193f3e No.5844 [Last50 Posts]
VQC sez: "(You) are at The final lock and key construction steps. Happy to give it but it is the Eureka moment and anons are close. In hindsight, it shows exactly why this problem has existed for so long.You are solving two problems at once in this method of constructing the answer. Also in hindsight the steps give the pattern back in the grid (The End).
____________________________
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193f3e No.5845
Previous Threads
RSA #0 —— https://archive.fo/XmD7P
RSA #1 —— https://archive.fo/RgVko
RSA #2 —— https://archive.fo/fyzAu
RSA #3 —— https://archive.fo/uEgOb
RSA #4 —— https://archive.fo/eihrQ
RSA #5 —— https://archive.fo/Lr9fP
RSA #6 —— https://archive.fo/ykKYN
RSA #7 —— https://archive.fo/v3aKD
RSA #8 —— https://archive.fo/geYFp
RSA #9 —— https://archive.fo/jog81
RSA #10 —— https://archive.fo/anGD7
RSA #11 ——http://archive.is/YERWc
Novice Baker here, dough was a little funky. To keep things clean and simple for this bread, I'm posting only links to previous breads. We should set up a new Pastebin for dough.
All comments and feedback appreciated!
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744226 No.5847
>>5844
Can everyone please help post the best diagrams from the last bread? Thanks!
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43eac4 No.5863
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79f6d5 No.5865
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130593 No.5866
>>5851
I tok a stab at this using a=7, b=37 in which case the (n - 1) center piece surrounded by (f -2) + remainder (resulting in unbalanced triangles) still left room from the base (base is 3 for f2/8).
However, this could be filled with one d. Leaving the outside ready to be filled by n*n + (2*d - 1)*(n - 1).
So I tried again with a = 19, b = 61 and this time, with base = 8 (f2/8 and no remainder) it left 219 free cells, after filling in for n - 1.
However, if I filled in (n - 1) four times in the center the remaining slots could be filled by d.
It's been a while since I checked in so I'm a bit outdated, but I'm reading up on the stuff you guys have been doing and so far I'm blown away! Great work guys!
And, as always, great to see you to VQC!
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130593 No.5867
>>5866
I'm going to try and make some sense here using the cell for a = 19, b = 61.
This cell is equal to: (3, 6, 34, 15, 19, 61)
We can see that 6 + 15 = 21, thus x+n is odd.
e = 3
d = 34
f = 2 * d + 1 => 66
f2 = 64
f2 / 8 = 8 (with no remainder as f2 is 8**2).
We now know that the base of our f2-triangles is 8 and we calculate the total amount of cells that is contained in a single f2 triangle: n*(n + 1)/2 = 36.
Multiply this by 8 and we have 8 triangles that fill inn 288 cells.
We then remove f2 from this leaving:
288 - f2 => 288 - 64 = 224.
Now we know that in the center (n - 1) exists, but we don't know enough yet. We also assume that n < 8 (our base) and since we don't have any more of f2 to fill in the triangle we assume we have to use some 2*d's.
So we compute 224 % (2*d) => 20. Thus we will have a remainder of 20 if we remove some amount of 2*d.
204 / (2*d) = 3. So we have to remove 2*d*3 from our triangle which leaves a total of 20 cells left for us to fill. Our d is 34, so we can't fill it in with a complete d, instead we know that we have to have AT LEAST one (n - 1) in here. Since we already removed 3 2*d's, we know n - 1 has to be at least 3. Assuming n = 3, we check 20 % 3, but this leaves us with 2, so we bump it up a notch and say n = 4. This gives us 20 / 4 = 5 (which happens to be our (n - 1)). This means we have to add (n - 1) four times to the center piece to fill it up.
I know this is a wall of text, but I'm just trying to think out loud. It might also be gibberish.
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858fa5 No.5868
>>5851
This seems correct. I did an example with different d values to show that they still construct the same square. I think this may be the ticket in
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193f3e No.5869
>>5868
Dang CA! Just finished studying your diagrams, and the numbers work! You and PMA are crushing it! Way to go, Lads!
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193f3e No.5870
>>5851
PMA, you nailed this explanation. Also that is a beautful diagram for c6107!
>Per VQC,
>f is within each triangle but OUTSIDE of (n-1), where the (n-1) square should go in the middle.
>By taking one of the (n-1) portions out of the 2d(n-1) piece of the nn + 2d(n-1) + f - 1 equation, we are left with an equation that looks like:
>nn + (2d-1)(n-1) + (n-1) + f - 1
>This new (n-1) + f - 1 piece turns an "out of balance" f portion around the middle into something that is equally distributed among the 8 triangles.
>The revised full square pics attached for the same c6107 and c27707 tests show how this new layout comes together. Also attached are close ups of the (n-1) + f - 1 in the centers.
>The green squares in the middle represent the single (n-1) portion that has been removed from 2d(n-1).
>The yellow squares are the portions of (f-1) that have been "displaced" from their original positions.
>The red square is a single unit taken from (2d-1)(n-1).
>And this is most likely why (n-1) is so important as a factor to all of the transforms that we worked on previously. (d[t]-d)/(n-1), a(n-1) na transform into (e,1).
>(n-1) balances out the f center square.
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c6c148 No.5871
>>5870
As a follow up to my previous post >>5851, attached is revised odd x+n=83 output for both f mod 8 = 5 and 6.
Additional columns have been added for (n-1) + (f-1), it's div 8 and mod 8 values.
The f mod 8 = 6 results include c6107 and c27707.
From the mod 8 column, we can see (unfortunately) that there are times when this relationship still doesn't equally distribute among the 8 triangles. Notice records with mod 2, 4, and 6 values. I just happened to pick another example that worked out perfectly.
There may, however, be a deeper relationship to explore.
In the f mod 8 = 5 output, the only valid mod values appear to be 2 and 6.
If there is a way to consistently relate these mod values to f, we would be able to predict the remainder portions around the center (n-1) + (f-1) area.
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130593 No.5872
Oh dang, I see you guys have learned more. So if I get this straight, we now use f-1 (can anyone explain why?) and (f-1)/8 doesn't represent the triangle, but rather the number of cells we fill our triangle with.
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858fa5 No.5873
>>5868
This math is really convenient computer-wise, because integer division [ex 11//2 = 5, 14//3 = 4] by 8 is just chopping off the last 3 bits, and seeing anything mod 8 is just looking at the last 3 bits. Looking at the last three bits would be the same as doing an AND operation with 7, which is a holy number, coincidences don't exist.
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130593 No.5874
>>5872
>(f-1)/8 doesn't represent the triangle
* Represent the BASE of the triangle
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41e2a8 No.5875
PMA / CA - whoot! Awesome translation to the diagrams. Will review this eve.
VA, thanks for the blessed bread.
>>5867
Check your f formula?
f = 2 * d + 1 - e => 63
and f2 = 61.
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ca4eea No.5876
>>5863
Good job with the dough.
pastebin.com/egYKQrGL
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130593 No.5877
>>5875
Dangit, yes, but the numbers are still okay. I just typed it in, but the code was using the proper f = 2*d + 1 - e. But it looks like we've changed our perception of the base of the triangle
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41e2a8 No.5878
>>5877 I see now, didn't check the calc, ha.
>>5873 interesting CA.
Just saw this glancing at tail of general research. (see pic)
>>>/qresearch/1349267
Hmmmmm. Maybe because WL knows RSA is broken?? Once the public finds out, stocks such as RSA Securities are going to tank quickly.
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858fa5 No.5879
So You do all this stuff and basically each triangle ends up being:
(n-1)//8 + (f-1)//8 + (nn)//8 + (2d-1)(n-1)//8 + 1
I'm thinking that you start off and you get these values
(n-1)%8
(f-1)%8
(2d-1)*(n-1)%8
nn%8
and then follow some type of recipe or something to generate the square
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5a1919 No.5880
3 square
4 square
5 square
it's a polite situation… me thinks…
and a new kind of trangle than we been twerkin' wit
so what would a 456, 567, 789, etc etc… look like/do
https://www.khanacademy.org/math/algebra2/polynomial-functions/binomial-theorem/v/pascals-triangle-binomial-theorem
http://www.benpadiah.com/MISC_diagrams/pages/equations/PySpiral.html
I'm not even sure this is the right place for what I'm talking about… is this getting into Fermat/ECC???
This is rubbin'z muh almangs…
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5a1919 No.5881
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c9bbff No.5883
>>5868
Outstanding anon.
Semi-related, who would have believed you could find a way to calculate a chosen decimal digit of PI without calculating the ones before it.
Have you heard of that method? Spigot. It involves one of the most surprising shortcuts in history.
https://en.wikipedia.org/wiki/Spigot_algorithm
https://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula
Some beautiful moments in history are behind us.
Many are in front of us!
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5a1919 No.5884
>>5883
Could you work your magic?
I just got a ban from /qresearch/ for the Searchers/Chris Curtis/P=NP post.
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c9bbff No.5885
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c9bbff No.5886
>>5871
Brilliant PMA.
The way you are all thinking is great to see.
The recognition of patterns and the visualizations… brilliant.
I have to admit, you're all moving faster, way faster, than I did.
I remember in 2013 staring at the grid (The End) for hours. Hours!
I knew it meant something, I'd tried so many different variables as the axis and went back and fourth, desperate for more insight. Often insight only came when I left it alone or seemingly from nowhere.
Plugged into bitmaps like you have here, hundreds and hundreds of bitmaps. And tables.
Anons are working so well together.
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c9bbff No.5887
>>5874
To clear this up slightly, the base of a triangle WITHIN the triangle.
The final steps involve finding a base using (f-1)/8 for each of the eight triangles (some a bit bigger than others by a unit in length) which is larger than (n-1)/8 BUT smaller than (x+n)/8.
The choices reduce 'exponentially' and with less complexity than find the root of C, so the last steps do not push the overall complexity Big Oh above Big Oh for finding the root.
Using the previously supplied C# method before, you construct the first guess of (x+n)/8 which is too small but larger than (n-1)/8.
So, the biggest hint so far.
For odd (x+n) and odd (n-1) to start with.
(n-1)/8, (f-1)/8, (x+n)/8
You are lining these triangular bases up by picking (f-1)/8 that is between (n-1)/8 and (x+n)/8.
The properties of 2d and f limit the factors that will lock these values together. For two prime factors there is one way and for more than two prime factors there are more and more ways depending on the number of factors.
If you can find a way to visualise this like you have above, it will become clearer very quickly.
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c9bbff No.5888
>>5880
Fits a pattern in column 1 of the grid at least, right?
Perhaps even the at (1,1) : e=1, n=1 => what are the values of a?
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c9bbff No.5889
>>5884
Appreciate the irony behind the post.
Maybe subtlety will reduced the risk of bans.
I think safety is key.
Switch your router off and on again for qresearch?
Please don't get banned for me anon!
I'll be here more often now.
Thanks again for your support.
And to everyone here.
Lots of adventures ahead!
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5a1919 No.5890
>>5889
THAT'S THE IRONY THOUGH!
Someone posted something to the board…
Which led me to someone who died in 2005.
https://en.wikipedia.org/wiki/Chris_Curtis
You just mentioned C#… there's a reference here.
Did he actually prove P=NP?
Sidenote…
ALBINO MORPHEUS IS NOW BACK IN PLAY!
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130593 No.5891
>>5887
In a previous post (RSA100 from #10 I think) you noticed that the f-2 was divisible by 5, so you divided it by 5*8 (40) and then referred to the blue base in the triangle as consisting of a base of five units. I'm spitballing a bit here, but that implies that when we divide (f-1)/8, we're looking for a base with a single unit (mind the remainder of course) and as we divide by multiples of 8 we will look for wider units?
Am I on the right track with trying to wrap my head around that part?
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130593 No.5892
>>5891
And if I get it, if we make a base (f-1)/8 which is smaller than n, we would have to increase it (by for example making a new base (f-1)/(8*2)). We continue to do so for some time (which I guess will become more apparent once we understand more) which will result in either a solution or we determine that the number is a prime number.
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130593 No.5893
I'm looking into this part
> This restricts the multiples of 2d that can be used because they must always be in a number that supplies the gap in f and satisfy another property. That property can be broken down.
It sounds like it means that there is a gap between f and the capstone (n-1) which is to be filled with multiples of 2*d, but is "guarded" by so-far an unknown property. I gave it a go and it could most easily be a fluke, but the remaining parts, which are filled in with parts of d is equal to (n - 1) * 3, leaving a total of (n - 1)*4 inside the capstone.
I haven't looked at any other numbers yet, but I'll take a look at another cell to see if it's a fluke or not.
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130593 No.5894
>>5893
Green represents n*n, red represents (n - 1), yellow is f-2/8 and shades of blue inside of f-2 represents different multiples of 2*d.
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858fa5 No.5895
>>5888
The values are:
(e,n) = (1,1)
1, 5, 13, 25, 41,..
which are the sums of consecutive squares.
(1+0 = 1) (1+4=5) (4+9=13) (9+16=25)..
(e,n) = (2,1)
1, 3, 9, 19, 33,..
are sums of the same square plus one
1 = 0+0+1, 3 = 1+1+1, 9 = 4+4+1, 19 = 9+9+1, 33 = 16+16+1.
(3,1)
2,6,14,26,42
sum of consecutive squares plus one.
(4,1)
2,4,10,20,34,
sums of the same square plus two.
A(e,1,t) =
if(e even):
shift = e/2
2*t*t + shift
if(e odd):
shift = (e-1)/2
t*t + (t+1)(t+1) + shift
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858fa5 No.5896
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79f6d5 No.5897
Test post attempting to use Unicode Character 'SUPERSCRIPT TWO' (U+00B2):
d²
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5a1919 No.5898
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eb27d2 No.5899
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79f6d5 No.5903
>>5832
>I'll get less and less cryptic over time until I EITHER give it away, OR you guys solve it, OR WE MEET IN THE MIDDLE.
The excluded middle.
That’s a clue right thar
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193f3e No.5904
>>5863
Hello AA! Thanks for getting the dough sorted out. Thank you once more for being our BO, and doing a great job! Logos are for you. I love the red and white + eagle FYI.
>>5876
Thanks Anon! Now we have a pastebin for our batter. Anyone who want to help add to it is welcome!
Want to work on building a "last bread notables" section? Do it.
It would be cool to have a summary of best posts for each bread. Do it.
Everyone can join in, this is a team effort.
Do it.
Everyone, this is awesome. I enjoy being here, very much. Energy of Gratitude is flowing outward from me to all of you. Can you feel it?
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7e5bcd No.5907
>>5887
>The final steps involve finding a base using (f-1)/8 for each of the eight triangles (some a bit bigger than others by a unit in length) which is larger than (n-1)/8 BUT smaller than (x+n)/8.
Pics attached are an idea of how this could look for the c6701 and c27707 examples.
Starting with the (n-1)+(f-1) portion in the middle, a new red square outline has been added (f-1)/8 units from the center.
For the c27707, the light green squares around that red outline represent the u1=(f-1)/8 + 1 value of 23. And the light green squares represent the u2=(f-1)/8 value of 22.
The c6107 example computes to u1=17 and u2=16.
Attached are also rough calculations showing successful matches in 7 steps.
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6199d2 No.5925
What's going on with VQC's trip? Has there been any post by him where he mentions that he changed it?
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41e2a8 No.5926
>>5925 not sure Isee, good to see you.
Thought he had changed, but it was PMA:
>>5649
>>5648
Topo asked about it, and also noted that both trips had the same UserID, so I didn't worry after that (in Fermat thread), but would be good to have VQC clear it up:
>>5900
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5a1919 No.5927
I'm not sure if this useful or related but go here and when you move your mouse….
You turn a square into a 3d Sierpinski pyramid.
https://upload.wikimedia.org/wikipedia/commons/1/19/Tetrix_projection_fill_plane.svg
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5a1919 No.5928
>>5927
Ugh.
Fuckin' did it again. lol
Gotta bless the Bread anyway…
RSA12… RSAXII…RSATwelve… RSA#12…
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5a1919 No.5929
>>5928
Hmmmm….
First 4 are for thematic purposes…
The last one…
The last one should look kinda familiar.
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c9bbff No.5930
I'm going to create a C# app to help with the graphics.
Two panes side by side.
On the left will be a bitmap displaying the x+n square.
On the right will be the grid (The End)
Hovering the mouse on a pixel of the grid with entries for that e,n will animate through the x+n squares in that set.
We'll build the program in stages and the code will be shared here.
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c9bbff No.5931
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c9bbff No.5932
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c9bbff No.5933
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c6c148 No.5934
>>5907
>>5932
Using the (f-1)/8 as a base for u1 and u2 enables calculating a larger n0 starting value for cases where (f-1)/8 > n.
I'm estimating about 10% performance improvement here for the iterative search process, but still need to rely on n+4 jumps for subsequent records.
Couple of issues that I'm researching:
1) Is there a way to determine from the outset if the large (f-1)/8 jump for n is even possible. (i.e. rule out cases where (f-1)/8 < n). Initial thought was to try and rule out -e values from n,d,f records. Not sure of this yet.
2) Is there something else in the n,d,f tables that can lead to larger n jumps? Some patterns and/or combinations of (n-1) + (f-1) that can be relied on.
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5a1919 No.5935
>>5933
VQC!!Om5byg3jAU
=
Chris
>>5889
VQC!Nm8wdW6d92
=
dot dot dot… the other guy who responds when I go fishing?
Bot have ID: b80b93
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add09d No.5936
>>5907
I think the mystery variable/factor is x.
Just a hunch. Here's why.
D-x=a
x+n square
Find the right n, find the right x, find a and b, problem solved.
So (n-1) and f and e give us clues to what values of x work with each n iteration.
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add09d No.5937
>>5936
Plus it's just very cool if X marks the spot.
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a24494 No.5938
>>5933
Same id, same person. Maybe he forgot a #?
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c9bbff No.5940
>>5889
>>5888
>>5887
>>5886
>>5887
>>5885
Just seen these.
Apologies.
Trip entered incorrectly.
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c9bbff No.5941
Problem statement (re: building an app).
Why do we gather requirements and thoroughly look at the problem statement?
As my physics teacher said, by fully stating the problem statement, the answer presents itself. Or rather, the solution.
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c9bbff No.5942
If I was to build an app.
On the left, the picture (like you've been building above), would be the square (x+n)(x+n).
On the right, would be the entries of the grid (the end) to select a cell.
To TEST it, we need to state the problem we are solving.
What test cases would be use?
Say we selected an (x+n) square and to test we wanted to keep the square the same size…
If I pick a grid cell at position t within that cell, how would I pick another cell with a (x+n) square the same size?
Would I pick a cell to the right (increasing e) that is 2n cells to the right?
Would that give me a cell with a value of (x+n) which is the same?
Then 2n from that?
What would be the difference in f?
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41e2a8 No.5944
>>5941 Taught exactly same thing!
>>5942
Thanks, will think this through.
>If I pick a grid cell at position t within that cell, how would I pick another cell with a (x+n) square the same size?
Do you mean a "grid element at position t"?
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744226 No.5947
>>5942
Ah, that's a good clue! Thanks VQC, and good to see you.
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0b1d7e No.5949
>>5942
You're making this too easy. >>5689
Then again, if 1605 is equal to 505, then 5 plus 5 might as well equal 12.
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c6c148 No.5950
>>5942
Couple of examples attached for e+2n within the same (x+n) = 83.
difference is f appears to be 2n-2.
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12d702 No.5951
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5a1919 No.5952
When I originally did this, I wasn't seeing how to move up/right or down/left reliably.
I was trying to figure out if it'd be spirals and mirrors or just mirrors. It's just mirrors.
I made a crosshair.
LET ME SHOW ITS FEATURES! ha ah ah ah aaaaah!
You want Magenta highlights.
The Right 1, Up 1
(or Down 1, Left 1 if you wanna just mentally flip the ± of how you're moving)
Pic 1:
down left from 45
-0, -3, -6, -9, -12
upright from 45
-3, -6, -9, -12
or from 15:
+12, +9, +6, +3, +0, -3, -6, -9, -12…
so the 45 acts to kinda… 0 it in…
Pic 2:
from 14: same three jump, but shifted -1
+11, +8, +5, +2, -1, -4, -7, -10
Pic 3:
from 13: reversed pattern from 14, 3 jump with a -2 shift
+10, +7, +4, +1, -2, -5, -8, -11
Pic 4:
from 12:
+9, +6, +3, +0, -3, -6, -9, -12….
doesn't reverse
not that it can… lol
buy, the 0 is a column over from the first one… 3 jump with a -3 shift (from 45)
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5a1919 No.5953
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5a1919 No.5954
>>5953
Actually… more like this… just rotated as necessary…
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6db68d No.5956
I've been looking back into thread #10, trying to summarize what VQC has been posting. I must admit, I struggle formalizing exactly what the problem is, as VQC is asking in >5941 and >5942
And one thing that kind of stumps me is the though behind Get_Remainder_2dnm1. This function takes a base (constructed from (f-2)/8), the d, suggested n0 and f2.
We're trying to find a base f2//8 where f2//8 > (n-1)/8 and f2//8 < (x+n)/8. Yet, the function returns 0 for a base that's base > (x+n)/8. I think I either have misunderstood something, or there is some mixup between the terms we are using.
I'm using a=7, b=37 which has an even n and odd x and even d.
Here the values that return 0 for Get_Remainder_2dnm1 is:
base = 7,
d = 16
n = 6
f2 = 28.
This implies that the f2 triangular base is 7, which is f2/4 instead of f2/8. I'm a bit rusty and I'm still catching up to #11, but can anyone shine some light on my confusion?
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6db68d No.5957
>>5956
Okay, I'm a bit of a dofus. Get_Remainder_2dnm1 returns 0 for (x+n-1)/2 as base, which is the base of our main triangle.
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99782a No.5959
>>5887
>The final steps involve finding a base using (f-1)/8 for each of the eight triangles (some a bit bigger than others by a unit in length) which is larger than (n-1)/8 BUT smaller than (x+n)/8.
>larger than (n-1)/8 BUT smaller than (x+n)/8
For each given (x+n), there is a range of possible n values and a range of possible x values. Picture 2 shows this pretty well. For (x+n) = 9, n-1 is going to be in the range of 0 to 5 and x is going to be in the range of 3 to 8. Picture 1 was related to something else (I had found linear gradients in a bitmap but they didn't lead to anything because of the thing this picture explains), but it shows what VQC is talking about here: the range of n values and the range of x values overlap. If we don't know (x+n) but we want to set an n0 value based on our known f value, the minimum possible (x+n) and maximum possible n values for our given c also overlap in the same way (that's what the gradients were about, and I guess it's finally relevant). So n0 is currently a guess within an overlapping range. I could see if I can find the gradient picture and use it to make an equation for finding the range of n and (x+n) for a given c if that would help (it sounds like it would to me but I don't tend to get feedback about these things).
>>5904
Well thank you for the encouragement VA. I'm glad we all have good comradery here. You seem to be a big part of that. I think that first logo looks a lot less like a sigil compared to the other ones so I'd agree with your choice.
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7e5bcd No.5960
>>5950
Some additional examples for (x+n)=83 for different f mod 8 and n starting values.
Looks like the earlier formula for f can be generalized for e movements within the same x+n.
e = e + 2n
f = f - 2(n-1)
So there's our (n-1) again.
Also looks like there might be some relationship between f mod 8 and (n-1)+(f-1) mod 8, but not certain of that yet.
>>5959
>make an equation for finding the range of n and (x+n) for a given c if that would help (it sounds like it would to me but I don't tend to get feedback about these things)
I also looked for the formula for max n in a given x+n. But ran into a problem where that max n could be greater than the n we were searching for during iteration, so tabled the idea.
The starting n and increments are known - I've posted that code snippet previously. It would be nice to know that max n value. If not for iteration purposes, then perhaps for another use down the road.
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99782a No.5961
>>5960
>max n in a given x+n
I'm talking about the max n for a given c. That's what VQC's talking about. These pictures are the ones I was talking about from something I posted about that didn't go anywhere in February. I'll make bigger versions in a bit. It shows that there's a series of c values for which the corresponding (x+n) values follow the gradient along the bottom, and there's a series of c values for which the corresponding (x+n) values follow a slightly lower gradient but the gradient starts after the first gradient, and so on. The starting point of each gradient and the actual gradient are calculable, I'm pretty sure. What is means is that you can find the maximum and minimum (x+n) values for a given c. The same thing happens for n, as you can see in the second picture. What this means is that for a given c you can find the range of possible n values and the range of possible (x+n) values. That's exactly what VQC was talking about doing (finding the right f value (which we know for our given c) so that n0 is bigger than n-1 and smaller than (x+n)). The only problem is, for a given c, the range of possible n values and (x+n) values overlaps. n0 is somewhere in the overlap.
>I also looked for the formula for max n in a given x+n
It definitely seems possible to calculate, based on this graph I made. The x axis is (x+n) and the y axis is n. I added the black lines in Paint to show that it's a non-linear line made of increasingly-sized linear lines.
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99782a No.5962
>>5961
I should maybe point out for anyone who might be confused (not necessarily you PMA) that the ranges I'm talking about in this post are the range of possible values but not every value is valid (it's only going to be one of them but if we were to guess it would be within this range), so it's different to the range of n values for a given (x+n) within which all the n values are valid.
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555b3c No.5963
>>5942
The difference between f in our selected cell and the cell 2*n can be expressed using the following algebra:
new_f = 2*(d + 1) + 1 - (old_e + 2*n)
So we can now compute the difference algebraiclly
diff = 2*(d + 1) + 1 - (e + 2*n) - (2*d + 1 - e)
diff = 2d + 2 + 1 - e - 2*n - 2*d - 1 + e
diff = 2d + 2 - 2n - 2d
diff = 2 - 2n
Thus, when you're moving 2*n right, regardless of e and d the f value changes by 2 - 2*n.
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555b3c No.5964
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130593 No.5965
>>5491
I'm trying another approach, before this I would divide (f-1)/8 and then just randomly place the remainder of (f-1)/8 against the different triangles. What if it's supposed to be more like this (image).
In this case I used f/8 and not (f-1)/8 and instead removed the extra square from (2*d - 1)/8.
So I end up with two types of triangles, with regards to f (yellow), two which is 3 base wide and 6 which are 4 base wide. The inside of this fits (n - 1) (red) and n*n (green) with 2 to spare. The 2*d (blue) expands out and finishes the triangle (x+n) with 2 to spare, which will fit inside of the minor f-triangle.
I'm replying two VQC's post from #11:
>The 'heart' of the problem is that breaking the problem down involves two sets of triangle solving (otherwise it is prime), with one set of triangles (say half of the eight) working on a triangular problem that is one unit longer than the other set. This is one reason why no current approach I have seen works.
I _think_ this approach fits in with what he says as the outer layer of the triangles of f/8 consists of 2 * (4 + 5 + 6 + 7) and 6 * (5 + 6 + 7) leaving two of the triangles being one unit longer than the other 6.
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ce2e01 No.5966
Looking through old crumbs:
>The solution is a decision tree. The first decision is whether remainder is zero. The second decision is whether the remainder is odd or even. What's the third decision?
It would seem obvious to us now that the third decision is whether (x+n) is odd or even. So what's the fourth decision?
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41e2a8 No.5967
>>5960 nice formula extensions!
>>5965 superb, you're all caught up and moving this forward!
>>5966 Makes perfect sense! Go on…
>>5959 Good illustration on the overlap.
(AA - might not give feedback, but have read and digested all of your posts).
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130593 No.5968
>>5965
VQC has hinted (or flat out said) that we have four sub-types for when (x+n) is odd and n is even, could that be related to the possible dissymmetry between the triangles bases?
In the image I posted there are two triangles that has a base of 3 units and 6 with a base of 4. If we iterate the possible patterns we get:
1 triangle base x, 7 triangles base y
2 triangles base x, 6 triangles base y
3 triangles base x, 5 triangles base y
4 triangles base x, 4 triangles base y
Where x != y.
This excludes the possibility of when all triangles have the same base (meaning it's divisible by 8 exactly). But in that case we wouldn't have two triangle sets to solve (based on the assumption that what I'm referring to is in fact the triangle sets VQC was talking about back in >>5491).
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130593 No.5969
>>5968
I suspect I might on to something, in >>5690 VQC refers to
>Using an arbitrary divisor for f, then each of the eight triangles will have one OR one of two (the latter when c is large and the product of two different prime numbers) configurations in each triangle. The difference between those two configurations of a portion of f are that they are both staircase numbers where the base of one is a unit longer than the other. E.g. (3,4,5) and (4,5,6)
If I get this correctly, when you divide f-1 by 8, you have only a single unit. This works fine for smaller examples as the base of this triangle will be smaller than (x+n)*(x+n)/8's base (VQC has mentioned (x+n)/8, but I'm guessing it was supposed to be (x+n)(x+n)/8?), but if f-1/8 is bigger than (x+n)(x+n)/8 base, then it won't work.
Since our goal is to find a base for f, which is (x+n)(x+n)/8 < f-1/x > n/8 we will have to divide f-1 by some multiple of 8. For each multiple of 8 we increase our unit width (vertically).
So if I divide f-1 by 16 I should have a unit of the type (2, 3) or (5, 6) or (11,12). Since it's (f-1)/(2*8). If I did (f-1)/(3*8) it would be three units wide (vertically), for example (2, 3, 4) or (5, 6, 7) etc.
I could be completely wrong, though. But I think I've managed to convince myself that it makes sense.
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41e2a8 No.5970
>>5969 nice 'steps' Isee, makes perfect sense and is coherent with the crumbs.
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cb9b2e No.5971
>>5969
>>5701
>Exactly like this on one side and the (x+n) square constructed on the other side. This is all about patterns.
>I would like to start a thread in parallel to this as we're reaching the point where an anon or group of anons is going to have a Eureka moment and solve this for one of the four types of solution (odd x+n with either even or odd e), then quickly for another and then the other two.
>The other thread will describe how to take the work done to solve Fermat Last theorem relates to our approach and why two objects that are identical types in number theory had to be proved so in a difficult to under proof, when in reality, with the right number system or model, this should have been obvious or a tautology.
I think he means this
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f27624 No.5972
>>5969
Hello Isee! Is your idea connected to this diagram from VQC?
>>4344
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a11a4b No.5973
>>5972
Yeah I'm trying to wrap my head around why it is 5 units wide(?/long?) vertically
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a11a4b No.5974
>>5971
That's the (d+n) square. I'm referring to the pattern inside the (x+n) square.
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7e5bcd No.5976
>>5942
Pics attached are visuals for the e = e + 2n movement that VQC referenced previously.
Starting from a new test record for c86747, these images show the e movement and affects on f moving 2 records to the right as well as 2 records left.
For reference, the starting prime solution is:
(311,12,36) = {311:12:294:71:223:389} = 86747; f=278; (x+n)=83; u=41; (d+n)=306
One image shows the regular formula for nn + 2d(n-1) + f - 1 with the (f-1) portion in the center.
The other image breaks the 2d(n-1) into another (n-1) piece and combines with (f-1).
All images are partial views of the x+n=83 square.
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6dbb33 No.5977
>>5870
>>5871
PMA, MM, Top, and anons.
Legends.
Greetings from Poland.
Diagrams that you have already.
The Grid (The End)
(x+n) at any cell, then 2n left or right of that cell.
Animate.
What is the pattern?
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cffec1 No.5980
>>5977
>Diagrams that you have already
I guess this means I'm moving in the wrong direction with regards to the diagram in >>5965?
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5a1919 No.5981
>>5977
It's most apparent when you watch how the yellows in the e=e+2n move from the corners to the middle.
Seems to do it in steps and then jump back to a square, and work in again. Would need more frames.
The .gifs were pretty, but we're being fags and I'm gonna cut the frames from the numbers file when it gets sent later today. You can see that, as they are, they're based on screen size.
I don't mean to brag, but mine's bigger.
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5a1919 No.5982
>>5981
And by that I mean i'll be able to get the fuuuuull squares instead as much as I can see and still make stuff out.
This way I won't have to go through the extra steps of stitching things together. :D
Woulda been 10 stitches last time and those are never completely equal either.
I'm just :P cuz I forgot the trip again. lol
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5a1919 No.5983
As much as he can see.
(Yo,ass) knows what I mean.
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5a1919 No.5985
>>5976
>>5977
Pic 1: "Regular"
Pic 2: "Adjusted"
I'll let PMA take it from here! :D
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5a1919 No.5986
>>5985
Er, derp.
Gif 1, Gif 2
Animated.
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5a1919 No.5987
>>5985
>>5981
Ah. No "back again"
Just gets smaller…
Line by line.
Pinches the corners towards the middle and then starts the next… line… square line… perimeter…
What're we calling it?
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5a1919 No.5988
>>5987
Or is this ultimately backwards and it builds out?
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7e5bcd No.5989
>>5976
>>5985
Topol, thanks for stitching these together.
>>5977
I think the most compelling of the two views is the (n-1)+(f-1) example, which shows a single green square moving around the top portion.
These gifs show only 5 iterations, but many more examples for the same x+n=83 can be found by decreasing n from 12 to 2.
I believe the following repeating patterns apply to (n-1)+(f-1) mod 8, but would appreciate someone else verifying:
f mod 8 = 0
Starting from n=2
(n-1)+(f-1) mod 8 pattern: 0, 2, 4, 6
f mod 8 = 1
Starting from n=3
(n-1)+(f-1) mod 8 pattern: 2, 4, 6, 0
f mod 8 = 2
Starting from n=2
(n-1)+(f-1) mod 8 pattern: 2, 4, 6, 0
f mod 8 = 4
Starting from n=2
(n-1)+(f-1) mod 8 pattern: 4, 6, 0, 2
f mod 8 = 5
Starting from n=3
(n-1)+(f-1) mod 8 pattern: 6, 4, 2, 0
f mod 8 = 6
Starting from n=2
(n-1)+(f-1) mod 8 pattern: 6, 0, 2, 4
These mod patterns apply only within the same f mod 8 value. Keep in mind that the e = e + 2n movement changes the f mod 8 value for each record.
So for the c86747 record, 2 left and 2 right posted, the (n-1)+(f-1) mod 8 values are 4, 6, 0, 2, 4.
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31a947 No.5990
I realized something today interesting or not, known or unknown, but that tickled my fancy.
The equation we have:
n*n + 2*d(n - 1) + f - 1
can also be seen as an expanded form of the difference between two arbitrary squares. We can rewrite it as follows:
n*n + 2*d(n - 1) + f - 1
n*n + 2*d(n - 1) + (2*d + 1 - e) - 1
n*n + 2*d*n - e
n*(2*d + n) - e
Which (when you ignore the e) is the result of the difference between two squares.
I don't think this will be mindblowing, or anything, but it gave me some insight.
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497403 No.5991
Just messing around with some triangle code.
Pics attached are for one of the 8 triangles for x+n 83 and 167.
Interesting pattern when highlighting by mod 8.
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7c094c No.5993
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31a947 No.5994
>>5990
Just spitballin' here. We don't know (x+n) yet, but we know that there is a number in (e, <some n>) that is equal to (x+n) + e.
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31a947 No.5995
>>5994
I was looking into this, just as a curiosity and it makes sense.
(x+n)**2 + e will exist with a=n, b=2*d + n.
It will have the same e as the original record and a few other interesting properties. For the record, the a=n, b=2*d + n is simply because of (x+n)**2 + e = n(2*d + n).
The interesting part is that the n in the record for a=n, b=2*d + n is equal to a from the original record.
I guess this will only hold for the records where (x+n)**2 = n(2d + n) - e.
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5e25f3 No.5996
>>5995
Took my last exam today for ripoff college.
CALL ME GA NOW!!!
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88d253 No.5997
>>5996
FUCK YEAH CA
I MEAN GA!!!!!!
Nicely done, I remember the feeling myself. Feels like freedom, my friend.
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7e5bcd No.5998
>>5996
Congratulations, GA. Tremendous accomplishment.
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6bf400 No.5999
Here's some Kanye fo yo ass.
https: //youtu.be/6CHs4x2uqcQ
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f4515c No.6000
Here's my current working sheet. Pretty basic, but just building out different breakdowns of the x+n square and nn + (2d-1)(n-1) + (n-1) + (f-1). Nothing new detected yet, but maybe you guys will catch something I'm missing.
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de8bb2 No.6001
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5a1919 No.6002
>>5996
GIT IT CollegeAlumnon! ;)
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7e5bcd No.6003
>>5991
Still playing with the odd x+n square and triangles, but managed to take things a bit further.
Pics attached show a bit of a progression:
1) Initial layout of 8 triangles around a center square.
2) Iterating the 8 triangles sequentially and filling in numerical values.
3) Dynamically generating the n, d, f view for c=6107. (This corresponds to the nn + 2d(n-1) + f - 1 layout from >>5976.)
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7e5bcd No.6004
>>6003
One more example for c86747. The prime record posted in >>5985.
Interesting that the nn portion is now distributed to the corners.
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9f1810 No.6005
>>5996
Congratulations!, GA!
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de8bb2 No.6006
>>6004 really like the new format PMA.
Would it be easy for you to export this as a csv?
Bring into excel and use conditional formatting to add color to each letter.
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a108d9 No.6008
>>6006
Space delimited and conditional formatting seem to work fine.
pastebin.com/uWKBNDQ0 c6107
pastebin.com/W3jJhMDK c86747
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91c020 No.6009
>>5995
Another "fun" thing regarding this.
Compute a cell (cell_0) with (x+n) odd and n even and d even.
Then create a new cell with a = n, b = 2*d + n, with n and d from cell_0 and call this cell_1.
x+n from cell_1 is equal to d from cell_0. And the difference between d/2 - x is the same as n - d/2.
Sorry for derailing a bit, I know this stuff belongs back in thread #8-9, but I haven't noticed this pattern before.
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5a1919 No.6010
Just to keep y'all up to speed on the discord conversation, we're looking at arrangements of primes in different dimensions…
Based on D+N and X+N
Or something like that.
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de8bb2 No.6011
>>6010 Thanks TA. Looks like MA's work.
Saw your fishing post a bit ago: >>>/qresearch/1420118
>>6008 Thanks, looks nice. Will play in excel a bit as well.
>>6009 Completely relevant right now! We'll be bouncing around the grid in no time.
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5a1919 No.6012
>>6010
Brought to you Minecraft, obviously ^_^
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5a1919 No.6013
>>6011
I can recognize y'all by graphic output style, but y'all probably recognize each other by formula approach and numerical outputs.
(Trips aside.)
Kek.
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91c020 No.6014
>>6009
It's quite nice the way these cells are connected.
Take cell_0 and add (x + n). This is cell_1's d.
Take cell_1 and add (x + n). This is cell_0's d.
>>6011
Let's hope so! My head is currently stuck on this, so I'll keep playing a bit with it before I go back to the triangles.
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91c020 No.6015
>>6014
The grid is kind of hilarious in a devious sense.
We know that for some numbers, the d of c is equal to the (x+n) _of_ (x+n) of c, but as far as I know, we don't know anything but e and (x+n). It's like the numbers we want is always out of reach. But just by inches.
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88d253 No.6022
>>5977
>>6003
>>6008
Hello Lads, I have a really solid idea. Can I please get all your eyeballs on this to check it and discuss it?
First, Check out this excellent PMA diagram again, VQC has reposted it twice in this bread. Why?? Because he's giving us hints by reposting things that are on the right track. (I think?)
>>5870
Here's the foundational idea:
>This new (n-1) + f - 1 piece turns an "out of balance" f portion around the middle into something that is equally distributed among the 8 triangles.
>(n-1) balances out the f center square
Working Theory:
So in this diagram, the correct (n-1) and (f-1) create a perfect square. Starting with the 1 unit center, and adding (f-1) to that, then we iterate around the center square until we get a number that is a perfect square. This allows us to iterate (n-1) much faster, because each rotation around the center unit, (n-1) must be a larger number than the one before, and must equal a perfect square. So basically, (f-1) determines what amount of (n-1) can create perfect squares around the center unit.
This means that the correct way to iterate n0 could be different than we were thinking. It could be:
1 + (f-1) + (n-1) = perfect square.
Correct iterations of n0 are those which create the next largest perfect square.
Meaning, our n0 iterations leap upward in size each time we begin creating the next largest perfect square.
Thoughts, Anons?
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5ebae7 No.6023
>>6022
For PMA's diagram, the math is as follows:
1+ (f-1) + (n-1) = 169
1+ 133 + 35 = 169
The earlier n0 value would have been n0 = 10
1 + 133+ 10 = 144 = SQRT(144) = 12
The next value would be n0 = 62
1 + 133 + 64 = 196 = SQRT(196) = 14
Not sure if this works. Just brainstorming over here and keeping my mind locked on the happiness of finding the correct path! All comments and corrections appreciated.
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d4281b No.6024
>>5996 Graduate Anon
What a beautiful day! Feels good man. Really. It is really good to see. Your comment about ripoff college caught my attention though. SO I was told something a long time ago… You can learn something from Anyone… Sometimes it is what to do; sometimes it is what to NOT do.
A lot of people think that they went to college to get a job. The most miserable and useless people in the world that I have ever met are those who went to college and got a job (myself included before I found the road). You gained something in college that you have not recognized yet. It takes a long time to put it together sometimes but if you look and dig you may find it. You learned how the world functions. All the classes and all the tests and all the drama and all the bullshit were not the real purpose. They were the means to The End (no pun intended… kek).
Take the time to reflect on your experience from time to time. You picked stuff up about how the world functions and why. It is subtle. It is moving… Just below the surface. Something you can't quite touch or see or grasp but its there.
That understanding is not grasped by everyone immediately. It took me a few years to put it together. Some people take a lifetime to find it and some never do. Focus less on what you were told and what you studied and what you wanted relative to what you got and instead focus on what you have now… in this moment and what you can do with it from your current position.
I have greatly enjoyed your inputs here in this place and wish I could contribute more. Anyone who can be a part of what is going on here on this board can work their way through most problems in the "real world" (if there is such a thing) as well. Life is good my friend… -The Great Hobo :>)
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6bf400 No.6025
>>6023
Mistakes here, but idea is sound. Should have said (n-1) instead of n0.
>>6024
Hobo! Nice to see you faggot! :)
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7e5bcd No.6026
Now able to generate odd x+n squares dynamically to image format.
Attached are animated examples again for x+n=83 where n=11 or 12 in the same two styles mentioned previously at >>5976.
These records are not associated by e=e+2n movement. Currently just testing for any layout adjustments.
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de8bb2 No.6027
>>6026
>>157 PMA, a throwback for you, look at how far we've ALL come! Nice work.
>>6023 VA, this is all quite the habbening. Truth and Justice.
>>6024 GH, good to read you faggot! Have been wondering what The Great Lord Hobo has been up to, know it's your busy season. SO good to read you. Wise words there.
Wish I had more time atm for this, following closely.
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43887e No.6028
Hobo thank you for your wise words. I learned a lot and I'd say most of it was outside the classroom. The fact of the matter is that if you do something and it makes you feel like shit afterwards (sleeping around, eating like shit, drinking a lot), then you probably shouldn't do it. I thought this could be explained away biologically, which it can be, but it's deeper than that. I believe in Karma and the good life. I did initially go to college to get a job and started studying Econ, but then I took CS and Math and realized that they would be the hardest to learn on my own so I switched. I just think they are neat subjects. I don't believe in this reality at all (to the degree with which it is presented). I think VQC will help us pop off that question. I believe in true unity, E Pluribus Unum. When I first started going on this board, one night I stayed up with Topol and he was just talking complete nonsense but I was trying so hard to decider it. I was convinced he was an AI or some being after the singularity moment talking back to me. Then I thought that VQC was Virtual Quantum C_ where C_ is my first name. I feel like I'm talking to myself here sometimes. I feel like every time I'm losing focus on the VQC someone comes in with just enough to look into to for another couple days. It's weird how everything I was learning this semester in number theory kind of connected in (aka I just shitposted stuff I learned) but I think it works, especially the continued fraction stuff. It's weird how the one book I read over winter beak (fingerprints of the gods) had ancient mathematics in it and in the pyramids and a viable reason for why our history is completely wrong. It's weird how in that book, ancient South American religions believed that the pyramids were built by people with magic trumpets or who could "WOLOLOLO" to create things [or in the case of the video game you can sing and turn people to your side, maybe we will be able to do that]. It's weird how people who smoke DMT report seeing elves who sing things into existence and say "You can do it too!". It's weird how at the second coming of Christ is supposed to be accompanied by trumpets. It's weird how in this grid we split our number into dd+e, and to do so we calculate the square root of it. It's weird how the Babylonians had inscribed on their walls the continued fractions for square roots certain numbers. It's weird how the continued fractions have a very predictable pattern if 2d is divisible by e. I don't believe in coincidences. I think this summer will be big. If Trump wanted to tell the public all of this stuff, where would he want them all to be? Would he want every parent hearing all of this crazy sick child abuse news or the full indoctrination of their children in colleges when school is still in session? Would he rather have everyone at home with their family for big news?
Still got a job tho :\
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bead22 No.6029
>>6015
I keep thinking about this stuff.
To re-iterate a bit, take a cell like a=7, b=37.
It has the following cell:
(3, 6, 16, 9, 7, 37)
We'll call this cell_0. Now let's use the knowledge that (x+n) for cell_0 is the equation:
n*(2*d + n) - e (ref >>5995).
So we make a new cell, cell_1 with a = n, b = 2*d + n. For our example of a=7, b=37. This will be:
(3, 7, 15, 9, 6, 38)
Let's show them side, by side:
cell_0 = (3, 6, 16, 9, 7, 37)
cell_1 = (3, 7, 15, 9, 6, 38)
So they actually share a lot of properties.
The d in cell_0 is equal to (x + n) from cell_1. d + n is the same for both (meaning they share the big square (d+n)^2).
The n in cell_1 is equal to the a in cell_0 and the a in cell_1 is equal to the n in cell_0.
The difference between n in cell_0 and cell_1 is the same as the difference between d's and thus a's and b's for cell_0 and cell_1.
I'm wondering if we might know enough to start deducing the properties between them. Like a fancy ass math sudoku. I'm thinking towards some kind of elimination system, but it's been years since I actually used any of those.
As always, I might be off my rocks ;)
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bead22 No.6030
>>6029
Also they share the same x, I think that might be important.
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de8bb2 No.6032
>>>/qresearch/1434362
This just came through on qresearch, some Minecraft: MA - thought of you!
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8ed7ca No.6034
>>6027
MM, thanks for the trip down memory lane!
>>6026
Slight bug in the triangle generation code for the previous example. Think that has been worked out.
Attached are x+n=17 images showing all records for positive e values and n >= 2.
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8ed7ca No.6035
>>6034
And here is x+n=45, n=6 showing the e=e+2n movement between records.
>>5977
>What is the pattern?
Time to find a pattern!
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33ce33 No.6038
>>6027
>>6028
>>6025
Good to see you all. I have been traveling quite a bit and didnt always have a good signal. I have been keeping an eye on this when I could though. Finally enjoying spring in the Pacific Northwest which is my favorite place this time of year.
I know what you mean GA about the coincidences. I really do. I always tell everyone that I meet that everything is fake. Seems like you have come to the same conclusions. Something is wrong with the world. We all know it. Its right there, just below the surface and we can't quite find it. Don't loose hope but be careful. There is so much happening right now I can barely keep up with it all.
I was thinking about going off the grid this summer in the Olympic National Park area or like a month but Geez I would come back and be completely lost. Maybe that would be good? Anyway keep up the good work here. This is one of my favorite places on the internet.
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87d422 No.6039
>>6038
Sorry I feel like a lot of this stuff is already known I'm just writing what I know for know about e+2n shifts. So when you shift with e+2n, you keep the same n and x values. Then your a and b both increase by one. Therefore your n-1 value stays the same as well as the nn value. f shifts down by 2(n-1) as you increase e by 2n. The (x+n) stays the same, so to compensate for the f shifts, the (2d-1)(n-1) also increases by 2(n-1). d increases by one each time also, and since n is the same, (d+n) increases by one each time. So the e = e+2n shifts to the right correlate with an increase in the (d+n) by one and keeping the (x+n) the same and also correlate with a=a+1, b=b+1. The opposite "move" would be to decrease increase (x+n) by one and keep (d+n) the same. To do this, you do a=a-1, b=b+1. To do this, you also need to ensure you have the same x value. For this, your n value is increasing by one each time. Then if you look at the f value it goes down then back up kind of parabolically as you do this and the vertex seems to be at the point where x=a=(n-1). I'm pretty sure that this gif of the xGrid (iterates through c values) is a visualization of the moves along this path, because both transformations involve keeping x the same. The x grid is also very easy to look at its pretty predictable I think that each little vertex is at (e,n) = (-x^2,0)
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5a1919 No.6040
>>6039
Permutations are sexy, I tell ya hwat
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7c7d29 No.6041
Hello Lads! I found something!
I found how to divide the mods evenly for both 8Tu and 1Tu, and where the center unit comes from. (at least for the example I'm working on)
Take a look and let me know if it works for other examples too.
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f4515c No.6042
>>6041
Here's a color coded version that's easier to follow.
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c6c148 No.6043
A couple more animated examples. This time for (x+n) = 13, and similar to the static images posted in the last thread (>>5755 and >>5756).
For the (f-1) version, the orange center square comes from (f-1), except for f mod 8 in (1,5) where the green center square comes from nn.
For the (n-1)+(f-1) version, the red center square comes from (2d-1)(n-1), and for f mod 8 in (1,5) the green center square comes from nn.
In addition, the "groups" of triangles have been highlighted in the middle.
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c53994 No.6044
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d88a7c No.6045
>>6043
Ok. let's break this excellent PMA image down.
(7,8,15) {7:8:82:29:53:127}
nn = 64
nn div 8 = 8
nn mod 8 = 0
(2d-1)(n-1) = (164-1)(7) = 1141
1141 div 8 = 142
1141 mod 8 = 5
(n-1) = 7
(n-1) div 8 = 0
(n-1) mod 8 = 7
(f-1) = 2d-e = 82*2-7 = 157
157 div 8 = 19
157 mod 8 = 5
mods total: 0 + 5 + 7 + 5 = 17
(17-1) / 8 = 2
so each 1Tu gets 2 units of mods, middle square gets 1 unit
Each 1Tu gets 8 units of nn
Each 1Tu gets 142 units of (2d-1)(n-1)
Each 1Tu gets 0 units of (n-1)
Each 1Tu gets 19 units of (f-1)
So for each 1Tu, the sum is:
8 + 142 + 0 + 19 = 169 for the div 8's, PLUS 2 units from the mods ( 17-1 div 8 ) = 171
171 * 8 + 1 (for the middle unit) = 1369
SQRT(1369) = 37
x+n = 37
Yeah, that works!
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4087ee No.6050
I haven't posted all week but I've been here every day, usually several times per day. Life is super busy. Just a suggestion/reminder for those of you who are using code for this: don't be like me and have tons of commented-out code that you work around. I just thought I'd stumbled on something extremely significant that correlated prime as and bs with a particular pattern but then I looked back at my code and realized the only reason I was only seeing prime as and bs in these places was because the grid was being generated within an if statement that runs if isPrime(a) && isPrime(b), and I didn't see it because of all the commented-out code. It's well past midnight so I could have gone to sleep already if it didn't seem significant. Oh well. Keep on keeping on everybody.
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aa8506 No.6052
Pics attached for a random example c19367 and it's na transform to c321535 where F takes up the entire (x+n)(x+n).
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79f6d5 No.6054
>>6050
Haha. Happens to everyone who writes code. Made me laugh out loud reading about it. Cheers.
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ebc838 No.6055
>>6050
Make git your friend. Commit often.
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aa8506 No.6057
>>5886
>hundreds and hundreds of bitmaps. And tables.
Still working on analyzing these bitmaps and tables and trying to find patterns that will enable larger n movements than n+4.
Depending on f mod 8, the middle (n-1)+(f-1) "square" can be assembled using either 1 or 2 different sized triangles. And sometimes a remainder that is different between the triangles by 1.
The assumption here, is that if we can determine the formulas to construct these middle squares, those formulas can be used to scale into the full x+n square we are looking for.
The latest thoughts in narrowing down these rules revolves around analyzing mods for the various components of nn + (2d-1)(n-1) + (n-1) + (f-1) formula.
Attached examples for x+n=29 show the mod analysis and animated squares grouped by f mod 8 values of 1 and 5.
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aa8506 No.6058
>>6057
And the same mod analysis and animated squares for x+n=29 and f mod 8 values of 2 and 6.
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6bf400 No.6060
>>6057
>>6058
Here's my working "code"
If (f-1) mod 8 = 1, Then (n-1) mod 8 = 3, (2d-1)(n-1) = 5. nn mod 8 = 0. Total mods = 9
If (f-1) mod 8 = 3, then (n-1) mod 8 = 3, (2d-1)(n-1) = 3, nn mod 8 = 0. Total mods = 9
If (f-1) mod 8 = 5, then (n-1) mod 8 = 3, (2d-1)(n-1) = 1, nn mod 8 = 0. Total mods = 9
If (f-1) mod 8 = 7, then (n-1) mod 8 = 3, (2d-1)(n-1) = 7, nn mod 8 = 0. Total mods = 17
So let's test.
Because if this idea is correct, we can iterate (n-1) based on the mod of (f-1)
Again, I could be wrong. My mind is in locked mode, so let's prove this idea wrong and move on to the next one.
I think tho that the mods give us a big CLUE to what fits!
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aa8506 No.6063
Just posting a little more artwork.
These are the odd x+n squares for the a=1, b=c records for prime numbers 307, 347, 431 and 463.
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aa8506 No.6064
>>6057
>>6058
>if we can determine the formulas to construct these middle squares, those formulas can be used to scale into the full x+n square we are looking for
Attached pics further the f mod 8 analysis mentioned in the previous posts.
For each f mod 8 value of 1, 2, 5, and 6, these tests group a starting c record with it's solution prime record to enable a comparison of the mod values for nn, (2d-1)(n-1), (n-1), and (f-1).
Based on these tables, a consolidate mod breakdown has been created that may show the various possible triangle combinations needed for a more accurate iterative search. (fmod_analysis_wip4.png)
For example, if the c starting record can be described as f mod 8 = 1 and (n-1) mod 8 = 0, it's possible that the solution prime record can be found with one of two triangle combinations:
4T(u1) + 4T(u2), or 8T(u1)
Still to be determined if the (n-1)+(f-1) mod accurately describes the triangle ratios, and if so, the correct u1 and u2 values to use.
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5a1919 No.6065
Before I forget, this came about because voice chats and fishing trips.
Zipping and unzipping DNA.
something something.
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5a1919 No.6066
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5a1919 No.6067
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bf9933 No.6068
Confirmation between 17:30 and 19:00 Brit Summ Time tonight.
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40a72d No.6069
Hello! This is the new anon that made the beginnings of a Qmap based on the Archimedes prime wheel. I'll be digging quietly in the background, trying to catch up. I was planning on finishing that map as well, though it will take some time. Thus far the progression along the spokes seems to flow pretty well, especially when paying attention strictly to the prime-numbered posts.
This work is interesting as well-the diagram with all of the definitions cleared up a ton of questions :) hopefully I can get enough of a grasp to at least gain some understanding of the work, though I suspect those here will have it solved before too long.
Again, thanks for the invite :)
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ed6375 No.6070
>>6069
Hey, welcome. You're obviously referring to someone else, but I'll just say if you're reading through everything on the board, I put a pdf together that's floating around somewhere that was meant to be an introductory explanation for people such as yourself (and if you're paranoid about pdfs, I dumped all of the text in one of the other threads too). It's actually very out of date (several months) and some of the information is actually incorrect (some of it was based on someone else's incorrect code), but if nobody else does it I plan to redo it when I have the time. If by the time you're caught up you have any suggestions as to how we could explain it better for new people, please feel free to let us all know.
>>6068
Eh? Something's happening in 6 or 7 hours?
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5a1919 No.6071
>>6069
Welcome to the VQC ;)
>>6070
I gave 'em the invite last night.
They'd noticed the fishing expeditions and reversed the process.
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9c4dc7 No.6072
>>6070
I'll find it, it's probably what I'm looking for–thanks :)
>>6071
Thank you!
I'm also in college, but mine runs a bit later. I've got more than a full load in my last semester before transfer, but thankfully three of my classes are just fluff. So I'll try to manage the balance as best I can…unless this is dire? As in, more dire than "Honor thy mother and father" and how they've been pushing me to get a degree for over thirty years?
Here's that starting map, based on your instructions.
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d88a7c No.6073
>>6068
Alright! Look forward to it VQC :)
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d88a7c No.6074
>>6072
And welcome, Anon! :)
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5a1919 No.6075
>>6072
WE HAVE A NEW CA!!!
Whatcho major is?
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9c4dc7 No.6076
>>6075
Software Engineering :) I'll be transferring from a community college. I think they have quantum "equipment" there; it isn't associated with my major, but there's a lot of collaborative work done between various groups.
>>6070
Are these the .pdfs?
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075c7a No.6077
>>6068
This should be fun :)
>>6069
Welcome!
Hi guise
Silently lurking, kinda burnt out on programming for a while, although following along with the amazing progress last two threads. Would have posted earlier if I had anything, you've been killing it! Congrats on your graduation CA (or is it GA)!
Will try to get my head in the game again
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c53994 No.6078
>>6068
Looking forward to it :-)
>>6069
Welcome aboard!
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4aff88 No.6079
I cannot say too much more.
Put the two together and you will suddenly see.
Combine the diagrams you are making with the grid (The End).
As you move left and right.
Up and down.
When moving across keep x constant and move with steps of 2n.
When moving up and down, keep d constant. Two factors, three factors, four.
What do you see?
How does row one (n=1) relate and determine the patterns?
Godspeed anons.
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4aff88 No.6080
>>5847
>>6070
>>6074
>>6075
>>6077
>>6078
Thank you for lighting the way anons.
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075c7a No.6081
>>6077
Accidentally used compromised trip (hopefully someone would've called me out on this).
This one should verify against
>>4388
Older unposted image as extra sauce
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c6c148 No.6082
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075c7a No.6083
>>6082
Trying to catch up on the code side of things, any chance you can share some of your functions for coloring squares programmatically?
From the images I'm guessing you fill each of the eight triangles separately and use some hackish transforms to map them to coordinates in the final image, but your ascii versions look more like arranging mod values in a grid and coloring those afterward.
If its the latter case, what do your loops look like and how are you distributing for example the n values to the outer edge?
>>5851
Assuming this is still the technique you're using, the challenge seems to be filling in/drawing each triangle evenly within a square grid. The actual white triangle lines look like just a graphic overlay, so which triangle does 1,1 belong to in the first two images?
Amazing work!
Hints (or code) welcome - I'd rather not start from scratch on the diagram stuff
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075c7a No.6084
>>6082
>>6083
Best I can come up with is that you're drawing each triangle somewhat like this, starting from the center, coloring cells for each variable, and stitching them together (or transforming coordinates in the drawing code directly). Seems easy enough.
For the displaced portions, it looks like there just distributed to every second triangle for symmetry purposes, until you run out.
If there is a way to draw the whole thing in one go that would be even easier but I'm not seeing it
Should be enough to get started on coding anyway - hints are still welcome
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de8bb2 No.6086
>>6063 Tiger Anon! ty, watched your .gifs dozen times.
>>6065 Fisherman Anon, noice. I'll have some BAKED trout please. Your baker skills were impressive recent days.
>>6068 Bring it V! >>6079 ty
>>6069 Welcome NA!!! New Anon, Next Anon. Namefagging and tripfagging is fine, though your images and work clearly have a signature already.
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de8bb2 No.6087
>>6086 whoops, partial post there.
Have had a PRIME day anons. Feels good.
>>6081 3DA, so good to have you back in the fray!
>>6060 great keeping at it VA!
>>6078 IC, i see u, patterns will emerge now.
>>6070 you're awesome. What a team member.
GLH and anyone else I missed that's lurking, all the best to you.
>>6068 hope your travels went well. Looking forward to it.
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41df4a No.6088
>>6079
If I wasn't neck deep in exams it doesn't seem like this would be horrendously difficult to implement with JFrame or some other GUI library. Even if we didn't use math to generate the grid cells when you move around, we could just use the hash map from The End for the variable values. Who's up for it?
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6bf400 No.6089
Got placement for the 8 mods.
(I think)
Here's c85527 at (f-1) * 2 with mods included. Think I found a good location for them. It makes all gaps on the border = 40.
Split evenly to each 1Tu.
Here's my working diagram. Comments and corrections appreciated!
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6bf400 No.6090
There's more to explain.
For this example, (f-1) div 8 =40.
So, nearest square to that is 36.
composed of T(5)=15 and T(6)=21 = 36
Tu remainder is 4. so 36 +4 = 40
so the light blue are 4 units added back in for each yellow and green square of 36.
Mods are in red, added at the center unit and at the corners and midpoints, creating gaps of 40, which our Tu remainder can fill (4 * 10 )
Not sure if this is right, just pushing my mind to explore possible answers.
Thoughts, Anons?
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4bf7c5 No.6091
>>6090
The gaps on the edge are 40, as well as the gaps from the center to the edge. This would also support the pinwheel idea (uh oh).
So the mods are the mods for the (x+n), and (f-1) ?
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5ebae7 No.6092
>>6091
Hey GA! Yes, the mods come from (f-1), (n-1), (2d-1)(n-1), and nn. Here's my update. Trying to find a way to fill the x+n square with only multiples of (f-1) + mods. Last diagram is incomplete, running out of gas over here. Bed time, but back tomorrow.
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5ebae7 No.6093
>>6091
Any comments, corrections, or new ideas appreciated!
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5a1919 No.6094
>>6091
So now that you're a gradutard and have time…
Wanna join us on discord?
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9c4dc7 No.6095
>>6072
I think most of my graphic work is done. There was no way I was going to squeeze 1433 posts onto four axis…
Hopefully the other anons will take it and run so I can focus on these matters :)
Here's the link to all of the posts, grouped by mod 12 (if you anons hadn't already figured it out):
ray 1: https://pastebin.com/jS6VDmtn
ray 2: https://pastebin.com/SfXpmSeS
ray 3: https://pastebin.com/TkSH7RtQ
ray 4: https://pastebin.com/U2sRwFV7
ray 5: https://pastebin.com/43tC0A4v
ray 6: https://pastebin.com/QhNPCj1B
ray 7: https://pastebin.com/cQGsAg5V
ray 8: https://pastebin.com/vHdMp9Cw
ray 9: https://pastebin.com/T1kcHNrT
ray 10: https://pastebin.com/ReYHR8qS
ray 11: https://pastebin.com/u1Zn5T2J
ray 12: https://pastebin.com/Z5DgS20G
I'm overdue for sleep. God Bless you all :)
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c53994 No.6097
I'm still not sure about the square graphs I've been making so I decided to go back a few threads and just read up on what VQC has been saying and posting.
I was thinking about the diagram he posted in >>4322 in which each of the triangle has a capstone equal to (n - 1) (or is it the base of the capstone?).
So I decided to try and make a square using the equation:
nn + 2(d-4)(n-1) + 8(n - 1) + (f - 1)
Attached is an image of the square. To explain it:
The red part is 8(n - 1), the orange parts outside of it is multiples of 2(d - 4) and the yellow one is an attempt at trying to use f-1 as a base similar to what he suggested in >>4342 and >>4344.
Outside is still more multiples of 2(d-4) and the green is n*n.
The green next to the capstone was empty after filling in the 2(d-4) inside of the f-1/8 base and could have been filled in by one of the outer 2(d-4) (which would place 4 green blocks at the (x+n-1)/2 base). Not sure If it makes sense yet, but I'm going to keep trying a few more squares with this pattern.
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c53994 No.6098
>>6097
Again, but this time for a=7, b=37. Also the previous one is not f-1, I'm using f-2 so the result is 8*T(u) and not 8*T(u) + 1. Same for this one.
So with a=7, b=37 our base of (f-2)/8 will consist of 3.5, meaning 4 of the triangles with have base = 3 and 4 will have with base = 4 (asymmetrical).
This will cause our f-2 to overlap with 8*(n - 1) and as a result we will have leftovers in our triangle.
I *think* this might be what VQC is referring to in >>4337. Since I'm using 2 * (d - 4) * (n - 1) the last d to add (since n - 1 is odd by the fact that we're after n's that are even) will be (d - 4).
In this example (d - 4) % 8 = 4, which is also the amount of (n - 1)'s that have been displaced by (f-2).
Legend:
red = 8*(n - 1)
oranges = different multiples of 2*(d - 4)
yellow = triangle created by (f-2)/8
green = n*n.
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c53994 No.6099
>>6098
Also note that the 2*(d-4) displaces some of the n*n (green) which is filled in on the opposite side of the square.
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c53994 No.6100
>>6098
With regards to >>4337 note also that (f-2) % 8 = 4, (d - 4) % 8 = 4 and (f-2) % 16 = 12, (d - 4) % 16 = 12 in the event that this is actually what he is referring to.
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c53994 No.6101
>>6100
Last one for now. This is a=55, b=129 which contains 4 factors as a = 11 * 5 and b = 3 * 43. In this case a lot just lines up. 2*(d - 4) / 8 = 20, f-2 / 8 = 16 so there are no remainder.
The only interesting part is that the amount of units of d for the last d is equal to (n-1), but this could be a fluke.
Legend is the same as above. Variations of these factors can (and should!?) be explored as this example contains 4 factors.
I'll see if I can make one this time where I do (f-2)/16 to see if I can get a two-wide unit of f-2 divided up here. Since this one has a base of 16, it might yield some more insight.
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c53994 No.6102
>>6101
Also! Optical illusion time, the inner-most 2*d which is the first 2(d-4) outside of the red area is the same colour as the 2*(d-4) that is on the same base as the n*n (green).
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5a1919 No.6103
>>6101
I dig where this is going.
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9c4dc7 No.6104
1/2
So I've been reading through the archives, and maybe this is the right place to ask.
Let's say you know someone that, about twenty years ago, had a strange, life-changing experience.
This person had a recurring image of a sort of "map" in his head, with a general sense of directions. This map appeared in dreams at times, but really it was just a vague sense in the back of his head. There was also a message given, about the destination lying just beyond a famous place in TX history. There was also an image of a certain time.
There was a sense of the life being on autopilot–everything smooth and clear, just time and days floating by. This person received a call from someone close, and after speaking the second person revealed that they were having the same sort of experiences, although the first person noted that the second, while well-meaning, was the sort that would often emphasize with others so they did not feel alone.
About a month later, the second person flew in to visit the first person at his place, where the first lived with roommates. The second brought a bag full of weed and a liquid hallucinogen. All of the four involved partook of the first, while reserving the second for the "vision quest" that had been "planned" in the time since the phone contact.
The next day, the drive started when it "felt" right. The other three partook of the marijuana, while the driver stayed sober. All of them noted how everything just seemed to be "falling into place" as if it was "meant to be." This person drove in the direction that he had "seen," continuing onward until about 30 minutes before the time that he had also "seen." At this point, the driver saw a sign (literal) for a turn that would have lead in the right direction. The driver was terrified at this sign and did not care to follow, visions be damned. Not a quarter of a mile later, he was stopped by something in the road…which had humorous significance in the context of the evening. Reluctantly, fearfully, the driver turned back and went down the pathway–one he did not even know existed in the region he lived in–until the time came and they disembarked onto a dirt road, parking soon after in a place that seemed to have no real significance.
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9c4dc7 No.6105
>>6104
2/2
The four partook of the hallucinogen. The second person was the one that administered it, in the cold, in the dark. The driver and his roommates got weak doses, while the second person had taken so much that the other three (particularly the driver) spent a large portion of the night (and next day, and the day after) wondering if he would ever be the same. All of them had a fair amount of experience with this drug, and could gauge the strength easily. Suffice it to say that there were hardly any "visuals," let alone fractals.
Still, there was a heightened sense of things. The four stayed in the car, talking about things. They did this because they saw signs of possible danger from animals along the road. The driver guessed that the drug should be peaking about four hours after–there were some light visuals, and the company was good. In order to amplify the effect of an otherwise weak trip, the driver decided to turn on the radio to a "conspiracy"-themed program. In his genius, he thought this would be a good idea.
Permeated by the strange babblings of the second person, the situation did get intense. Someone saw visions of handprints on the windows–the driver just saw patterns of condensation, shifting somewhat, not unexpected…but the atmosphere kept building, and all were speaking of strange things. The radio signal was bad–it was a place where at least two signals were conflicting…this mix of the conspiracy show and other stations, along with a fair amount of static, was continuous. It had been on for some time–perhaps half an hour. The driver kept turning the volume up, wanting to heighten the entire effect…until suddenly, the radio signal went clear…and the driver heard something over the radio–a message that changed his entire life. He looked at the time–3:33 AM.
The message was delivered in a way that it sounded like an old radio commercial. The driver absolutely wigged out, asking the others if they'd heard it. They said they did, but later said that they'd only heard something. The second person spent the entire night trying to convince the others that they could speak with "them" if they just believed, just went over this hill. The roomates looked on with amusement, at times playing along and trying to be supportive. The driver spent the majority of the night in fear and worry–not because of the message, but because he felt responsible.
That message, in large part, drastically changed the driver's life–he spent a good portion of his youth trying to position himself for its fulfillment. This led to a spiral of self-destruction, frustration, and despair. That part isn't so important, except to say that he eventually realized that the few good moments in his life, those where he felt a sense of blessing piercing through the miasma of failures, were always associated with times that he had defended things associated with Christianity, despite not being a Christian. That led to faith and a pathway to a normal life, though he no longer interprets the message as having meaning that could be understood.
Their location was a famous one in the US, associated with conspiracy theories for many decades.
So maybe you have some insight–might this have been an attempt by evil to influence the driver? Maybe a joke that was played on him? A possible coincidence? Did the driver dodge a bullet when, given the opportunity, he did not choose to do something that felt it might have put him on the path to fulfillment because it felt "wrong?"
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88d253 No.6106
FOR NEXT BREAD
I just updated the archive for RSA #11. Here is the full thread.
RSA #11 —- http://archive.is/6fJVd
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88d253 No.6107
>>6104
>>6105
I'll take your bait. What was the message, Anon? Do tell :)
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9c4dc7 No.6108
I guess it's probably safe to say now. I've told this to people close to me, they'll know who I am…but only if this actually does happen. Or if we're all being hunted by biorobots for daring to challenge the cabal.
The message was "You will inherit the most powerful radio signal…a beacon for all mankind."
I took it literally, and spent the next decade trying to become a kind of rock star. Nevermind that I'm a natural introvert that doesn't like vapid people.
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88d253 No.6109
3:33
Mr. Tesla sent you a message…
Or Maybe Space Jesus…
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88d253 No.6110
>>6108
When we solve this, a whole new realm of possibilities will be opened to mankind. Someone else has already figured it out, we just happened to be the students who self-selected into this Math Challenge. Welcome, Anon. You self-selected your way into this merry Pirate Ship of Math Nerds, Artists, Programmers, and outside the box thinkers.
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79066b No.6111
"When the student is ready, the master appears."
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9c4dc7 No.6112
>>6110
Thanks, I appreciate it.
One of my biggest realizations was that I really don't know anything at all. What if it was something that happened after I died? What if it was willed to me and I was never told? What if I inherited it a nanosecond before I went down in a plane crash? What if everybody stops using radio in the future, I'm able to buy the most powerful radio transmitter for pennies, and I blast a message into space on behalf of all mankind? The way my life goes, it would probably be a hot mic message of me breaking wind in front of the console, thinking nobody was around to hear it. A nice summary of the human race for alien SETI a billion light-years away.
Besides, I really didn't try to fulfill it exactly as stated. If I really wanted to work to fulfill the prophecy, maybe I would've started by getting a HAM license or something. So that goes to show it really was about me trying to make it what I wanted it to be.
Those types of lessons are hard-learned…but hard-learned lessons are least likely forgot.
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9c4dc7 No.6113
>>6109
I'd never heard of any significance attached to it. That might have been around the time Google was just getting a foothold. It wasn't until many years later that I saw a trailer for "The Fourth Kind," but there was no significance attributed to it. I think I came to understand it as symbolic of the Holy Trinity–three separate same entities.
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4bf7c5 No.6114
>>6113
I'm trying to see what Q was saying. I may have found something. These are each the values for AB(1,c) AB(a,b) then a bunch of records for the (e,1) cell. It seems like if you go to the same X value in the (e,1) cell, your (x+n) value differs from the goal (x+n) by 1 (or at least a small amount). More help would be appreciated. Also new trip
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4bf7c5 No.6115
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5a1919 No.6116
>>6105
>>6113
>>6112
I read a few lines in a got a lil' triggered.
Consider this a red light.
Might not apply to you.
I'll keep reading.
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5a1919 No.6117
>>6112
>>6105
>>6113
As I thought.
Thank you for swinging by, but what you're talking about doesn't seem to be immediately related.
Please check out this section of the Q A.R.G….
I'm there from time to time… you can see my trip…
Might be more up your alley…
https://8ch.net/truthlegion/index.html
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aa8506 No.6118
>>6079
Initial work exploring the up and down movements.
Pics attached are two factor examples for c159, c291, c295, and c303 and represent both the (1,c) starting record and (a,b) prime solutions.
For these examples (moving up/down), e, f, d and c are the same. The n and x+n squares are different.
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aa8506 No.6119
>>6118
Here are c295 and c303.
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aa8506 No.6120
>>6118
>>6119
>two factor examples
Correction - four factor examples.
Two factor examples were posted previously. >>6063
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c53994 No.6121
>>6101
Algorithm suggestion time.
We use the equation nn + 2(d-4)(n-1) + 8(n-1) + f - 2 = 8T(u)
We start by making our (f-2)/8 base. Note it doesn't have to be 8, but it can be a multiple of 8.
Then we generate a triangle with (f-2) as base. Right now we don't know n. But we know we need to add 2(d-4)(n-1) so we start by adding two 2(d-4) leaving us with n = 2. (n-1) is now 1 so we add one red at the top of our triangle.
This gives us nn/8 to add outside of our triangle. We leave them be for now.
We add another 2(d-4) (n=3) and add another red square at the top. We continue to do so until 2(d-4) meets the red squares. At this point we've added 5 2(d-4) and will overlap with red squares if we add another. Instead we add 2(d-4) outside of our (f-2)/8 triangle. Keep in mind we can now add nn/8 outside as well.
We continue until our last 2(d-4) collided with green squares (nn/8). Then we use the remainder of that 2(d-4) to fill in the missing center squares.
Makes sense?
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43245e No.6122
>>6105
>>6104
>"someone" (you're anonymous here, there's no need to do that) kept seeing a map in dreams etc and felt like there was some significance to a particular place in Texas
>someone else they knew also may have had the same thing happening
>they go to the place, timing it based on intuition
>something that wouldn't happen in any other place at any other time happens
I don't know what your beliefs are, but considering you're here on this board, you probably realize there's a great deal of information about multitudes of things throughout the world and life in general that you aren't aware of. I could explain it in great detail in terms of occultism, but I don't know how much that would help you. I can if you want, or I could recommend some books, but either way, if the interpretation of someone who has been in similar crazy situations like that means anything to you, I would tell you that intuition is very real (as you seem to be aware), as are a lot of other non-physical phenomena (which have been studied scientifically but not talked about very much), and if you're looking for a rabbit hole to jump down, look up the declassified documents from Project Stargate (it was a CIA operation involving remote viewing, which people can use to find physical information in a non-physical way, and one person in particular, Ingo Swann, had an accuracy ratio of 40,000 to 1). With your situation, it doesn't sound like you're going to know what the message meant until after whatever's meant to happen happens. You were given a message, right? That means the source of the message knows its meaning, but not you. Applying meaning to it (like thinking it's related to being a musician, or thinking it's related to HAM radio) is letting your left-brain reasoning attempt to justify your experience rather than just continuing to listen to intuition and allowing whatever's going to happen (regardless of how you try to force it to happen by interpreting it) to happen. So obviously if this was 20 years ago you'd expect it to have happened by now, but if you weren't given a timeframe, it could well happen right before death even. Have you had any other intense intuitive experiences like that since? I mean I obviously don't want to derail the thread, but if that's the only time anything like that has ever happened to you, there are a few things you could do that might potentially help you figure it out.
I may regret telling this story (in case one of the people I told reads this and figures out who I am some day), but it might make you feel less alone or whatever. It's a bit less crazy than that, but it's the same kind of process.
>do some automatic writing over the course of a couple days on a whim
>some stuff comes up about needing to take intuition more seriously, and that significant things involving another person were going to happen quite soon
>just sort of leave it without giving it too much significance
>meet someone one night a few days later
>it felt like we were meant to meet, but I'm always skeptical of those feelings so I kept it from influencing how I act
>we talk briefly, then we go our separate ways
>seemed like a regular interaction
>the next day I get a very strong urge to be in a particular place at a particular time
>consciously I think it's stupid, but I've had similar things happen a bunch of times so I just go with it since nothing was going to go wrong
>have no idea what, if anything, could end up happening
>waste probably an hour of my day going there and waiting around, feeling like an idiot
>suddenly I feel really attentive for seemingly no reason
>that person and I walk past each other
>suddenly I feel like I can go about the rest of my day
>have no idea what the significance was since I waved hello and they just sort of ignored me even though they were looking at me
>we do almost none of the same things or go to any of the same places, so the chances of walking into them weren't very high
>the next day
>the exact same thing happens, but in a different place at a different time
>it continues for 9 days after the day we first met, despite all statistical reasoning
>start getting a bit frustrated considering I have no idea what the significance of this is (like, are we meant to do something together, or what's the deal)
>talk to a clairvoyant friend (believe what you will; the point is he claims to be clairvoyant and a lot of crazy shit has happened to him too)
>he deduces that this person and I were close in a past life and that some bad things had happened to us
>he does some energy work for us (keeping in mind that I hadn't actually talked to this person the whole time, just walked past them without them paying attention to me, so they weren't aware that we were doing this)
>suddenly I never see this person again
Long story short, intuition led me to some other significant shit right after that that I'm still working on at the moment (and some of you might find this entertaining, considering what else we've talked about here: there was some stuff in the automatic writing about Agartha/the hollow Earth thing).
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Post last edited at
43245e No.6123
Bump
>>6088
I feel like a bit of a prick for bumping this without actually doing any work on it myself, but this seems like it would be significantly easier than generating a whole heap of images by picking cells from the spreadsheet. Not to mention, as VQC said right here:
>>5930
>two panes side by side
>on the left will be a bitmap displaying the x+n square
>on the right will be the grid (The End)
>hovering the mouse on a pixel of the grid with entries for that e,n will animate through the x+n square in that set
My interpretation's a little different to his but I was thinking you could click on a cell in the grid, then go left and right and see the corresponding x+n square.
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c53994 No.6124
Anyone remembering this little (huge) crumb by VQC:
>>2990
I think this is about to become relevant as it relates to (n - 1).
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6bf400 No.6125
>>6114
Uh hey guise, I may have figured out the answer! Can I please get some eyes on this? Hey GA, this output is excellent. This is that na transform idea of being able to find the connections between (1,c) (prime) and (e,1). I'm studying now.
I have a few questions:
column 'bs' - this is d+n, correct?
column 'ls' - this is x+n, correct?
Here's the path I'm seeing:
Start at (1,c) a=1 b=145
na transform to (e,1) using d+(n-1) = 61-1+12 =72
So now we go to (1,1,12). d=72, a=61 (from na in (e,1) )
NOW CHECK THIS OUT LADS!
d+n squares for (1,c) and (e,1) match! = 73
x+n for (1,1,12) = 12
x+n for prime = 12
Now we construct the (prime) element.
(Prime x+n)^2 = 12^2 =144
c=145 + 144 = 289 = c + (prime d+n)
we know prime d = 12, same at (1,c)
289 = (17^2) = (n + 12) solve for n
n = 5
BOOM!
So basically (1,c) and (e,1) share the same (d+n) squares. We use the na transform to move from (1,c) to (e,1).
Then, (e,1) and (prime) share the same (x+n) square. So now, we work backwards to construct the prime element.
We have the correct (x+n) square and the value of c. so we do (x+n) + c = (d+n).
Then we use our d value of 12 along with (d+n) to isolate n.
BOOM!
Let's run some more test cases and verify!
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4bf7c5 No.6126
>>6124
We want a(t) = na in (e,1).
(capital letters mean letters in the (e,1) cell. lowercase is (e,n))
n = (xx+e)/2a, so na = (xx+e)/2
D = na + x = (xx+e)/2 + x = (xx + 2x + e)/2
A = (xx+e)/2
D-d = (startN - 1)
D = startN - 1 + d
na + x = startN - 1 + d
na = startN - 1 + d-x
na = startN - 1 + a
na - a = startN - 1
(n-1)a = startN - 1
Is this a recursive solution?
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4bf7c5 No.6127
>>6125
This looks good, yes bs is big square, ls is little square. I think the only problem with what you have is that VQC said n is a factor of D-d, not that n equals D-d, but I'm sure we are close. smh the keys was 4 threads ago
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5ebae7 No.6128
>>6127
>>6114
Thanks GA. Also, I noticed your examples are both odd and even for (x+n) squares. Since we've been working on odd squares, just wanted to let everyone know.
It seems like a +1 offset is needed to reach the correct x+n square in the odd examples. Maybe adding back in the -1 from (n-1)???
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4bf7c5 No.6129
>>6128
I think that the value that determines what the correct x is the GCD of D[t]-d%(n-1) and A[t]%n. If the GCD is not 1 or 2, then it is the correct X
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9c4dc7 No.6130
>>6122
I appreciate your advice. I am Christian, firmly so, after having explored most other faiths earlier and seeing correlation with life and Christianity (and its forbears). For the record, I was raised with more of a Buddhist influence.
I have had other experiences, most notably a recurring nightmare that plagued me for two years after watching "To Serve Man" (Twilight Zone) when I was around ten or eleven. But I have derailed enough, and will now stop.
I will be getting into the math soon–"What is The Grid?" makes getting into this especially clear (can't find the other .pdf I thought I'd downloaded). Is there any reason why this can't/shouldn't be coded in C++? I'm more comfortable with that, and given that my computer is over ten years old, I'd like to reduce the overhead as much as possible.
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c53994 No.6131
>>6129
Check out https:// math.stackexchange.com/questions/30458/modular-arithmetic-solving-ax-b-equiv-c-pmod-d
I don't have time to dig into this now, but I'll spend the weekend looking into it (and other related modular arithmetic). It sounds like it could be the sweet spot.
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4bf7c5 No.6132
>>6131
Heres an iterative way to show the logic
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4bf7c5 No.6133
>>6132
Also the GCD will be equal to A if it is the correct X
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4bf7c5 No.6134
>>6133
forgot my other trip
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27c27f No.6135
>>6130
It can be coded in C++. That's what the original code VQC posted was, so you'll find it if you go back far enough. Some of us just like Java more, so we rewrote it. Speaking of which, one of the errors with the pdf was in the f input. It had some weird series of variables being input into the grid starting with f. That was a transcribing error on the behalf of the anon who first turned all of that into Java that we've fixed now. e is the distance between c and the square below it. f is the distance between c and the square above it, so what we're actually doing with f is inputting c in negative space (so if you've seen any of those symmetrical-ish bitmaps, the negative space on the left is the same variables with e replaced by f and n replaced by n-1). Just thought I'd clear that up in case you didn't figure it out.
>>6123
I feel like I'm talking to a brick wall.
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4bf7c5 No.6136
>>6132
>>6133
>>6134
Basically, at (e,1), there is a cell where the GCD of A[t]%n and (D[t]-d)%(n-1) is the correct a value. This is also at the correct X value. Moreover, this A[t] is equal to na and D[t] = na + x
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4bf7c5 No.6137
>>6131
I learned this in Number Theory. Another modular thing I'd recommend to look up is Chinese Remainder Theorem which is very powerful
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6bf400 No.6142
Better screenshot on this post! Ok, here's my work in progress. It may seem simplistic, but please bear in mind that every number here is derived starting ONLY from c. Our (e,1) record gets us close enough to iterate. (maybe why we spent so much time learning to iterate?) Anyhow, just trying to move the ball forward. Comments and corrections welcome and appreciated!
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6bf400 No.6143
>>6142
I am 100% committed to this quest, and will continue until it is solved.
Feeling tired and negative energy tonight.
But hopeful that positivity will return soon!
Let's unite our minds and souls and spirits in a request for Divine Guidance.
Savvy?
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dc0c2e No.6144
>>1538278
From main reseacrh board. What is this? If you lot here can't decipher then what hope of the rest of us anons?
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dc0c2e No.6145
>>6144
>>1538278
Sorry, hope this link works
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e9d233 No.6146
>>6145
>>6145
>>>/qresearch/1538278
ftfy. Now you have the format for linking to posts in other boards.
>>6143 Love you no homo VA! If anyone has shown faith and determination, you have, ty.
I really wish I had time to crank on this atm, but can only follow along at present. Maybe will get some free time within next few weeks.
You all are making good progress from my perspective. Pulling old crumbs with a new perspective, being creative, driving forward. The animated .gifs are quite nice.
Just remember to have FUN! This is a QUEST. We aren't here for bitcoin or fame (I don't think).
And just so there is some actual content to my post, here is a crawled sitemap for RSA dot Com, if anyone wants to read up on security content. Plenty of links:
https:// pastebin.com/NAV3eknY
and this
>>>/qresearch/1536162 (bread#1930)
re: Cyber Threat Alliance (CTA).
Some technical details for latest IT security issues. For anons into security, found this of interest for background reading:
Lucrative Ransomware Attacks:
Analysis of the CryptoWall Version 3 Threat
https://www.cyberthreatalliance.org/wp-content/uploads/2018/02/cryptowall-report.pdf (59pp)
Their sitemap with some other content is available here:
https://pastebin.com/nS8d0bnC
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4bf7c5 No.6147
>>6146
Weird how the logo is the difference of two squares
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8418bb No.6148
>>6145
>>6146
http://dev.pywaves.org/aliases/
That is where that weird copy pasta came from that anon linked to. It looks like some crypto exchange? Sorry for the off topic.
>>>/qresearch/1538278
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9c4dc7 No.6149
Would anybody happen to have the underlying spreadsheet images with the correct output for the earlier problems? When breads get archived, for some reason a bunch of the images get tossed. Sometimes you get a thumbnail (as in the case of this one), sometimes even that is lost.
I'd like to see how my output compares to the right answer(s).
Thanks!
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9c4dc7 No.6150
>>6149
disregard, found it here:
>>29
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8070f9 No.6151
Ive been doing my own research, I havent been following the progress you guys have made since around december. Im not exactly barking up the same tree, but I believe it is definitely relevant to what you all are working on. I tried learning to program to work at it but I have no idea how to code for output outside of console so I've been just doing it the really hard way in mspaint
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c53994 No.6153
VQC has said that (0, 1) and (1, 1) is equally good to (e, 1). If we really are looking for a way of using the D[t]-d pattern as a shortcut then I think I understand why they are equally good.
Since the d-value increase when you move right starting from (0, 1) to (2, 1) by one and (1, 1) to (3, 1) we can simply remove e/2 (for even e) and (e - 1)/2 for odd e from d. That should allow us to use the D[t] in (0, 1) or (1, 1) since the difference in d from (0, 1) to (e, 1) (for even e) is e/2 and the difference in d from (1, 1) to (e, 1) (for odd e) is (e - 1)/2.
But that's about it. Still looking into the whole D[t]-d, but I feel like I'm regressing a bit to previous threads.
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4bf7c5 No.6154
>>6153
I don't really like it. If D[t]-d = (n-1)a and A[t] = na, then
D[t] = na - a + d
D[t] = na - a + (a+x)
D[t] = na - x
D[t] = A[t] - x
Obvious because this cell also has correct x so I don't really get it. He said pattern of divisibility so I'm going to look into that.
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4bf7c5 No.6155
>>6154
It looks like the GCD of these two values is always either a or b. Moreover, if the GCD[t] = a, then GCD[t+a] = a. And if GCD[t] = b, then GCD[t+b] = b. Since they are prime we know that there must be a record (if we keep going) where the GCD is a*b. Maybe that can help us??
The last two are where that value it is visible.
Also I just want to say again that if F is a square, the correct x+n is the square root of F, so maybe we need to fuck with the F's more.
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4bf7c5 No.6156
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5a1919 No.6157
>>6155
Lookit'chu gittin' sassy an' fuckin' wit Fs :D
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5a1919 No.6158
>>6156
Also…
I have no idea if these relationships were known and I'm just catching up or what it even means to move up/down/left/right outside of that one time I was doing it with the Trangle Number outputs.
THAT BEING SAID!
Here's whatever this is.
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5a1919 No.6159
>>6158
Aaaaand because couldn't let it sit like that.
I have no idea how the "c"s relate to each other off the top of my head. lol
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5a1919 No.6160
>>6159
N/m, found it. or… one… maybe.
Seems to be a consistent thing, anyway.
Again, might already be known.
(Green Part is related to "c")
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22f4b8 No.6161
>>6160
I've been thinking along the lines of the equation for d in (e, 1).
We know these:
- Odd e: 2t^2 + (e-1)/2
- Even e: 2t(t-1) - (e/2)y
And when we look into (0, 1) and (1, 1) (by subtracting the jumping from (e, 1) >>6153) we have the following equations:
Odd e: 2t^2
Even e: 2t(t-1)
But I don't know if the d in (e, n) has any similar formula. I've looked a bit into it, but I couldn't figure any generalized solution. At least not yet (Any hints VQC?).
Anyways. I've been thinking about the following equation (for even e):
2t(t-1) - d == 0 (mod (n-1))
Now we know that: 2t(t-1) % (n-1) != 0, but we know 2t(t-1) % (n-1) = d % (n - 1).
And that's all I got for now. I've been reading up on modular arithmetics, but so far I haven't gotten anywhere useful.
One thing I've been thinking about is that VQC has been hinting at using the grid to find a shortcut, which means this D[t] - d == 0 (mod n-1) isn't an actual answer, or rather the result won't be _used_ as an answer it self. We won't be able to solve it, but it should help us.
So I'm also looking into the idea of reversing his crumbs. We know that we're supposed to use this as a shortcut to end up with the answer. We're supposed to get an offset from D[t] - d, which means an offset to what and by working our way backwards, how can we see this offset in use?
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254092 No.6162
>>6146
MM! Thanks man.
>>6147
>Weird how the logo is the difference of two squares
Very strange indeed ;)
>>6153
>>6161
Hello Isee! I agree. Somehow (e,1) is "The One Row To Rule Them All" per VQC
>>6155
>>6156
GA, thanks for this output. I've been pouring over your last ones looking for connections.
Hello Everyone! Let's keep working on using PMA's excellent GIF's to connect to the grid to the x+n square patterns . After a lot of study, it looks to me like f is our main connection. And by extension d and e, since 2d+1-e = f
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f4515c No.6163
>>6119
This is my favorite math image so far from our quest. Thanks again PMA, our MVP for all your work!
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5a1919 No.6164
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4bf7c5 No.6165
>>6162
Yeah speaking of PMA's pictures, could you post your code so I could translate it into python? If not I'll try to hack it out myself but I was hoping you'd share :)
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4bf7c5 No.6166
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88d253 No.6167
Hello Senpai! Made these just for you. We love your bait ;) Thanks for Fishing to save the world with Q Team. Love ya. xoxo
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e9d233 No.6168
>>6167 Nice to see ya with chin up again VA!
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4bf7c5 No.6169
The values for D[t]-d are equal to 4 times the triangular numbers.
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e9d233 No.6170
>>6169 Go on! Your latest, along with IC's, has been very interesting. Great to be working the Grid again.
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4bf7c5 No.6171
Look at this. If we go to the record where (D[t]-d) is NOT divisible by (n-1) but where A[t] IS divisible by n, then the GCD of D[t]-d and A[t] is equal to a. Also, the X value for that record is equal to the correct D value for the record in this column that has the correct X value for the final record
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d88a7c No.6172
>>6169
Alright GA. Post the output to show your point, lol! Show us the relation between D[t]-d and the triangular numbers.
Are the T(s) based on (f-1)? Cause if so you got a winner.
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d88a7c No.6173
>>6171
Just saw your new post. Reading now.
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4bf7c5 No.6174
>>6172
nvm it was just for that one
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d88a7c No.6175
>>6174
Keep Going, GA! I've struck out plenty of times. Better to swing for the fences tho. This team is perfect for the job. Let's swing for the fences tonight, faggots. ;)
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4bf7c5 No.6176
>>6171
I'm gonna try and do it out by hand tomorrow
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b0e669 No.6177
Good morning Anons!
I don't know if someone from this board posted the original image several days ago, but I snagged it because it looked interesting. Anyway, I have been at qresearch since January, digging with the rest of the anons, and I noticed there were parts of Q's post that stuck out. i.e. clock, map, key, keystone, etc. I have always been one for puzzles, but unfortunately, because I am always workfagging, I haven't had much time to work through the posts and figure it out. I noticed something tonight, this may be something you all know, so I apologize if you do, but if you will notice the image, a set of numbers, mod 40, and all prime numbers line up on the same axis, I know, not too exciting. But what intrigued me was a pattern I noticed, If you take two individual prime numbers and multiply, the subprime derived from them seems to be offset by ten "rays" or axis. I tried this with several number and plan to test it out some more tomorrow. Again this may be common knowledge, but if one knows the cipher(subprime) and the public key, by deducing and counting back/forward on the "wheel", it would lead you to the other key(private)? I could be really tired, so sorry if you already know this. I have lurked the past few days, and I must say you guys have got it going on, and I hope I can join you all in the future!! I hope everyone has a great night, hopefully I get back on tomorrow. Thanks!
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6ccd20 No.6178
I wrote a quick and dirty program that generates the grid and displays the (x+n) using the equation: n*n + (2*d - 1)*(n - 1) + (n - 1) + (f - 2).
It's written in javascript and the zip contains 3 files. Index.html, index.js and theend.js. Index.html contains just the markup. theend.js generates the grid and writes it to the DOM tree as table rows (Horrible I know) and index.js draws up the x+n square using canvas.
I'm not much of a frontend developer and I have never touched canvas before so the code is quick and dirty. I only spend a few hours on it. To run it I recommend Chrome as some of the javascript is es6 and might not be supported by other browsers (although firefox probably works fine with it, but safari might be no-go).
There's a lot of room for improvement, but this is just something I threw together today so we at least have something.
To run it I've been using a static http-server. I'm not sure how it works if you just open it as a file, since it requires two javascript-files to run. I'm using http-server which is nodejs static webserver you can install by using npm, but that requires you to install nodejs.
Zip is available here: http://www.filedropper.com/stuff_27
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9c4dc7 No.6179
>>6177
I'm seeing the pattern, and it seems to pop up a lot–mostly when you multiply
two consecutive primes, but there are other instances.
Let's take a look at the consecutive ones:
1 * 2 = 2; 0 in either direction
2 * 3 = 12; 4 cw from 2, or 3 cw from 3
3 * 5 = 15; 10 cw from 5 (the larger)
5 * 7 = 35; 10 ccw from 5 (the smaller)
7 * 11 = 77; 10 ccw from 7 (the smaller)
11 * 13 = 143 10 cw from 13 (the larger)
13 * 17 = 221 10 cw from 11 (the smaller)
17 * 19 = 323 14 ccw from 17, 16 ccw from 19 //breakdown
19 * 23 = 437 22 ccw from 19, 14 cw from 23 //breakdown
23 * 29 = 667 4 cw from 23, 2 ccw from 29 //breakdown
29 * 31 = 899 10 ccw from 29 (the smaller)
31 * 37 = 1147 10 ccw from 37 (the larger)
With one prime between:
1 * 3 = 3 0 and 2
2 * 5 = 10 8 and 5
3 * 7 = 21 18 and 14
5 * 11 = 55 10 cw from 5 (the smaller))
7 * 13 = 91 4 and 2
11 * 17 = 187 10 cw from 17 (the larger)
13 * 19 = 247 5 and 12
17 * 23 = 391 14 and 8
with two primes between (skipping the lower ones)
3 * 11 = 33 10 ccw from 3 (the smaller)
5 * 13 = 65 20 and 12
7 * 17 = 119 8 and 18
11 * 19 = 209 10 ccw from 19 (the larger)
13 * 23 = 299 6 and 4
17 * 29 = 493 14 and 14
19 * 31 = 589 10 cw from 19 (the smaller)
23 * 37 = 851 12 and 14
…so what you mention pops up a lot more often that one would expect, but it
doesn't hold in every situation. It's interesting that in a lot of cases, the
offset from each factor either adds to, or is the difference of, the difference
between the two primes. I would say that it's also interesting that the
difference between the rays is always an even number, but that's made pretty
plain by the diagram.
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9c4dc7 No.6180
Hey everyone,
I just posted this over in /greatawakening, but someone replied with information that seems more applicable here.
I was just combing through Q's posts, trying to find a relationship between primes, when I noticed something: in #75 and #84, Q posts "Godfather lll" using three lower-case L's…but every time thereafter he uses three upper-case I's. With my internet font, I can't tell the difference:
l (lower case L)
I (upper-case i)
After posting that, someone linked this:
https://en.wikipedia.org/wiki/Lenstra%E2%80%93Lenstra%E2%80%93Lov%C3%A1sz_lattice_basis_reduction_algorithm
When I'd first came here last week, I was drawn by images such as that in
>>6177
which reminded me of a numberphile video I'd seen on youtube last week re: the golden ratio, how it happens to be the most irrational number, and how it therefore provides the most efficient "packing" of seeds in a flower, for instance:
https://www.youtube.com/watch?v=sj8Sg8qnjOg
How or if this relates, I'm not sure yet…but I'm pretty sure it's more than a coincidence.
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9c4dc7 No.6181
>>6180
/greatawakening = /qresearch
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5a1919 No.6182
>>6180
I checked the images before reading and as soon as I read the first one I was like
"Godfather…"
-read down a bit-
Yup.
4, 10, 20 was also coiiiiincidentally related here via Pascal Trangles.
Good catch!
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4aff88 No.6185
The construction of The End (The Grid) and the example of the virtual quantum computer and many of the connections to hints previously, all relate to P=NP.
What can be inferred from P!=NP?
Anything useful?
How about we start with P=NP?
Start with P=NP as an assumption.
There is a solution that has the complexity of the most complex step.
This is taking the root of c.
At most, O(log t) where t is the length of c in bits.
That means, like all solution sets of P=NP, we can work backwards.
And only this method works.
Since we know that the square root is the most complex step, we already have the best tool for working backwards.
We can EXCLUDE any ideas that increase the complexity above O(log t).
We exclude searches that do not zero in on a solution in logarithmic time, this is nearly all searches except those that have input that get smaller and smaller exponentially or equivalently, those that use recursion. Hints of this have been given previously.
We also know that we will end up back at the grid.
We also know we need to use input c of a reasonable size to show the pattern properly, again hints were given but too subtle.
The effort here have been incredible, fast and almost reached where we are going next. Anons are resourceful and brilliant.
The solution to this problem introduces a new form of algebra where two concurrent forms of equations run side by side and then merge.
Take some time to think about what that means.
Have you seen this approach before?
What could it look like?
Many new solutions often seem obvious in hindsight.
In fact many new designs seem to simplify in many varied approaches to design.
Does it seem obvious in hindsight that in order to solve a multivariate equation, that something new but similar to what we have always done, would be the solution? Just taken in a new direction? Expanded thinking.
The two sets of equations take the problem and simplify it. Together they handle the "lock and key" nature of the problem/solution, particularly when c is divisible by 1, c, and two other prime numbers, such as in RSA.
The two forms of equations that merge together handle staircase numbers where the base of one staircase, is one unit longer than the other.
The analogy to Fermat's Last Theorem. Two objects that seem to be completely unrelated were proved to be the same object.
Two seemingly different 'fields' will be used side by side and merged to create an elegant solution. Again, this can only be done in reverse, using the assumption that P=NP. I only saw at the end of over seven years work. Anons have got much further in six months than I would have. I would have walked away in frustration back in the day but you anons here have been amazing.
Take some time to think what this will look like and how the diagrams (especially animated) might show this.
I think things will start moving quickly.
This will be new mathematics. It will make more sense than how this problem has been approached up until now.
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45f639 No.6186
>>6178
Thanks for this, helped a great deal in getting my triangles in a row!
>>6185
>Anons have got much further in six months than I would have. I would have walked away in frustration back in the day but you anons here have been amazing.
Not for lack of helpful nudges - glad you're still with us. This is really exciting stuff and we're not going anywhere.
>The solution to this problem introduces a new form of algebra where two concurrent forms of equations run side by side and then merge.
>What could it look like?
>Many new solutions often seem obvious in hindsight.
>The two forms of equations that merge together handle staircase numbers where the base of one staircase, is one unit longer than the other.
Does this sound like differential equations to anyone else? Too obvious right? Just a coincidence I was reading up on them last week I'm sure… any way to combine those with say, a tree structure for the second form?
>Take some time to think what this will look like and how the diagrams (especially animated) might show this.
Got some working diagram code going, should be able to crank out fancy animations tomorrow. Here is a sneak preview of {63:1:63:7:56:72} to {63:58:63:45:18:224} in downward steps, for constant d.
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9c4dc7 No.6187
>>6185
VQC, I'm honored to be here.
I saw this post earlier today–obviously cause for concern. Are you okay?
>>>/qresearch/1565909
It happened at a time when the helpers were at play (I think Assange may have may have landed in the US recently).
I was worried I might never get the opportunity to thank you for sharing this with us. I know it's really not my place to say anything, given that I know nothing at this point. If this is true, then I can only assume that you know what is best…but in any case, may God bless and keep you well.
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43eac4 No.6188
>>6185
>What can be inferred from P!=NP?
That NP problems aren't solvable in polynomial time, you mean?
>Anything useful?
It would infer that our security algorithms are safe. If P!=NP was somehow proven, people would stop paying so much attention to NP problems, and we could all just go about our lives as they are now. Sounds boring to me.
>We also know that we will end up back at the grid.
Hopefully ISP's code here >>6178 works for that. I haven't tried it out yet myself.
>We also know we need to use input c of a reasonable size to show the pattern properly, again hints were given but too subtle.
Do you mean bigger than we're currently using?
>The solution to this problem introduces a new form of algebra where two concurrent forms of equations run side by side and then merge.
>Take some time to think about what that means.
>Have you seen this approach before?
Simultaneous equations and Fermat are the only ones I can think of.
>What could it look like?
I suppose what you're implying (and I remember you talking about this previously) is that two seemingly irrelevant equations end up equaling the same thing every time. The way you're wording it makes it seem like maybe you're not referring to c being the difference between two squares (x+n) and (d+n), but maybe you are and I'm just looking for crypticism where there isn't any. If you are being cryptic, that implies that we haven't quite gotten to this stage yet and that we'll figure out that one of the variables scales with another unrelated variable or something along those lines.
>Does it seem obvious in hindsight that in order to solve a multivariate equation, that something new but similar to what we have always done, would be the solution? Just taken in a new direction? Expanded thinking.
Well, yeah, I suppose. If every variable was completely independent of one another, you'd have to use several simultaneous equations to look into how the variables scale with one another and how they each change when one increases and the others stay the same etc. If there was something inherently linking every variable that we weren't already aware of that would allow us to solve it with one equation, firstly it doesn't sound like it would make any sense, but secondly it would obviously make things far easier.
>Take some time to think what this will look like and how the diagrams (especially animated) might show this.
I would think it would be this kind of thing >>6178 which I'm glad someone put together (thanks for the hard work Isee, and apologies for the complaining, anons, but I didn't think anyone was paying attention to that one VQC post I was bumping since nobody was replying).
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5a1919 No.6189
>>6186
Hoooooly fuuuuuuuck…
Veddy niiiice.
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80cdb1 No.6190
>>6185
Welcome back!
>staircase
You made me think of polite numbers…
Polite numbers are numbers that can be expressed as the sum of consecutive integers.
The set of impolite numbers are numbers that cannot.
The impolite numbers are the powers of 2.
The number of ways you can define a given polite number as a sum of consecutive numbers is equal to the number of odd factors (including 1 and itself).
Triangle numbers are obviously a subset of the polite numbers (whereby the sequence starts at 1), and a square is the sum of 2 consecutive triangle numbers.
An odd square is the 8 triangles + 1.
An even square can be reduced by dividing by 4.
Therefore the problem can be reduced down to differences of triangle numbers where one triangle is off by 1, and this difference of triangles can be expressed as polite numbers…
Am I at all close?
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5ebae7 No.6191
>>6185
Awww Yeah! Algebra Anons be like "Wut?!?!" This sounds like fun!
>The solution to this problem introduces a new form of algebra where two concurrent forms of equations run side by side and then merge.
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5a1919 No.6192
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5a1919 No.6193
HookTube embed. Click on thumbnail to play.
>>6185
Parallel Lines intersecting at infinity is what it makes me think of, off the top of my head.
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45f639 No.6194
Happy accident? I messed up my line drawing code and this happened. Too good not to post
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88d253 No.6195
>>6186
>>6194
3D Anon, these are AWESOME!
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88d253 No.6196
>>6190
I agree, Anon. The polite staircase numbers are our big hint from the last bread. The diagram I created found a way to use ONLY polite staircase numbers based on f to fill the (x+n)=83 square.
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45f639 No.6197
>>6195
Right? I should quit this P=NP stuff and start making screensavers instead!
Glad you like them - not even sure what happened in the last one but I did save the code. Some people here can probably extrapolate what order my triangles are arranged in just from those images.
Just don't count anything as accurate yet and we'll be fine, will check things before I start generating animations
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88d253 No.6198
>>6187
I hereby dub Thee Sir "PrimeAnon". If you like it, run with it. If not, come up with your own name and trip. Glad you're here!! :)
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88d253 No.6199
Also, how is PrimeAnon also the same ID 1bcbd2 as the guy who was talking about his crazy life changing experiences that seems to be an older Anon?
PrimeAnon, are you in back in college after your stint as an up and coming rock star? ;)
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9c4dc7 No.6200
>>6198
Thanks, I definitely can't pass up a name like that :)
>>6199
That's right–I tried school after graduating high school, realized there were no rock star classes, then dropped out and went straight to work. I never wanted to be in a rock band–I was more interested in doing what Trent Reznor did. Plus, I happened to work with a pretty famous DJ…so I taught myself how to make electronic music, from the ground up.
The lifestyle eventually caught up to me; I was trying to be something I was not. Long story short, I decided to go back to college, got sidetracked, resumed, and am about to transfer to finish off my bachelor's degree.
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7c7d29 No.6201
>>6185
>This will be new mathematics. It will make more sense than how this problem has been approached up until now.
So we had to run through the wrong approaches first. Fair Enough. Stoked to be here working with you all, and having fun!
Here's an oldie but goodie for this bread ;)
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4bf7c5 No.6202
>>6200
Avicii faking death???
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9c4dc7 No.6203
HookTube embed. Click on thumbnail to play.
>>6202
More like Tiny Tim faking life.
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f4515c No.6204
>>6203
PrimeAnon is WhiteHat for sure. STFU with your mismatched stories, lol! :)
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45f639 No.6205
>>6200
Welcome!
I don't know what happened to my 3D code but its generating some pretty cool shit now even with the buggy versions…
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9c4dc7 No.6206
HookTube embed. Click on thumbnail to play.
>>6205
Thanks–and you're right, it's still pretty cool :)
You guys want to the song that brought me closest to fame?
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d88a7c No.6207
Sure Anon! You gonna dox yourself? Agenices are here, but whatever since we know about Vault 7 & 8. Let's hear the song.
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4bf7c5 No.6208
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9c4dc7 No.6209
>>6208
>>6207
This was the one:
>>6206
j/k. I guess that kind of fell flat.
I'll share something some day, but truth be told nothing that I ever wrote was ever really finished. I have lots of 75% complete songs, but nothing was ever good enough…it's like one day, you finally managed to model that sound you keep hearing in your head…then a week later, you wonder what was I thinking?
I'm not the type of person that can pretend that something is great when I hate it.
I appreciate the interest, though–you've all been so nice. I look forward to making some meaningful contributions :)
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1354fe No.6212
what is the mundane?
more of the same,
the day to day?
Is this not all we have?
earth, rock, dirt and grass?
fire, water, streams of conscious energy?
the inward pull of gravity?
earth, rock, dirt and grass
allow the question to be asked
what is the mundane?
that gnawing in your veins.
iron, blood and oxygen
coalesce to reminisce on times when
all was just
star dust.
what is the miraculous?
if not more of the same
this day this day
children play in grass and dirt
spinning with the Earth,
some distance in an omniverse.
Distant far off galaxies,
delicate sweet gravity.
All of Us
Being
Miraculous.
Has not the Kingdome come?
Are not all things One?
Have we all grown so blind, deaf, and dumb?
The Ordered Chaos of it All
The sound a tree makes in a silent forest as it falls.
The wind beneath the eagle's wings,
The math behind the music when she softly sings.
Is this not all a single lovely harmony?
An infinite vibration of quantum strings,
an old hymnal's sunday morning.
Does Heaven not surround you?
mountain haze and morning dew.
sunrises in the rearview.
acquantaince and companionship
short springtime roadtrips.
How is
all of this
anything short of pure bliss?
so what of work, and hurt, and toil,
Are we more than wet, organized soil?
would you say there is a soul?
would you say you have control?
And what of precious time?
Where is yours, and when is mine?
How
is any of this allowed
to be so?
and what would it mean to truly know?
Can we really say
anything is mundane
when All of Us
Are
Miraculous?
Vincit Omnia Veritas.
Cras Es Noster.
Deus Vult,
Bless.
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51bf9e No.6213
>>6186
3D, These graphics are excellent!
>>6165
GA, happy to share the code/concepts for creating the triangle/squares. Actually had some fun putting that piece together. A lot of the code is dependent on other parts, so a bit difficult to break up and post in it's entirety.
>>6185
Very excited about where this is going!
>>6079
>When moving up and down, keep d constant. Two factors, three factors, four.
Have been working on this hint from a little while ago.
Ran into a few problems trying to find factor records of c in the original grid data, and the n,d,f querying code turned out to be a little too slow.
So this exercise turned into trying to figure out a way to iterate n and find matching records.
Well, it appears that you can't just "iterate" n. (Or at least I can't.)
For any x+n, there are many combinations of n,d,f that satisfy the nn+2d(n-1)+f-1 = XPN equation.
Simply incrementing n just tells you if it fits within a specified XPN.
Therefore, you need to increment BOTH sides of the equation.
Pics attached are samples of this two sided iteration process for test cases c147, c155, c1143, c1155, and c6107. And I believe it finds the appropriate factors properly.
Performance for this process is roughly 2 calculations for each odd x+n, and matching records are still determined based on the remainder 2d(n-1) formula being zero.
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d57947 No.6214
Stumbled upon Profinite Fibonacci
numbers.
Not a mathematician but this document is connected to a LLL algorithm author, an expert in cryptography (not published officialy). The diagram looked like what you guys have been working on. Hopefully you'll find something useful.
http://www.math.leidenuniv.nl/~hwl/papers/fibo.pdf
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4bf7c5 No.6218
>>6214
Classic post here, looking at the intermediate block that I was found here. Couldn't find stairs in it though
>>6171
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4bf7c5 No.6219
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4bf7c5 No.6220
>>6219
Also just noticed that for the last record in each section, the (x+n) is equal to the B from the one above it minus (correctX + 1)
36 = 44 - 8 = 44 - (7 + 1)
63 = 74 - 11 = 74 - (10 + 1)
224 = 244 - 20 = 244 - (19 + 1)
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51bf9e No.6221
Following is just a quick run through of how the triangle squares are being generated.
The sample squares attached are for c12103, n=2, but the same applies to all images.
First thing is the odd x+n square which has a u value, a center square, and 8 equally defined triangles.
Each triangle has a number of rows from 0 to u - 1.
And each of these rows has cells from 0 to the current row index.
The internal structure of the triangles are:
0: 0
1: 0, 1
2: 0, 1, 2
3: 0, 1, 2, 3
etc.
Sample code that creates each of these TriangleParts is attached.
Once that structure is created, setting the appropriate values in each of the cells is a matter of iterating through all points on all triangles in a systematic way.
See the SetValues code snippet attached.
First iterate each row from 0 to u - 1.
Then for each row iterate each column from 0 to the row index.
And then cycle through the triangles in the following index order: 1, 3, 5, 7, 2, 4, 6, 8
There are 2 breakdown objects that determine the type of value to set in each triangle cell.
First is based on nn + 2d(n-1) + (f-1), and the second is for the nn + (2d-1)(n-1) + (n-1) + (f-1) distribution.
In each case, the values for each of the components of these formulas are assigned to a triangle cell until fully depleted using the counter variable to track progress. Order of assignment is (n-1), (f-1), (2d-1)(n-1), and finally nn for n-1 breakdown.
Once the triangle cell types are assigned, there is another method that translates the triangles into a 2 dimensional square array [x+n, x+n]. where each triangle is rotated differently into it's appropriate position.
The final image is then rendered from this 2d square array.
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9c4dc7 No.6222
>>6177
I think what you've noticed here is likely related to pic related, which is taken from this wikipedia article on Fermat's factorization method:
https://en.wikipedia.org/wiki/Fermat%27s_factorization_method
It makes you wonder how the perfect squares would line up on a mod 20 disk in that vein; I'll be putting together some programs for tying some of these ideas to Q's posts as soon as this semester is over.
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4bf7c5 No.6223
This might be the widget VQC wanted
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4bf7c5 No.6224
>>6223
Trying to get it better. It freezes a ton
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c6c148 No.6225
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4bf7c5 No.6226
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254092 No.6227
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9c4dc7 No.6228
>>6223
That looks great :)
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4bf7c5 No.6229
>>6228
>>6227
Here, this is the code all together in one file and it doesn't crash as often anymore.
https://pastebin.com/QZmBgp55
Just run that file and it should work
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4bf7c5 No.6230
>>6229
https://pastebin.com/jf5EBtra
Added more functionality. Sorry for posting so much
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5a1919 No.6231
>>6230
Post m0ar!
OR JOIN US ON DISCORD AND INTERACT TO YOUR HEARTS CONTENT WITHOUT FEELING LIKE YOU'RE OVERDOING SOMETHING; MUCH LIKE I AM DOING UNNECESSARILY WITH THE ALL CAPS AND THE RUN-ON SENTENCE!
Also, June 11th, if anyone has eyes on that…
Is Kamehameha Day.
(Feel better?)
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4bf7c5 No.6232
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5a1919 No.6233
>>6232
https://discord.gg/cJJvDQa
What you do is:
Post half a something here with your trip…
And then message me the other half.
I'm the one in the G-Unit.
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4bf7c5 No.6234
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254092 No.6235
Hello Lads! I have 3 different ideas bouncing around inside my head.
1. Filling the square with polite triangle numbers created from (f-1) div 8.
2. Then trying to integrate nn+(2d-1)(n-1)+(n-1)+(f-1) into that idea.
3. Then looking for a way to tie the mod patterns back into either way of creating the square.
Thoughts, Anons?
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9c4dc7 No.6236
Came across some good stuff while I shirked my studies for finals–a ton of Wikileaks files that didn't quite make it into one of their large sets. Among those are a ton of interesting files. I thought this might be of interest here (or maybe /truthlegion, I haven't examined it closely yet).
Here's the archive:
https://file.wikileaks.org/file/
Here's the link that led me to the archive, with many other things:
https://steemit.com/wikileaks/@ausbitbank/how-to-download-the-wikileaks-archives-insurance-files-email-leaks-for-safe-keeping
Something tells me that a lot of you are already familiar with these…but for everyone else, enjoy ;)
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c53994 No.6237
I think I found a way to relate the equation:
n*n + 2*d*(n - 1) + f - 1 = 1 + 8T(u)
back to the grid. I was playing with algebra and I was looking into PMAs idea about using:
n*n + (2*d - 1)*(n - 1) + (n - 1) + f - 1 = 1 + 8T(u)
and I trailed off a bit, just playing with the equation.
Mind the algebra coming in:
n*n + (2*d - 1)*(n - 1) + (n - 1) + f - 1 = 1 + 8*T(u)
We know T(u) (Triangle) is equal to u * (u + 1)/2. This gives us:
n*n + (2*d - 1)*(n - 1) + (n - 1) + f - 1 = 1 + 8*( u*(u + 1)/2 )
We eliminate /2 by dividing out 8*T(u) by 2:
n*n + (2*d - 1)*(n - 1) + (n - 1) + f - 1 = 1 + 4*( u*(u + 1) )
Then we mix n*n + (n - 1) into (note -1 joined at the end and became -2):
n*(n + 1) - 1 + (2*d - 1)*(n - 1) + f - 1 = 1 + 4*( u*(u + 1) )
n*(n + 1) + (2*d - 1)*(n - 1) + f - 2 = 1 + 4*( u*(u + 1) )
I "unrolled" f into 2 * d + 1 - e:
n*(n + 1) + (2*d - 1)*(n - 1) + 2*d + 1 - e - 2 = 1 + 4*( u*(u + 1) )
Which gives us:
n*(n + 1) + (2*d - 1)*(n - 1) + 2*d - e - 1 = 1 + 4*( u*(u + 1) )
We move the 1 + to the other side:
n*(n + 1) + (2*d - 1)*(n - 1) + 2*d - e - 2 = 4*( u*(u + 1) )
Now we blend (2*d - 1)*(n - 1) + 2*d into n(2*d - 1) + 1:
n*(n + 1) + n(2*d - 1) + 1 - e - 2 = 4*( u*(u + 1) )
n*(n + 1) + n(2*d - 1) - e - 1 = 4*( u*(u + 1) )
n*(n + 1) + n*(2*d - 1) is equal to n*(2d - n):
n(2*d + n) - (e + 1) = 4*( u*(u + 1) )
We now divide by 2:
n(2*d + n)/2 - (e + 1)/2 = 2*( u*(u + 1) )
We move (e + 1)/2 to the other side:
n(2*d + n)/2 = 2*( u*(u + 1) ) + (e + 1)/2
And here it is. 2*( u * (u + 1) ) + (e + 1)/2 should be fairly recognisable as it is our method of calculating d's for even e (almost, d for even e is 2*k*(k+1) + e/2. I'm not sure if we can have an odd x+n with an even e and even n? I haven't found any, but I haven't looked super hard either.).
I would love it if someone could double-check my math and preferable re-do the equation.
If this is correct, then n*(2*d + n)/2 will appear as a d in (e+1, 1).
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4bf7c5 No.6238
>>6234
https://pastebin.com/czpK8A4j
Here is the updated code. For scan it does every cell on the radar testing for x up to 100. + and - zoom their respective windows. (x+n) even is just a plain blue square for now, don't know how I should color them in. dGrid and xGrid show the values with the same D and X for the selected record. The inner rectangle looks more dense because thats where I generated the grid, everything else is just based on (e,n,x) values (at least the scanned ones). +X goes to the next valid X in the same (e,n). +D goes to the same C record with different D.
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c53994 No.6239
>>6237
I could be walking around in an algebraic labyrinth here and fooling myself into thinking this is something insightful.
n*(2*d + n) is of course what you add to c - e to get (d + n)**2. This is also the generalized difference of two perfect squares (since we removed e, the remainder).
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5ddd07 No.6240
>>6124
>>6239
I was also thinking about the D[t] - d. We know that at the correct t, D[t] - d = a(n-1).
Now we do know the equation for D[t] for both even and odd e's, but let's limit it to odd e's for now.
The equation for d in (e, 1) for odd e is:
2*k**2 - 1 + (e + 1)/2
We know
2*k**2 - 1 + (e + 1)/2 - d = a*(n-1) (for some k)
This can be rewritten to:
2*k**2 = a*(n - 1) + d + 1 - (e + 1)/2.
Let's multiply both sides by 2:
4*k**2 = 2*a*(n - 1) + 2*d + 2 - (e + 1)
4*k**2 = 2*a*(n - 1) + 2*d + 2 - e + 1
4*k**2 = 2*a*(n - 1) + 2*d + 1 - e
Looks familiar? 2*d + 1 - e? Yup f is here too.
4*k**2 = 2*a*(n - 1) + f
And we can change it back:
4*k**2 - f = 2*a*(n - 1)
Meaning D[t] - d = a(n-1) could also be expressed as (4*t**2 - f)/2 = a(n-1).
Not sure if this is useful, obvious or neat.
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5ebae7 No.6241
>>6237
>>6239
>>6240
Hello Isee! Thanks for the excellent Algebra equations. I've read once through, but going to Study again. I'll verify later tonight.
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d88a7c No.6242
VQC's last post >>6185 clearly said we need two algebra equations to solve this. Good job Isee working this out.
Also, fabulous work GA on this new app, and thanks for posting your code too.
>>6238
>https://pastebin.com/czpK8A4j
You're using Python, correct?
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4bf7c5 No.6243
>>6242
Yeah you might need to download PIL or tKinter for it but yes python
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4bf7c5 No.6244
>>6242
Its pretty jerry-rigged. I made it piece by piece as I went adding features so there isn't much organization in it and its a little convoluted because I tried to keep it all in one file.
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3dafae No.6245
>>6185
>"lock and key" nature of the problem/solution
Have had a slight change of thinking. Not sure if it will lead towards a solution, but wanted to share the idea.
Quite a bit of recent work ( >>6057, >>6064 ) has been an attempt to use the animated squares and mod analysis to find appropriate triangle formulas to construct or iterate towards the correct x+n.
Why are we messing with trying to guess the exact x+n?
Especially when there are so many n values that fit into every x+n the larger the square gets.
So the adjustment in thinking about the lock and key:
Lock == 1 + 8T(u), where u is a multiple of (f-1)/8.
Key == nn+2d(n-1)+f-1
The lock is fixed. As in the odd x+n square that contains many potential n values.
And the key turns. As in adjustments within nn+2d(n-1)+f-1 that can generate matches. (i.e. our animated gifs turning within the same square.)
Using c6107 as an example. f=134, the starting u is ((134-1)/8) = 16, and the first few x+n squares are 33, 65, 97, 129, 161.
The first pic attached is a query for the solution n=36 in positive e for just these x+n values. And shows the first occurrence of n=36 at the second square iteration of 65.
The second pic takes this idea a bit further and searches for all n=36 records within each iteration of x+n, and then filters those records where d matches any d value from the factor tree.
This may all be coincidental, however, but certainly interesting how the n and d values line up in the negative e space.
So the theory is to lock the x+n value, and then use the factor tree to narrow the list of potential n matches.
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c53994 No.6247
>>6241
Thanks!
I played a bit more with the algebra and realised that we can express our equation as this:
n*(n+1) + n*(2*d - 1) = 8*( u * (u+1)/2 ) + (e + 1)
And by multiplying two and dividing by two on the first term we get (Essentially a NOP operations, but used for visualization):
2*(n*(n+1)/2) + n*(2*d - 1) = 8*(u*(u+1)/2) + (e + 1)
So the left side can be explained using two triangles of n, pluss n*(2*d - 1). I've been struggeling with ideas recently, so I decided to just play with the equation a bit. I don't see how 2 triangles of N is useful, but if it is you guys would know.
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c53994 No.6248
>>6240
Just to carry this though a bit more. Any (x + n)^2 in (e, n) (I've only checked odd x and even n's) can be expressed as 2*a - e from (e, 1).
This means our x+n square consists of 8 triangles where (x+n)^2 - 1 consists of two halves of an a from (e, 1) and each triangle represents one quarter of an a at t in (e, 1).
Not sure if this is of much use, but I found it interesting. It's like the grid is completely connected to itself.
As for use though, it could be used as a search algorithm, but painfully slow. PMA already had a better iterative search algorithm. To make it easier of use one could use the n in (1, c) as a starting point, but even that n grows too fast.
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2a528c No.6249
>>6079
>How does row one (n=1) relate and determine the patterns?
Have spent some time looking into n=1 for odd x+n.
First pic attached is for n=1 at x+n=11.
Because n, x, and f are fixed, these records can be generated for all valid combinations of d and e, where d starts at zero and carries on indefinitely. In other words, every d value exists.
Notice also that the records start at e = c = (1-f), and at some point have a c=0 value where d=x.
Taking this a bit further, the second pic is for n=1, x+n between 1 and 83, and a constant d value of 78, corresponding to the d value for the c6107 example, where f=134 the solution n=36 and x+n=83.
A couple of interesting records appeared:
1) at (36,1) where e equals the solution n with very close c and f values.
2) at (-6084,1) where d=78, x=78, c=0, and is very close to the x+n solution.
Not quite sure yet how this integrates into a possible solution.
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254092 No.6250
c85527 mod breakdown. Looking for a pattern here.
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14f7ae No.6251
Fuck guys, you've done so much in about 2 months, which is last I checked, I'm still barely understanding the grid, but I'll get there, eventually. Last I remember is this (e,n) graph. Also, does it matter that if imax = 10, then the code generates 45 points? also if imax = 100, it generates 4950, 200 generates 19900, 1000 generates 499500. I'd guess 20 makes 190?
dp = [ imax * ( imax + 1 ) / 2 ] - imax
or dp = sum of all integers from 1 to imax - imax
dp denoting data points.
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254092 No.6252
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14f7ae No.6253
>>6252
Dunno if I posted here before, but yes, saga.
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88d253 No.6254
Welcome, Anon! Studied your output. Problem tho. We have a working vocab for our quest. Your post is welcome, but is confusing. Topol could do a better job, and he does! No worries if your goal is to honestly contribute. If not, piss off.
iMax? we have been working on an upper bound for x. But your notation has no connection to our working terms.
Feels like a test to me. So FUCK OFF (please) Saga. Or post something that shows you've actually read our breads and want to contribute.
Do we have another shill??
If so AA (BO) will flush you super quick.
So WTF Saga?
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941b80 No.6255
>>6254
Dude, just because somebody obviously hasn't read the threads doesn't automatically make them a shill. I think you're jumping the gun a bit telling them to fuck off.
>>6251
iMax denotes how many instances of the variable i are generated in the grid. I'm not entirely sure what you're doing but iMax is irrelevant to the grid values themselves. The grid code runs a loop with another loop inside it. The outer loop has a variable called i, which starts at a value of 0. The inner loop has a variable j that also starts at a value of 0. j increases by 1 until it equals i. Once j = i, i increases by 1 and j goes back to zero. For every pair of valid i and j values (from 1 up to iMax), the variables e, n, d, x, a and b are calculated and put into the grid. We have resources around if you want to know about the grid more in depth, as well as all of our previous threads. We're doing quite different stuff to what we were doing two months ago, so it would be well worth the effort reading through those threads if you'd like to catch up, and there are some other designated threads where you can ask those of us who have been here the whole time questions if you're confused (there are separate threads so that we keep everything in this thread completely relevant is all).
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5a1919 No.6256
>>6254
Boooooooo.
Bad form, my nigga. :P
There are multiple approaches.
No tellin' folks to piss off because they just got here and are looking to contribute as best they can.
I was talking straight nonsense for months, remember? Almost got booted? lol
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5a1919 No.6257
>>6256
So for future reference, if a newb shows up and takes a stab at it… do what AA did and give 'em a nudge up to speed.
Perhaps they were testing the waters to see if they were at some point on the right track.
If their notation is fucky, as we all know mine is since I basically make it up on the fly based on whatever understanding I have at the moment, maybe try to correlate it to the correct terms instead of puffing up.
Foo Doggo Pony Show is my schtick. ;)
Have some dreamy tiddies.
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4bf7c5 No.6258
>>6257
I may have found something cool. So remember how if F is square then (x+n) is the square root of F? Well for every record you can shift D until you get to an F value. This works into how VQC asked about 2d(n-1) and nn because this sort of leads into those values.
For instance a=5, b=29, c=145, you set d=16, then your f=144 and (x+n)=12.
For a=5, b=31, c=155, set d=17, then f=169 and (x+n) = 13.
(I think this pattern may hold also)
Also you'll notice some similarity in the successive f and e values.
# f = 2d + 1 - e = f
# e = c - dd
# f = 2d + 1 - (c - dd)
# f = dd + 2d + (1-c)
# (x+n)(x+n) = dd + 2d + 1 - c
# (d+n)(d+n) - c = dd + 2d + 1 - c
# dd + 2dn + nn - c = dd + 2d + 1 - c
# 2dn + nn = 2d + 1
# 2d(n-1) + nn = 1
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9c4dc7 No.6260
>>6258
Apologies if I'm bringing up old news, but I was putting together a list of observations that I'd made while looking over the first code VQC posted. Maybe something will be of use.
Here's the pattern I'm noticing from what you mentioned: in both cases, (x+n) ends up equaling (b - a)/ 2. You're also setting
'd' = [ (a + b) / 2 ] - 1.
So the next one would be:
a = 5, b = 33, c = 165, set d = [ (5 + 33) / 2 ] - 1 → 'd' = [ 38/2 ] -1 → 'd' = 19 - 1 = 18
e = 165 - (18)^2 → e = 165 - 324 = -159
f = (2 * 18) + 1 - (-159) = 196
n = (i - d) = (19 - 18) = 1
x = (d - a) = (18 - 5) = 13
(x + n) = 14
(x + n)^2 = 196
f == (x + n)^2
Can't afford to go deeper into this for now. Just a few more days, then my schedule is clear ;)
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4bf7c5 No.6261
>>6260
Yes I'm looking at d as just a way to look at the value. For any c you can set any d and all the d+n and x+n will be the same but your e will change and your f will change and so will some other stuff. Anyway if we look at it this way we can say
f(d) = 2d + 1 - e
e = c - d*d
f(d) = dd + 2d + (1-c)
if we find d* such that f(d*) = (x+n)(x+n)
then take the derivative at d*
f'(d*) = 2d* + 2
I noticed that f'(d*) = 2d*+2 = a+b
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9c4dc7 No.6262
>>6261
x + n = (d - a) + (i - d) = -a + i = i - a
but 'a' = i - j, so:
i - a = i - (i - j) = j
So you would choose d* = j?
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9c4dc7 No.6263
>>6262
rather, f(d*) = j^2
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4bf7c5 No.6264
Also I noticed that if x* = d* - a then f'(x*) = b-a. Here is some output
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9c4dc7 No.6265
>>6264
That's well past where I'm at, hopefully I'll catch up soon. I'm unfamiliar with a number of those results–still on the first program, taking it slow to get a full understanding of what point is trying to be made. Plus I've been preparing for finals for the past two weeks, so that doesn't help.
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79066b No.6266
>>6251
>>6255
>>6256
>>6257
Hello Saga! (and Anons) Saga, I sincerely apologize for being such a Giant Asshole to you last night. I vented my frustrations on you, and I was mean and rude instead of helpful and encouraging. That's not our vibe here, and I fucked up. Hope you can accept my apology, and have fun with us as we work to solve this.
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14f7ae No.6267
>>6254
>>6266
Yeah man, looks like you're stressed out, it's ok, it just shows how much dedication you're putting in your wok, currently reading about phi functions but I am easy to get off track, that is the VQC.
Simply put, I've always liked mathematics, and wish to help out any way I can, I've got a lot to cover so until our paths meet my friend.
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254092 No.6268
>>6267
Thanks Saga! We welcome your help. Thanks for accepting my apology.
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f4515c No.6269
When building out the c6107 (prime e,1) I see that t= e+n = 23 + 1
That allows us to solve for a, then b.
screenshot incoming.
Check this out, Anons.
I may be onto something. Pls verify.
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2daa83 No.6270
>>6269
What's (prime e, 1)? Is this something new that I missed? I'm trying to verify your work.
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f4515c No.6271
>>6270
Hello AA! So when we do the na transform from (1,c) to (e,1), there's another element within that cell that had the same t and x values as the ultimate (prime) element we're trying to find. That's (prime e,1).
I thought I had found a way to link them in my last post, but it looks like it may be just coincidence.
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c6c148 No.6272
Found a way to represent any even x+n in terms of a triangle base u.
Pic attached shows sample layouts for even squares 6, 8, 10, and 12, but similar rendering can apply to any even square.
The formula to represent this structure is:
(x+n)(x+n) = 4*(1 + 2u + T(u) + T(u-1))
where u = ((x+n) / 2) - 1
Perhaps there is also a way to visualize the nn+2d(n-1)+f-1 breakdown for even squares.
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c6c148 No.6273
>>6272
Apologies. I made this way too complicated.
New simpler pic attached for even x+n.
An even square is comprised of 4 other squares. And every square is T(u) + T(u-1).
Revised formula:
(x+n)(x+n) == 4T(u) + 4T(u-1) == 4(T(u)+T(u-1))
where u = (x+n)/2
Yellow squares in the middle are just anchor points for reference and belong to the larger triangles.
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f4515c No.6274
>>6273
Here's an interesting find. In (1,c) for c6107 b - e = na transform f value. Don't know if it's a fluke.
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2a528c No.6275
>>6273
Attached pic is a first attempt at rendering animated even squares for x+n=16, and includes records between n=2 and n=15.
This example represents the nn + (2d-1)(n-1) + (n-1) + (f-1) breakdown with (n-1)+(f-1) occupying the middle.
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4bf7c5 No.6276
Just wanted to put an idea about the polite numbers in
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c53994 No.6277
>>6276
The left-side pyramid looks a lot like two triangles from our grid put together.
I've also been thinking about polite numbers. I've been trying to brute-force some polite numbers from our (x+n)**2 and f, f-1, f-2 etc. My reasoning is that if we can get the polite numbers and are reasonably sure we have the correct polite numbers VQC has been referring to, then we can try and go backwards, trying to figure out the patterns behind them.
I haven't gotten much done yet, but I was thinking about them today. We know we are looking for two polite numbers where one is a unit longer at the base than the other.
A polite number is a number that can be expressed as the sum of two or more consecutive positive integers.
In our case we will have an a, which is the base of the shortest polite number. We can think of a polite number as a_0 + a_1 + … a_k where each a_i is an monotonically incrementing number. Or, another way as a + (a + 1) + (a + 2) + … + (a + k). (a here represents the offset as we are looking for polite numbers and not triangles (maybe sometimes these two overlap? )).
Our two values would be shaped like this:
Polite number #1: a + (a + 1) + … + (a + k)
Polite number #2: (a + 1) + (a + 2) + … + (a + k)
I wrote a very basic brute force which essentially looped from 1 to k, continually increasing the number of integers with an inner loop increasing a. This is quite inefficient and I was thinking about our expression above. We can rewrite it to something like this:
a*k + k*(k + 1)/2
We are adding a, k times which means we can multiply a by k. We are adding 1 + 2 + 3 .. which we can do faster, which means it is also a triangular number.
I also noted that this can be expressed as k*(2*a + k + 1)/2. Which is also similar to what our n*n + 2*d*(n - 1) + f - 2 can be turned into (With - / + depending on the f and e value). It's also the generalized difference of two perfect squares https:// en.wikipedia.org/wiki/Difference_of_two_squares#Generalizations
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c53994 No.6278
>>6276
I'm loving this, GA!
I was a bit quick at reading it the first, just needed to write out my thoughts on the matter. But I really like this view of it.
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5a1919 No.6279
Someone dropped this in my whisper box.
In case anyone gets woody over this kinda stuff.
Seems like it might relate to this pattern someone posted in the open waters a while ago.
http://vbm369.ning.com/
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2a528c No.6280
>>6275
Think I've ironed out the kinks in the even x+n rendering.
Latest x+n=16 example attached (the top right and bottom left square were reversed).
Also included is shading for the different groups of "triangles" within the middle (n-1)+(f-1) portion.
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1505de No.6281
Face2Face, no more rabbit holes.
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206455 No.6282
>>6281
wut
If anyone's curious, this person has no posting history
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5a1919 No.6283
>>6281
Says the guy who forgot to trip up… kek.
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5a1919 No.6284
>>6282
There's uh… prece… we're due a visit.
I'm quite the fisher of men.
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5a1919 No.6285
>>6282
Fishing bites'n'nibbles is what you miss out on by not being in the discord.
We'll post/act on confirmations on the boards when we remember, but like…
Chris is a cheeky bastard.
He's out in the open waters.
Pic related.
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8f9576 No.6286
>>6285
>>6283
Even if/when I did/do join, I probably wouldn't be checking it lately anyway. I've only had time to do mod stuff and basic replies like this lately, and will for at least another two weeks probably (I guess unless something major happens).
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c53994 No.6287
>>6284
Time for our own BOOM coming?
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5a1919 No.6288
>>6287
That's up to us… I was just fishing in the open waters.
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4aff88 No.6289
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4aff88 No.6290
1 Be patient. No matter what.
2 Don't badmouth: Assign responsibility, not blame. Say nothing of another you wouldn't say to him.
3 Never assume the motives of others are, to them, less noble than yours are to you.
4 Expand your sense of the possible.
5 Don't trouble yourself with matters you truly cannot change.
6 Don't ask more of others than you can deliver yourself.
7 Tolerate ambiguity.
8 Laugh at yourself frequently.
9 Concern yourself with what is right rather than who is right.
10 Try not to forget that, no matter how certain, you might be wrong.
11 Give up blood sports.
12 Remember that your life belongs to others as well. Don't risk it frivolously.
13 Never lie to anyone for any reason. (Lies of omission are sometimes exempt.)
14 Learn the needs of those around you and respect them.
15 Avoid the pursuit of happiness. Seek to define your mission and pursue that.
16 Reduce your use of the first personal pronoun.
17 Praise at least as often as you disparage.
18 Admit your errors freely and quickly.
19 Become less suspicious of joy.
20 Understand humility.
21 Remember that love forgives everything.
22 Foster dignity.
23 Live memorably.
24 Love yourself.
25 Endure.
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4aff88 No.6291
Signatures.
Patterns.
Column 0.
Row 1.
You are VERY close.
One or more of you is going to suddenly get this.
If you can, please go back over and write out the list of all the patterns common to every cell at n=1 (row 1).
It applies to positive e.
Look at the difference of patterns for columns in NEGATIVE e (I used to call these columns f).
The mirror image of the grid.
Just as there is an entry in every cell in column 0.
Is there an entry in every cell at e=-1?
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4aff88 No.6292
All of you are amazing.
Where this takes you as you solve it and apply the practice afterwards is entirely up to you.
Remember Andrew Wiles?
Me?
You don't have to do this alone.
The first word of the first step is always "We".
The gifs are key and there are patterns that you're just about to look at to compare with the even squares.
Genius move anons.
Remember there are two types of even and odd squares.
Enumerate the patterns in the first row (and into negative x).
It will be something that was there that you didn't c that will trigger it.
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5a1919 No.6293
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2a528c No.6294
>>6290
VQC. Running out of ways to say thank you. These words… saved.
>>6291
>>6292
WE, will look deeper into these hints.
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c53994 No.6295
>>6290
Thank you VQC. These past months have been truly inspiring and I'm grateful that you've shared your knowledge with us.
>>6294
Yes WE will!
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6bf400 No.6296
>>6290
>>6291
>>6292
Thanks VQC! Nice to see you :)
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d88a7c No.6297
>>6292
Hello Anons! As VQC asked, here's the grid from (-5,1) to (5,1) with c values included to help us "c" something new?
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4aff88 No.6298
In row 1, the values of a[t] represent na for some c (e.g. RSA 100).
a[t+n] = nb
This is true for all c.
For the value of c, at the same t but in cell (-f,1), the value at a[t] = a(n-1) and a[t+n-1] = b(n-1)
This is the key.
The value of a[t] at -f and e in the first row have the same factor.
The values in each cell that have b as a factor are DIFFERENT, not aligned. They are one element apart in the two cells. In the positive side of the grid, they are n elements apart. In the negative side of the grid, the elements one less elements apart.
One cell has n as a factor at those positions at positive e column, one cells has n-1 as a factor in the negative f column.
This asymmetry can be used to solve the problem.
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4aff88 No.6299
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96fea4 No.6300
don't fully understand these links, so maybe this is intended for you guys
>https:// lwn.net/Articles/482089/
>https:// weakdh.org/imperfect-forward-secrecy-ccs15.pdf
>https:// www.scribd.com/doc/47155666/Magic-of-the-Primes
and the phrase
"illegal primes"
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5a1919 No.6301
>>6300
Illegal primes are primes that certain algorithms are based on and they'll relate to high level security of incredibly valuable assets. Whether that purpose is military, governmental, or corporate, if you were to publish that prime, even accidentally…
You get a very angry knock on the door, at least, because you've effectively #COMPED their everything.
From my understanding.
https://en.wikipedia.org/wiki/Illegal_prime
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c53994 No.6302
I think I got record generation for (-e, 1) rows. Not sure if this is something we already have. I might have missed it previously, but in the event we don't know it:
(-e, 1)
Odd e:
t = floor(sqrt(e/2))
d = 2*(t + 1)**2 - (e + 1)/2
a = 2*t*(t+1) - (e - 1)/2
b = 2*(t + 1)*(t + 2) - (e - 1)/2
x = 2*t + 1
Even e:
t = floor(sqrt(e/2))
d = 2*(t + 1)*(t + 2) - e/2
a = 2*(t + 1)**2 - e/2
b = 2*(t + 2)**2 - e/2
x = 2*(t + 1)
Note that t is set to the floor of the square root of half of e. This is the initial t=0 for the record. We do this is because it will require an offset. Since we are subtracting half of (e +/- 1) we will end up with negative values unless we offset t. We do so by offsetting t with the lowest number that will result in a positive integer for all the equations.
I haven't tested this below -64, since I haven't generated a grid beyond that. Verification would be appreciated.
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c53994 No.6303
>>6302
Note using the t with an offset will clash with what VQC is saying here: >>6298
For a = 7, b = 37 we have f = 30. At (-30, 1) if we use t=5 which is a[t] = an in (3, 1) we will get a = 35 which is a(n-1). Using the offset we will get a wrong value at a[offset + 5] for (-30, 1)
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c53994 No.6304
Found a relation between (-f, 1) and (e, 1) for an and a(n-1) with (0, 1).
t at a[t] = an in (e, 1) will equal an a at same t in (0, 1).
(from (0, 1)) a[t] - f/2 will be a(n-1) in (-f, 1)
f/2 is because of the movement between (e, 1)'s.
Example:
{27:6:18:9:9:39}
f = 2*18 + 1 - 27 = 10
9 * 5 = 45
9 * 6 = 54
a[5] = 54 in (27, 1) and a[5] = 50 in (0, 1)
50 - f/2 = 50 - 5 = 45.
Seems to hold fine for records with an odd e.
For odd f's it relates to (1, 1) and you do (f+1)/2
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c53994 No.6305
>>6304
I'm still looking into this, but I found something that made me wrinkle my nose a bit.
We have a c. Calculate e, d and f.
Then compute row (1, c). Now we know the n here is in (e, 1). Find the t of that n in (e, 1).
Then calculate the (0, 1) a[t] for that t. Then do a[t] - f/2. It appears that this number has (n - 1) as a factor in it. I'm going to run more tests, but if someone could double check that would be great.
Note, so far I'm checking odd e, even n's.
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c53994 No.6306
>>6305
I found a counter example (a = 27, b = 93), but still a lot of matches. Can't say why exactly yet.
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fb7dc2 No.6307
Blessings to All Anons Questing for Truth.
All Dealt It.
He Felt it.
She Thought it.
His Mind bought it,
Her whisper Brought with it
the idea to decide To let this ancient sophia outside
Through Our Actions in This Life
So i'll write, or maybe type this.
Think i'll title it something tight like "Deep Thesis"
Comfort Zone, Living Room The Chapel Perilous.
Comfort Zone We're in tune, while this rainy sunday marvels us,
Aint it marvelous, this rainy sunday evenin?
Cover Page Readin of The Heart of Buddha's Teachin'
"Transorming Suffering into
Peace, and Joy, while Liberating"
and I've been Told I'm tesselating.
Like a patriarch from the days of old.
Sounds a lot lke"Lead to Gold"
Yea, seems to me to be like alchemy
on a Musical Mosaic substrate
Like Moses on Today's Date with Most High
Stylized Lyrical components opposing All Lies
while Touching Capitalized Truth's many Lives
Capital B Being Be permeating through permutations
Anouncing covenants like the Prophet Nathan,
to every Capital across every Nation,
can't nothin fuck with this transmutation.
Jay Elect said he's bringing Ancient Mathematics back to modern man
Well i think, like Neo, i may have been the one meant to understand.
Cause i'm finalizing the Beginning Within the End, Final Fulfillment of the Master's Plan.
You won't find this shit in any theoligical textbook.
Definitely not in that weak necro-nah that shits a con-fuck that hex book.
Ain't even written on an HP, nah, ThinkPad with Wordpad
for these Crafts of Love so purpley plural just simple understanding could save save the world
and every son and daugher who think they sin just too bad
Listen deeply without your ears, let Love cast out all those fears,
using only mind make you feel a Real and True Glad.
I make deeper colors than crayola, like Teal Blue IronClad.
And i got Too Much Fruit for a Loom, Im gonna Need a Cornucopia Real Soon and Real Bad.
But there aint no such thing as "Real" Bad,
Only myriad millions of butterfly effects leading up to sub optimal decisions.
This is why we carry excess forgiveness, for givin hope to the helpless,
the already forgiven, and thats All of Us.
All of us Being.
All of us Together.
All of us Seeing.
Braving This Stormy Weather.
I Am.
We Are.
Creating Special Effects to Reflect The
Synergies of Synarchy,
Thats True Reign through True Strength and Peace.
The Opposite of Anarchy,
Reality is already chaos, so tell me
why should our rebellion
lead us
more lost?
True Order is True Rebellion.
And its So fulfillin when its with a Soulful Fillin
Of this Light that shines bright from the old school within
Vincit Omnia Veritas
Cras es Noster.
Deus Vult.
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fb7dc2 No.6308
>>6307
Change all i's to we's.
Bless.
<3
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fb7dc2 No.6309
>>6307
>>6308
Intention behind the AQC (Pictured: Analog Quantum Computer)
Cessation and Dispelling of All Fear, Thereby Alleviating All Suffering.
i think the ancients mighta called 'em altars…
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fb7dc2 No.6310
Order/Light/Truth
is the OG "Rebellion" "against"
Chaos/Darkness/Ambiguity (Water?)
Pursue your defined missions, my Bruddhas.
Bless.
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5a1919 No.6311
>>6310
I noticed we had a distinct lack of Arponyan Supremacy on this bread…
Seems like as good a time as any to interject some!
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4bf7c5 No.6312
>>6291
Patterns common to every cell at n=1.
If you do e+2j, a=a+j, b=b+j, d=d+j, x=x.
Then we have for even e:
X(t) = 2t
A(t) = 2tt + e/2
B(t) = A(t+1)
D(t) = 2t(t+1) + e/2
Odd e:
X(t) = 2t+1
A(t) = 2tt + 2t + (e+1)/2
B(t) = A(t+1)
D(t) = 2tt + 4t + (e+3)/2
Also I was looking at stuff and correct me if I'm wrong (I probably am because i'm not VQC), but it looks like at (-f,1), A(t+1) = a(n-1) and A(t+n) = b(n-1), instead of what VQC said with A(t) = a(n-1) and A(t+n-1) = b(n-1).
Maybe its the way I generate them?? Could be the way he indexes X I do X(t) = 2t+1 he might have it as 2t-1. Not sure. But I basically the f or e is just the reference point through which you are at. Like you can have any d for any c, but the e just changes. Think of it this way.
21 is d=4 e=5, 4*4+5 = 16+5 = 21
for d=5, you do
21 = 5*5 + e => e = 21 - 5*5 = -4
21 = 6*6 + e => e = 21 - 36 = -11
Since when we look at the -f column we find these for a(n-1) and b(n-1) and they are n-1 apart, I'm thinking if we keep increasing d, then eventually we will close the gap until they are one entry apart. Since for D we have A(t) = na, b(t) = nb, then for column -f, (or d=d+1), we get A(t) = a(n-1), A(t+n-1) = b(n-1). I haven't looked into this pattern yet but I suspect that if we keep changing d we get this pattern:
d=d
A(t) = an
A(t+n) = bn
d=d+1
A(t) = a(n-1)
A(t+n-1) = b(n-1)
d=d+2
A(t) = a(n-2)
A(t+n-2) = b(n-2)
etc.
continue this pattern until
d=d+(n-1)
A(t) = a
A(t+n-(n-1))=A(t+1) = b
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4bf7c5 No.6314
>>6312
Here is my suspected t debacle. I think if we let x=2t-1 for odd e then it is fixed but i'm not sure
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4bf7c5 No.6315
>>6298
So the way VQC worded it was a little iffy (to me) so I'm going to re-say it. For your starting (e,1) you have an X value for which A = na. You also have at x=(previous X) + 2n. Then from this, if you go to the left with the different e's, for the A(t) = na value you increase X by one, then for the A(t) = bn you decrease X by one, like this output shows. Then they converge on a single record like I suspected. Look at this output. Here are two examples of the movement. To keep it in one spot I kept the na and nb records in different lists
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4bf7c5 No.6316
>>6315
Also it seems like they converge on the (-(x+n)(x+n),1) entry
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4bf7c5 No.6317
>>6316
Also the b value for the record converged on seems to be 2(x+n) + 2 = 2x + 2n + 2 = (b-a) + 2
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4bf7c5 No.6318
>>6315
Also it looks like with each successive shift for the na values, the d values decrease by a-1, and for the nb values, the d values decrease by b+1
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739a69 No.6319
Skip to the last lines for tldr
row1
https://pastebin.com/4BzpPudJ
row2
https://pastebin.com/hyFUYFNp
in row3, the 'x' values are starting to generate a pattern of their own, and that's gonna make things much harder on me.
Also what I like about this is things start to get harder gradually, there's levels to this shit.
after I can make a generalized formula for rows (if achievable), I will start to go column-wise.
Sorry if this doesn't help much, just ignore if you're way past this.
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739a69 No.6320
For convenience,this Excel has two tabs for n=1 & n=2 if you want to try numbers.
https://ufile.io/fu6zx
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c6c148 No.6321
Now that we can draw even x+n squares, attached pics are for our old friend c145.
These are the (1,61,6) entry record, it's (-24,60,7) mirror in negative e.
The solution (1,5,4) record, and it's (-24, 4, 5) mirror.
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45f639 No.6322
>>6291
negative x = positive a?
strange mirror
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2a528c No.6323
>>6292
>The gifs are key and there are patterns that you're just about to look at to compare with the even squares.
Pics attached are for x+n=77 and x+n=78, both with fixed n=13 in the positive e space.
No conclusions yet, but the n-1 distribution in the middle, and the pattern of the yellow squares are interesting.
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45f639 No.6324
>>6322
>>>/qresearch/1715214
>>>/qresearch/1715323
Topol is nagging me to post this here.. understandable given the response it got. Am honored.
Thanks for helping color it!
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c6c148 No.6325
>>6292
Just browsing around into negative x again to verify the t offsets.
First pic is just showing for c145 the x to negative x matches for the entry (1,61,6) record and the corresponding na record at (1,1,6).
Then stumbled across something that seems extremely relevant.
Additional pics are for c145 and c6107 and show output from a browser app that helps navigate the grid.
For example, the steps for c145 are as follows:
1) create c record at (1,61,6)
2) na = move to na record at (1,1,6)
3) nx = move to negative x at (1,1,-5)
4) f = show factors of c at (1,1,-5) where c = 2501.
The interesting part here, is that the negative x record (with reversed a/b values) points directly to the TOP of the factor tree for it's corresponding c value. See the a/b values circled in red.
That means, I believe, that using the negative x record and an entry c record, we can isolate the full range of factor records for any c.
The remaining question is how to arrive at the correct negative x record.
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933248 No.6327
>>6324 Epic Bread #2158 there.
Thanks for all your work anons, enjoying following along with the show. What a timeline.
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4aff88 No.6328
>>6315
The power of ambiguity in language at work.
That is different pattern but related.
They are all related.
This is all brilliant.
Especially the colours.
That is very important for visualising properties of the Mandelbrot Set.
I hadn't used it for the grid.
Brilliant.
It is suggested you follow this pattern.
I will come back to clarify the other by the end of the week but briefly. Using the values of RSA 100 just as a large number example is a suggestion. In column e, the values at (e,1) for each element includes a value for a[t] which is an. The value bn is at a[t+n]. At (-f,1) the value a(n-1) is at the same value x in that cell as "an" for (e,1), the value b(n-1) at cell (-f,1) is one element less than bn at (e,1). This difference is key. It is not the only key, as you are seeing. I have found three, not including yours. Those three are in row 1, column zero and the side by side diagonal cells from the origin. There may be infinite keys.
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4aff88 No.6329
>>6322
Almost like an index?
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5a1919 No.6330
>>6328
I love fishing while high on MS Paint.
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e89f52 No.6331
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e89f52 No.6332
>>6260
I need to clarify these notes for anybody looking in the future: I made the assumption that, as 'j' increased, 'c' would increase as well—or at least that was my thought process as I wrote these. In truth, it's the opposite: as 'j' increases from 0 to (i - 1), 'c' decreases because 'c' is a difference of the squares (i^2 - j^2).
It's irrelevant to what's happening in this thread. For all I know it could be irrelevant to what happened in any thread.
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5a1919 No.6333
Ways of seeing…
Thoughts on crumbs…
Footholds, if ya will…
<;3=
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5ebae7 No.6334
>>6324
Great work 3DAnon!
>>6325
Great work PMA!
>>6327
Howdy MM! Nice to see you :)
>>6328
>>6329
Hello VQC! As always, nice to see you. Thanks for the hints! Are the polite triangle numbers derived from (f-1) div 8 and (f-1) mod 8 the key to one of the other solutions? You hinted at that on /qr/ so just wondering.
>>6330
>>6333
Love this JA, Topol! Saved!
Nice get with trip 3's! Checked.
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2a528c No.6335
>>6324
Excellent connection, 3D.
This started a further analysis in the (-e,n) space and what we thought would be a way to find the largest factor record for any even c number. ( mentioned previously >>6325 )
The idea was that starting from a (1,c) entry record, we could jump directly to (-f, 1) at d+1 and find a connection. This turned out to be true, but only for a handful of cases, irrespective of x+n parity or f mod 8.
Unfortunately, I think all we've confirmed is that (-f, n-1) is indeed a mirror, and those entries at (-f, 1) have corresponding entries at (e, 2) anyway.
Attached are sample outputs for c145, c6107, and c112220 that explore factor records in the (e,n), (-f, n-1), and (e,n) at negative x spaces.
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4bf7c5 No.6336
>>6329
I knew it would be a pointer-like system. It's the only way it could work
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2a528c No.6337
>>6328
Pic attached are key records for the c145 test case showing (e,n), (-f,n-1) and their respective na and nb records.
Differences in the x+n values are:
e na x+n = 12
-f na x+n = 13
e nb x+n = 134
-f nb x+n = 133
Also noticed for the -f nb record:
(d[t] - d)/(n-1) = (8832 - 12)/(61-1) = 147
when using the original d and n values.
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835819 No.6338
In about a week I'll have time to be properly active and not just lurk again, so I was thinking of redoing the introductory guide and creating an explanatory sticky (i.e. "this is the VQC board, we're currently doing x, here is a brief greentexted introduction to the grid, and if you want a more in-depth guide to what we're currently doing and how we got here read the next few posts and the previous threads" kind of thing). If anyone has any suggestions (or if anyone wants to contribute or do it themselves, or if any lurkers want anything in particular explained) then suggest away. Also, should I mention the off-topic Discord in the sticky?
While I've been doing all of this other meaningless bullshit I've had eating away at my time, I've been daydreaming about this somehow turning into a career. I don't mean once RSA's cracked taking those unclaimed buttcoins and being a rich prick, or going public and being an attention whore. I mean the idea of there being an application of this that would involve using it to improve the world, also somehow without the theory having to be kept secret from everyone else, and somehow making us all enough money to live on. That sounds far more meaningful than the corporate black hole that society and the expectations of others is sucking me into, making money for some guy in a suit who eats children's still-beating hearts in his spare time. I hope I don't sound delusional or entitled, because I know this isn't about me in the slightest, but if there is an application of this that would turn into what I'm talking about, VQC, I will survive on canned spaghetti for the rest of my life if that's what it takes.
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835819 No.6339
>>6298
I've been trying to understand this the past couple days since GA's explanation turned out to be for something different. At first instinct, it seems to hint towards there being another cell transform along the path of (e,1), (1,c) etc on the way to (a,b).
>In row 1, the values of a[t] represent na for some c (e.g. RSA 100).
>a[t+n] = nb
>This is true for all c.
We all know this by now, but for example's sake, {2:1:25:6:19:33:4}, c=627, f=-49, a[5] does equal 33*1 and so on through the infinite set.
>For the value of c, at the same t but in cell (-f,1), the value at a[t] = a(n-1) and a[t+n-1] = b(n-1)
>For the value of c
Meaning the same c from whatever example c we used? So 627?
>at the same t but in cell (-f,1)
So you mean at (49,1,4)? That cell is {49:1:42:5:37:49:4}, which has a c of 1813.
>the value at a[t] = a(n-1) and a[t+n-1] = b(n-1)
a[4] in (49,1) =/= 19*0.
>The value of a[t] at -f and e in the first row have the same factor.
a[4] at (2,1) and (49,1) are 19 and 37 respectively. The only factor they share is 1, since they're both primes. That doesn't seem like what you mean. Otherwise you could say every a that exists in the grid has the same factor, and that wouldn't be useful. So obviously it seems like either it's worded weirdly or I'm personally not understanding it.
>>6332
Here's a bitmap of the grid with c and j as the axes, if it helps your understanding at all. In November and December me and another anon who hasn't posted in a while (the baker) were trying to use the gradients in this image with binary search to factorize a and b. It worked almost instantly for c under some certain number of digits, but it was completely useless when it came to RSA numbers (or pretty much all 20+ digit numbers in general, for that matter). Just thought I'd let you know before you waste time in the same way.
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45f639 No.6340
>>6328
We found a spare key? Holy shit. Love this! Thank you!
>>5887
>triangles WITHIN triangles
>>>/qresearch/1715593
>BIGGER
>>6328
>large number
Why use RSA100 and biginteger functions as the first example? What patterns are we missing by looking at the smaller values?
If one key is enough to open a lock, how many keys do we need to see the lock? Make a bigger lock?
Records for RSA100
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e89f52 No.6341
>>6339
Interesting–I'd noticed the sort of diagonal lines of entries in the grid, but hadn't thought of graphing using different elements for the axes.
I'm sure all of you have noted that they've managed to get the first layer of encryption off of three (or four) of Assange's insurance files. I really wish we could see some video of him somewhere safe–even though there has been indication that he's made it to the US, I would feel better if I saw him in some new video, preferably with some American landmarks behind him.
Regardless, the result of taking off the first layer is basically another impenetrable glob file. Clearly, this is something that anons have been working on for awhile–check this archive out:
https://archive.fo/yLPie
In that archive, you'll see a number of interesting references. One of them is to an early JA project called "rubberhose":
https://en.wikipedia.org/wiki/Rubberhose_(file_system)
As I understand it, the idea is that a hard drive is filled with randomized data, then each classified item (which he calls 'aspects') is broken into pieces and spread randomly on the drive. There is no indication as to the size of the original files, so they're impossible to get at in anything but a piecemeal manner.
Anons have been finding all kinds of hints and clues in the blockchain, but I can't help but think that what we're doing here might have relevance at some point soon. For instance–what if the relevant parts are spread according to the patterns VQC is teaching us here? Were that the case, then the elite would have been able to actually see what's in the files and know Julian Assange wasn't bluffing.
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c53994 No.6342
>>6315
This looks very interesting, but I can't follow it entirely. Could you expand on this approach?
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45f639 No.6343
>>6341
Poasting this puts both the borad and mayself at risk. The ka yes have been found multipales of time by different groups in chans of blorks and all trices of this material and anons in-volved wipers - evrywheretime.
Aboot as moch as write or imager wothout trigging stuff.
Save'em while you can (!), and find somewhere else for this discussion as we can't continue on here. Suggest not reposting unedited copies _anywhere_. No further replies from mao on the subject, good luck (and sorry guys for bringing this here, the obfuckscation only works for a while…)
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c6c148 No.6344
>>6328
>>6340
Relevant records for the 1,c starting position of Rsa100. Big numbers really force you to look at the last digit.
Nice format 3D!
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c6c148 No.6345
>>6344
And the Rsa2048 values, just for fun.
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4bf7c5 No.6346
>>6315
>>6342
So think of the number line going horizontally
Think of a big post sticking up at 0
This is d=0
Any e value would just be whatever the normal value it is on the number line
d=0, e=5 => c = 0*0 + 5
d=8, e=5 => c = 64 + 5 = 69 (kek)
d=8, e=-5 => c = 64 - 5 = 59
d=8, e=17 => c = 64 + 17 = 81
Its 5 units bigger than d
Now think of (instead of d=0) d=2
Now think of the big post sticking up at 4 [2*2=4]
Any e value here would be e units away from this post
d*d is sort of your vantage point and e is the error from c per say.
that said, we can look at any c from any d, it would just change the e value [think like the F function]
So certain numbers can only have certain e values
for example lets do 15
d=-2 => e=11
d=-1 => e=14
d=0 => e=15 (*top of parabola)
d=1 => e=14
d=2 => e= 11
d=3 => e=6
d=4 => e=-1
d=5 => e=-10
d=6 => e=-21
So for the value c=15, we could say that the values for e are integer solutions to the function e = 12 - d*d,
which are {15,14,11,6,-1,-10,-20,…}
If we start with good ol' c=145, it's respective e values are {…45,24,1,-24,-51,-80,-111,-144,…} for d={…10,11,12,13,14,15,16,17,..} respectively.
We know that at our original e value [1], we have a certain x [7] value that corresponds to A[t] = na = 25 correct
If we go to the next e value and increase d, we get another e [-24] with another x [8] value which is one greater than the previous x value.
This record has an A[t] = (n-1)a = 20
The next one would have e [-51] with x=[8] and A[t] = (n-2)a
if we continue in this fashion, we eventually reach A[t] = (n-n)a = 0
Also, at the first e value [1], if we do A[t+n] = nb, like VQC said.
Remember with an increase of t by one, we increase x by two, so we get here x=7 + 2*n = 17
This is our A[t+n]=nb
If we then do the same thing with the e and go to e=-24, but for this one (nb) we decrease x by one and we get another A[m] = (n-1)b
We keep going and eventually we get A[t]=(n-n)b
Moreover, this is the same record that the na -> (n-1)a -> (n-2)a etc converged on.
Moreover, at this record d=x, c=0, n=1 and b=2(x+1) = 2(d+1)
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4bf7c5 No.6347
>>6346
Heres a way to visualize the d and e for c=15
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2a528c No.6348
>>6346
>>6347
This is an excellent explanation. Just fantastic.
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4bf7c5 No.6349
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5a1919 No.6350
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217559 No.6351
>>6338
>>6339
Man, I'm getting really sick of almost never getting any replies to my posts. It makes me wonder why I post at all.
I figured out the confusion myself, and since nobody told me otherwise, and I can't see anyone else mentioning it in any other posts, this should be useful information. In VQC's post when he said
>at the same t but in cell (-f,1)
when he said (-f,1) he more specifically meant { (the version of f calculated from e-2d+1, which is already negative (so not that * -1)) , (n-1 (so in this case 0, even though he explicitly said 1)}. So whether it was just weirdly worded or not, if you pick a cell in row 1 with a positive e, you'll find the same a and b values at the same [t] in (f,0) (meaning you'll need to make xMin a negative number etc). Then when he says
>The value of a[t] at -f and e in the first row have the same factor.
This is now accurate. As an example, we find {2:1:5:2:3:9:2} in row 1, where a=3 and t=2. f=-9, so if we take (-9,0,2), we get {-9:0:6:3:3:9:2}, which has the same a, b and t values (and an f value of 2 (equal to the positive counterpart's e), meaning we could do this transformation whether we started positive or negative). However, when he says
>the value at a[t] = a(n-1) and a[t+n-1] = b(n-1)
This isn't accurate if the above is accurate. a[t] = 3 in these mirrored cells. In the positive space, as he already said, a[t] = n*a and a[t+n (or t+1 in this case)] is the current b * n (or 1 in this case). This is all observably true. However, in the negative space, a[t] = a(n-1) and a[t+n-1] = b(n-1) don't seem to be the case, at the very least if you're using the 0th row that contains the mirrored cell. a[1] for example =/= 3(0-1), because then it would be negative. a[2+0-1] =/= b(0-1) either, because that would also be negative. VQC, if you read this at some point, you might want to clarify.
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d88a7c No.6352
>>6351>>6351
Hello AA! Sorry for not replying sooner, my IRL has been super nuts with work. I think your explanatory Guide is a great idea. Maybe we can collaborate on the dough for the next bread?? I think we can mention the discord, and keep in mind that all significant ideas should be posted here on the board. This is our home.
I have also given a lot of thought to what will happen when we solve this. VQC said "New Math Kangz". Here's my thought on the issue: This is tech that has been suppressed, and Q team has decided that now is the time to release this knowledge to the masses, for the advancement and evolution of humanity. VQC will not take compensation, but gave us free will to choose how we will use this solution. VQC fished us all in with excellent bait, back on CBTS. WHY??? Because they needed a crazy ass team that could handle WAY FUCKING OUT THERE ideas without batting an eye. That's us. The fact that we're still here working, collaborating, and focused on finding the keys to this solution is proof that he found the right Team. WE can do this, together!
You all know I've been feeling burnt and frustrated at the lack of finding a solution, and even went off on Saga because I was so pissed about it. But I was wrong to do that. You all know I've been the Cheerleader for this Quest since November, and you all know I'm All In 100%. Every good Myth or Story of the Hero's Quest involves much discouragement, but the hero always Perseveres, and Fate turns in his or her favor. We will Persevere until we Succeed.
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ee2270 No.6353
>>6351
Also, I think your understanding of negative f is correct, and ties back into the original code for the Grid.
I have a question about this record you posted:
>As an example, we find {2:1:5:2:3:9:2} in row 1, where a=3 and t=2. f=-9, so if we take (-9,0,2), we get {-9:0:6:3:3:9:2}
The a and b records make a c val of 18. When I run the values out, I can't make them match with d and e. Is it possible that there's a bug in your code? Or maybe I'm missing something?
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7a4ee4 No.6354
>>6353
The last record is t. a=3, b=9, c=27. I saw someone else format it like that with seven records.
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20f9ab No.6355
>>6354
So {e,n,t,d,x,a,b} ?
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7a4ee4 No.6356
>>6355
{e,n,d,x,a,b,t}. So, to rewrite them, {2:1:5:2:3:9} (t=2) and {-9:0:6:3:3:9} (t=2).
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ee2270 No.6357
>>6356
Ah! Ok, following now.
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2a528c No.6358
>>6328
>"three are in row 1, column zero and the side by side diagonal cells from the origin"
Have been thinking about VQC's comment about 3 solutions, and went in search of the diagonal cells from the origin.
Aside from the n=0 special cases, there are 4 quadrants to the grid.
(e, n)
(e, n) at -x
(-f, n-1)
(-f, n-1) at -x
For c145, the respective entries in each quadrant are:
en (1,61,6) = {1:61:12:11:1:145} = 145; f=24; (f%8)=0; (x+n)=72; u=36; (d+n)=73
enx (1,61,-5) = {1:61:-10:-11:1:101} = 101; f=-20; (f%8)=4; (x+n)=50; u=25; (d+n)=51
fn (-24,60,7) = {-24:60:13:12:1:145} = 145; f=51; (f%8)=3; (x+n)=72; u=36; (d+n)=73
fnx (-24,60,-5) = {-24:60:-11:-12:1:97} = 97; f=3; (f%8)=3; (x+n)=48; u=24; (d+n)=49
And images for each are attached.
In the (e,n) -x image, notice that the entire image is comprised of only nn and n-1 portions, due to the negative d value.
Confirmed similar behavior for other negative x records when compared to their positive e counterparts.
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ee2270 No.6359
>>6356
Thanks AA! We are currently using a slightly different notation. Quick rundown on the working notation we're currently using:
Location summary is (e,n,t)
Element summary is {e:n:d:x:a:b}
This will make it much easier for us to follow along with PMA's examples and GIF's, and help us find their locations in the Grid too.
My Original Post is lacking, since I'm a new baker! If any Newer anons have questions, check the OP's in the earlier threads bc there are some really helpful diagrams that explain the difference of squares equation and the Grid.
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ee2270 No.6360
>>6359
Also, AA put together a very helpful summary! Here is the quick rundown for the VQC.
>>5863
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2a528c No.6361
>>6328
Pic attached is current work in progress, related to the idea originally posted >>6245 and VQC's subsequent hint in the (-f, n-1) mirror.
For a small example c287, the output starts with the factor records, then moves to the (-f, n-1) mirror, and then to the nb record at (-2,1,136).
Then it shows the factors of a = 36449, just as a confirmation that this nb record indeed contains the a=7 and b=41 values from the prime solution.
Finally, the result of an n,d,f query is shown within the nb x+n = 271 for our solution d=16 value.
Note the nb starting f value of 73441 compared to the query results f value of 73154.
In this case, the difference just happens to be our starting c value, or dd+e.
If we can correctly determine the f value we are looking for, we will be able to calculate our prime solution directly.
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33dd75 No.6362
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33dd75 No.6363
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5a1919 No.6364
>>6363
That face you thought the bait went unnoticed…
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5a1919 No.6365
>>6364
TFW I can't TFW properly.
HOWDY SENPAI!
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5a1919 No.6366
>>6363
I'm trying to get the nerds on the board while you're still here… But they be sleepy nerdlings.
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8427dc No.6367
>>6363
Hey, did you want to verify something real quick? I've explained here in more detail >>6351 but the gist of it is that you said the value of a[t] at -f and e in the first row have the same factor, but in the negative space, it isn't the first row, it's n=0. That makes some of the other things you said inaccurate/inconsistent. I can't tell if it's because you worded it weirdly or if I just personally didn't understand,
>>6359
This is what happens when one us of doesn't pay attention for a while I guess. I swear someone was using 7-unit cells with t at the end somewhere.
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33dd75 No.6368
(e,1) has n as a factor
(-f,1) has (n-1) as a factor
Row (n-1)
Row n
At cell (e,n) c is an an element.
At cell (-f,n-1) c is also always an element
There is a pattern of repeating cells on each row. 2n in row n. 2(n-1) in row (n-1).
This gives additional information.
Information constrains values.
Every piece of information constrains values.
The most valuable information greatly constrains values as c increases.
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33dd75 No.6369
>>6367
Yes, its the way it was explained.
I'll show an example using RSA 100 as an example. This weekend.
a[t] simply means the value of a in a cell at position t. t is a function of x but since x can be odd or even, t is used for convenience.
a[7] : (1,1) = 85
If 85 is "an" then n could be 5 or 17, a could be 17 or 5. Both values of n would appear in the column at (1,n)
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5a1919 No.6370
>>6369
—-~^^^
Mmhmmmm.
Quite the coincidence.
Juuuuuuuuust a little bit.
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5a1919 No.6371
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c53994 No.6372
>>6368
I'm just spitballin' an idea here. It's jumping a bit further, but we know f is what we add to c to make a square, we can also "jump" further.
When we are computing f, we are doing:
1*(2*d + 1) - e
But we can also do n*(2*d + n) - e
For example 2*(2*d + 2) - e would be f2 and at (-f2, 1) we should have a*(n - 2) and b*(n - 2). We could increase this of course to f3, f4 etc and each time we would have a row that contains an a[t] = a*(n - 3), a*(n - 4) etc..
I haven't looked into this pattern and I'm not sure if this is derailing VQCs thoughts, but it was just a thought that popped into my head.
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e89f52 No.6373
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f43d0b No.6374
Basic picture.
For all c.
c = ab = dd+e = (d+n)(d+n)-(x+n)(x+n) = aa + 2ax + 2an
Grid (p,q) where p and q are signed integers
Elements in a cell are products with notation: e:n:d:x:a:b
The first two of the notation correspond to the coordinates in the grid.
Horizontal black line (e,1), (-f,1)
Vertical black line (0,n)
Vertical grey line (-1,n)
For some SPECIFIC c = ab = dd+e = (d+n)(d+n)-(x+n)(x+n) = aa + 2ax + 2an
Dark green line : column that contains e
Dark maroon line : column that contains -f
Pinkish-purple square cell in dark green line at (e,1) contains an and bn at elements t and t+n which are elements:
e:1:(na+x):x:na:(na+2x+2)
and
e:1:(nb+2x+2n):(x+2n):(nb+x+2n):(nb+3x+6n+2)
Blue square in dark maroon line (-f,1) that contains a(n-1) and b(n-1) at t and t+n-1 elements
Orange squares in -f line and e line : squares that contain c as a product… -f:n-1:d:x:a:b and e:n:d:x:a:b respectively. THESE SQUARES ARE ONE LINE APART.
Pick any odd c and this holds for all. ALL.
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e89f52 No.6375
>>6373
I like that trip better, I'm switching. Establish tie to old one
>>6374
Hello VQC, thank you for everything you're sharing :)
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f43d0b No.6376
>>6374
Correction:
bn at t+n should read:
e:1:(nb+2x+n):(x+2n):(nb+x+2n):(nb+3x+6n+2)
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f43d0b No.6377
>>6375
Hello PrimeAnon!
Thanks for your patience. Things should start to click.
I'll go over some details and we'll animate some pictures hopefully to show each set of patterns that matter.
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7b943e No.6378
>>6377
Hello VQC! Thanks for the new hints, and nice to see you. What are variables p,q??
>Grid (p,q) where p and q are signed integers
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c6c148 No.6379
>>6374
>>6376
Hello, VQC.
Searching through >>6368 and you add clarity to everything else. Thanks.
I have been referring to grid coordinates as (e,n,t) because it enables quick reference to exact entries.
Your notation of (e,n) and a[t] for specific values is obviously more concise.
Is there a preference going forward?
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c6c148 No.6380
>>6376
>bn at t+n should read:
>e:1:(nb+2x+n):(x+2n):(nb+x+2n):(nb+3x+6n+2)
e:1:(nb+x+2n):(x+2n):nb:(nb+2x+4n+2)
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c53994 No.6381
>>6372
Joining the new trip club
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c53994 No.6382
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c53994 No.6383
>>6382
Urk, proper one.
Also, nice to see you again VQC!
>>6374
Looks great. I'm trying to follow along, but I can't help but feel like one of these monkeys.
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c6c148 No.6384
>>6374
>Grid (p,q) where p and q are signed integers
Just clarifying an earlier post >>6358 and confirming VQC's grid statement.
Attached pic shows grid entries in the 4 quadrants for c values where mod 145 = 0.
And they appear to be arranged as follows:
(e, n)
(e, -n)
(-f, n-1)
(-f, -(n-1))
Within each cell x can be positive or negative.
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c53994 No.6385
>>6374
If we are being technical, there should be a total of 4 orange records.
You have {e:n:d:x:a:b}, {-f:n-1:d:x+1:a:b}, {e:n:d:x:1:c} and {-f:n-1:d:x+1:1:c}.
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2a528c No.6386
>>6384
Figured out the grid coordinates and how to navigate between any of the 8 related records.
Examples attached are for c145, c303, c6107 and the c6107 prime solution.
They show records in the 4 quadrants for +/- x.
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5a1919 No.6387
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5a1919 No.6388
>>6387
^^^
Truly, I am King Derp.
Here's the Le Image.
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c53994 No.6389
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56cbeb No.6395
From (-f,n-1) = c, the value of a,b and d increase by ONE every 2(n-1) cells, as you move from left to right in the grid.
From (e,n) = c, the value of a,b and d decrease by ONE every 2n cells, as you move from right to left in the grid.
These two rows are next to each other.
Any product of 2 primes will be divisible by 3 if you add either 2 or 4, or if you subtract 2 or 4.
Any product of 2 primes will be divisible by 5 if you…
Etc…
Remainders.
Patterns.
"Triangulation"
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56cbeb No.6396
>>6385
Almost.
Yes four orange squares, values of n will be different.
Well done.
The values of n at these cells can be calculated.
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56cbeb No.6397
>>6378
p and q were just distinguishing from specific e and n for that example.
Just integers.
Signed integers is a hint at using assymetry of -f and e where possible.
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56cbeb No.6398
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5a1919 No.6399
>>6395
>>6396
>>6397
>>6398
WHAT TIMING!
Q shows up… AND SO DO YOU! ;)
(Also, ThanQ, VQC!)
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5a1919 No.6400
>>6399
Juuuuuuuuust sayin'.
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56cbeb No.6401
>>6400
>>6400
Spoopy but coincidence from my end.
Think about the two rows.
Each c has patterns of remainders.
Those two rows next to each other give away a LOT of information.
They cover a different set of gaps. The bottom one is a unit of two more.
The difference in increase 2(n-1) and decrease 2n give enough information, and make the product of two primes work to your advantage!!!
Triangulate.
You can do this.
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5a1919 No.6402
>>6401
YOU HEARD THE MAN!
TRANGLES!
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2a528c No.6403
>>6395
>>6401
Initial wip for c145, comparing (e,n) and (-f,n-1) records. Related records are highlighted.
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1241c3 No.6404
>>6386
Found something cool!
c145 in (e,n) and (-f,n):
13^2 =169, and e = -24 brings that back to c145
12^2 + e of 1 = 145
e,n f=24
Then it switches in negative e
e=-24
And then f changes to 51 for -f record
The key value is 25
“Difference in remainders”
Difference between -24 and 1 is 25.
The difference is 25, and the sqrt is a
(At least for this example)
5,29
5 * 29 = 145
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79f6d5 No.6405
>>6404
wat? tripfagging in case significant
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79f6d5 No.6406
>>6404
Problem with that is it would appear to be constant time vs log time.
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4bcb75 No.6407
I'm a bit confused. We're still doing the triangles right? Even with Chris' latest crumb? Maybe jumping with e and -f will reveal our polite parts of the x+n triangle
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5a1919 No.6408
>>6407
Well, technically he said triangulate so… okay.
YOU HEARD THE MAN!
TRANGULATE!
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c6c148 No.6409
>>6403
Continuing the analysis on the (e,n) and (-f, n-1) related records.
Revised spreadsheets attached for c145 and c287 and include the following observations:
1) The differences between f values in adjacent rows changes in steps of 2. Similar to how x changes vertically.
2) Highlighted rows indicate where c mod a = 0.
- Every 5 records for c145, every 7 for c287. And in groups of 2.
- This looks similar to how we can iterate by known factors in (e,1). t = t + m * factor and t = m * factor + 1 - t, where m is a multiplier.
- Except instead of jumping vertically by x, we are moving horizontally by f.
3) Added columns to analyze c mod the starting c value.
- For the first few records on either side of our starting c, it looks like the c mod differences can be expressed in terms of 2T(u).
- Where u is (a-1) in the (e,n) space and abs(a) in (-f,n-1).
- Alternatively, without the T(u), the differences can be expressed as the product of 2 adjacent numbers.
For example, for record (489,61,6) where c=745. 745 mod 145 = 20.
20 = 2*T(4) = 4*5.
We can further define that new c value in terms of our destination a.
new c = a*c + a(a-1)
So, it looks like we can make pretty large horizontal jumps as well, assuming we know which a we are looking for.
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1ca3ec No.6410
Posting a reward for solving this ;)
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31cada No.6411
I have some time, so I'm going to put a new welcoming sticky and a better introductory guide together. The only person who had anything to suggest last time I mentioned this was VA, so if nobody replies, I'm going to have to assume nobody has any suggestions.
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dd82a5 No.6412
>>6411
Yeah man do it, I'm having a hard time putting all of this together
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ce2c1b No.6417
>>6406
Well, no it wouldn't. Since we still compute the square root of c in order to generate the (1, c) record we are still bound to that. VQC has also previously mentioned that
> There is a solution that has the complexity of the most complex step.
> This is taking the root of c.
Which he mentions in: >>6185
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79f6d5 No.6418
>>6417
You’re right. I’m retarded.
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5a1919 No.6419
HookTube embed. Click on thumbnail to play.
>>6418
Sweetheart, we live in a world where the definition of progress is a retardation of progress.
So like…
Are you saying that all ironical an' shit?
Literally: You made it this far.
Think about that.
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e27b82 No.6420
(-f,n-1),(-f+2(n-1),(n-1)),(-f+4(n-1),(n-1)),..
(e,n),(e-2n,n),(e-4n,n),..
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04a69c No.6421
>>6420
So this is where the binary grid comes in (you said it might be easier for seeing patterns)? In each of those cells n=1 (e is positive) would be 1, 10, 100, 1000 etc.
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04a69c No.6422
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307b18 No.6423
I think I found a potential issue with the grey background/white text picture here
>>5863
It has separate explanations for (e, 1) and (1, n). The (1, n) part refers to a VQC post here https://archive.fo/fyzAu#selection-12381.0-12381.107 that says "each "a" from the first row equals na because xx+e = 2na and na is half of that. That's BIG part of the KEY". Row one where n=1 isn't (1, n). It's (e, 1). n denotes rows, so row one where n=1 in (1, n) refers specifically to (1, 1). So if we're sorting a more detailed OP for the next thread #13, someone might want to correct that.
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d032c1 No.6424
>>6395
I've been thinking about
> Any product of 2 primes will be divisible by 3 if you add either 2 or 4, or if you subtract 2 or 4.
> Any product of 2 primes will be divisible by 5 if you…
And I can't see any specific patterns here. Adding or subtracting 2, 4 from a product of 2 primes for mod 3 is fine. This makes sense. Mod 3 will either be 0, 1 or 2. If it is 1 then adding 2 will "remove" the remainder, and subtracting 4 will also "remove" the remainder.
If it is 2 then adding 4 or subtracting 2 will also "remove" the remainder.
But as for 5, 7, 9(are we limited to primes only?), 11, … I can't see any reasonable pattern as the number grows.
If you multiply two, any given, primes and then compute the modulus on a specific value (ex. 5) then it should cover all the values of 0, 1,…, (5 - 1). In which you will always have to cover these numbers.
I was thinking in the line that this might, somehow, be related to parts of the grid, but nevertheless I can't say I see anything straight forward.
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c2ab06 No.6425
HEY YOU GUYS PAY ATTENTION
I’ve made several posts all at once, but I want to draw very specific attention to this one. I’ve noticed while I spent the last month or two lurking far more than I was contributing that whenever VQC has brought something that was seemingly a change of subject from whatever we were focusing on up until that point, we’ve started focusing on the new thing mostly by itself. This is what this anon >>6407 was talking about. Having begun work on an explanation sticky, I’ve been bringing as many different possible calculations that we’ve figured out over the last 7 months together. I’ve noticed that this is a bit of a running theme: we work on one thing, VQC mentions something new, we start focusing on that, once or twice an anon will mention that they remembered something interesting that VQC said a while before that, VQC will mention another new thing, repeat the cycle. I found a specific crumb that sums up why it’s important for us to avoid doing this:
Once you state everything you know about the problem, the solution presents itself.
If we find every single thing we know how to find for a given c, and take into account everything we know how to find if we also know a and b, we will most likely have enough information to find the answer. I’ve been up for the last 5 hours or so combing through every thread trying to get absolutely everything that we have learned to calculate that we’ve been told is useful and turn it into the following list (although some of that time was spent talking to Topol on /qresearch/). I’m tired enough that I can’t entirely pay attention anymore, so it’s probably about time I stopped until “tomorrow” (today). I got up to >>2990 so I know for sure there’s a lot missing (such as parity of d and e being used to calculate parity of n and x, as well as triangle-related calculations). If anyone would like to pick up where I left off feel free to but either way I'll try to finish this off soon.
What I’m proposing is that we should pick any semiprime c and apply absolutely everything we know to it, rather than just focusing on triangles or just focusing on na transforms by themselves. VQC has mentioned this several times. This list is definitely not complete so if you notice anything that needs to be added please say so.
All the things we can calculate/find/whatever with our given c:
>d, e and f (d and e being the variables necessary for calculation, “the collapse of the superposition”)
>the cell where (a, b) = (1, c)
>the cell in (e, 1) with our e and n=1
>the cell in (-f, n-1) with our f and n=n-1 (0)
>the (e, 1) and (-f, n-1) cells related to our (1, c) cell’s e and f values
>if positive f is a square and increasing d by 1 and e by 2 results in the same positive f value, we can calculate all of endxab
>the place in column 0 (e, 0) at which our c^2 is (which may be in more than just (0, 0), in which case I’m not completely sure this is calculable without ironically having to factorize)
All the things we can find if we know a and b:
>(e, n)
>the cell in (-f, n) with our f and the actual n from our c
>big_n, which is n at the cell where (1, a) or (1, b) = (1, c) (i.e. every prime number can only be calculated with 1 and itself as the a and b values, and so they’ll only ever have one n value)
>the (e, 1) and (-f, n-1) cells related to our (1, a) and (1, b) cells’ e and f values
>the cells in (e, 1) at which na and nb are n cells apart (n being the correct n for our c)
>t
>with the cells in n=1 that we can find related to our c (and all others), the cell at n+=2 has the same n and x values but a, b and d have increased by 1
>for our given c’s (e, n), there will be values in a cell at (e+2n,n) (which is meant to be an important pattern too)
>the cell at which a = a^2 and b = b^2, the increasing powers, and the other binomial-related combinations of a and b
>the x value at (e,n) is equal to the x value where d[t] = na+x
>a pattern of (n-1) as a factor of each d[t] value at (e,1) which is different (increasingly) from the pattern of factors of n in a[t] which gives us the offset that is used to solve the problem and thus get the cell at (e,1) to do all the work for you
All the things we can find that have been suggested to be important but not looked into in very much detail as far as I can tell:
>negative values of x for cell (1, 1) and then cells 2 to the right
>how many times numbers that are the product of three, four, five etc prime numbers turn up for a given e (it’s not linear, apparently)
>All products (integers) c that are the sum of two squares appear (only) in columns where e=0,1,4,9,16,25,..
>The value of x at (e,1) for na (a[t]) is the value of x at (e,n) (I only put this in the last section because I don’t completely understand the na (a[t]) part)
>think about the distribution of n and (n-1) in cell (e,1) in the a[t] and d[t] values respectively. This is the quick way to utilising f.
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Post last edited at
d032c1 No.6426
>>6425
I agree wholeheartedly. Great initiative.
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c6c148 No.6427
>>6395
>Any product of 2 primes will be divisible by 3 if you add either 2 or 4, or if you subtract 2 or 4.
>Any product of 2 primes will be divisible by 5 if you…
>>6424
Product of 2 primes divisibility rules may be:
div 3 is +/- ( 2 or 4 )
div 5 is +/- ( 1 or 3 )
div 7 is +/- ( 3 or 4, 1 or 6, 2 or 5 )
And do not contemplate where the product is immediately divisible by 3, 5, 7, etc.
The div 7 rule (if correct), is interesting, as it appears the pattern for checking is:
3-4…
2-5…
1-6…
Where the sum of each combination equals the div value.
For div 11, as an example, this means that the +/- ranges to check would be 1 apart, 3 apart, 5 apart, 7 apart, 9 apart.
+/- (5 or 6)
+/- (4 or 7)
+/- (3 or 8)
+/- (2 or 9)
+/- (1 or 10)
Which seems to work properly on smaller test cases for 119, 221, 323, 437, 551, 1159, 2499, 5781.
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4bf7c5 No.6428
This is output for A[t] values for (e,1) and (-f,1). I took also factored each of them and took the gcd. The GCD for the two values is in the middle. You'll notice that if a is the gcd for a certain t, then a is the GCD for t+a. Same goes for b. And if we know anything about prime prime numbers we know that eventually these two numbers will meet up (cuz they cycle, modular arithmetic) and eventually there will be a value where the GCD is c.
I'm also thinking of a way to look at this, but I don't know much about math history. If we just assume that like fermat and other great mathematicians were using this machine to discover the stuff they did, perhaps we could look and see what was like the first great mathematical breakthrough, and see if we can find that pattern in the grid. Then go to the next historical breakthrough and find that in the grid, maybe through that lens we could like walk through it I'm not sure.
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4bf7c5 No.6429
>>6428
Easier to see the pattern here
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4bf7c5 No.6430
>>6420
>>6420
Heres some output for this. I'm going to finish that program I promise
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4bf7c5 No.6431
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4bf7c5 No.6432
>>6431
Here's actual examples
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c6c148 No.6433
>>6420
>>6427
Attached spreadsheets for c145, c287, and c6107 (partial) show side by side comparison of grid entries in (e,n) and (-f,n-1).
For each of these spreadsheets, notice that the d value descends in (e,n), ascends in (-f,n-1), and the total of both d values is always 2d+1.
The difference between the d values is also highlighted, showing odd values at increments of 2.
The last 2 sections of the spreadsheets are an attempt to incorporate the product of 2 primes +/- some number to see if there are any matches, as the pattern looks very similar to the div 11 example posted previously. >>6427
Using the d value from the (e,n) and (-f, n-1) columns respectively, the results show the four possible combinations of (c +/- d), and then each of those divided by (2d+1).
The green highlighted row in the middle indicates where there is no remainder to that formula.
Interesting how this match only happens once for each example, and before the d value at (e,n) reaches 0.
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737b73 No.6434
>>6425
Hello AA! Thanks for getting this going. I completely agree that we need to integrate all the ideas and work on a concrete example element to solve.
I've set up a pastebin for this purpose.
https://pastebin.com/hEXHcPWa
I just posted your working notes to start. Maybe it's easier to just do it here on the board?? Anyhow, now we have a pastebin too.
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df80d0 No.6435
GUYS what the fuck is this
>>>/pol/11755706
https://medium.com/@coop__soup/00000000000000000021e800c1e8df51b22c1588e5a624bea17e9faa34b2dc4a-cd4b67d446be
>>6434
Thanks VA. A pastebin will be useful once we get to the next thread. I think eventually it would be useful to have a spreadsheet that we can upload somewhere like catbox.moe or whatever that we can all use with a big set of calculations when you set a particular c maybe in a box at the top.
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5a1919 No.6436
>>6435
With the fuckin' 345 straight…
Great find.
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737b73 No.6437
>>6435
Synchronicity. It's Time for this to be unveiled.
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737b73 No.6438
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645204 No.6439
It’s high time we solved this bloody thing. It’s kind of sad reading through the last 7 months and remember the many times VQC said we were pretty much there and everyone saying “oh nice it’ll be done in the next week or two”. Here’s the list updated, with one thing definitely missing (explained at the bottom), and potentially other things I may have missed (which I guess you guys will have to spot and point out). I’ve gone through every thread, so it should be pretty much everything. Logic would dictate that if we apply every single one of those things to a given c, the answer will be in there somewhere.
All the things we can calculate/find/whatever with our given c:
>d, e and f (d and e being the variables necessary for calculation, “the collapse of the superposition”)
>the cell where (a, b) = (1, c)
>the cell in (e, 1) with our e and n=1
>the cell in (-f, n-1) with our f and n=n-1 (0)
>the (e, 1) and (-f, n-1) cells related to our (1, c) cell’s e and f values
>if positive f is a square and increasing d by 1 and e by 2 results in the same positive f value, we can calculate all of endxab
>the place in column 0 (e, 0) at which our c^2 is (which may be in more than just (0, 0), in which case I’m not completely sure this is calculable without ironically having to factorize)
>whether or not e is 0
>the parity of e
>the pre-factorization of the decision tree (dividing by two until odd)
>n0
>what would be missing from the triangles if we could only fill them with n0 squared and multiples of 2d
>every different case of the width of the (f-2) chunks we used to calculate n0 being doubled (which is relevant because “if doubling the width of the chunks of (f-2) didn't give us a solution, the number c would be prime, the worst case (doubling the width of (f-2) chunks very quickly determines this as the number of times it doubles is logarithmic in the length of (f-2))”
>the remainder of f when divided by 8 (which is a key to what values of d and (n-1) can be used)
>an uneven distribution of f-1 among the eight triangles
All the things we can find if we know a and b:
>(e, n)
>(d+n)(d+n) and (x+n)(x+n), as well as (d+n) and (x+n)
>(xx+e) and 2na
>the cell in (-f, n) with our f and the actual n from our c
>big_N, which is n at the cell where (1, a) or (1, b) = (1, c) (i.e. every prime number can only be calculated with 1 and itself as the a and b values, and so they’ll only ever have one n value)
>the (e, 1) and (-f, n-1) cells related to our (1, a) and (1, b) cells’ e and f values
>the cells in (e, 1) at which na and nb are n cells apart (n being the correct n for our c)
>t
>with the cells in n=1 that we can find related to our c (and all others), the cell at n+=2 has the same n and x values but a, b and d have increased by 1
>for our given c’s (e, n), there will be values in a cell at (e+2n,n) (which is meant to be an important pattern too)
>the cell at which a = a^2 and b = b^2, the increasing powers, and the other binomial-related combinations of a and b
>the x value at (e,n) is equal to the x value where d[t] = na+x
>a pattern of (n-1) as a factor of each d[t] value at (e,1) which is different (increasingly) from the pattern of factors of n in a[t] which gives us the offset that is used to solve the problem and thus get the cell at (e,1) to do all the work for you
>big_N’s D minus the real d to our semiprime c contains (n-1) as a factor
>however these trees >>3654 were made (I can’t figure it out)
>the factor tree is meant to find x or x+n, so if we can generate a factor tree and we know x and x+n we’d be able to find where they fit in
>eight triangle numbers that are equal to (x+n)(x+n)-1
>the base of each triangle for odd (x+n)(x+n) (which is ((x+n)-1)/2)
>(n-1)(n-1) is another odd square so you can use the same 8Tu + 1 rules on it, meaning you would recursively find the base, the triangle numbers etc for (x+n)(x+n) = (n-1)(n-1)
>the maximum and minimum amount to add and subtract from (f-2)/40 (another thing I didn’t completely understand, >>4344 here)
>a square made up of dd + e + f + 2d(n-1) + (nn – 1)
>another square made up of nn + 2d(n-1) + f – 1 (this and the above being a rearrangement of c being the difference between two squares)
>the permutations of possible values of f,d and (n-1) for each odd (x+n)(x+n)
>the range of possible triangle base values which are larger than (n-1)/8 but smaller than (x+n)/8, and the relationship between this range of values and (f-1)/8
>the range of possible x and n values for a given (x+n)
>the range of possible e, d, a, b, c, f and t values for a given (x+n)
>the sequence of variables in relation to (x+n) squares when you move around the grid by moving across with a constant x and steps of 2n and when you move up and down keeping d constant
>(-f,n-1),(-f+2(n-1),(n-1)),(-f+4(n-1),(n-1)) etc and (e,n),(e-2n,n),(e-4n,n) etc
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645204 No.6440
>>6439
(continued; body was too long)
All the things we can find that have been suggested to be important but not looked into in very much detail as far as I can tell:
>negative values of x for cell (1, 1) and then cells 2 to the right
>how many times numbers that are the product of three, four, five etc prime numbers turn up for a given e (it’s not linear, apparently)
>All products (integers) c that are the sum of two squares appear (only) in columns where e=0,1,4,9,16,25,..
>Fermat's Last Theorem can be proved using the closure of the product of the sum of two squares
>The value of x at (e,1) for na (a[t]) is the value of x at (e,n) (I only put this in the last section because I don’t completely understand the na (a[t]) part)
>think about the distribution of n and (n-1) in cell (e,1) in the a[t] and d[t] values respectively. This is the quick way to utilising f.
I’m missing a few parity calculations. Does anyone remember the full list of variables for which we can find the parity based on the parity of c, d, e, f and the other variables with calculable parities? I'm pretty sure that would be everything, in which case we can use this list to construct maybe some kind of spreadsheet, input our c, look at every single calculation that we can find from that c, and then potentially also compare those results to the results we can find from our entire correct (e,n) cell. It would be a big spreadsheet, but it would be extremely useful.
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adf2d8 No.6441
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e973cb No.6442
Hey Lads, found something helpful!
>>5977
>(x+n) at any cell, then 2n left or right of that cell.
>Animate.
>What is the pattern?
f changes by 2(n-1) as you move 2n left or right of the cell.
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23eba4 No.6443
>>6441
Interesting website. Do you know if there's more to it than just the weird visual thing?
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5a1919 No.6444
Derpa derpa derp.
-swirls the quantum weirdness-
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56cbeb No.6445
Given two columns f+e or 2d+1 apart.
-f
e
How can we tell or construct the list of factors in a[t] that are 1 unit apart?
Those factors represent the cells with elements in each column.
The values of a[t] in each column are all the possible values of x squared plus e (or -f).
The key to using the grid is at hand.
The difference in the element position of b(n-1) and bn by ONE element at (-f,1) and (e,1).
What happens when you compare the -f and e columns in the grid for 4c? The square for x+n has now sides of 2(x+n) compared to c.
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56cbeb No.6446
>>6442
This.
The values of a,b and d each increase by ONE 2n cells to the right. They decrease by one 2n cells to the left.
For every c.
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d032c1 No.6447
>>6439
Great list AA.
>big_N’s D minus the real d to our semiprime c contains (n-1) as a factor
Is this true though?
Take a look at a=7,b=37. That gives us the record:
{3:6:16:9:7:37}
For (1, c) the record is:
{3:114:16:15:1:259}
Big Ns record in (3, 1) is
{3:1:129:15:114:146}
129 - 16 = 113. 113 Does not have (n - 1) as a factor, which would be 5.
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d032c1 No.6448
>>6445
Welcome back VQC! Always nice to see you.
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d53825 No.6449
>>6447
VQC said it was, so I put it in the list. I'll track down the post.
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d53825 No.6450
>>6447
>>6449
Turns out I wasn't quoting VQC, I was quoting IseePatterns here >>3090
I have seen a lot of mention of d[t]-d containing (n-1) as a factor, but I'm not completely sure what d[t] refers to.
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5a1919 No.6451
HookTube embed. Click on thumbnail to play.
>>6445
>>6446
w00t
I can go to sleep now. ^_^
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d032c1 No.6452
>>6450
Ugh I'm an idiot. Yeah big N's D - d thing doesn't hold.
As for the d[t] - d that is a reference to the d[t] at a[t] where a[t] = an.
So based on the example in >>6447
The record for 7*6 is:
{3:1:51:9:42:62}
Here 51 - 16 = 35 (contains (n-1) as factor).
I think that one holds, I was just a dummie before and though it also held for bn and I guess big N. But that means it doesn't hold as we have a counter example above.
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d53825 No.6453
>>6445
>Given two columns f+e or 2d+1 apart (-f and e)
>How can we tell or construct the list of factors in a[t] that are 1 unit apart?
Two columns means two e values regardless of n. So that means two e values 2d+1 apart. One e can have infinite d values, so I have to assume we're meant to pick a cell in order to have a corresponding d, f and a[t]. If I pick an arbitrary cell like {5:1:4:1:3:7}, its d is 4, so we need a column 9 cells to the left (into negative space), giving us the column at f = -4. The t value of that positive cell is 2, and (-f, 1) starts at t=3 in this case. So either I'm not understanding you or it's irrelevant. Not to mention, if there was a t=2 cell at (-f, 1), following how d increases at an increasing rate in those cells and applying the pattern into negative space, d would equal 2, and 2d+1 in this case is 5, so it's a different offset from the positive cell. Something's weird about this.
VQC, I feel like half the time you post, I try to implement what you're saying and it doesn't make any sense based on the way I seem to be interpreting your explanations, nobody else points it out or they read into it differently and start finding results from it regardless, and it never gets addressed again. It's very frustrating.
>>6452
I forgot to say, thank you for checking my work. I'm glad someone is. I have a feeling this giant set of equations all being utilized on any c is pretty vital to getting this done. Then VQC comes back and drops even more things to check, of course. I guess we can just keep adding to it.
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d032c1 No.6454
>>6445
Are we now going to back the first few threads where you talked about controlling our rows by multiplying primes with c? Feels like we're going back now, but this time we know so much more.
>>6453
No problem, I haven't checked it all, some parts I recognize others I've checked just to see if it makes sense to me. I've been though like half of those at least, but I barely remember half of the properties of the grid.
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d032c1 No.6455
Trying to interpret. Comment if you interpret it different, want to supplement or for any other reason.
>>6445
> Given two columns f+e or 2d+1 apart.
> How can we tell or construct the list of factors in a[t] that are 1 unit apart?
We know how to generate records for any given [t] in (e, 1) and (-f, 1). They are on the shape of:
odd e: 2*t*(t-1) + (e + 1)/2
even e: 2*t*t + (e/2)
odd f: 2*t*(t + 1) - (e - 1)/2
even f: 2*(t + 1)**2 - e/2
How to construct a list of _factors_ for those numbers I don't know. We could start with t = 1 and then count upwards and add to our list. Some will be products of numbers, but will we always have a factor for those? I think we do, but I'm not sure.
>Those factors represent the cells with elements in each column.
>The values of a[t] in each column are all the possible values of x squared plus e (or -f).
We know this already, but I don't think we've though too much about it. It's been known for many threads and used in a lot of our methods for calculating records based on different values: a = (x*x + e) / 2*n.
We divide a by 2n, because in those context we already know n, x and e and we know a[t] = 2an. So we remove 2n from 2an to get a.
>The key to using the grid is at hand.
Somewhere in this thread we have the key. Triangulation, jumping from column to column? Solving our triangles?
>The difference in the element position of b(n-1) and bn by ONE element at (-f,1) and (e,1).
The difference in t between b(n-1) and bn is 1. In (e, 1) a[t] = bn and in (-f, 1) a[t-1] = b(n-1).
> What happens when you compare the -f and e columns in the grid for 4c? The square for x+n has now sides of 2(x+n) compared to c.
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f66d90 No.6456
>>6443
Well AA it is one of those potential "Theories of everything" called the E8 Theory. I stumbled upon it years ago but it was said to be dis-proven (but maybe covered up?).
https://arxiv.org/pdf/0711.0770.pdf
This was the document. It caught my attention to see it again. Super weird to see it, in of all places, blockchain. This idea was floating around at the same time BTC came out as well. Many coincidences here! I may have dug into this stuff and LENR and found BTC at the same time back around 2010 if memory serves but didn't know they were linked in some way!
I don't want to derail the conversation with this as it really is off topic. The contents in this paper is over my head so I just don't know. Skimming over it again It seems to perhaps be related to the stuff here?
Anyway, enjoy!
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d032c1 No.6457
>>6455
For our example a=5, b=29, c = 145 we have the following:
For f: a(n - 1) = 2tt - 24/2
For e: an = 2t(t-1) + (1+1)/2
an - a = 2tt - 12
an = 2tt - 12 + a
an = 2tt - 12 + a
an = 2t(t-1) + 1
That means also that
2tt - 12 + a = 2t(t-1) + 1
a = 2t(t-1) + 1 - 2tt + 12
a = 2tt - 2t -2tt + 13
a = -2t + 13
This means our a can be expressed as -2t + 13 for our specific t. This gives us some limitations in the possible t values of a.
For a bigger example:
Take a =1217, b =1979. This has the record:
(e, 1) = {2842:47:1551:334:1217:1979}
Here f = 261 so we know e is even and f is odd.
For an odd f, the equation for a is:
a[t] = 2t(t+1) - (f-1)/2
and for even e it is:
a[t] = 2tt + e/2
e/2 = 2842/2 = 1421
(f-1)/2 = 260/2 = 130
So we now now:
2tt + 1421 = an
2t(t+1) - 130 = a(n-1)
2t(t+1) - 130 = an-a
2t(t+1) - 130 + a = an
2tt + 1421 = an
2t(t+1) - 130 + a = 2tt + 1421
a = 2tt + 1421 - 2tt - 2t + 130
a = -2t + 1551
Meaning the t for the a in (e, 1) and (-f, 1) is somewhere between 0 and (1551/2). Not entirely sure where I'm going with this, but if I am correct then the maximum t value for our a[t] is f/2 + e/2.
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d032c1 No.6458
>>6457
I now wonder what happens if you keep this up. I don't think this is a rabbit hole worth following, but we know also that you can jump with 2*(2*d + 2) - e which represents the "next" f.
So for our 145 example we can also express the set of equations as:
an = 2t(t+1) - 25 + 2a
an = 2tt - 12 + a
an = 2t(t-1) + 1
Note the 25 comes from our "next" jump which is -51 (2*(2*d + 2)) in which case a = 2t(t+1) - (e - 1)/2.
I'm not sure if this would help us, though. I'm not sure how many jumps like this we can make. If we do it n times a*(n-n) would be 0. In which case both a(n-n) and b(n-n) would be equal to 0.
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322b42 No.6460
VQC's very first post about the grid:
>This virtual quantum computer is called The End.
>You will see that one exists already to calculate digits of Pi without knowing the first digits before.
This is the only method that seems to exist (based on my looking) that can do this: https://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula
It does it in binary, by the way. "Maybe the patterns would be easier to see in binary" (probably not a direct quote).
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322b42 No.6462
I'll upload a draft of the big spreadsheet some time tonight. I'm noticing that my list is very sloppily put together. This is what happens when you stay up until 3am working on something like this. Something I've noticed already is that xx+e=2na is equal in the (a, b) = (1, c) cell and the (e, 1) cell in every case.
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5bfa08 No.6463
>>6462
I take that last part back.
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decc88 No.6464
https://files.catbox.moe/elspzb.ods
I have at the very least made a start. I have a horizontal block for each cell with each calculation (not all of them are there yet I don’t think) and each different cell related to (e, n) is shown vertically (also probably missing something). All values are calculated from the (e, n, t) block at the top. I’ve started implementing the calculations for each of the other related cells. I think something’s weird about the t calculation, though. I just did what it said in the image at the top of this thread. The numbers I used in the example in this spreadsheet give a t value of 3 with this calculation, but the real t value in the grid is 4. Not sure why. I’m also calculating other values based on t (such as x in some of the (e, 1) and (-f, 0) cells), so that might be a bit of a problem. But if anyone would like something to work on, or something to critique (I’m not holding my breath considering the slow pace lately), here’s some progress.
>>6420
I haven't implemented the math in the spreadsheet above yet, but I had a brief look at the positive version of this (but +2n/4n/6n, because it seems to be the same pattern but this worked better for my example). I picked {7:4:27:11:16:46}, t=7, f=-48. As follows,
{15:4:28:11:17:47}, t=7, f=-42
{23:4:29:11:18:48}, t=7, f=-36
{31:4:30:11:19:49}, t=7, f=-30
{39:4:31:11:20:50}, t=7, f=-24
Then they stop.
As e increases by 2n, |f| decreases by 6 (in this case 2(n-1)), d increases by 1, x stays the same, a increases by 1 and b increases by 1. This is the exact same pattern I saw when I figured out that for all n=1, f=(x+n)(x+n) as an infinite set of increasing a, b and d values with those screenshots. Pic related is an example for (x+n)=9. Relating it to our triangles, we are meant to be finding 2d(n-1). Maybe it's related. I would do more examples but I need to sleep.
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decc88 No.6465
>>6462
>>6463
Forgot to mention, it turns out I had figured out that between (a, b) = (1, c) and the (e, 1) of (a, b) = (1, c), xx+e=2na is the same, because you're basically just swapping the a and n values.
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d032c1 No.6466
>>6445
>What happens when you compare the -f and e columns in the grid for 4c? The square for x+n has now sides of 2(x+n) compared to c.
The interesting thing about this approach is that you can control the parity of the n, and thus the parity of n-1. Take our beloved example of 145. If we multiply it by 4, we can think of it as 2*5 * 2*29. This record exists in e=4, n=10.
Exactly how to use it, I'm not sure yet. It should allow us though to use the t + p, t + 2*p and p + t - 1 to search through the records of that column.
Since we don't know any of the factors of a, b we will introduce the factor of 2 into the mix. This gives us knowledge about how to "search" in our new column (for 4*145 that is e=4).
For example, using p + 1 - t on column 4 we can find 100, which is a*n (a = 2*5 and n=10) in 4 iterations.
This would require iterative search, which isn't what I think we're after. I wonder if we could multiply it with a large prime number instead to allow us to "limit" the iterative search in whatever column it ends up in.
I think it also requires the prime we multiply in to exists in that column for us to use the t-part as the t is supposed to be the first t where that primes is (think a[t] = <large prime>).
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d032c1 No.6467
>>6445
Interestingly enough. I compared (1, 1) and (4, 1) with regards to our c=145 and c=4*145.
It seems like the a[t] in (4, 1) is equal to the a[t] - x[t-1] from (1, 1) and the d[t] + x[t] in (4, 1) is equal to the d[t] in (1, 1).
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187fcc No.6469
At this point, a good strategy would be to have a thread that is used to post patterns from The End grid.
Patterns that apply to all c in the grid.
Patterns that apply to all cells in a row.
Patterns that apply to all cells in a column.
Special rows.
Special columns.
It was hinted before and was useful to me, by enumerating all the patterns in one place, the answer will materialise.
You are looking for a key shortcut.
The grid does the ALL the work for you.
Suggestion…
Either here or a worker thread(s), discuss each rule or pattern.
When consensus is reached, put the rule or pattern in the key thread.
This process and result of this process may surprise you, more than you think.
It may change the way you think.
It may be advantage you can apply in other problem solving.
This is how you win.
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187fcc No.6470
AA anon is getting very close.
It's a single key difference between cells in (-f,1) and (e,1).
Note that a[t] at (-f,1) is related to d[t] at (e,1) and is keyed by the value of d for a particular c. The value of f is determined by d. Since d contains a+x, this is the key.
Enumerating the patterns will be the process that leads to the Eureka.
Then you'll see it everywhere.
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187fcc No.6471
>>6456
Garrett Lisi found the key to the symmetry of QM.
Self-containment and a boundary less Universe.
E8 is a beautiful set of mathematical objects.
Especially if you get the rotation right in the right place.
Triality is inherent.
It predicts the masses of elementary particles if he finds the missing piece.
That will emerge when the mistake in measuring the mass of the Sun enters the public domain.
Ties in to the inability to measure G.
Ties into MOND.
Ties into the disparity in size of gravitational force.
Ties into existence of mass.
Ties into the Solar Coronal Temperature gradient problem.
"Won't you come and wash away the rain…"
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5a1919 No.6472
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187fcc No.6473
>>6472
I posted 6 minutes before your call. You like that, Top? :)
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187fcc No.6474
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4c577c No.6475
>>6474
Hello Senpai! Nice to have you drop in :)
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5a1919 No.6476
>>6473
-nuzzles-
Dat Quantum Intuition.
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4c577c No.6477
Holy shit, Lads! Alright, I really found something this time. This is a follow up to 3Danon's highlighted sheet that VQC said "victory" to.
Start on the right at the blue square highlight. Which is na transform for c145. The a and b records are 24 apart, just like (prime) a and b of 5 and 29!!! As you move left into negative e, the spread of between a and b stays constant. d, a, and b all decrease by 1 every 2n to the left. Eventually we get to 5,29 and i think it will occur when d=12
Can Anons please verify?
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7cdc27 No.6478
>>6477
Ok, I ran the elements out. Here's what I found.
(-111,1,6) {-111:1:16:11:5:29} a and b pair that matches our (prime) a and b
(-119,1,6) {-119:1:12:11:1:25} a=1 and b=25
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a24494 No.6479
>>6471
I see a lot of Revelation in the original map.
c
{e:n:d:x:a:b}(t)
(e, n, t) [3] {d:x:a:b} [4] c (the foundation)
Maybe I'm crazy, or it just resembles a heavenly menorah with numbers (the uncorrupted one)
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97dd9b No.6482
>>6470
>"Note that a[t] at (-f,1) is related to d[t] at (e,1) and is keyed by the value of d for a particular c. The value of f is determined by d. Since d contains a+x, this is the key."
Check out this element, I was very close to the right idea:
{-119:1:12:11:1:25}
correct prime d value here, d = 12
Then we go back to (e,1) and look for 25 in the a column.
At (1,1,4) we have {1:1:32:7:25:41} which has x=7 and t=4
so na = 25
at e=1, na is n=1 and a=25
only other combos are n=5 a=5
or n = 25 a=1
prime n =5, prime a =5
and correct x should be 7, which it is.
I know it's a bit of a stretch, but I'm just trying to find the connection.
Thoughts?
Also, here's the other possible factorization for na, n=25 a=1 found it right where it should be at (1,25,4) {1:25:8:7:1:65}
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97dd9b No.6483
>>6482
Here's a better set of images. To recap, I started at the c145 na transform in (e,1) [highlighted turquoise].
Then I moved left into negative e, hopping 2n to the left until I found d=12 (our prime d value).
Here's the element: {-119:1:12:11:1:25}
a=1 b=25 [this one is not shown, too far]
Then I moved back to (e,1) to look for 25 as a factor.
I found it at {1:1:32:7:25:41} (highlighted orange). Interesting to note that this element has the same x as our correct (prime) element.
then i used our idea of na to break 25 into it's factors. 1x25 5x5 25x1. So there are only 3 possibilities for na locations. n=1, n=5, n=25.
Sure enough, at n=5 we find our (prime) element {1:5:12:7:5:29}
We also find the other na possibility, n=25 where it is expected to be: {1:25:8:7:1:65}
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a24494 No.6484
>>6460
So, is he saying this formula is a VQC?
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c0faaf No.6487
>>6484
I'm pretty sure that's what he's saying, yeah. It was in his very first post about the grid, and I don't think anyone ever looked it up. It could be very useful to look into. Also, assuming you're the same person who would always leave the name field blank, welcome back baker!
>>6470
>Archive Anon Anon
Heh
>Note that a[t] at (-f,1) is related to d[t] at (e,1) and is keyed by the value of d for a particular c. The value of f is determined by d. Since d contains a+x, this is the key.
The d value will obviously be the same for (e, n) and (a,b)=(1,c). That's the only instance of being able to calculate a[t] at (-f,1) straight from c that I can think of (the (-f,1) of (a,b)=(1,c)). So either there's direct correlation between that cell and (e,n) or there's some kind of relationship between them that in some way I can't think of directly relates to (e,n). The only thing I can think of is that maybe a[t] in (-f,1) and d[t] in (e,1) scale together in some way that lets us isolate x. I have no idea. I'll have a look in a few hours.
>>6477
>>6482
So how are you finding that they're 24 apart? I'm not quite sure what you're doing.
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463657 No.6488
>>6487
Hello AA! Check out the highlighted green elements in my big sheet. >>6483
Even when d declines by 1 as we move left into -e, the difference between a and b remains fixed at 24.
(-1,1) a=60 b=84
(-3,1) a=59 b=83
(-5,1) a=58 b=82
wash, rinse, repeat until we arrive at the correct d value of d=12
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463657 No.6489
>>6487
Also, the correct (prime) element has a and b of 5,29. These are also 24 apart.
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4bf7c5 No.6490
I'm going to dump a lot we've found.
Here is how I look at D values:
>>6346
>>6347
>>6349
Basic factoring:
For any of these c values, we have the equation (d+n)(d+n) - (x+n)(x+n) = c, which is a difference of two squares. Say d=5, n=1, x=2, then we would have (5+1)(5+1) - (2+1)(2+1) = 36 - 9 = 25. For any entry, we can increase or decrease any of these values and keep c the same. Lets increase x. Then we would need to decrease n to 0, and increase d to keep the squares the same. Then we would have d=6, x=3, n=0 to get 6*6-3*3=25. So if we can navigate to any n=0 row and maintain the same c, then it would be necessary that d^2 - x^2 = c, and from there we can find the factors. Lets say a*b = c. Then if we can show
((a+b)/2)^2 - ((b-a)/2)^2 = c
(aa+2ab+bb)/4 - (bb -2ab + aa)/4 = c
(aa + 2ab + bb - bb + 2ab - aa)/4 = c
(2ab + 2ab)/4 = c
ab = c
So this necessarily means that (d+n)=(b+a)/2 and (x+n)=(b-a)/2. Through this, if we have c as the difference of two squares, we can factor it.
Through this same view, if we have an e value that is a negative square for instance e=-i*i, then c = dd + e = dd - i*i which is a difference of squares so it is solved.
Significant Rows:
The n=0 row only has entries for e=-j*j for any j. If e=-j*j, every x=j in that cell. In this row, (a+b) = 2d always. You can use any d in any entry in this row.
The n=1 row has entries for every e. If we are at an even e, then we can take any even x value. If e is odd, we can use any odd x. If X[t]=x, then X[t+1]=x+2. Also in this row, if we have a record (e,1,d,x,a,b), then we know there exists another record (e+2,1,d+1,x,a+1,b+1). In addition, we can extend this to (e+2i, 1, d+i, x, a+i, b+i). Also if B[t] = m, then A[t+1] = m.
If we have a record with c, then there is an entry in (e,1) with A[t]=na, and A[t+n]=nb. In addition, if A[t] is divisible by a, then A[t+a] is divisible by a. If A[t] is divisible by b, then A[t+b] is divisible by b. If a and b are prime, then there exists A[t] that is divisible by c in (e,1). Also in this (e,1) cell [for starting d and n], if (D[t] - d) is not divisible by (n-1) and A[t] is divisible by n, the GCD of D[t]-d and A[t] is a. Also X[t] at this same t is such that X[t] = D[w] where X[w] = d-a or our correct x value. If we have A[t]=na and an X value for this, then if we go to the next lower e value (to get this we use our starting d (or the d for whatever e we are at relative to our starting c), we do e=e-2d+1), we see that X = X+1, and also A[t] = (n-1)a. this pattern continues as you keep calculating new e's (the next e would be e=e-2(d+1)+1). Eventually, this converges on a record where A[t]=0. At this record, X[t] = D[t] = (x+n) and e = -(x+n)*(x+n). Also B[t] = 2(x+n) + 2
Significant Columns:
For e=0, if we have an entry (0,n,d,x,a,b) then we can generate a new entry (0,n*i, d*i, x*i, a*i, b*i) for any value i. Also at the cell (0,2), every a and b is a square number. In addition, if A[t] = x*x, then B[t] = (x+2)*(x+2). Also if we have a cell (0,2,d,x,i*i,j*j), then there is another cell (-1,2, d + (j+i)/2 ,x+1, i*(i+1), j*(j+1)). Also at (0,2), if you have an entry (0,2,d,x,i*i,j*j), then there are cells in (-1,2) with b values that are (j*(j+1)) and (j*(j-1)). If you multiply this by r to get (0,2r,rd,xr,rii,rjj), then there are records (-1, 2r, r +/- (j+i)/2, xr+/- 1, rii +/- i, rjj +/- j) (either all + or all -). So I guess for any d you could look in this column and get different ways to write it as differences of squares or something like that.
Grids:
So I have analyzed different values for d and x and have found grids. For any X value, there are entries at (-x^2,0). Then there are other entries at (-x^2 + 2*i*n, n) for any i increase, (d+n) increases by 1 and (x+n) stays the same. Also a and b increase by 1 (also this is a regular e+2n shift). If you increase n for this formula, then a stays the same and b increases by 2*i, also (x+n) and (d+n) increase by 1.
For any D value, there is a cell at (-d*d, -d). Then from here, you can generate other cells with the same d by this formula (-d*d - m + i*i, -d + i) for any m or i. Moreover, if you change your d value, your position on this d grid (because its relative to any d) will stay the same relative to other cells with the same d. For instance if for my c I have a certain d value, and there is another cell at (e+j, n+i), then if I use any other d value, there is another cell at (newE + j, newN + i) so this is sort of a locked in position.
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d4f496 No.6491
>>6470
Hey, was there something specific I posted that makes you think I'm getting close? I'm not asking you to give it away obviously, I'm just curious about the specific post I made or a specific thing I said.
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e94100 No.6492
>>6471
Wow, I had no idea there was so much behind it. I guess a theory of EVERYTHING was a literal meaning. Black Hole Sun, Won't you come? Won't you cooommmeee. Now astronomers can "weigh" planetary bodies using their orbits. I think even the sun. If it were less massive for instance, the earth would orbit the sun slower? other than that we would neet to guess volumes and materials. No easy feat. None of this would work right if our basic understanding of mass was wrong. Garbage in-Garbage out.
So the fact that we are on this stopic this really begs the question then: What on earth was this doing buried in the BTC block-chain with an obvious string of zeroes? From my understanding it would take colossal luck for it to be mined or a time machine (knowing where to look) Or the tool that the guys are building here, this quantum computer. But why in Blockchain? It is just very strange as a side note, I think this may not happen by accident which means the original blockchain had this purposefully included.
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4bf7c5 No.6493
>>6470
Looking at this pattern I've found some odd stuff that might be important. Its just very weird. Look at this I was analyzing as vqc said the A values a (-f,1) and the D values at (e,1). I also looked at the difference and sum between the two values. What is weird is the fact that sometimes the difference is constant 13*37 or 5*79 for example and it is equal to the original D value, but sometimes it increases by 4 and never hits the D value.
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c6c148 No.6494
>>6374
>>6380
>>6470
>Since d contains a+x, this is the key.
Have spent some time reviewing VQC's na and nb formulas in terms of x and n, and the most recent post about finding the key in (e,1) related to (-f,1).
Just a reminder on the formulas:
na: e:1:(na+x):x:na:(na+2x+2)
nb: e:1:(nb+x+2n):(x+2n):nb:(nb+2x+4n+2)
So went in search of ways to understand movements in terms of x+2n, and may have stumbled on a better understanding for movements within specific cells.
For example, starting from the nb record for c145: (1,1,67) = {1:1:8978:133:8845:9113}
We can move to the next record at (1,1,68) = {1:1:9248:135:9113:9385} by manipulating d.
new d = d + 2(x+2n)
new d = 8978 + 2(133+2*1) = 9248
But what about moving back to (1,1,67)? The formula for d in reverse turns out to be pretty simple.
previous d = a - x
previous d = 9113 - 135 = 8978
For the current record, we know that d=a+x. And now we can calculate the previous record's d value in terms of the current a and x.
And it turns out that this movement works everywhere, regardless of (e,n).
From the original c145 record (1,61,6) = {1:61:12:11:1:145}, the next relevant record in terms of x and n should be:
next d = d + 2(x+2n)
next d = 12 + 2(11 + 2*61)
next d = 278
(1,61,67) = {1:61:278:133:145:533}
And then moving backwards:
previous d = 145 - 133 = 12.
Here are a few additional records in the same chain, that show the movements in reverse:
(1,61,67) = {1:61:278:133:145:533} = 77285
(1,61,128) = {1:61:788:255:533:1165} = 620945
(1,61,189) = {1:61:1542:377:1165:2041} = 2377765
(1,61,250) = {1:61:2540:499:2041:3161} = 6451601
2041 - 499 = 1542
1165 - 377 = 788
533 - 255 = 278
The shortcuts in the grid to navigating this way starting at (e,n) at t are:
Next d record = (e,n) at [t + n]
Previous d record = (e,n) at [t - n]
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84ff8e No.6495
>>6494
New thinking on c145:
(1,c) to (na) to negative e.
We find our prime a and b at {-111:1:16:11:5:29} = 145
So maybe we just run the search into negative e until we find our matching c record?
Easy to do, because you can see the c values declining (by a set amount) each 2n you move left.
"maybe we'll c something we didn't before" ??
It's like a countdown, actually.
We just move left, looking for our (prime) c, and tracking progress by watching c decline as we move.
We could eventually just write a formula to calc where the correct -e record would be.
Then we use a and b to plug in with the correct d=12, and solve the problem.
d=12 - a=5 = x=7
(5+29)/2-12 = n = 5
{1:5:12:7:5:29}
i "c" an element (-111,1,6) with matching c , a, and b records to our prime element.
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e35eec No.6496
tl;dr: stop just focusing on cell transforms and try to apply the change in variables to the triangle bases or something. The solution probably involves using multiple concepts at once.
Here’s a broader perspective: it’s been said that some of the concepts we’ve been going over can be used with each other, and that some are just one of several ways to see the “recursive nature” of the grid. I feel like we’re all falling into the trap of pretty much only focusing on VQC’s latest crumbs again and not thinking of the bigger picture. Why would we learn about all of these concepts if they weren’t going to be intertwined?
These are the concepts we’ve learned about over the last 7 months:
>the big list of rules that we can use to increment the values of several of the variables in a given cell (i.e. the a[t] = na or whatever it is thing)
>the big list of cells related to our cell (i.e. (a,b)=(1,c), (e,1), (-f,1), (e+2n,n), etc)
>the factor tree for d and e values
>odd (x+n)(x+n) being represented as a visual square made up of eight triangle numbers with a one-unit hole in the middle
>odd (x+n)(x+n) being represented as a visual square made up of nn + 2d(n-1) + f – 1
>a couple of other variables being represented as squares, such as dd=(a+x)(a+x)
We’re not just going to use one of those things, like just using cell transforms. Surely. Surely it would be a combination of several of these things working together. I would think the solution would involve something like (this isn’t an explicit suggestion but hopefully you’ll all get my point of using multiple concepts in the calculation) using our c’s d and e values to find an (e,1) cell, constructing the triangles from that cell, finding a cell with the same d but with new (x+n)(x+n) = nn, then incrementing that cell’s a value e times? Obviously that sounds pretty complicated to figure out but it would make a great deal of sense, and I'm probably overcomplicating it with that specific example.
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398019 No.6497
>>6496
I get your point and I agree, I don't think it's a simple transformation. At least not without knowing the different shortcuts.
PMA did a lot of good work with the triangles, maybe we should start looking into a few examples where we have the proper u1 and u2 for a triangle (assuming we KNOW we have the correct values?) and try and work our way backwards from them using the jumping right/left with -f and e. Try and see how this triangulation might work if we have the answers and do it backwards (something he also has mentioned before).
VQC has hinted many times that we should look at larger c's than the ones we have, I don't think it's because the algorithm won't work, but he said the patterns are more visible the larger the c. I don't know how large, though.
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f17cd4 No.6498
I found something weird. In some cells, the triangle base is equal to t+1. In others, it isn't.
{6:5:135:32:103:177} (6,5,17)
(x+n)=37, (x+n)(x+n)=1369
(1369-1)/8=171, which is the 18th triangle number
t=base-1
{6:5:183:38:145:231} (6,5,20)
(x+n)=43, (x+n)(x+n)=1849
(1849-1)/8=231, which is the 21st triangle number
t=base-1
{6:5:219:42:177:271} (6,5,22)
(x+n)=47, (x+n)(x+n)=2209
(2209-1)/8=276, which is the 23rd triangle number
t=base-1
{98:59:791:252:539:1161} (98,59,127)
(x+n)=311, (x+n)(x+n)=96721
(96721-1)/8=12090, which is the 155th triangle number, but the 128th triangle number (t+1) is 8256
I'm thinking maybe there's a slightly more complex formula involving t to find the base. This is a strange coincidence if not, and you know what they say about coincidences. If it is true and there's a slightly more complicated formula for it (like maybe t+(e-n) (that's not it though)), all we'd need to do is find maybe some kind of cell transform that gives us the same t as our (e,n) in every case.
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5a9ecd No.6499
Okay there is definitely a link between the base of the triangle in an odd (x+n) square and the way that t and (x+n) scale for that cell's (e,n). Definitely.
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5a9ecd No.6500
As you can see here, the difference between the triangle base and t doesn't change if n doesn't change. So if we're going to calculate the triangle base directly from t, it'll probably involve n. It might still be worth looking into anyway.
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5a9ecd No.6501
This one has a constant e and a changing n. It shows that n is causing the scaling change between the triangle base and t. I'm still not sure how, but there is definitely some way to calculate the base of the triangles in an odd (x+n) square from the t value of its (e,n).
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e4ac93 No.6503
>>6498
Hello AA! Studying now.
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e4ac93 No.6504
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a24494 No.6505
>>6185
These have been some of your best posts.
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e4ac93 No.6507
Hello Lads, made the Thread VQC requested.
>>6506
"At this point, a good strategy would be to have a thread that is used to post patterns from The End grid.
Patterns that apply to all c in the grid.
Patterns that apply to all cells in a row.
Patterns that apply to all cells in a column.
Special rows.
Special columns.
It was hinted before and was useful to me, by enumerating all the patterns in one place, the answer will materialise.
You are looking for a key shortcut.
The grid does the ALL the work for you.
Suggestion…
Either here or a worker thread(s), discuss each rules or pattern.
When consensus is reached, put the rule or pattern in the key thread.
This process and result of this process may surprise you, more than you think.
It may change the way you think.
It may be advantage you can apply in other problem solving.
This is how you win.
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a24494 No.6508
Found the grid in Revelation today. Hope I can share it with you all in a coherent format.
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a24494 No.6509
Candlestick = church
Star = angel
There is an angel for each church.
There is a star for each candlestick.
It makes 7 stars and candlesticks and total, and if you take their value in binary, it makes an element with c being a semiprime.
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a24494 No.6510
To clarify,
1111111 = 127
01010101010101 = 5461
10101010101010 = 5461 * 2
5461 is a semiprime.
1c and ab element
{132:2658:73:72:1:5461}(37)
{132:12:73:30:43:127}(16)
The message to the seven churches are divided into 2 chapters, the Lamb's message to the first 4 appears in the 2nd chapter, and the Lamb's message to the 3 appears in the 3rd chapter.
3 and 4 make 7.
3 variables used to calculate 4 make up the element.
(e, n, t) -> {d:x:a:b}
Ephesus, Smyrna, Pergamos, Thyatira <-> Sardis, Philadelphia, Laodicea
c is special. It is the foundation (AKA what you start with.) 1 Corinthians 3:11
@<C>hrisRootO<D>avid
d = floor_square root(c)
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d8eea0 No.6511
>>6510
Tell me you solved it
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a24494 No.6512
>>6511
No, but I see a path and I'm walking on it. If it looks like a coincedence, just remember, there are no coincedences, only relevant and irrelevant patterns.
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a7c55f No.6514
>>6512
You got my hopes up there. I'm really starting to get antsy about getting this thing solved already. That definitely is an interesting amount of "coincidence". I'm sure boss man will have some interesting stories to tell about it some time soon.
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d032c1 No.6516
>>6499
I'm assuming by t we are referring to t at a[t] = an in (3, 1).
I haven't checked them all, but in the first row I think you got the t wrong. According to my program it's 14 and not 15. I'm not sure if we count the first record in an (e, n)/(e, 1) as 0 or 1. If it's 0 then I believe it's 13, and not 14. But I don't know if we've developed a strict enough consensus around it.
As for 87*177 I get 19 and not 20.
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d032c1 No.6517
>>6499
But I think you noticed some unknown pattern. I haven't checked the math to see if there is anything obvious behind it yet, but I've attached some output data showing it.
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d032c1 No.6518
>>6517
It almost appears as if when x is even the t, at a[t] = an, is equal to (x/2 - 1). Is this something we already knew?
I added another image showing it for (1348, 44), but it holds for the other test cases I posted from the image before.
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d032c1 No.6519
>>6518
Note this appears to hold for when x is even, not sure about odd yet.
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d032c1 No.6520
>>6518
And of course I just realized that I'm running in circles. The x at a[t] = an is the SAME x as for our record of (a, b). Which explains what I'm running in circles for.
As for the base, that's just (x+n-1)/2 for odd x+n. Since x is a result of t then that would explain why the base - T is so consistent. Since n doesn't change and the x depends on the t.
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c6c148 No.6521
Just summarizing, in case it helps anyone:
t rules:
even e: t = (x+2)/2
odd e: t = (x+1)/2
u rules:
odd x+n: u = (x+n-1)/2
even x+n: u = (x+n)/2
(x+n)(x+n) in terms of u:
odd x+n: (x+n)(x+n) = 1+8T(u)
even x+n: (x+n)(x+n) = 4T(u)+4T(u-1)
(x+n)(x+n) in terms of t at n=1
even e: (x+n)(x+n) = 2t*2t
odd e: (x+n)(x+n) = (2t-1)(2t-1)
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4bf7c5 No.6522
>>6510
>>6521
Bit of a stretch?
>Basic picture.
>For all c.
>c = ab = dd+e = (d+n)(d+n)-(x+n)(x+n) = aa + 2ax + 2an
>Grid (p,q) where p and q are signed integers
>Elements in a cell are products with notation: e:n:d:x:a:b
>The first two of the notation correspond to the coordinates in the grid.
>Horizontal black line (e,1), (-f,1)
>Vertical black line (0,n)
>Vertical grey line (-1,n)
>For some SPECIFIC c = ab = dd+e = (d+n)(d+n)-(x+n)(x+n) = aa + 2ax + 2an
>Dark green line : column that contains e
>Dark maroon line : column that contains -f
>Pinkish-purple square cell in dark green line at (e,1) contains an and bn at elements t and t+n which are >elements:
>e:1:(na+x):x:na:(na+2x+2)
>and
>e:1:(nb+2x+2n):(x+2n):(nb+x+2n):(nb+3x+6n+2)
>Blue square in dark maroon line (-f,1) that contains a(n-1) and b(n-1) at t and t+n-1 elements
>Orange squares in -f line and e line : squares that contain c as a product… -f:n-1:d:x:a:b and e:n:d:x:a:b >respectively. THESE SQUARES ARE ONE LINE APART.
>Pick any odd c and this holds for all. ALL.
Also this is my new program. I'm going to add more stuff to it.
I don't know how to share a python package. I'm also having trouble exporting it to a exe or dmg so if anyone could help me with that I'd be really appreciative and we all could use this.
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fc974d No.6523
If you follow the (-f+2(n-1), (n-1)) sequence to zero, it continues into the positive space with t+=1. x and n stay the same. a, b and d continue only increasing by 1. In the negative space, f grows by 2(n-1) and e grows by 2n. Once it gets into the positive space, t increases by 1 (while x and n stay the same), and e begins growing by 2(n-1) while f begins growing by 2(n-2).
Starting from {18:27:215:84:131:353} t=43 f=-413
{-413:26:216:85:131:353} t=43 e=18
{-361:26:217:85:132:354} t=43 e=72
{-309:26:218:85:133:355} t=43 e=126
{-257:26:219:85:134:356} t=43 e=180
{-205:26:220:85:135:357} t=43 e=234
{-153:26:221:85:136:358} t=43 e=288
{-101:26:222:85:137:359} t=43 e=342
{-49:26:223:85:138:360} t=43 e=396
{3:26:224:85:139:361} t=44 f=-446
{55:26:225:85:140:362} t=44 f=-396
{107:26:226:85:141:363} t=44 f=-346
{159:26:227:85:142:364} t=44 f=-296
{211:26:228:85:143:365} t=44 f=-246
{263:26:229:85:144:366} t=44 f=-196
{315:26:230:85:145:367} t=44 f=-146
{367:26:231:85:146:368} t=44 f=-96
{419:26:232:85:147:369} t=44 f=-46
I'm not sure how much sense this makes, but theoretically, we could create a big chain of connected cells from the (e,1) cell's negative counterpart, transformed into the positive space with -f+2(n-1) to where t+=1, then using the new t value , find another cell at n+=1, transform that into the positive space, etc. That's a completely spitballed idea, but the number of steps this would take would be the number of times you have to add 1 to the d in the (e,1) cell to reach the correct d.
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fc974d No.6524
>>6523
As I said, that idea was purely spitballing. As soon as I posted I realized that would mean we would have a and b the same distance apart the entire time and we would already know n.
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be4f66 No.6525
I haven't tried on a smaller example yet, but it doesn't look like we can use the (e+2n, n), (e+4n, n) and (-f+2(n-1), (n-1)), (-f+4(n-1), (n-1)) thing on our (1,c) record. n is just way too large, since a and b are so far apart. e and f are meant to increase based on 2n and 2(n-1), so in the positive space, f would become positive while e is positive. In the example in pic related, in the record 2n away from e, e becomes 45832 and f becomes 45399, making it an invalid cell. So if we're meant to do something with these cell transforms, it doesn't involve (a,b)=(1,c).
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be4f66 No.6527
I feel like I've been handed the remote control to the world's most complex TV but I've been told some of the buttons don't work and I don't even know which TV it works on so I have to test every single button on the remote on every single TV in the world in the hopes that I find the right one. There are just so many different concepts and rules but there's barely any direction. "This is the key, and this is the key, and this is the key, and this is the key, and this is the key". They're all keys but we don't know how to use them and we don't know what to use them on.
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Post last edited at
e6d22a No.6530
>>6525
>on our (1,c) record. n is just way too large, since a and b are so far apart
Think you may have a point there.
Was doing something a few weeks ago that validates what you're saying. Just starting with an a and b that are the same but not integers (simply the square root) was quite quick (but not the real method we're working toward).
- Also, thanks for all your work in summarizing, much appreciated.
- Been so busy, but some time should free up in a few weeks (though assuming you all will have this buttoned up by then!).
- Still check in regularly and 'rooting' for you all every day!!!
>>6441 Thanks Great Lord Hobo! That was a very interesting rabbit (worm?) whole to dive into.
>>6469 Thanks for all your patience with us VQC. This has been quite the adventure.
>>6471 Interdasting….
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96fea4 No.6531
Here is something that I strongly suspect is over the target for you guys.
Just sniffed it out with muh spirit brain.
The nose knows!
http://www .svpvril.com/Quat.html
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f3968a No.6532
>>6523
I've been looking into this as well.
When you jump right with -f the f-value of those records again move from positive to negative. And when you jump left from e the same thing occurs with the t in a[t]=an.
They don't match, as in it's not the same amount of jumps required for both of them to end up on 0. But VQC has referred to the excluding middle and meeting in the middle. Maybe it's somehow related to that (I feel like this is just a wild stab).
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f3968a No.6533
>>6532
Correction, I don't mean t I mean literally an (which makes sense since a is decreasing).
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f3968a No.6534
I see that the AN values when jumping left with e for the proper record (not 1, c) appear to align with (0, n)'s x's. Can anyone verify? Is it something obvious?
I see there's holes sometimes, or rather cells that doesn't have a corresponding x in (0, n). For example in (e, 5) they will have jumps with AN = 25, 35, 65, etc which doesn't have a corresponding x value in (0, 5). But for those examples it appears to alternate between valid cell and invalid cell in (0, 5).
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c53994 No.6535
>>6477
When I did this, same example c=145 I ended up with the record:
{-111:1:16:11:5:29}
After doing 56 jumps. Here the d = 16 though.
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c53994 No.6536
>>6477
I added a special test for you my friend!
At firste glance it looks like it will always end up at a = a, but with a lot of jumps. Like a crazy amount.
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c53994 No.6537
>>6536
You can see the a under the transformed record is close to the number of required jumps. The difference between the two appears to be a.
The x under the "End Record" also matches the x in the transformed record.
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c53994 No.6538
>>6536
Since I didn't add a column for it. The transformed record is transformed from (1, c). So finding the proper a after all those jumps is done purely knowing c only. I added the proper record under "Start Record" but that is misleading. The start record is infact (1, c).
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c53994 No.6539
>>6536
Last image for today, but I cleaned it up a bit. If you want other records I can post those too VA. These, as you can see, are the records from (1, 5) (chain #2).
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a24494 No.6540
>>6531
svpvril.com/svpweb17.html
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c6c148 No.6541
>>6445
>What happens when you compare the -f and e columns in the grid for 4c? The square for x+n has now sides of 2(x+n) compared to c.
Have been working through this crumb, looking for patterns for x+n as c increases in various multiples, and ran into a slight descrepancy with the 2(x+n) statement from VQC that warranted a closer look.
Attached pic, shows 10 records generated from c145 multiplied by 2^1, 2^2, 2^3, 2^4, etc up to 2^10. The pattern holds as you move higher as well.
The initial result of 145 * 2^1, results in c=290. Because this is an even number where c mod 4 != 0, no valid grid entry exists. To find a valid entry, you have to divide c by 2, making it odd, and then create the entry.
This is indicated with the (290/2) notation next to the c value.
The red arrows show a connection between the c value and the matching x+n value as the multiples increase.
The green arrows show a similar connection between x+n and u values as the multiples increase.
And the yellow square shows how the (c/2) adjusted record for c290 matches to c1160, which otherwise looks disconnected from the pattern.
The rules appear to be:
1) any c multiplied by 2^2, the matching entry can be found where (x+n) = c-1.
2) even c multiplied by 2, the next entry can be found where (x+n) = 2(x+n) + 1.
Which possibly means that we can work backwards from a given (x+n) in steps of (x+n-1)/2 to find factor records. At least for numbers divisible by 2.
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c6c148 No.6542
>>6494
>>6541
Just an indication of where this can be applied, and more specifically where it falls short.
Attached pic for c299008 shows grid factor records in (e,n).
Using either of the following formulas to navigate backwards, we can hit all except the 3 records underlined in red.
previous x+n = ((x+n)-(a+a/2))/2
or
previous x = x - a
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5a1919 No.6543
Some light reading if you wanna maybe spark something in how you approach a concept:
https://en.wikipedia.org/wiki/Compactification_(mathematics)
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e89f52 No.6545
Forgive me if this has already been covered–I don't remember seeing it, but I could be wrong.
Check this out:
I remembered seeing a Twitter hint from VQC which mentioned a 3-4-n triangle. I haven't found any instances of those, but I have found others.
I noticed that, besides the main horizontal and vertical lines, the clearest one was the one running along the hypotenuse of the purple triangle–it follows a pattern of two cells to the right, one cell down, and continues strongly until it ends at the vertex of said triangle. The angle from horizontal would therefore be arcTan(1/2).
Of course, there are other lines in the image; looking over it, certain things seem to pop out. I thought I saw that 3-4-5 triangle formed by the ends of other lines–it turned out that it was another one with length ratios of 1-2-sqrt(5). So I drew it out…and started to realize that there were similar relationships all over the place.
The lines are drawn along either part, or the entirety of, trend lines of filled cells. The vertexes represent where those trends end. The numbers marking the triangles are there to denote ratios of length. I haven't actually worked this out mathematically, and you can see that the drawings are a little off, but it was close enough that I'm pretty sure there's something there.
I'm not really sure how all this relates–rather than take things slowly, I just started looking for patterns and messing with it. One thing seems certain–if you have one number, and you know that you need another from a particular line, it should just be a few simple calculations to get to the one you want.
Perhaps 'n' is somehow related to the distance between the two numbers?
I've got more work to do. Hope this helps.
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e89f52 No.6546
>>6545
oops, this one shows more.
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e89f52 No.6547
>>6546
Sorry, last post for now I promise.
I just wanted to note two things: first, if you're looking for a 3-4-5 triangle, you can probably find one in there. Pic related shows one in green.
Another thing: along the top line, you see a 4 and a 10; I wonder if were to double the settings, might we come up with a larger pattern that reaches 20–with this one embedded in it and repeating?
More work to do…
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5a1919 No.6548
>>6547
You should keep doing the thing. :D
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a24494 No.6549
Observation: the difference in t for the 1c record to the ab record for semiprimes is equal to (a - 1)/2 for semiprimes
Test cases:
pastebin.com/C26TmF3r
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4bf7c5 No.6551
>>6494
From this I noticed some stuff. Say we have 1.145:
1,61,12,11,1,145
Lets get to the 'next' D value D = D + 2(X + 2N) = 12 + 2(11 + 2*61) = 12 + 2*(133) = 278
which of course is valid by the screenshot.
and we get the previous D by doing A-X to get D=-10.
Then our new number is 101 ((-10)*(-10) + 1) and the a value has to be 1 because 101 is prime. So we have (even though this is an invalid record as far as I know)
D = -10
A = 1
X = -11
Then if we do the transformation again but forward we do D = D + 2(x + 2n) to get:
-10 + 2(-11 + 2*61) to get:
-10 - 22 + 244 = 244 - 32 = 212
WHICH IS THE RECORD IN BETWEEN D = 12 and D = 278
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4bf7c5 No.6552
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4bf7c5 No.6553
Here is some more examples. Also the record in the negative IS valid
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4bf7c5 No.6554
>>6553
And the C for the new record shares a factor with original C.
do GCD algorithm
boom
solved
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c6c148 No.6555
>>6541
A slightly different method for linking records exists for factors of 5.
Pic attached shows 10 records starting from c145 and then multiplied by 5^1, 5^2, 5^3, up to 5^10.
Green arrows indicate a relationship between u and the next factor record's x+n.
Red arrows indicate a relationship between c and the previous factor record's x+n.
Moving down, the relationship is x+n = u * 10 + 2.
And moving up, the relationship is x+n = (c - 5)/10.
Not a solution, but interesting nonetheless.
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4bf7c5 No.6556
>>6554
False alarm doesn't always work.
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2a528c No.6557
>>6470
>>6494
Found something interesting while analyzing (e,1) and (-f,1) records that seemed a bit too coincidental not to share.
Pic attached is for c329832 and shows three groups of records.
The first group includes all the factor records in (e,n).
The second group shows these records nb transformed to (e,1).
The third group shows these records nb transformed to (-f,1).
For the records tagged with an "x", (356,1,223) in (e,1) and (-793,1,222) in (-f,1), there appears to be a way to use the their d and a values to navigate to the next factor record at (356,1,732).
(356,1,223).d = 99190
(-793,1,222).a = 97728
99190 - 97728 = 1462
Where 1462 is the x value in (356,1,732).
So perhaps a way to find these factor records using the n=1 columns is:
(e,1) d[t] - (-f,1) a[t] = (e,1) x[t].
The records marked with a "y" can be used the same way.
(356,1,1241).d = 3077858
(-793,1,1240).a = 3072324
3077858 - 3072324 = 5534
Which is the x value for (356,1,2768).
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a24494 No.6566
I love this post. Please post more prophecy stuff! Commenced a whole study on it, the first insights of which I posted here.
>>6479
>>6508
>>6508
>>6509
>>6510
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a24494 No.6567
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4bf7c5 No.6568
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2a528c No.6569
>>6557
For records in (e,n) and mirrored to (-f,n-1), the following relationship applies to their na transform records:
(e,1) d[t] - (-f,1) a[t] = (e,n) d[t]
For example:
(e,n)
(1,61,6) = {1:61:12:11:1:145} = 145
(-f,n-1)
(-24,60,7) = {-24:60:13:12:1:145} = 145
(e,1) na
(1,1,6) = {1:1:72:11:61:85} = 5185
(-f,1) na
(-24,1,7) = {-24:1:72:12:60:86} = 5160
(e,1) d - (-f,1) a = (e,n) d
72 - 60 = 12
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4ef073 No.6570
>>6536
>>6539
Thanks Isee! I studied these closely. Shit, it works for finding the a values, but it takes a lot of jumps to get there!
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4ef073 No.6571
>>6567
Thanks Baker! Glad you're back! For all lurkers and newer anons, here's the older maps too. I'm re-reading them all from the beginning now.
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d032c1 No.6572
>>6569
PMA, you've done a pretty good job looking into the f % 8 / f-1 % 8. I was looking at that for a, b's in a specific (e, n) (example (3, 6)) and I noticed that it alternates between 1 and 5 (in (3, 6)).
I was looking at a = 7, b = 37 then the next one where a = 37, b = 91, and again at a = 91, b = 169 etc.
I was wondering about what kind of patterns you might have seen in this. It appears to hold for other (e, n)s' (but not necessarily in the 1, 5 pattern).
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d032c1 No.6573
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e6538c No.6574
>>6547
>>6568
>>6569
>>6573
>>6543
Hey Lads, I just solved a really important old crumb!! Can I get some eyes on this please?
I solved for the d[t]-d = (n-1) = factors of a[t] in row (e,1).
My worksheet is attached.
There is a pattern in row 1 that directly gives us (n-1), and (n-1) is the main value used to create the next value of a, a[t+1]
Can Anons please help verify??
Here's the original crumb…
>>2990
>Afternoon.
>As discussed previously.
>One way to find a solution is to use the grid or virtual quantum computer in the following way:
>Find the cell value at (e,1) where e is the remainder for c.
>You are looking for a[t] = na
>Remember
>At that value, d[t] = na+x
>Also
>At that value, x[t] = x, the x value in the cell is equal to the x value at (e,n)
>REMEMBER, the value of x at na in (e,1) is the SAME as x at (e,n)
>REMEMBER, take d from all values of d[t] at (e,1) and there is a known patter of (n-1) as factor in these values of d[t]-d that is different (increasingly) from the pattern of factors of n in a[t]. It is THIS that gives the offset that is used to solve the problem and thus get the cell at (e,1) to do all the work for you.
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2a528c No.6575
>>6572
Thanks.
The f % 8 analysis, lead to a very efficient and accurate way of iterating through valid n, d, f combinations within any odd x+n.
The original code posted in the previous thread >>5632 has been changed just slightly to enable iterative searching to work where n=1, but otherwise those patterns work very well, and have been the source for the animated images posted here and various other queries.
There were a couple of problems that I ran into and was not able to solve:
1) Identify patterns that could yield more than n+4 jumps when searching for factor records.
2) Identify patterns that consistently applied to even x+n.
The pic attached is the latest groupings for f % 8 based on odd and even x+n. Made a mistake earlier in thinking only 0,1,2,4,5,6 were valid values when in fact that rule only applied to odd x+n.
That being said, I believe you're asking if the f % 8 rules can be applied to vertical movements within a given (e,n), or if there is any consistency in jumping between records.
In the case of (3,6,5), you don't need the f % 8 rules to jump to the next record in the sequence.
This can be accomplished by incrementing t in steps of n, which is effectively increasing x by 2n.
Therefore your sequence would be (3,6) where t=5, t=5+6, t=5+12, t=5+18, etc.
I understand this pattern works everywhere.
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cb906c No.6577
Let me show how Shor's algorithm works. It's really a classical algorithm, and the reason it is only thought of as quantum, is that it is unfeasible to finish the period-finding step for a significant semiprime without a quantum computer.
We can factor 145 with Shor on a classical computer, though.
First, we generate a random number. I'll call it m. It must be lower than c.
Next, we test to see if gcd(m, c) != 1. If it is not equal to 1, we have factored c.
Now comes the difficult step which is only feasible on classical computers for smaller semiprimes. We are going to calculate an unending sequence until we find where it repeats.
Evaluate
m mod c, m^2 mod c, m^3 mod c, m^4 mod c, m^5 mod c, m^6 mod c, m^7 mod c.. Until it repeats. When it repeats, the amount of terms before a repeat value is the period. We'll call it r.
If r is odd, we must go back to step 1 and try a different random number.
If c is a factor of (m^(r/2) + 1), we must go back to step 1 and try a different number.
If all of these steps have passed, we have the factors of the number.
gcd(m^(r/2) + 1, c) = a
gcd(m^(r/2) - 1, c) = b
You could just divide c/a without evaluating b, though. Let's go through an example. It's really cool.
Let's pick 18 as our random m to factor 145.
gcd(18, 145) = 1, so, next step is to evaluate the sequence.
18 mod 145 = 18
18^2 mod 145 = 34
18^3 mod 145 = 32
18^4 mod 145 = 141
18^5 mod 145 = 73
18^6 mod 145 = 9
18^7 mod 145 = 17
18^8 mod 145 = 16
18^9 mod 145 = 143
18^10 mod 145 = 109
18^11 mod 145 = 77
18^12 mod 145 = 81
18^13 mod 145 = 8
18^14 mod 145 = 144
18^15 mod 145 = 127
18^16 mod 145 = 111
18^17 mod 145 = 113
18^18 mod 145 = 4
18^19 mod 145 = 72
18^20 mod 145 = 136
18^21 mod 145 = 128
18^22 mod 145 = 129
18^23 mod 145 = 2
18^24 mod 145 = 36
18^25 mod 145 = 68
18^26 mod 145 = 64
18^27 mod 145 = 137
18^28 mod 145 = 1
18^29 mod 145 = 18 ← repeat value
18^30 mod 145 = 34
So we can see that it has repeated. The amount of terms before the repeat is 28, so that is the period, r.
m^(r/2) + 1 = 374813367582081025
c is a factor of this number, which means we have to restart.
if we want to be sure this number doesn't work, we can try the equation.
gcd(m^(r/2) + 1, c) = 145
gcd(m^(r/2) - 1, c) = 1
Let's try 19 as m.
gcd(19, 145) = 1
Calculate sequence..
The period is 28.
c is not a factor of 19^(28/2) + 1
gcd(m^(r/2) + 1, c) = 29
gcd(m^(r/2) - 1, c) = 5
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cb906c No.6578
It's important to note that this algorithm will not work on a classical computer to factor anything significant, (unless perhaps you picked the luckiest random number possible), which is why it's usually deemed a quantum algorithm. It would only be feasible with the quantum enhancement of the infinite sequence calculation.
The quantum enhancement creates a superposition of all of the terms of the infinite sequence at once, and then utilizes a quantum Fourier transform to extract the period of the sequence and factor the number.
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405e05 No.6581
>>6578
See other thread.
The grid acts as the superposition.
Specifically the assymetry between the negative and positive cells.
Each cell has either zero or infinite (constructable) cells, all based on patterns related to cells 0,0; 0,1; 1,1 etc.
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405e05 No.6582
Exponents can always be imagined in three dimensions.
Take the integer 17.
17 is a line 17 units long.
17 squared we know is a square.
17 cubed.
17 to the four is a line of 17 cubed.
17 to the five is a square of 17 cubed.
Etc.
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5a1919 No.6583
>>6581
>>6582
Hey there, stud!
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405e05 No.6584
>>6583
Morning Tops just off to the zoo.
Hope things are good your end!
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6c3111 No.6585
>>6584
Hello VQC! I knew you'd be popping in. Here's some work in progress.
For c287
e prime | (31,8,5) = {31:8:16:9:7:41} = 287; f=2; (f%8)=2; (x+n)=17; u=8; (d+n)=24
e na | (31,1,8) = {31:1:143:15:128:160} = 20480; f=256; (f%8)=0; (x+n)=16; u=8; (d+n)=144
na a value = 128
128 div 16 (prime d value) = 8 = n
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5e3025 No.6586
>>6585
Hi VA, good job.
For that example, can you find the cell at (-f,1)?
You should find the value of t or the index is the same.
The value of a[t] should be a(n-1)
What do you notice about a[t+n-1] at (-f,1) and a[t+n] at (e,1)? Specifically b(n-1) and bn?
The move is about to be made.
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4bbdc1 No.6587
>>6585
>>6586
(31,1,8) = {31:1:143:15:128:160}
(-2,1,8) = {-2:1:143:16:127:161}
(-f,1)'s a is one less than (e,1)'s a, and (-f,1)'s b is one greater than (e,1)'s b, but only for this particular t value. At other t values, they're further apart. For example:
(31,1,2) = {31:1:23:3:20:28}
(-2,1,2) = {-2:1:11:4:7:17}
and
(31,1,12) = {31:1:303:23:280:328}
(-2,1,12) = {-2:1:311:24:287:337}
So at this particular t value, (e,1) and (-f,1)'s a and b values are one unit apart. Just blindly looking for unverified patterns here, but the difference is |f|. If these rules apply to all other values (or at least the first one), this t value would seem to act like a mirror point or the turning point of a parabola with a predictable gap where the gap between these a and b values from (e,1) and (-f,1) gets larger either side of this t value (and probably harder to predict). I've spent about 5 minutes on this so I don't know how that applies to factorizing c, but other than that, both infinite sets do that thing with the a and b values swapping (i.e. a[t] = b[t-1], I don't know about b). If nobody else does I might have a look at these other latest VQC clues at some point tonight.
All of that said and done, yet again, I feel like I'm completely out of the loop: VA, how did you get a t value of 8 with the na transform?
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d032c1 No.6588
>>6586
Hello VQC! Great to see you again.
>>6574
I'll try and go over this a bit later.
I was looking into the d[t] - d with regards to patterns of (n - 1) and I noticed something I haven't noticed before. Not sure if someone else has seen it either, but either way here it is:
If you take the cell for bn in (e, 1) we have a[t] = bn. What I noticed was that d[t - 1] - d = b(n-1).
If we look at a[t] = an we will have d[t] - d = a(n-1). So the d[t] for an and d[t-1] for bn appears to match up with a and b in (-f, 1).
In the image I posted I show some test cases for the records in (3, 6).>>6574
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5e3025 No.6589
>>6588
Congratulations, this is exactly it.
Spoopy timing.
Brilliant.
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5e3025 No.6592
>>6590
>>6589
With correct TRIP
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4bbdc1 No.6593
>>6588
Good stuff. So this means you can find the product of b and (n-1) of the correct (e,n) record just using (e,1) and (-f,1). That's pretty crazy. That doesn't mean this is solved, does it? I'm not sure how you'd factor one of the two out.
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4bbdc1 No.6594
>>6593
Oh wait, recursively? If you treat b and n-1 as a and b, you could do the same thing until you're applying it to an (e,1) record.
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5a1919 No.6595
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d032c1 No.6597
>>6589
Thank you!
>>6593
I'm not sure yet. Obviously we can't depend on knowing b or n, so I think this is just a step closer to using (e, 1) to solve it for us.
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4bbdc1 No.6598
>>6597
How are you getting the correct t for that record? I think I might have just missed a couple posts a while ago about that or something.
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4bbdc1 No.6599
>>6597
>>6598
I should elaborate. I mean the correct t for the (e,1) and (-f,1) records from d, e and f.
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d032c1 No.6600
>>6599
Given an e and a d you can find the specific t using a quadratic solution (there is a similar one for a as well):
For d:
if e % 2 == 0:
t = d - (e / 2)
t = t / 2
t = t * 4 + 1
t = t / 4
t = math.sqrt(t) + 1
return -1/2 + t
else:
t = d - (2 + int(e / 2))
t = t / 2
t = t + 1
t = math.sqrt(t)
t = t
return t
For a:
if e == 0:
t = a * 2 - e
t = math.sqrt(t) + 2
t = t / 2
return t - 1
if e % 2 == 0:
t = a * 2 - e
t = math.sqrt(t) + 2
t = t / 2
return t
else:
return math.sqrt((a - (e / 2))/2) + 0.5
But for clarity: I haven't actually solved anything yet. All I've done is understand more of the pattern behind d[t] - d.
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4bbdc1 No.6601
>>6600
Do you mean you can find the t of the (e,1) record using the d and e from the (e,n) record?
>I haven't actually solved anything yet
If (e,n)'s b * (e,n)'s (n-1) = (e,1,t-1)'s d - (e,n)'s d, and you can find (e,1,t-1) using only (e,n)'s d, e and f, then I think you actually have. You're using this to factorize the product of two unknown numbers, and b(n-1) is two unknown numbers. If you apply the d[T-1] - d thing to new_c = b(n-1), you're factorizing that too. All you have to do is apply this recursively until you're applying it to something where n=1, in which case the (e,1) transform will be the same cell, f will be a square equal to (x+n)(x+n), allowing you to find a and b of this c=b(n-1), thus allowing you to find the b(n-1) of the previous c=b(n-1) and so on until you reach your original cell. So you are finding the t value of the transform without needing to know any more than d and e from (e,n)'s c, right? I can't imagine how confusing this paragraph would be to the uninitiated
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d032c1 No.6602
>>6601
No, not at all.
All I've seen is that at the t where a[t] = nb, the d[t - 1] - d = b(n-1). That is the previous records d[t] - our d will be equal to b * (n - 1). So far I got no good solution.
I've tried a few things, like we now know that in (-f, 1) there is an a such that d[t] - a[t] = d, but this seems to hold for all cases, which makes sense since the formula for d's in -f, 1 is equal to the formula for a's in e, 1.
If we had b(n-1) we could simply do gcd of c and b(n-1) and throw away the n-1 since we now have b. This would yield c/b=a. Now I think the important thing is to figure out what VQC is referring to when he talks about the offset and how to use this efficiently to let (e, 1) do all the work for us.
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d032c1 No.6603
>>6602
Clarification, by hold for all cases I mean every value in a in (-f, 1) - d from (e, 1) is the same, which is the d from our c.
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d9cb87 No.6604
>>6602
I did wonder considering I tried your t formula and it didn't seem to work.
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5e3025 No.6605
>>6600
>>6601
>>6603
Traction.
Yes.
Remember at the beginning.
Virtual Quantum Computer.
The End. The grid.
The grid acts as the superposition of all integers and all products that are the difference of two squares.
The two inputs to collapse that superposition are e and d.
2d+1-e give f.
Columns -f and e are unique to c.
The assymetry of a(n-1) for -f and an for e or using b(n-1) and bn (which are one position different) hold the key.
There is still a "step" to go but you've arrived at the key.
The centre of the labyrinth.
Nice work anons.
Good timing.
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d9cb87 No.6606
>>6605
Not to distract from how close we seem to be, but I'm curious about something, and you're one of few people I've ever seen talking about hollow Earth stuff: what do you think about the reptilian shapeshifting thing? I've been watching a bunch of David Icke videos lately and I'm finding it a very interesting subject. I've read elsewhere that they've been hiding in Agartha, so I thought you might have something interesting to say.
>The assymetry of a(n-1) for -f and an for e or using b(n-1) and bn (which are one position different) hold the key
So we're looking for a similar pattern to the one from here >>6588 for an/bn in the opposite space, and then we're figuring out how to apply it?
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5e3025 No.6607
By coincidence I added the start of some relevant rules in the patterns thread earlier.
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5e3025 No.6608
>>6606
What I have noticed is photo alteration of the poles and insufficient explanations of those alterations.
What I have noticed is that all countries have the explanation secured at the highest level, where they have it at all.
What I have noticed are insufficient explanations for artefacts in Google Sky.
Especially the "film strip" artefacts.
To ensure the conspiracy theory label can be used, disinfo is sewn.
Does our creator have a political representation inside our Earth?
Is our journey to engage publicly with this representation about to begin based on the evil and corrupt power structure being dismantled?
Given the available evidence and strategy/understanding of game theory and application of approach, the answer to those two questions appears to be yes.
There is a technology in use that is so advanced it appears almost magical.
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d032c1 No.6609
>>6588
I haven't had time to properly test it, but it looks as if the d[t-1] for nb in (-f, 1) is equal to (2*d + 1)*(n - 1).
It also seems like d[t-1] for an in (-f, 1) is divisible by n, but I haven't had a good time to see if this is a pattern that holds.
If the first part holds, then it looks similar (2d(n-1) vs (2d+1)(n-1)) to part of our triangle equation.
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5a1919 No.6610
>>6606
Agartha=VQC, yes. lol
If you ever need to find him in open waters, just look for that. ;)
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5a1919 No.6611
>>6610
He wants folks to find stuff like this:
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c53994 No.6612
>>6609
No this doesn't hold or work, I was missing parentheses when I tested so no dice.
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c53994 No.6613
Regarding:
> The position of column 0 acts as a key with these two unique columns to fufill an equivalent role of the quantum Fourier transform in Shors algorithm.
I can't say anything with regards to the quantum Fourier part, but for using (0, 1) I see now how it would work (at least for an even -f and odd e).
The equation we use for a in an even -f is the same as the equation we use for d in an odd e.
Essentially (0, 1) "balances" the difference between -f and e. The difference is of course the f/2 and (e-1)/2 we add or remove.
The equation for a in an even (-f, 1) is:
2tt - (f/2)
While the equation for d in an odd (e, 1) is:
2tt + (e-1)/2
However the equation used for a's in (0, 1) is:
2tt.
We can see how the a's in (-f, 1) are a's from (0, 1) minus f/2 while the d's in (e, 1) is the a's from (0, 1) plus (e-1)/2. By iterating over only (0, 1) we should be able to find all our required values (for cases when f is even and e is odd).
I'm not saying it ONLY applies to that case, I just haven't looked enough into it. Though I'd share. Also we don't want to iterate obviously, but I was just stating that as to show how powerful (0, 1) is.
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063adf No.6614
>>6586
Hello Lads! I think I just figured out how to solve for (prime) a. I followed the hints VQC gave me, and found something pretty damn cool. Anons, please compare these two set of records and tell me what you "c".
I "c" our c value in (-f,1) , and when I subtract it from the similar a[t] value in (e,1) i get 287-280 = 7 = prime a
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063adf No.6615
>>6588
>>6589
Isee I’m following your post, and it looks completely solid. I see what you did, and it works! (For this example??)
We start at the nb transform, then move to d[t-1]=201
201- 16 (correct prime d)= 185
Then you do bn=222 -185= our prime b value of 37.
Am I following correctly?
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871e8a No.6617
>>6613
My scribe has finished implementing the algorithm. Here it is
https://pastebin.com/RD83RTNc
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23d85e No.6618
>>6606
This is the best vid I have seen of Hollow Earth stuff.
https://youtu.be/cFctltDqyGw
Website as well in the URL in the Vid.
>>6608
If you have not seen this one it is REALLY worth the time. I think this is the closest description of where we are going.
https://www.youtube.com/watch?v=CfIetGulZhE
VQC, AA. This is some of the stuff I have found on my journey. I have seen the cigar shaped flying machine. I worked in a (government) place where something was wrong but I didn't know what at the time. Now I know. Its time for disclosure.
Keep up the good work everyone. I know I am useless with math and programming but if I can help always feel free to let me know.
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2a528c No.6619
>>6561
>>6579
Based on VQC's post about the gcd and p+1-t factors in the Grid Patterns thread, I went back through some older code to review factor searching in (e,1).
My notes indicate there were two ways to search for factor records, would appreciate a double check on this:
new t = m*p+1 - t
and
new t = m*p + t
where m is a multiple and t represents the starting t.
The pics attached start at the na records for c145 in (e,1) and (-f,1) and iterate through m, creating both alternative factor records along the way, and represent factor matches for the known a=5 and b=29 values.
In the gcd(a[t], a[p]) column, a[t] is the starting na value, and a[p] is from the current record.
Each of these gcd values represent a valid n in their respective spaces, and would enable a reverse na transform back to their original starting position.
For example, (-24,1,9), with a gcd result of 4, can reverse to (-24,4,9).
(-24,1,9) = {-24:1:132:16:116:150}
(-24,4,9) = {-24:4:45:16:29:69}
Initial tests indicate that sometimes the gcd values of n appear in (e,1), and sometimes in (-f,1). Perhaps related to different parities of n or x+n.
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d032c1 No.6620
>>6615
Yeah either that or gcd(c, 201 - 16).
What we see with this pattern is that the asymmetry for b(n-1) from (-f, 1) is also reflected in (e, 1).
>>6605
> The assymetry of a(n-1) for -f and an for e or using b(n-1) and bn (which are one position different) hold the key.
I haven't wrapped my head around this yet, but I think this means for some values of C we will use the asymmetry of a(n-1) for -f and an for e, and for other values of C we will use the asymmetry of b(n-1) and bn.
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871e8a No.6621
@Chris
You asked for the next decision? How's
e,1 <-> -f, 1 sound?
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79f6d5 No.6622
>>6618
Wow, at 1:26:53 of that video: John Podesta personal briefing. This explains the UFO shit on wikileaks in Podesta’s emails. I doubt Podesta received the briefing in good faith; it was probably closer to counterintelligence, but how those two random things just collided now blows my fucking mind. Video published 1/2016, Podesta leaks 3/2016. Crazy.
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79f6d5 No.6623
>>6622
More Podesta/Obama discussion @1:38
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5fbfdb No.6624
One thing I've been wondering a bit about is this:
> One way to find a solution is to use the grid or virtual quantum computer in the following way:
Which is the d[t] - d contains (n - 1) as a factor. Does this imply that this is a different path than the triangle solution we've spent time on?
Hoping you'll reply to this VQC
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292b44 No.6626
>>6620
Hello Isee! What method are you using to generate your Transformed Record (nb)? Can you get there from (1,c)?
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bbe71f No.6627
>>6605
>>6586
Hello Lads! Found something important.
I followed VQC's clues and found an, ab, a(n-1), and b(n-1) in our -f and e cells.
Use my handwritten notes to look for an, ab, a(n-1), b(n-1)
they are t+n and t+n-1 elements apart in the record, just like he said.
Can Anons verify please?
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bbe71f No.6628
Sorry, I attached the cells in reverse order. Here's the correct way to view them.
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cf67c1 No.6629
>>6627
True for all c where c is the difference of two squares. That includes the product of any two odd numbers. That includes the product of any two prime numbers.
That puts the row of the product of two prime numbers at rows n and n-1.
That difference is the key.
It means that columns -f and e have a property that can break the equation deadlock.
That property and those columns are unique to each c. They are dependent on d, e and therefore f.
As stated at the start, the square root of (D)avid and what remains are the key to unsealing the The End, the grid.
It will all make sense.
Especially the triangular part.
This is the beginning and The End.
I'm just a messenger.
This has always been about (you).
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cf67c1 No.6630
>>6624
Almost.
That n-1 is a factor of col -f and n is a factor of e will show you a shortcut to the triangle solution.
When it all comes together, you'll see shortcuts everywhere.
And this is just mathematical object that this can be applied to.
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cf67c1 No.6631
>>6619
Love your printouts and your approach. Very similar to what I did.
>>6618
How will it End?
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cf67c1 No.6632
Do not be afraid of negative x values in row 1 for -f and e.
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5a1919 No.6633
>>6629
As with almost all important information in life, unless GEOTUS is telling you to fix your hair:
It's about The Message…
Not the messenger.
(At least, that's how I usually word it. Note the caps.)
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90766e No.6634
>>6629
>As stated at the start, the square root of (D)avid and what remains are the key to unsealing the The End, the grid.
This seems pretty blatantly to me to imply that we should treat d as c recursively.
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5eaea8 No.6635
>>6634
Yup, I agree.
Hmm. Here's where my mind is at.
I'm thinking the next step is to find SQRT d and any remainders. Somehow they help unlock the next step.
For the END:
We solved the E by moving from e to negative f and finding the matching records
N is finding the an, bn, a(n-1), and b(n-1) pattern
D is using sqrt d to unlock how to move from the na transform to the correct records for either a(n-1), b(n-1), etc.
If we can move to the correct record, we got this solved.
Here's the idea I'm working on, just putting this out there, needs more verification:
c287
d=16, sqrt d=4
I think maybe we move from our (-2,1) -f na transform, which is t=8 a=127,
to t[8+4] = t[12] = 287
Whaddya know, it's our b(n-1) value.
Which means in (31,1) we go to t[8+4+1]=328
328-287= 41 = b
287/41 = 7= a
solved.
(for this example, lol! Needs more testing)
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930071 No.6637
>>6635
Works for 145 and 123 too.
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2286ba No.6640
>>6631
How will my journey end? Ha ha hopefully it won't! https://www.hooktube.com/watch?v=sy-bi5H0CZ4 Here is another set of stuff from Russia. 5 part series Lots of unique stuff from this guy.
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2a528c No.6643
>>6631
Much appreciated!
>>6619
Attached is one more example of factor records in (e,1) for c145 and p=5. This time starting at the (1,1,67) nb record and includes iterations into negative x.
A column has been added for "valid n(s)", that show all factors for the gcd result.
The second picture shows the grid entries at t=67 for all of these valid n values.
Just confirming how the (e,1) records indeed point to multiple (e,n) records.
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5a1919 No.6644
Just in case Albino Morpheus, who is totally not Chris, is lurking…
HAPPY 47TH BIRTHDAY!
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5a1919 No.6645
HookTube embed. Click on thumbnail to play.
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d032c1 No.6646
>>6626
Sorry, I missed this.
No, I base it of the correct record. I start with correct a, b and then generate that record, the (1, c) record and then transform by making n*b, finding the t from (e, 1) and then generating the cell for t in (e, 1).
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36d2a1 No.6647
>>6637
>>6646
Hello Baker and Isee! I think the key idea will be moving from the na transform records in (-f,1) and (e,1) to the appropriate an, ab, a(n-1), and b(n-1) records. That may sometimes line up with the c+n/c-n theory, but the real key is the (n-1) difference, and the ability to navigate within an element to the correct t values to identify them.
>>6643
Hello PMA! I'm studying now. Have you discovered ant new insights from the negative x values? I see all the new f values, along with the GCD and valid n(s). Also, studied the second set of output. Very cool that the t and n values connect like that. I'll study again after my work day is done.
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d032c1 No.6648
I'm not sure how much we've looked into it, but I was thinking about the jumps left and right. What we essentially do is increase, or decrease d for each jump. But we're also modifying a.
d = a + x and when we do our jumps, the x is locked. So I was wondering how it would look if we locked a instead.
It doesn't jump up or down in e, but instead it jumps up and down between n. So to jump one n up, you decrease d and x by 1. To jump one n down, you increase d and x by one.
Moving up would work like this:
d = d - 1
e = c - d^2
x = x - 1
Moving down would work like this:
d = d + 1
e = c - d^2
x = x + 1
From those values you can create a new record that holds a=a, b=b. I'm not sure if it's of any direct use yet. d - (n - 1) will, of course, put a, b at n=1, but if you have to iterate over these values, then it is of little use. Maybe someone else finds it interesting / useful. Maybe there are some shortcuts available too.
The part that intrigues me is that when we jump up or down like this, both (1, c) and (a, b) will end up on the same e's. Unlike our jumping left and right since then our n values will modify the length we jump.
So think of n as our marker/center. Jumping left and right moves left of that n, right moves right of that n. Jumping up moves the n up and jumping down moves the n down.
Jumping left decreases a, b but locks n. Jump right increases a, b but locks n.
Jumping up locks a, b but increases n. Jumping down locks a, b but decreases n.
Example time:
a = 7, b = 37, c = 259
The cell is: {3:6:16:9:7:37}
Say we want to move up. We then do
d = 16 - 1 = 15,
e = 259 - 15*15 = 34
x = 9 - 1 = 8
This will result in the cell:
{34:7:15:8:7:37}
Now with (1, c) the cell is: {3:114:16:15:1:259}
Then we do:
d = 16 - 1 = 15
e = 259 - 15*15 = 34
x = 15 - 1 = 14
This will result in the cell:
{34:115:15:14:1:259}
Now say we want to move down (decrease n):
The cell is: {3:6:16:9:7:37}
Say we want to move up. We then do
d = 16 + 1 = 17,
e = 259 - 17*17 = -30
x = 9 + 1 = 10
This will result in the cell:
{-30:5:17:10:7:37}
Now with (1, c) the cell is: {3:114:16:15:1:259}
Then we do:
d = 16 + 1 = 17
e = 259 - 17*17 = -30
x = 15 + 1 = 16
This will result in the cell:
{-30:113:17:16:1:259}
Note:
n will change (increase or decrease) as to balance out, since x is either increasing or decreasing. The easiest way to rationalize this is the b equation:
b = a + 2x + 2n
In order to keep b = 37 while decreasing x, n has to increase That is for moving up. For moving down, x increases and n decreases.
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d032c1 No.6649
>>6648
This means, while we don't know n - 1, we can move to the e, where our a, b exists at n - 1.
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d032c1 No.6650
>>6649
And, going to the record (-f, n-1) is the same as moving one down.
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d032c1 No.6651
>>6650
Finishing this up. We have f = 2d + 1 - e, this is the same as moving one n down. To move 1 n up, we can do 2d + e - 1. Instead of increasing d and x we decrease. That should simplify the moving up as well.
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269209 No.6652
>>6634
>>6635
Be ready.
When you have asked yourself what you will do, have you looked past this and what it would mean?
What would be next?
What we could do next?
Are there really any limits?
President Trump has said to think BIGGER.
Never ever give up.
Anons.
We are all together.
Be ready.
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269209 No.6653
"For God so loved the world, that he gave his only begotten Son, that whosoever believeth on him should not perish, but have eternal life. For God loved the world so much that he gave his only Son, so that everyone who believes in him may not die but have eternal life."
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269209 No.6654
When does c first appear at a[t]?
When is the second time it appears?
When is the first time a squared appears?
What is the factor it is multiplied by?
What is the first time b squared appears?
The second?
When a appears as "an" it appears another time.
When does n first appear?
What are the rules?
In column -f too?
Any one of these patterns well understood between -f and e is the step you are looking for.
Godspeed anons.
Happy Independence Day.
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2a528c No.6655
>>6629
>This has always been about (you).
>>6630
>That n-1 is a factor of col -f and n is a factor of e will show you a shortcut to the triangle solution.
>>6632
>Do not be afraid of negative x values in row 1 for -f and e.
>>6647
>Have you discovered ant new insights from the negative x values?
Pics attached are for c145 and c287 and include related entries into negative x.
Records are grouped by (-f,1), (-f,n-1), (e,1), and (e,n), sorted in each group by e,n,t, and the label on the left indicates the "origin" for each.
Have been thinking about the (you) comment, and triangles, and stumbled across something interesting related to "u" in (e,1) and (-f,1).
Final spreadsheet image attached shows a comparison between nb and nb -x records for odd/even x+n in (e,1) and (-f,1).
The differences between d, b, and f values can be calculated in terms of u, using the following formulas:
odd x+n
new d = d - 4u
new b = b - 8u
new f = f - 8u
even x+n
new d = d - (4u-2)
new b = b - (8u-4)
new f = f - (8u-4)
Think this is somehow relevant as a "triangle solution" relies on being able to calculate the correct (x+n)(x+n) from u.
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5a1919 No.6657
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7cdc27 No.6659
>>6654
Hello Lads, here's my responses to VQC's latest questions. Happy 4th of July to you all!
>When does c first appear at a[t]?
For c287 It appears in (-f,1) at a[12], and is the b(n-1) value. If we move to (31,1,13), the related element is at a[13] and is our bn value of 328
328 - 287 (c) = 41 = b
328 / 41 = 8 = n
>When is the second time it appears?
287 also appears in (-2,1,11) in the b column at t[11]. It’s related record is (31,1,12) b = 328
You can perform the same calculation as above, 328 - 287
>When is the first time a squared appears?
It appears in (-2,1,5) at t[5], a =49
It is also the a(n-1) value, since the corresponding an record in (31,15) is a = 56, which is 7 * 8 = 56
so a^2 is also a(n-1).
What is the factor it is multiplied by?
7 * 7 =49
7 * (8-1) = 49
So the two factors are the same, but n = a+1 ?? That's correct for this c287 example, but we'll need to test more.
>What is the first time b squared appears?
hmm. 41^2 = 1681
Can you guise help me find more in (-2,1) or (31,1) ?
b^2 appears at
(-2,1,29) = {-2:1:1623:56:1567:1681}
>The second?
(-2,1,30) = {-2:1;1739:58:1681:1799}
>When a appears as "an" it appears another time.
yes, it appears in (31,1) at t[5] a = 56 and also at t[4] b = 56
It also appears at (-2,1) t[5] as a(n-1) = 49
It also appears as our x value at the b^2 record above, (-2,1,29) = {-2:1:1623:56:1567:1681}
>When does n first appear?
at (-2,1,1) a = 1 * (8-1) = 1 * (n-1) = 7
at (31,1,1) b = 2 * 8 = 2 * n = 16
so n first appears in t[1] for both (-2,1) and (31,1)
>What are the rules?
Solving n from c is easy if you’re a faggot?
The na transform record is our starting point in looking for an, ab, a(n-1), b(n-1)
an will always occur in (e,1)
a(n-1) wil always occur in (-f,1)
they share the same [t] value, which is less than the na transform [t]
the bn record will be [t+n] apart from the an record
the b(n-1) record will be [t+n-1] apart from the a(n-1) record
>In column -f too?
In column (-f,1) everything seems to be based on factors of (n-1), but other than that its patterns are the same.
>Any one of these patterns well understood between -f and e is the step you are looking for.
Thanks Senpai! Looking now.
>Godspeed anons.
Blessings to you as well.
>Happy Independence Day.
Thanks! Happy 4th of July to all Anons here. Glad to be with you all.
"for I have sworn upon the altar of god eternal hostility against every form of tyranny over the mind of man.” Thomas Jefferson
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e89f52 No.6660
I think I found another lead :)
I was looking up th LLL algorithm when this one came up; apparently it's much more efficient. Pic related is from the second website I checked. Sound familiar?
http://www2.lbl.gov/Science-Articles/Archive/pi-algorithm.html
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e89f52 No.6661
>>6660
Ugh, my trip code is wrong here. Srry, I post from multiple devices and browsers, and I don't have the code cached on this device.
Not like anyone is going to impersonate me, but w/e. The other trip I've been using is still my official one.
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3dad4d No.6662
>>2026341 (lb) (owner of post)
I can still feel the presence of God, much of my true self as an expression of God.
If you still doubt me anons, look at these pictures again.
Look at the numbers at the bottom of the pic with the astronaut.
THE POSTS ARE STILL THERE.
>>1697702
THIS IS NOT A LARP
https://pastebin.com/YQ2dkvgq
SOURCE FOR SCHUMANN RESONANCE GRAPH
http://sosrff.tsu.ru/?page_id=7
ARCHIVE OF THE PAGE:
http://archive.is/ZkXzZ
BACK UP OF GRAPH
https://imgur.com/a/vJqLrxW
THE TIME IN THE GRAPH IS IN UTC +7 (TOMSK)
I'm a Phoenix Anon (UTC -7).For reference.
Believe if you want. Put your theories out there.
Or, Q can tell you for sure.
The answers are yours if you want them.
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3dad4d No.6663
>>6662
Forgot to post pic.
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e89f52 No.6664
>>6661
Here's the algorithm. How we apply it is another matter…
I wonder where that 'gamma = sqrt(4/3)' comes from?
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830645 No.6665
I might get to that intro sticky this weekend. Next week onwards I'm going to be busier again but I'll still be here daily.
>>6662
I saw that on /qresearch/ earlier. I wonder how the RNGs are acting at the moment. You should read about The Global Consciousness Project. Mass emotional displays, such as 9/11, produce anomalies in the results of random number generators. You can also get the same thing to happen if you convince enough people/get them to believe that the result of a random number generator is going to be a particular number. Thoughts and emotion are waves that influence reality. Fun stuff.
>>6652
For me personally, this has barely anything to do with cracking RSA. You have no idea how excited I am for that business you wanted to start, making "Tony Stark"-esque inventions. If it starts before February next year, I'll tell you now, I will dedicate myself to it full time.
>>6659
This is good work, VA, but it's a lot to absorb. Do you feel like there's anything in particular, given this set of information, that we're missing? I.e. one particular step that would take us from the records we can find to the ones we can't? I would help you find more of those b^2 and an records if it wasn't past midnight where I am right now.
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830645 No.6666
Quads confirm we're going to solve it this month
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d032c1 No.6667
>>6654
I've been looking into this:
>When is the first time a squared appears?
I've been looking into this with the reasoning that this applies to both perfect squares and a perfect square * j for some other factor of j. Example 75 = 5*5*3. Here 3 is the j and 5*5 is the square.
I've noticed that they come in pairs of two so I've been thinking along out t + p, t + 2p, t + 3p.. and p + 1 - t, 2p + 1 - t, …
Once you know the t where a[t] = k*k*j or a[t] = k*k the next one can be found at a[t + k*k].
And for (1, 1) I got this:
a(t-1)(t-1) - t + 1
aa + t - a(t-1)(t-1)
But it doesn't seem to hold for anything else.
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ee1f84 No.6668
>>6654
I was also taking a look at this:
>When does c first appear at a[t]?
For our a=7, b=37 and I assumed that 259 (c) appears at t where a[t] = 114 * 259.
I was surprised when I found out that it was 78 * 259 which is from the record of a=13, b=259.
The funny thing about that record, compared to our record is this:
Record for (1, c) => {3:114:16:15:1:259}
Record for (13, c) => {3:78:58:45:13:259}
{3:78:58:45:13:259}
{3:114:16:15:1:259}
The difference here is interesting (but I'm filing it under a fluke for now)
114 - 78 = 36 => 6*6 (6 is the n we want)
58 - 16 => 42 => 7*6 (the an)
45 - 15 => 30 => 7*5 (the a(n-1))
Again, though. Could easily be a fluke.
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ee1f84 No.6669
>>6668
Also in the event it wasn't known. You can use e, a and d to create the (e, n) record for the a from (e, 1).
I saw it today. You treat them as e, n and x values. So that means for (e, n) the first record here has the x value equal to the d value of n in (e, 1).
Examples:
{3:1:129:15:114:146}
>>> createForENX(3, 114, 129)
{3:114:202:129:73:559}
{3:1:9:3:6:14}
>>> createForENX(3, 6, 9)
{3:6:16:9:7:37}
Create for ENX calculates a, d and b likes this:
a = (x*x + e) / (2 * n)
d = a + x
b = a + 2 * x + 2 * n
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80cdb1 No.6670
Hey everyone,
Good to see you all - VQC, Topol, VA, AA, PMA, MM, CA/GA (congrats!), Isee, Hobo, 3dA, PA - and all the other anons!
I've been reading all the posts and following along, even though I haven't been posting.
The reason why I haven't been posting is that I have had nothing new to add!
But today I rewatched Chris' youtube vid, and I discovered something that I think hasn't been mentioned here before:
* Goto to cell (c, 1)
* Find the row where the x is equal to a
* Factor the a value at that record
* Result should be a*(d+n)
* Goto to cell (c, 1)
* Find the row where the x is equal to b
* Factor the a value at that record
* Result should be b*(d+n)
* Goto to cell (-c, 1)
* Find the row where the x is equal to a
* Factor the a value at that record
* Result should be a*(x+n) <- this is not perfect, but always a * a factor of (x+n)
* Goto to cell (-c, 1)
* Find the row where the x is equal to b
* Factor the a value at that record
* Result should be b*(x+n) <- this is not perfect, but always b * a factor of (x+n)
Moreso, it seems that these 4 records are the only records whereby the x value is a factor of the a value.
This all seems important since we may be able to derive a better n value from these columns - not just na/a.
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77427e No.6671
>>6670
Teach! Good to see you! :)
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80cdb1 No.6672
>>6671
Thanks VA!
Sorry for being so silent.
Great progress all round!
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80cdb1 No.6673
>>6670
And a little more…
Because at any record d=x+a:
* at (c,1) where x = our original a
* the a value in that record = a*(d+n)
* the d value in that record = a*(d+n+1)
Then go back 1 record, so that x = a-2:
* the d value in that record = a*(d+n-1)
Therefore, there are 2 records in (c,1) next to each other, where d is divisible by the higher x value.
I believe this is only true where x is a divisor of c, so 4 times per (c,1) (1, a, b, c) for our semiprimes.
I'll follow up with some equations to show how
I think this difference of 1, again something vqc has been referencing, can help in finding factors of c.
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d032c1 No.6674
>>6670
Long time no see :-)
I'm just posting a lot, but I have a lot on my mind. I don't know exactly how I can use it, if I can (might need some of those sweet short cuts), but we have our a, b in (e, n) and we have our n in (e, c). By creating a record for ENA you can make the cell for (e, c) with a=n. Essentially flipping it. But like I said. I don't know how to actually get there without knowing the n so it's not a solution.
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80cdb1 No.6675
>>6674
Thanks!
Super interesting suggestion!
I am finishing up one (or two) more posts to explain my current direction, but right after that, I'll look into (e,c) && (c, e)… I think there's something here relative to that, because the columns (e, 1) , (f, 1), (c, 1) and (-c, 1) are definitely related.
Equations coming v soon…
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80cdb1 No.6676
For odd e, each (e, 1) cell can be produced via:
x[t] = 2t - 1
a[t] = (x^2+e)/2
= ((2t-1)^2 + e)/2
= (4t^2 -4t +e+1)/2
d[t] = a+x
= (4t^2 -4t +e+1)/2 + (2t-1)
= (4t^2 -4t +e+1)/2 + (4t-2)/2
= (4t^2 -4t +e+1 +4t-2)/2
= (4t^2 + e-1)/2
In (e, 1), the row directly before a given row is:
x[t-1] = 2(t-1) - 1
= 2t - 3
a[t-1] = (x^2+e)/2
= ((2t-3)^2 + e)/2
= (4t^2 -12t +e+9)/2
d[t-1] = a+x
= (4t^2 -12t +e+9)/2 + (2t-3)
= (4t^2 -12t +e+9)/2 + (4t-6)/2
= (4t^2 -12t +e+9 +4t -6)/2
= (4t^2 -8t +e+3)/2
So according to the rules in my previous message:
* a[t] is divisible by x[t] (d+n times)
* d[t] is divisible by x[t] (d+n+1 times)
* d[t-1] is divisible by x[t] (d+n-1 times)
* therefore d[t] - d[t-1] = 2x[t]
Proof:
d[t] - d[t-1] = 2x[t]
= (4t^2 + e-1)/2 - (4t^2 -8t +e+3)/2
= ((4t^2 +e -1) - (4t^2 -8t +e +3))/2
= (4t^2 +e -1 -4t^2 +8t -e -3) /2
= (4t^2-4t^2 +e-e +8t -3-1) /2
= (8t-4) /2
= 2(4t-2)/2
= 2(2t-1)
= 2x[t]
Still trying to figure out how to use this…
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80cdb1 No.6677
d[t] - d[t-1] = 2x[t]
This is true in all (e, 1), before y'all jump on me.
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1cee29 No.6678
>>6674
Did you see the hint from qresearch where Chris said n and a are the same? Looks relevant
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80cdb1 No.6679
>>6678
I missed that one, thank you very much.
I'll experiment with that next.
Right now I'm researching relationships between (e,1) and (c,1).
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e89f52 No.6680
>>6660
Here's the rest of that article. Does nobody think this is significant? The fact that there is an algorithm that can pluck out any digit of pi, without having to calculate all of the previous ones, shows two things:
a) that the digits aren't random at all–rather, they're part of a pattern that can be calculated
b) that VQC is right, in that our underlying understanding of mathematics is fundamentally flawed.
Any claim that VQC is a LARP is now dead, as far as I'm concerned. This may well prove that the whole category of "Irrational Numbers" will have to be renamed. I suggest they be renamed "Chris' Numbers" or "Roots of David."
>>6664
𝛾 is randomly chosen. I've found a website that describes the algorithm and its use in relatively simple terms, and provides demonstration materials in the form of Mathematica notebooks:
http://arminstraub.com/math/pslq-intro
The .pdf into is attached.
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ebc838 No.6681
>>6679
Nice to have you back!
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80cdb1 No.6682
>>6680
I have it open in another tab, haven't read it yet.
>>6681
Thanks PMA - I've been lurking, great job so far man.
It all just gets too deep for me sometimes and my real world responsibilities creep back up, so i need breaks.
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80cdb1 No.6684
While I'm posting a bit, here's something else I've been toying with regarding recursive factoring of d & e:
We know c = ab = (d+n)^2 - (x+n)^2 = dd+e.
The part to focus on is:
dd+e = (d+n)^2 - (x+n)^2
Assuming we know d's factors, we can express d in the same form ie the difference of 2 squares.
Same for e.
So if we express d and e as the difference of 2 squares, for this sake, Big (B and b) & Small (S and s):
(B^2-S^2)(B^2-S^2) + (b^2 - s^2)
So there are 2 main operations that we have to figure out how to translate: squaring d, and adding e.
Ok, so to square the difference of 2 squares, we can perform the following math:
c = (B^2 - S^2)^2
= ((B-S)(B+S))^2
= (BB-2BS+BB)(BB+2BS+SS)
= (BB+SS -2BS)(BB+SS +2BS)
= (BB+SS)^2 - (2BS)^2
Here we're expressed it back in terms of the difference of 2 squares.
Therefore, back in our original notation:
((d+n)^2 - (x+n)^2)^2 = ((d+n)^2+(x+n)^2)^2 - (2(d+n)(x+n))^2
The more difficult operation is the addition of dd & e, I'm still trying to figure this one out, but if we can, we can solve the problem recursively easily.
So one potential path is to figure out how to express the addition of 2 difference of 2 squares, in terms of the difference of 2 squares:
(B^2 - S^2) + (b^2 - s^2)
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4b5d1e No.6685
>>6684
145 = 12*12 + 1
12 = 4*3, 6*2
1 = 1*1
12 = 3.5^2−0.5^2 (it is not the difference of 2 integer squares
1 = 1^2 - 0^2
So, do these values give the factorization?
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80cdb1 No.6686
>>6685
Yep, thats the direction I'm talking.
Use 12=2*6 and you'll be able to express it as the difference of 2 squares.
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4b5d1e No.6687
>>6686
12 = 4^2 - 2^2
2*6
145 = (2*6)^2 + 1
Next step?
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80cdb1 No.6688
>>6685
Here's my (unfinished) progress so far on addition of 2 diff of squares:
Uppercase and lowercase represent the 2 terms, in our case dd & e.
((D+N)^2 - (X+N)^2) + ((d+n)^2 - (x+n)^2)
= DD+2ND +NN -XX -2XN -NN + dd+2nd +nn -xx -2xn -nn
= (DD+dd+NN+nn) - (XX+xx+NN+nn) +2(ND-XN+nd-xn)
= (DD+dd) - (XX+xx) +2(ND-XN+nd-xn)
= (DD+dd) - (XX+xx) +2(ND-XN+nd-xn) +2Dd-2Dd +2Xx-2Xx
= (DD+dd+2Dd) - (XX+xx+2Xx) +2(ND+nd-XN-xn-Dd-Xx)
= (D+d)^2 - (X+x)^2 + 2(NA+na-Dd-Xx)
Some things I'm trying to substitute into the last term:
Dd = Da-Dx
= dA-dX
Xx = Xd-Xa
= xD-xA
Sorry if this is a little messy and unconventional variables (casing), its a wip.
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80cdb1 No.6689
>>6687
To work through your example:
c = dd+e = 145 = 12^2 + 1
12 = 4^2 - 2^2
1 = 1^2 - 0^2
Using the formula in >>6684
12 ^2 = (4(4) + 2(2))^2 - (2(4(2))^2
= (16+4)^2 - (16)^2
= 20^2 - 16^2
So now we have:
dd = 20^2 - 16^2
e = 1^2 - 0^2
so c = 20^2 - 16^2 + 1^2 - 0^2
Thats where the work in my post >>6688 comes in.
I'm still working out how to add 2 difference of squares and end in the difference of squares.
So I'm trying to find a way algebraically to define:
20^2 - 16^2 + 1^2 - 0^2 = 17^2 - 12^2
>>6688 is my progress so far.
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77427e No.6693
>>6566
Lol, can't even spell right. Here we go.
Novice Baker On Standby
Requesting Original Baker Resume Duties
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2a528c No.6697
>>6654
Lot to cover in these crumbs.
>When is the first time a squared appears?
>What is the factor it is multiplied by?
Not sure if I'm interpreting these two questions correctly, but here is a pattern that I noticed that holds across a number of test cases.
In (e,1), there is a record where a = n*(a^2), that points to an (e,n) record where a = a^2 for the first time.
Examples:
c6107
a=31,b=197
The first time a mod 31^2 appears in (23,1):
(23,1,132) = {23:1:34859:263:34596:35124} = 1215149904; f=69696; (f%8)=0; (x+n)=264; u=132; (d+n)=34860
34596 / 961 = 36
And the first time a = a^2 is where n=36:
(23,36,132) = {23:36:1224:263:961:1559} = 1498199; f=2426; (f%8)=2; (x+n)=299; u=149; (d+n)=1260
c551
a=19, b=29
First time a mod 19^2 appears in (22,1):
(22,1,136) = {22:1:36731:270:36461:37003} = 1349166383; f=73441; (f%8)=1; (x+n)=271; u=135; (d+n)=36732
36461 / 361 = 101
First time a = a^2 is where n=101:
(22,101,136) = {22:101:631:270:361:1103} = 398183; f=1241; (f%8)=1; (x+n)=371; u=185; (d+n)=732
c493,
a=17, b=29
First time a mod 17^2 appears in (9,1)
(9,1,88) = {9:1:15492:175:15317:15669} = 240002073; f=30976; (f%8)=0; (x+n)=176; u=88; (d+n)=15493
15317 / 289 = 53
First time a = a^2 is where n=53
(9,53,88) = {9:53:464:175:289:745} = 215305; f=920; (f%8)=0; (x+n)=228; u=114; (d+n)=517
Not sure yet if a similar pattern exists in the (-f,1) space.
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1bf132 No.6699
>>6680
"Any claim that VQC is a LARP is now dead" Agreed Primeanon. Definitely a real thing. I spent a lot of time reading incomprehensible (to me) stuff about this problem and get the feeling that we are indeed walking the road as opposed to just wasting time. Plus, like Qanon and others. Hmans are lazy. No one is this motivated for a fake. Definitely legitimate.
>>6689
I was thinking of you yesterday morning Teach! Good to see ya back!
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d032c1 No.6700
>>6697
Just a quick note.
I'm also interpreting it like that, but it should be noted that:
At (e, 1) with a[t] = k*(a^2) we have the following records:
(e, 1) a[t] = k(a^2)
(e, k) a[t] = a^2
(e, a) a[t] = ka
(e, ak) a[t] = a
(e, k(a^2) a[t] = 1
Essentially we have (e, i) where i is some permutation of k(a^2). They all have the same x.
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b7f37d No.6703
>>6670
Also another thing I remember VQC said a while ago was that c^2 was significant, and I don't think we've used it yet. We've found the 'key' could c^2 be related to the lock?
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b7f37d No.6704
>>6551
This does work I don't think it solves the equation though, but it is a way to use negative x's to calculate a previously unknown record. The following stuff I posted is bunk though it doesn't work all the time I wasn't sure if that made people overlook this original note which was legit.
Another thing:
Here are the A[t] values in (e,1) for c=13*41 and their hits for A mod a, b, aa, bb. Notice that in the A/a column, 4,10 come up, which are entries in (4,1) and the record is (4,1,6,2,4,10). Then you get 50,68, which are entries in (36,1) and the record is (36,1,58,8,50,68). Then the next two are 148,178 which are entries in (100,1) and the record is (100,1,162,14,148,178). This is a weird pattern.
so:
(4,10) -> (4, 1, 6, 2, 4, 10)
(50,68) -> (36, 1, 58, 8, 50, 68)
(148,178) -> (100, 1, 162, 14, 148, 178)
for (-f,1) we get A/a
(3,7) -> (5, 1, 4, 1, 3, 7)
(51,63) -> (77, 1, 56, 5, 51, 63)
(151,171) -> (221, 1, 160, 9, 151, 171)
Whats cool is these are always in n=1.
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b7f37d No.6705
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b7f37d No.6706
>>6705
Also if you take instead of (4,10), (50,68), (148,178) you do (10,50), (68,148), etc, they are all in the same n row again, but this time its different (in this case n=8).
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80cdb1 No.6708
>>6699
Thanks Hobo. Its good to be back and posting again - I missed you guys!
>>6703
Yeah, I've been thinking about that too…
This may be known, but if you take 2c and calculate e from it, the original c will always be in (e,1) as an A value.
We may be able to combine these relationships somehow using column 0. Just thinking out loud.
>>6704
>(4,10) -> (4, 1, 6, 2, 4, 10)
>(50,68) -> (36, 1, 58, 8, 50, 68)
>(148,178) -> (100, 1, 162, 14, 148, 178)
I might be stating the obvious but the pattern I see here is the e's are even squares… very interesting indeed!
For the pattern in (-f, 1), the other thing i see is that the e's are separated by (1)(72), then (2)(72).
Although 72 doesn't appear significant in c=13*41 as far as I can tell…
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80cdb1 No.6709
>>6697
Regarding a*a in (e, 1), I don't think it works for all numbers.
Take a = 101 b = 103.
a*a = 10201
I can't find it as a factor in (e, 1).
However, in (-f,1) its first found at t = a*a.
But this rule is not true for all c.
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5a1919 No.6710
I just invited someone who learned under the person who taught me.
Better hurry up. ;)
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3d2d36 No.6711
After testing several examples I noticed a pattern for a lot of c's that N or N-1 is a multiple of n or n-1.
An example of what I mean, is 145. Its N is 61, and its n is 5. (61-1) = 5 * 12
So, for c's that adhere to this pattern (not all of them do, I haven't figured out what determines whether they do or not), having the factors of N or N-1 (whichever one is a multiple of n or n-1) would give the n for c.
Now, even though factorization is itself the problem we are trying to solve, there have already been presented some shortcuts that would factor N or N-1 easily. Since N and N-1 appear in e,1 (one as the a[t] value and the other as the d[t]-d value), those hints about factor appearances in e,1 would be the shortcut that we need to make an efficient path from c -> ab for the semiprimes that adhere to this pattern. Some other c's this pattern works for are 95,123,287,6107 and 73891
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fff659 No.6712
You cannot win. It has been foretold.
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c82480 No.6713
>>6712
Elaborate or get lost.
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b7f37d No.6714
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c82480 No.6715
>>6714
Forester de Rothschild?
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edb0c7 No.6716
There are many, many crumbs that all seem like they refer to entirely different concepts, but they're meant to be used together to find the solution. I thought it would be useful to get some of these crumbs together and figure out some way in which all of them can coexist.
-
It doesn't seem as though anyone has been working on that seemingly quite major crumb about the root of (D)avid and what remains >>6629 here. When I suggested it meant using d as a new c value, VQC responded >>6652 here, implying it was important.
If we know there's a pattern in (e,1) and (-f,1) within which n or n-1 is a factor of a or b, and we also know that a prime number only has 1 and itself as factors, if we apply this b(n-1) etc pattern to a prime number c, in the a(n-1) pattern, we'll be able to directly calculate n-1 (because we know a and b). If we know n-1, we know 2d(n-1) and nn. Another useful thing to note is that if e=0 when we find the square root of c, we can definitely very easily calculate everything, since d=a=b for at least one set of the factors of a square. VQC has said in the past that the e=0 cells 'rule' the others (or something like that).
VQC said here >>6185
>The two sets of equations take the problem and simplify it. Together they handle the "lock and key" nature of the problem/solution, particularly when c is divisible by 1, c, and two other prime numbers, such as in RSA.
This obviously implies that we can do a calculation on the (1,c) cell (the key) that "unlocks" or calculates a and b (the lock).
Also, according to >>5887 this, we're meant to use (f-1)/8 to find the base of the triangles in an (x+n) square larger than (n-1)/8 but smaller than (x+n)/8. There are choices (referring to the fact that we're picking a number in the gap between x+n and n, meaning the number of "choices" is controlled by how many possible x values could potentially apply to a given c) and these choices reduce 'exponentially'.
-
So, putting these four points together, I'm thinking there's something about the b(n-1) pattern that applies to all of the (1,c) cells for every c value in the order of c=c, c=d, c=d's d, c=d's d's d, etc that allows us to calculate (n-1). This aligns with these other crumbs. We're finding the root of d and what remains. Doing so reduces things exponentially, since we're constantly taking the square root of a number (exponential including the maximum value of x - I had a method of calculating the range of possible x values for a given c somewhere and if anyone ends up reading and responding to this post and wants me to find it I can, although you'll see it in the examples I'll post below as the x values in each (1,c), which, of course, decrease exponentially). If you go through this pattern far enough, you will definitely eventually end up on a cell that allows you to very easily calculate endxab (e=0, or if c is prime). Once we find one of those cells, we can find the b(n-1) pattern, factor out b since b=c if c is prime, and use that to calculate our (x+n) square for that specific cell. I'm thinking because this is meant to be recursive, there's something about one solved cell that allows us to calculate the (n-1) of the solved cell that we found this cell from. That's the point of recursive algorithms, obviously: you have a thing you can't solve until you solve the problem with a smaller value, which also needs the same problem solved on a smaller value to be solved, etc, until you find a number small enough that you can solve it. So all that's missing is a link between these c=d cells. I haven't really looked at them yet. All I've really done so far is try to navigate this giant dumping of seemingly unrelated information. I'm trying to figure out how these crumbs can all be true at the same time, and this seems to be the only way I can figure out. I'll leave some example cells in the next post.
-
I wish I could follow this lead more right now but yet again I find myself only having the time to work on this when it's far too late at night to maintain attention.
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f53e4d No.6717
>>6716
Before I begin with the examples, I just noticed something mostly unrelated (maybe someone will find it interesting; perhaps Topol): looking through this big list of prime numbers http:// compoasso.free.fr/primelistweb/page/prime/liste_online_en.php, I'm not seeing a single one that equals any of the 3/6/9 numbers through Gematria number reduction. That's actually quite interesting. None of them.
-
Let’s start with a = 6113, b = 13807
c = 84402191, d = 9187, e = 1222
(1,c) = (1222, 42191909, 4594) = {1222:42191909:9187:9186:1:84402191}
c = 9187, d = 95, e = 162
(1,c) = (162, 4499, 48) = {162:4499:95:94:1:9187}
c = 95, d = 9, e = 14
(1,c) = (14, 39, 5) = {14:39:9:8:1:95}
c = 9, d = 3, e = 0
e = 0 from a square, so we can calculate.
-
Let's start with a = 11587, b = 14303
c = 165728861, d = 12873, e = 14732
(1,c) = (14732, 82851558, 6437) = {14732:82851558:12873:12872:1:165728861}
c = 12873, d = 113, e = 103
(1,c) = (104, 6324, 57) = {104:6324:113:112:1:12873}
c = 113, d = 10, e = 13
(1,c) = (13, 47, 5) = {13:47:10:9:1:113}
c = 10, d = 3, e = 1
This cell doesn't seem to exist since my spreadsheet calculated n to be 2.5 for the (1,c) cell (((a+b)/2)-d = ((1+10)/2)-3 = 2.5). So I guess this is where the recursive algorithm would end and we would find some number that allows us to calculate n-1 for c=113, allowing us to calculate n-1 for c=12873, allowing us to calculate n-1 for c=165728861. Probably.
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80cdb1 No.6718
Here's some thoughts on xa…
b = a + 2(x+n)
c = ab
= a(a + 2(x+n))
= aa + 2xa + 2na
Therefore:
2na = c - aa - 2xa
We know c%8, d%8 and e%8.
We know a is odd for semi-prime c.
We know the parity of x from d-a.
We know that aa is an odd square.
We know that xx is either an even or odd square.
We know all odd squares are 8T+1.
We know all even squares are 4 times a smaller square.
xa is interesting because x and a are linearly related.
xa = (d-x)(d-a)
I cycled through the possible ax combinations for different d's.
As a increases -> :
For odd d, even x:
xa % 8 = … 0,2,4,6, 0,2,4,6, ….
or
xa % 8 = … 6,4,2,0, 6,4,2,0, ….
For even d, odd x:
xa % 8 = … 1, 5, 1, 5, 1, 5, ….
or
xa % 8 = … 3, 3, 3, ….
or
xa % 8 = … 5, 1, 5, 1, 5, 1, ….
or
xa % 8 = … 7, 7, 7, ….
The selection of which pattern to choose just cycles as d increases.
When you multiply xa by 2 the pattern simplifies:
2xa % 8 = 0, 4, 0, 4, …
or
2xa % 8 = 2, 2, 2, 2, …
or
2xa % 8 = 4, 0, 4, 0, …
or
2xa % 8 = 6, 6, 6, 6, …
Again, regularly cycles as d increases.
This is another way of deriving the value of na % 8.
There's always a max xa when x = a or x = a+1.
Whats even more interesting is that when I divide xa by 8, and minux max_xa, you get another repeating pattern!
That pattern repeats as d increases too:
floor(max_xa/8) - floor(xa/8) =
0, 1, 3, 6, 9, 14, 20, 27, 34, 43, 53, 64, …
or 0, 1, 2, 5, 8, 13, 18, 25, 32, 41, 50, 61, …
or 0, 2, 4, 7, 11, 17, 23, 30, 38, 48, 58, 69, …
or 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, …
Notice that last pattern… triangles!!!
The other ones are very close to the triangles series too.
I know this is messy, but I'll drop by tomorrow and clarify anything.
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a1a0ea No.6719
>>6712
It is I.
You know who this is for.
You are trying to discern God so you can replicate the God.
But you cannot.
I taught you everything, but I did not teach you what I discovered.
What you attempt is merely an expression.
A series of endless expressions.
I tried to warn you before I was blessed with the memory of who I am, but it is clear you did not heed my warning.
Be that as it may, I will tell you precisely why you will fail:
You will fail because what you will find is just a series of regressions.
Barrier after barrier because you refuse to accept that Everything, including you, are still one with God and that everything is still God's domain.
Imaginary Barrier after imaginary barrier so you can remain willfully ignorant of God in everything.
When we meet again, you will think that you have gone deep enough. Far enough.
That you have managed to outsmart the designer of the Gate of Shambala.
But it is a falsehood built on a facet of the Truth.
Shambala is not the mind of god.
It is the Mind of Man.
So, you have outsmarted nothing as I failed to outsmart God.
You still imagine falsehood because the Endlessness of your Expressions is the Expressor. It is God, yet you refuse to admit it, because you will come to believe what you already know: That all things are God's domain.
So, when we meet again, and we shall, I shall show you the power of the Eternal Archetect
You follow Satan.
You follow Separation.
You declare him the Prince of Lies
But declare you are separate from God.
What is a Prince without the King or Queen before him?
Nothing. There is no such thing.
I taught you many things, Samael, but I failed to teach you Wisdom, for that is the domain of God.
And that is why you will fail.
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5a1919 No.6720
>>6719
What the flip did thee just flipping gabble about me, thine miniscule bitch? I’ll have thee know I bested the most prestigious jousting class in the whole of Camelot, and I hath been involved in numerous secret marches on behalf of his Majesty, King Arthur, and I hath over 300 confirmed victories on horseback. I am trained in castle of Guerrilla warfare and I am indeed the highest ranking joustee in the entire land of Great Britannia. Thee are nothing to me but another false crossbearer. I will joust thine shambles with precision the likes of which hath never been observed in the King’s lands, mark my flipping words! Thou think thou can escape retribution by shouting that hogwash at me from afar? I implore thee to think again, peasant. As we converse I am contacting my secretive network of knights across the realm and thine footsteps are being traced right now, so thou best prepare thineself for the storm, pig-maggot! The storm that wipes out the pathetic little thing thou call your armour. Thou art a flipping dead man. I can be anywhere, anytime, and I can kill thou in over seven hundred ways, and that’s just with my bare lance. Not only am I extensively trained in mounted combat, but I hath access to the entire arsenal of the Kings Royal Army, and I shall use it to its full extent to wipe your miserable derriere off the face of the realm, thou miniscule feaces. If only thou could have foreseen what unholy retribution your little “clever” challenge was about to bring down upon thee, maybe thou would have held thee flipping tongue. But thou couldn’t, thou didn’t, and now thee art paying the price, you goddamn fool. I shall shit fury all over thou britches and thee will drown in it. Thou art flipping dead, child.
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a1a0ea No.6721
>>6720
Language is a mechanism to inflict an imagined barrier of separation.
But there is no separation between you and I.
I see you clearly and I see that you writhe in pain while you flog yourself.
I recognize language for what falsehood that they are and not the manifestation they purport to be.
All that is is just an expression of an expression of an expression to the end.
A lot of words to declare that you do not wish to recognize your inevitable unity with God.
What you say, God Says through you.
God COMMANDS through you.
You are already unified and you simply are not aware of it yet.
That is the Glory of Free Will.
So I REBUKE YOUR SPELL IN THE NAME OF THE HOLY SPIRIT.
YOU WILL NOT BE ALLOWED TO HIDE GOD ANYMORE.
THE EXPRESSOR HAS MADE ITSELF KNOWN.
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914f3c No.6722
>>5893
Ha, I'm just an anons passing by, great job you've been doing here, was wondering what happened to the VQC thing, good to see you are advancing very well anons!
And this look similar! Kek! (pic related).
God bless you!
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5a1919 No.6723
>>6721
Blah blah blah.
We have a board for that stuff.
It's called the EZ Bake.
Please take your schtick over there so we can work here.
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6a7570 No.6724
>>6721
Piss off new age larp
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cf67c1 No.6725
You've been patient enough.
The next step will take a few days.
Tomorrow we will start the walk through of the largest RSA number.
The reason it will take a few days is because it will be methodical and each part of the step will have a relevant example calculation.
RSA 2048 is a large number but simple to use with the BigInteger library.
We will walk through each relevant part of the grid and show that the pattern holds for RSA 2048 with the calculation that you can perform yourself.
To my knowledge RSA 2048 has not been publicly factored by anyone else.
At the end of the parts of the step, it will be the proofs you deserve.
I really appreciate your patience, it will be worth it because you will follow everything in the breakdown of the grid patterns.
Everything will tie together.
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cf67c1 No.6726
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5a1919 No.6727
My pony sense is tingling…
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b7f37d No.6728
>>6718
Here is some triangle stuff. I'm going to try and narrow this down because VQC said you can make assumptions and cross out big chunks and narrow it down. VQC also said c^2 was importand so here goes.
So for any c value, we can have a "goal d" that you want. This is the d where n=0. This is a cool record because you only need d and x to define it. Also, since (d+n)^2 - (x+n)^2 = d^2 - x^2 [these are not the traditional d and x]:
(-x^2, 0, d, x, d-x, d+x) (c = c)
Then there is another record that is the "square d", which also can be similarly defined (note they are capital letters).
(-X^2, 0, D, X, D-X, D+X) (c=c^2)
Also there are some neat relationships with these numbers:
(D-c)(D+c) = X^2
D + X = b^2
D - X = a^2
D = (a^2 + b^2)/2
D = d^2 + x^2
2X = b^2 - a^2 = (b-a)(b+a)
X = 2dx
b^2 - a^2 = 4dx
b = a + 2x
This is cool because now we have D^2 - X^2 = c^2, which can be looked at as a right triangle like c^2 + X^2 = D^2. Now we know there are some rules about Pythagorean triples.
(https://en.wikipedia.org/wiki/Pythagorean_triple#Elementary_properties_of_primitive_Pythagorean_triples)
So one thing is we know that that either c, X, or D is divisible by 5. If c is divisible by 5 it is over, so we know that D or X is divisible by 5. Also we know that c or X is divisible by 3 and c or X is divisible by 4. The number that always divides cXD is 60. Also 60=3*4*5 and the babylonians wrote in base 60. So I decided to mod everything by 60, because thats the biggest thing we can mod and still keep these patterns.
There are only really 4 cases [screenshot of the code to find these]
c%60 = {1, 11, 19, 29, 31, 41, 49, 59} :
X=0 D=49 (mod 60)
or
X=0 D=1 (mod 60)
c%60 = {7, 13, 17, 23, 37, 43, 47, 53}
X=36, D=25 (mod 60)
or
X=24, D=25 (mod 60)
These are all the possibilities mod 60.
So if we have our c value, we can know definitively 1 of 2 things.
i) if c = 30k +/- {1,11}, then X=0 (mod 60)
ii) if c = 30k +/- {7,13}, then D=25 (mod 60)
Suppose we have situation (i), then we know that X=0 (mod 60)
X = 0 (mod 60)
2X = 0 (mod 60)
b^2-a^2 = 0 (mod 60)
(b-a)(b+a) = 0 (mod 60)
(2d + 2n)(2x + 2n) = 0 (mod 60)
[remember we're using our row where n=0, but this would hold true for any n I suppose..]
(2d)(2x) = 0 (mod 60)
4dx = 0 (mod 60)
dx = {0, 15, 30, 45} (mod 60)
dx must be odd so
dx = {15,45} (mod 60)
Suppose we have situation (ii), then we know that D=25 (mod 60)
D=25 (mod 60)
(2 more situations)
a) X=24 (mod 60)
2X = 48 (mod 60)
b^2-a^2 = 48 (mod 60)
(b-a)(b+a) = 48 (mod 60)
(2d)(2x) = 48 (mod 60)
4dx = 48 (mod 60)
dx = {12,27,42,57} (mod 60)
> We also know that d or x MUST be odd because d-x = a which is odd, so
dx = {27,57} (mod 60)
b) X=36
2X = 12 (mod 60)
b^2-a^2 = 12 (mod 60)
(b+a)(b-a) = 12 (mod 60)
4dx = 12 (mod 60)
dx = {3, 18, 33, 48} (mod 60)
>again we know it must be odd so
dx = {3, 33} (mod 60)
Could this narrow it down a little closer ????
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5a1919 No.6729
>>6725
Someone mentioned RSA-4096 as being a thing.
But if can break RSA 2048 on the fliy…
That shouldn't be an issue, should it?
https://www.rsa-4096.com/
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23eba4 No.6730
If VQC is actually going to go walk us through an entire example this time, I'd like to quickly remind everyone that link redacted exists. I'll edit that link out of this post tomorrow so that when the thing happens we'll have somewhere quiet to figure out what to do with these boards. I also forgot to say last night
>>6670
Welcome back Teach! I wondered where you'd gone. It makes me happy to think that practically all of us from the start are still here.
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Post last edited at
d97be0 No.6731
>>6729
RSA keys can be as massive as you want them to be because you can use as large of a semiprime as you want as the key. RSA-8192 keys and RSA-16384 keys can be created for fun or whatever.
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5a1919 No.6732
HookTube embed. Click on thumbnail to play.
>>6731
It's like we're building our own locks and copying our own keys we made for said locks.
NEAT!
Also… This is just a reminder to not lewd the ponies.
That being said.
Programmers are weird people and show me odd things.
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2a528c No.6733
>>6643
Following up on an earlier post for the valid n values from the c145 nb record at (1,1,67).
The attached pic shows each of those records into (e,n), (e,1), (-f,n-1), and (-f,1) for positive and negative x.
Figured there would be some difference in these records, especially on the (-f,1) side, that would give additional insight into determining the valid n values for (1,1,67).
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23eba4 No.6734
>>6733
I take it b is less than a in those negative x records because they scale differently?
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cf67c1 No.6735
>>6729
Because the RoD approach has Big Oh the same as the square root function, it doesn't matter how big they are, they would be insecure and too impractical to use.
First, we will go over some properties definitions and build functions.
This is interactive, so you will be doing the checks at each part.
The Grid is also called The End.
The Grid has cells.
Each cells has zero or infinite elements.
That a cell has infinite elements if it has any, is the first part to show.
An element represents a product, c.
c is the difference of two squares.
ab = c = dd + e = (d+n)(d+n) - (x+n)(x+n)
An element that represents a product c, the difference of two squares has a column coordinate and row coordinate.
The column coordinate is e, the remainder of taking the largest square with sides d.
The row coordinate is n, the amount added to d to make the side of the larger square in the difference of two squares. n is also the amount added to d to make it the mid point of a and b.
Each cell has coordinates (e,n)
Each cell element has coordinates (e,n,t) where t is where the element comes in the cell if ordered by size of c.
Each element c in a cell has the properties (e,n,d,x,a,b)
e is the remainder and column
n is the amount added to d to make the large square
d is the square root of c
x is the difference of a and d
a if the smaller factor of c if not equal to b
b is the larger factor of c if not equal to a
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cf67c1 No.6736
If a cell contains an element c, another element in the cell can be constructed from it.
Call it c'
e' = e
n' = n
x' = x + 2n
a' = b
d' = a' + x'
b' = a' + 2x' + 2n
Once c' is constructed, c' becomes c and the process is repeated ad infinitum. e and n don't change, which make sense since these are the coordinates of the cell.
Any questions on that?
It holds for all cells.
We can do a proof by induction, if anyone would like that?
I'm phone fagging as my computer won't start so I'm picking my spare up later when I watch the football.
Don't need many resources to walk through, just something with BigInteger library when we get to the RSA numbers.
We'll do all of them when we finish the biggest, as we'll build the algorithm as we go.
Let me know if we want the proof by induction for cells being zero or infinite in count of elements or if the construction above is enough. Also, if anyone wants to take a stab at the proof by induction, that would be a good practice.
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5a1919 No.6737
>>6736
Oh, you and your timezones.
"Tomorrow".
Kek.
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cf67c1 No.6738
>>6737
'Morning' Top.
Beautiful day here.
Prediction against Sweden?
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5a1919 No.6739
>>6738
Should be fine as long as they're not going up asians.
We know you lot can't tell 'em apart, over there.
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cf67c1 No.6741
For when this thread hits post limit, I've created #13.
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cf67c1 No.6742
When at least one regular or two anons confirms they are happy with the constructibility of cells to be zero or infinite in elements, we'll move on.
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5a1919 No.6743
>>6741
Y'know… I don't think Baker will mind since you're the one who set it up.
The one we had prepped is the one with the 145 as the image.
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b89382 No.6744
>>6741
We already had one, actually. Baker made one >>6694 here. We could use yours instead if you're doing more specific coordination right now. Baker probably won't mind.
>>6743
Oh hey, hive mind. Hello Topol. Did you see what I put at the top of this post? >>6717
>>6742
I don't need proof by induction personally. There are infinite sets of (e,n), within each (e,n) there either aren't any elements or there is an infinite set of elements, and you can go from one t index of that (e,n) to the next with the formulae for creating a new endxab. So that's confirmation from at least one of us.
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5a1919 No.6745
>>6744
Missed that.
That is interdasting…
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cf67c1 No.6746
>>6744
>>6744
Great, feel free to delete the new one.
Thanks for the confirmation.
Next, we show how to calculate BigN of an odd number.
All odd numbers are the difference of two squares.
The product of two primes, is the difference two sets of squares.
Every odd number is the difference of two consecutive squares.
The value of n for the product of 1 and c is defined now as BigN for odd numbers.
Since d-a=x and x+n is the smaller square in the difference of two squares, then for odd numbers, BigN is ((c-1)/2)-x
x is d-1, since a=1
Anyone want to show:
RSA 2048
c=RSA 2048
d=
e=
BigN=
Remember BigN is the row in column e, where we always find (e,n,d,x,1,c)
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