77656e No.12463 [View All]
The virtual quantum computer (VQC) is a grid made of constructably infinite elements that follow a known pattern. It is indexed using e, n, and t, where e is the column, n is the row, and t is the specific element in the cell.
The grid in its entirety serves as the superposition. The required input parameters to collapse the superposition are d and e, which are trivial to calculate for all c that is the difference of two squares. When the integers that are the difference of two squares are arranged into the grid and their corresponding properties are shown, a pattern emerges that shows a path to calculate the factors of c instead of searching for them. It can be understood using only the basic operations of arithmetic and sqrt. All currently-known patterns can be found within one thread here >>6506
C# VQC generator - pastebin.com/XFtcAcrz
Java VQC generator - pastebin.com/2MPYrJVe
Python VQC generator - pastebin.com/NZkjtnZL
Glossary
Look-up
A pattern used to calculate the factors of c, like a value look-up table.
Column
All cells for a given e
Row
All cells for a given n
Cell
All entries for a given e,n (not to be confused with an entry itself.)
Entry; record; element
A set of variables corresponding to a factorization for a given c. The legend to read entries is {e:n:d:x:a:b} (e, n, t) = c
Example: {1:5:12:7:5:29} (1, 5, 4) = 145
ab record; nontrivial factorization
The element that contains the factorization of c that is not 1*c, hence, nontrivial.
1c record; trivial factorization
The element generated from setting a=1 and b=c
Mirror element
The element in -f corresponding to an element in e, in the context of a given c.
Variables
a and b are, to reiterate, the factors of c. a is the smaller factor of c, and b is the larger one.
d is the integer square root of c.
e is the remainder of taking the integer square root of c. Unless c is a perfect square, a remainder will be left over.
i is the root of the large square. It is equal to (d+n).
j is the root of the small square. it is equal to (x+n).
n is what you add to d to be exactly halfway between a and b, and it is the root of the large square, so it takes you from d to the large square.
x is what you add to a to make d. When added to n it makes the root of the small square.
f is what you add to c to make a square. (e is what you subtract from c to make the square below it, f adds to make the square above c.)
t is the third coordinate in the VQC, it is a function of x.
q is a product created by multiplying successive primes until the product is above d.
u is the triangle base of (x+n)^2. 8 times the triangle number of u plus one is (x+n)^2 for c with odd x+n.
ab = c
dd + e = c
(d + n)(d + n)-(x + n)(x + n) = c
a + 2x + 2n = b
a = d - x
d = a + x
d = floor_sqrt(c)
e = c - (dd)
b = c / a
n = ((a + b) / 2) - d
d + n = i
x = d - a
x = (floor_sqrt(( (d+n)*(d+n) - c))) - n
x + n = j
j^2 = 8*T(u) + 1
f = e - 2d + 1
u = (x+n) / 2
if (e is even) t = (x + 2) / 2
if (e is odd) t = (x + 1) / 2
Past threads
RSA #0 - archive.fo/XmD7P
RSA #1 - archive.fo/RgVko
RSA #2 - archive.fo/fyzAu
RSA #3 - archive.fo/uEgOb
RSA #4 - archive.fo/eihrQ
RSA #5 - archive.fo/Lr9fP
RSA #6 - archive.fo/ykKYN
RSA #7 - archive.fo/v3aKD
RSA #8 - archive.fo/geYFp
RSA #9 - archive.fo/jog81
RSA #10 - archive.fo/xYpoQ
RSA #11 - archive.fo/ccZXU
RSA #12 - archive.fo/VqFge
RSA #13 - archive.fo/Fblcs
RSA #14 - archive.fo/HfxnM
RSA #15 - archive.vn/59GwR
RSA #16 - archive.vn/F49fw
RSA #17 - archive.vn/u2Tu6
RSA #18 - archive.is/FDVP9
RSA #19 - archive.is/6mJhe
____________________________
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d756e2 No.12848
>>12846
>>12845
>Today is the day.
Sweet.
>Multiply c by 4.
Done.
>Take the new e',1 at 4 x c.
>That cell at new e',1 will have a value that is double a value at e,1.
Not sure which "a value that is double a value in e,1" value you mean, but a quick exploration turned up the value of 2c as an a[t] value at the locations x = 2d (e',1) and x = 2d +1 (-f',1) for our two small classic values. Please see the attached work.
>The value at e,1 is na or (x^2+e)/2 and the value in e',1 is 2na.
>If we have two cells at e,1 and e',1 how do we find values that are twice the other?
Hmmm. OK. Going back to the original (-f,1) (e,1) to reference the new (-f',1) (e',1)
We need some kind of concrete reference point to begin, and so finding 2c at a[t] in (-f',1) (e',1) is a helpful discovery. It's 2 * a * b (of course) so perhaps that could be useful.
Gotta keep investigating the patterns. I'll be back when I have something useful to report.
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d756e2 No.12849
>>12847
>WE BRINGIN' ANCIENT MATHEMATICS BACK TO MODERN MAN
Boom. What's up my bruddah!
No doubt this idea was known to previous thinkers.
Been watching/reading too much Graham Hancock over here.
Definitely several advanced civilizations have been wiped out.
"We are a species with Amnesia"
>>12846
>Have fun.
Aye aye, Senpai!
>If you need any help with finding the double value between cells, I'll be back later.
Is the double value 2c?
>Let's do the biggest RSA number together?
>End of an era, beginning of a new one.
Sure, sounds good. Let's have fun and get some work done!
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2dc828 No.12850
>>12849
Blessings to you & your mind, and your family. Let us take care of those we love and make sure they know that we cherish them and hold them near to our hearts.
Blessings to All. Past, Present, & Future.
And now, as our savior taught us, we are bold to say:
Our Father, who art in Heaven,
Hallowed be Thy Name…
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bca452 No.12851
>>12845
>That cell at new e',1 will have a value that is double a value at e,1.
>The value at e,1 is na or (x^2+e)/2 and the value in e',1 is 2na.
For a=13, b=43, na=65. This implies there should be a value of 130 somewhere in (e',1), which pic related shows there isn't. The only way would be to make t equal 130, but then you would have just said to do that. t=130 for this (e',1) is (27,1,130) = {27:1:33813:259:33554:34074}, f=-67600, c=1143318996, u=130, i=33814, j=260 and I don't see anything useful in this element.
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bca452 No.12852
>>12851
I only thought to check variables, now I'm checking gaps between variables within elements and between adjacent elements. It does exist as a gap between adjacent d values here:
(27,1,32) = {27:1:2061:63:1998:2126}, f=-4096, c=4247748, u=32, i=2062, j=64
(27,1,33) = {27:1:2191:65:2126:2258}, f=-4356, c=4800508, u=33, i=2192, j=66
2191-2061=130, which is 2na. The higher of the two x values is na. I don't know if that actually matters though because for every element in (e,1) the gap between two adjacent d values is two times the higher element's x value, so that's implied algebraically. I still don't see anything useful about these elements that would help us find them, and all the patterns that involve gaps between d values in (e,1) either involves taking away our original d value from d[t] or d fitting between the two d values, so I don't know if any of this is useful. Also you would need to know unknowns to find these elements regardless of anything else.
The only way I can make sense of this post is if the 2na he's talking about in (e',1) is the gap between d values. Can't find any use for this or patterns that could be used to find these elements though.
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bca452 No.12853
>>12852
By the way, pretty sure today is the 7th anniversary of the "my hand is forced" post, what better time to finally post the fucking solution already
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10ec0b No.12854
>>12853
Happy 7 Year Anniversary Faggots! Love you all, and Happy Thanksgiving.
Found another double value, printed out the (e,1) and (e' ,1) columns to do some old school pencil and calculator work.
Please see attached work.
I found a double(d) value.
d = 29 in (e,1)
d = 58 in (e' , 1)
b = n = 36
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bca452 No.12855
>>12854
Multiplying c by four is the same as multiplying a by 2 and b by 2, which doubles x+n and doubles d+n (n' is 2n-1, d' is 2d+1 and x' is 2x+1). That also means you can use variables from one element to create the squares for the other element. Given (x+n)(x+n) = nn + 2d(n-1) + f - 1, and given xx+e=2na, I would assume there's a way to rearrange the variables that make up the square so that you have xx in the corner with e divided into two equal parts (depending on parity of x) surrounding it so that when you divide it by two you end up with two right-angled triangles with na at the tip of both (since na = (xx+e)/2). That would mean if you had these (unknown) elements you could use one square to find the other's square with geometry. None of this is useful information right now though, and if you just keep multiplying c by 4 nothing useful seems to happen, as far as I can tell (pic related).
Some grid patterns posts relevant to what he said too
>>7639
>>7624
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bca452 No.12856
>>12855
Should clarify, in that first linked post he specifically mentions "gaps between the square remainders" (i.e. square remainders are e so maybe relevant to gaps between d values, which is the only place 2na exists in (e',1) that I could find other than forcing it as a t value or something) and the fractal thing is relevant given the fractal nature of one element's squares being made up of double another's variables and so on upwards as you multiply by 4. I think these latest two posts are related to the n0 triangles.
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Post last edited at
c29784 No.12857
>>12845
Assuming this has anything to do with >>7423 solely because the only way I've found that makes this post make sense is a gap between d values in (e,1) that equals 2na, I put together these tables showing where n-1 is a factor of d[t]-d vs where n is a factor of a[t]. I know we looked into this a little bit years ago and didn't find anything but I don't think anyone ever applied it to multiples of c. Still doesn't really seem like anything is obviously relevant.
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c00ae2 No.12858
>>12857
Thanks AA, reviewed your output in detail.
{30:5:23:10:13:43} c=559
I think the only advantage of multiplying c x 4 is that we get more n locations?
>Still doesn't really seem like anything is obviously relevant.
Yeah, not sure exactly how this helps.
Doing work here is a welcome daily break when IRL duties are done, so I'll just keep thinking and working. Nice to have a hobby I enjoy, and friends to work on it with.
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723dd6 No.12859
Huh.
I don't know what to make of the new crumb.
At face value, it doesn't take any advantage of the Offset between (-f,1) and (e,1).
That's fine, since I love being here and this is my happy place lol.
Also, a couple finds…
I found 4(an) but not 2(an)
For 4 x c(6107), the 4(an) element is located at (92,1,48) with 4(an) being 4464 = 1116 * 4
I also found 1/2 * (an) = 558 at the location (92, ,1, 17)
Not sure what this means.
Just another day tending the rows and columns at the VQC farm.
"In all labor, there is profit. But idle chatter leads only to poverty."
Proverbs 14:23
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6c8ff0 No.12860
AA you lurking?
Did we actually find solutions for every semiprime c up to 1M?
I recall that you said we did, using the binary idea.
For example, if the a[t] values in (e,1) are even, we moved to (-f,1) to use the odd a[t] values to look for the binary tag.
Same with this multiply c x 4, 9, 16 idea.
We need a column where a[t] is odd.
For example, in c6107 the (e,1) column is even a[t] values
But when we move to (-f,1) then we have odd a[t] values to examine.
So we need movement to the place where a[t] values are odd.
>>12856
> gap between squares and remainders
xx+e / 2 = an
(square + remainder) / 2 = an
This is why there is a flow, or repeating pattern of remainders in the a[t] column for any (c) value.
The squares cycle, but the remainder is fixed based on (e).
A wave pattern.
That is embedded in the data.
Perhaps by moving to 4c we can observe this wave pattern more easily.
I haven't tried a binary analysis on 4c yet.
There should be 4x as many binary tags for (a)
There should also be 4x as many (n) locations
The key is that we need a multiple that delivers odd a{t] values, as that is where our prime value for (a) will be found.
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6c8ff0 No.12862
Whoops. Copied part of my last post, here's the new part.
The wave pattern is (x squared + e) / 2
It's the part that doesn't fit that's left over.
For example, 2975
101110000000 = 2944
011111 = 35
101110011111 = 2975
So what happens with this binary stuff when we multiply to a column that has MORE of the (a) value we need?
It just has to be a (c) multiple that gives us odd a[t] values I think.
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cc88ed No.12863
>>12860
>Did we actually find solutions for every semiprime c up to 1M?
>I recall that you said we did, using the binary idea.
I don't remember where it stopped (the fact that it stopped alone shows that it's not the solution) but it was inconsistent and it would be unpredictable which variable appeared where. A year or two later (I've lost track of time in this context) I wrote an entirely new program that checked the same thing except I tried to see if every possible combination of parities among the variables and other types of groupings brought out any patterns and there weren't any. If the solution is based on finding an unknown in the binary of a known, I've exhausted every possible way of doing that that I could and I didn't find a way to do it that wouldn't require waiting thousands of years for bigger numbers (since it would be O(n) where n is the length of each variable in bits, rather than O(log n) (and even then it wouldn't find a solution in all cases)).
>For example, if the a[t] values in (e,1) are even, we moved to (-f,1) to use the odd a[t] values to look for the binary tag.
I was already just generating every known and every unknown and checking all the unknowns (including recursively through the factors of d, e and f) to see if they existed in the binary of the knowns, so I already did that.
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62dc3c No.12864
>>12863
Hello AA. I appreciate all the effort you've put into our quest. Thank you.
Thanks for answering my question, and confirming that this path has already been explored.
>I've exhausted every possible way of doing that that I could and I didn't find a way to do it that wouldn't require waiting thousands of years.
Lol.
Yeah, so many dead ends.
My archives are full of paths that didn't work.
Since this is my favorite hobby, I'll just keep thinking and working.
Edison tried 10,000 times to achieve the electric bulb.
Worst case I get to hang out with you excellent faggots some more.
It hurts my sense of personal self-confidence the most:
>"Maybe I'm just not smart enough. I've failed so many times."
Maybe I just won't give up.
>"Maybe I need a higher IQ to solve this."
I took an IQ test online 2 months ago because I had never done it. 135 is fine.
>"WTF is wrong with me?"
Uhhh. Nothing.
>"Why can't I find the pattern underlying the crumbs?"
Not sure, but that's no reason to be ashamed or mad at myself
>"Look how much time you've wasted on this."
Made some great friends, and continue having fun every evening. Learned a ton.
>"Now you're definitely on every Fed list."
Fuck the FedBois, hopefully the White Hats have us on their nice list.
>"What a fkn joke this whole thing is."
The only joke would have been not to accept the Mission, and persevere until it's solved.
All my personal angels & demons come into play when working on this.
Which makes me more sure that I should continue working on it.
This is the Quest I was born to be a part of.
Let's keep going.
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62dc3c No.12865
>"WTF is wrong with me?"
Uhhh. Nothing.
False.
(upon retrospection)
Just another normal human being in need of forgiveness and Divine blessings.
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75cb83 No.12866
Hello everyone! Reviewing crumbs and relaxing,
IRL duties are complete.
Glad to be here with you all.
>>12853
Back To The Future.
Set the computer to 11/30/2017….
Back when we all responded to the craziest faggot on CBTS.
VQC.
It doesn't seem like that long ago to me.
My life has changed dramatically since this quest started.
I lost a woman and a family unit due partly to Cue and VQC.
Gained a better woman 3 years ago.
I don't even go into detail with her about us, I just tell her "I'm working with my Math boys."
She just chuckles and says, "Ok, whatever."
On the upcoming solution and forward progress:
Some good hints would be very welcome, Chris!
You have a tried and tested team who have worked hard to follow your bread crumbs.
We've learned a ton.
It's gotta be something simple we've overlooked.
Senpai, teach your disciples something new pls.
It should be obvious by now that we will persist until we succeed.
Drop us a good crumb to work on.
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362d91 No.12867
>>12866
>Glad to be here with you all.
Just checked in - guess will work a bit this eve.
>>12853
>7th anniversary of the "my hand is forced" post, ..
and just over 6 yrs:
>>8176 (VQC in #14)
>Last post.
ALL indications are that was the last post by Chris.
>>12864
>Hello AA. I appreciate all the effort you've put into our quest. Thank you.
SAME!
>False.
>(upon retrospection)
Same same.
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362d91 No.12868
>>12859
>I found 4(an) but not 2(an)
>For 4 x c(6107), the 4(an) element is located at (92,1,48) with 4(an) being 4464 = 1116 * 4
>I also found 1/2 * (an) = 558 at the location (92, ,1, 17)
Confirmed those. Interesting relationship.
>>12855
>Multiplying c by four is the same as multiplying a by 2 and b by 2
Right, and for 9c, a, b * by 3 and 3.
So the "trivial" "Big N" (as '1' will produce half-integer n with 0.5 remainders for 4c, ..) for c9 would use a=2, ..
One thing will be reviewing more, is the difference in n=1 between the "d_min" rows, above and below.
To review "d_min" is the minimum 'd' value for a given 'e' column. As e grows, so does d_min.
Consider e=23. The minimum d=12.
d^2+e = c = 144+23 = 167
Record (23,1,6) {23:1:83:11:72:96:6912:84:12}
Here starts perhaps a unique set of BigN, where each an relates to an N where a=1.
For t=1:5, the 'c' are valid, but the a,b are not of simplest form.
So t=5:
(23,1,5) {23:1:61:9:52:72:3744:62:10}
a=52, b=72
=2^2*26*36 = 2^4 * 13 * 18
In 'f', 'c' and 'i' again positive at t=6.
We look at the "related" 'c' values in the 'e' column at d_min and above.
d increases by 2 for each successive row. For e=23, d is even.
Also noted that the 'd' for the related c = 2*t.
Consider trivial case for c=6107.
t=39: *2=78 = d.
for the 6107 solution row, related c=2327:
t=24, related c is 2327 with d=48.
It has a BigN: (2327+1)/2-48 = 1116
BigX "happens" to be 47 (solution x for 6107).
(23,1,24) {23:1:1163:47:1116:1212:1352592:1164:48}
(23,1116,0) {23:1116:48:47:1:2327:2327:1164:1163}
And using that element as the BASE for the sub-series Generator, and looking at gcd patterns for 6107 (instead of 2327 as would typ do), 31 and 197 emerge:
b for:
(23,1116,-2) {23:1116:4324:-4417:8741:2139:18696999:5440:-3301}
= 2139 = a for:
(23,1116,-1) {23:1116:-46:-2185:2139:1:2139:1070:-1069}
gcd = 31.
at t=14, d and i have gcd = 31
(23,1116,14) {23:1116:470084:31295:438789:503611:220978967079:471200:32411}
Which is right in middle of repeated t=-2 and t=-1 pattern:
(23,1116,29) {23:1116:1944614:64775:1879839:2011621:3781523609019:1945730:65891}
(23,1116,30) {23:1116:2078628:67007:2011621:2147867:4320694362407:2079744:68123}
gcd(2011621, 6107) = 31
the 197 first appears at t=51 (b) and t=52 (a):
(23,1116,51) {23:1116:5924106:113879:5810227:6040217:35095031899259:5925222:114995}
(23,1116,52) {23:1116:6156328:116111:6040217:6274671:37900374443607:6157444:117227}
where gcd(6040217, 6107) = 197
also at 52, mod 'x':
mod(116111, 6107) = 78
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362d91 No.12869
>>12868
A comment on this "related c" and "d min" with the 4*c and 9*c.
ALL of the "related c", if large enough, move to the same 'e' column.
You noted for 6107*4, e' = 92. and 6107*9, e''=207.
The related 'c' all jump to the same (see pic).
What was interesting, is the 4*c is another way to quickly ID the d_min (had another way), as that's when the e values are then equal (see pic).
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362d91 No.12870
>>12869
>What was interesting, is the 4*c is another way to quickly ID the d_min
Incorrect statement. Meant to simply show with the base calcs (c*1) that e=23 starting with t=6 (d_min = 12, c=167)
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362d91 No.12871
>>12868
>So the "trivial" "Big N" (as '1' will produce half-integer n with 0.5 remainders for 4c, ..) for c9 would use a=2, ..
just read that, which states the a for c0 a=2, should state 3, as a=2 is for c*4
In case c*9 a=3, c*16 a = 4, …
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a3a3b7 No.12872
>>12867
Hello MM and everyone!
Checking In for some banta and work.
I'm gonna review crumbs for the moment to look for a good place to dig.
If anyone wants to hang out, I'll be here for about 2 hours before I gotta get some rest.
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a3a3b7 No.12873
*bantz lol.
Too tired from all full day of work in the field.
Autocorrect obviously couldn't help.
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a3a3b7 No.12874
*a full day
Pretty beat over here.
Glad to check in with you my bruddahs.
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a3a3b7 No.12875
Going to read some crumbs.
I'll check the board while I work if anyone wants to hang out.
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420f67 No.12877
Hello everyone!
Hope you are all well.
Reading crumbs and relaxing this evening.
Also reviewing diagrams and spreadsheets to look for somewhere to dig.
The Offset is still stuck in my mind.
Every time I review crumbs, I am drawn to think and strategize to make it work to our advantage.
We have two pairings in the (-f,1) and (e,1) columns.
a(n-1) and (an) (at the same [t] value)
b(n-1) and (bn) (offset by [t-1] value)
Somehow being able to split these is our key.
At any given [t] location, we can perform these simple operations to split the (an) and (bn) apart.
(an) - a(n-1) = a
(bn) - b(n-1) = b
(an) / a = n
a(n-1) / a = (n-1)
(bn) / b = n
b(n-1) / b = (n-1)
So at the correct [t] location for (an), we should find equal (a) values, along with (n) and (n-1).
There's gotta be a way to use this to calculate or intelligently search.
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260571 No.12878
Here's some highlights from this evening's crumb review.
"We are using f-2 as a guide to how to construct the square."
Having fun playing with my calculator, wondering if a clue exists using (f-2) / 2
For our classic,
f = 134
(f-2) / 2 = 66
Using the pythagorean theorem, if this is c, then c^2 = 66 ^ 2 = 4356
4356 = a ^ 2 + b ^ 2
4356 / 2 = 2178
sqrt(2178) = 46.667 = 47 = x
lol.
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d219c7 No.12879
For c145 it works also
24 - 2 =22 = (f-2)
22 / 2 =11
11 ^ 2 =121
121 / 2 =60.5
Sqrt(60.5) =7.778 = x = 7
So the idea is using (f-2)/2 as the hypotenuse (C) in the Pythagorean theorem calc.
And solving for A or B
“We are building a method that uses (f-2) as a guide to constructing the square”
8:20 PM
Hmm.
Well it’s a new idea at least.
Sqrt(((f-2) / 2 ) ^ 2 / 2) = x
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d82066 No.12880
Hello been a while but I've gotten back into the problem and I've found a couple patterns worth looking into.
First off, if we look at the A grids, we can parabolically jump to another record in the E column with the same A value, for any A in an E,N cell, and I theorize there is a way to jump to the desired record, but I haven't found a solid way to do so yet. You can do it for 145 with A=101, then you jump to 41, shift to A=65, jump to 25, shift A to 37, jump to 17, then shift A to 17 and then jump to 5 and you get the correct N, but this is not consistent for other starting points so TBD. pic related is the 17 to 5 jump visualized
Secondly, for any E,N, if you go to the cell E*i*i, N*i*i, for any i, they will contain the exact same A values. Not sure how this will help us yet though. What would be interesting is if you have an E,N that are both divisible by a square, you could jump in to E/(square), N/(square) and I'm not sure how that would help us. Would be worth it to inspect some more of these patterns though. Just found this out so still trying to figure out how to use it
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d82066 No.12881
>>12880
Also another, and I'm not sure how helpful this is since just calculating massive factorials is intensive, if you shift E + 2N you know that you'll get another cell, since the valid -Es have to be quadratic residues mod 2N (or something like that), if you shift over by N! you will have the same exact cell layout relative to your starting cell, but again this is probably not worthwhile
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c197ce No.12882
>>12880
Hello GA! Good to see you, and Happy New Year to you and your loved ones.
How's the post college life going so far, my brother?
I hope you've found work that you find enjoyable and fulfilling!
I'm studying your post closely and working to understand your ideas.
First, Is this output specifically for c145?
I can see the parabolic patterns on the left side of your output, so I'm following you there!
With no headers or column notation for the right side, it's a little tough to follow
Also, there's some notation I don't yet know.
Can you please explain your idea of A Grids?
> I theorize there is a way to jump to the desired record, but I haven't found a solid way to do so yet.
Glad to see you here my man.
There is a simple and elegant way to solve this that we are about to discover.
Also love that our faggot Elon is making Pepe famous today.
But fuck his hiring of a bunch of subtard India residents.
America First, Elon.
We know you're lurking here anyhow.
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c197ce No.12883
>>12882
To be fair, Elon is probably hiring only the best.
But the H1-B thing needs reform.
Too many jobs given to non-citizens.
And too many of our American Industries outsourced to cheap ass countries.
For who?
Global Corporations to make moar profit?
Fuck that.
Tariffs on all that, and on foreign workers.
We protect our own.
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d82066 No.12884
>>12882
Family is well and I'm working in DevOps now very happy with where I'm at in life. Thanks for asking! And here's a little write up about the A grid.
For a given A, there is a grid for which there is a line, for a given i, with an origin at (e=0, n=0) and continuing at (e=2Ai, n=i).
Then on this line there are parabolas aimed to the left, with origins at (e=2Ai + A^2, n=i+A). There are more parabolas with a larger vertical gaps below.
The way it is set up, is that the Ath value going towards -N is the rightmost value on another ray going to the left.
For any given cell, (e,n) and A, and i, there exists cells at (e+2Ai, n+i)
These cells are also on infinitely many parabolas facing the left with increasing vertical gaps as you go down.
To idenfity which position on the parabola we are on, say we want the parabola with vertical gap sizes of R, then for a given cell, X - R*A is the vertical difference from the origin cell for the parabola.
Examples
Moving down the A grid to the right, move_a_diagonal the E value will increase by 2*A, the N value will increase by 1, the D, X and A values will remain the same, and the B value will increase by 2.
Switching to a new A grid, switch_a_grid, the E value will increase by 2*N, the X and N values will remain the same, the D, A, and B values will increment by 1.
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d82066 No.12885
>>12882
Run this python code in your terminal (you may need to pip install something) but then type "!hl a 7" to higlight the A 7 grid. Then WASD will move around the view and J/K will scroll up and down in the panel on the right. Still working on a better tool but stil isn't as featureful as this one (and its not in one easy to run file)
https://pastebin.com/H56HTXmy
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d82066 No.12886
Line 240 and on shows the help messages which contain all the controls
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d82066 No.12887
Another pattern found.
For any cell (e, n), you can calculate any other cell with ALL the same D values which is located at
E = (n*n)*i^4 + (e - n*n)*i^2
N = n*i^2
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d82066 No.12888
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d82066 No.12889
I tried factoring random primes from a list of primes less than 100, did 1000 tests among all of them and all of them passed
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d82066 No.12890
I think its a false alarm I'm looking further into it
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c92e58 No.12891
>>12884
>>12885
Thanks GA!
I appreciate the explanation, and I'll see if I can run your code this evening.
Still need to check that calc for R100 for the idea above >>12878
>For our classic,
>f = 134
>(f-2) / 2 = 66
>Using the pythagorean theorem, if this is c, then c^2 = 66 ^ 2 = 4356
>4356 = a ^ 2 + b ^ 2
>4356 / 2 = 2178
>sqrt(2178) = 46.667 = 47 = x
Trying now for R100:
f =
16822699634989797327123095165092932420211999031886
(f - 2) / 2 =
8411349817494898663561547582546466210105999515942
((f - 2) / 2) ^ 2 = a ^ 2 + b ^ 2
70750805752271465055608860340573027765685157371017713344822767273620415511976554537019938312147364
(((f - 2) / 2) ^ 2) / 2 =
35375402876135732527804430170286513882842578685508856672411383636810207755988277268509969156073682
Sqrt(((f-2) / 2 ) ^ 2 / 2) = x
5947722494882871763235375286825560268388553286177
actual x =
1045343918457591589480700584038743164339470261995
Eh, bust!
What's new lol.
That's fine, I'll go read some crumbs and look for a new path.
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d82066 No.12892
So when we solve this what's the plan? Claim RSA money? Sell it to the US government? Release it for free?
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bba8ed No.12893
>>12892
Let's sell it to the Laotian government instead
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71f7e7 No.12894
>>12806
Hello everyone!
Circling back to a couple months ago, there were some good ideas we were working on.
"once you c i[t]"
Every new location of c contains the factors we need to solve.
Maybe at these new a[t] values, we need to do simple operations to help reveal the factors.
For example,
If a[t] is even, divide by 2 until odd.
If a[t] ends in 5, divide by 5.
etc.
For this example, I was able to simplify a=42195 and get a prime factor to pop out.
42195 / 5 = 8439
8439 GCD 145 = 29 = solution(b)
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71f7e7 No.12895
>>12805
>the big idea is that moving to a related (e) column gives us the opportunity to explore a bunch of new a[t] values that have the factors of a, b, n, and c that we are looking to factorize
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a29984 No.12896
>>12892
>>12893
I love the idea of giving it to We The People.
It’s like Heath Ledger’s Joker lighting the pile of money on fire and watching it burn, but in a good dark to light way.
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47f3cb No.12899
Hello everyone!
I hope you are are well on this fine Friday evening.
Reflecting on our quest, I see a few things more clearly now.
"Ready Player One"
Great movie, and great explanation of this mission.
How was it that Chris showed up to post our breadcrumbs?
Could be Govt, Higher Celestial Beings, etc.
No doubt, the fish were caught. That's us.
The Matrix is breaking in real time all around us.
This forum is just one place where this miracle is occurring.
Our wildest dreams are coming true, and yet….
We could be the crowning achievement of excellence for this new era of existence.
We could earn fame and glory, not for ourselves, but for our cause.
We will change the future of maths.
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d82066 No.12900
I was thinking what would the reason for the F value be? If we do ABD(1, c, d+1), the new E value is the negative f value and N is N-1. From these two records, we know that
one N will be even.
Then I found that for each factor of N, (call it J), there exists a cell ENX(E, J, (start X)). This means if we factor N we can hope directly up the E column. Since we have two N values for C, one positive and one negative, we can assuredly find one which N/2 is a valid record. Then we can continue that pattern and go down into the N=1 row.
I haven't found anything definitive here yet, but for my testing of products of primes (factors<1000) we get roughly 80% of the records using GCD with each value.
Also in terms of instant solutions, if F is square, we have a solution, and if ENX(e, 1, x).C is our start C we have a solution. I've tried this algorithm with RSA100 and it didn't generate a solution, but it's another pattern
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d82066 No.12901
>>12900
>one positive and one negative
I mean odd and even
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