2e6c6f No.5042 [View All]
I knew this would make a swastika eventually.
____________________________
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7b70c2 No.5806
>>5803
Some Observations:
only the first and second rows seems to matter, oving in increments of 1's and 2's. LET'S TRY 3's!
Whatever number you're starting with can always be tracked back Up and Left.
Moving Up/Right or Down/Left makes things get weird. (For me, at least)
—————————————————
"right 3, down 2" is an increment of +3 after you find the base value of the increment ({Row# * 3}+2).
If you're starting with the first column,
Right 3, Down 2…
You base increment is ({Row# * 3}+2)
And then it increases in +3 per step
Example:
15*3=45, 45+2 = 47
Right 3, down 2 from 15:
62
now add +3 to the increment and that's your next step
62+50=112
+53
165
Also worked when I did it for 7, 21, and 23
Try it out fer yer own bedazzlement!
>>5805
Preeeeeetty :D
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9c4900 No.5807
>>5805
CA we need a higher res shot of your second diagram!!
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7b70c2 No.5808
>>5807
Open it in a new tab :P
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fcff8c No.5809
>>5808
New tab didn't work, but downloading it and opening it did. Thx!
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ffd9d7 No.5810
Ok, lads. I have a match. Still don't quite understand how it all works. But additional iterations of 16 filled the gap. Take a look, can't confirm it's 100% correct, but the symmetry / asymmetry is pretty cool.
The border needed corners and midpoints of 40/8=5. So each corner and midpoint has and odd number. The 8 remaining gaps to be filled were all 36 units. Found a way to make them all fit.
Yellow = 6
Green = 10
They all fit.
But shouldn't there be extra mod units to add in? I'm stoked but at the same time confused. Any ideas welcome. Posting a work in progress for group discussion.
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bc568f No.5811
>>5807
Sorry the background wasn't white. This has all the gaps filled out
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ffd9d7 No.5812
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9c4900 No.5813
>>5810
Update:
Properly filled borders.
10 * 18 = 180
6 * 18 = 108
= 288, our remaining units to be filled.
Still confused about additional mods related to the 288, but here is the diagram for your consideration and analysis.
Thoughts, Anons?
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1ea3a1 No.5814
>>5813
>>5798
Based on VA's new square layout, and the pyramid numbers posted earlier, attached is another test case for c6107 that attempts to bring these ideas together.
The solution takes a few more iterations than posted in >>5728, and includes the following changes:
1) u1 and u2 are now incremented by pyramid staircase values offset by 4 (meaning each iteration is descending the pyramid).
2) There are 4 squares made up of 2 triangles each that are themselves 1 staircase unit apart.
3) The estimated XPN formula has been adjusted to:
1 + 4*(T(u1) + T(u2) + (2d + 1)) - (2 * f mod 8)
4) the 2d+1 portion represents the border around the 2 adjoining triangles and fills the remainder portion. The d value in each iteration is calculated as (u1+u2)/2.
5) Subtracting the (2 * f mod 8) resulted in each iteration returning a perfect square.
This method hasn't been tested on any other cases and is highly depended on identifying the correct starting u1, u2, and the "polite" way to descend the pyramid.
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1ea3a1 No.5815
>>5814
Pics attached show two possible layouts for the first iteration of the c6107 example previously posted.
Left square is the f-1 starting position.
Right squares represent the T(u1), T(u2), 2d+1 and f mod 8 portions of the formula.
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fcff8c No.5816
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17115d No.5817
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7b70c2 No.5818
Ooh la laaaaa…
Timing and events…
Baker?
Jan?
Other unaccounted Id?
;)
We're doing this.
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89e146 No.5819
Nice following along with you all.
Found this old crumb by mr. curtis. Might be relevant now! (see pic)
Also, no breakthrough patterns, but just another dig back from couple months ago. Did those plots of EvenOdd x+n, D, n, etc in excel.
Attached is a matrix of (f Mod 8) for the product of the first 751 primes (5701 is the 751st prime). So the matrix is from 1x1 to 5701x5701.
Set color f=0 Red, 3=Yellow, 7=Green
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4ba6ca No.5822
>>5766
Absolute pleasure.
This was WAY better the moment it could be shared.
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4ba6ca No.5823
>>5769
The realisation that something is missing in our mathematics, and can be found starts to arrive doesn't it!
Thanks for being part of this.
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4ba6ca No.5824
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7b70c2 No.5825
>>5822
Right on time!
I was about to go fishing with this bait ;)
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4ba6ca No.5826
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4ba6ca No.5827
>>5825
Good to see you Tops.
Thank you for fishing, always good to get an invite.
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1302cc No.5828
>>5827
Glad to catch you online.
Seems like we are close, but missing a critical piece.
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4ba6ca No.5829
OK, so looking at the case of (x+n) is odd.
If the unit square is thought of as the middle and then there are eight triangles.
Remember that (f-1) will be distributed among those eight triangles UNEVENLY.
Some triangles will have more of f than others.
How does this work?
Especially at scale? (Getting very large)
The size of f and d restrict the values of n and n-1.
If you draw out some odd (x+n) squares, relate them graphically side by side with the grid and see any patterns that f make and each 8 triangle, especially if portions of f are outside the (n-1) square in the middle.
Summary:
(x+n) is odd.
f is divided amonst the 8 triangles (some have a bit more than others).
f is within each triangle but OUTSIDE of (n-1), where the (n-1) square should go in the middle.
This will join up with what you were doing before.
This is for cases where f/8 > n
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4ba6ca No.5830
>>5828
Good to see you too anon!
Yes. 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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7b70c2 No.5831
>>5830
Here's a question,
You said there are four "problems"….
RSA/PGP, ECC… P=NP… what's the 4th?
Fermat's Last Theorem?
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4ba6ca No.5832
>>5828
Essentially, as I was showing with RSA100, you take a stab at the number by creating a triangle base out of f div 8. The remainder (f mod 8) is divided among the eight triangles. 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 is easier to break down the less factors there are as there are less solutions.
The answer is easier to come to the more factors there are as there are more solutions.
You get the best of both worlds, which is the ideal for this problem.
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.
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4ba6ca No.5833
>>5831
Four sub types for this one. Two key types (x+n) = odd or even
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e64041 No.5834
>>5830
No. Don't give away the Eureka moment. I think it will be more meaningful once earned.
>>5832
Here's hoping that we solve it first.
The (n-1) square in the middle is throwing me for a bit of a loop. I believe all the previous pictures have had the 2d(n-1) portion surrounding f-1.
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7b70c2 No.5835
>>5832
>The remainder (f mod 8) is divided among the eight triangles.
Like this?
I'm not sure about the 5th in the centers…
But… maybe?
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7b70c2 No.5836
>>5833
Also, VA is out'n'about and can't 8chan at the moment.
"Ask him this: what about the (f-2) portion? Do we add that back in?"
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89e146 No.5837
>>5830 Awesome.
>>5820 Thanks for starting this thread.
>>5834 agreed PMA
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7e1d47 No.5839
>>5832
EITHER = XOR
OR = OR
OR MEET IN THE MIDDLE = ???
>The excluded middle (xor??)
VQC please wait until at least 8:30 pm I have a numerical analysis exam at 6!!
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7e1d47 No.5840
>>5839
Middle of XOR and OR would be.. (a XOR b + a OR b)/2 (average of the two)
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7e1d47 No.5841
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9c4900 No.5843
>>5817
new baker on standby
Novice Baker here, but have my cookbook from /qr/ handy. Topol haz Dough? I'll bake in 2-3 hours if nobody more qualified steps up.
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ffd9d7 No.5846
>>5844
>5844
>>5844
>5844
FRESH BREAD
Let's fill this one up!
I kept it clean and very simple for my first bake ever. No time for fixing funky dough. We can work on getting all the code links up later.
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6fc1ab No.5848
>>5819
Holy shit that first key seems HUGE.
This equation works for A,B and D values.
A/B/D(t) = T(t)*4 + t*const + start
where T(t) =(t*(t+1))/2
I figured out a way earlier to calculate the constant and start for each (e,1) record I just don't have it on hand. These are known values.
an = A(t) = T(t)*4 + t*const + start
= 2*t*(t+1) + t*const + start
bn = A(t+n) = T(t+n)*4 + (t+n)*const + start
= 2*(t+n)*(t+n+1) + (t+n)*const + start
b = a + 2n + 2x
so
bn = an + 2nn + 2xn
2*(t+n)*(t+n+1) + (t+n)*const + start = 2*t*(t+1) + t*const + start + 2nn + 2xn
(2t+2n)(t+n+1) + t*const + n*const + start = 2*t*(t+1) + t*const + start + 2nn + 2xn
[2tt] + 2tn + [2t] + 2tn + [2nn] + 2n + [t*const] + n*const + [start] = [2tt] + [2t] + [t*const] + [start] + [2nn] + 2xn
4tn + 2n + n*const = 2xn
4t + 2 + const = 2x
4t + 2 + const = 2d - 2a
Idk if this helps but I think the algebra is correct.
I forgot what the calculation for t was.
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6fc1ab No.5849
>>5848
If x is odd (d even), then t = (x-1)/2,
so 4(x-1)/2 + 2 + const = 2x
2x - 2 + 2 + const = 2x
const = 0
FUCK
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1ea3a1 No.5850
>>5829
>>5830
>>5832
>(x+n) is odd.
>f is divided amonst the 8 triangles (some have a bit more than others).
May have an idea of how to proceed.
Pics attached are the (x+n)=83 small squares for c6107 and c27707 arranged according to the nn + 2d(n-1) + f - 1 formula.
nn around the edge, then 2d(n-1), and finally the f-1 portion in the middle including the center square.
c27707 is a new test case, so for reference, here are the relevant details from the grid representing the prime solutions.
{23:36:78:47:31:197} = 6107; f=134; (x+n)=83; u=41; (d+n)=114
{151:20:166:63:103:269} = 27707; f=182; (x+n)=83; u=41; (d+n)=186;
Notice the f-1 distribution for both cases is uneven with respect to the 8 triangles.
(1/2)
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1ea3a1 No.5851
>>5850
And here is (I think) the find.
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.
(2/2)
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9c4900 No.5852
>>5851
Damn. Nice work PMA! Studying in detail now.
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8bba68 No.5853
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8bba68 No.5854
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fcff8c No.5855
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8bba68 No.5856
>>5855
Just a little bit of inspiring OC for the group in the final posts of this thread
The third pic of this
>>5851
reminded me the artwork/plans when halfpol was larping on and on about invading Tuvalu lol
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8bba68 No.5857
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ffd9d7 No.5858
>>5844
>5844
>>5844
>5844
>>5844
>5844
>>5844
>5844
FRESH BREAD HERE
Simple loaf, but delicious!
My first Bake.
Please post your favorite diagrams, or links to code. Batter was funky. I chose to err on the side of simplicity. Let's set up a pastebin for batter.
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ffd9d7 No.5859
Also, every Anon archive everything offline. I've got most of it, but need to save the maps first. Baker did a great job with that.
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8bba68 No.5860
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8bba68 No.5861
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7b70c2 No.5862
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