2bf75e No.4140 [View All]
The only thing you need to do to solve this is to organize.
Enumerate EVERY rule.
Global rules.
Row rules.
Column rules.
They are a finite set and RELATED.
This is your MAP.
You will be able to use the MAP to find n from c.
Enumerate the rules.
Win.
____________________________
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5d00f1 No.5009
(e,n,d,x,a,b)=
(a, (CC - a + aa)/2a, C, C-a, a, (C*C+a)/a)
Does this mean this?
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8298a2 No.5010
>>5007
Pic attached for c=6107 and Rsa 100, showing the breakdown between nn, 2d(n-1), (f-2), and their respective mod results.
Included is an analysis of the various triangle formulas 8T(u), T(u), and T(u)/5 - which equates to div 40. T(u) represents the area for one of the 8 triangles that makes up the odd x+n square.
This was an attempt to explore the relationship between the mod values and how they affect the estimation required for an iterative solution.
In these pictures, u represents the triangle base - the blue bar in VQC's images.
The reason the iterative solutions are failing so far is because of the u % ((f-2)/40) result at the bottom. Finding something to quickly fill that gap is the problem.
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1cda4a No.5011
>>5010 great PMA, am also looking at some mod values, and the f, see examples. Next going to look at how the 2d-1 fills in to take space for the n-1 block.
>>4681
>Try this group:
>odd e, even d, odd (x+n)
>From the smallest upwards incrementally.
>Notice any patterns with f?
For this group:
The (1,c) and prime records are in the same group (odd e, even d, odd (x+n))
n always EVEN, X alway ODD,
f is identical for (1,c) and prime records
d is identical for (1,c) and prime records
e is identical for (1,c) and prime records
fMOD8 and (f-2)MOD8 are identical for (1,c) and prime records
the (1,c) and prime records are in the same group (odd e, even d, odd (x+n)) with same mod value
fMOD8=2,6
(f-2)MOD8=0,4
>>5008 Very interesting CA. Seems you've found something of interest!
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8298a2 No.5012
>>5011
Spreadsheets look good.
Have you given any thought to the relationship between the large square and the small square?
in >>4678
>We are creating a method that USES f as a guide to find how to construct the square.
Is there something in the large square that directs the use of f further into estimating 2d(n-1) or nn ???
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7a77e2 No.5013
>>5010
>>5011
>>5012
>The reason the iterative solutions are failing so far is because of the u % ((f-2)/40) result at the bottom. Finding something to quickly fill that gap is the problem.
>Have you given any thought to the relationship between the large square and the small square?
Maybe f is a measurement for both the small square and large square.
>Is there something in the large square that directs the use of f further into estimating 2d(n-1) or nn ???
Yup, gotta be a connection.
CA is very close to tying everything back to the number trees!!
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1cda4a No.5014
>>5012
Haven't seen a match/pattern yet. Was working through the (n-1) triangle cap piece.
>>4678
Did get the "fill area" to check out. Left side has the pieces, just by taking the chunks visually and comparing to the 2d(n-1):
(d+n)^2-d^2-n^2+1-e-f=2d(n-1)
Then, the box with the red 2d(n-1), f, and n^2-1 was tallied, and this is indeed (x+n)^2 for the odd group. We can see some fractions for the Even cases, so will have that to address in part 3b.
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f3b045 No.5015
>>5013
The theory is that each pair of numbers points you to a different prime (maybe not even a prime idk) as you recurse up the tree and it should end at the right factor.
Also my idea is a way to generate prime numbers.
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fa8576 No.5016
>>5015
CA - I wrote a test around your "toOther" and "web" methods from >>4998.
Called them using the a and b values from our c and p records and noticed something VERY interesting.
Pic attached is for c=6107 and c=9874400051. But this holds true for all my other test cases including RSA100.
web.one = toOther(a, b*a)
web.two = toOther(b, b*a)
I then determined the d and e values for each of these results. And for giggles, added them together to get d+e.
For a and b values from the original c record, d+e is the same for the web.one and web.two.
But, for the a and b values from the prime record:
- 1/2 of the difference between d+e for web.one and web.two equals the (x+n) value we are looking for.
- the d value for both web.one and web.two always equals the x+n value from our starting c record.
There is a link here.
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fa8576 No.5017
>>5016
Or to simplify
(web.two - web.one) / 2 == solution (x+n)
where the d for both web.two and web.one equals the original (x+n).
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d0d511 No.5018
>>5016
Yeah I actually think this may be the secret underlying link between all the prime numbers. There is probably some criteria for the p and q that go into the web function to generate primes that I haven't discovered yet.
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fa8576 No.5019
>>5018
Any thoughts as to how this might assist with the (f-2) div 40 iterative search?
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d0d511 No.5020
>>5019
Not that I can think of. Check this output:
>>> makeTree(85)
85
9
3
1
[[[1, 2], 0], [1, 0]]
>>> web(1,2)
(1, 2) (1, 2)
1, 2 2 [2] 2
2, 2 3 [3] 3
(2, 3)
>>> web(3,0)
(3, 0) (3, 0)
3, 0 3 [3] 3
0, 0 0 [] 0
(3, 0)
>>> web(3,1)
(3, 1) (3, 1)
3, 3 5 [5] *5*
1, 3 3 [3] 3
(5, 3)
>>> makeTree(7*19)
[[[1, 2], [1, 0]], [1, 2]]
>>> web(1,2)
(1, 2) (1, 2)
1, 2 2 [2] 2
2, 2 3 [3] 3
(2, 3)
>>> web(1,3)
(1, 3) (1, 3)
1, 3 3 [3] 3
3, 3 5 [5] 5
(3, 5)
>>> web(1,5)
(1, 5) (1, 5)
1, 5 7 [7] 7
5, 5 11 [11] 11
(7, 11)
>>> makeTree(13*61)
[[[1], [1, 2]], [3]]
>>> web(1,2)
(1, 2) (1, 2)
1, 2 2 [2] 2
2, 2 3 [3] 3
(2, 3)
>>> web(1,3)
(1, 3) (1, 3)
1, 3 3 [3] 3
3, 3 5 [5] 5
(3, 5)
>>> web(5,3)
(5, 3) (5, 3)
5, 15 61 [61] 1
3, 15 59 [59] 59
(61, 59)
So for some of these if you extrapolate out you can generate the correct factor through this. Some examples don't work but I'm not completely deterred because maybe my trees are terminating too quickly or something idk
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fa8576 No.5021
>>5020
Some other nifty coincidences from these formulas:
( (d+n)(d+n) + (x+n)(x+n) ) / 2 - web.one == solution a
( (d+n)(d+n) + (x+n)(x+n) ) / 2 - web.two == solution b
web.one e - (x+n) = a
web.two e - (x+n) = b
(cc / 4) - web.one == a
(cc / 4) - web.two == b
I think you should come up with a better name for these methods!
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fa8576 No.5022
>>5021
Revised test cases showing the additional connections.
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6133d1 No.5023
>>5021
>>5022
CA and PMA tearing it up! I've studied all your guys' output, and it looks good. I'm working to understand the underlying ideas, I've got about 90% of it down.
>The theory is that each pair of numbers points you to a different prime (maybe not even a prime idk) as you recurse up the tree and it should end at the right factor.
>Also my idea is a way to generate prime numbers.
What are the algebra formulas for web.one and web.two? Or how to calculate web1 and web2? I understand this idea:
>for any a,b, where a or b is even (so that a*b is even), we can make a continued fraction [(a*b)/2; (a, a*b)] which evaluates to the square root of an integer value. The integer value would again be dd+e which is (a*a*b*b)/4 + b.
Can CA or PMA give a quick explanation of the formulas and how they work? Almost there in my understanding.
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fa8576 No.5024
>>5023
VA - If I'm understanding this correctly, there seems to be a direct relationship between the x+n value for the entry c record, and the squares of these values as you go higher up the tree.
For example, for a = 31, b = 197, c=6107
web.one = 31 + (6107*6107)/4 = 9323893
web.two = 197 + (6107*6107)/4 = 9324059
The interesting parts here are that the sqrt of both numbers is x+n = 3053.
And the web.one number 9323893 is prime.
I have also checked a=5, b=29, c=145, and the web.one method also produces a prime number. 5+(145*145)/4 = 5261.
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2343be No.5025
>>5024
Thanks PMA! Studying now. Also, I agree with you here:
>I think you should come up with a better name for these methods!
I propose "CAprime.one and CAprime.two :)
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2343be No.5026
Lost my trip. Baker, you there? I'm looking forward to some excellent new bread! New loaves? Here's some loaves for you lads.
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8298a2 No.5027
>>5026
Seems appropriate to respond to this one.
So a prime number can sometimes be created by adding a prime to the product of two primes squared divided by four?
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6133d1 No.5028
>>5027
Alright, I've found an appropriate shitposting niche! If PMA is commenting on it, it's gold.
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0f243d No.5029
>>5024
>The interesting parts here are that the sqrt of both numbers is x+n = 3053.
The sqrt of both numbers is 3053, but where does x+n = 3053? The x+n of 31*197 is 83 and the x+n of 6107*6107 (or any square) is 0.
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8298a2 No.5030
>>5029
That’s the x+n from the original c record.
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0f243d No.5031
>>5030
What's the original c in this case then if it isn't c=31*197=6107 or c=6107*6107=37295449? I'm not seeing it in any other post.
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8298a2 No.5032
>>5031
Original c record is where a=1 and b=c. In the case a=1, b=6107.
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0f243d No.5033
>>5032
So what this means is that floor(sqrt((c^2)/4)) = (x+n) iff a = 1? I took the real a and b out of it because it still seemed to work (it floated around 3053.5). Maybe there's a link between the remainder from the square root of ((c^2)/4) across all values of c that allows us to change the equation around to calculate a and b.
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f7dce4 No.5036
Looking at you, MandelBaker…
(Trap may or may not be included. See site for details.)
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f7dce4 No.5037
>>5036
C'mon Nerds… if you can arrange a prime (points to it, -rimshot-) you can arrange a board.
We'll need some solid dough to bake some tasty bread.
(The pics may or may not get better… maybe.)
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a60c59 No.5038
>>5036
Verging on lewding the ponies there, Topol. ;). Should I report you to you?
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f7dce4 No.5040
>>5038
Bro, if you think anyone is lewding that pony; we have a bigger problem here.
>>5037
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a60c59 No.5041
>>5040
I hadn’t updated yet. Yikes!!! I’ll take the first pony girl. Ivanka pony girl would be the best.
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2bf75e No.5044
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2bf75e No.5046
Something that I completely didn't notice when I calculated the inverse triangle function is that it involves calculating the odd square, just like in VQC's methods
public static BigInteger TM1(BigInteger T) {
return sqrt(eight.multiply(T).add(one)).subtract(one).divide(two);
}
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8298a2 No.5048
>>5046
Very interesting observation.
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8298a2 No.5051
Attached diagrams are attempts to further understand the relationships between e, f, dd, nn-1, and 2d(n-1), the small and big squares, triangles, and how this might lead to an iterative solution.
These examples represent the prime solutions for c=115, c=259, c=287, and a partial view of c=6107.
The 2 light blue blocks from (f-2) can always be positioned one in the center and the other to fill the nn-1 square.
The (x+n)(x+n) square is bordered by e and some dd blocks represented in light grey. I was looking for a way to define a boundary for the small square that could be calculated.
Also notice how the light blue f squares can start directly after the green e squares. Together they sometimes match the x+n side or get pretty close.
In the c=6107 example, there are 2 lightly colored vertical lines that represent (f-2)/2 and (f-2)/2 + (e-1)/2.
This was an attempt to use the e and f relationship to find a better starting position for the (f-2) div 40 iterative search.
An estimated triangle base can be calculated as:
u = ((f-2)/2 + (e-1)/2) / 2
In some cases, this formula will find an exact match. In others, it can be used as a starting position for a quicker search.
Unfortunately, there are quite a number of cases where this completely misses the mark. RSA values and small values of n in particular.
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2bf75e No.5052
>>5051
So, an iterative solution would be calculating a fraction of the odd square, and filling the rest with 2d multiples?
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8298a2 No.5053
>>5052
Yes, I believe we estimate the base of one triangle, compute it's area, then the square, account for the 2d multiples and mods, and check for a match.
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bf3baa No.5054
>>5051 we're on the same wavelength! I was playing with the colored geometry earlier, came on and saw your post. Here's what I came up with, in the middle of it and was going to look at how the (1,c) record looks, or another ODD case such as this.
Thought the 3rd one here was the most interesting configuration. Would like to try for another case and see if that shape holds.
The comment >>4680
>Those who are good with their grid diagrams, can you show any examples of how all the pieces fit together for smaller examples with the triangles?
>The only piece missing before is the left hand big square with the pieces added to the side.
>Should make for a good diagram. Particularly with respect to the contribution of 2d(n-1) and how the symmetry of that looks.
..has been hanging there and been intending to come back to it.
For each of the 3 pics:
2d(n-1)=32*(5)=160 units (red)
160 / 8 = 20 (area of each triangle, highlighted in yellow)
n=6, so n2-1 = 35 units (purple)
f=30 (blue)
Landing on the 3rd pic, the symmetry felt right, as it fits with both the (x+n) square, and also the larger squre of (d+n)=22, with=9.
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f7dce4 No.5055
>>4795
I found where this image came from.
Meru Foundation.
Is pretty interesting.
https://youtu.be/midn6ABiZMA?t=10m43s
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bf3baa No.5056
>>5055
Interdasting. We have a "legitimate BRIDGE" as well:
https://youtu.be/midn6ABiZMA?t=1656
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bf3baa No.5057
>>5054
Checked for c=287, and a similar pattern can be constructed. f=2.
The (1,c) show some interesting patterns, and coalesce around squares with some configurations, but nothing ready to share yet.
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2343be No.5058
>>5051
>>5054
Excellent diagrams, PMA and MM! Studying now.
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bf3baa No.5059
>>5058
The symmetric fill patterns on those last couple were lucky. Did a couple more and the f and n don't fill the small square so cleanly, there is a d remaining that needs to fill in.
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4cf918 No.5060
>>5059
Hey lads! It's been kinda slow here, but that's fine. Just checking in, I'm thinking and working over here. Reviewing diagrams and crumbs. Q is going off too. Let's keep up the good work. Hopefully VQC will pop in soon.
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8298a2 No.5061
Still stuck on an iterative solution, but sharing some images with the hopes that someone else will have a better insight into this.
Attached are pictures for c=259 at various stages during the iteration process while determining n0.
Examples are for n0=2, n0=3, n0=4, and the final solution where n=6. Darker colored squares are the values after n0 has been determined.
The f-2 calculation on the right of each image shows the iteration step, f-2 div 40 and f-2 mod 40 results, and the triangle formula. Currently using just the iteration*(f-2) div 40 result as the triangle base. Not quite sure yet how the mod 40 factors in.
One thing that has become a bit clearer in doing this exercise is the meaning of the remainder 2d(n-1) calculation.
It's the difference between the 1+8T(u) formula and the nn-1 + 2d(n-1) + f formula for the small squares.
Also, just a guess that the "All your base are belong to us" comment somehow relates to the triangle base (u) in the small square formula.
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8298a2 No.5062
>>5061
My apologies. Revised pics attached indicating the contributions from the 2d(n-1) remainder. Indicated by "r".
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4cf918 No.5063
>>5061
>>5062
Check Here? ;)
> "All your base are belong to us" comment somehow relates to the triangle base (u) in the small square formula.
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8298a2 No.5064
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819f3a No.5065
PMA, we love you man. Just keep up the good work! Here's a good video for tonight.
https:// youtu.be/xFntFdEGgws
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8298a2 No.5066
>>5065
Thanks, VA. I think we all have made that leap of faith.
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1bfa60 No.5067
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