add73d No.3361 [Last50 Posts]
RSA General
Crumbs to focus on
In the (e,1) solution, d[t] = na+x
and a[t] = na
Therefore, if you can find the t value of this element, you can factor all semiprimes.
>Which column has NO gaps? [0]
>What is the pattern in COLUMN ZERO?
>One Row to Rule them All.
>A column that contains ALL and is the KEY to ALL the patterns.
>What are the patterns in COLUMN ZERO?
>If c is a semiprime, how many times does its square (c^2) appear in column zero? [5]
Why are there gaps?
The reason for the gaps may be because the VQC does not include decimal numbers. That is, because cells with gaps are the cells where the midpoint of a and b would not be a whole number. Or, worded another way, the difference between a and b would be an odd number.
There is a repeating pattern of gaps as you move horizontally from (0, n) to the right.
The pattern of gaps repeats every 2n if n is even.
The pattern of gaps repeats every n if n is odd.
>If first appearance of factor p is element t, second appearance will be at (p+1-t)
>p will be a factor of a in elements: t+p, t+2p, t+3p,…
as well as in elements:
>p+1-t, 2p+1-t, 3p+1-t,…
take d from all values of d[t] at (e,1) and there is a known pattern 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]
Why do the values in row one determine all the values in the columns below them?
Let me break that down.
Find every value of d - d[t]
There is a pattern of (n-1) being a factor.
Find every value of a[t].
There is a pattern of n being a factor.
The pattern in the first group of values is "increasingly different" from the second. We must catalogue these patterns.
____________________________
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add73d No.3362
Code
C#
BigInteger Square Root —— https://pastebin.com/rz1SdACZ
Generate Bitmap within original code —— https://pastebin.com/hMTtJF6E
More on generating a bitmap with the original code —— https://pastebin.com/JUdtehb4
Generate the large square for e and t —— https://pastebin.com/nbjs2kz4
Original VQC code —— https://pastebin.com/XFtcAcrz
How to run VQC code on Linux —— https://pastebin.com/6HnN7K5X
Unity Script —— https://pastebin.com/QgAXLQj3
Unity Script 2 —— https://pastebin.com/Y38nVWgT
Java
VQCGenerator w/ Bitmap —— https://pastebin.com/Dgu9aP1h
VQCGenerator —— https://pastebin.com/VMRnkXFP
Traverse the VQC cells in real-time —— https://anonfile.com/TeH6q3d8bd/VQCGUI_v2.7z
NodeJS
BigInteger Library and Sqrt —— https://pastebin.com/y8AXtFFr
Python
College Anon's code —— https://pastebin.com/d8xZZnm0
Create the VQC —— https://pastebin.com/NZkjtnZL
3D VQC —— https://pastebin.com/vdf8SpYt
3D VQC (v2) —— https://pastebin.com/wZM5Thzu
Fractal cryptography —— https://pastebin.com/XuN4U7Dv
Generate cells for a (and more) —— https://pastebin.com/iAizgLFF
Generate any cell in (0,1) and (0,2) —— https://pastebin.com/gRTYpdMU
Generate genesis cell —— https://pastebin.com/GKzcCpMF
Generate positive AND negative genesis cells —— https://pastebin.com/9ixjRyxt
Calculate variables based on e and t —— https://pastebin.com/4s6McdbN
Get A and B from C and N example —— https://pastebin.com/s0SZ9BNF
VQC + t —— https://pastebin.com/Lgufk0db
Rust
Check if a number is prime —— https://huonw.github.io/primal/primal/fn.is_prime.html
Create Bitmap using the VQC Generator —— https://play.rust-lang.org/?gist=c2446efeec452fe14e1ddd0d237f4173&version=stable
Create Bitmap using the VQC Generator [V2] —— https://pastebin.com/zGSusyz5
Additional VQC code —— https://play.rust-lang.org/?gist=50def916ad48400bc5d638fbf119ae85&version=stable
Generate the VQC —— https://play.rust-lang.org/?gist=6b6beb372b6b931f1abd30642a35a80c&version=stable
Factorization methods (Java)
Binary search for i —— https://pastebin.com/TAt5bDsR
Count down from t of 1c element —— https://pastebin.com/xxYa946V
Mirrors 1c until e=(-x+n^2) —— https://pastebin.com/WJBqPM4P
Calculate factors using -x jumps —— https://pastebin.com/gKX9GW9r
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
Videos on cryptography —— https://pastebin.com/9u3hwywe
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add73d No.3363
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7fa95d No.3381
>>3363
Posting this again since you forgot again
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add73d No.3382
>>3381
The best thing I envisioned for keeping track of crumbs was making this but for the newer crumbs. Just never got around to it.. I'll see if I can work on it tonight
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7fa95d No.3383
>>3382
A picture will probably look better than unformatted text, but until then at least this means you don't have to wait for 8 751-post threads to load.
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7fa95d No.3390
Repost
>>3385
>"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?"
When did he say that? Link?
>What's the third decision?
At the top of the decision tree, if e = 0, it's a square, so it's not an RSA number (right?). If e != 0, we need to find n or x, which apparently relies on whether it's odd or even. In the second decision, whether it's even or odd determines what t is, but then to calculate t you need to know x, and if you knew x you'd know a. So do we know any other calculations that require knowing if e is even or odd? Otherwise, should we be using this to find relationships?
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add73d No.3391
>>3390
Yes, you need to analyze odd and evens. Notice how, in the VQC, a lot of places are a sequence of odd and even numbers?
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ae181d No.3392
>>3390
We can already find X it is the first entry where the GCD of (d(e,1) - d(e,n)) and a(e,1) is not equal to 1. Moreover this GCD is a factor of the original number.
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ae181d No.3393
>>3392
The even E and odd E have to do with where you start your X value. For even E, you start with X=0, and with odd E you start with X=1. You calculate your first D by the equation:
D = (X*(X+2) + E)/2
Also this E even or odd has to do with how much you shift D by each successive entry. If E is even, then your shift starts at 4, if you're at an odd E, it starts at 6. With each successive entry in the cell, you increase the shift by 4.
Ex: (odd E)
D1 = D0 + 6
D2 = D1 + 10
D3 = D2 + 14
So those are your decisions
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add73d No.3394
>>3393
Is it still a search algorithm?
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ae181d No.3395
>>3394
You just iterate through the values in the e,1 cell so I guess it's a search. The thing is it works with every D value (including D=0) so I've been thinking that we may be able to cancel out D entirely from the equation or generating a system of equations for multiple D's to then simplify it but I haven't figured out exactly what I'm doing with that
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add73d No.3396
>>3395
That's similar to what I just found.
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add73d No.3397
Notice how all d - d[t]'s are even and all a[t]'s are odd.
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aeeab0 No.3398
>>3397
Look at the first time where they have a gcd that isn't one
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add73d No.3399
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aeeab0 No.3400
>>3399
Which X is that? (the correct one)
Also what is the GCD? (the correct factor)
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add73d No.3401
>>3400
{1:1:32:7:25:41} (1, 1, 4)
↓
{1:5:12:7:5:29} (1, 5, 4)
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add73d No.3403
Nvm, got it to work. Does your implementation work for 123?
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aeeab0 No.3404
>>3402
def getGCD(A, B):
if B == 0:
return A
else:
return getGCD(B, A % B)
def factor(number):
C = number
D = int(math.sqrt(C))#can be any number s.t. d^2 < C
E = C - D*D
newX = E%2
if(newX == 1):
dShift = 6
else:
dShift = 4
newD = int((newX*(newX+2)+E)//2)
newA = newD - newX
GCD = 1
dCand = newD - D
while(GCD == 1):
newX += 2
newD += dShift
dCand += dShift
dShift += 4
newA = newD - newX
GCD = getGCD(dCand, newA)
print(D-newX, GCD)
#return GCD
#newA = D - newX
#If we switch up the D value, one of these might be negative but it will still be a factor
return newA
This is all the Python code. I switched up the loop to make it fewer calculations but there may be a better way.
>>3403
Yes
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aeeab0 No.3405
>>3404
Also I've tested it for products of the first 50 primes and it solved them all
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add73d No.3406
>>3404
GCD is the only one this is always a factor of c.
D-NewX and NewA at the end are only sometimes factors.
So, got it to work for every number, but it may not fast enough.
Took too long on
856005528343335436096708147408503565008227
and
111086395036901246938296321
Factored 4447786245211932229 in a few seconds. Still have it running on rsa300 if it makes you feel better.
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2ae1e5 No.3408
>>3406
Are these results significantly faster than normal methods for a number of this size?
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add73d No.3409
>>3408
No.
Here is the (fixed) code: https://pastebin.com/70GJSMrv
Again, I think you are going on the right path, since this code was derived directly from the a[t] vs d[t] - d hint.
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add73d No.3410
>>3409
Even with that cold hard fact that we haven't beaten GNFS (though with proper use of our research one could most likely develop a faster algorithm), I still have faith that we'll find the solution, because once we make the jump from search algorithm to complete solution, it will go from exponentially slow to O(1).
Calculation would have the same number of operations for every c, or just about. Amazing thought. (Because the only time complexities lower than log n are log log n and O(1)), and Chris said LOWER than log n.
It's going to be a miraculous jump when it goes from searching to calculation. So hold out.
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ed19bd No.3411
>>3410
You are one of the most faithful people I have ever seen.
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7fa95d No.3413
>>3393
>you increase your D by 4
Are you sure that 4 is constant or do you think maybe the reason it's not working yet for big numbers is that the offset (maybe this is what Chris was talking about) changes depending on the cell? How did you get 4? I noticed last thread when I was analyzing c^2 in the 0 column (nobody else seemed to care about what I was doing so I don't know if anyone will know what I'm talking about) that when c is the square of a semiprime (so the square of what we want to factor), the infinite set starting from (a, b) = (1, c^2) has a beginning d value and it increases at a constantly increasing rate throughout the infinite set (i.e. for (0, 200), d = 21, 44, 69, 96, 125 - this is +23, +25, +27, +29), but the start values and increase values change depending on the cell. The value of x in these sets increases just with the one increase value for each set (same example, x = 20, 40, 60, 80 etc) and the start value and increase changes from cell to cell too. I don't know how the beginning value or the increase values are determined, but if we're using a decision tree instead of iterating through an infinite set's values, perhaps the next decision in the decision tree relates to finding the beginning value and increase values or either d or x. Of course I say this without knowing how you found those original values. If I'm right about this, I have a feeling it has something to do with the relationship between (a, b) = (1, c) and (a, b) = (1, c^2) and something to do with the relationship between t and x. Maybe it would be worth graphing a few. That's a little bit of an educated shot in the dark though.
>>3411
>(1)
Looks like you have some secret admirers, baker. I'm actually personally surprised at how many of us are still here. I thought more of us would have given up. As unrelatedly fitting as Q's message a couple hours ago about nobody being asleep tonight is considering how much progress we seem to suddenly be making, I'm off to bed.
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add73d No.3414
Found this while ACTUALLY FINISHING VQC MAP 2
What the actual fuck, kek
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316b0b No.3415
I'm not making heads or tails of these latest crumbs from VQC. So I've been looking at something else.
I noticed how we can "climb" down the n's from a given record. As we increase the a and b monotonically we will jump between different e's and the n will at certain times lower by one until it hits 1.
For (e, 2) (where applicable) this follows a simple equation of d - floor((e - 1)/2).
Meaning to go from (3, 2, 3), which is {3:2:12:5:7:21} you add 11 to both 7 and 21.
d here is 12, 12 - (3 - 1)/2 = 11. The record for 7 + 11, 21 + 11 is (0, 1, 24, 6, 18, 32) which is also the first record that has n = 1 for 7 + k, 21 + k for k < 11. For these records the x in the record (e, 1) will always be the (e, 2) x + 1.
No idea if this is useful, but it was interesting. For n > 2 I haven't spotted an obvious pattern yet.
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f077f5 No.3416
>>3413
Yeah I'm pretty sure It's constant.
>>3414
I don't get this at all lol
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2ae1e5 No.3417
>>3409
>>3410
Thank you for the replies. Thanks to the glorious US education system I find it hard to follow along sometimes. This is the coolest thing ever though and I encourage you all to keep up the good work!
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d6f59c No.3418
>>3416
Good luck with an RSA sized number and GCD
Here is an example of two miniscule numbers relative to RSA.
D(t)-d E N D X A B
28 32 61 38 23 163
47 28 1 14 0 14 16
43 28 1 18 2 16 22
35 28 1 26 4 22 32
23 28 1 38 6 32 46
7 28 1 54 8 46 64
-13 28 1 74 10 64 86
-37 28 1 98 12 86 112
-65 28 1 126 14 112 142
-97 28 1 158 16 142 176
-133 28 1 194 18 176 214
-173 28 1 234 20 214 256
-217 28 1 278 22 256 302
-265 28 1 326 24 302 352
-317 28 1 378 26 352 406
-373 28 1 434 28 406 464
-433 28 1 494 30 464 526
-497 28 1 558 32 526 592
-565 28 1 626 34 592 662
-637 28 1 698 36 662 736
-713 28 1 774 38 736 814
-793 28 1 854 40 814 896
-877 28 1 938 42 896 982
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d6f59c No.3419
>>3416
Try again
D(t)-d E N D X A B
28 32 61 38 23 163
47 28 1 14 0 14 16
43 28 1 18 2 16 22
35 28 1 26 4 22 32
23 28 1 38 6 32 46
7 28 1 54 8 46 64
-13 28 1 74 10 64 86
-37 28 1 98 12 86 112
-65 28 1 126 14 112 142
-97 28 1 158 16 142 176
-133 28 1 194 18 176 214
-173 28 1 234 20 214 256
-217 28 1 278 22 256 302
-265 28 1 326 24 302 352
-317 28 1 378 26 352 406
-373 28 1 434 28 406 464
-433 28 1 494 30 464 526
-497 28 1 558 32 526 592
-565 28 1 626 34 592 662
-637 28 1 698 36 662 736
-713 28 1 774 38 736 814
-793 28 1 854 40 814 896
-877 28 1 938 42 896 982
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950189 No.3420
for CONTEXT of >>3414
>>>cbts/42019
https://8ch.net/cbts/res/34072.html#38114
Anon wrote:
▶Anonymous 12/05/17 (Tue) 23:30:18 f03740 No.41094>>41110 >>41997
Or the golden ratio is like, our denominator. Like our universes calling card or something?
A % Golden Ratio = Prime is fucking crazy!!!What is up with this equation?! My mind is being continuously blown! What the hell is this? VQC! Tell me! Tell me my duderino famalam!
A % Golden Ratio = Prime
>A % Golden Ratio = Prime
A % Golden Ratio = Prime
>A % Golden Ratio = Prime
A % Golden Ratio = Prime
>A % Golden Ratio = Prime
A % Golden Ratio = Prime
>A % Golden Ratio = Prime
A % Golden Ratio = Prime
>A % Golden Ratio = Prime
A % Golden Ratio = Prime
>A % Golden Ratio = Prime
A % Golden Ratio = Prime
>A % Golden Ratio = Prime
Whhhhhhaaaaaaaaaaaaaaaaaaaaat theeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee fuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuckkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
which was a response to VQC:
▶VQC!!Om5byg3jAU 12/06/17 (Wed) 04:58:10 ecb64b No.41997
>>41094
This is where the fun really begins.
"I'm not a Monster, I'm just ahead of the CURVE"
Well done.
VQC responded to this >>3414
with:
▶VQC!!Om5byg3jAU 12/07/17 (Thu) 04:19:54 ecb64b No.47800
>>42019
Yes and only a small, finite number of dimensions are required to utilise the properties of the whole.
It's all about CONTEXT.
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950189 No.3421
re: Decision Trees
>>3386
>Is this something VQC has said?
▶VQC!!Om5byg3jAU 12/06/17 (Wed) 15:59:38 12c5d9 No.44171
Integers are also related into families that you have noticed.
Like fractals.
They repeat.
They GROW.
Sea shells grow.
Establishing which patterns govern which level of the fractal and which part of the branch of that fractal, will help identify the factors of c.
Think decision trees too.
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950189 No.3422
>>3385
>>3387
So Divisible by 5 likely the 5th decision, so what would be the fourth, assuming no to #'s 2&3? A quaternion?
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fc4336 No.3423
>>3420
>>3421
>>3422
WHOA!! A%Golden ratio = prime? That means we could work backward from d to find a and x? Nice find on the crumb!!
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7fa95d No.3424
>>3416
>Yeah I'm pretty sure it's constant
But my other question was how did you find exactly 4 and exactly 6?
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0f37bf No.3425
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0f37bf No.3426
>>3423
Quick Question: how do you do this if golden ratio isn't an integer. Is it just an estimation?
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7fa95d No.3427
>>3425
Where did you notice it, so I can look into it?
>>3426
Whoever that was did it when we were blindly looking for patterns by changing the colours and criteria for bitmap generation. It isn't necessarily going to help exactly the way it looks, but obviously Chris did reply to it saying that it was "where the fun really begins". Isn't there some formula for calculating the golden ratio exactly instead of just rounding to 1.618?
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079d4f No.3428
>>3427
>>> from math import sqrt
>>> sqrt(5) / 2 + 0.5
1.618033988749895
>>> from decimal import *
>>> getcontext().prec = 100
>>> Decimal(5).sqrt() / Decimal(2) + Decimal(0.5)
Decimal('1.618033988749894848204586834365638117720309179805762862135448622705260462818902449707207204189391138')
>>>
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bd26b6 No.3429
>>3427
You can find a lot of it by just looking at the values in the N=1 row.
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add73d No.3430
I'm working on the new VQC map.
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add73d No.3431
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3a9068 No.3432
>>3426
Good Question CA! I didn't know the formula for the golden ratio. Glad someone else did!
>>3428
Thanks Anon!
I have to say, I've taken a few days break from intense work here, and I feel refreshed, rested, and ready to dig in again. If anyone else out there is feeling worn out, make sure to take good care of yourself, rest, and get comfy. We will work more efficiently over the long haul if we make sure to care for our bodies. I exercised today, and am now drinking a beer and getting even more comfy. Glad to be here, fellow MathFags.
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add73d No.3433
>>3432
_Conf_goProgMathAnon_P_build_that-shit#_
RED1_RED2_College_Anon_build_solve.
Q
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7fa95d No.3435
>>3434
Oh, you were the one who was emailing me in December. I wasn't sure which one of you it was.
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add73d No.3436
Establish comms for when the VQC takes the internet with it.
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3a9068 No.3437
>>3433
_All_wings_Report_IN#go_Faggots_Math-Time
RED5_RED6_Anons_op_MEGAMATH
D_CsTBA_YES[AUTH_H7^pZBVTZ7302-]
##FLY##
[OWLS] and_Reptilians_GO-BTFO.
HOT-1_pre_D
AS THE WORLD TURNS.
HAPPY HUNTING.
P_PERS: WRWY [N1LB][FG&C]
Jeremiah 29:11
Q
>>3436
Bantz + Shitz?? :)
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7fa95d No.3438
>>3436
>for when the VQC takes the internet with it
>email
?
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add73d No.3439
>>3438
It's like this:
Anon msgboard -> Private
Private -> Uncomped comms
Also, the new VQC map is MASSIVE.
I'm going to put the rest of the crumbs in it.
14999x10310
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5d1335 No.3440
>>3433
With a call out like that, I’ll be sure to post my limited progress on dt/at later.
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fc4336 No.3441
>>3433
>>3440
Nice job whoever is working on the archive and VQC Map! Baker or AA? This stringer was back in RSA #2!
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7fa95d No.3442
>>3441
I made the pdf but I'm not the one making the image version.
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c14c81 No.3443
>>3434
>>3439
HA! Good to see you baker! Thanks for the delicious new bread. I DID thank the anon who posted the GR formula. "Looked up first Anon, I did."
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add73d No.3444
I found a lot of instances of Chris triggering people when I searched the archives.
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c00f3b No.3445
Where's Topol? He should piss on the fire hydrant too and drop some LSD math memes. I'll go say hi over at EZ Bake.
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7fa95d No.3447
>>3444 (nice trips)
He did say some pretty crazy shit. He said Agartha is real and that 9/11 was meant to be worse. That was fun.
>>3445
He seems to leave for several days at a time so he'll probably be back soon.
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8e44f0 No.3448
>>3445
Chris put me time out for being to about "the chatter", though this reminds me I was gonna deep dream something. ;)
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add73d No.3449
>>3447
That just gives it more credibility if HE believes that. Hitler was the one who originally researched Agartha and that is why I refuse to dismiss it.
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32324c No.3450
This meme is one of Topol's finest, created on the fly back in RSA 3 or 4. Posting for posterity.
>>3447
It was SUPER fun! When you start questioning everything you've ever been taught, Agartha, hollow earth, etc. doesn't seem implausible. At this time, I just want facts to validate truth. A is A. Cause and Effect. Since I have no facts (personally validated) on those specific issues, my mind remains open to new solutions.
>>3448
What's up Topol!
"That's a nice global government you have there.
Would be a shame if someone
broke the conditioning."
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7fa95d No.3451
>>3449
>>3450
As someone who has always thought Flat Earth was bullshit but could see there was some weird shit going on (like planes not being allowed to fly over the poles, the heavy protection around the poles and Joe Biden etc going there after the US election) I think Agartha makes perfect sense. Since something like this would be provable with video etc when the time comes, and Q said the 40,000ft view would be too much for 99% of people who would just outright deny it if it was told to them, imagine how crazy the 40,000ft view really is and where this would even fit into it.
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add73d No.3452
>>3451
https://archive.fo/8upab
2018's your lucky year.
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add73d No.3453
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8e44f0 No.3454
>>3453
That's both brilliant and fucking retarded.
*FTFEveryone
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add73d No.3455
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82ada7 No.3456
Still working on the d[t] and a[t] analysis. I have been looking at quite a bit of data, trying to make heads or tails of this. Would appreciate anyone taking a look and letting me know if I have gone too deep down the rabbit hole. (or not deep enough.)
Latest version of d[t] and a[t] analysis with a new "summary" at the top is attached. This is again for c=145.
The negative t records have been removed to save space. The results I am showing are mirrored in negative t.
The summary at the top is an attempt to condense relevant information and make it easier to review multiple test cases.
Which will follow in separate posts.
The summary columns are as follows:
(d[t]-d)/(n-1): values where n-1 is a factor.
(d[t] diffs): difference between values.
(x diffs): difference between x values where n-1 is a factor.
(t diffs): difference between t values where n-1 is a factor.
(d[t] diff)/t: Only shows where t is a factor of d[t] diff. (An attempt to find a common basis.)
(a[t]/n): values where n is a factor.
(a[t] diffs): difference between values.
(x diffs): difference between x values where n is a factor.
(t diffs): difference between t values where n is a factor.
(a[t] diff)/t: Only shows where t is a factor of a[t] diff. (An attempt to find a common basis.)
(1/4)
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82ada7 No.3457
>>3456
Pics attached are d[t] and a[t] summaries for different values of e, odd and even n.
c=441
an example at e=0 for a=3^2, b=7^2.
c=143 and c=145
examples at even and odd e where n is odd.
c=115 and c=481
examples at even and odd e where n is even.
(2/4)
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82ada7 No.3458
>>3457
Pics attached are side by side comparisons of the d[t] and a[t] analysis for various values of c and their prime solutions.
Samples are for c=145, 533, 785, and 901.
Results are run over the same list of records. All starting from (e,1,1) and up to t=c.t + c.n.
(3/4)
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82ada7 No.3459
>>3458
If you've made it this far, following are my current thoughts:
1) We can accurately identify 3 records in each of the (d[t]-d) and (a[t]) sequences. This applies only to the initial 1,c record. Don't know the value of this yet, but it is there.
For (d[t]-d): 1, xx-f, and c.
For a[t]: 1, xx-f+4, and c.
2) For the initial 1,c records, some records have t as valid factors of the d[t] diff or the a[t] diff. When these factors exist, the pattern appears to be:
For even n:
d[t] diff / t are even (2,4,6,etc)
a[t] diff / t are sequential (1,2,3,4,etc)
For odd n:
d[t] diff / t are sequential (1,2,3,4,etc)
a[t] diff / t are even (2,4,6,etc)
Don't know if this is important or not.
3) One of the sequences of values in a[t]/n uses x.
4) One of the sequences of values in (d[t]-d)/(n-1) uses (x+1).
Still have no clue what offset we are looking for.
(4/4)
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7bfaf5 No.3460
>>3459
Check out
>>3413
Haven't had the time to test that out.
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7fa95d No.3461
>>3460
I'm the one who posted >>3413 and I think it would pay to mention that I looked at the places CA was talking about in (e, 1) and they seem to always follow the pattern he mentioned (for whatever unknown reason). That said, I actually don't completely understand what CA is doing, so what I said is possibly still important, just in a different context than calculating an infinite set of ds and xs given (e, 1). We just don't know what that context is yet, and I don't really have the time to do anything other than lurk for the rest of the day. If we're using a decision tree rather than iteration then it potentially becomes important when we link (e, 1) to (e, n). Currently, from what I understand (and I might be wrong), CA's method iterates through (e, 1) to find the correct x for (e, n).
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add73d No.3462
I was comparing a[t] and d[t]-d values for 91, 123, 133, and 145 and some profound discoveries hit me.
Since d[t]-d values always seem to have the opposite parity (odd or even) of the a[t] values, then that means that for any e, we already know the parity of the d[t] - d values.
The n values for a given c appear to all have the same parity. EX:
n for 145 = {5, 61}
n for 147 = {2, 62}
n for 287 = {8, 128}
It'd be great if we could confirm these, because if we knew the parity of n that would be a huge improvement to our algorithms. Remember when Chris said Decision Tree? DECISIONS IMPLY A BINARY CHOICE. ODD AND EVEN!
True or false tree.
Boolean tree.
1 — is e = 0
2 — parity of e (is odd or even)
3 — ???
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add73d No.3463
>>3459
>>3458
It has to be parity. That's what jumps out to me.
But I haven't yet related the parity of a[t] and d[t]-d's with e (not an obvious pattern), so I haven't yet been able to come up with a good idea of what the third decision is. Though if we are able to know for certain the parity of the n we want that would be HUGE.
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316b0b No.3464
>>3462
Decision Tree's aren't strictly binary, though.
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8e44f0 No.3465
Hey nerds, I've been doing my thang over at the red triangle group of also coding nerds.
I have a feeling that the RR from the Vortex Math is the same RR that's leading them through an ARG. Also kinda points to the Zodiac Killer murders being the "cover" for the Skull and Bones initiatory sacrifices. They're supposed to flaunt it because the occult likes to brag… hidden in plain site… thus the letters. "Come on! Catch me, peasants/pissants!"
https://en.wikipedia.org/wiki/List_of_mesons
https://www.thefreedictionary.com/meson
Also a photo of ATLAS
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7fa95d No.3466
Some of Chris' clues directly relate to (e, 1), and since we're working on that at the moment, I'll point a few out:
>At the correct element in the grid at (e, 1) where the value of a at that element equals the na we want, if you subtract 2d+1 from na then you are in the negative half of the grid in terms of e and the value at the same element in the first row will be (n-1)a
So does that mean either the next or one of the next parts of the decision tree relates to na?
>You're going to use column e=1 to show whether the conjecture about Fermat primes is true during this
Has anyone looked for the Fermat numbers in (e, 1) yet?
>A prime appears once in a column. This is SIMPLE to calculate. This calculation is important. I call n for a prime number big_n or N
>The product of two primes appears twice. Once the same as the prime, which is easy to calculate and the second time, this is the n that we are after.
We already know most of this with (a, b) = (1, prime) etc, but this sounds like it's part of the decision tree based on how sequentially Chris describes it.
>How many times does the product of three distinct primes appear?
I looked up the phrase "three distinct primes" and I found this: https://en.wikipedia.org/wiki/Sphenic_number
He wouldn't have mentioned it if it wasn't important, so does anyone know anything about Spehnic numbers? I haven't seen anyone discussing them. He said this right after the above two lines about the product of two primes, so maybe if we can find one prime and we can find the product of three primes we can use those methods to find the product of two primes.
>The relationship between cells 2n to the left or right in (e, 1)
Has this come up in relation to the decision tree yet?
>Extend the cells at (e, 1) into negative x
Has anyone done that yet? He's said a bunch of times that listing everything we know (every rule, every positive and negative value of every integer, etc) would make the answer obvious.
>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.
This seems like the most important of all things to focus on based on where we are right now.
Also this might be a stupid thing to consider but has anyone tried contacting Motti Milgrom and asking if he remembers emailing with a guy called Chris Curtis 10 years ago?
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add73d No.3467
>>3466
Glad you read the crumb map. I don't think I've ever felt this close to the solution. I feel like I'm only lines of code away from it. If you've read all the crumbs you'll see he said he might not even exist (which I'm guessing is referring to his name), as well as when he asked "who is CC" as if it refers to someone other than himself.
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add73d No.3468
>>3464
Yeah, but the 2 decisions he listed are booleans. One or the other. And I've been iterating through so many instances of binary in my head. And like I said, knowing the parity of n AT LEAST halves your search space.
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316b0b No.3469
>>3468
Of course, my intention wasn't to discourage, but simply to point it out. It's also a good thing to keep in mind, if some branches might require more than binary splits.
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316b0b No.3470
>At the correct element in the grid at (e, 1) where the value of a at that element equals the na we want, if you subtract 2d+1 from na then you are in the negative half of the grid in terms of e and the value at the same element in the first row will be (n-1)a
I'm trying to wrap my head around this one. I'll try with an example and if you guys understand it in another way let me know:
We use c = 145. This is a=5,b=29 so the correct na in our case is 25.
The d is 12, so we do 25 - (2 * 12 + 1) which equals 0.
We are now supposed to be in the negative half of the grid with regards to e, which I then assume, we are going after: e - (2 * 12 + 1) => -24. The first record in -24 is {-24:1:28:8:20:38}.
Here a is (5 - 1)5 which is correct, but the crumb states "the value at the same element in the first row will be (n - 1)a". I interpret that as the same record at t (1, 1), where a[t] = 25 should also be the t in (-24, 1) where a = 20. However, this doesn't match, since 20 appears in t=1 in (-24, 1) and 25 appears at t = 4 in (1, 1).
So does that mean that na - (2*d + 1) doesn't reference the same e, as e - (2 * d + 1)?
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316b0b No.3471
>>3470
Now, I'm not sure if we *know* but from what I've seen, if we do e - (2 * d + 1) then (n - 1)a will appear in (e - (2 * d + 1), 1) at some point.
Could it be that we have to use these two together? Somehow combine (e, 1) and (e - (2 * d + 1))?
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316b0b No.3472
>>3471
And, of course(?) shouldn't then (n + 1)a appear in e + (2 * d - 1) as well?
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316b0b No.3473
>>3466
>>You're going to use column e=1 to show whether the conjecture about Fermat primes is true during this
Has anyone looked for the Fermat numbers in (e, 1) yet?
I haven't looked into it myself, not sure if someone else has, but we know that they will always be in (1, n) for greater than 3, and this makes sense since Fermat Primes are all perfect squares + 1. All perfect squares will occur in (0, n) and since we add 1, we will give it a remainder of 1 shy from a perfect square.
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316b0b No.3474
>>3473
This probably won't work for large numbers, but what we do when we compute f (2*d + 1) is that we are actually increasing by one square.
Now, I'm thinking out loud again. We can do e + (2 * d - 1) and here we will find (n + 1)a. But we can generalize it to k * (2 * d - k), where k is the number of squares to add to e.
This means that the (n + k)a will be in (e + k * (2 * d - k), 1) and it will occur at a lower t. I don't know if we can guarantee it to exist as t = 1, though. I mean if we could then we would have found (n + k)a which would be amazing. Then we would be even closer to the actual answer.
Take our beloved example of c = 145. We know the record we want is (1, 5, 12, 7, 5, 29).
If we then do 1 + (2 * d - 1) we will get 24. Here we will find 5 * 6 (30) at t=4 (same t as an btw, but probably not something that holds true for every number).
But let's add more squares to it: e + 4*(2 * d - 4) => 1 + 4*(2 * 12 - 1) => 81.
This is the same as (n + 4)*5 which is (5 + 4)*5 which is 45.
For e = 81, we have the following records:
(81, 1, 42, 1, 41, 45)
(81, 1, 48, 3, 45, 53)
(81, 1, 58, 5, 53, 65)
(81, 1, 72, 7, 65, 81)
(81, 1, 90, 9, 81, 101)
Thus we've "moved" a(n+4) up to t=1. This works fine for c = 145, though. It probably won't work smooth for big numbers.
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316b0b No.3475
Someone want to take a look at this and see if it make sense?
def genesis_block(e):
if e == 0:
return (0, 1, 4, 2, 2, 8)
x = e % 2
a = int(e/2) + x
n = 1
b = a + 2*x + 2*n
c = math.fabs(a*b)
d = math.floor(math.sqrt(c))
return (int(e), int(n), int(d), int(x), int(a), int(b))
def factorize(c):
e, n, d, x, a, b = rowForAB(1, c)
dd = d/2
if dd % 2 == 1:
dd -= 1
ee,nn,dd,xx,aa,bb = genesis_block(e + dd * (2 * d - dd))
print(math.gcd(c, aa*bb))
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316b0b No.3476
>>3475
It only works for small numbers though. But maybe there is a way to get it faster?
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316b0b No.3477
>>3476
Actually, thinking more about it I just got caught up with low numbers. The big numbers multiplied by (n + k)*a will diverge too far from the genesis cell. I don't think this is a the direction to the solution.
Although I am interested in what you guys think about the (n + ..) angle, though.
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8cd29a No.3478
I will look into negative X. I believe it may be very helpful because (xx+e)/2a doesn't care about the sign of x.
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a67aa4 No.3479
On -X, it is not a perfect mirror. Here is a view you guys are familiar with, the end in 3d plotted by E, X, and N.
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a67aa4 No.3480
-X is much smaller and compact. A different pattern.
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a67aa4 No.3481
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a67aa4 No.3482
The line across E0, X0 is a bug, but can anyone else confirm that no cells exist with N= -1 for +E?
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a67aa4 No.3483
>>3482
Sorry, can anyone else confirm that cells don't exist where X = -1?
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316b0b No.3484
Aren't all negative x's simply the next records x * -1?
For (1, 1) we have the following records:
(1, 1, 2, 1, 1, 5)
(1, 1, 8, 3, 5, 13)
(1, 1, 18, 5, 13, 25)
(1, 1, 32, 7, 25, 41)
(1, 1, 50, 9, 41, 61)
(1, 1, 72, 11, 61, 85)
Isn't the negative x for (1, 1) these:
(1, 1, 2, -3, 5, 1)
(1, 1, 8, -5, 13, 5)
(1, 1, 18, -7, 25, 13)
(1, 1, 32, -9, 41, 25)
(1, 1, 50, -11, 61, 41)
(1, 1, 72, -13, 85, 61)
Which are the same records, except the x refers to the next records x multiplied by negative 1 (or simply negated)?
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68a051 No.3485
Hey guys, sorry I've been busy for a bit with work and haven't been able to post.
But this problem has still been on my mind, and I've made some more observations in patterns:
ab = dd + e
2an = xx + e
a is a factor of both dd+e and xx+e. This is the tree growth I believe that we were guided with.
2ab = 2(dd + e)
In this case our n for c = c becomes our original b.
4ab = 4(dd + e)
4ab = (2d)(2d) + 4e
In this case n = 2b.
I think these could be the key to unlocking this thing.
—-
1 more pattern i've observed that I think is linked:
a + b = 2(d+n)
a - b = 2(x+n)
Still trying to put this all together. But I think with these 2 tips we should be able to get it done.
How's everyone else?
There have been many posts since I've been here last, anything in particular I should look into?
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a67aa4 No.3487
>>3485
a + b = 2(d+n) is correct, but a - b = 2(x+n) is not. In the original grid, A < B is always true, therefor a - b < 0, and 2(x+n) > 0.
Also been gone a bit, safe to ignore everything other than VQC posts.
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68a051 No.3488
>>3487
>a + b = 2(d+n) is correct, but a - b = 2(x+n) is not. In the original grid, A < B is always true, therefor a - b < 0, and 2(x+n) > 0.
sorry, b-a.
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d6f59c No.3489
New useless solution record corrected.
For {E, N, D, X, A, C}
Increase N by X
Decrease D and X by X so X=0 and D=A
A and B stay the same
Record is:
{ 2*(NA+XA), N+X, A, 0, A, B}
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a67aa4 No.3490
>>3488
Checked, I assumed that after a minute, and it seems like a property of all valid cells, but it also exists for some invalid cells. Not sure how to apply it.
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68a051 No.3491
>>3490
Can you give me an example of it working for invalid cells?
In our c, a and b are both odd. So they're separated by an even distance. That's the 2 in 2na.
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69c043 No.3492
Negative X would mean that the factor A is greater than D, which means that the other factor would need to be less than D. This is how you have B < A.
If you have a record in (E,1)
D X A B
then there is another cell
D -X-2 B A
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7fa95d No.3493
>>3467
It would surprise me if he was using his real name to give people on 8ch the knowledge necessary to crack RSA. I tried looking in various places on the internet for someone called Chris Curtis from Auckland, New Zealand out of curiosity at one point and while I didn't put much effort in, I didn't find anything.
>>3485
>>3487
CA found a way to calculate any d and x in (e, 1) cells >>3393 although it doesn't work as it's meant to because to find the correct x value at the moment we're iterating.
There's a newer version of the map >>3453
Chris hasn't been here and said anything in a while. Someone messaged him on Twitter, though, and he pointed out some things about the decision tree. The first question in the decision tree is whether e = 0. The second is whether e is odd or even (which relates to CA's thing). We're looking for the third decision at the moment. There are a lot of things you'll see in the map that seem like they relate to decisions on a decision tree, possibly some of the ones I pointed out >>3466 here, but we don't know what the third decision is yet. Other than that, there have been lots of discoveries that didn't go anywhere. I might have missed something but that's what happened last thread and this thread so far off the top of my head.
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68a051 No.3494
Also, is everyone aware that the distance between each 2 consecutive squares is the set of odd numbers?
Another way to think about what we're doing is that we're summing a sequence of consecutive odd numbers.
Example for c=145.
c = 25 + 27 + 29 + 31 + 33.
The middle of the sequence is b, the number of terms is a. And:
25 = 13^2 - 12^2
27 = 14^2 - 13^2
29 = 15^2 - 14^2
31 = 16^2 - 15^2
33 = 17^2 - 16^2
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68a051 No.3495
>>3493
For n = 1, the series of d's, a's, and x's are easily calculable from t. Here's my function, all in terms of t:
if (e == 0) {
x = 2*t;
a = 2*t*t;
d = 2*t*t + 2*t;
} else if (e % 2 == 1) {
x = 2*t - 1;
d = 2*t*t + (e-1)/2;
a = 2*t*t + (e-1)/2 - 2*t + 1;
} else if (e % 2 == 0) {
x = 2*t - 2;
d = 2*t*t + e/2 - 2*t;
a = 2*t*t + e/2 - 4*t + 2;
}
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a67aa4 No.3496
>>3494
Renders done with two different isValid checks. Both use A != 0.
First image if (c - (d * d) == e && d - a == x){return true;},
second image if (a + b == 2 * (d + n) && b - a == 2 * (x + n)) {return true;}
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68a051 No.3497
>>3496
Could you explain your axis's? I think I get what you're saying, but I'm not sure exactly.
You're multiplying a * b in the x and y i'm guessing, but what is z?
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a67aa4 No.3498
>>3497
For X,Y,Z I'm using E,X,N. E extends away, N extends right, X extends up/down.
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68a051 No.3499
>>3498
Gotcha, so you're trying all combinations of e x and n?
And in the grid where there are empty cells, the formula still holds is what you're saying?
If so, then yes, that makes sense, but it doesn't hinder us by holding true for non-grid cells.
The a's & b's in the grid are always the same parity, otherwise, n would not be a whole number. And as I mentioned before, a & b in our case for semi-prime c will always be odd.
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a67aa4 No.3500
>>3499
Checked again. Yes, this is all combinations of E, X, and N using my bruteforce cell creation method.
And correct again, I'm saying formula applies in cells that aren't valid but I can also say the formula always applies in cells that are valid. At least where E>=0, but for any X (positive or negative) and any N < 0.
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a67aa4 No.3501
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68a051 No.3502
>>3500
Checked!
>>3501
Cool cool :) same page.
A pythagorean triplet can be made by:
hypotenuse = b+a
side1 = b-a
side2 = 2ab
This all seems connected.
b+a = 2(d+n)
b-a = 2(x+n)
2ab = 2c
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ab147f No.3503
1st pic, the front of the end, with +E left, +F right, +X up, -X down.
2nd, at E=0, looking towards +E.
3rd, at E=0, towards +F (-E).
4th, at E=0, X=0, looking up towards +X.
5th, at E=0, X=0, looking down towards -X.
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ab147f No.3504
>>3503
I'm a little shaky on the -X values, I don't have anything to check them against, and I lost my code to put the signs on the front of the blocks. Will validate -X next by re-coding the sign process.
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ab147f No.3505
Pretty sure those rays aren't supposed to be there across F, -X.
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ab147f No.3506
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add73d No.3507
Topol was right. My phone could hardly render that whole image and it took some major tweaking to work. Perhaps I could split it into parts or something.
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ab147f No.3508
There are patterns in plus X that seem to be mirrored but rotated in minus X. -X appears to be much more compact across N space. I think the signs will show something interesting.
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68a051 No.3509
>>3502
My pythagorean triplets equations are wrong, sorry.
hypotenuse = b^2+a^2
side1 = b^2 - a^2
side2 = 2ab
This makes it a little trickier.
But it may be related to the mod 4 = 2 set of numbers.
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ab147f No.3510
Here's something else, this is the same bruteforce view, but plotting by D instead of X. -X has positive D's as well, so they're inverted for the purpose of the render. It would seem that -X is a perfect mirror in terms of E,D,N.
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add73d No.3511
So, it looks like we can know the parity of the n we want for semiprimes who do not contain 2 as a factor. (Basically the numbers we need to know how to factor.) I have attached both n values to a list of these numbers. You will see that both n values have the same parity.
c——{N, n}
9——{2,0}
21——{7,1}
25——{8,0}
33——{12,2}
35——{13,1}
39——{14,2}
49——{18,0}
51——{19,3}
55——{21,1}
57——{22,4}
65——{25,1}
69——{27,5}
77——{31,1}
85——{34,2}
87——{35,7}
91——{37,1}
93——{38,8}
95——{39,3}
106——{43,17}
111——{46,10}
115——{48,4}
119——{50,2}
121——{50,0}
123——{51,11}
129——{54,12}
133——{56,2}
141——{60,14}
143——{61,1}
145——{61,5}
155——{66,6}
159——{68,16}
161——{69,3}
166——{71,30}
169——{72,0}
177——{76,18}
183——{79,19}
185——{80,8}
187——{81,1}
209——{91,1}
221——{97,1}
247——{109,1}
253——{112,2}
287——{128,8}
So, what is n for rsa617? I don't know but it is an even number!
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7fa95d No.3512
>>3511
Sounds like another decision for the decision tree. It also cuts the time in half if we were to iterate. Nice work. If n is even and d is even, that means (a+b) % 4 == 0, right? Because it means (a+b)/2 is an even number too. That means if n is even and d is odd that (a+b) % 4 != 0. Vice versa for odd n values. Chris said something about mod 4, right? I don't remember what it was so I don't remember if it applies to this situation.
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add73d No.3513
>>3512
c mod 4 tells you whether or not c is a difference of squares. If c%4 == 2, c is not the difference of squares. If it is any other number, it is.
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68a051 No.3514
>>3513
c mod 4 will never be 2 or 0.
Otherwise c is divisible by 2…
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ab147f No.3515
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7fa95d No.3516
>>3513
>>3514
a+b, not a*b. I'm not talking about c, I'm talking about (a+b) from the equation n = ((a+b)/2) - d.
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68a051 No.3517
>>3516
Gotcha, sorry for teasing you.
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add73d No.3518
n's for rsa100 (same parity theory):
761302513961266680267809189066318714859034057480651323757098846024549192056136964455981270339102876
14387588531011964456730684619177102985211280936
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add73d No.3519
I find it interesting that his purported initials are CC and he referred to c^2 as cc.
"What did Einstein say about co-In(c)iden(c)e???"
God's way of staying anonymous is what he said.
"I might not EVEN exist."
"No fear for EXPRESSION."
Even, expression? Math words.
"C root of D"
Read THIS carefully.
IT is important and will be my LAST message on 8ch but NOT my last message.
You can follow ME on Twitter.
What you have ABOVE and HERE is over 95% of what You need.
Over 95%
The rest of what MATTERs is You and (you)
My name here is VQC.
Victory means Peace,
Questions mean Release,
Chris is my name,
God+/-speed
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add73d No.3520
>Yes, Fermat's Last Theorem can be proved using the closure of the product of the sum of two squares. If higher dimensions existed, the product of the sum of two squares would not be closed.
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add73d No.3521
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8e44f0 No.3522
>>3519
Frickin' double c again…
Sight beyond Site!
Blasting off faster than the speed of light.
c^2… c squared… a squared open circle…
Planes of Light?
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32324c No.3524
>>3455
Thanks for putting the new Map together Baker! Read it and expanded my thinking again.
>>3494
Hey Teach! Good to see you, and thanks for this great explanation.
Decision Trees:
I think step 3 is finding t and x for na (1,c) in (e,1).
Thoughts?
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7fa95d No.3525
>>3524
How do you turn that into a decision?
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8e44f0 No.3526
Last one before I pop over to the EZ bake with variations and anything else that might pop up (for a bit)
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add73d No.3528
>>3523
>>3524
>>3525
Look at these. I think this may be related to the decision.
if (e is even) t = (x + 2) / 2
if (e is odd) t = (x + 1) / 2
I don't know who derived these, I had them in the batter at one point.
if (e is even) n = (floor_sqrt(c + (2 * t)^2)) - (d)
if (e is odd) n = (floor_sqrt(c + (2 * t - 1)^2)) - (d - 1)
if (e is even) (d + n)^2 = (2t)^2 + c
if (e is odd) (d + n)^2 = (2t - 1)^2 + c
if (e is even) (x + n)^2 = (2t)^2
if (e is odd) (x + n)^2 = (2t - 1)^2
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c00f3b No.3529
>>3525
Wondering the same thing, Anon. It's not binary tho. Here's what I have so at the moment: "Choose Your Starting Point." Find na (1,c) in (e,1) and complete the element (x,t now known). Then use that knowledge to find the element that has matching d[t], x[t], and has the "Prime Solution". LOL, Baker should make us some Hitler memes with the Prime Solution.
>>3528
I derived those, with PMA's help. Pic relevant.
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c00f3b No.3530
PMA found that in (e,1) the small square was tied to t, this part was all his work:
if (e is even) (d + n)^2 = (2t)^2 + c
if (e is odd) (d + n)^2 = (2t - 1)^2 + c
if (e is even) (x + n)^2 = (2t)^2
if (e is odd) (x + n)^2 = (2t - 1)^2
T equations verified by Chris, see pic.
if (e is even) t = (x + 2) / 2
if (e is odd) t = (x + 1) / 2
These were my work, based on using t to solve for the small square:
if (e is even) n = (floor_sqrt(c + (2 * t)^2)) - (d)
if (e is odd) n = (floor_sqrt(c + (2 * t - 1)^2)) - (d - 1)
The quadratics are derived easily by subbing out PMA's t solution for small square, and then using algebra to isolate n.
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add73d No.3531
>>3530
Chris explicitly said binary tree. What decision happens after?
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8e44f0 No.3532
>>3531
Tree have ROOTS, a main trunk/body (the genesis cell?), and branches/twigs/leaves that GROW toward THE LIGHT.
So what's being mapped out with the VQC may be all the points where the roots, trunk, branches, twigs, and leaves "are". Are those the primes?
Wasn't there talk about - and going backwards to move forward or something? If y'all know where to find m'discord, you can see all the direction I've been pulled in throughout the Q A.R.G. Phenomenon.
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8e44f0 No.3533
Hmmm… this is where the square root of primes and… something someone hit me up with but I'm not finding the conversation…. maybe it was this?
https://en.wikipedia.org/wiki/Splitting_of_prime_ideals_in_Galois_extensions
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c14c81 No.3534
>>3531
Sheeeit. Good question, Anon. Trying to figure out how to put it in binary decision form. (I'll take Step 3 in Binary for $1000, Alex?)
>>3532
Great visual projection Topol! This is where I have to keep my spiritual and emotional focus. If I let myself get frustrated by "lack of progress", I lose my joy. TREES ARE GOOD AND BEAUTIFUL! I love trees and math. This is an abstract math tree that exists in the source of all knowledge. We have been called here (self-selected perhaps?) to help reveal this beautiful tree to the world. We never give up, we just keep going. How do you say that in Latin?? That shit should be our motto here.
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add73d No.3535
>>3534
I generated this. It's something Chris said to make.
>>3521
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c00f3b No.3536
>>3535
It's beautiful! The patterns are all laid out to see. Interesting that a+b seems to highlight specific patterns that flow in lines from the origin, like branches.
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add73d No.3537
>>3536
It's 3 maps put together. Orange is where e is even, green is where e is odd, and the last is both of them together.
Never realized it, but yeah, 2*i = a+b.
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32324c No.3538
"Perseus Usque Semper Victoria"
"Perseus Usque Victoria"
"Usque Fidelis Victoria"
Vote for your favorite motto, faggots.
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add73d No.3539
>>3538
Hitler nihil mali fecerit
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7fa95d No.3540
>>3538
"Autismus Possum Telum" it's the closest Latin I could get to "weaponized autism"
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c00f3b No.3541
>>3539
"Solutio Primo Oportet Te Auris Laxant, Padawan"
Prime Solution Unlock You Must, Padawan.
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8e44f0 No.3542
>>3538
Hang on…
Perseus was coming up elsewhere in the Q…
…and the Gorgon.
Now was it Red Triangle or NROL/SpaceX related… hmmmm…
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c00f3b No.3543
>>3540
Weaponry Of The Mind
"mens illa armamentarium"
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8e44f0 No.3544
>>3543
>>3540
Autismus illa armamentarium my Bruddahs.
(Pic related to Autism Possum)
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c00f3b No.3545
>>3542
Topol, you wonderful faggot! We all know you are Chris' shitpost account. You and Chris "Had a little talk" (between yourself(s)) about not blessing us with TOO many Math LSD memes. Fess up.
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8e44f0 No.3546
>>3545
I legit was looking at the page to tell y'all to take this to the EZ Bake :P
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add73d No.3547
Just wanna say these are probably erroneous. Google can hardly get living languages right so it's probably taking a shot in the dark when it comes to a dead language.
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fba0a4 No.3549
>>2974
>>3031
>>3131
Let's focus again on c=145 for a minute.
We never made the connection between c and p records. This is where we gotta focus. na=a at (e,1) for c solution. Then d[t] x[t] for n are the same in the prime solution. The only (e,1) difference was the lesser t value for the prime solution.
Side Note: There's gotta be some Golden Ratio / Fib Sequence to how d,x,a,b tie together. All Nature and Growth are governed by these same rules, We're just helping uncover them.
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c14c81 No.3550
>>3549
Error Correction:
>Then d[t] x[t] for n are the same in the prime solution.
Should be:
Then d and x for (prime n) are the same in the prime solution as they are in (e,1)
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add73d No.3551
>>3549
For any semiprime the t value of the c record is bigger than the t value of the prime solution in e,1. So the straight path to factorization is to generate (e, n, t) for 1*c and then go to (e, 1, t) and iterate downwards (by lowering t) until you reach the solution record where x is the x that gives the factors of c.
So, a solution would be finding a way to make that fast enough/skip it entirely.
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fc4336 No.3552
>>3551
Thanks for understanding my idea, Baker. It's so nice to be with your minds who can do the things I like to do. Most people around me can't even begin to understand how excellent this forum is.
Just doing Math(s) is/are? improving my life.
The relationship between d, x, and a is amazing. I'm wondering if there's a pattern between [delta]t for (1,c) and t for (Prime Solution). Can we run those numbers, programAnons? Let's take known c's from the basic grid so everyone can follow. Let's do the (1,c) and (Prime) for each c value. If willing, PMA, would you please pick some c values starting at c=145 and greater? Just a few examples needed to examine this [delta]t idea between na=a for (1,e) and (e,n). Thoughts welcome.
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06387b No.3553
>>3551
>So the straight path to factorization is to generate (e, n, t) for 1*c and then go to (e, 1, t) and iterate downwards (by lowering t) until you reach the solution record where x is the x that gives the factors of c
Exactly. But if [delta]t is known (or patterns of [delta]t are known) can we then lower t to the proper integer using a known formula? Asking for a friend.
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add73d No.3554
>>3553
Am gonna generate a list right now of t diffs.
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add73d No.3555
Iterating t is a smaller search space for factorization than iterating n, but it still is probably going to take several years, so it needs a massive improvement.
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316b0b No.3556
>>3511
So…?
1. Check if e is zero
2. Check if e is odd or even
3. Check if big N is odd or even
…?
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316b0b No.3557
>>3556
4. Check if d is odd or even?
Parity of d will give us parity of x, right?
We can't deduce parity of d from e alone, as you can see in (4, 2).
Of course, I'm now assuming we CAN figure out the parity of x by looking at d. Not sure if we can / haven't looked enough to see if it's possible.
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fc4336 No.3558
>>3555
Checked!!
Agreed. But if we can go from (1,c) to (prime) using a formula based on t in (e,1), then that gives us our x and t values, which are supposed to be the same in (e,n)?
>>3556
>>3557
Looks like this could be next steps for decision tree?
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7fa95d No.3559
>>3557
If we know the parity of d and n then we know the parity of (a+b) too, as explained >>3512 here. a = d - x. If d and x are both odd or both even, a is even. If d is either odd or even and x is the opposite, a is odd. If a is even and (a+b) is odd, b is odd. If a is even and (a+b) is even, b is even. If a is odd and (a+b) is odd, b is even. If a is odd and (a+b) is even, b is odd. If (a+b) is odd then c is even (odd*even or even*odd). If (a+b) is even then c could be odd or even (odd*odd=odd or even*even=even). So if we know the parity of x, we know the parity of every variable. I'm too tired to remember if we know how to find the parity of x yet.
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7fa95d No.3560
>>3559
Okay I don't know if we actually do have the parity of d and n yet just rereading my post from earlier. As I said I'm a bit sleep-deprived. But if we do, all the logic in this post still checks out.
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add73d No.3561
What follows is a similar list, but of their t values, to attempt to illustrate our pattern. Similar to N vs n, I'll refer to the t value from our start point as T, and the one we want as t, and add their differences. The starting point and factors will also be added, so a pattern can be deduced.
Example:
(1,1,6) starting point for 145
(1,1,4) solution element
c——{T, t, T-t} {E:N:D:X:A:B} a*b
9——{2, 1, 1} {0:2:3:2:1:9} 3*3
21——{2, 1, 1} {5:7:4:3:1:21} 3*7
25——{3, 1, 2} {0:8:5:4:1:25} 5*5
33——{3, 2, 1} {8:12:5:4:1:33} 3*11
35——{3, 1, 2} {10:13:5:4:1:35} 5*7
39——{3, 2, 1} {3:14:6:5:1:39} 3*13
49——{4, 1, 3} {0:18:7:6:1:49} 7*7
51——{4, 3, 1} {2:19:7:6:1:51} 3*17
55——{4, 2, 2} {6:21:7:6:1:55} 5*11
57——{4, 3, 1} {8:22:7:6:1:57} 3*19
65——{4, 2, 2} {1:25:8:7:1:65} 5*13
69——{4, 3, 1} {5:27:8:7:1:69} 3*23
77——{4, 1, 3} {13:31:8:7:1:77} 7*11
85——{5, 3, 2} {4:34:9:8:1:85} 5*17
87——{5, 4, 1} {6:35:9:8:1:87} 3*29
91——{5, 2, 3} {10:37:9:8:1:91} 7*13
93——{5, 4, 1} {12:38:9:8:1:93} 3*31
95——{5, 3, 2} {14:39:9:8:1:95} 5*19
106——{5, 4, 1} {6:43:10:9:1:106} 3*51
111——{5, 4, 1} {11:46:10:9:1:111} 3*37
115——{5, 3, 2} {15:48:10:9:1:115} 5*23
119——{5, 2, 3} {19:50:10:9:1:119} 7*17
121——{6, 1, 5} {0:50:11:10:1:121} 11*11
123——{6, 5, 1} {2:51:11:10:1:123} 3*41
129——{6, 5, 1} {8:54:11:10:1:129} 3*43
133——{6, 3, 3} {12:56:11:10:1:133} 7*19
141——{6, 5, 1} {20:60:11:10:1:141} 3*47
143——{6, 1, 5} {22:61:11:10:1:143} 11*13
145——{6, 4, 2} {1:61:12:11:1:145} 5*29
155——{6, 4, 2} {11:66:12:11:1:155} 5*31
159——{6, 5, 1} {15:68:12:11:1:159} 3*53
161——{6, 3, 3} {17:69:12:11:1:161} 7*23
166——{6, 5, 1} {22:71:12:11:1:166} 3*81
169——{7, 1, 6} {0:72:13:12:1:169} 13*13
177——{7, 6, 1} {8:76:13:12:1:177} 3*59
183——{7, 6, 1} {14:79:13:12:1:183} 3*61
185——{7, 5, 2} {16:80:13:12:1:185} 5*37
187——{7, 2, 5} {18:81:13:12:1:187} 11*17
209——{7, 2, 5} {13:91:14:13:1:209} 11*19
221——{7, 1, 6} {25:97:14:13:1:221} 13*17
247——{8, 2, 6} {22:109:15:14:1:247} 13*19
253——{8, 3, 5} {28:112:15:14:1:253} 11*23
287——{8, 5, 3} {31:128:16:15:1:287} 7*41
7463——{43, 35, 8} {67:3646:86:85:1:7463} 17*439
93801——{153, 152, 1} {165:46595:306:305:1:93801} 3*31267
940683——{485, 484, 1} {1722:469373:969:968:1:940683} 3*313561
2611447——{808, 2, 806} {3222:1304109:1615:1614:1:2611447} 1613*1619
44231413——{3325, 544, 2781} {8913:22109057:6650:6649:1:44231413} 5563*7951
173648257——{6589, 740, 5849} {14928:86810952:13177:13176:1:173648257} 11699*14843
6867225401——{41434, 4805, 36629} {119977:3433529833:82868:82867:1:6867225401} 73259*93739
10967535067——{52363, 2, 52361} {209442:5483662809:104725:104724:1:10967535067} 104723*104729
867072408247——{465584, 22940, 442644} {426358:433535272957:931167:931166:1:867072408247} 885289*979423
15778598254603——{1986114, 257299, 1728815} {2970619:7889295155074:3972228:3972227:1:15778598254603} 3457631*4563413
81311001417221——{4508631, 64212, 4444419} {5475100:40655491691350:9017261:9017260:1:81311001417221} 8888839*9147539
215303158862641——{7336606, 385003, 6951603} {8465697:107651564758109:14673212:14673211:1:215303158862641} 13903207*15485863
237051716747561——{7698242, 19017, 7679225} {27978272:118525842977298:15396483:15396482:1:237051716747561} 15358451*15434611
Added some huge semiprimes just to give an idea of how far t is from T in a practical setting.
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add73d No.3562
>>3561
If the factors are close together, then t is closer to t=1.
If the factors are far apart (like 3*313561) then t is closer to T.
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add73d No.3563
So, as T-t increases, b-a decreases.
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add73d No.3564
I've found the crumb we need to analyze.
>>3462
Review the pictures here with these formulae in mind.
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add73d No.3565
So, in the starting point in (e,1), the d[t]-d value is (N-1)*1, and the a[t] is N*1.
And at the solution record, the d[t]-d value is (n-1)*a, and the a[t] val is n*a.
We have a new way to iterate.
Find first appearance of factor of the factor of our N.
Calculate formulae for appearances of its multiples.
Iteration from t=1 becomes a tree instead of blind iteration.
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add73d No.3566
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add73d No.3567
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70f5db No.3568
>>3564
This essentially states that 5 as a factor accumulates over t's.
And since this holds for 5 and 25, it also holds for 85, 145 etc. So 5 will continue to grow as a factor of a in (1, 1). So will 25, 85 etc.
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add73d No.3569
Working pretty nice so far will improve tomorrow. If you want the algorithm it basically calculates d[t]-d for (e,1,1) (then takes absolute value of it) and then iterates over the formulae described in the crumb, and tests the x values produced.
End result are negative values of a and b.
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add73d No.3570
>>3568
Yep, it GROWS exactly like he said. It's still a search algorithm but it's exactly the kind of improvement I was looking for - it's able to skip many of the elements we don't want when iterating t.
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70f5db No.3571
>>3511
Take a look at >>3090
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70f5db No.3572
>>3571
Err, i meant >>2920
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15411f No.3576
>>3361
>>3564
>https://archive.fo/RgVko (RSA#1)
VQC:
Anonymous 12/05/17 (Tue) 09:34:37 ID: ecb64b No.36154
I'll work through RSA100. If you're using C# the BigInteger namespace is handy.
c = 1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139
d = 39020571855401265512289573339484371018905006900194
e = 61218444075812733697456051513875809617598014768503
n = 14387588531011964456730684619177102985211280936
x = 1045343918457591589480700584038743164339470261995
a = 37975227936943673922808872755445627854565536638199
b = 40094690950920881030683735292761468389214899724061
In the element at (Rsa100e,1) you find that the element where x = Rsa100x, you will find a = Rsa100n x Rsa100a
If you find a that is Rsa100n elements further down, then that a will equal Rsa100n x Rsa100b
Then at grid cell (Rsa100e, Rsa100n) which is obviously in the same column, you'll find the product c.
In summary, the factors in any cell in the first row govern ALL the behaviour in the column below in a pattern or, if you perfer, in MANY patterns.
A member of Pink Floyd said in an interview during a documentary on MUSIC PRODUCTION that the GAPS are absolutely as important as where there are NO gaps.
Let the VQC take the strain. The more you c it's beauty, the more it will give you (back).
You all amaze me.
Can't find the crumb, but VQC mentioned we can take a prime of our CHOOSING. The crumb you pulled is an example of this, using 5.
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add73d No.3578
Still kinda wondering why this works.
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add73d No.3579
>>3578
It gets all semiprimes in a few steps up until around 7463 and fails after that. I don't know why.
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5f2245 No.3580
Thank you for your patience.
The decision tree has parts.
The decision tree terminates upon finding a factor.
Pre-factorisation, trim trailing zeros (in binary) or equiv is to divide by two until odd.
This is 1 and c for primes.
This is the square root for squares.
This is the pair of factors that are not 1 and c for the product of two primes that are not the same numbers (not a square).
This is the pair of factors closest to the square root for products of more than two factors (not including 1 and c as factors).
The first part.
Determine if a square.
Find the square root d and remainder e.
First decision, if e is 0, return d.
Second decision, if(GCD(e,d)!=1) return GCD(e,d);
That is the end of part one.
Part two.
Part two is recursive (since the grid is effectively an overlay of the fractal nature of integers).
A recursive part should immediately make sense in hindsight or will come as less of a shock in due course, considering that the complexity claim was at most big oh of the natural log of the length of c in bits.
The inputs into part two are to factorise d and e.
The end result of part two is a tree, descending from c with branches d and e, then branches descending from d and e or each of their square roots and remainders.
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15411f No.3581
>>3579
What changes? Is there a branch in the binary tree? Does the growth pattern remain the same?
>>3580
Oh, hello VQC! Wow, that is a lot to digest, but makes perfect sense. Please provide a moment for the 'in due course' bit to hit. Feels YUUGE. Much gratitude.
>New inputs! factorise d and e!
Excellent, was digging into the d-primes. Hadn't thought about the e-primes. ok!!
Question please? With the Einstein crumb.(and 2 c's), was going down the path of Eisenstein and Conway and Gaussian and Eisenstein integers. Wrong turn for Part 1?
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d1075e No.3582
>>3580
Thank you for this direction.
>or will come as less of a shock in due course
A bit surprised, yes. Will get over it! :-)
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dda169 No.3583
>>3580
Welcome back! These hints are great
First stab at part one (using gmpy2 for fast bignums)
from gmpy2 import mpz, isqrt_rem, t_div, gcd
def one(c):
# Determine if a square
if t_div(c, 4) == 2:
# Make it square?
c = t_div(c, 2)
# Find the square root d and remainder e
d, e = isqrt_rem(c)
# First decision, if e is 0, return d
if e == 0:
return d
# Second decision, if(GCD(e,d)!=1) return GCD(e,d)
tmp = gcd(e, d)
if tmp != 1:
return tmp
return None
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add73d No.3585
O(log n) is still less than a second, or a few seconds, so it doesn't really matter.
So, I'm probably going to make a replica of this tree soon. Until then, here's the code I had. I really have no idea why it worked for the numbers it did.
https://pastebin.com/nD4F7AFa
This tree is actually an amazing idea, though.. Just keep going through the remainders and square roots until you hit a factor.
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fc4336 No.3586
>>3561
>>3562
>>3563 Factor Trees Crumb
>>3565 (N-1)*a and n*a combined with Factor trees crumb? How to combine?
Thanks Baker for crunching the numbers for [delta]t. Seeing some very interesting patterns. You're right about the distance from T and t being too large in the RSA setting. Good progress on the (N-1)*a and n*a ideas tho, and how they combine with the factor tree crumbs.
>>3576
Great find on this crumb!
>>3580
All right! Just saw this, VQC is back! Good to see you.
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add73d No.3587
Let's see if we can get to a factor of 145 by going down this tree shape.
145 = 12^2 + 1
12 = 3^3 + 3
3 = 1^2 + 2
1 = 0^2 + 1
Hmm, add 12's remainder and its remainder's remainder? That would give you five. Let's try another.
287 = 16^2 + 31
16 = 4^2 + 0
31 = 5^2 + 6
6 = 2^2 + 2
2 = 1^2 + 1
1 = 1^2 + 0
41 = 31+6+2+1+1, so this COULD work..
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add73d No.3588
>>3587
Try making this into a recursive program.. see if it reaches factors of c. Gotta tend to my business.
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d6f59c No.3589
What did Q say about something being necessary?
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316b0b No.3590
Some preliminary work:
def rowForAB(a, b):
c = a*b
d = int(math.floor(math.sqrt(c)))
e = int(c - d*d)
n = int(((a + b)/2) - d)
x = int(d - a)
return (e, n, d, x, a, b)
def recurse(c):
e, n, d, x, a, b = rowForAB(1, c)
if e == 0:
return [d, c//d]
if gcd(e, d) != 1:
return [gcd(e, d), c//(gcd(e, d))]
return [[e, recurse(e)], [d, recurse(d)]]
This can be used to study the branching.
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316b0b No.3591
>>3580
Of course, welcome back!
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dda169 No.3592
>>3590
Thanks. Reminds me how much I hate recursive programming too :)
If you're lazy like me and want to see the tree immediately you can cheat a bitprint(json.dumps(recurse(c), indent=4))
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15411f No.3593
>>3589
Disinfo is necessary and used.
anon thinks can at times be related to requirement for plausible deniability.
>>3592
please drop a screencap of that!
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dda169 No.3594
>>3593
Filtering out some useless formatting, this is what I get for RSA100 61218444075812733697456051513875809617598014768503
13204334350125320769126278
2
6602167175062660384563139
7824221627472775882440665
1915100302336
1357695
5
271539
1383871
895
54
5
1
1
1
2
1
1
1
1
1
1
7
3
2
1
1
1
1
1
1
1
1
1
2
1
1
1
1
1
1
29
4
2
2
5
1
1
1
2
1
1
1
1
1
1
1176
2
588
2797181014427
1698898
1089
33
33
1303
7
3
2
1
1
1
1
1
1
1
1
1
2
1
1
1
1
1
1
36
6
6
1672477
628
3
2
1
1
1
1
1
1
1
1
1
25
5
5
1293
68
4
17
35
5
7
39020571855401265512289573339484371018905006900194
2
19510285927700632756144786669742185509452503450097
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add73d No.3595
So, got a prototype running. It's not fast enough yet, but it's getting there.
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68a051 No.3596
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68a051 No.3597
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15411f No.3598
>>3597
Good to see you Teach!
Exactly! Been picturing this, including ordering of the factors.
When we look up at the tree, we see primes. If there is something behind, obstructed, then what we are seeing can factor that integer.
>>3594
Nice. Thanks.
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15411f No.3599
>>3597
>>3598
Wasn't quite picturing everything you put down, that was helpful. Good notation.
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d1075e No.3600
>>3597
teach, why d_2? Shouldn't they all be sqrt(d) and e?
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06387b No.3601
>>3597
Great diagrams, Teach! Thanks.
>>3600
>teach, why d_2? Shouldn't they all be sqrt(d) and e?
I was wondering the same thing. Chris did say you could pre-factorize by dividing by 2 until odd. Then you take d and e and repeat until those branches terminate in a factor.
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dda169 No.3602
>>3597
I see your graph viz and raise you ete toolkit
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15411f No.3603
>>3602
Keep going anon!!
Was looking at your 'screencap', >>3594
playing a bit in Wolfram Alpha. Then you drop this.
What a big week.
etetoolkit.org
Title:ETE Toolkit - Analysis and Visualization of (phylogenetic) trees
Description：The ETE toolkits is Python library that assists in the analysis, manipulation and visualization of (phylogenetic) trees.
>>3601
>I was wondering the same thing. Chris did say you could pre-factorize by dividing by 2 until odd. Then you take d and e and repeat until those branches terminate in a factor.
That recursive part is starting to make sense, using the power of the VQC (on d and e).
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dda169 No.3604
>>3603
Looks way more like a fractal when chopping off the trailing zero bits
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d1075e No.3605
>>3602
>>3603
Pretty sweet results.
Anyone have a better square root function for biginteger in c# (other than what was originally posted) that works well at smaller values?
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dda169 No.3606
>>3604
I screwed up the filename on the last one, both of those are RSA100. The 2048 one is apparently too large to post. You can have 1536 instead.
>>3605
I love these, just need to figure out how to change the background color! Tons of possibilities for layout too
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8e44f0 No.3607
bee tee dubz, this is "c root of d" according to wolfram:
http://www.wolframalpha.com/input/?i=c+root+of+d
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68a051 No.3608
Good to see you all too!!!
Hello VA, Topol, PMA, CA, MA, Baker, (am i missing anyone).
>>3605
d_2 is the divide repeatedly by 2 like VA assumed.
I have one in js you could port:
function sqrt(c) {
if (c.isZero()) {
return c;
}
if (c.lesser(4)) {
return BigInt(1);
}
var n, p;
var high = c.shiftRight(1);
var low = BigInt.zero;
while (high.greater(low.add(1))) {
n = high.add(low).shiftRight(1);
p = n.square();
if (c.lesser(p)) {
high = n;
} else if (c.greater(p)) {
low = n.equals(1) ? n.add(1) : n;
} else {
break;
}
}
if (c.equals(p)) {
return n;
} else if (c.equals(high.square())) {
return high;
}
return low;
}
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68a051 No.3609
>>3602
I haven't heard of ete before, I'll check it out.
Unrelated kinda, but I've seen the collatz conjecture represented using that circular tree diagram. Great stuff.
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8e44f0 No.3610
Sure does look familiar….
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dda169 No.3611
>>3609
Hadn't heard of it either till your post inspired me to look for better options. There is an interactive viewer too!
The code I used (after changing dicts in IseePatterns code to tuples) is this. There are probably better waysdata = (c, recurse(c))
t = Tree(str(data) + ";")
ts = TreeStyle()
ts.mode = "c"
ts.arc_start = 0
ts.arc_span = 360
nstyle = NodeStyle()
nstyle["shape"] = "sphere"
nstyle["size"] = 3
nstyle["fgcolor"] = "red"
for n in t.traverse():
n.set_style(nstyle)
t.show(tree_style=ts)
#t.render("image.png", tree_style=ts)
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d1075e No.3612
>>3608
Thanks teach. Things are getting exciting now!
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d1075e No.3614
>>3612
Forgot image. You guys with the gcd call were spot on. Good job!
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68a051 No.3615
>>3612
Sure is! I freaked out when I saw that Chris had posted.
Even in the tree diagrams there are patterns.
Considering we can ultimately express any c breaking it down to its most smallest pieces, then what are those pieces… 1 & 2 I think.
Example:
c = 145
c = 12^2 + 1
c = (2*2*3)^2 +1
c = (2*2*(1^2 + 2))^2 + 1
c = (2*2*(1^2 + 2*1))^2 + 1
Trying to figure out what this means now. But any number can be expressed as a recursive equation in the for:
2^j * (recursive part) + (recursive part)
where each recursive part starts from the root of 1,1.
Moving backwards towards a (or x) though, still unclear.
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d1075e No.3616
>>3615
>we can ultimately express any c breaking it down to its most smallest pieces
I spent quite a bit of time over the weekend breaking down the (0,1) space for d[t] and a[t] changes. Didn't post the results as it seems a bit unnecessary, and I might be stating the obvious, but all the movements between records look like they be described by prime numbers.
At it's most basic, the numbers seem to work together in very obvious patterns.
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68a051 No.3617
>>3616
Soooo many patterns…
When this is all over I'm going to post my notebook… I need to go buy another one in fact, filled up a grid book.
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7fa95d No.3618
I might put a program together that takes in a public key and outputs the equivalent private key. I'll have a gap where the decision tree code will eventually go. That way, once we're there, we don't have to figure out how keys work in code at the last minute (and so far it seems quite complicated). Nobody has any objections to Java still, yes?
>>3608
>(am I missing anyone)
hi
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add73d No.3620
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68a051 No.3621
>>3618
Sorry AA! I knew I was missing someone, duh.
Regarding the rsa cracking algorithm, sounds like a good idea. I have something similar ready to go ;)
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7fa95d No.3622
>>3621
Would you mind posting it then? I've figured out most of it but if you already have something working I won't need to waste my own time.
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68a051 No.3623
Will do, let me fix it - might be a couple of hours sorry (gotta check in with the real world occasionally).
In the meantime, here's some slides from my security lecture, not that its nothing you won't already find on wikipedia or stackoverflow.
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d1075e No.3624
>>3617
>When this is all over I'm going to post my notebook
Thanks Teach. I'd be interested in reading.
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68a051 No.3625
RSA:
Steps to produce private key (n,d) from public key (n, e).
1) factor n into 2 primes, p, q
2) compute phi = lowest_common_multiple(p-1,q-1) // euclid algo
3) compute d = modular_inverse(e, phi) // use ext. euclid algo
Private key is (n, d).
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7fa95d No.3626
>>3623
Oh I already know that stuff. I was the one who wrote those guides for the lurkers who didn't know anything several threads ago and then stopped because I didn't understand the grid well enough at the time myself. I'm just talking about how you're meant to import keys into Java and get the modulus and exponent out of them. If you've already got that working I won't need to, but I think I can figure it out if you haven't.
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dda169 No.3627
>>3618
>>3621
>>3622
>>3626
Ok so I've had this ready to go since december… Good time to drop! Please keep it safe and don't go destroying the internet
Presenting sonic_rainboom.py
https://pastebin.com/vNqnPRJR
Full support for PEM certificates and certificate requests, armored PGP public keys, SSH public keys, and verification of results before outputting private key. And ECC of course.
Ponies not included
Have fun
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7fa95d No.3628
>>3627
This is fantastic but where does the math go? In the VQC class? This is a great contribution but I'm still tempted to rewrite it in Java at some stage just so I understand it better.
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add73d No.3629
>>3628
You don't have to bother with that. Download a python java linker
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dda169 No.3630
>>3628
Yep. Solution goes in the factors function.
If you generate your own keys you can test it with the private keys like this
class FakeVQC:
def _load(self, key):
f = open(key, 'rb')
key = serialization.load_pem_private_key(f.read(), None, default_backend())
return key.private_numbers()
def factors(self, c):
priv = self._load('rsa/key.pem')
return priv.p, priv.q
def ecpoint(self, curve, x, y):
priv = self._load('ecc/key.pem')
return priv._private_value
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add73d No.3631
>>3630
That's very useful. So it works for PGP?
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dda169 No.3632
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add73d No.3633
In regards to implementing the tree, it's not so easy. I wrote some methods that added up the remainders like in >>3587 but it's too slow for higher numbers.
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add73d No.3634
Also, holy shit, the House Intelligence Committee voted to #ReleaseTheMemo.
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226bb3 No.3635
>>3627
Wait so you solvedd the grid and this is it all solved? What's the algorithm can you explain it?
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f2af50 No.3636
>>3634
Yeah, just saw that now!! Crazy! #ReleaseTheMemo
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dda169 No.3637
>>3635
No, this is just boring wrapper code to directly read the most common types of public keys, and output correctly formatted private keys. For some reason most crypto libraries have no way to just add private numbers to a public key (since its considered impossible, who needs functions for it). Just so we don't have to code it later!
The final ingredient is missing
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add73d No.3638
>>3637
Glad I didn't have to implement that. And you're right, it's so asinine that they wouldn't have functions for that. There are also ways to crack public keys due to noobs not configuring stuff right. If they have a bad RNG you can test some factors from similar keys that have been cracked against them to see if they are a factor. There was a professor who did this and he cracked a ton of them.
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dda169 No.3639
>>3638
Yeah and its tedious to do by hand even when you have the factors (especially PGP, the keys this spits out can be used right away). I wanted this available for when shit hits thef fan, and its a good way to verify our factoring algorithm against real keys as well (this could be considered illegal if they are not yours, fyi). This is the most obvious proof I can think of, right after solving all RSA numbers in a few minutes.
The "keep it safe" bit was serious too, please create backups!
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add73d No.3640
>>3639
How so? Mathematics is not illegal. Academics have always tried to crack it. Though they never actually get anywhere.
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dda169 No.3641
>>3640
I am not a lawyer. Hence the "could be". I'm betting someone will make the argument that cracking the Verisign root key shows malicious intent or something, DMCA laws and all. Would rather not be the one to defend that in a courtroom claiming "but mathematics!". Not disagreeing, just saying.
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add73d No.3642
>>3641
I have no problem with starting some shit. Cryptocurrency needs to be shown as the house of cards that it truly is. Do you think bitcoiners will become distrustful of the genuine security of crypto when we get a working algorithm of this tree, or do you think they will go even further into their delusion?
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dda169 No.3643
>>3640
>>3641
Just to clarify, cracking RSA is completely fine, and can be proven without using keys you do not own as proof. Posting "someone elses" private keys, even if they are just numbers is a grey area imo. What you do is none of my business, but am pointing it out anyway.
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dda169 No.3644
>>3642
Obviously the proof will spread like wildfire and everything based on broken crypto will have a bad time. Nothing can stop that.
The tree we're working on now is not gonna put a dent in any cryptocurrencies until we get the crumbs on how elliptic curves are the same as prime numbers (or something). When that happens and ECC is broken as well, _everyone_ will shit their pants, including bitcoiners, as there are no implemented alternatives ready to go, anywhere.
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7fa95d No.3645
>>3644
I can't wait for the crazy shit Chris said we would do when we're ready to turn this into a company or whatever he called it, but, I've got to be honest, causing mass world hysteria sounds quite entertaining too, even if we're also affected ourselves. I bet CNN etc will try to act like they have any idea what they're talking about and act like they're super smart while saying absolutely nothing of substance (like the "hacker known as 4chan" thing), and all those YouTubers will try to make sense of it while the ones who use their accounts to teach people math will fumble around with it for a while.
Actually, now that I think about it, people might actually kill themselves over this when crypto goes under. A few did when it tanked recently. Fuck. I'm definitely still on board because this is inevitable and if anyone should do it it should be people like us who have good intentions, but I point this out because it should hopefully discourage anyone from taking credit for it publicly.
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7fa95d No.3646
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c14c81 No.3647
>>3602
So I have a question for everyone: I'm looking at these factor trees (nice job, Anon!) and seeing all the smaller factors (the leaves?) medium size factors (branches) and larger factors (main trunks). Are these larger factors all prime?
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dda169 No.3648
>>3647
Thanks! RSA100 with primes colored red
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add73d No.3649
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fc4336 No.3650
>>3649
I see where you're going Baker. Sum of (1,c) factors can be used to create all the remaining factors? However, checking your math wouldn't it be 12/2 = 2 and 6. Then 6 = 3 and 2? So the termination points would be 1, 2, 2, 3? Could be wrong here, just trying to make sure I understand what you're hinting at. Can you elaborate on your idea?
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add73d No.3651
>>3650
2^2 + 2 = 6
The tree was recreated using 12/2 instead of 12 because 12 as the d branch does not reach a factor.
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add73d No.3652
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add73d No.3653
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add73d No.3654
Getting it yet? We just have to build a tree smart enough to automatically do this.
If rules for when zeroes are to be trimmed can be ascertained, 7463 in particular would take about 3 steps.
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add73d No.3655
>>3648
This is an amazing graphic.
If we master that tree we can crack all the RSA numbers!
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fba0a4 No.3656
>>3655
Yeah, it's beautiful. Looks like the cross section of a giant sequoia, where you can see the growth rings.
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dda169 No.3657
>>3655
Checked! I will cook up some more tomorrow if the whole thing isn't broken already. Seem to have made more art during this adventure than actual breakthroughs, hope they're inspirational
>>3653
>>3654
You're just killing it with these, did you make them by hand? I might steal your color scheme. Awesome stuff
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add73d No.3658
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fba0a4 No.3659
>>3654
So are you saying we can find one of the prime factors by adding up the "leaves" or "outer ring" of smaller factors and then use that to find the other prime factor? That would be cool!
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add73d No.3660
>>3659
If you find one factor, you win.
c/a = b
But yeah, generally it seems adding up the leaves on the bottom gives a.
Algorithm would basically generate the entire tree, and if it doesn't find a factor, it needs to trim the zeroes in different places until the leaves at the end add up to the factor.
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add73d No.3661
>>3660
Given how fast square roots are, I'd say there's hardly an optimization issue in generating the tree.. so once the algorithm is smart enough, it's over.
>>3648
Say, just how long did it take you to generate the rsa100 tree? And what language?
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7fa95d No.3662
>>3660
What so we've figured it out? I was away for the last couple hours.
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add73d No.3663
>>3662
The general structure of how the solution would be is known. It needs to be automated and perfected.
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dda169 No.3664
>>3656
Just for you
>>3661
Its stupidly fast, even with unoptimized python code on a single core. Generating the images takes way longer
RSA100 tree without removing trailing zeroes (binary) is 0.3ms
RSA100 tree with all trailing zero bits removed on every branch is ~1ms (can probably cut this in half compared to my current code)
RSA2048 tree with zero bits removed is about 2.4ms…
The full RSA2048 circular image which is 18673x18723 and too large to post takes about half a minute.
Will see what happens if we need to remove one bit at a time and try all combinations of every branch though
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7fa95d No.3665
>>3664
That looks like it could be turned into a really cool album cover.
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add73d No.3666
>>3664
So, with optimization & implemented in Java it would be stupendously fast to generate the tree, and could be done hundreds of times in a second. Looking pretty good so far.
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fc4336 No.3667
>>3664
Thanks Anon!
>>3666
Interesting Malicious Digits, Checked!
So we sum all the "leaves" (termination points) in the whole circle/tree to find one prime factor? That should be easy to verify for you programAnons. Is that your idea Baker?
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add73d No.3668
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dda169 No.3669
>>3667
>Java
>Malicious Digits
Right on!
>That should be easy to verify for you programAnons
>should be
Did I mention I hate recursive programming?
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add73d No.3670
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dda169 No.3671
>>3665
Gonna have to make an album now…
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dda169 No.3672
>>3670
Yeah but thats even more annoying when dealing with recursive problems. Easier to write it with recursion and refactor it with tail-calls and accumulators and shit afterwards, which says a lot.
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7fa95d No.3673
>>3671
I'm a musician but as cool as I think it looks I wouldn't want to out myself publicly as a part of this.
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dda169 No.3674
>>3673
Good point. Have to wait till its public knowledge or obscure it really well. Tree-based procedural generation of landscapes from random RSA keys or something.
Wouldn't call myself a musician, I just mess around with synths tbh
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68a051 No.3675
>>3660
Is this leaf addition thing true? If so, its easy to code. I'll test it out gladly if its known to be true.
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7fa95d No.3676
>>3666
Man, you keep getting trips all the time. Were you working on a Java version or did you want someone else to?
>>3674
Yeah, I don't know if you've thought about it much, but consider >>3645 this. Once it's all over the media and everything people might be able to get away with it. I've been trying to think of ways to use the grid to create music, too.
>Wouldn't call myself a musician, I just mess around with synths tbh
There are people who consider themselves musicians who don't even know how to mess with synths, so I hope you don't say that out of pessimism. I'm a little more invested in it than that. I have a thing recorded and mixed professionally, and one of the song names is a subtle reference to Q. It's not out yet, but hopefully it's subtle enough that all 5 people who will ever hear it don't think I'm an evil internet hacker nazi who masturbates to anime.
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82ada7 No.3677
>>3675
Teach. Looking into this now also.
Sample for c=481 attached. I added (x+n) just as conceptual for which nodes get included.
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dda169 No.3678
>>3676
There is all kinds of applications for this that we can't even imagine yet. Guess its called the tree of knowledge for a reason… can't wait to se where we're going, now that the tree is taking shape my mind is already going crazy with possibilities
No pessimism here, just a fun hobby I hope to get good at someday, if not its still fun! Good luck with your album/EP!
>evil internet hacker nazi who masturbates to anime.
Did I forget how deep this rabbit hole is again? Yay for reminders
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82ada7 No.3680
>>3677
Also added prime factors for d[t] and a[t] analysis. See pic attached for c=481.
The d[t]-d primes column really clarified for me where the values are we are searching for.
Also, if you look at the (40,1,209) record, the d[t]-d value can be calculated by (xx-f)*(n-1). Perhaps the tree factors are a way to simply remove certain factors from the value?
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add73d No.3681
>>3675
That's what it showed me for the numbers I tested. It's not going to be a piece of cake because I don't know the rules behind when you are supposed to divide by two.
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8e44f0 No.3682
>>3641
Yeah, I'm so screwed on that also.
BUT THERE WERE POOOOOONIIIIIIIES!!!!!
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dda169 No.3683
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68a051 No.3684
Btw, here's my algorithm for calculating private key from 2 primes, written in js, not using BigInts yet. Its very simple.
var p = 61;
var q = 53;
var e = 17;
var p1q1 = (p-1)*(q-1);
var lcm = (p1q1)/xgcd(p-1, q-1)[2]
var mi = xgcd(e, lcm);
var d = lcm + mi[0];
console.log(d)
function xgcd(a,b) {
if (b == 0) {
return [1, 0, a];
} else {
temp = xgcd(b, a % b);
x = temp[0];
y = temp[1];
d = temp[2];
return [y, x-y*Math.floor(a/b), d];
}
}
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68a051 No.3685
>>3681
That's where I'm stuck too. Playing with the tree now.
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dda169 No.3686
>>3685
>>3681
My only thought so far is adding every possibility (no division, divide once, twice etc.) as a separate branch and summing the leaves for every case. It should still be pretty quick considering how small the trees are, but a pain to code.
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add73d No.3687
>>3686
How would that be implemented, and can you send your code?
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dda169 No.3688
>>3687
Yeah, about that, I don't have any yet. Recursion with multidimensional trees is throwing my brain for a spin. Will get back to you
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add73d No.3689
>>3688
How did you generate >>3648 then?
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8e44f0 No.3691
>>3683
You see, your honor, if I may call you that, I was LARPing so hard as someone with a clue that I ended up getting one. I didn't mean to have such a raging clue and showing it to the other boys, who then got super excited about my clue but then got their own clues so we followed theirs. Also, why are there two of you and who took my rum?! WHY IS IT GONE?!
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dda169 No.3692
>>3689
The credit goes to >>3590. The only addition is removing all trailing zero bits (divide number by two for as many times as needed until it is odd). This is one tree. Using the numbers as is and not dividing at all is another (original code). I posted both. There are lots of possibilities in between.
If >>3654 is right, we don't know which numbers have to be divided, and how many times, so this a bounded logarithmic number of possible trees. Since the trees are small (computationally speaking), generating all of them and checking if the sum of all leaf nodes are a factor of c should still be pretty quick (seconds.. ish). One of these trees is the solution. This is the part that is hard to code.
If we do that, we may find ways to decide how many times and when to divide each number, and not have to try every tree.
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add73d No.3693
Ever notice that any number whose d value has the same parity as its e value is even?
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82ada7 No.3694
>>3692
Try just removing trailing zeros from the d values.
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dda169 No.3695
>>3694
So… my code has a bug, I have been removing zeros from only the e values so far.
Here are trees with trailing zeros removed from both e and d, just e, and just d (third image is the one you asked for). I haven't added up the leafs and checked these yet
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add73d No.3696
>>3695
How are you making those images?
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dda169 No.3697
>>3696
Combination of >>3590 and >>3611. With some divide by two code added (actually just stripping off zero bits in binary by shifting right) for e and d. And some coloring code mixed in (primes have a red dot)
Best way I found to visualize trees so far
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add73d No.3698
Which branch you are on is important.
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add73d No.3700
Finally factorized something with this bloody tree.. I'm happy now
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add73d No.3701
I created a program that generates the d branch separately because I recognized a pattern where, whichever variable is even (d or e) is the branch where everything is trimmed.
This truly is log n if I ever saw it. Instantly generates the tree even for gigantic RSA numbers.
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666a85 No.3706
Am I the only one skeptical that this 'solution' doesn't use the original VQC at all, that it essentially boils down to 'factorize the square root and it's remainder recursively'? And it all just neatly adds up to a massive prime number? And goes from "This is my last message on 8ch" to "disregard the last 2 months, this is the answer"?
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8757e1 No.3707
>>3706
MA - I concur. Still looking into this latest crumb.
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7fa95d No.3708
>>3706
The grid is meant to have many uses, so it might be used to connect the logic of factoring numbers with elliptic curve cryptography. Also, one of the early crumbs said something like "you factor a number with two input variables d and e". The whole time this was meant to revolve around d and e.
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add73d No.3709
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7fa95d No.3710
>>3709
So I guess that means there's an easier way than the way we have right now? Maybe it reduces the number of leaves to involve n or x in the math somehow.
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add73d No.3711
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d1075e No.3712
>>3709
btw, VA pointed out that you can construct f and x from d and e assuming n=1.
>>3711
So that means twos are removed for both d and e?
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add73d No.3713
>>3712
Depends which one is initially even or odd.
I need that tree code.
Mine isn't very good.
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add73d No.3714
>>3713
One universal pattern is this, d and e will never have the same parity for a semiprime, and the branch that starts with an even number is the one that needs to be trimmed.
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d1075e No.3715
>>3713
relevant code posted here
pastebin.com/CZy8f0HD
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d1075e No.3716
>>3714
>>3711
from your example, doesn't look like you completely trimmed the 40. What's the logic there?
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add73d No.3717
>>3716
I don't know. When I was doing it by hand I just kept remaking it until it gave a factor. That's why it isn't solved yet.
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fba0a4 No.3718
>>3709
Baker, thanks for your good work, and thanks for keeping connected with Chris! Can you give a quick explanation on why you stopped at 10 for the right branch, leaving d=3 and e=1? Am I reading that correctly? Just want to make sure I'm following correctly.
N, or a or b for (e,n) has gotta be one of those large primes in the factor tree. We should take an RSAc from a solved example, run the factor tree program, and see if n or a or b pops up in the (1,prime). Thoughts?
>>3712
Thanks for remembering PMA! F can be constructed from c,e,d. In row one it helps derive x. In n>1 still not working. But if x in (e,n) is the same as in (e,1) maybe it can help us locate the prime x.
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add73d No.3719
>>3718
I made tree generation stop when every number on a level of a branches is less than or equal to 3, because the sum of their square roots and remainders is always the same if you branch out any more.
E.G
3 = 1^2 + 2
2 = 1^2 + 1
1 = 1^2
The sum is the same as the original number, so it can't branch any more after that.
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d1075e No.3720
>>3717
Perhaps his response means that we should remove 2's until we hit a triangular number?
Is there a way to detect a triangular number?
>>3718
VA - important discovery!
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add73d No.3721
>>3718
For the 481 example I just did what worked. If all the zeroes were trimmed it would start with 5 and not 10 (40/2²)
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fba0a4 No.3722
Here's a big picture idea:
Right now we're working on part 2, the factor tree.
Next, part 3: VQC meets factor tree.
At (1,c) in n is going to be the largest possible, because a and b are farthest apart.
At (prime solution) n will always be less because a and b are closer to d.
Factor tree allows us to see all prime factors, correct? (Could be wrong here)
(1,c) has highest n, a and b max distance.
(Prime) always has lower n, closer to f for a and b.
Factor tree allows us to quickly identify primes and test them for compatibility with c.
Thoughts?
And we can use (e,1) to help us find x?
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fba0a4 No.3723
>>3722
Should read
>(Prime) always has lower n, closer to d for a and b.
>>3719
Good explanation, thanks!
>>3720
I like triangles. /_\ especially when they are broken down like this:
+++
++
+
0
Bye bye scumbags! Our math triangles helped break your evil triangle HAHAHA
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68a051 No.3724
>>3711
I'm telling you guys… pythagorean triples…
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add73d No.3725
{e:n:d:x:a:b} + c makes the VQC a 7-dimensional grid. We've been analyzing an object with many dimensions that has been reduced to a 3 dimensional object.
Now if only I was a 7-dimensional being.
hooktube.com/watch?v=Q_B5GpsbSQw
hooktube.com/watch?v=eyuRLmCphHc
hooktube.com/watch?v=S-yRYmdsnGs
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add73d No.3726
>>3725
And you may not notice, but all 7 coordinates can be encoded into 3 coordinates.
(e, n, t) -> {e:n:d:x:a:b:c}
So you have 7 dimensions in 3.
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1d5a78 No.3727
>>3725
>>3726
I was gonna say, "Baker you forgot x's little brother t. Don't mess with our little brother, homie."
>{e:n:d:x:a:b} + c makes the VQC a 7-dimensional grid.
Plus little brother t = 8 dimensions?
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1d5a78 No.3728
>>3724
Hey Teach! Can you give an example of the pythagorean triples as they relate to the number tree or the VQC? I'm interested over here to see your ideas.
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add73d No.3729
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06387b No.3730
>>3719
>>3720
>>3724
Quick question: is n supposed to be prime too?
>>3729
Checked it out Baker! WTF does it mean, LOL? Found the cool X tho. Check out our tattooed posse…. Teach is in the middle, FYI.
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82ada7 No.3732
>>3730
don't think so. prime solution for 481, n = 4. (40,4,5) = {40:4:21:8:13:37} = 481
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8e44f0 No.3733
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c00f3b No.3734
>>3733
Checked!! 3,7,3,3 All very good numbers. How does 7 fit into Tesla numbers?
Post some triangle images here too Topol. Visited EZBake, me need some math memes here too.
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c00f3b No.3736
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7fa95d No.3737
>>3736
Unfortunate that there was nothing about the memo but I suppose it's coming out regardless.
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add73d No.3738
>>3737
The fight really isn't over. They're going to try to make it a redaction fest.
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c14c81 No.3739
>>3738
>>3737
Watch the full speech. Mastery of PR.
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add73d No.3740
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c14c81 No.3742
>>3740
Back to work lads!
>Pythagorean Triplets
>Number Tree w VQC
>Let's Melt the Establishment
>Topol's triangles
>>3741
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1d5a78 No.3743
I'm here and able to work for the next 120 minutes. Let's work faggots. I'll check in with VQC too.
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06387b No.3744
Many Questions:
Can the factor tree help us find (1,c) and (prime)? Or match them to number tree?
Can we factor (1,c) and find primes to match for a and b using the tree?
Can we use factor tree and (e,1) to get what we need?
We need to know the factors AND how to use the grid.
STEP 3. Tree + VQC
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add73d No.3745
>>3744
As he stated, it gets even more complicated to factor larger numbers. When you try to sum the last levels for large numbers as I did in the examples, the number is too small to be a factor.
7463 = 228^2 - 211^2 (17*439)
7463 = 3732^2 - 3731^2 (1*7463)
7463 appears in (67, 3436, 43).
7463 appears in (67, 142, 35).
It's interesting that the t value it first appears in is 43, and the first branch of the right factorization tree starts with 43 (86 / 2). The t difference is 8, which also appears in the tree.
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add73d No.3748
I'm going to contribute to putting together what we know about the grid tomorrow.
There are almost a hundred different concepts this touches upon.
Hyperspace, difference of squares, square roots, golden ratio, platonic forms, etc.
Something interesting. Ever wonder why diff of squares is always an odd number?
Every square is a sum of odd numbers.
4 = 1 + 3
9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7
25 = 1 + 3 + 5 + 7 + 9
36 = 1 + 3 + 5 + 7 + 9 + 11
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
So, when you are taking the difference of squares, you are really just taking elements out of this summation.
9 - 4 = 5
16 - 9 = 7
25 - 16 = 9
36 - 25 = 11
49 - 36 = 13
Perhaps it is here where a solution lies. To solve RSA one needs to master the number of line of squares, and thus, the set of odd numbers. It doesn't sit right with me that after all these years, this hasn't been done yet.
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add73d No.3749
The RSA problem is basically to find the two different combinations of squares whose differences make the number. The amount of combinations of squares that can be made corresponds to how many factorizations a number has, for odd numbers.
For 145, it's 73^2 - 72^2. These are easy to calculate for any number, becayse 1 is always a factor. And because these two squares are adjacent, the combination of them simply makes 1 * c. So, when you look at a different factorization, the squares are never going to be adjacent.
145 = 17^2 - 12^2.
17^2 = 1 + 3 + 5 + 7 + 9 … + 33
12^2 = 1 + 3 + 5 + 7 + 9 … + 23
So, again, when you take the difference of squares you are just removing elements. Thus, factorization can be restated as finding what two odd sums' differing elements.
To factorize, all you have to do is find the sum pictured. To calculate the smal square, simply calculate the sum all the way to the first element, and calculate the sum all the way to the last element to make the big square.
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add73d No.3750
>>3749
Finding what two sums' differing elements make c*
If we can put all of this together, the problem can be solved.
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d6f59c No.3751
Found pattern in (e,1) and (f,1) records. Using /_\ for delta for example:
/_\ d(e) = d( e, n, t) - d(e, 1, 1)
Then
/_\d(e) = /_\ a(e) + /_\x(e)
/_\d(f) = /_\ a(f) + /_\x(f)
After testing large number of combinations both A and X exhibit occasional negative /_\. I was working on way to try and solve for x or a using differences between e and f but grid continues to hide its secrets.
New hint from QVC.
Recursion gives the pattern of triangle numbers used to construct the square that added to c makes the square with sides d+n.
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add73d No.3752
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d6f59c No.3753
>>3752
Awesome!
Math memes!
Great explanation of QVC hint!
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12fd55 No.3754
>>3751
Secrets secrets, #NoMoreSecrets!
Thanks anon, that crumb is yuge!
>>3752
Nice!
So many crumbs to reconcile still…
>>1099
>The pattern GROWS in a certain way as n increases in column zero.
>This pattern is related to the pattern across row one (e,1)
>>1447
>In the meantime, consume anything you can on what drew mathematicians to believe elliptic curves and modular forms were the same object.
That was an interesting dive, still digging.
>>12
>Let's have some fun.
And, for a little VA enthusiasm injection: anons, time to get the triforce gloves on!
And speaking of triforce, note the dollar. We will be the change! (just my ¢¢)
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add73d No.3755
>>3754
The (e, 1) pattern you referenced is what I used to make a factorisation prototype algorithm. I posted the source earlier, only worked for small numbers.
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d6f59c No.3756
>>3754
Thanks for crumbs and triangle symbols.
>>1099 seems very appropriate.
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d1075e No.3757
>>3751
excellent observation!
>>3752
nice!
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316b0b No.3758
The triangle numbers are interesting.
To generate the A's for an odd e, you compute
(e + 1)/2 + 2 * (t * (t - 1)).
And the function for even e D's its:
e/2 + 2 * (t * (t - 1))
Which is (e + 1) / 2 + 4 * T_(n - 1) (A for odd e)
and e / 2 + 4 * T_(n - 1) (D for odd e)
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06ef9d No.3759
Okay I may have found a type of solution for 145.
145 = (12)^2 + 1
= (3^2 + 3)^2 + 1
= 3^4 + 2*3^3 + 3^2 + 1
= 81 + 54 + 9 + 1
= (81 + 9) + (54 + 1)
= (90 + 55)
GCD = 5.
or = (81 + 54) + (9+1)
= (135 + 10)
GCD = 5.
If we keep the terms separate maybe we can get it to work. Moreover I've looked at a bunch of trees and I think we could terminate the trees at either 1, 2 or 3.
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ed48a1 No.3760
>>3755
Right! I'm mostly caught up with you all at this point, still chasing down some _bits_ here and there.
Oh, this crumb too has been on mind…
>>699
>One Row to Rule them All.
>A column that contains ALL and is the KEY to ALL the patterns.
>What are the patterns in COLUMN ZERO?
Did we complete this one? I know we did (0,0). Not sure I have that question fully answered yet.
And for the Column 0, it appears from the first few D and A grow in linear fashion without interleaving, will look more this evening. Using excel images such as attached to start, and PMA's output.
>>3756
You bet! So many crumbs. Was looking at the one about imagining the grid coming up at you, and this being another dimension (the complex I think it was). So much just about sq rt of -1.
Last night, was going back to some work by Hagen Kleinert, a German professor. Was going to start a separate thread, but will just point here. A 2001 paper:
https:/ /arxiv.org/pdf/quant-ph/0106095.pdf
Towards a Simulation of Quantum Computers by Classical Systems
..and if you want to listen to a video by Prof. KIeinert, this is tough quality but super. His energy is infectious:
San Marino 2011
Lecture 21: The Purely Geometric Part of Dark Matter. A Fresh Playground
for String Theory
Prof. Hagen Kleinert, Germany
http://www.world-lecture-series.org/san-marino-2011-purely-geometric-part-dark-matter
or download the mp4:
www.world-lecture-series.org/san-marino-2011
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d1075e No.3761
I know we're currently working on trees, but if anyone has an interest in reviewing d[t] and a[t] output:
c145 pastebin.com/mJcE0CDy
c481 pastebin.com/yx1ym1C0
c533 pastebin.com/U4KDsUnv
c785 pastebin.com/Kthbi32S
both (1,c) and (a,b) records run against (e,1), (0,1), and (f,1).
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add73d No.3762
>>3761
This is hard to read. There is too much data in one place.
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d1075e No.3763
>>3762
Yes, and that was the condensed version! Someone may find it useful. Or not.
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68a051 No.3764
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fba0a4 No.3765
>>3754
Triforce gloves at the Ready, Lads! Thanks for the enthusiasm injection, MM! Gannon is going down with those wooly gauntlets. Link related, old school Legend Of Zelda final boss.
https:/ /youtu.be/O33Xybjegtc
>>3762
>>3763
>>3764
Lol, Wut? There is a LOT of info there, thanks PMA! I'll look for patterns in it a bit later tonight.
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68a051 No.3766
Ok guys, so it doesn't work for all numbers, but it does for a lot how about this…
recurse through the d and e tree, at some point, a d value passed in will be equal to x in our original problem.
so each iteration, calculate gcd(d*d + e, orig_c) for each branch of the tree.
Like I said, it doesn't work for all values of c, but it relies on the xx+e = 2na & dd+e = ab common factor of a.
I'm trying to figure out the cases when it doesn't work, and trying to solve those cases individually.
I'm also going to write up a big int version and test on some larger numbers.
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add73d No.3767
>>3766
I'll try that out. How do you know when it's the correct x value?
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add73d No.3768
>>3766
Also, did you trim the d values or just leave them as is? (Trimming = divide by 2 until odd)
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68a051 No.3769
>>3766
Correction, gcd(d*d + orig_e, orig_c), sorry math is hard
>>3767
You'll know because gcd will be greater than 1.
>>3768
At the moment, i'm missing several cases, since i'm not testing the 2's we're extracting, or the root of the perfect squares - going to do that next.
But I'm testing each and every d/2 value on the way down, yes.
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add73d No.3770
>>3769
What numbers did it work on?
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68a051 No.3771
>>3769
Actually, this isn't looking so good on bigger numbers.
Sorry, back to the drawing board.
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7fa95d No.3772
>>3771
It's not the end of the world if we don't figure this out right away. Chris did say he would pop in before the end of the week, remember.
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add73d No.3773
>>3697
Whoever this poster was, could you generate that tree image (the long one that shows the whole thing) for 2^61 - 1 and 2^521 - 1? They are Mersienne primes.
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fba0a4 No.3774
>3131
>3580
Still looking at c=145. Don't know why, but have the gut feeling the answer is starting us in the face with this example. Combine the factor tree knowledge (leaf sum?) with n=na and we can somehow get x or a.
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fba0a4 No.3775
>>3774
Sorry lads, green links, typing in by hand. Here are the correct ones.
>>3131
>>3580
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add73d No.3776
>>3774
I just did some research into General Number Field Sieve and I realized something..
When we finish the log n algorithm,
There are gonna be a LOT of pissed off number theorists, mathematicians and cryptographers.
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add73d No.3777
>>3776
There are people that have spent probably a large portion of their entire life on sieving algorithms.
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fc4336 No.3778
>>3776
>>3777
Checked 666 and 777? WTF Baker is on a roll!!! Here's a prophecy for the occasion:
"I see great and malicious confusion upon you, Baker. But the 777's of YHWH will deliver you and lead you to peaceful living space"
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add73d No.3779
Here's a method for generating triangular numbers: https://pastebin.com/jPTTpgjE
For some reason Wikipedia made it look complicated by using Sigma Notation but T(n) is just (n(n+1)) / 2
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add73d No.3780
>>3778
I'm not malicious, but I'm probably participating in this because I think bitcoiners are faggots.
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7fa95d No.3781
>>3776
>>3777
This is part of why I keep saying that anyone who thinks they should take credit and come out publicly about this is really stupid.
>>3780
This too. People have put their entire life savings into crypto.
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33fd02 No.3782
>>3779
Hey fags, popping back in. I'm the guy from >>934 back in RSA #6. I think I was the first to bring up triangular numbers and I demonstrated a simple transform on them.
I don't think anyone realky paid much attention to the posts, which I spent a fair amount of time making as straightforward as possible, so if any of you don't have much to do right now, maybe take 15 min and work through my example.
Good to see that faggot VQC finally show back up again.
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fba0a4 No.3783
>>3782
Will do! Thanks Anon!
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f2af50 No.3784
>>3780
>the 777's of YHWH will deliver you and lead you to peaceful (((living space)))
You missed my funny reference to Lebensraum. I made that joke just for you, Baker. Geez, Anon. Not even a Hitler fan over here.
BTW, good work faggot.
Weaponized Autism at work.
8Chan Autists On The Job.
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68a051 No.3785
>>3782
Hey anon, will you do a walk-through with a non-perfect square this time?
I re-read your post, and I'm in agreement with everything you said, just want to see a non-perfect square example, and if it works, i'll code it up.
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add73d No.3786
>>3782
Thanks for doing the research. This might be related to the pattern Chris described. Did you look into it any more beyond that?
>>3784
But the Lebensraum was faked.
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65f089 No.3787
>>3752
Hey yea that is pretty cool. You have transformed a*b into a sum of consecutive odd numbers where a is the number of terms and b is the midpoint:
5*29 = 25 + 27 + 29 + 31 + 33 = 145.
The dots you drew can also be drawn similary to triangular numbers, but you are only adding the odd rows:
_____
__e__
_eee_
eeeee
_eeeeeee_
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32324c No.3788
The trippy thing is that the difference of squares = triangles. >>3751
What a cool connection!! VQC is correct that our language and understanding of numbers is lacking currently. Squares relate to triangles relate to circles relate to trees.
>>3580
Fractals:
>(since the grid is effectively an overlay of the fractal nature of integers).
Part One:
>First decision, if e is 0, return d.
Second decision, if(GCD(e,d)!=1) return GCD(e,d);
So the GCD between d and e is the end goal of Part 1? What do we use this for? Ideas?
Part 2:
>The inputs into part two are to factorise d and e.
The end result of part two is a tree, descending from c with branches d and e, then branches descending from d and e or each of their square roots and remainders.
So the terminating factors are a big key. How so? We are working to understand. This is working from the opposite way we were doing with the VQC. Instead of building factors for c, we're breaking c into factors. Gotta be able to go both ways? Build or Break down?
Part 3: >>3709
We will combine factor tree knowledge with VQC knowledge. Then solve this.
Summary up to this point, let me know what should be added.
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add73d No.3789
New hint from VQC:
Column 0 is just as good as row 1.
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3a9068 No.3790
>>3752
Is the correct squares = triangles reference.
>>3789
Wut?! If we can finally figure out (0,n) with (e,1) combined we can do it.
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33fd02 No.3791
>>3785
I believe VQC called direct attention to the e == 0 row for 1 x c^2. So no real point with a non-square number because c^2 is (obviously) always square.
>>3786
Each n calculated using the triangular number approach is also a perfect square * 2.
It's a fucking miracle when you see it working. For those following along at home, you can calculate the height (I call it index) of the triangular number of the n value in e == 0 for c^2 by typing the following into your desktop calculator:
Using our trusty 5 x 29 == 145, c^2 == 21025, first we calculate n. Calculator prompt is >:
> 21025 + 1 =
> / 2 =
> - 145 =
n = 10368
> 2 - 10368 =
> * 8 =
> * -1 =
> +16 =
> sqrt
> - 4 =
> / 4 =
height = 71
(on my PC now instead of mobile)
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33fd02 No.3792
>>3791
> height = 71
The reason I call this an index is because it's zero-based like programming languages, so the first element is element 0.
72 might be an interesting number for c = 145 as well because of this (i.e., normal counting based where 1 is the first element).
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32324c No.3793
>>3791
>>3792
TriangleAnon in da house. Let's go! a^2+b^2=c^2 "Pythagoras, we summon your mighty power to help us combine triangles with squares and unfuck the universe."
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65f089 No.3794
>>3788
>Second decision, if(GCD(e,d)!=1) return GCD(e,d);
>So the GCD between d and e is the end goal of Part 1? What do we use this for? Ideas?
If GCD(e,d) is not 1, that means e and d have a common factor, but then this common factor divides c because c = d*d + e. So this line of pseudocode tests for this possibility, where you will happen to get factor of c right away.
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a9b066 No.3795
Think of it like this
d= a*b
e = a*c for some a, b and c.
Then d*d + e is equal to a*b*a*b + a*c.
This can be changed to a*(a*b*b + c). Thus c is divisible by a.
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a9b066 No.3796
>>3795
Poor variable name choosing here though. The c in e = a*c isn't the same as our c (d*d + e). I mean the equation holds it might just be a bit confusing
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add73d No.3797
>>3795
So how does that help?
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a9b066 No.3798
>>3797
I was just referring to >>3794. Explaining why the GCD(e, d) works.
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ed48a1 No.3799
>>3782
Your insights in RSA#6 were superb, thanks for popping back in! And what a co-In(c)iden(c)e, was working through column 0 more this eve (and Eve will see all Bob and Alice have been chattering about soon!).
>>3789 Noted!
VA listed a few (0,n) patterns here >>1542 and here >>1544.
And in #6 from TLA there's >>1548
Had missed the 'closed under multiplication' and 'Gaussian integers' back then. Making much sense now, understanding this as a HIGHER-dimensional problem (old VQC hint).
scalar > vector > bi-vector > tri-vector…
Not sure if we'll need it for the first part, but it's coming, or going back anyway, to Grassman et. al.
Hestenes celebrated this:
http:/ /geocalc.clas.asu.edu/pdf/GrassmannsVision.pdf
as related to Grassman. I mentioned Clifford Algebra in last bread, well it just keeps unfolding. Also dig into Alan Macdonald's Geometric Algebra / Geometric Calculus and the Conformal model. And on it goes, a unifying model for mathematics. Recall other VQC drop in an anon's DM >>3139 "All your base belong to us". Boom! There it is. And, given this includes Blades, and Philip Dick was on my mind earlier, a Blade Runner graphic!
- What's SO cool about this, is the variety of applications! Read: https:/ /arxiv.org/pdf/1305.5663.pdf New forms of neural nets and neural computing (AI/ML anyone?), computer graphics, motion tracking, FLAT space (flat earth, haha, get it now VQC!), crystal structures (for you Topol). Oh, and bumping into bitcoin along the journey. Yuuge! I feel like the guy in the attached image, a slice of light across the eyes looking at a blinding future.
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add73d No.3800
>>3793
Just found this. ;>
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23f2bb No.3801
>>3792
Thats cool I get what you're saying there. I just checked and saw that sqrt(10368/2) = 72 also.
What do you mean by "triangular" number though. Like is it the c for a^2+b^2=c^2? I googled it and it doesn't match the definition google gave me.
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316b0b No.3802
>>3801
He means https ://en.wikipedia.org/wiki/Triangular_number
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23f2bb No.3803
>>3801
Also noticed that if you (for c=145) take 72 and subtract your original n, you get your original x.
goal (1, 5, 12, 7, 5, 29)
start (1, 61, 12, 11, 1, 145)
sqare (0, 10368, 145, 144, 1, 21025)
n = 10368
sqrt(n/2) = 72.0
sqrt(n/2) - start[n] = 11.0
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23f2bb No.3804
>>3802
>The sequence of triangular numbers (sequence A000217 in the OEIS), starting at the 0th triangular number, is
>0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666…
I don't see 72 or 71 here so thats why I was confused.
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316b0b No.3805
>>3804
>>> rowForAB(1, 17 * 53)
(1, 421, 30, 29, 1, 901)
>>> rowForAB(1, 17 * 53 * 17 * 53)
(0, 405000, 901, 900, 1, 811801)
>>> s(405000/2)
450.0
>>> getTFromA(1, 421)
15.0
>>> generateNthRowFromGenesis(1, 15)
(1, 1, 450, 29, 421, 481)
So it appears the n in c**2 is equal to the corresponding d[t] where t is the t that a=1,b=c's n appear in (e, 1).
Not sure if this will hold though. But if it is, then it would explain why sqrt(n/2) - n = x. Since d - a = x.
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23f2bb No.3806
Here we have
(e,n,d,d-1,1,c)
Then if we do our c^2 record we get
(0,(c^2+1)/2 - c, c, c-1, 1, c^2) N= (c^2+1)/2 - c
Then we can see that sqrt(N/2) - n = x
sqrt( ((c^2-2c+1)/2)/2 ) - n = x
sqrt( ((c-1)(c-1)/2)/2 ) - n = x
(c-1)/2 - n = x
c - 1 - 2n = 2x
c - 1 = 2x + 2n
c - 1 = 2(x+n)
c = 1 + 2x + 2n
c = a + 2x + 2n
nothing new
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80a284 No.3807
>>3773
Here are some Mersenne trees for you. These are without trimming bits / dividing by two.
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80a284 No.3808
>>3773
>>3807
Same, with all bits trimmed. These are 61, 521, 607, 1279 and 2203 in case you missed the filenames. Not sure this is useful though but maybe one of you spots something
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316b0b No.3809
>>3808
Well, if I understood the VQC decision tree, we should trim d and e.
So we are making a function f, that will return a factor of the input parameter. So far we know the following:
Step 1:
Trim input c until it is an odd number.
Calculate e, d from trimmed c
Compute GCD(e, d)
If GCD(e, d) != 1 then return GCD(e, d)
Step 2:
We now have to factorize d and e, which essentially means we have to call our function f again, two times, one for e and one for d.
Step 3 is still unknown, but somehow involves the VQC and triangle numbers.
So assuming I wrapped my head around it, trimming d and e are necessary (and thus also what makes it a recursive function).
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316b0b No.3810
>>3809
Step 3 also involved a tree from our d, e factorizations.
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316b0b No.3811
Actually, thinking more about it. I don't think we are supposed to use f(d) and f(e). Given the idea that f is supposed to return a factor, I think we need two functions.
f which takes our c and g which takes our e and d. We use f to factorize a number and g to generate a tree.
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80a284 No.3812
>>3809
True, and the resulting tree makes more sense when trimming. Still, >>3653 gets the right answer without completely trimming the 40. So did we just get lucky with small numbers or are there rules for how many bits to trim? Maybe this is too simplistic anyway, given the hints about larger numbers being more complex, and triangular numbers.
I tried just coloring triangular numbers instead of primes but there are to few of them to really do anything on their own… Can't figure out how to relate the recursive tree to our grid so I'm a bit stumped at the moment
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316b0b No.3813
>>3812
>>3811
If f is supposed to just return a single factor, then how will we be able to build a tree at all?
Assuming the flow of the code is something like this:
function f(c)
d = floor(sqrt(c))
e = c - d * d
if gcd(d, e) != 1:
return gcd(d, e)
factor_d = f(d)
factor_e = f(e)
…
How will f(d) and f(e) be used to build a tree, assuming this function will return a single factor?
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d1075e No.3814
>>3789
Just a reminder regarding (0,n), the a value from the prime solution can be easily calculated from the cc and aabb records:
a = sqrt( cc.d - aabb.x )
We just never found a way to navigate to the aabb record without iterating.
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80a284 No.3815
>>3813
Yes, we need to split into branches (return two or more results) for each recursion, otherwise there is no tree. The rules for how to manipulate numbers (trim all bits, only trim bits if number is not triangular, ect.) for each step are up to us. But VQC was pretty explicit on trimming all zeros, not trimming _some_ zeros. So assuming the tree we can generate now is right, step 3 has to be walking it and summing numbers in some way?
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316b0b No.3816
>>3815
I agree. I find it a bit hard to reason about, though.
As for step 3 I have no real idea on how that might be. There's a lot of crumbs that wouldn't fit into it. Like why was f (2 * d + 1) important if it's not used somewhere?
He also keeps reminding us about the x*x + e = 2an. Maybe step 3 will somehow give us x, so we compute x*x + e and then find 2an? In which case gcd will help us out. So far I'm still trying to make sure I got the step 2 part correct.
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80a284 No.3817
>>3816
Wish I was better at recursive programming. I got pretty stuck trying to add dimensions to a binary tree in addition to branching…
Just summing up the leafs for a completely trimmed RSA100 tree and dividing by two gets us to
40364365001756683226800431241044541563293164124405
which is pretty close to our factor
40094690950920881030683735292761468389214899724061
but not close enough. Maybe we can use the grid on this.
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316b0b No.3818
>>3817
I'm just thinking out loud here. We know that we will get a tree of d's and e's. Maybe we need to look at the records of the leafs and somehow build up?
Like, we generate a big tree, then we go down to the bottom and start working our way up. We do <something> with the leafs to build up and keep doing that until we "bubble" up to the top and by then we will have a factor(s)?
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80a284 No.3819
>>3818
This is what I meant by "walking" the tree. I've been considering it but the only thing I can come up with is starting with an untrimmed tree and trimming one more bit for every possibility, and regenerating the lower nodes. Then checking the leafs again for a factor.. this is when I figured I could just generate a complete multidimensional tree in the first place. Still doable computation wise, not nearly log n but if we get a solution tnat may give more hints.
Then I realized you have to try all permutations of the multidimensional tree, not just walk it once, so we're back to exponential…
So maybe something like walk up the tree until you find a prime, triangular number, or some grid variable, and choose your next direction based on that (up or next branch down)… thinking out loud as well
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d1075e No.3820
>>3818
>>3819
VQC previously mentioned regarding the grid:
>How many times does 105 with three factors appear in one column?
>How many time for a number with four factors?
>It's not linear.
Perhaps the tree we are building is a way to query these numbers. And not necessarily to provide the prime solution.
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316b0b No.3821
>>3820
Well I tried generating the tree for 145. It yields (among others) 3 and when you compute the same function with 145 * 3, you'll get 5, 87. But if this holds up for more numbers I'm not sure about yet.
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80a284 No.3822
>>3820
Trying something new. RSA100 with 1xcc for each number. Let me know if anyone has ideas for what functions to use
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8e44f0 No.3823
>>3822
What happens if you did 2xcc? or -1, or d√1xcc…
or 1xtcc… I have no idea what this means; 1st pic related.
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80a284 No.3824
>>3823
I may have posted the wrong image, that was 1xc, not 1xcc. Or I misunderstood everything completely, whatever. So here they are with bonus 2xcc. I am not sure how to add cats yet.
And apparently bioinformaticsfags don't like having dark backgrounds in their programs so I had to hack the shit out of it to get this.
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65f089 No.3825
>>3811
>>3813
Right, the factor has to come in part 3. We have seen that we can sometimes add up 'leaves' on this tree and get a factor of c, but no consistent pattern has been put forth and VQC hasn't actually said yeah, add them up that's the secret. So depending how you define the function f(c), if you want it to return factors then either that's part 3 after the branching business; if the recursive part returns a factor then you are getting a factor from every subtree so you have to do something with that, could be worth messing around with, but seems more likely coming in part 3..
Also, for everyone's interest, the part 2 recursive tree has a time complexity of its own which we can compute. If this part is slower than log(bits of c) then VQC has misspoken, which I wonder about. It could still be fast enough to be revolutionary without being that fast.
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add73d No.3826
>>3825
Time complexity is small. Tree generates in milliseconds for rsa2048.
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8e44f0 No.3827
>>3824
¡Muchas Gracias, My Bruddah!
I have no idea if what you just did is in any way useful but apparently whatever I said could be done after all!
HAVE AN EBOLA DASH!
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316b0b No.3828
>>3825
It shouldn't be too hard to estimate the time complexity of step 1 and 2. Best case we do one squart root (to estimate d and e) and return d (e = 0).
This should be O(log n).
If e != 0 we compute GCD(e, d) which also has a time complexity of O(log n).
This can be seen as the time complexity of O(log n + log n) which is the same as O(log n).
For the recursive parts, we still only compute O(log n) so overall I think step 1&2 should have a time complexity of O(log n).
It's been years though since I did this formally at uni, so I'm only using what I remember. I might have some old textbooks I can go over at some time, to double check. Maybe CA can chip in on this?
But without knowing the remaining steps, it's impossible to say anything with regards to the algorithm as a whole.
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7fa95d No.3829
>>3828
So far it is O(log n) in the worst case but that's based on the number itself and not the bits in c. Chris said we had 95% of the information we need, didn't he? If he did then I really have no idea how we get to log of the number bits in c if he has somehow told us that. Hopefully we're up the point at which he can tell us that and we'll understand (it seems like it).
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65f089 No.3830
>>3824
Anon I like your work. Is this adapted from philogenetic software? I was trying to interpret your tree. Sometimes two sibling nodes have a "d^2 + e" relationship to their "uncle" node (sibling of parent node), and sometimes they are the factors of their uncle node.
examples pictured: 307*3=921 or 5*5=25, vs 633 = 25^2 + 8
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add73d No.3831
>>3807
Thanks. Haven't came up with a use for it yet, but we've just started studying these types of trees.
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65f089 No.3832
>>3828
> If e != 0 we compute GCD(e, d) which also has a time complexity of O(log n).
Yeah this is already more than log (bits of c). As many of you know, bits of c ~= log c, example a 4 bit number is between 1000 and 1111 (binary), or between 8 and 15.
log (2^3)<=log c < log (2^4)
3 <= log c < 4.
So generally bits(c) - 1 <= log c < bits(c).
Then a time complexity of log (bits (c)) is log log c, and we are already slower than that. Like I said, log c is still very fast, it's an arrow instead of a bullet, when now the on the books factoring speed is a knife strapped to a turtle.
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add73d No.3833
>>3832
Enough of that. You're ruminating on how a slow a bullet is compared to a turtle.
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65f089 No.3834
>>3833
ok fine, no analogies or ruminations. strict math for you then. this is not log (bits c). read the math don't worry about the turtle
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65f089 No.3835
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7fa95d No.3836
>>3832
>>3833
>>3834
It might be useful if we theorize about what we could do with the current decision tree in order to turn it into an O(log log n) algorithm.
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82ada7 No.3837
>>3830
Pic attached is a tree example of c=921 with and without trimming values.
In the top example, 30*30 + 21 == 921 as expected.
The bottom example, the d node is created trimmed in a recursive call, and that trimmed value is used for the d and e assignment and so on.
Not sure if we have an answer if the nodes should always be in balance with the formula.
In other words, should be trim before or after the sqrt(d) ?
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f2af50 No.3838
Time to have some fun working on Math! Persistence will win the day. I'm sitting over here relaxed, peaceful, and with an open mental state. It's really nice to check in after a good day of work to read your new ideas, Anons. Love you all no homo.
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add73d No.3839
>>3838
God's with us. Through Him I am able to form patterns from chaos. Through Him I am able to help all of you.
If you love God, he'll solve this in the most beautiful way and he'll do it with your hands.
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316b0b No.3840
>>3837
I haven't had time to really dig into this, but I've been trying to play with primes.
We know that this function will return a factor, so for a prime number that means it will return either 1 og the prime itself (I'm assuming). This we can also use the function as a primality tester.
If that's true, then somehow, the tree will be used to show that it has no factors other than 1 or it self.
No idea on how though, but maybe it will be easier by starting to look at a few smaller numbers where the branches can't be big.
Just an idea, maybe I'll take a stab at that during the weekend.
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65f089 No.3841
>>3580
>The decision tree terminates upon finding a factor.
>Pre-factorisation, trim trailing zeros (in binary) or equiv is to divide by two until odd.
>This is 1 and c for primes.
>This is the square root for squares.
>This is the pair of factors that are not 1 and c for the product of two primes that are not the same numbers (not a square).
>This is the pair of factors closest to the square root for products of more than two factors (not including 1 and c as factors).
>>3840
>We know that this function will return a factor, so for a prime number that means it will return either 1 og the prime itself (I'm assuming). This we can also use the function as a primality tester.
Based on VQC quote above, yes it looks like it brings back the two factors closest to sqrt (c). When c is prime you'll get 1 back, and c, so you'll know it's prime.
If your c might have more than two factors, and it spits back out c = a*b, you can further put a and b into the algorithm. Like if you put 360 in, you should get: 18*20 (not 10*36 or anything else). Then 18 = 6*3 (not 2*9); 20=4*5 (not 2*10) etc.
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add73d No.3842
>>3841
Logarithmic primality test.
2 birds; one stone.
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65f089 No.3843
>>3842
>Logarithmic
For a long time I have felt that there is something really beautiful and mysterious about this word
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add73d No.3844
>>3843
Want to see something neat?
Draw the d,e tree for some c.
Then redraw the tree with each number replaced with its triangle number.
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65f089 No.3845
>>3844
Ok but what's the triangle number, of another number?
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add73d No.3846
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add73d No.3847
And the inverse function =
(sqrt(8*T+1) - 1) / 2
Where T is your triangular number.
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add73d No.3848
Now repopulate the tree with the inverse function of every value in the original c tree and find where a factor appears.
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add73d No.3849
……………287
…..16………………31
……………23
…..5……………….7
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add73d No.3850
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add73d No.3851
I'm not saying a factor is going to appear every time.
But they do happen to just "show up" in the tree.
So we need to find every possible way that makes them show up to see the underlying pattern.
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65f089 No.3852
>>3851
>>3849
ok gotcha you are taking the floor of the inverse triangular number. 287 gave me 23.463… interdasting
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ed48a1 No.3853
>>3841
Good callout. Had been playing with the 'closest factors' and you stated clearly.
>>3848
Like the way you're thinking anon!
Also, we haven't used Mod in our decision tree. Recall VQC's crumb to dig into Mod and Elliptic curves, I think it was? Mod could be helpful too (3, 6, 9 rule) and also divisible by 5. Will post up reference in a bit, but the tree might just handle all that anyway - three birds…
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ed48a1 No.3854
>>3840
This is a fun site for playing with primes: http:/ /www.primesdemystified.com/
Had pulled this paper up a while back, very interesting. Can post pdf if you all want, but here's a link:
https:/ /papers.ssrn.com/sol3/papers.cfm?abstract_id=2706936
Notes on Prime Numbers, Their Numerical Statistics & Patterns I: Modular Arithmetic and the Eight Fold Period 24 Model
25 Pages Posted: 22 Dec 2015 Last revised: 18 Jan 2016
Fredrick Zia Michael
Agathos Scientific and Education
Date Written: November 24, 2015
Abstract
We have not yet obtained a simple, complete and accurate method for determining if an integer number specifically an odd number that does not digitally sum to 3,6,9, whether it is prime, and that does not involve trial & error and/or as by a sieving method or factoring. This after thousand(s) of years of searching for such. Alternatively great strides are made towards understanding the behavior of numbers, fields of numbers obtained by some method or another, statistics probabilities & entropies of such numbers, & so forth, & it seems that it is assumed that a better method (but not 'full') for determining primality will appear as an afterthought or byproduct. In this letter we wish to revitalize the interest of the intrepid professional & amateur mathematician & scientist towards seeking the 'full' or 'complete' method of determining primality, or rather towards obtaining the method by which a prime odd integer can be determined completely, simply and efficiently & with applicability to small to very large numbers with a minimum of effort eschewing supercomputing & other recent methods however by utilizing methods resembling pencil & paper approaches.
The caveat that we wish the 'method(s)' to be efficient and simple means specifically that we are referring to the utilization of methods such as digital sums & digital roots, modulus methods, combinatorial & progressions & recursions patterns easily applied, these connected to geometries, polynomial equations and relationships, discernible patterns in sieves and doodles and spirals & other topologies investigated in one context or another yet often deemed curiosities when viewed from the focused area of number theory as pertaining to identifying primes.
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ed48a1 No.3855
>>3854
This was another paper that was interesting:
Carlo Cattani, 2010
One of the main tasks in the analysis of prime numbers distribution is to single out hidden rules and regular features like periodicity, typical patterns, trends, etc. The existence of fractal shapes, patterns and symmetries in prime numbers distribution are discussed.
Fractal Patterns in Prime Numbers Distribution https:/ /www.researchgate.net/publication/221433042_Fractal_Patterns_in_Prime_Numbers_Distribution
and check out the Modulo 2pi plots by H. Sabelli (image attached):
The Biotic Pattern of Prime Numbers
Supports the Bios Theory of Creative
Evolution from Radiation to Complexity
http:/ /www.necsi.edu/events/iccs7/papers/3114216e0aebd76eb51ebe74680c.pdf
for those who love biology, Hibbs, 2008
The Double-Helix Pattern of Prime Number
Growth
http:/ /www.iaeng.org/publication/WCECS2008/WCECS2008_pp278-283.pdf
and on the fractal side, S. Woon 1994:
Riemann zeta function is a fractal
https:/ /arxiv.org/PS_cache/chao-dyn/pdf/9406/9406003v1.pdf
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d1075e No.3856
>>3854
>>3855
Thank you or sharing these.
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ed48a1 No.3857
>>3856
You bet, I have many now, just posting a few relevant ones. I think the Fredrick one could be used when this is cracked to spread disinfo. "Using techniques adopted from this article, a team…"
Oh, and here's another link to 5 pages!!!
https:/ /intelligence.house.gov/uploadedfiles/memo_and_white_house_letter.pdf
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f690f0 No.3858
>>3857
Have we decided how to make the tree? Does it go
145
12 1
then does 12 turn into 6 then 3 or does it turn into 9 and 3 etc?
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f690f0 No.3859
>>3649
>>3652
>>3653
>>3654
What was your stopping criteria for these? Did you go until they were less than 5?
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add73d No.3860
>>3859
My general criteria was stopping when every value on the level was less than or equal to 4. But sometimes reached a factor before that.
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316b0b No.3861
Do we have any knowledge of how to use f (2*d + 1) yet?
Has anyone looked into making a map over the, still, not solved crumbs from VQC?
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add73d No.3862
I'm mad at myself for not finding this 2 months ago.
It literally didn't show up in Jewgle search.
Chris Curtis - Wikipedia
https://archive.fo/UJy1M
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add73d No.3863
And now there's random people on 4 mentioning Chris.
What is happening
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7fa95d No.3864
>>3862
I had already found that but it obviously isn't him so I didn't say anything. Are you saying he took that name as some kind of analogy (since this guy was apparently in a band called The Searchers)? Based on his video, he seems to be from New Zealand.
>>3863
Link?
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d1075e No.3865
>>3863
???
>>3861
I did spend some time looking at f movements. Specifically, enumerated all values in the original grid, created a corresponding record at (f,1,t), and then tried to analyze the differences.
c is the original (e,n,t) record from the grid.
cf is the (f,1,t) record
For odd values of e, found the following:
cf.e == c.f
cf.n == 1
cf.x == (c.x - 1)
cf.d == (c.d + (c.n - 1) * c.a + (c.a - c.e))
cf.a == ((c.n + 1) * c.a - c.e + 1)
Wasn't able to fully describe the movements for even values of e, only figured out the x:
cf.x == (c.x + 1)
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add73d No.3866
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7fa95d No.3867
>>3866
There's no other context that suggests that they're talking about our Chris.
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06387b No.3868
>>3839
Agreed, Baker. I've prayed and continue to pray for the success of our quest. Perhaps our generation will be one of few to see Evil overthrown and defeated in a Great Awakening.
Here's an idea:
Part one ends with finding the GCD of e and d. Kind of a cliffhanger moving into part two.
Part Two: Maybe we should sum the factor tree leaves that match the GCD? Or only sum termination points that are >GCD? Thinking out loud here. Thoughts?
Why did Chris make so much of a point that Pt 1 ends with finding GCD? Must be pretty important for Pt 2?
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add73d No.3869
>>3868
Come on. You already have hints that will lead you to the solution.
THESE ARE YOUR HINTS.
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f2af50 No.3870
>>3869
Cool! Thanks Baker. Seems like you're getting close or have figured it out. Love the clues, thanks for posting. I'm enjoying the quest as we go, and continuing to work, think, and pray.
Clues: So, more than one way to solve it, but the recursive nature of solutions is key.
Squares are triangular numbers, as your cool graphic demonstrated.
And Column one again. We've solved row 1, let's solve column 0. PMA did a lot of good work on that.
Well, Baker seems to know the way lads! Baker, what's up? Did you solve it? Do tell.
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add73d No.3871
>>3870
No, I haven't, I just went in the right direction with these triangle functions.
https://en.wikipedia.org/wiki/Square_triangular_number
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add73d No.3872
This article demonstrates how squares are made from triangle numbers.
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add73d No.3873
The square of a number is equal to its triangular number added to its last triangular number.
Namely, n^2 = T(n) + T(n-1)
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add73d No.3874
If you were to attempt to apply this to our equations, you would have something like this.
d^2 + e = c
T(d) + T(d-1) + e = c
i^2 - j^2 = c
(T(i) + T(i-1)) - (T(j) + T(j-1)) = c
Interesting to note, that n-1 appears in the hints,
and T(d+n) + T(d+n-1) would be the large square.
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3a9068 No.3875
>>3873
Funny. I was just outside thinking about this exact equation, based on your diagram from earlier. Then I come back in and you've posted it.
>>3874
So c is the difference of two squares, which means that difference is also the difference between two triangular numbers. This would allow us to scale down (for a) and up (for b) starting from d. The triangular numbers should help us solve for a or x, by helping us calculate steps up or down from d?
Very interesting that n-1 appears in our records. A larger n means that the two squares (and their triangular numbers) are farther apart (from d). There can be multiple possible combinations I think, but only one solution.
>>3752 for example reference Anons, very helpful graphic by Baker.
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add73d No.3876
Anyone else try to apply the VQC to different problems lately?
If you forget the equations for each variable then what you're left with is a geometrical object that
-has coordinates
-has variables derived from each coordinate
-has 1:1 correspondence from element to set of coordinates
And then if you fill those in with other mathematical functions, you could apply it to anything.
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add73d No.3877
So, really all you'd have to do to apply it to other cryptography and math problems is map coordinates to variables with equations.
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33fd02 No.3878
>>3872
This sounds completely fucking gay, but one observation is
n^2 = 2/_\(n - 1) + n
Using the example of n = 7, if you remove the center stripe (7 black squares) you are left with 2 identical /_\6, one white, one black.
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add73d No.3879
>>3878
Yes, that observation works as well.
n^2 = 2*T(n-1) + n
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add73d No.3880
I feel like this can be used somewhere..
145^2 = T(144) * 2 + 145
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7fa95d No.3881
>>3880
Which is also to say c^2 = T(d^2) * 2 + c, or T(d^2) = c*(c-1). Is that in every case or just this one?
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add73d No.3882
>>3881
No, those identities don't work. They only work for perfect squares. You could add the remainders into the equation.
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7fa95d No.3883
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add73d No.3884
>>3883
Just a fancy word for equation. It means that
"Function a = function b + this and that"
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add73d No.3885
>>3884
Basically that the equation is always right. So like many of the equations we use for the VQC.
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7fa95d No.3886
>>3884
Oh okay, I learned a new meaning of the word identity then.
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add73d No.3887
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316b0b No.3888
I've been looking more into the triangle numbers, nothing new to report, but just another way of finding c**2 (0, N).
Calculate the triangle number of d**2 and subtract the d[t] in (e, 1) where big N is a[t].
Example:
Record for 145 is (1, 61, 12, 11, 1, 145). At t=6 in (1, 1) we have (1, 1, 72, 11, 61, 85).
Compute the triangle number for 144 and subtract 72. You're left with triangleN(144) - 72 = 10368
Which is also the record for a=1, b = c*c.
Doubt it's useful for step 3, or any steps, but it is interesting.
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316b0b No.3889
>>3887
x or x + n. So this can be interpreted in a few ways, either that we have no clue if we'll find x or if we are going to find x + n. However, there are no coincidences, right?
I'm speculating that maybe it depends on the parity of e (or d). Maybe if e is even, we're going to find x, if e is odd, we're going to find x + n.
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2fe852 No.3890
Internet is super shit, can't post images today. Looking into triangular numbers, I've discovered cell (15,4), where D is a triangular number for every single entry.
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cf736f No.3891
What's his twitter? Just made one
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add73d No.3892
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82ada7 No.3893
>>3869
>>3887
Baker - thanks for these additional clues. Still plugging away.
I've taken your idea of triangles and inverse triangles and incorporated them into a parse tree. Tried to find combinations that will get us back to any solution.
Pic attached is just latest attempts adding all d's, e's, and just the d triangles that are direct descendants from c.
Working in (0,1) space for c=145, I was able to match an x value of 120 by 4*(24 + 6). Also managed to find the factor of 5 by adding triangles sqrt( 2*(1+6+78) - 145 ).
Of course, this doesn't work in too many places.
Interesting that things keep coming back to difference of two squares.
For example: 2*78 - 2*6 = 144 + 1 = 145.
If we are searching for x or x+n (x+n is just x+1 in (0,1) space), then we're looking for the correct level in the tree, not a factor.
Does this parse tree tell us what level we are at or the level we are looking for? Does the triangle value or difference in triangles between branches of the parse tree tell us this?
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80a284 No.3894
>>3888
Checked, nice trips
>>3893
Good work both of you, so far any correlations I've found appear to be just luck and work for one number only.
In other news I wrote a variant of the tree-generation algo in C and ran some tests:
RSA100 0.07ms
RSA2048 0.4ms
RSA4096 0.7ms (self generated key)
RSA8192 1.5ms (self generated key)
RSA16384 3ms (self generated key)
And just for fun, M74207281 * M77232917 (two largest known primes multiplied by each other) takes… ~42 seconds. Somehow I don't think anyone will be using RSA keys that large. More surpised my computer didn't blow up trying that!
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add73d No.3895
>>3894
Nope, not at all. Takes hours to create keys that large.
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316b0b No.3896
>>3895
Not with this algo it won't. Finding Mersenne primes should be a simple game.
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32324c No.3897
Hey lads! Working over here, just plugging along. Doing factor trees for (1,1) old school style, pencil and paper. Seeing interesting patterns, nothing to report yet.
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e93cea No.3898
In the same way -E and -X exist, it seems -N also exists. The patterns are mirrored or rotated or something. Will try to show something that makes sense.
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82ada7 No.3899
>>3896
I wrote a test a little while ago to see if there was any pattern to prime numbers in the grid. Obviously haven't figured that one out, but it certainly looks possible.
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e93cea No.3900
Here's a view of (E,X,N) to the right, (E,X,-N) left. The glass blocks are where D-A=X is true, the red; false. Still figuring out what part of -N is "real".
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e93cea No.3901
>>3900
Basically -N appears to be a mirror the same way -X is. A and B trade places and are inverted (multiplied by -1)
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e93cea No.3902
This is the start of my proof that -N is a mirror of -D or -X
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33fd02 No.3903
>>3900
>>3901
>>3902
Keep rockin, MA, this shit is interesting
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e93cea No.3904
Last picture was (E,D,N), this picture is (E,X,N). It's possible I have some flaw with how I'm generating this, but I think it's correct.
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8e44f0 No.3905
>>3904
I'm diggin' where your head's at.
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e93cea No.3906
Confirming this holds true on -E as well.
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7fa95d No.3907
I'm not aware of anyone having thought about this yet but in this image a point is plotted whenever c is a square. So the line in the middle is (0, n) and every other point has a relationship in terms of e and n with (0, n).
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7fa95d No.3908
>>3907
Another interesting thing: this is the same idea (looking for relationships between the e=0 column and other cells) but with x as the x axis instead of e and n still the y axis. Is this maybe the start of an elliptic curve before it diverges? The second image is to make the shape more obvious.
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68a051 No.3909
I've been building a tool to help me visualize the trees and test different theories…
Its code is not ready to share yet, but I'm going to clean it up and share it when its ready.
Using electron + node.
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b5b90b No.3910
>>3909
Teach - this is pretty sweet!
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68a051 No.3911
>>3910
So far its on small ints only.
I just added traditional slow factoring to find a & b and calculate the grid values given a & b.
So I've added the grid to the view now.
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68a051 No.3912
>>3911
I'm also currently developing a ruleset to test the climbing back up the tree.
There are some interesting things I've seen so far, such as when you divide by 4.
c = ab = d^2+e
4c = (2d)^2 + 4e
It appears that you can also calculate movement by 2, but you end up with n ending in 0.5
It seems for calculating x for combining the perfect squares is pretty simple too, as is gcd, since you get a factor out which gives you x and n.
The last remaining rule is the dd+e.
So far, I'm seeing patterns around if e is even or odd, and if so, x is an addition of d+e in some cases. Something like that. Simple d+e addition is the point i'm making.
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add73d No.3913
>>3912
Add triangle number functionality to the application. Chris implied it had something to do with finding x+n
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68a051 No.3914
>>3913
I've been developing a grid app too… hope to turn squares and rectangles into triangle numbers.
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68a051 No.3915
Also, I wanted to show a drawing a made of 2 primes multiplied together as differences of squares.
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add73d No.3916
>>3914
If you think about it, c is a rectangle.
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68a051 No.3917
Looking at my own drawing… is na always a triangle number?
c = (d+n)^2 - (x+n)^2
c = (d^2 + 2nd + n^2) - (x^2 + 2xn + n^2)
c = (d^2 + 2nd) - (x^2 + 2xn)
c = d^2 - x^2 + 2n(d-x)
c = d^2 - x^2 + 2na
>>3915
In this diagram, d is the small dotted square. d+n is the large dotted square.
The area above d+n fits into the area between d and d+n on both the top and the side.
Hence 2 * na.
It seems na is going to be a triangle number.
Double check my thinking guys?
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68a051 No.3918
>>3917
This would make sense as to why we only see a and b of the same parity…
Ok, so I think this ties back to d[t] - d = (n-1)a at n=1
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add73d No.3919
>>3918
If na is a triangle number, which triangle number is it?
E.g if T(n) = n(n+1) / 2
what is n
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68a051 No.3920
>>3918
And I'm pretty sure this would work for L shapes (diff of 2 squares) of any size. It would just scale up.
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68a051 No.3921
>>3919
It will be the a'th triangle number… let me double check that…
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68a051 No.3922
>>3921
This isn't correct…
give me a sec… it has to do with ((b-1)/2) -a
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68a051 No.3923
Sorry, yeah, I've got my math wrong… still correct on concepts, wrong variables…
I there is something here though.
Will update soon.
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add73d No.3924
>>3865
I was trying to replicate your code and I had no clue what was going on regarding the string manipulation.
Could you clarify lines 108, 130 & 134?
https://pastebin.com/CZy8f0HD
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68a051 No.3925
Not sure if this was already said before, but:
ab + △a + △b = △(a+b)
Example:
c = 145
a = 5
b = 29
△a = 15
△b = 435
ab+△a+△b = 595
solve quadratically:
595 = n(n+1)/2
n = 34
△34 = 595 = △(a+b)
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add73d No.3926
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68a051 No.3927
>>3926
This is what I've got so far, very interesting…
c, the difference of 2 squares, is also the difference of 2 triangles.
Our equations in terms of triangle differences is:
c = △(((b-1)/2)+a) - △(((b-1)/2)-a)
Another example for clarity:
c = 145
a = 5
b = 29
(b-1)/2 = 14
(b-1)/2 + a = 19
(b-1)/2 - a = 9
△19 = (19*20)/2 = 190
△9 = (9*10)/2 = 45
△(((b-1)/2)+a) - △(((b-1)/2)-a) = 190 - 45
△(((b-1)/2)+a) - △(((b-1)/2)-a) = 145
△(((b-1)/2)+a) - △(((b-1)/2)-a) = c
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add73d No.3928
>>3927
Fascinating. I wonder what relation triangle numbers have to actual triangles.
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add73d No.3929
>>3928
Besides the fact that they are the result of arranging integers in an equilateral triangle.
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68a051 No.3930
>>3928
This is all bloody beautiful.
Squares can be added an subtracted via their associated triangles.
You can define a triangle number as either a square - a smaller triangle, or as a square + 2 smaller triangles.
This is the missing piece I've been searching for for weeks.
For a while I've had a hunch that we're representing c as dd+e because its like a new number system. Just like we can add i to the integers or rationals or irrationals even, and get complex numbers.
I was looking for a way to add, subtract and multiply squares and errors (dd+e), and its possible via triangles!!!
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68a051 No.3931
Another comment, since I'm on a roll tonight.
I've been playing with using an alternative to f.
So instead of using d+1 to get f, I've been using d-1.
I'm doing this whenever e is odd and d is even, so that e will be even when d-1 is odd.
The reason for this is so that I have enough error to put on either sides of my square.
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33fd02 No.3932
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68a051 No.3933
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33fd02 No.3934
New private message from VQC:
>VQC!!Om5byg3jAU
>Thank you for your patience.
>The decision tree has parts.
>The decision tree terminates upon a set of conditions.
>The first part.
>Determine to stick dick in or not.
>That is the end of part one.
I'm going fucking loony. Great shit tonight, Teach.
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68a051 No.3935
>>3934
Great night man.
I feel so close.
I can't sleep just yet.
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8e44f0 No.3936
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add73d No.3937
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add73d No.3938
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65f089 No.3939
>>3938
this is good to know. no more trying different possibilities.
>>3934
>Determine to stick dick in or not.
Depends if even or odd
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add73d No.3940
>>3939
Yeah. Now, the FINAL piece necessary for our tree algorithms to work is figuring out what we have to do to each value on our tree to make it spit out x or x+n.
If any of you are frustrated that he didn't come back onto the board, don't be. It's important for us to step through us ourselves. It is a million times more rewarding and meaningful to go through the steps of solving this on your own. I'm already learning how to apply the VQC to other maths, and before this I didn't even like math.
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316b0b No.3941
>>3940
There is probably some pattern behind it either being x or x+n. If it is something as simple as e being odd or even, I don't know. Maybe it also depends on which branch you choose?
Like if you use the e branch you will find x, if you use the d branch you will find x+n. Of course, I don't understand the tree yet.
Take our example c = 145. If you use the d-branch, do you have to stick to the d branch through the tree, or does it depend on the different records in the sub tree?
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65f089 No.3942
>>3941
good question. the result would have to in some way be informed by both branches it seems to me. the factors of a number can't just depend on e, or on d alone… something else has to be going on.
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316b0b No.3943
>>3942
Hmm, yeah that makes sense. The question is are we going to traverse the whole tree, or just different branches (Has VQC said anything regarding this)?
I'm looking into the tree for 3^k for odd k. I figure if there is a pattern here, it should be easier to figure it out by observing smaller trees where we know what to expect and preferably in a predictable way (Like just 3 as a growing factor).
I also think we are still missing pieces right? Like how the f works, I know PMA has been looking into it, but I haven't had time to go over his results yet.
Do we also know the underlying pattern of d[t] - d and the growth of (n - 1) as a factor? I'm not sure if we need to know there before we look for patterns in the tree, or if the tree will help us find these patterns (and how to apply them).
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65f089 No.3944
>>3943
>Do we also know the underlying pattern of d[t] - d and the growth of (n - 1) as a factor?
I personally have still not looked at this. I don't even know what growth of (n-1) as a factor means. One thing we still don't exactly do on these threads is define what we mean :)
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d16bcc No.3945
>>3917
Because no one is double checking, N * A = △ is false. Between E,X,N 0->255, there are 1,431 valid cells where N*A = △, and 177,158 valid cells where N*A != △.
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68a051 No.3946
>>3945
Yeah, thanks MA.
Not quite right, but theres something there.
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d16bcc No.3947
>>3946
Flexing my programming skills, I have this for you.
A*B 1066 triangular, 177523 not triangular.
A*C 142 triangular, 140574 not triangular.
A*D 1577 triangular, 177012 not triangular.
A*E 4488 triangular, 174101 not triangular.
A*N 3907 triangular, 174682 not triangular.
A*X 2517 triangular, 176072 not triangular.
B*C 34 triangular, 139377 not triangular.
B*D 830 triangular, 177759 not triangular.
B*E 2842 triangular, 175747 not triangular.
B*N 2309 triangular, 176280 not triangular.
B*X 1328 triangular, 177261 not triangular.
C*D 240 triangular, 139792 not triangular.
C*E 1782 triangular, 155938 not triangular.
C*N 400 triangular, 178189 not triangular.
C*X 420 triangular, 154549 not triangular.
D*E 3244 triangular, 175345 not triangular.
D*N 2823 triangular, 175766 not triangular.
D*X 1870 triangular, 176719 not triangular.
E*N 11106 triangular, 167483 not triangular.
E*X 5440 triangular, 173149 not triangular.
N*X 8336 triangular, 170253 not triangular.
I'm skipping the calculation if it overflows past java's MAX_INT (2147483647), which is why all the total's aren't all the same, and the original N*A numbers are different because I changed the sieve to add 0 as a triangular number.
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89ff50 No.3948
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d16bcc No.3949
Ran addition as well because the hard part of the code is already done.
A+B 3426 triangular, 175163 not triangular.
A+C 1098 triangular, 177491 not triangular.
A+D 4533 triangular, 174056 not triangular.
A+E 5156 triangular, 173433 not triangular.
A+N 5710 triangular, 172879 not triangular.
A+X 5713 triangular, 172876 not triangular.
B+C 1072 triangular, 177517 not triangular.
B+D 3326 triangular, 175263 not triangular.
B+E 3506 triangular, 175083 not triangular.
B+N 4088 triangular, 174501 not triangular.
B+X 3960 triangular, 174629 not triangular.
C+D 430 triangular, 178159 not triangular.
C+E 441 triangular, 178148 not triangular.
C+N 455 triangular, 178134 not triangular.
C+X 402 triangular, 178187 not triangular.
D+E 6097 triangular, 172492 not triangular.
D+N 4902 triangular, 173687 not triangular.
D+X 4885 triangular, 173704 not triangular.
E+N 11695 triangular, 166894 not triangular.
E+X 10210 triangular, 168379 not triangular.
N+X 11511 triangular, 167078 not triangular.
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d16bcc No.3950
Also tried (Var1 + Var2) ^2, and Var1 * Var2 * 2 with similar numbers. Any other ideas?
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68a051 No.3951
>>3947
>>3949
>>3950
Thanks MA.
Taking the c = 145 example, I'm not sure if this works for all numbers, or only 145 because n=a, but assuming the answer:
c = 145, a = 5, b = 29
d = 12, e = 1
n = 5, x = 7
d^2 - x^2 + 2na = 12^2 - 7^2 + 2(5)(5)
so focusing on 2na = 25, and using the diagrams I've shared above, you can know some things about na.
I think its something like:
2na = 2n + △a
or
2na = 2a + △a
(not sure yet since n = a)
Using our example:
△a = △5 = 15
25 - 15 = 10 = 2(5)
10 is a triangle number too, but I'm not sure if this is a coincidence.
I'm working now on describing x+n in triangles too. I think there's a relationship.
Think about this…
c = (d+n)^2 - (x+n)^2
(d+n) = 17
(x+n) = 12
(d+n)^2 = △17 + △16
(x+n)^2 = △12 + △11
c = (△17 + △16) - (△12 + △11)
c = (△17 - △11) + (△16 - △12)
Any time you have △-△, you also have square - square, so the pattern repeats.
Thats what I've got so far.
I'm also working on a process for adding 2 triangle numbers, if you know anything about that, I'm all ears!
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68a051 No.3952
>>3951
>so focusing on 2na = 25
* 2na = 50
I always make typo mistakes.
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68a051 No.3953
>>3951
Also
>2na = 2n + △a
* na = 2n +△a
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d16bcc No.3954
>>3953
All cells where 2 * N * A = 50. Format is {E:N:D:X:A:B} : C
{1:1:32:7:25:41} : 1025
{1:5:12:7:5:29} : 145
{1:25:8:7:1:65} : 65
{14:1:31:6:25:39} : 975
{14:5:11:6:5:27} : 135
{14:25:7:6:1:63} : 63
{25:1:30:5:25:37} : 925
{34:1:29:4:25:35} : 875
{41:1:28:3:25:33} : 825
{46:1:27:2:25:31} : 775
{49:1:26:1:25:29} : 725
{50:1:25:0:25:27} : 675
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d16bcc No.3955
>>3954
All cells in +E,X,N** If you want ALL cells, they're here: https:/ /pastebin.com/w1E6jWcu
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82ada7 No.3956
>>3924
The only important thing to note with those code lines is the recursive call with the prefix string. That's what gives the indentation in the output.
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d16bcc No.3957
My interesting observation is the
{1:1:32:7:25:41} : 1025
cell. This has the same 2*N*A as our answer cell, and shares the same E and X as our answer cell. I played with this solution idea for a while at the end of December. This was the path I was trying to work out,
{1:61:12:11:1:145} : 145
-> {1:1:72:11:61:85} : 5185
-> (Unknown logic) {1:1:32:7:25:41} : 1025
-> {1:5:12:7:5:29} : 145
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8e44f0 No.3958
>>3953
So…
nanna, not nanners? ;)
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add73d No.3959
>>3944
>>3943
The growth of n-1 as a factor in d[t]-d is similar to the growth of n as a factor in a[t]
Use the equations in >>3564 with both.
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add73d No.3960
>>3948
It'd be great at solving any unsolved problem in mathematics you can think of.
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68a051 No.3961
(d+n) - (x+n) = a
(d+n) + (x+n) = b
Not sure if this has been stated before.
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4cc89d No.3962
>>3961
Looks familiar, also validated true for +E,X,N. Here's a new formula for ya, related to some work I'm doing with T right now. Given a valid cell (E,X,N), to predict the next X in it's "chain",
next x = sqrt((2 * n * b) - e)
This let's you step up X, where first cell's B becomes the new cell's A. There are many cells that have multiple "chains", going to work on that tomorrow.
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4cc89d No.3963
>>3962
Not so much predicting, as it is deriving. Not sure how you would get the previous cell's X other than,
previous x = x - (sqrt((2 * n * b) - e) - x)
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80a284 No.3964
>>3961
Found it here >>2709
Shouldn't we be able to use this to get x or n from d and x+n?
If the tree gives us x we can just a = d - x, but what about the x+n case? There are some good crumbs on it in RSA #7 and you guys saying its possible but nobody posted the formula and I suck at algebra
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4cc89d No.3965
I'm quite sure that this negates some of my previous statements on T. If you can find one valid cell in an (E,N), you can search between the valid cell and the next cell in it's chain, which should reveal all of the other chains, thus unlocking the entire cell.
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add73d No.3966
>>3961
This has been stated from the beginning. The roots of the squares added together is b, and their difference is a.
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add73d No.3967
>>3964
How behind are you? Finding x+n solves c. (x+n)^2 + c = (d+n)^2
x+n = j
d+n = i
i+j = b
i-j = a
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add73d No.3968
You check if a number is x by subtracting it from d and checking if it is a factor of c. You check if a number is x+n by squaring it and adding it to c, if the sum is a perfect square, it is the correct value of x+n. You are looking for one or the other in the tree. You are to trim every even value until it is odd by repeated divisions of 2.
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80a284 No.3969
>>3967
>>3968
Thanks, this is what I was looking for. Just couldn't figure out how without having d+n first. My math skills aren't always able to keep up with you guys, more of a codefag.
n = sqrt((x*n)^2 + c) - d
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80a284 No.3970
>>3969
Screwed that up when converting code to post formula. Supposed to be x+n, not x*n. So if you have x_plus_n:
n = sqrt(x_plus_n**2 + c) - d
x = x_plus_n - n
a = d - x
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82ada7 No.3971
>>3909
Thanks for the very clear example, Teach. I've updated my tree parsing to incorporated the /2 step for clarity.
VQC has hinted that the tree will lead to an x or x+n solution. Still searching in the dark for this, but ran some sample code against a list of prime numbers.
Pic attached is for prime numbers 509, 521, and 523.
If you isolate the d branch in the tree and calculate the triangle values for each d node, the total is really close to the (x+n) value.
Well, close enough to be interesting.
These examples share the same d branches and the e branches vary slightly, perhaps there is something to learn here.
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68a051 No.3972
>>3971
I'm working on similar things.
I reworked my tree code, added x+n and d+n.
There are so many patterns again, so confused today.
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add73d No.3973
>>3972
I find it VERY interesting that 14 and 11 show up and the x+n you're looking for is 13.
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68a051 No.3974
>>3973
Want me to run any numbers?
I'm down to share my code too. I think its ready to be packaged up, just gotta figure out where to post it to share. Its really really big.
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add73d No.3975
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7fa95d No.3976
>>3711
>I'll be back on VQC board by the end of the week
So much for that.
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add73d No.3977
>>3976
Do you want to be babied through this or something?
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7fa95d No.3978
>>3977
Of course not (that would defeat the purpose), but this definitely isn't the first time he's said he'll be here and he didn't turn up.
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add73d No.3979
>>3978
If that's the purpose, don't you see the reason why he would underestimate our progress and then want to let us go farther ourselves?
It's not hard to post a hint on a board of something you know forwards and backwards; he worked on it for ~7 years. He clearly values the importance of us working throughout it ourselves.
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add73d No.3980
So, of course he would go back on his word in giving us a hint that could potentially complete the log(n) algorithm.
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7fa95d No.3981
>>3979
>>3980
All I'm saying is that he's done this a bunch of times now so I question why he would keep saying it. I know he knows what he's doing. I don't have anything to add to what we're doing at the moment that anyone else hasn't said at the moment.
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add73d No.3982
I also believe that the fact that he walked us through the VQC instead of giving us this tree which doesn't have an (immediately) clear connection, (though you can see how it was derived from the VQC if you note the appearance of fractals) is because he devised a specific plan for people to work through it exactly as he did. Except with a VQC that was improved from the original version.
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add73d No.3983
I just don't see why anyone would want any more hints. There's a million things to discover from these mathematical devices, they've probably never been studied in this way.
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add73d No.3984
>>3983
And you don't discover much if the solution is spoonfed to you.
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33fd02 No.3985
>>3983
>There's a million things to discover from these mathematical devices, they've probably never been studied in this way.
Studied in what way? Studied by a bunch of degenerate, unqualified faggots? (I count myself in these ranks)
Maybe that's the point? Who knows. Also, maybe VQC counts twitter DMs as being on /vqc/. I agree VQC is a complete faggot, but he's /our/ faggot.
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cdc778 No.3986
The tree thing is probably very important, but maybe we should take a step back and review the patterns we already know.
Get a list of everything and then double check it.
VQC has hinted that triangular numbers are related to this. Maybe we should reevaluate our methods of calculating D's and A's to see if there is a more elegant triangular method?
We also know he said the pattern was complete in a tweet during thread #7(?). Maybe we should revisit that thread and go over the results. We might have found something important and overlooked it.
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7fa95d No.3987
>>3986
Link to the thing about the pattern being complete? Or you could point it out in the baker's big image or my pdf if it's in one of those and it's easier to find there.
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add73d No.3988
>>3987
From the beginning he has stated that he had provided all that was necessary to solve it.
Guessing we didn't put 2 and 2 together as easily as it looked. But it's a lot of 2's and 2's to put together to surmise that the solution to this would be fractals and triangular numbers.
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cdc778 No.3989
It's not in the big picture and I'm not opening anyones PDF.
>>2469
Chris has since deleted the tweet but it said something along the lines of the pattern being complete.
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add73d No.3990
Replicated PMA's recursive tree algorithm in Java. It's self-contained and ready to run.
https://pastebin.com/VZnQQR2i
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add73d No.3991
>>3990
I changed the order of it to e and d instead of d and e so you can just go up the tree and see that 3^2 + 3 = 12, 12^2 + 1 = 145, etc.
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13fa49 No.3992
Y'all cianiggers talk too much and don't do enough math.
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d1075e No.3993
>>3972
Teach - this is great work. Makes it real easy to follow along.
A couple of the records don't look like valid grid entries. Specifically the even nodes where n's are fractions.
Also, instead of treating the d nodes as "c" values and trying to calculate their "prime" solutions, perhaps just treat them as "d" entries, and resolve to a valid record based on e=0, d=2d, a = d. Don't know if that will add any clarity, but certainly something to look into.
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add73d No.3994
I programmed in a version of the tree that checks for x and x+n values, as Chris has initially described the solution would work. It doesn't work for large ones (probably because they have to be extracted out of the tree in a different way than just checking values) but if you enter some random values from 1-400 you'll see it does find the factorization for quite a lot of them. It needs to be optimized a bit more before it's able to instantly check the rsa-sized trees.
https://pastebin.com/0jPr3RrE
>>3992
>(1)
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80a284 No.3995
>>3994
I have been doing something similar on the RSA100 tree trying to find close matches for x or x+n using various formulas for every combination of the numbers in the tree, up to ~10 different numbers. Summing nodes in any combination isn't any good. Neither is summing triangular numbers, or trimmed versions of triangular numbers. Initial runs on multiplication aren't looking too hot.
for i in range(1, 42):
print(i)
#for comb in combinations(traverse(tree), i):
for comb in combinations(map(lambda n: (n*(n+1))//2, set(traverse(tree))), i):
result = sum(comb)
...check how close result is to targets...
I think the C version counts as instant though, just need to figure out what to checkclang -O3 -march=native -mtune=native -lgmp tree.c -o tree
0.000493 seconds - create_tree
0.000018 seconds - traverse(tree_count)
2587 nodes
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80a284 No.3996
>>3995
Those timings are for RSA2048, not RSA100. Sure beats python.
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13fa49 No.3997
>>3994
>(178) posts
>0 insights
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add73d No.3998
>>3997
Let's make that 179 just for you
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80a284 No.3999
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316b0b No.4000
>>3995
I think we're being given the crumbs in a disordered fashion.
I don't know if we have everything in place for the tree right now. We might be missing crumbs, still, rendering the tree useless.
Unfortunately VQC is sporadic in his hints, so it's hard to tell. At the same time, I don't want to be spoon fed this stuff either, it's a fun challenge. However, I do question how much we've been able to solve ourselves.
It seems like we're just hitting our heads against the wall until he somehow drops a few crumbs (or parts). And then we jump on it until we're stuck. Rinse repeat.
Now we know we're talking about fractals here, right? Instead of going the other way around, why don't we try and understand why we need to recurse down the tree. Try and study the patterns at the core of the nodes, see if we can understand why the tree is necessary in the first place.
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68a051 No.4001
I don't mean to distract from the triangle numbers, and our current tree theories, but I was going back through the original crumbs, and found this one:
>At the correct element in the grid at (e,1) where the value of a at that element equals the na we want, if you subtract 2d+1 from na then you are in the negative half of the grid in terms of e and the value at the same element in the first row will be (n-1)a
Re-reading this crumb, and looking at the grid, at (f, 1), there is an a that = (n-1)a.
For 145, this is our value 20, and when we add d to 20, in (e,1) we'll find d=32 & a=25.
These two numbers are connected via x.
The x for f is 1 more than the x for e.
More than that too, it seems that the x at (f, 1) shares a common factor with the original d.
I'm going to try to use this info to roll up the tree.
We know how to generate d, x, and a for (e,1) or (f,1).
We know the relationship between x's and factors of x in (f,1), and so hopefully there's a way to mathematically solve the a's for e and f using simultaneous equations.
Also, regarding sharing my code. I fell asleep last night so I'll package it up and share it today.
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316b0b No.4002
>>4000
We know we are talking about a fractal. And we've been told previously that we are going to try and figure out which branch we are in of the fractal.
Once we know that, we can solve it. So maybe the tree represents a part of our fractal, and we are going to use this tree to figure out where we are in the fractal.
If we can figure out how this relates to a fractal, somehow generate and study it, then maybe then we can understand the tree more.
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68a051 No.4003
>>3993
You're right. These numbers are not officially in the grid, I was just trying to understand patterns.
One thing that Chris said was that this process finds the 2 factors that are closest together, ie, smallest n.
For this reason, I've made my slow_factor method return precisely these factors.
I like your idea of treating d as 2d. I'm not quite there yet as to understand why and precisely how. But I see your point.
>>4001
If you get a chance today PMA, I know you can generate d,x,a for n=1, check out this post and tell me what you think.
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80a284 No.4004
>>4000
I have to agree, we're not really solving much on our own, just collectively understanding the crumbs. Seems almost designed that way.
The fractal part has been obvious since day one imo, and is one of the main reasons I'm here. They just got way more obvious in tree form than looking at a flat grid. I think the tree is only neccesary for factoring numbers though, and not related to navigating an infinite fractal grid, although recursion still applies. Think of it as a tool, not a solution.
The worst part is I'm not really interested in solving RSA, I want to see what we can come up with using this. A software quantum computer ffs, the possibilities are ridiculous. And given what we've already seen its pretty fucking fast too.
I'm not sure what the strategy behind this slow drip of info is since Chris (or the "Chris entity") is obviously smart enough to have one, and its kinda working well (we're still here).. Trying to drop this knowledge any other way would probably be suicide and it'll get wiped from the internet, so that may be part of it.
A darker twist would be that some group already has this math (the elite hoards knowledge) and are already using it in secret. We know the NWO plan depends on chaos and the financial system is already set up to fail, so if (((someone))) didn't want to get the blame, what better way than having a bunch of channers release it in a timed way that blows shit up good…
^^ The above may be a bit out there and I'd like to think my intuition is good enough to avoid such traps, but knowing what we're up against I'm not so sure. Think its worth considering at least.
Now lets get back to math!
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80a284 No.4005
>>4004
Sooo.. didn't mean to kill the thread.. too weird? too real? too crazy?
Everyone working on math? Or upgrading their VPNs?
Already know where I stand on the spoiler part, and will continue anyway but assumed it'd start a discussion at least. Post some trees or something!
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d1075e No.4006
>>4003
Thinking more about f.
In the original grid code, the f record was used to illustrate a mirror in the negative half of the grid.
The (n-1)*a transform, which takes us from (e,n,t) to (e,1,t), was explained as being a shortcut for an f movement. (I believe VQC stated this).
These are both big movements in the grid. We can do individual record movements (by adjusting x, t, or even computing the next d), but unless we know a factor, we can't move around any better than this.
While creating the tree, the only values we can compute definitively from e and d are c and f.
Perhaps the point of making the tree is not to figure out the x or x+n directly, but to create a structure that allows medium sized movements within the grid itself (phase 3).
What if the f value we can generate at each d node in the tree can be used to navigate backwards in bigger steps towards the correct solution?
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33fd02 No.4007
>>4005
>>4004
Just for the record, I agree with you 100%
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a89af7 No.4008
>>4005
Gonna have to better than that to kill the thread. There's just not that many people here, which is why every contribution is helpful.
>>4003
I've gone over that crumb a few times, but I don't know what to do with it because we still don't have a way of getting the N*A we want. We know the answer is 25, and that the (N-1)*A is 20, but we don't have a way to reach that answer. Maybe you could do some kind of Fourier transform but I don't know.
>>4006
If I had to guess, the medium sized moments would be an exponential growth formula which intersects with the factor cell.
The only note I have about F, is that any (E,D) with any values for (N,X,A,B) all have the same F. The inverse is true for any (F,D) with any (N,X,A,B) - they all have the same E.
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1d5a78 No.4009
>>4006
I like your thinking here PMA! I'm working over here, and will post some work in progress and questions a bit later, just nothing major to report. Here's a quick fun question:
for all n>1, Floor(SQRT(Abs(f))) + what? = x
You know we gotta start from c.
"what?" could be a piece from the factor tree.
>>4005
I could care less about RSA too, just enjoying the challenge of solving a math problem. I think NSA already has this secret math, they just want release to be crowdsourced. Also, it will increase GEOTUS' Anons to Legend status and ALL THE WORLD WILL TREMBLE at our autistic Reeeeeeeeeeeeee! Think about it, it's partly a PR campaign for us.
"We the people forgot how to Play"
"We the people have the Power"
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a89af7 No.4010
>>4009
X ≈ floor(sqrt(abs(F)*(N)))
See my posts >>1929 >>1930 >>2346
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a89af7 No.4011
>>4010
One of the problems with that approximately X formula is solved with this >>3962 >>3963 formula, i.e, working your way quickly down to the root cells.
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d1075e No.4012
>>4008
>If I had to guess, the medium sized moments would be an exponential growth formula which intersects with the factor cell.
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d1075e No.4013
>>4012
sorry. forgot pic.
c=6107 and c=7463 samples attached.
Look at the f values in corresponding d branches.
MA - I think you are spot on here.
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68a051 No.4014
>>3962
I didn't pay enough attention to this MA.
How confident are you in the accuracy, because if its true, it could be super useful.
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d1075e No.4015
>>4008
>>4013
Add f values for the Rsa100 tree at d branches just to show growth.
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1d5a78 No.4016
>>4010
>>4011
Hey MA! Agreed.
Only problem is we don't have n starting from c. Can we find n in the factor tree and plug it in? We get c,d,e,f to start and that's it. So the missing piece for the formula is an n approximation?? Let's find it.
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a89af7 No.4017
>>4014
>exact for N2 (or F's N1), then X's variance sets in. Off by ~128 around ~N500 and grows, but was still less than 2000 at N5000.
It's more accurate for longer on higher E/F's because it's a prediction for that slope, not the A=1 cells.
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a89af7 No.4018
>>4014
Sorry, didn't read which post you were replying to. It's a 100% accurate derivation of the next X in the chain.
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add73d No.4019
>>4004
You chec the DOW lately? Whichever Jew it is they're going all in with the sabotage, I can't see a reason for the market to be crashing like this.
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a89af7 No.4020
>>4016
That formula is better suited for a way to brute force if an N contains any entries for any given E/F.
>>4015
I still haven't even started on the tree thing, I'm still working through all of the ideas I had while I was away. Honestly it doesn't make any sense to me at all right now, still processing it subconsciously I guess.
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d1075e No.4021
>>4020
Still - great insights, thanks.
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06387b No.4022
Hello Lads! Solid idea over here! Lets factor f.
Just looked over my formulas and realized that d of f solves for x in row 1. If we factor Floor(SQRT(abs(f))) we get d and e for f. The d portion solves for x in row one, but the remainder e is still there!
But what about the remainder (e) portion of f? Can this help us bump up our value so that we can solve for x in n>1?
Maybe program anons can run the numbers. I'm working over here but wanted to share now bc my preferred programming language is Excel, lol.
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a89af7 No.4023
Finally debug'd my next big render. Here's a view of E to the right, N down. Red block, N is invalid. Yellow block, (E,N) has one X chain pattern. Blue block, # of X chains not equal to one.
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a89af7 No.4024
Yellow block, X chain = 2. Blue block, else. Red block, invalid cell.
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a89af7 No.4025
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a89af7 No.4026
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a89af7 No.4027
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a89af7 No.4031
Pay no attention to the man behind the curtain. X chain = 6.
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7fa95d No.4032
>>3992
>>3997
I don't see you giving any useful insights, (2). Starting arguments is only going to cause the conversation to go off-track even more. Not all of us have any useful ideas right this second. A couple slightly irrelevant posts isn't stopping anyone from working and collaborating.
>>4019
>DOW
Did you see it was at 666 points the other day?
>>4023
The gradient of each of those lines increases by 1 each time (rise/run = 2/2, 4/2, 6/2, 8/2, etc). What's an X chain pattern?
>>4022
>d of f solves for x in row 1
Here's the bitmap with e and f either side of the zero line on the x axis and d on the y axis.
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7fa95d No.4033
>>4032
>>4022
Here's e and f on the x axis and d on the y axis when c is a perfect square.
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3a9068 No.4034
>>4032
I'll take a triangular square with rectangles for $5000, Alex. That's fucking cool as shit, Anon!! Patterns everywhere!
That's considered a chevron symbol, correct? Did a quick search. Baker will die and go to heaven if you can find a swastika. ;)
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a89af7 No.4037
>>4032
{1:1:2:1:1:5}
{1:1:8:3:5:13}
{1:1:18:5:13:25}
{1:1:32:7:25:41}
{1:1:50:9:41:61}
{1:1:72:11:61:85}
{1:1:98:13:85:113}
{1:1:128:15:113:145}
{1:1:162:17:145:181}
{1:1:200:19:181:221}
{1:1:242:21:221:265}
{1:1:288:23:265:313}
Here, B leads directly into the next cell's A. X Chain = 1.
{3:2:2:1:1:7}
{3:2:6:3:3:13}
{3:2:12:5:7:21}
{3:2:20:7:13:31}
{3:2:30:9:21:43}
{3:2:42:11:31:57}
{3:2:56:13:43:73}
{3:2:72:15:57:91}
{3:2:90:17:73:111}
{3:2:110:19:91:133}
{3:2:132:21:111:157}
{3:2:156:23:133:183}
Here, 2 different chains of cells lead from their B to their next cell's A. So X Chain(s) = 2. I'm calling them X chains because with the formula here, >>3962 (You) you only need one entry from each chain to unlock the entire chain.
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f2af50 No.4038
>>4034
>triangular square
I have no clue what I meant there. Except that trapezoids are combinations of squares, triangles, and rectangles. Just too many shapes at once, got all excited.
>>4033
Looks like a nice right triangle to me, Anon. WTF, I love this challenge. How do we use it? Any ideas?
>>4036
Love this X chain work, MA! I was thinking of it like Jacob's Ladder, but X chain is badass too.
For (e,1) that starting point is in t=1.
For odd e, (e+1)/2 = a for t=1
For even e, (e/2) = a for t=1
We're all working to unlock n>1, and we'll find it soon. Appreciate all your excellent ideas and graphics.
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7fa95d No.4039
>>4034
I hope this is actually useful and not just pretty pattern shit again, but I did e&f by x both for square cs and not based on your post and found something interesting. In the non-square one, you can kind of see a diagonal grid in the f space within the grid we're already looking at.
>>4037
It's a little irritating that we can't just edit our posts on this website. I can but that's only because I control the BO account. I get what you mean about the x chain thing. Do you think there's any way to mathematically predict these points? Just looking at this image, there are obvious patterns in the valid/invalid n cells (the first one being two valid, two invalid over and over, the second being one invalid, two invalid etc) but they seem to get a lot more complicated the further down you go. If we could find a way to predict the pattern of a particular line, I don't know that it would help with the tree but we'd be able to find something.
>>4038
>How do we use it? Any ideas?
I'm not sure that the thing I posted was entirely useful just since all of the variables used in the image come from c (if c's a square, as well as d, e and f). It shows a linear relationship between d and f, but that doesn't give us n or x.
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a89af7 No.4041
>>4039
If you look at X Chain = 3, the first (E,N) is (0,9). Then (9,9). Then (18,9). Then (27,9). Basically (i*9,9) for any i. The next row down is (27*i,27) for any i. Then (0,45) (45 = 27+18). Every coord is a factor of 9, but not every factor of 9 is a coord.
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3a9068 No.4042
>>4039
AA? Nice work! I see the diagonal grid.
Baker, can you please repost your idea about e and f being two mountains to the sides of a prime valley? That was a great explanation.
Guys, I'm feeling it. It's like when you know somebody is about to get
7777777777777
Trump will win.
Maybe not tonight, but I can feel our combined minds going up a level right now. Seriously listening to "Chariots Of Fire' Vangelis Theme right now. Also, I just smoked a toke. Choose Joy tonight, Anons! It is possible by choice alone, we control our own VQC's: Our mind, emotions, and spiritual essence. The conversation today is up another level of understanding! All Anons On Deck!!
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add73d No.4043
>>4042
Sure thing. Your semiprime exists between two squares, they're the two mountains, and every number between that is differentiated by it's remainder. You can reference a number by the first square, d^2 and e, it's distance or you can reference a number by it's 2nd adjacent square, (d+1)^2 and it's distance from that, equal to f.
So all your numbers exist between two squares.
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add73d No.4044
>>4043
You can denote c as the distance from any square. I haven't looked into whether that gives any insight.
145 = 1^2 + 144
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fc4336 No.4045
>>4043
And you can locate every position in the valley because c, d, and e. Like GPS. You have enough coordinate info to locate in more than 2D. E and f are the balance between two adjacent squares?
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add73d No.4046
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3a9068 No.4047
Alright, so what should we focus on tonight? We have a lot of good ideas going on. Maybe everyone just keep going and see what we come up with? Post your shit, even if you're not sure about it. Let's get some constipated ideas out.
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82ada7 No.4048
>>4047
VA. I added f into all the d nodes in the tree, and am reviewing against (e,1) entries to see if anything obvious jumps out.
This has to relate back to the d[t]-d hint at some point. But not seeing anything.
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06387b No.4049
>>4048
PMA, nice to see you. I feel like I'm in the LOTR at this point. Hero's quest, we're doing the walking in between all the cool parts. Long walk, but maybe we'll get our math asses in shape.
>"There is a simple and elegant way to solve this shit." - PMA
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32324c No.4050
Is Bantz and Shitz lurking here? Anon had some great advice back in RSA#4: loose paraphrase…
>If you don't have shit to say, Bantz and Shitz till someone smarter like Teach says something good..
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82ada7 No.4051
>>4049
Still believe in the simple and elegant. Trees are pretty elegant, after all.
Unfortunately, still walking around in the dark and hitting walls at every turn. Guess that's what makes it fun. Learning something that hasn't been taught before.
Would love to hear the back story to this one day. It must be fascinating.
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7fa95d No.4052
>>4042
>AA
Yeah, I'm still here every day even if I don't always have ideas. This is an important board so someone has to keep it from being overrun by shills.
Have we considered using f in the tree somehow? That might be a stupid idea but I'm just thinking out loud. Obviously it revolves around input variables d and e, like Chris said, but if the tree either terminates with x or x+n, maybe the thing that determines whether it's one or the other is another variable that comes from input variables d and e (i.e. f).
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fba0a4 No.4053
>>4052
Thanks for defending us against shills, AA! F may hold an important key, not sure yet. Starting from c, f is one of the few we can immediately derive.
Let's factor f like so: d of f AND e of f using the tree formula. For the following cells:
Format {e,n,d,x,a,b} c, f, d of f, e of f.
First, (1,1) t=1-100
Then (1,2) t=1-100
Then (1,3) t=1-100
Can you anons please check my idea? Thinking that e of f can help make x for n>1.
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7fa95d No.4054
If when this is implemented we just plug numbers in and get numbers out, how are we going to know whether the number it spits out is x or x+n?
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fba0a4 No.4055
>>4052
AA, please post some good esoteric reading about Meme Magick, and how we create our own reality. Topol should chip in too. Math is nice, Math + Mystery = Sexy.
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7fa95d No.4056
>>4055
I've been meaning to keep going with that thread explaining the occult but nobody seemed all that interested and my life has been quite hectic. Plus, a lot of you keep using Christian themes in the things you say, and the Bible says occultism is inherently wrong (you're all entitled to your opinions but as someone who used to be Christian I think that's complete horseshit), so I don't want to step on anyone's toes for no good reason. If you read what I wrote about astral projection in that other thread and you are interested I guess I could keep going.
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3a9068 No.4057
AND I'm down to work too, faggots. To prove it, here's some epic ponies. Meme magick waiting to happen. Topol?? PLS help?
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8e44f0 No.4058
Oh yeeeeeeah…. I was doing something…
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3a9068 No.4059
>>4056
Thanks Baker! I am open to all ideas, because they are the currency of life. Confirmed Jesus Fag, who is high and drinking beer now here.
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32324c No.4060
>>4058
Love you, Topol. Plenty homo, my faggot. How do you do what you do when you do?
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7fa95d No.4061
>>4032
Noticing the part of this image at the top where it seems like there's a solid pattern - do we have a maximum number at which the tree doesn't work anymore? Maybe that section of the image shows where it works and lower is where it doesn't work. This is a blind guess, though, since I don't know what value of d it is when that pattern stops in this image.
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32324c No.4062
>>4058
Chevron Ponies? ChevRonies? Quick Draw Topol wins the gunfight.
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8e44f0 No.4063
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3a9068 No.4064
See it's way more fun like this, guise. Math + real life + memes + equations + honesty + dreams = VQC. This board isn't just for math. It's for life and world improvement.
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fba0a4 No.4065
Prepare for acceleration?
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899548 No.4066
Been researching factor pairs, Pythagorean triples, prime factorization, hexagonal numbers, various rules, etc.
Maybe read this page for some ideas:
https:/ /findthefactors.com/tag/centered-triangular-number/
Couple example snips:
Prime factorization: 976 = 2 × 2 × 2 × 2 × 61, which can be written 976 = 2⁴ × 61
The exponents in the prime factorization are 4 and 1. Adding one to each and multiplying we get (4 + 1)(1 + 1) = 5 × 2 = 10. Therefore 976 has exactly 10 factors.
Or the 901 example, 901 is the sum of two squares two different ways:
30² + 1² = 901
26² + 15² = 901
901 is the hypotenuse of FOUR Pythagorean triples:
60-899-901, calculated from 2(30)(1), 30² – 1², 30² + 1²
424-795-901, which is (8-15-17) times 53
451-780-901, calculated from 26² – 15², 2(26)(15), 26² + 15²
476-765-901, which is 17 times (28-45-53)
Two of those were primitives. That can only happen because ALL of 901’s prime factors are Pythagorean triple hypotenuses.
901 is a composite number.
Prime factorization: 901 = 17 × 53
The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 901 has exactly 4 factors.
Factors of 901: 1, 17, 53, 901
Factor pairs: 901 = 1 × 901 or 17 × 53
901 has no square factors that allow its square root to be simplified. √901 ≈ 30.016662
694 is a composite number.
Prime factorization: 694 = 2 x 347
The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 x 2 = 4. Therefore 694 has exactly 4 factors.
>>4065
oooh, thanks VA! It's about Time (something VQC has said a few Times). Perhaps with the market fuckery, it IS time, now finally! VQC has been holding and tweeting, and I think this may be the case!
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fc4336 No.4067
>>4066
Exactly! Thanks MM. (Mr. E. Melange?)
There are multiple answers for a given C? But only one is prime? What about the remainder of f? Factor f into d and e.
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899548 No.4068
Also took a turn down the biblical / gematria route, and Star of David stuff.
example:
http:/ /www.whatabeginning.com/Misc/Tess2/P.htm
and (seems less exact, more 'fishing' for numbers, but could provide some inspiration)
Structural Numbers of the Bible
https:/ /www.biblewheel.com/Collaboration/Paardekooper2003_09_09.php
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82ada7 No.4069
>>4065
Ready!
>>4066
Very interesting. Thank you.
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899548 No.4070
>>4067
Yes VA, the Melange here. Been into this for a while. Mostly reading academic papers, but tonight dumping more of the 'image' type stuff, as sensed your call for some new tangents to think along.
>>4031
Spent a good while looking at your outputs MA. Vahddy interdasting!!
Examples of some basic rules:
A number with three or more factors is a
composite number
Divisibility rules can be used to factor a number and to test the primality of a number. Some divisibility rules:
• All numbers are divisible by 1
• Any even number is divisible by 2. A number is divisible by 2 if it ends with a 0, 2, 4, 6, or 8.
• A number is divisible by 3 if the sum of its digits is divisible by 3.
• A number is divisible by 4 if the last two digits (tens and ones) are divisible by 4.
• A number is divisible by 5 if it ends in a 5 or 0.
• A number is divisible by 6 if it is also divisible by 2 and 3 (see these tests above).
• To test a number for divisibility by 7
o Take the last digit in a number.
o Double and subtract the last digit in your number from the rest of the digits.
o Repeat the process for larger numbers.
• A number is divisible by 8 if the last 3 digits are divisible by 8.
• A number is divisible by 9 if the sum of its digits are divisible by 9
• A number is divisible by 10 if it ends in a zero.
A prime factorization is a factor string expressing a number as the product of only prime factors. Every number has exactly one prime factorization. This prime factorization can be written using exponents if any of
its prime factors appear more than once in the string.
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899548 No.4071
>>4069
Hey PMA, your work has been killer!! You and Teach are really inspiring one another. Nice to follow along with you fine anons.
Has this been brought up? Star of David Theorem. (Given RootODavid and all).
http:/ /mathworld.wolfram.com/StarofDavidTheorem.html
Generalizations to Large Hexagons of The Star of David Theorem with Respect to Gcd
https:/ /link.springer.com/chapter/10.1007/978-94-011-5020-0_4?no-access=true
The Star of David theorem with respect to the greatest common divisor states that GCD(A 1, A 3,A 5) = GCD(A 2, A 4,A 6) for the six coefficients A 1,A 2,A 3,A 4,A 5 and A 6 in order surrounding any element X in Pascal’s triangle. This property was discovered by Gould [4] and first proved by Hillman and Hoggatt [5]. Subsequently a number of alternative proofs and various interesting generalizations have been presented.
https:/ /7daysand8nights.wordpress.com/2010/02/26/star-of-david-theorem/
This blog has same info. Enjoyed this bit:
According to Wolfram’s Mathworld, the Star of David Theorem was first stated by H. W. Gould in 1972, and there were several generalization in the years immediately following. Apprently the association with Pascal’s triangle wasn’t noticed until 6 years ago, however, by B. Butterworth in this article (which is originally about using Pascal’s triangle to illustrate the song “The Twelve Days of Christmas”).
https:/ /threesixty360.wordpress.com/tag/pascals-triangle/
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899548 No.4072
Learning about different types of numbers.
https:/ /www.fq.math.ca/Scanned/9-2/bauer.pdf
Numbers that are both Triangular and Square - their Triangular Roots and Square Roots. R. L. BAUER. (3 pg pdf, 1971)
This popped up including 'recursive' in search.
And, patterns in these triangle numbers. An email exchange with a mathematician regarding The Triangular Squares:
The purpose of this note is not Pell’s equation, but rather to give a more elementary solution, once we propose that the triangular squares obey a linear recurrence. The recurrence I gave needs only two consecutive values, say 0 and 1, to produce all the rest.
http:/ /www.gosper.org/triangsq.pdf
That all stemmed from this guy Hans, and researching Pascal's triangle. It's where I stumbled on the Star of David. He has an interesting perspective.
http:/ /hans.wyrdweb.eu/about-the-number-nine/
About Pascal’s Triangle
When a number represents a Geometric Structure it is called a Figurative Number.
Every possible figurative number is generated by the Triangle of Pascal.
The Fractal Sierpinsky Triangle is the Triangle of Pascal Modulo 2.
The Triangle of Pascal was known long before Pascal (re)discovered it.
It was known in Ancient India as the Meru Prastara and in China as the Yang Hui. Meru Prastara relates the triangel to a Mystical Mountain called Mount Meru. Mount Meru is also implemented in the Sri Yantra.
The Triangle shows the Coefficients of the Function F(X,Y))= (X+Y)**n. If n=0 F(X,Y)=1 and if n=1 F(X,Y)=X+Y so the Coeffcients are (1,1).
Pascals Triangle is a 2-Dimensional System based on the Polynomal (X+Y)N. It is always possible to generalize this structure to Higher Dimensional Levels. 3 Variables ((X+Y+X)N) generate The Pascal Pyramid and n variables (X+Y+Z+….)**N generate The Pascal Simplex.
The rows of the Pascal’s Triangle add up to the power of 2 of the row. So the sum of row 0 is 20 and the sum of row 1 is 21 =2.
The Sum of the rows of the higher n-dimensional versions of the Triangle is nN where n is the Amount of Variables and N the level of expansion. So the Sum of Pascal’s Pyramid (3 variables X,Y,Z) is 3N.
The most interesting property of the Triangle is visible in the Diagonals.
The First Diagonal contains only 1′s. The Ones represent Unique Objects. They are the Points in the Tetraktys.
The Second Diagonal contains the natural numbers. These Numbers are used to Count Objects that are The Same. The Natural Numbers are the Lines that connect the Points. The Natural Numbers are the Sum of the previous Ones.
The Third Diagonal contains the triangular numbers. The Triangular Numbers are the Sum of the previous Natural Numbers.
This pattern repeats itself all the time.
The Fourth Diagonal contains the tetrahedral numbers (Pyramid Numbers) and the Fifth Diagonal, the pentatope numbers.
Fermat stated that Every Positive Integer is a Sum of at most three Triangular numbers, four Square numbers, five Pentagonal numbers, and n n-polygonal numbers.
The Tetrahedron with basic length 4 (summing up to 20) can be looked at as the 3-Dimensional analogue of the Tetraktys.
The Diagonals of the Triangle of Pascal contain every Possible 2-Dimensional Figurative Number (and Structure).
These Numbers are Projections of Higher Dimensional Numbers and Higher Dimensional Structures.
The Higher Dimensional Versions of the Triangle (the Pascal Pyramid, The Pascal Simplex) contain these structures.
The Rows of the Triangle Sum to the Powers of Two (2 Dimensions). These Powers control the Levels of Expansion.
Every 7th step the Fractal Pattern of the Triangle repeats itself on a higher Level.
The Figurative Numbers are the Geometric Shapes that are created by the Lines of the Natural Numbers that are connecting the Points of the One.
Pascal’s Triangle also contains the numbers of the Fibonacci Sequence (“The Golden Spiral“).
When we take the Modulo 9 (the Digital Root of Pythagoras) of the Numbers of Fibonacci a repeating patterns of 24 steps shows itself that can be represented by a Star Tetrahedron or Stella Octangula. The Star Tetrahedron is a Three Dimensional Star of David.
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ceb3da No.4073
>>4065
Welp, I don't have much time today to work on it, but I'll be ready.
I hope he will stick around long enough afterwards to teach us how to make more virtual quantum computers.
This shit is too good.
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899548 No.4074
^^ uggh, sorry for blacking out and reddit formatting with that last copypasta.
>>4073 Hey Isee, good to see ya!
Agree, it's interesting, and given it's last thing I do before crashing, in my dreams. I often wake grasping at some concept. This morning it was a series in the e that was *just* there, if only a bit more lucid of a dream…
So Hans sends me down the path of
Hexagonal numbers, which are a form of pyramidal numbers. Didn't VQC mention pyramids at one point or am I losing it?
https:/ /en.wikipedia.org/wiki/Centered_hexagonal_number
The sum of the first n centered hexagonal numbers is n3. That is, centered hexagonal pyramidal numbers and cubes are the same numbers, but they represent different shapes. Viewed from the opposite perspective, centered hexagonal numbers are differences of two consecutive cubes, so that the centered hexagonal numbers are the gnomon of the cubes. (This can be seen geometrically from the diagram.) In particular, prime centered hexagonal numbers are cuban primes (more new terms! Like pronic numbers, and polygonal numbers, etc.).
To bring this back to Pascal:
When all the odd numbers (numbers not divisible by 2) in Pascal's Triangle are filled in (black) and the rest (the evens) are left blank (white), the recursive Sierpinski Triangle fractal is revealed (see figure at near right), showing yet another pattern in Pascal's Triangle. Other interesting patterns are formed if the elements not divisible by other numbers are filled, especially those indivisible by prime numbers. Go here to download programs that calculate Pascal's Triangle and then use it to create patterns, such as the detailed, right-angle Sierpinski Triangle at the far right.
(http:/ /ptri1.tripod.com/)
Reminds me of CONWAY, and the patterns that grow. When I see this image of the blacked out Pascal triangle and Sierpinski's triangle (attached), and relate to some of the Chain work MA and PMA highlighted, makes me think that maybe that's what we're seeing. A form where the path switches back and forth black to white (chain shifting). And along the edge, it's solid, the ROOT. The rays that shoot off are like the internal edges of the more major features, within the (fractal) structure.
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33fd02 No.4075
>>4074
This GIF is relevant to squares
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33fd02 No.4076
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8e44f0 No.4077
Hey nerds…
I have an epic fuck load of math noms for you.
Green and Blue:
https:// discord.gg/qZtpAPc
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add73d No.4078
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8e44f0 No.4079
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fba0a4 No.4080
Hello Lads! Working on factor trees over here for f. Anyone have ideas or progress today?
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82ada7 No.4081
>>4080
Running out of ideas with the tree.
Went back to the grid and added prime factor output to various test cases for d[t]-d and a[t] values. Just further researching triangle numbers. c145 example attached.
Also, the hint "if a number at position t has a factor s at (e+1)" makes more sense when you break down the values into their prime factors. Sample for e=2, d, and the d prime factors also attached.
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add73d No.4082
>>4081
There are plenty of ideas. Have you analyzed (0,n)? The significance of n and n-1 becomes clear.
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7fa95d No.4083
>>4081
I'm curious about something. Any chance you could post the results from c = 221 (13*17)?
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7fa95d No.4084
>>4083
>>4081
Oh wait, that's the same (e, n). Whoops. Any semiprime that isn't the same (e, n) and also has an a of larger than 5.
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add73d No.4085
>>4084
What do you mean?
221 is 14^2 + 25 and n value is 1
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7fa95d No.4086
>>4085
I just wanted to look at something is all. There's a pattern that probably doesn't exist in any other cell but, I mean, if it magically does then it'll be useful.
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7fa95d No.4087
>>4085
>>4086
Oh okay, I was looking at d. Jesus Christ I need to sleep or something.
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add73d No.4088
c will either appear in (0,n) or (0,n-1). That's the pattern I've seen
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8e44f0 No.4089
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179b2d No.4090
I honestly think this tree solution is extremely hard to find just due to the sheer magnitude of the possibilities. We may need to wait for our next step from VQC. I've played with this for a while and I cannot figure it out.
I devised a strategy to find it though. So we know that for each split in the tree, we should use both branches to get the correct X value for the source of the branches. If we analyze the primes, we should have that the prime branches should return us with a x value of (a=1, d-a = x) d-1. This way we could just try for those and maybe extrapolate out. Idk I'm just thinking we shouldn't try to do 145 right off the bat because the pattern must work for smaller numbers
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8e44f0 No.4091
>>4090
Honestly I think you're a goober.
You're not looking for every individual leaf and each strand of root.
They'll grow on their own.
All you need is the seed.
The Genesis Cell.
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8e44f0 No.4092
>>4091
Get it?
Goober?
Peanut?
Seed?
Nyuck nyuck.
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8e44f0 No.4093
I know what's going on.
Y'all are thinking like the wrong kind of asian.
Chakra reset ;)
https:// youtu.be/2Gfr-nVKnhs?t=1m55s
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d1075e No.4094
Found a connection between the parse tree and the grid. Not sure if relevant or what to do with this, but wanted to share.
c=6107 tree attached with an "a" formula for d nodes.
From the d value in the tree, and using the e value at the same level, you can calculate a valid "a" in the tree using the formula:
a = 2*d*(d+1) + (e+1)/2
This formula works only for odd values of e.
Also attached is a spreadsheet analysis of "a" values in (23,1). At small values of a the formula doesn't work, but above that it's spot on.
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c761fd No.4095
>>4091
We know that this is solved recursively, which means that each node can be solved (for x or x+n as VQC says) with the solution of the previous two nodes. For 145, the correct X value we want is 7. The nodes for this are 12 and 1. We know that for 1, the X value must be 0, because d=1, a=1, x=d-a. 12 is even, so we chop 12 down until it is 3. Then for 3 we have factors a={1,3}, d=1, so x={0,-2}. So somehow from these two sets:
{0,-2} and {0}
we need to get this: {7,-22}
We may be able to do a function on {0,-2} twice because of the 'chopping' of 12.
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b92360 No.4096
>>4094
Not sure what to make of this. I broke out the grid and ran your formula on some trees. This can't be a coincidence right?
13*37 = 927
23*55 = 2523
Doesn't seem to hold for evrything. Are we just proving some relationship we already know? Sooo many patterns
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b92360 No.4097
>>4096
One more
9*11 = 185
11*17 = 369
17*27 = 929
27*41 = 2249
Is this a new way to navigate the grid? Holy shit the fractals are killing me
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b92360 No.4098
>>4094
Great work PMA, still scratching my head over this one. What kind of magic causes your a formula to be the d value for higher cells in columns left of a cell in the same row?
I still fear this is just proving some obvious relationship in reverse… Am I just being a faggot or is this pattern really here? Mind blown for now
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d1075e No.4099
>>4098
Thanks for looking into this.
VQC hinted about triangles.
T(n) = (n*(n+1))/2
I bit strange that the formula for a puts 4 of them together, then adjusts for e.
>a = 2*d*(d+1) + (e+1)/2
At worst, this is a coincidence in the tree, and we've learned a bit more about the "a" value and how it works for odd values of e.
Alternatively, we need to figure out a formula for even values of e, and see if walking the tree produces anything of value.
The other hint was that the tree leads us to an x or x+n. I don't believe we know which x or x+n it is. Could very well be higher up.
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d1075e No.4100
>>4098
Apologies. I'm so lame.
Of course Teach has already figured out these formulas for a in terms of t.
>>3495
I guess all I'm trying to do is use them from the tree by substituting d for t and seeing what happens.
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b92360 No.4101
>>4100
I see, missed that post. Should read the whole thread again every so often. No need for apologies
Isn't the pattern I highlighted a different series? I don't see how we would move between columns using x, d, a without having the new e or n first. Note the first image is n=4. There are variations of this for all rows.
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b92360 No.4102
>>4101
e or b*. n is obviously constant since its the same row.
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b92360 No.4103
>>4102
Answered my own question here. Having x*n and d*n is enough.
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d1075e No.4104
>>4101
for your n=4 example at {40:4:35:12:23:55}, it does seem pretty interesting that 2*(35*36) - 4 + 1 == 2523.
Perhaps there is a "triangle" way to navigate between columns.
Also would be interesting to see if there is any factor shared between these records. Could just be numbers thrown together.
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1d5a78 No.4105
Ok guys, I'm gonna take another swing at the big picture.
So, we start with c,d,e,f (known). n,x,a,b (unknown) What about a basic scan down the e column for matching c,d? We already know there are multiple possibilities, (1,c) [which we can calculate n for], (prime a b) and we know that they have different n values. What if there are more possibilities than just (1,c) and (prime a b)? How do we find all difference of squares that fit c and d for a given e column? Thinking out loud here. Can we write code to find all matching a b for c d? Wouldn't be log n tho :(
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3a9068 No.4106
>>4105
Basically, Is there a Pattern or Golden Ratio or Fibonacci sequence to the appearance of a and b that make c? And can we then analyze them to find the (prime a b) pattern? Thoughts??
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b92360 No.4107
>>4106
>Wouldn't be log n tho :(
Undestatement of the year, any kind of searching (binary, fibonacci numbers, whatever) is completely useless. We already have algorithms to get pretty close to a factor in seconds, but pretty close doesn't cut it with numbers this huge
I know you're trying to inspire us, and since the grid is already infinite, we should assume there are infinite possibilities. Even VQC has said there are multiple ways / paths to factor a number.
Not meaning to be harsh on the internet but without an underlying pattern we're shooting in the dark, which we've been doing for months like a bunch of (really smart) idiots!
We've made progress but analyzing patterns is still what we're doing
I think we must be missing something since we've been given enough hints to solve everything and are still fumbling around, if the VQC #1 thread was 95% of it must be some kind of fractal percentage math I don't know about. All you need it addition and multiplication huh? What are we doing with all this square root and division stuff. Re-reading everything tomorrow!
Needs more triangles
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3a9068 No.4109
>>4107
Yeah, Agreed! I have faith. We haven't made quantum leap yet. What are we missing? I'll re-read all crumbs too. We gotta go to archives. Here's the links.
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
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b92360 No.4110
>>4109
Nice timestamp digits, faith is not the issue! Break your links. Adding this one
VQC+++ postings —— https://archive.fo/XmD7P
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b92360 No.4111
>>4110
RSA #0 eh, oops. I should learn to check. And gonna stop spamming the thread until I have something!
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b92360 No.4112
>>4111
Not spam, images from first VQC thread since the archive doesn't have them
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b92360 No.4113
>>4112
More relevant spam, the internet fogets you know
Save these
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1d5a78 No.4114
>>4110
Faith = happy confidence in the eventual solution of the problem under consideration.
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b92360 No.4115
>>4114
Palindrome digits, fuck yeah - loving it! Re-reading the early VQC posts now is a trip, future really does prove past
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65f089 No.4116
>>4112
>>4109
I would like to see our results consolidated, with a list of claims / theories and results. The first part of the list would include vqc crumbs (interpretations and results) and the second part would include leads that anons are following (and could reference the first list). There are old crumbs and new ones which I think haven't satisfactorily been settled or maybe even understood.
example here
>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
a= na because n=1 in that row, that's obvious. so what's the other, non obvious meaning? If it was stated, I missed it. I'll look at this old nugget now, and post the result. but if we can work as a group on an organized list, that looks a little like the baker's post but longer and with results beside crumbs… or some other way… I think it'd help.
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f2af50 No.4117
>>4113
Don't get us started, LOL! VQC was tearing up the CBTS boards with this, and we are all pretty much open to the idea here. I can't find a non-shopped photo of NP or SP. And there are plenty of cool ass rare photos like this with a hole. All the first VQC maps were mind bending and mind opening. Huge part of why I'm still working on this math quest. New Mysteries to explore here. Opened my mind to all truth, found a bunch of cool new friends, and get to work on math, while traversing the infinity of mind with you all. I'm just fucking honored to be here with all you excellent faggots. May Truth And Justice Prevail. Veritas Aequitas. May the ancient Serpent be smashed down by the heel of Yeshua. May our math success contribute to this WAR of AGES.
"The Kingdom Of Heaven Is Inside You."
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b92360 No.4118
>>4116
Me too, wish we had a wiki, lol. Can't set one up without compromising ourselves. Since we're confined here maybe BO could have a thread for it with the list consolidated in main post if he has time for that.
I don't think we've gone down the road of double meanings and coded words in a meaningful way. Will help out if we can get a useful format going
>4117
>Opened my mind to all truth, found a bunch of cool new friends, and get to work on math, while traversing the infinity of mind with you all
Love you! Is this the deepest rabbit hole on the internet right now? Go anon!
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65f089 No.4119
>>4118
>I don't think we've gone down the road of double meanings and coded words in a meaningful way.
I really mean, what hint is this? Not double meanings.
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b92360 No.4120
>>4119
Not something from VQC :)
I think I lost a zero in my post number just now
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65f089 No.4121
>>4120
>I think I lost a zero in my post number just now
?
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b92360 No.4122
>>4121
I was drunk enough to think my post would be >>4200. Yeah. Lets pretend I made a binary joke or something
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7fa95d No.4123
>>4118
That was the original point of the board: to more eloquently organize everything, have a list of thread archives, etc. Then /cbts/ turned to shit. We've just sort of been winging the structure of threads and everything so far. Of course you're more than welcome to put something like that together, and I could sticky it if everyone would like. I don't really have the time to go through everything myself at the moment. My life is quite busy and stressful lately (not that anyone needs my life story but one of the things is that I'm moving out of my parents' house quite soon) so while I can easily lurk and maybe make some bitmap images occasionally, someone else is going to have to get all of that information together. I did put some pdfs together early on for the people who knew nothing about programming or how the grid worked. There's a thread somewhere. Maybe that information would come in handy too.
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65f089 No.4124
>>4123
Yes you are right, I intended to do it earlier but I didn't get around to it. I'll have to chip away at a few of them, and maybe do a mini list. Expand if useful; hard to know which loose ends are most important right now.
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b92360 No.4125
>>4123
Winging it is fine but organization has its merits too. I will reread everything and cherry pick some stuff but all I can do is post that as a new thread and it'll be outdated when anons contribute. Editing the top post would be nice but its not like we can just ask for mod rights.
Thanks for all you do, its appreciated by all of us! And good luck with your move/rl stuff!
Any anons with organization skills are welcome to step up
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b92360 No.4126
>>4123
>>4125
Just to add, the foresight in getting the hell out of /cbts/ early really saved us. Thanks!
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06387b No.4127
>>4120
>>4119
You almost got the 4120 for your post of the earth smoking a bowl, lol. :)
>>4123
Thanks again AA for planning ahead!
Thought: VQC is further along the road of enlightenment, so he doesn't party for good reason. We disciples, on the other hand, were probably attracted here in part by his self which used to party. You gotta like expanding your mind for this kinda work. Pretty cool that he's doing it sober now. Much respect.
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06387b No.4128
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8e44f0 No.4129
Rules? As in measurements or modes of behavior?
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06387b No.4130
>>4129
Rules? As in "nature, in order to be ruled, must be obeyed" ?
Dance Of The Cosmos?
Numbers That Grow Like A Tree?
Patterns That Traverse Infinity?
Underlying Order Of The Universe?
Sign Me Up, I wanna be there for that Habbening.
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65f089 No.4131
>>4127
>You almost got the 4120 for your post of the earth smoking a bowl, lol. :)
What if my post was actually a hint, because inhaling smoke into the pole creates a toroidal vortex?
I just realized the pdf here is as useful as anything.
>>3381
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65f089 No.4132
ok still gotta work on the hint I was asking about, but… reading the pdf and getting excited here by the imagery:
>Integers are also related into families that you have noticed.
>Like fractals.
>They repeat.
>They GROW.
>Sea shells grow.
>Establishing which patterns govern which level of the fractal and which part of the branch of that fractal, will help identify
the factors of c.
>Think decision trees too.
————————————————————————————————————————
>If in doubt go back to the original (e,n) grid and look for MORE patterns!
>The strange is, the more you enjoy this the more it SINGS it's beauty.
>It IS beautiful.
>Like the Mandelbrot Set.
>Similar rules.
————————-
What kind of logarithmic spiralling fractal mushroom trip are we looking for here? Rules of mandelbrot set: The Mandelbrot set is the set of complex numbers c for which the function f_c(z) = z^2+c does not diverge when iterated starting at z = 0. We know that VQC's part 2 factor tree iterates taking roots, but for all sub-branches. The mandelbrot set looks at sequences. But it is self-similar (download / write yourself some fractal viewing software for inspiration).
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7fa95d No.4133
>>4125
>>4126
>>4127
Do keep in mind, the reason we left /cbts/ is because a human started doing dumb human stuff that interfered with the function and culture of the board, and I'm also a human. Maybe I don't think I'm going to start doing anything shitty, but I can't imagine anyone who controls this kind of thing ever does, and sometimes they do. You all have the responsibility of keeping me in check.
>>4131
I didn't think to put the Twitter DMs that some of us have posted screenshots of into that pdf, but the images in the archived webpages don't work. It would be a good idea to get those messages together (maybe with their context in time too if possible), if everyone who has been messaging Chris on Twitter is reading this and could post them again.
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7fa95d No.4134
>>4131
>>4132
By the way, if you didn't already notice, there's an image version here (featuring some extra posts and possibly missing a few) >>3453
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b92360 No.4135
>>4132
>What kind of logarithmic spiralling fractal mushroom trip are we looking for here?
Would love to know. If I weren't already fucked up in the head, getting a My Little Pony set from the toy store (they sell these right?), adding a random generator and some fancy rotation code to the 3D VQC and taking too much acid could provide insights. Maybe make a slideshow out of Topol's art, not sure I would ever come back from that though
>>4133
Those damn humans…
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a8eda6 No.4136
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5cbb4c No.4137
So, re-read a few crumbs and had a breakthrough this eve!
Finally understood the problem: big square, little square, to the point I can picture the whole thing in mind clearly now.
That enabled an excel model to be put together to brute force it, which is working. Not efficient, but a step on a path.
It takes about 1 min to factor a 10-digit "c". Lots of extra functions and lookups being done that aren't needed for display purposes, and about 5000x more computationally intensive than needed (see comments below). Ironically, the our beloved 145 case takes about the same time, as it does what it does, isn't smart.
Attached are a couple examples (see pics):
c=2,171,394,821
e=71,980
a=42,043
b=51,647
Small square = 42,043
Large square = 4,802
and
c=9,806,432,891
e=111,893
a=91,583
b=107,077
Small square = 7,747
Large square = 99,330
Some of that time was to calculate "e" which wasn't even being used.
But, knowing the "e" column, thus providing the d-pattern(s), would drastically reduce the table created (this proto just did linear D, still works).
So look at pic in post >>4097 for n=1, in any column, the D value is already around 2250 by the 33rd row (element). That's a 70x reduction in the search space. And matching 2 columns as done in this proto, that means it's 70x squared, or a 5000x reduction. Get it out of excel, this thing should humm.
Hey VA - you could try this in excel, screencapped the formula for you.
Here's the algorithm if someone wants to code it, should be straightforward. One column in the table is D, next column is D^2, and next is C+D^2.
Basically, just using the INDEX_MATCH function. Return row in COL_A (the "D"), where D^2 (COL_C) = C+D^2 (COL_M). That is the Large Square, returned in COL_N (have extra columns hidden not needed for this from when exploring how e=0 grew). Then, COL_O is the D for that row. Once you have Large and Small square, just a matter of plus and minus to get "a" and "b" (and use the "" if Cols N and O are blank).
In the excel model, Columns R, S, T, U have the same info in first row, just using a MIN function to find them, as it's hard to scroll through 100,000 rows to find where the match occurred.
Big night with Q too - takedown in China (with juicy intel re: the window and techniques) and big pharma treadmill vs cures. YUUGE! Feels like a winning day all round!!!
Thanks for the motivation everyone!
Think we've got this now!!!
Also believe VQC let us go down the triangle and other paths, as even if we don't need that for step 1, we're going to need it soon enough, and we're learning math, and that's one of the goals.
BTW, everyone should read about the singer Chris Curtis as listed here: >>3862
That VQC is one clever Kiwi!!!
Not only a band called "The Searchers", read the whole archive, it's fun. Here's a snippet - note the tin-foil story, perfect for us conspeericy folks!
"Curtis's concept was a band with a core of three members: Curtis, Lord and Robbie Hewlett. The other musicians would be engaged whenever the core felt like it. "They would jump on and off the roundabout. But I left that party in a new band, Roundabout." said Lord.
Curtis would arrange for Daimler limousines to taxi him about and was charging the cost to Tony Edwards. Edwards realised that he had made a mistake agreeing to manage Curtis but he liked what he saw of Jon Lord. And Lord was also having problems with Curtis, who had started to use LSD. Lord returned from a few days away with The Flowerpot Men to find the entire flat covered in aluminium foil. Everything; even the furniture and the light bulbs.
Curtis moved out soon after this but he did tell Lord that he had a guitarist in mind for Roundabout. Curtis arranged for Ritchie Blackmore and his girlfriend to fly over from Hamburg, Germany and meet Tony Edwards. The meeting was a success for Blackmore, Edwards and Lord but they had no room for the erratic Curtis. They changed their name to Deep Purple and their first single was Joe South's "Hush", which Curtis had been playing in Lord's flat for months."
>>4117
>>4136
HERE HERE HERE!!!
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5cbb4c No.4138
>>4137
oops, late, types that one wrong, should be:
c=2,171,394,821
e=71,980
a=42,043
b=51,647
Small square = 4,802
Large square = 46,845
Attached 2 crumbs that led to this, as listed in the MAP above. One was a DM - please do share your DM's anons!
"Where we go one, we go all!"
"The Kingdom Of Heaven Is Inside You."
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add73d No.4139
>>4137
You don't need to search. Calculation is possible
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5cbb4c No.4141
>>4139
Right.
And in:
[71980,1]
where exactly are the 4802 and 46845, which should be 42,043 elements(?) apart.
And for:
[111893, 1]
the 7747 and 99330 values, which are 91583 apart (providing a).
^^^
I think the e's listed may be the f's, not to the floor (easy to adjust). But need to find in the cells.
Can use 115 (5x23 with e=15) to validate.
That's next, then back to the calcs.
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add73d No.4145
Good morning to all! Fresh bread.
I included my self-contained recursive tree generation code.
>>4140
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add73d No.4146
>>4141
na and nb for any c are n elements apart.
You might find my traversal code useful.
https://anonfile.com/TeH6q3d8bd/VQCGUI_v2.7z
The source code is included, I only put it in a JAR because there are a lot of classes and it has a neat theme
Example:
145 = 5 * 29
5*5 is 5 cells apart from 5*29 in (1,1).
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add73d No.4147
>>4116
Will do. We've discovered a lot. For example, the parity of n is known. Plus we've invented a ton of new algorithms, though none (yet) have beaten sieving. Another is that if gcd(e,d) isn't 0, it is a factor of c.
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add73d No.4148
>>4147
1* I mean. I'm going to love explaining to the cryptographers how it works when we find the solution.
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7fa95d No.4149
>>4148
That makes me think; how are we going to explain this to the world when it's done? Should we try to make some kind of long and concise text-based explanation, or should we turn it into a video with graphics and someone brave's voice, or what? I kind of just went with the text-based thing when I was trying to help the lurkers who can't code etc however long ago that was but it would be better to discuss and organize it rather than just having someone like me decide that they should do it all.
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add73d No.4150
>>4149
Yeah, I think so. There's gonna be a LOT of heat, probably in the news too.
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add73d No.4151
>>4149
We need to say it loud and clear, ANYONE can be a GENIUS!
I will definitely make an explanation of the VQC, fractals, the tree, and the final piece that ties it together when we find it.
Needs to be a picture too, so it spreads far and wide :-)
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8e44f0 No.4152
>>4149
You make it an open class with video series and different levels for people who comprehend this stuff vs people like me who have no earthly idea what you're talking about. Programming? Primes? Making the Matrix? The hell is a VIRTUAL Quantum Computer? Etc etc.
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7fa95d No.4153
>>4150
It's going to be really funny when mainstream news anchors try to explain how it happened. Every so often when a big tech thing happens I look for the reaction from TV news places just to watch them bumble around not knowing what the fuck they're talking about, e.g. when they talk about darknet drug markets, or the hacker known as 4chan.
>>4152
>people like me who have no earthly idea what you're talking about
This has been my thought since the start. The whole point of this disclosure is to make people aware that we have a fundamental misunderstanding of how mathematics works and to hopefully change the world for the better. When it is ready, it'll most likely seem like something only scholarly PhDs can grasp to everyone who hasn't been following along like us, at least at a surface glance. If we could explain everything from the ground up as simply as possible, it would greatly help in making this common knowledge instead of just being something that gets studied at high levels in universities behind literal and figurative closed doors. It's just a question of whether it would be enough to type it all out or if we'd need to put videos together (and if we put videos together instead of typing it all that implies that there wouldn't be a lot of reading involved, which implies that one of us has to record their voice, potentially de-anonymizing themselves), or if we could have a big infographic (it would have to be huge and there would likely be a fuck ton of text anyway). Any ideas?
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8e44f0 No.4154
>>4153
For starters, I'm about as anonymous as the sun, so if need be I can do the voiceover.
Second, i'm not completely clueless considering that I've gotten ahead of you guys on multiple occasions… not that I understood why or how but my head was in the right spot. This isn't rocket science. It's SecondLife Science.
As for how to essplain to the laymen, that's going to be infinitely easier when the equation/Genesis Cell/algorithm is finalized. You start by showing them how the Genesis Cell expands… make it a self contained download package/zip/rar with everything needed to operate the VQC for even the most computer illiterate. Plug'n'play.
Then we'll have TL;DR videos for the basic gist of it, then intermediate videos for the Math Savvy, and then advanced courses for the Programmers. Like I said… make this a class.
Show people what REAL education looks like.
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add73d No.4156
>>4154
We're educators now? I love it
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8e44f0 No.4157
>>4156
Been sayin' that for a while… lol
>>865
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a8eda6 No.4158
>>4154
That's a really cool idea man. I like the video idea. This is probably the most positive and up beat room on the internet. :)
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add73d No.4159
>>4157
You can create (e,n,t) w/ pen and paper.
You can calculate (e, N, T) w/ pen and paper.
You can calculate (e, 1, T) w/ pen and paper.
No programming necessary. I'll write down the equations in a sec.
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1d5a78 No.4161
>>4137
Thanks MM! I'll run your formulas later today!
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1d5a78 No.4162
Lol, Meme of the day so far over on /qresearch/
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add73d No.4164
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8e44f0 No.4165
This is just a reminder that we'll need new bread in about 10 posts or so.
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8e44f0 No.4166
OOOH! OOOH! I FOUND ANUDDER ONE!
https:// en.wikipedia.org/wiki/Centered_heptagonal_number
AND FUCK IT! I FOUND MORE ON TOP OF THAT!
https:// en.wikipedia.org/wiki/Category:Figurate_numbers
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3a9068 No.4170
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8e44f0 No.4171
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d1075e No.4172
>>4170
Lol. It works. Except when it doesn't!
VA - thanks for posting these and keeping in touch with Chris.
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add73d No.4173
>>4171
Open the catalog.
>>4170
I've already seen somewhat of how 0,n and e,1 are the same. You'll see that (0,n) or (0,n-1) contains c, and e,1 of course has the pattern of factors of n and factors of n-1.
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f2af50 No.4177
NEW BREAD
In case anyone missed Baker's earlier post:
>>4145
Thanks Baker!
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7fa95d No.4180
Don't mind me, just posting 6 more times so the thread's full and I can sticky the next one.
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7fa95d No.4181
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7fa95d No.4182
Different words so I don't trigger flood detection
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7fa95d No.4183
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7fa95d No.4184
Text text text text text text text text text typing typing typing typing typing typing words words words
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7fa95d No.4185
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