2dbaf3 No.2555 [Last50 Posts]
Virtual Quantum Computer
Definition
The virtual quantum computer (VQC) is a grid made of infinite yet constructable sets that follow a known pattern. Like a quantum spreadsheet.
The grid is the superposition. The collapse of that superposition will be two input parameters, d and e which can be calculated easily for all integers, c, where c is the difference of two squares. Its purpose and our goal is to be able to factorize large semiprimes, all the way into the hundreds of digits, which would break the RSA cryptosystem.
When the integers that are the difference of two squares are arranged into the grid and their corresponding properties are shown, a pattern emerges that shows calculation instead of searching is possible.
Variables
The map's legend is {e:n:d:x:a:b}, where d is the result of removing the largest square from c AKA the square root,
e is the remainder,
n is what you add to d to be exactly halfway between a and b,
and x is what you add to a to make d.
c is any number that is the difference of two squares, so odd numbers are included. It is the number you want to factor.
f is what you add to c to make a square.
t is the third coordinate in the VQC..
Rules
Each cell of the grid (e,n) has infinite elements or ZERO elements.
Each cell with one value has infinite elements, since every element can make a new one.
By induction, a cell only needs one value to make infinite values, that's part of the power of this and is why it is a virtual quantum computer as a whole.
The t variable is what will allow you to walk across these infinite elements.
If a grid cell has elements, all elements are constructable from a finite set of root elements.
The grid is indexed using e, n, and t, where e is the rows, n is the columns, and t is the specific element in the cell-group.
Thus, only three variables are required to identify an element: e, n and t.
All products of odd numbers and all products of pairs of even numbers are the difference of two squares.
The x-intercept of the line that goes through the point containing the factors of c is (a + 1).
(1, 1) - the key
The values of a and b at 1,1 are related to the length of the longest side in right angled triangles.
The values here can be used to create the entire grid.
The values here determine the values of the rows to the left and right, which determine the values of the whole column.
Columns
Each cell at n=1 contains the roots of products in the column.
If c is a prime number, it will appear in one column exactly once.
If c is the product of two prime numbers that do not equal eachother, c will appear in two cells of one column.
All products (integers) c that are the sum of two squares appear (only) in columns where e=0,1,4,9,16,25…
All factors in a column are factors of the elements of the first cell in their column.
All Fermat primes (except) 3 appear in column one.
(e, 1)
(e, 1) = the genesis cell.
If a number at position t has a factor s, then s is a factor at (t+s), (t+2s) and so on for a at (e,1).
Also, if a number at position t has a factor s at (e+1), then s is a factor at (s+1-t), (2s+1-t), etc for a at (e,1).
n*a and n*b for any c can be found n places apart in the cell at (e,1).
(1, n)
The cells in row one where n=1 have a relationship with the cells 2n to the right and 2n to the left.
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
Each element in a cell can be generated by moving up (t-1 = x-2) or down (t+1 = x+2). Other variables can be generated from x.
Useful Equations and Notation
ab = c
dd + e = c
(d + n)(d + n)-(x + n)(x + n) = c
a + 2x + 2n = b
a = d - x
d = a + x
d = floor_sqrt(c)
e = c - (dd)
b = c / a
n = the difference between the square root d and the larger of the two squares
n = ((a + b) / 2) - d
d + n = number that is exactly halfway between a and b
d + n = i
x = d - a
x = (floor_sqrt(( (d+n)*(d+n) - c))) - n
x + n = j
f = e - 2d + 1
t = the variable that lets you traverse the infinite elements in for a given (e, n) that has values.
if (e is even) t = (x + 2) / 2
if (e is odd) t = (x + 1) / 2
____________________________
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2dbaf3 No.2556
Code
C#
BigInteger Square Root —— pastebin.com/rz1SdACZ
Generate Bitmap within original code —— pastebin.com/hMTtJF6E
More on generating a bitmap with the original code —— pastebin.com/JUdtehb4
Generate the large square for e and t —— pastebin.com/nbjs2kz4
Original VQC code —— pastebin.com/XFtcAcrz
How to run VQC code on Linux —— pastebin.com/6HnN7K5X
Unity Script —— pastebin.com/QgAXLQj3
Unity Script 2 —— pastebin.com/Y38nVWgT
Java
Create a Bitmap using the VQC Generator —— pastebin.com/Dgu9aP1h
VQCGenerator —— pastebin.com/VMRnkXFP
Traverse the VQC cells in real-time —— anonfile.com/W44cofd6b6/VQCGUI.7z
Traverse the VQC cells in real-time [V2] —— anonfile.com/TeH6q3d8bd/VQCGUI_v2.7z
NodeJS
BigInteger Library and Sqrt —— pastebin.com/y8AXtFFr
Python
College Anon's code (VERY USEFUL) —— pastebin.com/d8xZZnm0
Create the VQC —— pastebin.com/NZkjtnZL
3D VQC —— pastebin.com/vdf8SpYt
3D VQC (v2) —— pastebin.com/wZM5Thzu
Fractal cryptography —— pastebin.com/XuN4U7Dv
Generate cells for a (and more) —— pastebin.com/iAizgLFF
Generate any cell in (0,1) and (0,2) —— pastebin.com/gRTYpdMU
Generate genesis cell —— pastebin.com/GKzcCpMF
Generate positive AND negative genesis cells —— pastebin.com/9ixjRyxt
Calculate variables based on e and t —— pastebin.com/4s6McdbN
Get A and B from C and N example —— pastebin.com/s0SZ9BNF
VQC + t —— pastebin.com/Lgufk0db
Rust
Check if a number is prime —— huonw.github.io/primal/primal/fn.is_prime.html
Create Bitmap using the VQC Generator —— play.rust-lang.org/?gist=c2446efeec452fe14e1ddd0d237f4173&version=stable
Create Bitmap using the VQC Generator [V2] —— pastebin.com/zGSusyz5
Additional VQC code —— play.rust-lang.org/?gist=50def916ad48400bc5d638fbf119ae85&version=stable
Generate the VQC —— play.rust-lang.org/?gist=6b6beb372b6b931f1abd30642a35a80c&version=stable
Solutions (Java)
Binary search for i —— pastebin.com/TAt5bDsR
Calculate factors using -x jumps —— pastebin.com/gKX9GW9r
Previous Threads
RSA #0, or the VQC thread —— archive.fo/XmD7P
RSA #1 —— archive.fo/RgVko
RSA #2 —— archive.fo/fyzAu
RSA #3 —— archive.fo/uEgOb
RSA #4 (not finished, but dead) —— archive.fo/eihrQ
RSA General (#5) —— archive.fo/Lr9fP
RSA #6 —— archive.fo/ykKYN
Videos on cryptography —— pastebin.com/9u3hwywe
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2dbaf3 No.2557
Chris' posts
(RSA #2 was shoahed)
RSA #3
>>>/cbts/87168
>>>/cbts/87234
>>>/cbts/87300
>>>/cbts/87378
>>>/cbts/87414
RSA #4
>>>/cbts/98492
>>>/cbts/98560
>>>/cbts/107338
>>>/cbts/107342 rt >>>/cbts/107256
>>>/cbts/111903
>>>/cbts/111975 rt >>>/cbts/111942
>>>/cbts/111983
>>>/cbts/112148
>>>/cbts/112422
>>>/cbts/112425
>>>/cbts/112429 rt >>>/cbts/112172
RSA #5
>>11
>>12
>>17
>>18
>>19
>>20
>>21
>>23
>>24
>>25
>>26
>>27
>>28
>>29
>>30
>>31
>>32
>>33
>>495
>>699
>>709
>>710
RSA #6
>>1099
>>1380
I'm working on the new VQC crumb map.
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6f6ea0 No.2561
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ff4dbe No.2564
I'm only going to shill this once since it isn't RSA-related, but I know some of you might be interested.
>>2563
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87160d No.2567
Tasty Bread! Thanks Baker!
>>2506 Made this for ya'
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312ec3 No.2568
>>2567
Its 0+0 right? fuck
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ab8fb8 No.2573
>>2568
The solution? Lol.. we got this.. just gotta think trees.
Trees.
Trees.
Trees of numbers.
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87160d No.2574
Hello PMA! Can you update me on your ideas for how we can use a=1b=C, CC, AA, BB, etc to jump records? Also if anyone else has working formulas or programs for using these to make the jumps between records, jump in too!
On a separate note, what's the latest on mirroring x, lads?
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ab8fb8 No.2575
>>2574
Run my program. I'm working on the fact that it's n! speed right now, lol.
Any ideas? I'm going to try looking for other patterns like the x jump. I was completely unaware walking down the tree was as simple as it is.
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87160d No.2576
>>2575
Lol! Thanks for the deep laugh, Topol! That was just what the doctor ordered.
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ab8fb8 No.2577
I just realized something.
The amount of jumps is n*a.
Like VQC said.
We can use it because we know n*a.
Oh.
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3ed644 No.2578
>>2574
Was enjoying the -x discussion earlier and tinkering in the background.
I have made a little bit of progress, but nothing that I'd call finished.
Pictures attached are of the (0,1) and (1,1) -x space for c=65.
In the left columns, you'll note (abc), (a), (b). These represent where the a, b, or c values from our prime solution are factors of the a value in the record.
Interestingly, if you navigate directly to (e,n,t) at (0,1,-c), this is the first occurrence of an a value that has abc as factors.
Based on these pictures, I don't think we have to do much searching at all. The a factors are literally everywhere. Pulling them out is a different story.
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ab8fb8 No.2579
>>2578
If you take any c, and generate the 1*c cell, you can go n down and find a number whose second factorization contains the a you want for c.
145:
start with {1:61:12:11:1:145}
go down n (61) entries in the cell
and you get {1:61:278:133:145:533}
145*533=77285
all you have to do is find out how to get the second factorization of 77285, and you have a:
5*15457 = 772851
145 / 5 = 29
5*29 = 145
Second example: 533
{4:244:23:22:1:533}
getting to this cell is trivial. now go 244 cells down.
{4:244:1043:510:533:2041}
533*2041=1087853
now all you have to do is get the other factorization of 1087853 to factor 533
1087853 = 13 * 83681
533 / 13 = 41
13 * 41 = 533
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ab8fb8 No.2580
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455aad No.2581
>>2580
Based on that I wrote this function. It takes in c and number of steps, then it iterates down n and uses gcd to attempt to find the factors.
def goDownSteps(c, steps=2):
e, n, d, x, a, b = rowForAB(1, c)
bb = b
for i in range(steps):
ee, nn, dd, xx, aa, bb = rowNegX(n, -(2 * i * n + x), bb)
g = math.gcd(c, bb)
if g != 1:
print(g, c / g)
break
Unfortunately it's still way too slow.
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ab8fb8 No.2582
>>2581
Can you write a function that just gives me the n values down jump? I'm not good with t.
A function that will give me 145*133 when I input 145, because all you have to do is jump n down from 1*145.
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455aad No.2583
>>2582
You mean 145, 533?
That's what rowX already does.
>>> rowForAB(1, 145)
(1, 61, 12, 11, 1, 145)
>>> rowX(61, (2 * 61 + 11), 145)
(1, 61, 278, 133, 145, 533)
The code for rowX is:
def rowX(n, x, a):
b = a + 2*x + 2*n
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)
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455aad No.2584
I've been thinking about what f means and the (-e, n) records and it's quite obvious.
I'm just writing this as I'm thinking loud, but of course (-e, n) records contain the same info, but with d + 1, n - 1 etc.
It's because it's the big square (d + 1)^2.
e is the remainder after removing d^2 from c. So negative e is the "remainder" after removing the upper square (d + 1)^2, or rather the number of empty parts.
n - 1 makes sense because we are now 1 square closer to the n we need to find the factors. Increase of d again makes sense because d represents the square.
Same as x + 1 because n is now (n - 1) so we need to increase x by 1 to maintain the same number (n + x).
I don't see any obvious way to make use of f.
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455aad No.2585
>>2584
So if we extend the idea of f, we can do
e - 3*(2 * d + 3) which should give us n - 2, d + 2, x + 2.
e - 4*(2 * d + 4) which should give us n - 3, d + 3, x + 3 etc..
This is because we are increasing the squares. Note though: This isn't how we can factorize as this is the original fermats factorization.
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ab8fb8 No.2586
>>2585
>>2584
We ARE doing Fermat's factorization.
Just a super advanced, instant version.
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455aad No.2587
>>2585
Wait, I think that's wrong. e - 3*(2 * d + 3) should give us n - 3, d + 3, x + 3 since f = e - 1*(2*d + 1).
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455aad No.2588
>>2586
Well yeah, but the e - 3*(2*d + 3), e- 4*(2 * d + 4) etc is the slow, original method.
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88565a No.2589
>>2579
These have more than 2 factorizations, but good idea if you can find how to do it
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ab8fb8 No.2590
Somebody tell me how to mirror a cell.
I think that's the solution. Every cell on the positive e side has a counterpart on the negative e side.
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ab8fb8 No.2591
>>2590
Found it, I'll work on it tomorrow.
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455aad No.2592
It feels like we're so close. All we need is one last nudge from VQC.
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ab8fb8 No.2593
>>2592
I believe the answer is in (e,1).
Look at my picture of the x jump factorization of 145 and compare to (1,1).
You can calculate it, I bet. Brainstorm. I need some sleep.
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99eff8 No.2597
>>2592
VERDE QUANTUM CODESMONKEY!
WHERES DIDS YOUS GOES?!
(Granted… he did this… you all can do it….)
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51b282 No.2598
Wasted a lot of time on (e,n).
I think that the solution is back on (e,1)
Exploring relation between (2c,1) and (2Nc,1)
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455aad No.2599
>>2598
Given the tweet from vqc yesterday Im sure he thinks we're close.
Maybe we found the last pattern, but we continued instead of studying it?
Maybe we should look over the past two days and recheck the patterns discovered?
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51b282 No.2600
>>2599
I'm sure there is value in (e,N) but it take massive iteration to find records. I have not found a computational pattern.
Looking at VQC's hints they remain focused on (e,1) on (n,0) and on factor trees
a(t=1) equals c ….
for (2c, 1) and 2c-1,1)!!!!!!!!!!!!!!!!!!
.
The a(2c-1, t) factor tree has twice as many a,b as factors as does a(2c, 1, t) tree.
I'm doing side by side looking at different multiples of e = 2c and (2c-1)
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87160d No.2601
>>2593
Nice work last night on n and na, Anon! I also think (e,1) is the key. How do we USE (e,1) to get what we need in higher values of n? Working to solve that over here, still haven't made the breakthrough. Hopefully we'll find it soon.
>>2592
Hello Isee! Thanks for all your work, anon. We are getting close. Let's keep going!
>>2578
Hello PMA! Thanks for the clear explanation and great output as always. So the left column is used in this example to search for factors of c=65? Cool. Then you can change your search to whatever factors you want. Pulling them out will be the next step?
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ab8fb8 No.2602
>>2601
→is current cell
↓is the cell after a lower -x jump (see defined method)
↑is the cell after a higher x jump
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51b282 No.2604
>>2603
Quiz
for what two values of e
does a(e, 1, 1) = c??
Is this useful in analyzing factor trees?
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51b282 No.2605
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87160d No.2606
>>2604
>>2603
Hmmm... Quiz, Ok!
Q1: (1,1) and (1,2).
At (1,1) we have a,b factors of 1,5 and c=5
At (2,2) we have a,b factors of 1,3 and c=3
Q2: Seems like a repeat of Q1?
Yes, this is very useful for analyzing factor trees. Not sure what you're hinting at. Have you found something new to report, Anon?
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3ed644 No.2607
>>2604
>>2606
a(e,1,1) = c
can be found at both:
(2c,1,1)
(2c-1,1,1)
>>2603
(e,1,1) = c I'm having a bit more trouble finding.
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51b282 No.2608
>>2606
>>2607
Sorry, deleted first post forgot the a.
I was hoping to generate some interest in analyzing factor trees. VQC gave a few hints regarding them.
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51b282 No.2609
Deleted post for first time found myself in alternate universe full of deleted posts. Found a VQC post and got excited until I realized it was a deleted post!
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3ed644 No.2610
>>2608
shit… any here I was thinking all my code was buggy.
So you're confirming that the relevant records are at (2c,1,1) and (2c-1,1,1) correct?
Do you have any additional insight into how they are useful?
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ab8fb8 No.2611
>>2610
Knowing how the factor trees work in the VQC lets you factor it.
When you make an xjump from a 1c cell you get a number which has a for c as a factor.
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87160d No.2612
>>2608
Definitely interested over here in factor trees. I've been working for weeks to understand the crumbs about this. Kinda stuck.
>>2610
Can you guys quickly explain your notation? Are we using (e,n,t) as usual? If so, what does a(e,1,1) =c mean? I'm working to follow.
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3ed644 No.2613
>>2612
my understanding: a(e,1,1) means the a value for the record (e,n,t)
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ab8fb8 No.2614
>>2612
>>2613
As if it would kill you to type "a of (e,1,1)".
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3ed644 No.2615
>>2614
dude… wasn't me. But I like your notation better!
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87160d No.2616
>>2613
Hmm. Ok, that's what I thought. Thanks!
>>2614
No worries over here. Just wanted to make sure I could follow along.
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ab8fb8 No.2617
>>2616
Just keep working on it. We're closer than we've ever been. I finally understand the na is n apart crumb.
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3ed644 No.2618
quick question. if we have multiple records with a values that share factors a, b and c, all with different values.
Then divide the c out, we're left with some combinations of a*b.
With 2 records, with different remainders, can we solve for a?
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ab8fb8 No.2619
>>2618
Give me an example. I think this is how you factor it.
Each cell at n=1 contains the roots of products in the column.
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ab8fb8 No.2620
All factors in a column are factors of the elements of the first cell in their column.
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
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3ed644 No.2621
>>2619
>>2620
ok. Just an example:
for c=145
a of (290,1,-289) = 168345
a of (289,1,-290) = 168925
both share a,b and c as factors.
can we discern anything from this?
There are so many ways of getting to records that share a,b,c as factors.
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51b282 No.2622
>>2618
It could but as t gets bigger more extraneous factors get added.
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ab8fb8 No.2623
>>2622
>>2621
Yes, there's somewhat of a wall when you try to factor it by moving around and ignoring the e,1 cell, because you can't "go back." You can go up the branch of the tree and get more numbers who share prime factors of the c you are using, but you can't get the factors. I really think we just have to pay attention to e,1. e,1 contains the factors for every x jump from 145.
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3ed644 No.2624
>>2622
I understand, I expanded the search to 10*c*n just to pick up more data.
>>2623
yeah, screen shot was testing cc, 2c, 2c-1, and various other (e,1) combinations.
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455aad No.2625
>>2618
Gcd will find the greatest common divider
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ab8fb8 No.2626
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455aad No.2627
>>2625
Given two numbers a, b where a factorized is k, j and b factorized is j, l and k, j and l are primes gcd(a, b) will return j. It's also in On(log n)
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51b282 No.2628
>>2621
Any multiple of c will contain at least the same multiple of a and b. Since you still don't know a and b we are no better off.
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455aad No.2629
>>2627
O(log n) stupid auto correct
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99eff8 No.2630
I might be tardy but I thought I noticed that in one of the VQC crumbs on the map said something about only needing addition and multiplication?
I could have misunderstood.
I think it was in one of the twitter grabs.
Also, gimme some shit to fuck wit.
Where'd the pretty visuals go?
HOLIDAY'S OVER!
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3ed644 No.2631
>>2628
thanks. back to digging.
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ab8fb8 No.2632
>>2628
I disagree that we are "no better off." Have you learned nothing from Fermat? And yes, that's a very important rule, that multiples of c will contain a and b.
>>2629
>>2627
O(log(n)) or O(log(n)*n)? If it's in O(logn) we can use it.
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99eff8 No.2633
>>2632
I keep seeing Oolong.
Tae, anyone?
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51b282 No.2634
>>2620
I thought a lot about this hint. In row one n = 1 so na is just a. There was also a hint about subtract d from a to get n-1. Well at t=1 a = d and n=1 so we get a=d=n=0. If we subtract 2d we get a n = -1.
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51b282 No.2635
>>2623
I was looking at some of the records you were generating. Didn't follow all of them but they tested. Do you think you give me a couple (30, 98) records
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3ed644 No.2636
is it possible that the records we are searching for in (e,1) have a values equal to either a^2 or b^2?
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3ed644 No.2637
>>2636
scratch that. doesn't apply everywhere.
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ab8fb8 No.2638
>>2636
>>2637
Just because it doesn't apply everywhere doesn't mean we can't make it apply everywhere. What did you learn from factoring c^2 instead of c?
That you can make e = 0.
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51b282 No.2639
>>2636
theoretically we know where the aa and bb records are. I've looked at odd next to even. I've looked at 2Nc versus 2c records but if you just know c the a and b remain hidden. I think Chris has some magic sauce that will seem easy when someone figures it out.
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3ed644 No.2640
>>2638
>>2639
ok. here's what I'm looking at, still for c=145.
starting position at (e,1,1) where a = n*c
(1,1,67) = {1:1:8978:133:8845:9113}
factors of a are a,b,c
negative x move from original (1,61,4)
(1,1,-66) = {1:1:8712:-133:8845:8581}
factors of a are also a,b,c
These search boundaries can be created for any (e,1).
Between these t=67 and t=-66, there are 2 instances where a = a^2 (at (1,1,-3) and (1,1,4)), and a = b^2 (at (1,1,-20) and (1,1,21)).
Not sure what to make of these, as I didn't find them in other examples.
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ab8fb8 No.2641
Just to give you an idea how little the world understands about factorization, How RSA 155 was solved:
pastebin.com/8BnqvvtT
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51b282 No.2642
Ghost in our machine??
For e=2c-1 and factor tree A of (e,1, t) and c=a*b
Using A for value of a in tree because its bigger than c=a*b
a is factor of A when t/a is integer or has remainder 1
b is factor of A when t/b is integer or has remainder 1
c is factor of A when t/c is integer or has remainder 1
here's the weird part when a is a factor of t but b has a remainder of 1 (or vice versa) c is
still a factor so c is a factor more often than a or b
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d86709 No.2645
>>2633I love Oolong Tea! Is Teach still working here? Haven't seen him back in a while. Iv been lurking. Will do another chapter and some cleanup tonight if I can think of a gripping topic. There seems to be a major lull in the action here and on the CBTS/Thestorm boards. Loving what you guys are making! Sad that I can't contribute more. Keep up the good work!
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87160d No.2648
>>2645
>>2633
Thanks Topol for the fun as usual, and thanks Hobo for updating your project. Sorry we don't have any cool new visual output for you guys! The vibe here has been off too. We're missing MA, CA, Teach, Baker, AA, and the BruteForce anons! I know I just got over being sick for almost three days, so maybe a few people are under the weather.
In the attached pic, note the value for var e, remainder. It's freaking huge!!! Baker, I know you love running RSA thru your shit. Is it normal to get huge e values like that?
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ab8fb8 No.2649
>>2648
Yes, they're huge. If e wasn't large enough the number would be vulnerable to primitive versions of Fermat's factorization. You don't really have to worry about how big RSA numbers are. Just solve it for humanly-understandable c's, program it and let the computer do the work.
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87160d No.2650
Thanks for the quick reply, anon! Is this Baker?
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ba51d0 No.2651
>>2648
I'm not missing, computer problems, life problems. Hope to pick it back up this weekend.
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ab8fb8 No.2652
>>2650
Yes.
To go into more detail, when the prime factors of RSA numbers are calculated, they have to be a certain amount apart, so that programs like our binary search program are made to fail. If abs(a-b) is too small, you can factor the number using the normal Fermat factorization method.
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ab8fb8 No.2653
Patterns.. oh so beautiful patterns.
f+c of 145 = 169
625 = 5^4
145 = 5 * 29
225 = 5 * 45
225*841 = 145*1305
225*841= 5*37845
290*1090=145*2180
f+c of 95 = 100
190*706=95*1412
190*706=5*26828
95 = 5 * 19
100*358 = 5*7160
bottom of picture.
{1:25:8:7:1:65}
c = 65 appears at (1,1,2).
(1,1,2) = {1:1:8:3:5:13}
See the pattern to crack it yet? :x
Am I walking up and down the branches now?
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ab8fb8 No.2654
>>2653
Forgot this.
29 - 58 - 87 - 116 - 145
841 - 3364 - 7569 - 13456 - 21025
29^2 = 841
116*29 = 3364
261*29= 7569 (29*9 = 261)
464*29 = 13456 (29*16 = 464)
725*29 = 21025 (29*25 = 725)
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3ed644 No.2655
>>2653
>>2654
playing along. I'm with you on the up step. not so much on moving down. Are you manipulating something other than x? If so, can you share that formula?
also, your 290 example isn't a valid record.
c != ( d + n ) * ( d + n ) - ( x + n ) * ( x + n )
21025 - 20736 = 289.
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ab8fb8 No.2656
>>2654
There's so many adjacent squares, they're everywhere. Seems as if the tree of numbers + tons of squares = simple factorization of all numbers.
All products (integers) c that are the sum of two squares appear (only) in columns where e=0,1,4,9,16,25…
0^2 = 0
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
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ab8fb8 No.2657
>>2655
I must have not checked that. I'm using this method. If I want to go down, I pass in n,x,a for 1c
If I want to go up, I pass in n, -(2n+x), c
>>> def rowNegX(n, x, a):
... b = a + 2*(-x) + 2*n
... 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)
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ab8fb8 No.2658
>>2657
Also, I think it says it's an invalid record because 290 is not the difference of two squares because it's even, but if you divide it by two it becomes the difference of two squares.
I wonder if that's important.
Passing in -(2n+x) as the formula makes it make the na jump like the crumb says.
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3ed644 No.2659
>>2657
Thanks.
fyi, an alternate to moving up:
BigInteger newA = ter.b;
BigInteger newD = 2 * (ter.b - ter.d) + ter.d;
TheEndRecord up = TerFactory.CreateForNDA( ter.n, newD, newA );
I got similar exceptions with other even values.
And what logic are you using to resolve a and b?
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3ed644 No.2660
>>2658
I think it might be.
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ab8fb8 No.2661
>>2660
>>2659
What do you mean resolve a and b?
This is the whole code for the "tree" code I was taking pictures of.
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907488 No.2662
Evening all.
So as you may have noticed, I haven't posted much lately, because I haven't made any progress.
But I think I just had a mini breakthrough.
I can't explain it better than this at the moment:
We have the grid organized by (e,n). Any number that is e bigger than a perfect square will be listed in column e.
What about f though? For each entry in the grid, what is the f value and how does it change in the column e?
Our semiprime value c has a constant e & f. There are many many entries in column e, but only 2 entries will share the same f.
So lets look at f in (e, 1).
If e is odd, the series of f is:
f = 4, 16, 36, 64
f = 2^2, 4^2, 6^2, 8^2
If e is even, the series of f is:
f = 1, 9, 25, 49, 81
f = 1^2, 3^2, 5^2, 7^2, 9^2
This is super interesting, since we should be able to find our axb record by simply searching for a matching f.
I'm trying to figure out a number of things as next steps:
* How does this pattern change as n grows?
* This seems to be related to x+1, more analysis needed for n>1
* How do we use the na cell in (e,1) to locate our axb for n>1?
A little more info. This came out of analyzing pythagorean triplets. They can only be 3 even numbers, or an even and 2 odds. We only care about the second case.
In the c=145 example, the na records in (e,1) both have the even side of the triangle = f (24), and the hypotenuse = a value in (e, 1).
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3ed644 No.2663
>>2661
I thought you had taken it a bit further.
I've playing a little in the (e,1) space trying to find more patterns and/or relevant records.
pic related show output for c=65, 145, and 785 and only includes records that meet the criteria:
1) a = n*a, n*b, or n*c from the c entry record (indicated as na, nb, nc); or
2) a = n*a, n*b, or n*c from the p solution record. (indicated as pa, pb)
as we know, the t values for these records are n apart. And we can see that extend into negative t.
I've also indicated the relationship between a and the prime a, prime b, and c values.
What's interesting is that none of the "prime" records are divisible by c. So unless the search scope needs to widen beyond n*c, we can't rely on c to find our answer.
Was looking into some relationship between f and c+f to see if there was another way to navigate here. No such luck yet.
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ab8fb8 No.2664
>>2662
But if you are able to find a matching e,f in (e,1) you don't need to analyze (e,n) do you?
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907488 No.2665
>>2664
Correct, this is in the case that you have f = a perfect square, also n = 1.
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ab8fb8 No.2666
>>2665
But I mean even for a c with n!=1, if you find the matching f in (e,1) you've already got the factorization. Right?
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907488 No.2667
>>2666
Sorry, maybe i said it wrong. It wont match in (e,1). It will just match elsewhere in e. In the (e,1), all f's are perfect squares.
There are patterns in f too for higher n. I think I'll have an equation soon. Once we have that we can calculate n from f.
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ab8fb8 No.2668
>>2667
Can you provide some examples? I haven't tried jumping with f.
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907488 No.2669
>>2668
Take our example c=145.
d=12
e=1
f=-24 (or 24 if you use absolute value)
If we look at (1,1), we get the following elements:
{1:1:2:1:1:5} (f=-4)
{1:1:8:3:5:13} (f=-16)
{1:1:18:5:13:25} (f=-36)
{1:1:32:7:25:41} (f=-64)
{1:1:50:9:41:61} (f=-100)
{1:1:72:11:61:85} (f=-144)
{1:1:98:13:85:113} (f=-196)
{1:1:128:15:113:145} (f=-256)
{1:1:162:17:145:181} (f=-324)
In each cell -f = (x+1)^2. This equation works for any (e,1).
I don't have the full process yet for jumping to the correct n, thats the piece thats missing.
But what I'm testing out right now, is to jump to the n = a record for the largest square under our f.
Then traverse that list.
Its obvious that thats not all there is to it, because c=85 doesn't work.
But if we figure out how f grows with n, we should be able to find the correct n given f.
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907488 No.2670
If you have code to generate the grid, add f to the output and you'll see the patterns i'm talking about.
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ab8fb8 No.2671
>>2669
Just use geometry if you want to figure that out.
Reminds me, I need to draw a few things.
Got a lot on my plate right now.
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ab8fb8 No.2672
So, in other news, /cbts/ has been shoahed. (((BaruchTheScribe))) (The BO) lied about many things:
>having a private conversation with Q
>not being able to verify Q's IP
>Q's second trip being cracked
He's also a big enough dunce to get screenshots of the Discord posted to the top of plebbit for all to see. He also insulted Q. He is very Weak Minded.
/thestorm/
and
/qpol/
Are coming in quite handy.
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ff4dbe No.2673
>>2672
This is why it's good that those of us here on this board (as far as I'm aware) don't want to be known for this (we're not famefags, as one you said). I have a feeling since most of the /cbts/ people came from other places on the internet and haven't always necessarily been part of anonymous imageboard culture that they got in over their heads and their egos took over when they became part of a 'super awesome secret club for special people'. That might be why some of them talked to Infowars etc, and it's obviously why so many 50+ year olds are making YouTube videos about it constantly. I have a feeling it's part of why the mods are being dicks. Either they're overwhelmed or they're compromised. I wouldn't want to run a board like that.
Why doesn't Q just make his own board? That way he could lock and sticky all of his own posts and everyone would know it could only be him.
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ab8fb8 No.2674
>>2673
Either way, I love Q even more for BTFOing Zion agent Baruch. Lots of new ideas in /thestorm/
Maybe they'll even convince Q to verify himself with GPG..
but oh dear.. One problem.. Kek.
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51b282 No.2682
>>2669
When we square c there are 5 possible factorizations. They all have different n's but d=c so f is the same for all 5 records including target record {0,n,c, c-aa, aa, bb}. We use f to move to between n's.
If we look at records with different f like moving down the (e,1) factor tree we have to be careful that we know how to get back to original f.
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51b282 No.2683
>>2672
Disinfo necessary.
Loop capital legit crumb!
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51b282 No.2686
>>2663
all the e's equal 1?
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51b282 No.2687
>>2661
I have a couple of questions regarding your and other transformations I see all over this board.
Some review first.
We start with c Looking to find a and b such that a*b =c
Original record {e, n, d, x, a, b}
VQC suggested squaring c.
New record {0, (cc+1)/2-c, c, c-1, 1, cc}
Are a and b factors of new c?
Questions about your transformations
Is original c a factor of new c's?
Are both original a and b factors of the new c's?
Are we looking to determine a and b?
If a or b or both aren't factors of new c how do we find a and b????????????????
If c isn't factor of new c is transformation legit???????????????
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51b282 No.2689
>>2687 (You)
After testing it turns out that
(e = 2c, 1, 1) = {2c, 1, c, 0, c, (c+2)}
(e = 2c-1, 1, 1) = {2c-1, 1, c+1, 1, c, ((c+1)^2+2c-1)/c}
for both new records a of (e,1, 1) = c
So new.c = old.c * new.b
Is old.c a factor of new.c??
Are these transformations legit??
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87160d No.2690
>>2682
Interesting! So from the algebra approach, and starting only from C we'd need to use your ideas to factor backwards to one or more factorizations of C.
To find the n value for a=1,b=c I took the basic formulas and just solved for n. Anyone able to verify that a cell exists at this location?
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51b282 No.2691
>>2690
That's a (1,1) record but I think you're right it works for (e,1) also. I'll put some work into it. Could be what we're looking for because we know how to get to (e,1).
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6cbda8 No.2692
>>2690
VA - not to be a downer here, but isn’t that just our standard n = (a+b)/2 - d formula??
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5167b5 No.2693
I haven't made any progress yet, but I'm glad to see you guys are spurring with ideas!
I made this functions for generating the content of (e, n).
def generateN(e, n, rows=10):
t = getTFromA(e, n)
initX = 2 * t - 1
e, nn, d, x, a, b = rowX(n, initX, 1)
print((e, nn, d, x, a, b))
for i in range(1, rows):
e, nn, d, x, a, b = rowX(n, 2*i*n + initX, b)
print((e, nn, d, x, a, b))
I'm going to see if I can make another similar function.
As you'll see, it only generates the one's that are 'native' to that n. Take for example (1, 5):
>>> generateN(1, 5)
(1, 5, 4, 3, 1, 17.0)
(1, 5, 30, 13, 17.0, 53.0)
(1, 5, 76, 23, 53.0, 109.0)
(1, 5, 142, 33, 109.0, 185.0)
(1, 5, 228, 43, 185.0, 281.0)
(1, 5, 334, 53, 281.0, 397.0)
(1, 5, 460, 63, 397.0, 533.0)
(1, 5, 606, 73, 533.0, 689.0)
(1, 5, 772, 83, 689.0, 865.0)
(1, 5, 958, 93, 865.0, 1061.0)
It doesn't include a=5, b=29. The reason for this is that a=5 is the result of 25 / 5 (or n*a / n yielding a).
This only uses the rowX method for finding the next values in the chain. As a result it will not be a complete (e, n). Nevertheless it's a starting point.
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5167b5 No.2694
>>2693
I don't think it's actually possible to generate complete (e, n) record. Every (e, n) record should potentially grow infinitely, right?
Every number in an (e, n) is there because there exists a number in (e, 1) that is a*n. If that a is a prime, then it should start a new "chain" in the n records. Take for example (0, 8). It contains 4 chains based on the initial map generation program by VCQ. It's only 4 chains, because we have a limited generation of numbers in the initial map program. If you extend this further then you should see even more chains in (0, 8)
In my function I had a limit of 10 rows, so for a given amount of rows it should be possible to generate the entire (e, n), but since every (e, n) contains these chains, it's not necessarily useful to do so.
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5167b5 No.2695
>>2694
I hope I'm making sense.
To explain the term chain look at these records:
{0:8:5:4:1:25}
{0:8:12:8:4:36}
{0:8:21:12:9:49}
{0:8:32:16:16:64}
{0:8:45:20:25:81}
{0:8:60:24:36:100}
{0:8:77:28:49:121}
{0:8:96:32:64:144}
{0:8:117:36:81:169}
{0:8:140:40:100:196}
{0:8:165:44:121:225}
{0:8:192:48:144:256}
A "chain" is something that links a row that contains a, b to b, c where a != b, b != c.
In (0, 8) we can see 4 different chains.
# Chain 1
1, 25 -> 25, 81 -> 81, 169
# Chain 2
4, 36 -> 36, 100 -> 100, 196
# Chain 3
9, 49 -> 49, 121 -> 121, 225
#Chain 4
16, 64 -> 64, 144 -> 144, 256
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5167b5 No.2696
>>2695
This rowX seems to hold a lot of potential.
We can generate every instance where a = c or b = c using it.
>>> rowBX(0, 0, 145)
(0, 0, 145, 0, 145, 145)
>>> rowBX(1, 0, 145)
(286, 1, 143, 0, 143, 145)
>>> rowBX(2, 0, 145)
(281, 1, 142, 1, 141, 145)
>>> rowBX(3, 0, 145)
(274, 1, 141, 2, 139, 145)
>>> rowBX(4, 0, 145)
(265, 1, 140, 3, 137, 145)
>>> # ...
...
>>> rowBX(72, 0, 145)
(1, 61, 12, 11, 1, 145)
Below is two functions, one for rowX which takes in three parameters (n, x, a) and rowBX which takes in three parameters (n, x, b).
def rowX(n, x, a):
b = a + 2*x + 2*n
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 rowBX(n, x, b):
a = b - 2*x - 2*n
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)
The difference between rowX and rowBX is that rowX generates a record for n, x and a while rowBX generates the same, but for b.
If you ignore one of the variables, for example x = 0, and just adjust n it will still generate records and will fill in for n (as x increases).
So ignoring the second (x) parameter and just adjust n you'll see that rowXB(72, 0, 145) is the highest n will go. This is because it will generate the record where a=1, b=145. n = 0 will generate (0, 0, 145, 0, 145, 145). So you can think of n as decreasing the a-variable.
The limit is (b - 1)/2 so for b = 289, the highest n will go is 144. (289 - 1)/2 = 144.
By using rowX you can find every (e, n) where a is equal to whatever value you want.
I haven't fully wrapped my head around it, but if you modify the code to only output e = <some value> you can see VQC's reference to:
> Take the number 71. It appears once in the grid. The number 15, twice. 71 is prime. All prime appear once. 15 appears twice, in the same column. 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. If you multiple c where it is the product of two primes by say, 105, you're going to have a product, a new c, that appears quite a few times in the same column… This is where we are heading.
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51b282 No.2697
>>2695
For (0, n) when n= 2mm for integer m there are multiple chains the rest of (0,n) you can develop from the first record.
These are records which a and b are perfect squares!
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51b282 No.2698
>>2690
I tried your formula on some (e,1) records based on various combinations of c and N.
Unfortunately those records have already been solved. It just confirms what we already know. Your formula doesn't get us to the goal record. At least so far.
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5167b5 No.2699
I don't think this is *the* solution, but is it possible to generate a record for (e, n, x, c)?
that is, use e, n, x and c to generate all possible solutions for c?
VQC said we only need 3 variables to solve for c. It could yield interesting results.
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87160d No.2700
>>2691
>>2692
>>2698
Hey lads, thanks for your replies. So the first part is just the (e,1) factorization we already know. The newer highlighted part is just setting ab to a=1 b=c and then using the basic equation (you weren't being a downer, PMA) to solve for n when ab are set to 1,c.
What I should have taken more time to be clear about is that starting from only C, this is a way to generate two valid cells for C, one at: {1,1,d,x,a,b} which we know using f, and a new one at {1,BigN,d,BigX,1,c}. This gives us an upper bound for n and x. All other factorizations for c should be in lesser n values.
At a conceptual level, I guess I'm trying to take our knowledge in (e,1) and figure out how to expand it. This 1,c bigN may have no value at all, just keeping the mental oven baking over here.
Baker mentioned that he had been able to get the f formula to generate factors for the RSA numbers, although they were in n=1. Baker, any chance you could post some examples of RSA numbers that you've been able to factor in n=1?Thanks!
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ab8fb8 No.2701
>>2700
Lmao, ok. I'll post the ones for RSA-240 since this is the lowest that hasn't been factored. I didn't factor the number, just set n = 1. Also, ab != c because there are only 2 factorizations for semiprimes. 1c and ab
I have to rewrite the method because I lost it. Can you repost the formula?
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87160d No.2702
>>2662
Teach, nice to see you! I'm following all your ideas here. Var f seems likely to hold some more goodies for us to discover.
>How does this pattern change as n grows?
>How do we use the na cell in (e,1) to locate our axb for n>1?
>>2651
Hello MA! Hope to see you over the weekend!
>>2701
Sure Baker. Also, thanks for clearing up a question I had. Only two factorizations for semiprime c? 1c and ab. perfect. Here's the formula.
x= floor(SQRT( abs(f)))
then use x to create a and b.
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ff4dbe No.2703
>>2699
>VQC said we only need 3 variables to solve for c
The only variables they could be are e, n, d and f, right? If we knew any others we'd already be able to do the factorization, and that would be O(1).
>>2702
>x = floor(sqrt(abs(f)))
Am I missing something here? Does it only work for some cases? Pic related.
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ab8fb8 No.2704
>>2703
>>2702
It assumes n=1 and generates a cell like that.
The c created doesn't equal the original c.
It's like creating a factorization for (e,1).
Might be more useful than I realized if we can find other cells containing same e and f.
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ff4dbe No.2705
>>2704
I don't know if this is useful but I haven't really contributed to this thread yet.
Where n = 1, x = floor(sqrt(abs(f)))
Where n = 1, i - d = 1, which means i = d+1
Where n = 1, x + n = j, which means j = floor(sqrt(abs(f))) + 1
Where n = 1, i scales up linearly with d and j scales up linearly with floor(sqrt(abs(f))).
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87160d No.2706
>>2703
It works for ALL (e,1), but only for (e,1). However, within row 1 it's very powerful. Just start from C, derive d,e,f,x, then calculate a and b. Boom, done. Try it out. grab your grid, pick a random element, multiply a and b to get c, then feed your c var into the calculator. It will solve any c value in row 1, and with basic math. Somehow f is always related to x. Like a "golden ratio" but different. It's pretty incredible.
>>2704
Cool, thanks for looking into it Baker.
>>2705
Thanks for contributing, Anon!
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ab8fb8 No.2709
>>2705
Thanks. Yes, j = x+n
all odd numbers can be represented by
i^2 - j^2 = c
(d+n)^2 - (x+n)^2 = c
(d+n) - (x+n) = a
i - j = a
(d+n)+(x+n) = b
i + j = b
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ab8fb8 No.2711
Here are the values if you assume n=1 for rsa240.
It is not the factorization of rsa240.
rsa240c = 124620366781718784065835044608106590434820374651678805754818788883289666801188210855036039570272508747509864768438458621054865537970253930571891217684318286362846948405301614416430468066875699415246993185704183030512549594371372159029236099
if you assume n=1 (it isn't), these are the values you get:
pastebin.com/W40p18aG
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51b282 No.2712
I can beat it. But I need to buy a computer and some software. Any recommendations for a windows fag.
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be63a1 No.2713
>>2712
AMD laptop from eBay with Windows 7 key. Use Windows 7, 10 if you must. You can still get free Windows 10 from a 7 key, but it's a bit of work. Or install and learn linux, you're gonna need it.
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51b282 No.2714
>>2713
Thanks MA. Can I have it run both?
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be63a1 No.2715
>>2714
It's a little more technical, but of course. It's a computer, it'll do whatever you want it to if you try long enough.
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87160d No.2716
>>2711
Thank you Baker!! Appreciate your work. Checked them out carefully in the pastebin link. Is this RSA unsolved?
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ab8fb8 No.2717
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87160d No.2718
>>2717
So t+na works within the same cell?
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ab8fb8 No.2719
>>2718
The ↑ I was using in my previous was t+na
I changed it to t+na,
and now the ↑ stands for -x jump with parameters
(n, -(2n + x), a)
t+na uses the parameters
(n, -(2n + x), c)
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5167b5 No.2724
I've been looking at the d for (1, n) records and I have the following:
(1, 5, #1) = 2(5 t^2 - 2t - 1)
(1, 5, #2) = 2(5 t^2 + 2t - 1)
(1, 13, #1) = 2(13 t^2 - 8t - 1)
(1, 13, #2) = 2(13 t^2 + 8t - 1)
(1, 17, #1) = 2(17 t^2 - 4t - 4)
(1, 17, #2) = 2(17 t^2 + 4t - 4)
(1, 25, #1) = 2(25 t^2 - 18 t - 3)
(1, 25, #2) = 2(25 t^2 + 18 t - 3)
Here I use (e, n, #chain).
I was wondering if it's possible to get the t from d only as a point of exploration.
So the t is squared and multiplied by n, but I don't understand the pattern behind the second part.
Although it's interesting to note that the calculation for these d's for t in the different chains are so similar.
I don't think it will be possible to get a way of calculating t of d without knowing n.
I checked a few other (e, n)'s and they appear to follow a similar pattern.
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ff4dbe No.2726
When was the last time Chris posted here?
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5167b5 No.2727
>>2726
It was in the last thread post >> 2388
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e28e44 No.2729
Hello lads! Pretty mellow here today. I spent yesterday looking to expand the formulas for f into n=2. Some success, but nothing solid to report. Seems like we're still looking for a way to tie (e,1) and column 0 together.
It would be nice if Senpai could pop in and drop a few tasty crumbs like he promised! VQC, you there? We've made it this far, let's finish this.
In other good news, CodeMonkey got shit sorted out with Q (as Baker mentioned) and there are a bunch of good new Q drops if you haven't seen them. Q is no longer on /cbts/. Might as well read up on Q posts and keep an eye here till VQC pops in. Good updated Q map over there too. New location for Q:
>>>/thestorm/
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51b282 No.2730
Saw something in factor trees yesterday that wasn't there. Have to be careful doing math and weed together.
I thought factor trees worked because factors were seeded into the tree and would pile up at their appointed times . It doesn't work like that.
The algorithm for generating the factor trees is simple to follow. The initial record for the a b generating factor tree is {2c, 1, c, 0, c, c+2}
generating new records is simple
next.d = d+4t
next.x= x+2
next.a = b or next.d-next.x
next.b = (next.d*next.d + e=2c)/ next.x
Somehow this algorithm that knows nothing about a or b produces records containing a and b as factors without c many times more often than c records.
Somehow the knowledge to produce records with a and b factors exists in the initial record cause it isn't in the algorithm. Ghost in the machine!
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ab8fb8 No.2731
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ab8fb8 No.2732
>>2731
Chris says we are close to discovering the pattern to the number tree of factorization.
If you believe that we create our reality, then faith is more important than the most powerful man in the world.
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51b282 No.2733
>>2731
>>2732
Quantum computer teaching quantum reality
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51b282 No.2734
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e28e44 No.2735
>>2731
>>2732
Love the number trees, Baker!! Also, I agree that faith in realizing our goal creates that reality. That's why I'm here, I've already seen the end result achieved. This is the persistence part, where winners keep going until they win. Did VQC say that to you on DM?
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99eff8 No.2736
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3ed644 No.2737
>>2735
VA - still trying…
I've put a lot of output together, still trying to find any pattern that links to any potential prime match. No such luck yet. Notwithstanding the -x hint earlier. All I've been able to confirm is that the pattern holds for na, nb, nc in negative x and f spaces.
I had a thought earlier that I've been trying to work through in the (e+1) space. Perhaps someone can assist or tell me if I'm barking up another wrong tree…
VQC mentioned that every factor for c exists in (e+1). I originally thought that meant (e+1,1), but I'm leaning now towards (e+1,n). And we also know that the a value at (e,n) moves to the d values at (e+1,n). (I believe this is correct).
For example:
for c=145 (1,61,6), the factor 5 exists as the d value of (2,1,2).
for c=901 (1,421,15), a factor of 17 exists as the d value of (2,33,8).
I thought that if there was some way to link records to e+1, we would have a simpler way to calculate d?
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e28e44 No.2738
>>2736
ThanksTopol! Are you ready to help FIRE UP SPIRITS?!?
>>2737
PMA, you have worked so hard, man. It'll click and fall into place!! Like the last tumbler on the lock.
Come on, fags, we can do this! Here's your inspirational Literature for this evening. We need to get our levels up to 9999!!
KING HENRY V:
Once more unto the breach, dear friends, once more;
Or close the wall up with our English dead.
In peace there's nothing so becomes a man
As modest stillness and humility:
But when the blast of war blows in our ears,
Then imitate the action of the tiger;
Stiffen the sinews, summon up the blood,
Disguise fair nature with hard-favour'd rage;
Then lend the eye a terrible aspect;
Let pry through the portage of the head
Like the brass cannon; let the brow o'erwhelm it
As fearfully as doth a galled rock
O'erhang and jutty his confounded base,
Swill'd with the wild and wasteful ocean.
Now set the teeth and stretch the nostril wide,
Hold hard the breath and bend up every spirit
To his full height. On, on, you noblest English.
Whose blood is fet from fathers of war-proof!
Fathers that, like so many Alexanders,
Have in these parts from morn till even fought
And sheathed their swords for lack of argument:
Dishonour not your mothers; now attest
That those whom you call'd fathers did beget you.
Be copy now to men of grosser blood,
And teach them how to war. And you, good yeoman,
Whose limbs were made in England, show us here
The mettle of your pasture; let us swear
That you are worth your breeding; which I doubt not;
For there is none of you so mean and base,
That hath not noble lustre in your eyes.
I see you stand like greyhounds in the slips,
Straining upon the start. The game's afoot:
Follow your spirit, and upon this charge
Cry 'God for Harry, England, and Saint George!'
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99eff8 No.2739
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3ed644 No.2740
new avenue to explore, perhaps.
Records at c and ab both share f and c+f values.
They also share a common record at (e,1, (c+1)/2) which has a different f value.
If you move the 3 records to (e+1, n, t) they each now have different f values. Is there a way to find a difference between these 3 records?
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e28e44 No.2741
>>2740
Nice PMA! This looks promising, f is part of the key I think. I'll check out the record movement you're describing. Need some sleep, but on it tomorrow!
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ab8fb8 No.2742
Pay attention to b for all the cells pictured.
Subtract b of ↑d from b of ↓d
Subtract b of ↑e from b of ↓e
Subtract b of ↑n from b of ↓n
Subtract b of ↑x from b of ↓x
Subtract b of ↑f from b of ↓f
Subtract b of t+na from ↓c
Always the same number.
Always a multiple of 2.
Let's call it.. o for now
Because why not.
o for 991 = 1980
o for 533 = 1064
o for 169 = 336
o for 145 = 288
o for 54 = 104
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ab8fb8 No.2743
>>2742
2*(c-1) works for most instances tried.
but not c=54, where o is actually 104
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ab8fb8 No.2744
>>2742
I am using this method.
>>> def rowNegX(n, x, a):
... b = a + 2*(-x) + 2*n
... 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)
Here are the parameters for the method for each element derived from c in the pictures:
→ = 1*c cell
↓ = (n, x, 1)
↓c = (n, x, c)
↓d = (n, x, d)
↓e = (n, x, e)
↓x = (n, x, x)
↓n = (n, x, n)
↓f = (n, x, f)
↑ = (n, -(2n + x), 1)
↑d = (n, -(2n + x), d)
↑e = (n, -(2n + x), e)
↑n = (n, -(2n + x), n)
↑x = (n, -(2n + x), x)
↑f = (n, -(2n + x), f)
t+na = (n, -(2n + x), c)
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ab8fb8 No.2745
>>2744
In addition, you can generate cells for the real factorization of c if you replace 1 with a. (and know a)
Try subtracting ↓d on top from ↓d on bottom.
Try subtracting ↑d on top from ↓d on top.
Always multiple of 2.
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ab8fb8 No.2746
>>2745
Try subtracting ↑d on top from ↑d on bottom*
Lots of patterns here.
Maybe you can find more.
I've provided the way to generate these.
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51b282 No.2747
Wow >>2745
Some excellent work. I need to spend some time to digest it.
Include >>2744 in bread
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51b282 No.2748
With no visits from Chris I thought I would try to imitate him. I hope i don't sound like a [D]umb [M]oron
Any factor of any a in (e,1) will be the value of all n at (e,n)
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51b282 No.2749
That was fun. Here's another [D]umb [M]oron idea
Patterns. Focus on a[t] and d[t] in (e,1) subtracting d from d[t] gives (n-1)a
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51b282 No.2750
If I get any more [D]umb [M]oron ideas I'll be sure to share them!
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5167b5 No.2751
>>2750
Okay guys. So I've been thinking.
If we do a, b = 5, 29 we have c = 145.
If you "scale" the number by 5 on a, b such that a,b = 5*5, 5 * 25 then c = 25 * 145.
I noticed that when we do that, we can predict the e it will land in. e = 25. Both a and b will contain 5 as a factor => (25, 25, 60, 35, 25, 145).
It doesn't predict the 'n', but shit. We know that we can find a number a % 25 == 0 in (25, 1). Then just use 25 * i + t and 25 * i + 1 - t to search for a, b.
Again, we simply multiply c with a square of our choosing. Then we find the first a in (e, 1) that contains our number (non-squared) as a factor.
This is our prime, we can then use this to factorize c by generating more a's that fit p * i + t, p * i + 1 - t and using gcd.
def solveTest(c, rows=20):
e, n, d, x, a, b = rowForAB(1, c)
e, n, dd, xx, a, b = rowForAB(1, x**2 * c)
li = generateGenesis(e, rows)
t = 1
for cell in li:
e,n,d,xx,a,b = cell
if a % x == 0:
break
t = t + 1
for i in range(1000):
p1 = x*x * i + t
p2 = x*x * i + 1 - t
pp1 = int(getAFromT(e, p1))
pp2 = int(getAFromT(e, p2))
gp1 = gcd(c, pp1)
gp2 = gcd(c, pp2)
if gp1 != 1 and gp1 != c:
print(t,i, pp1, gp1, c, c / gp1)
break
if gp2 != 1 and gp2 != c:
print(t,i, pp2, gp2, c, c / gp2)
break
Here I use x as an example. It doesn't have to be x, it could even be a chain of primes squared (I think).
I've also been thinking that maybe we can use e again. e will be the k * c, where k is some squared number.
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5167b5 No.2752
>>2751
> If you "scale" the number by 5 on a, b such that a,b = 5*5, 5 * 25 then c = 25 * 145.
I'm an idiot, here I meant a, b = 5 * 5, 5 * 29
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5167b5 No.2753
>>2751
This is heavily non-optimized. I don't even think we need to search for the first a that contains our factor.
If we for example use e as our prime (the number we squared) then we could simply use getTFromA(e, e). Then iterate on that to find the number that gcd(c, somenumber) != 1 and gcd(c, somenumber) != c.
I haven't tested that, but I think it should work. We should then be able to skip the first generation of genesis cells and go straight to the prime iteration process.
In my code I added a range(1000), but if we *know* what we are looking for we can just iterate indefinitely until we hit the proper i.
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5167b5 No.2754
>>2753
> In my code I added a range(1000), but if we *know* what we are looking for we can just iterate indefinitely until we hit the proper i.
Correction, if we *know* that e * i + t, e * i + 1 - t will yield a number where gcd(c, somenumber) == a or b, then we don't need a range-limit.
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5167b5 No.2755
>>2751
Correction, we can't "predict" e. It appears it doesn't always equal the number we square. But I think we can still use this approach.
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ab8fb8 No.2756
>>2755
That is an unacceptable solution.
Read the pictures I sent and try taking the difference between two ↑d's or two ↓d's
It's predictable based on a. I'm working on whether you can derive it from c and work backwards.
Always multiple of 2.
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08dd7b No.2757
>>2748
I think I know what this one means.
Look at records where e=n
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ab8fb8 No.2758
For simplicity sake I am going to refer to o as the difference between a ↓ element and a ↑ element, and I'm going to refer to the difference between the ↑ element of the a*b element and the ↑ element of the 1*c element as p
Two↓ elements also works here.
If you don't recognize this notation just look at the pictures I sent.
This difference, p, may be directly related to a and maybe can be derived from c.
I don't know the linear formula, but most of the values of p for a can be found when you put a into the formulas for all the elements.
For example, you get 8*22 when you type in 8, and p is always 22 when a=8
For example, you get 4*10 when you type in 4, and p is always 10 when a=4
Full list of relationships:
a = 1, p = 0
a = 2, p = 4
a = 3, p = 6
a = 4, p = 10
a = 5, p = 12
a = 6, p = 16
a = 7, p = 18
a = 8, p = 22
a = 9, p = 24
a = 10, p = 28
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ab8fb8 No.2759
>>2758
It might be congruent to this sequence.
oeis.org/A006093
Try entering d for any c into the element deriver that I sent the formulas for. I noticed the value of p is always there, but I haven't confirmed this. If it is, then you'd be able to factor the number by finding what term p is of the sequence above. (E.g 16 is the 6th term of that sequence if you ignore 1).
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99eff8 No.2760
lookin' gooooood!
(Pic related. I have no idea where to plug stuff in. Tech support didn't want to talk to me, and Jakarta was afraid I'd take over.)
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3ed644 No.2761
>>2757
interesting. thanks.
(1,61,6) = {1:61:12:11:1:145} = 145
(1,61,56) = {1:61:212:111:101:445}
(1,61,67) = {1:61:278:133:145:533}
The following formula works for generating these next records for (e,n).
a=a+2*(x-n) will generate (1,61,56) - this one can also come from (e,n,t+n-x)
a=a+2*(x+n) will generate (1,61,67) - this one can also come from (e,n,t+n)
My previous move by d formulas stepped over (1,61,67).
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ab8fb8 No.2762
>>2761
The way to generate the 101*445 element was the puzzle piece I needed.
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3ed644 No.2763
>>2762
not sure if this is of any value, but I found a few more records for (1,61) at -t.
(1,61,-66) = {1:61:12:-133:145:1} = 145; f=24; (x+n)=-72; f+c=169; -x: 11
(1,61,-55) = {1:61:-10:-111:101:1} = 101; f=-20; (x+n)=-50; f+c=81; -x: -11
(1,61,-5) = {1:61:-10:-11:1:101} = 101; f=-20; (x+n)=50; f+c=81; -x: -111
(1,61,6) = {1:61:12:11:1:145} = 145; f=24; (x+n)=72; f+c=169; -x: -133
(1,61,56) = {1:61:212:111:101:445} = 44945; f=424; (x+n)=172; f+c=45369; -x: -233
(1,61,67) = {1:61:278:133:145:533} = 77285; f=556; (x+n)=194; f+c=77841; -x: -255
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ab8fb8 No.2764
>>2763
Please take a look at my method.
If you can get p from c you just know what term of the sequence it is to find a.
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ff4dbe No.2765
Has anyone figured out how O(log n) factors into the grid navigation? I noticed you're mostly all working on calculations and things. Surely if we knew how the idea of this being done in O(log n) time makes sense it would help immensely.
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ab8fb8 No.2766
>>2765
log(n) means it's instant.
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3ed644 No.2767
>>2764
if you are referring to the rowNegX method, yes, I'm familiar with what you're doing. I have some sample code that follows along.
I'm not quite understanding what you mean with p. How is it different to the x or t values that we already know, or how we can navigate between records by factors?
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ab8fb8 No.2768
>>2767
It's all there. Value of p seems to be close to d.
a is equal to the number of the sequence I posted that p is, minus one.
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ff4dbe No.2769
>>2766
What are you talking about? O(1) means it's instant. Also what does that have to do with anything? VQC said it would be in O(log n) time. I'm not aware of many ways in which an algorithm can work in O(log n) time. Wouldn't that be a constructive thing to think about? It might more clearly show what we should be looking for in these relationships.
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ab8fb8 No.2770
>>2769
You have a good point, but you can't draw many conclusions from it being log(n), because that n can be different things.
O notation is just meant to analyze how fast something scales with input.
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ff4dbe No.2771
>>2770
How many different O(log n) algorithms can you list? What n is is irrelevant to what I'm trying to say. Even if we don't apply any existing O(log n) algorithms to the grid, what we do with the grid is meant to fit the criteria of an O(log n) algorithm according to Chris. Does that mean there has to be some variable or relationship between variables that allows us to halve n's search pool every time? Does that mean doing something weird with powers? If we know what criteria are necessary for something to be O(log n), we'll know more about what kind of mathematical relationship we're looking for, so that whenever we find an interesting relationship, we can figure out if we're going in the right direction.
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88565a No.2772
>>2769
There is a linear relationship between log n and the number of digits or bits of n. (Using base 2 or e or 10 would just give a different scale ). So a log n algorithm would have solving routine that might:
-narrow down (magnify) search area by 10 times per step, so step 1 tells you first digit, step two tells you second digit, and so on
-narrow down by 2 times per step, e.g a binary tree. If there is a Tree of Knowledge, and the first step is whether e is even or odd, and the following steps iterate some similar decision, then this would be log n
-being able to decide first bit, second bit, third bit in constant time each
One point about what VQC said, either the VQC works in log n time (really we mean log c, since it is c which is given in our problem), and VQC misspoke by saying log n where n is the size of the digits of the input. If VQC meant that literally and wasn't being sloppy then since the number of digits is already log of the number, the VQC's speed is log (log c).
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51b282 No.2773
>>2764
Baker these r legit but maybe [D]ont [M]ention
Up to you.
>>2748
>>2749
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99eff8 No.2774
>>2772
2 7 7 2
I see log log c
measure measure lightspeed
tree tree lightspeed
that's one tree faster than light speed.
(I'm actually the couch. Or am I the wall? The… photogr… hold on…. I'm not actually in this picture… yet… B{D Pic Related.)
http://www.infinitelooper.com/?v=wzZzR-7lskg
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3ed644 No.2775
>>2773
back to understanding (n-1)*a movements.
Pic attached is of c=156. Record starts at (1,61,6), I've moved the analysis to t + n at (1,61,67), and included the factor records at (1,29,67), (1,5,67) and differences between each and differences to (1,1,67).
As expected, they all follow a pattern. d and a move together. The f difference is always 2*d difference. And (n-1)*a applies when moving to (1,1).
There is also a pattern to moving from (1,61, 67) to factor records.
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3ed644 No.2776
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e28e44 No.2777
Hey guise, I'm here working away too. Nothing to report yet tonight. Checking out all your N and -x patterns.
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ab8fb8 No.2778
I'll be busy until Wednesday or Thursday. Tell me what you all find regarding the prime sequence and the -x jumps.
I'm not 100% sure the sequence of p for increasing a is congruent to the sequence I sent but it was very similar:
oeis.org/A006093
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ab8fb8 No.2779
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51b282 No.2781
>>2775
Excellent! I couldn't figure out what that meant.
>>2749
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5167b5 No.2782
>>2775
Why did you select (1, 29)?
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5167b5 No.2783
>>2782
Nevermind re-read.
Btw great work! This looks very interesting
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5167b5 No.2784
>>2775
8700 - 160 = 8540
8540 - 1464 = 7076
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3ed644 No.2785
>>2784
yes.
I also thought that the d difference between (1,29,67) -> (1,5,67) equals (1,61,6) n * 2d or n*f. But that formula doesn't apply to different test cases.
So for these records, if we can figure out the a or d formula…
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5167b5 No.2786
>>2785
Yeah, this shit feels soooo god damn close. If only we can get to one of those records, we got it.
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3ed644 No.2787
>>2786
I've also looked into a formula for b. Haven't found a combination of x+n, x-n, multiple of n, etc. that will work.
Interesting thing about these records is the shared e and x. and the 2an is all the same - as you would expect.
Pity we don't yet understand moving between n's other than to 1.
So the (n-1)*a takes you to the n=1 value and is a shortcut for the "f transition" (whatever that means).
What's the formula to take you to another n value other than 1?!???
Guess that's the Integer Factorization secret…
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3ed644 No.2788
>>2786
important formula I believe:
n = (xx + e) / (f - 2x)
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5167b5 No.2789
>>2788
Based on (4, 34, 161, 76, 85, 305)
n = (xx + e) / ((f + a) - 2x)
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5167b5 No.2790
>>2789
Damnit, clicked reply too fast:
n = (xx + e) / ((2 * (d + a)) - 2x)
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5167b5 No.2791
>>2790
Doesn't work for c=145, so might be a fluke, but then again there might be a different equation for odd e
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3ed644 No.2792
>>2791
my equation worked for c=145. I'm trying to get a formula for n in terms of known variables. e, x, d movement we have some formulas for already.
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5167b5 No.2793
>>2792
I know, but for (4, 34, 161, 76, 85, 305) I calculated the following rows based on the factors of 34 * 85:
(4, 1, 2966, 76, 2890, 3044)
(4, 2, 1521.0, 76, 1445.0, 1601.0)
(4, 5, 654.0, 76, 578.0, 740.0)
(4, 10, 365.0, 76, 289.0, 461.0)
(4, 17, 246.0, 76, 170.0, 356.0)
and
(76 ** 2 + 4) / (2 * (161 + 85) - 2*76) = 17.
I was thinking maybe that's the key.
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3ed644 No.2794
>>2791
>>2793
works for c=65 also.
more testing required for e other than 1.
n = (xx + e) / (f - 2x)
n = (xx + e) / (2d + 1 - e - 2x)
if this holds, we have a formula for n in terms of known x and e variables. And now we need to understand better the d jump.
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5167b5 No.2795
>>2793
(76 ** 2 + 4) / (2 * (161 + 85) - 2*76) =17.0
(76 ** 2 + 4) / (2 * (161 + 85 + (85 + 34)) - 2*76) = 10.0
(76 ** 2 + 4) / (2 * (161 + 85 + (85 + 34) + (34 + 85 + 85 + 85)) - 2*76) = 5
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5167b5 No.2796
>>2795
Adding more 34 + 85.. terms doesn't get to the (4, 2) record though.
However, the (4, 2) record, when divided by 2 is equal to 85 * 34 (2890)
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51b282 No.2797
Banged through a bunch of e=n records.
Discovery!
Guess z in c terms?
a of (z, z, 1) = c if c semi prime
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e28e44 No.2798
>>2787
Hello PMA! This is exactly what I've been working on. How to get out of row 1 starting only from c.
>>2789
Where'd you derive this? I'll check it out, could be very useful.
>>2797
Interesting, Anon! PMA, remember "Use the z on y at x" ??
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e28e44 No.2799
>>2787
I think the secret of getting out of row one could be multiples of f. An expandable search pattern for a simultaneous lock on x,a,b, and n that works outward from d. A very restricted search algorithm using f as its base. Thinking out loud here.
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3ed644 No.2800
>>2798
>PMA, remember "Use the z on y at x" ??
VA - I still have no idea what this means…
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e28e44 No.2801
>>2800
Lol, PMA! Neither do I. However, it was in relation to your idea about solving for the small square using t. We have been one variable away from success for weeks!
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ab8fb8 No.2802
>>2794
Elaborate. You evaluate that formula with x from 1c to get n from ab?
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3ed644 No.2803
>>2802
I've been trying to get further along with the (n-1)*a hint.
I'm starting with a record for c at (e,n,t). I move up to a record at (e,n,t+n). This record contains all factors of a,b,c.
It is also similar to a record at (e,1,t+n) in that e and x are the same.
Then if you find records at (e,a t+n) and (e,b,t+n) where a and b are the values we are searching for, you'll notice that they also share e and x values. And 2an are the same value for all of these records.
We know how to take our (e,n,t) for any c and translate it to (e,1,t). That's the (n-1)*a transformation.
So I've been researching relationships between these records to try and learn how to transform the n value to a or b instead of 1.
In analyzing those differences, and searching for a formula to define the change in a or d, I stumbled upon a way to represent that change in terms of known variables for e, x, and d.
Those are the formulas posted at >>2794.
Unfortunately, they only appear to work where e=1.
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ab8fb8 No.2804
>>2803
What is the point of that formula? It gives you n for 1c
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ab8fb8 No.2805
>>2804
>>2803
Wait.. it gave me 2*5 for 10..hmm
but 1*145 for 145
Looks like it can be improved
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3ed644 No.2806
>>2804
no. it gives you any n from any (e,1) and it may provide a way to get to n for a,b in terms of e, x and a change in d values.
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ab8fb8 No.2807
>>2806
It gave me the factorization of 10 which has n of 0.
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3ed644 No.2808
>>2807
I may not be explaining myself. That formula doesn't solve for n yet.
It is a step I am using to break down variables into smaller pieces to try and analyze the jump to a valid d that will solve for n.
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ab8fb8 No.2809
>>2808
But you might find it interesting that I was unable to find the factorization of 10 in the VQC output and when I factored 10 using that formula it give me a non-existent element
{1:2:3:1:2:5}
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3ed644 No.2810
>>2809
sure. I'll look into it.
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ab8fb8 No.2811
There are a lot of things I missed in my -x jump brute force factorization program. Try calculating change between each b it prints out. It also sometimes gives a instead of n. I think there is something valuable there. I just noticed it but I have to go.
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3ed644 No.2812
Continue some research into records calculated from c into (e,n,t+n).
These are labeled nt in the attached pic.
Can someone please tell me why (b-1)/x is always equal to 4?
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87160d No.2814
>>2810
Hello lads! Here's what I'm working on: took the formulas and built a sheet that calculates all vars and displays them visually for the difference of squares equation. Looking for ways to solve for x when n>1. Row one is solved, as you anons know. Seeing patterns in n>2, but still nothing yet.
I'm honing in on x being the key. X increases in a linear pattern. Can we link X to the c^2 patterns? I'm wondering if the stable growth of x can be tied to the exponential growth of C somehow. Maybe this ties into the -x patterns? Just keeping my head down and looking for patterns over here.
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3ed644 No.2815
>>2812
And if you compare the b values between t=n+t and t=(1-(n+t)) records, the difference also appears to be 4x.
Example
(1,5,67).b - (1,5,-66).b = 2045 -1513 = 532 = 4x133;
(1,29,67).b - (1,29,-66).b = 629 - 97 = 532 = 4x133;
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ab8fb8 No.2816
>>2814
That's some good shit. Send the spreadsheet!
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ab8fb8 No.2817
Add System.out.println(v);
after line 46 in my x jump code to see what it actually does
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5167b5 No.2818
>>2787
What do you mean "move between n's other than 1"?
While right now we are a bit limited in the moving regard, but we can, for example, move from (1, 61) to (123, 61).
>>> rowX(61, 2 * 61 + 11, 145)
(1, 61, 278, 133, 145, 533)
>>> createForENX(1 + 2 * 61, 61, 133)
(123, 61, 279.0, 133, 146.0, 534.0)
>>> createForENX(123 + 2 * 61, 61, 133)
(245, 61, 280.0, 133, 147.0, 535.0)
>>> createForENX(1 - 2 * 61, 61, 133)
(-121, 61, 277.0, 133, 144.0, 532.0)
>>> createForENX(-121 - 2 * 61, 61, 133)
(-243, 61, 276.0, 133, 143.0, 531.0)
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5167b5 No.2819
>>2818
It even keeps the logic behind xx + e = 2na
>>> rowForAB(1, 13 * 19)
(22, 109, 15, 14, 1, 247)
>>> rowX(22, 2 * 22 + 14, 247)
(40, 10, 317, 70, 247, 407)
>>> createForENX(40 - 2*10, 10, 70)
(20, 10, 316.0, 70, 246.0, 406.0)
>>> createForENX(20 - 2*10, 10, 70)
(0, 10, 315.0, 70, 245.0, 405.0)
>>> 245 * 10
2450
>>> getTFromA(0, 2450)
35.0
>>> generateNthRowFromGenesis(0, 35)
(0, 1, 2380, 68, 2312, 2450)
>>> generateNthRowFromGenesis(0, 36)
(0, 1, 2520, 70, 2450, 2592)
>>> createForENX(0 + 2 * 1, 1, 70)
(2, 1, 2521.0, 70, 2451.0, 2593.0)
>>> createForENX(2 + 2 * 1, 1, 70)
(4, 1, 2522.0, 70, 2452.0, 2594.0)
>>> createForENX(38 + 2 * 1, 1, 70)
(40, 1, 2540.0, 70, 2470.0, 2612.0)
>>> (70**2 + 40)/2
2470.0
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5167b5 No.2820
>>2819
Oh boy I messed that code up. I was trying to look at the next row for 247, but I used e instead of n and it jumped from e=22, to e=40. I don't trust the result from the code above. Need to verify
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5167b5 No.2821
Wasn't there an equation for moving around (e, n)'s?
(e + 2n, n) or (e - 2n, n)?
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5167b5 No.2822
>>2812
I can't tell you why, but I did some minor algebra on it:
b - 1 = 4x
b = 4x + 1
b = a + 2x + 2n
4x + 1 = a + 2x + 2n
4x - 2x = a + 2n - 1
2x = a + 2n - 1
x = (a + 2n - 1)/2
x = a/2 + n - 1/2
For n:
n = -a/2 + x + 1/2
For a:
a = 2x - 2n + 1
Appears to check out for multiple records:
c = 145
c = 1313
c = 3905
I don't know if these equations will be helpful, but at least here they are.
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3ed644 No.2823
>>2821
yes. for moving right or left.
>>2822
thanks.
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5167b5 No.2824
>>2823
I played some more with the numbers
We know x = a/2 + n - 1/2 for some c's (at least the ones I've been making by taking two primes and multiplying them).
We also know x*x + e = 2*a*n. I ran this through wolframalpha:
So by substituting x*x we get:
(a/2 + n - 1/2)^2 + 1 = 2 * a * n (for (1, n))
This yields:
n = 1/2 (a - 2*sqrt(a - 1) + 1)
Integer solutions:
a = 4 m^2 + 8 m + 5
n = 2 m^2 + 2 m + 1
m element Z
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5167b5 No.2825
>>2824
So for c = 145,
a = 4 * 5^2 + 8 * 5^2 + 5 = 145
n = 2 * 5^2 + 2 * 5^2 + 1 = 61
(1, 1, 5) = {1:1:50:9:41:61}
c = 901
a = 4 * 14^2 + 8 * 14 + 5 = 901
n = 2 * 14^2 + 2 * 14 + 1 = 421
Again, not sure if this is of any actual use.
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3ed644 No.2826
>>2824
Look at the attached image in >>2815
The mirrored (e,n) at x and -x follow a pattern where the difference in x is always 2x, and the difference in b is always 4x.
The equations make this obvious. Just restating that there are 2 possible ways to get to a solution, and 2 hints about movement in b.
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87160d No.2827
>>2816
Thanks Baker! Here's the spreadsheet, anons. Couple of quick notes:
1. The formula for x works for all (e,1) but not yet for higher values of n. Replace the x value if you're inspecting an element that has n>1. It will auto fill the visual and re-calc all values.
2. The third box on the right is actually just the top left box (a * (x+n)) moved down to more easily see a*b = c in rectangular form. This is where b = a+2x+2n comes from. I left both boxes for accuracy, but you only use one.
3. Hope you lads find it helpful for visualizing the problem at hand!
https://anonfile.com/H4Qawed4be/DiffOfSquares.xls
https://anonfile.com/J1Qcw5dab4/DiffOfSquares.xlsx
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3ed644 No.2828
>>2827
>"Use the z on y at x"
this hint was from VQC, correct?
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5167b5 No.2829
One thing I've been thinking of.
How would these patterns look if we look at numbers we know are primes?
For example, take a,b = 1,17. We know 17 is a prime. So patterns should diverge from when the case it's not.
My point is, have we discovered a better primality test than the ones that exists so far? If so, then it lends creedence to the truthability of the whole VQC.
Another thing I'm curious about, VQC said he was going to show us a lot of things. This grid is just one virtual quantum computer. He never said anything else about the second one which we are supposed to use to show us the language of math.
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ab8fb8 No.2830
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3a91e0 No.2831
>>2830
>>2828
Baker, any ideas on who you think it was? I thought it was Chris for sure. Trips were disabled on cbts main, and it sure sounded like VQC's style. Whoever it was also claimed to have set parameters within the VQC itself. (pic #1 related) I thought it was Senpai showing up for work all baked. Any thoughts? I'll go dig through the archives, CBTS has deleted all our RSA breads.
PMA, I'll try to find the reference. Worst case is that it's an irrelevant hint. And the accurate quote is "Use the Y on C at X." Pic related. There's a way to tie x to c, I think! Linear x, exponential c. We gotta write a function for it.
No matter what, this interchange is highly entertaining and badass, Chris or not.
RSA #2 —— archive.fo/fyzAu
ID: aaa2ec
Ctrl F is your friend.
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ab8fb8 No.2832
>>2831
I honestly thought it was an AI.
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3ed644 No.2833
>>2831
thanks. I've reread. Still searching…
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ff4dbe No.2834
>>2831
>use the y on c at x
Are you sure that's meant to mean the variable x? We don't have a variable y, so x and y might refer to e and n, meaning "use the n on c at e".
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3ed644 No.2835
>>2834
>"Use the Y on C at X."
this is the correct quote. it also mentioned the boiling point of vinegar. 100.6c or 213f. Mentioned 1,1,2,3. Some other stuff I didn't understand.
and this gem also:
112115 is also 5510101020
which I think means {1:1:2:1:1:5} is also {5:5:10:10:10:20}?
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3ed644 No.2836
>>2835
>{5:5:10:10:10:20}?
doesn't look like a valid record. no clue.
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3a91e0 No.2837
>>2834
Not sure, Anon! It could mean exactly what you think, that's definitely one possibility. I thought it could be that as well.
>>2835
Lol, PMA there are some gems in there for sure. At least it's a fun read. There's a bunch of stuff I don't understand either.
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3ed644 No.2838
Guys… A little bit of progress.
pic related. See "connection?" record.
Only works in 3 of my test cases. Perhaps we can take it further.
Process:
1) create c record.
2) create record at (e,1,1-(n+t)). Only need this for the negative x value.
3) create new (e,n) record with EDA where:
newd = (n-d)*3 + 2*d + 1
newa = newd - xFromStep2.
4) the n value in the record from step 3 is the b value in the prime result. Solve using EDB.
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51b282 No.2839
Interesting records to investigate:
For c = a*b prime
(2*c-1, 2*c-1, 3*c-1, 2*c-1, c, b)
(2*c-1, 1, c+1, 1, c, b)
(2*c, 1, c, 0, c, b)
(0, 2*c, 3*c, 2*c, c, b)
For all t (0, 2*c, t) a and b are multiples of c
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ab8fb8 No.2840
>>2838
What formula are you using to calculate t?
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3ed644 No.2841
>>2840
t comes from the c record at (e,n,t)
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ab8fb8 No.2842
>>2841
Yes, what formula? And what is EDB, EDA?
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3ed644 No.2843
>>2838
>>2842
Create the c record normally. Using the values from there to create another record via ent:
e = c.e
n = c.n
t = (1-(c.n+c.t))
This will give you a -x value.
Alternatively, create a record via ent with t equal to n+t, and just use the x value as negative.
>EDB, EDA
just some helper functions to create records.
public static TheEndRecord CreateForEDA( BigInteger e, BigInteger d, BigInteger a ) {
BigInteger c = d * d + e;
BigInteger b = c / a;
BigInteger n = ( a + b ) / 2 - d;
BigInteger x = d - a;
return new TheEndRecord( e, n, d, x, a, b );
}
public static TheEndRecord CreateForEDB( BigInteger e, BigInteger d, BigInteger b ) {
BigInteger c = d * d + e;
BigInteger a = c / b;
BigInteger n = ( a + b ) / 2 - d;
BigInteger x = d - a;
return new TheEndRecord( e, n, d, x, a, b );
}
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51b282 No.2844
>>2843
excellent!
Thanks PMA
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3ed644 No.2845
>>2844
you're welcome!
By the way, we can save a calculation at the beginning (if this ends up being a correct approach), by creating our entry record with a,b values of c,1.
This gives us the same e, n, d values, but with the -x we are looking for.
example:
145=5x29
created from a=1, b=c
c = (1,61,6) = {1:61:12:11:1:145} = 145;
created from a=c, b=1
-c = (1,61,-66) = {1:61:12:-133:145:1} = 145;
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ab8fb8 No.2846
>>2845
Wow. It was that simple.
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5167b5 No.2847
What's the row for a, b = 2, 7?
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5167b5 No.2848
>>2847
14 mod 4 = 2. So ignore it or divide it by 2.
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3ed644 No.2849
>>2848
There are similar entries in the -n space:
(-4,0,2) = {-4:0:4:2:2:6} = 12
(-9,0,2) = {-9:0:5:3:2:8} = 16
I don't see one for 2, 7 though.
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3ed644 No.2850
>>2849
>There are similar entries in the -n space:
-e space.
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3ed644 No.2851
The formula to go to matching -x records depends on e:
even values of e
(e,n,t) -> (e,n, (2-t))
odd values of e
(e,n,t) -> (e,n, (1-t))
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3ed644 No.2853
Still working. Rereading crumbs.
>A prime number 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.
The simple to calculate comment was with regards to the entry point at a=1,b=c.
Anyone made any further progress?
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99eff8 No.2855
This is likely unrelated, or not, or… maybe…
Does… is this looking like it makes sense?
(0, {[0.0…n]~[0.4999…n]}, 0.5, {[0.5000…n]~[0.999…n]}, 1)
and somehow ø and "universal set" are involved or something.
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ab8fb8 No.2856
>>2853
No, it wasn't. According to what he wrote, "only appears in one column" was simple to calculate.
This would be a very important discovery, because it would mean being able to instantly identify a prime.
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51b282 No.2858
What's in a name?
ChrisrootODavid?
[D]oes it [M]ean anything?
Root of d = {0, 2*d, 3*d, 2*d, d, 9*d}. All a and b of ( 0, 2*d, t) are multiple of d for all t.
Yes and that pattern can be used elsewhere.
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fe8a56 No.2859
>>2853
>>2856
Hey Baker and PMA! I’ve got some good ideas on how to find all the primes. We should work on solving that big N crumb. Even had some verification about my ideas from Stony Senpai back in RSA #2. I’ll post it up later today, away from my notes at the moment.
>>2858
Good idea Anon! Worth exploring for sure. Have you found any patterns in roots of d?
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51b282 No.2861
>>2859
>>2858
I thought that record might be root of d.
He verified it!
Plenty of patterns but no tie back to our goal yet.
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5167b5 No.2862
>>2853
Sorry, I'm still at this project, just real life that is catching up now that the new year is upon us.
I'm not going to give up this easily. I don't have anything new, I've been working for the past few days and after work I have other commitments, but I still read over crumbs and think every day.
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ab8fb8 No.2864
So, we can factorize d?
If we can factorize d with (0, 2d), can we factorize c^2, with (0, 2c)?
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51b282 No.2865
>>2864
>>2839
Unfortunately it isn't factorizing. Use the pattern for c, d, e, n or whatever. For example e. For any t … e is a factor of every a and b of (0, 2*e, t) using the pattern.
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87160d No.2866
>>2835
What's up Topol!! Nice to see you. Thanks for the inspiration as always!
>>2862
>I'm not going to give up this easily. I don't have anything new, I've been working for the past few days and after work I have other commitments, but I still read over crumbs and think every day.
Thanks IseePatterns! I think we're all feeling pretty much the same. Let's keep going. I'm seriously falling asleep thinking about the grid at night, and going over crumbs every day too. The solution will present itself if we keep adding to and writing down the rules like VQC asked us to.
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ab8fb8 No.2867
>>2865
So, instead of going down the tree, you're going up the tree, eh? Factorizing numbers with e instead of finding the factorization of e?
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51b282 No.2868
>>2867
Yes its moving up the factor tree. There were a number of hints in that regard in terms of where factors would turn up. For example (2*c-1, 1, t) a and b turn up as factors almost twice as often as c but no way to find them without iteration. But it doesn't help move down from N.
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ff4dbe No.2869
I don't say this to be a demoralization shill but I've really lost enthusiasm for this lately. Chris hasn't even posted for two weeks. I know it's real, and I haven't contributed since I gave up looking for a linear relationship to use in binary search so I have no idea if any progress has really been made. I just wish Chris hadn't kept saying "oh I'm going to tell you how it works now, oh wait I'll tell you in a week, oh wait I'll tell you in a month" etc.
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3ed644 No.2870
>>2858
>Root of d = {0, 2*d, 3*d, 2*d, d, 9*d}. All a and b of ( 0, 2*d, t) are multiple of d for all t.
>Yes and that pattern can be used elsewhere.
See pic related of c=145 where d=12 (both c and prime result records), for (0,2a), (0, 2b), (0,2n), (0,2c)
p e=0 n=2a a=a => (0,10,6) = {0:10:15:10:5:45} = 225;
p/c e=0 n=2d a=d => (0,24,13) = {0:24:36:24:12:108} = 1296;
p e=0 n=2b a=b => (0,58,30) = {0:58:87:58:29:261} = 7569;
c e=0 n=2n a=n => (0,122,62) = {0:122:183:122:61:549} = 33489;
c e=0 n=2c a=c => (0,290,146) = {0:290:435:290:145:1305} = 189225;
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3ed644 No.2871
>>2870
and c for these records is d^2.
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51b282 No.2872
>>2870
for all t of (0, 10, t)
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3ed644 No.2874
>>2872
Here are valid (0,10,t) records from t=-10 to t=50.
They only exist 5 t apart.
(0,10,-9) = {0:10:0:-20:20:0} = 0; f=1; (x+n)=-10; f+c=1; -x: 0; (x-n)=-30
(0,10,-4) = {0:10:-5:-10:5:5} = 25; f=-9; (x+n)=0; f+c=16; -x: -10; (x-n)=-20
(0,10,1) = {0:10:0:0:0:20} = 0; f=1; (x+n)=10; f+c=1; -x: -20; (x-n)=-10
(0,10,6) = {0:10:15:10:5:45} = 225; f=31; (x+n)=20; f+c=256; -x: -30; (x-n)=0
(0,10,11) = {0:10:40:20:20:80} = 1600; f=81; (x+n)=30; f+c=1681; -x: -40; (x-n)=10
(0,10,16) = {0:10:75:30:45:125} = 5625; f=151; (x+n)=40; f+c=5776; -x: -50; (x-n)=20
(0,10,21) = {0:10:120:40:80:180} = 14400; f=241; (x+n)=50; f+c=14641; -x: -60; (x-n)=30
(0,10,26) = {0:10:175:50:125:245} = 30625; f=351; (x+n)=60; f+c=30976; -x: -70; (x-n)=40
(0,10,31) = {0:10:240:60:180:320} = 57600; f=481; (x+n)=70; f+c=58081; -x: -80; (x-n)=50
(0,10,36) = {0:10:315:70:245:405} = 99225; f=631; (x+n)=80; f+c=99856; -x: -90; (x-n)=60
(0,10,41) = {0:10:400:80:320:500} = 160000; f=801; (x+n)=90; f+c=160801; -x: -100; (x-n)=70
(0,10,46) = {0:10:495:90:405:605} = 245025; f=991; (x+n)=100; f+c=246016; -x: -110; (x-n)=80
Maybe this means that (0,2d,t) records only exist where t = d apart.
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3ed644 No.2875
>>2874
not quite correct. Examples where d = 12 at (0,24). Valid records are 6 apart.
(0,24,-23) = {0:24:0:-48:48:0} = 0; f=1; (x+n)=-24; f+c=1; -x: 0; (x-n)=-72
(0,24,-17) = {0:24:-9:-36:27:3} = 81; f=-17; (x+n)=-12; f+c=64; -x: -12; (x-n)=-60
(0,24,-11) = {0:24:-12:-24:12:12} = 144; f=-23; (x+n)=0; f+c=121; -x: -24; (x-n)=-48
(0,24,-5) = {0:24:-9:-12:3:27} = 81; f=-17; (x+n)=12; f+c=64; -x: -36; (x-n)=-36
(0,24,1) = {0:24:0:0:0:48} = 0; f=1; (x+n)=24; f+c=1; -x: -48; (x-n)=-24
(0,24,7) = {0:24:15:12:3:75} = 225; f=31; (x+n)=36; f+c=256; -x: -60; (x-n)=-12
(0,24,13) = {0:24:36:24:12:108} = 1296; f=73; (x+n)=48; f+c=1369; -x: -72; (x-n)=0
(0,24,19) = {0:24:63:36:27:147} = 3969; f=127; (x+n)=60; f+c=4096; -x: -84; (x-n)=12
(0,24,25) = {0:24:96:48:48:192} = 9216; f=193; (x+n)=72; f+c=9409; -x: -96; (x-n)=24
(
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ab8fb8 No.2877
>>2875
>>2874
Chris says the solution is meant to be hidden in plain sight
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51b282 No.2878
For say d = 5 then these are the proper t.
t e n d x a b
1 0 10 15 10 5 45
2 0 10 40 20 20 80
3 0 10 75 30 45 125
4 0 10 120 40 80 180
5 0 10 175 50 125 245
6 0 10 240 60 180 320
7 0 10 315 70 245 405
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3ed644 No.2879
>>2878
>>2877
I have no doubt the solution is in plain sight. And something very simple and elegant.
But those are not valid t values for d=5, unless t needs to be redefined based on column (0,n). Which I think is an incorrect approach.
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3ed644 No.2880
>>2878
Starting looking at (0,2d,c) and (0,2d,n) for c=65 and c=145.
Interesting that they both exist.
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51b282 No.2881
I disagree with your t's.
What you call (0, 10, 1) isn't a valid record. If a = 0 everything is 0.
{0 10 15 10 5 45} is the first record.
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51b282 No.2882
>>2880
Every positive integer t will exist for (0, 2d, t) and x/2*d will give you t.
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51b282 No.2883
>>2879
This is (1, 1, t) first 10 records. Notice d pattern of two times perfect square starting with 1. t=1 d=2*tt. Lots of patterns for t math when you use this for
t = 1
t e n d x a b
1 1 1 2 1 1 5
2 1 1 8 3 5 13
3 1 1 18 5 13 25
4 1 1 32 7 25 41
5 1 1 50 9 41 61
6 1 1 72 11 61 85
7 1 1 98 13 85 113
8 1 1 128 15 113 145
9 1 1 162 17 145 181
10 1 1 200 19 181 221
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51b282 No.2884
>>2877
I'm also convinced it will be relatively simple but I don't remember a hint quite like it would be in plain sight.
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3ed644 No.2885
>>2871
d = {0, 2*d, 3*d, 2*d, d, 9*d}
I think the c for these records is 9 * d^2.
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3ff2ea No.2886
>>2884
>>2879
Often times really good equations are simple and elegant!! I don't remember a crumb about "Hidden in plain sight." PMA, can you provide sauce?
Important Question!: Ok anons, been thinking long and hard about variable t: Row one has ascending t, in 1-infinity order. As n increases, t jumps upward in patterns. Does this matter at all for calcs? Maybe t naturally skips as n>1, and is still useful in higher values of n? maybe some values of x don't exist at higher levels of (e,n)?>>2879
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3ff2ea No.2887
>>2877
Oops! Sorry, it was Baker. Can you provide sauce for the "hidden in plain sight"?
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ab8fb8 No.2888
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70de64 No.2889
Okay I think I have something.
Any cell generates an (e,n,d,x,a,b). From this if we have the entry (e+d*d, n+d) we have our position on the D grid. The D grid can be generated in two different ways.
Start at (0,0) and move to the left by square amounts (ie (0,0) -> (-1,0) -> (-4,0) etc) and from here generating parabolas up and down (or just down for positive a and b) to get
(0,0): (1,1), (4,2), (9,3), (1,-1),(4,-2) etc
(-1,0): (0,1), (3,2), (8,3), etc
This way you have a bunch of parabolas that are basically parallel.
Or you can generate it another way by starting at (0,0) and making the parabola that goes (w^2, w). Then for any cell (w^2, w) you can generate lines that go (w^2+2wt, w+t) so you get a bunch of rays going out from a parabola. Lets focus on this way to generate it and talk about the rays. We can identify these Rays from the bottom up by 1,2,3, etc. This is because one is generated after w=1, then w=2 etc.
If you generate values for (a,b) = (1,c) and place them on this grid, they are always on the bottom-most line. Then if you want any other factors of C they are on the same vertical line on the D grid. (ie they have the same E value). If you notice where these are placed on the D grid, these entries are always either ON the D grid, or they are halfway between two points on the D grid and have the same N value as the first value on the ray (or the one with the smaller N). Also from our D value, we know that any solution cell is on a ray (or exactly between 2 points on a ray) and the ray must be of a value less than D. So basically we need to get our value and check if there is a solution to one of our lines (or halfway between one) with the same E. Then that N value is the correct N value
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70de64 No.2890
>>2889
this is the D grid and the two lines are the coordinates of the D value of the cells
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70de64 No.2891
>>2890
Also this is a cool pic I forgot exactly how I made it. I think it is the C values with their respective values in each D grid (up until a certain amount, then I did the F function on each of them. This looks pretty neat but I think its worthless.
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ab8fb8 No.2892
>>2891
>>2890
This is fascinating. It shows that it's all linear, somehow.
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70de64 No.2893
>>2889
Basically for this type of algorithm you would get your number and get the E value for the record and then transform it into a D coordinate by doing E + D*D (which is actually just C, neato!)
So actually from the D grid we know that the E coordinate is just the C value. If we check the lines we can only be on the d lines from 1 to D, starting from the bottom line. Then we need to check for each d (from 1 to D) if (C - d^2) is divisible by d [d^2 is the origin of the line which is tangential to the parabola]. If we get this IS divisible, then we have the correct N and it is at whatever value that would calculate to. I don't know it off the top of my head. This would involve factoring a bunch of numbers, but we could probably do it recursively which could be simple.
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70de64 No.2894
>>2893
>>2893
damn this is worthless
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ab8fb8 No.2895
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70de64 No.2896
>>2895
if you're checking to see if C-d^2 is divisible by d, you already know d^2 is, so checking if C is would be presumably worthless because you could just check C anyway. There would be no point in going past D for the little d's because thats just common sense
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ab8fb8 No.2897
>>2896
But your original post is worth looking into?
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70de64 No.2898
>>2897
Yeah I think the Gif and the coordinates will help. Just the specific algorithm is bad. My idea is that if we can transform into the D grid and somehow navigate that grid to find the right coordinate and then transform it back. I'm also thinking it might be worth it to (idk exactly how this would be done) but to basically stretch everything into parallel lines instead of the parabolas, then we would have a normal looking grid which could potentially be even easier to navigate. Maybe then the pattern on that grid would be worth noting? I'll try and render an image of that now. I also have some other stuff I was working on and I think that the parabolas on the left which start at (-d^2,0) each produce all products of d. Also I reflected the image over the y axis to to do negative factors, but idk if that is useful yet.
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ab8fb8 No.2899
>>2889
Just noticed the first bit. Did you forget d^2 + e = c?
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70de64 No.2900
>>2899
The last paragraph is the important part. Download the gif and look at the lines when the red dots shift. The red dots are either on a line or between two lines
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70de64 No.2901
>>2898
This is what happens when you transform it into the grid where the parabolas are actually just straight lines and the vertical is based on the T value which is (b-a)/2. This T value also corresponds to which entry it is from the origin of the ray. Basically if you look at the D grid, I ripped out the prabola on the right and turned all the lines into vertical lines and closed the holes. idk if that helps. Also I generated the same for negative factors. For this, we have a hyperbola. If we can find the two lines for the parabola, then we can solve integer solutions for it and those would all be valid entries for C.
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ab8fb8 No.2904
>>2903
You lost me a while ago. What is w?
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70de64 No.2905
>>2904
Just an input variable.
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70de64 No.2906
Here is all the values colored in a gradient
>>2901
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70de64 No.2907
>>2906
lmao could we then transform this into another, more linear graph? We'll see next time. Also on a more serious note, I am posting this stuff because I find doing this stuff fun and I figure that it can't hurt to have more info. If this has already been discovered or if you guys think this is irrelevant I won't post it anymore.
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ab8fb8 No.2908
>>2906
I really don't understand any of this, so I won't condemn you by saying it's not relevant. Your brain seems to be on max CPU usage though, so think about the crumb that the solution is hidden in plain sight.
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ab8fb8 No.2909
Found some pretty patterns. Go to (1,1) and take the difference of (d+1)^2 and d^2 for each cell.
Very beautiful pattern, and it's exactly what my program prints out when I factorize 145
Each value is also only 1 integer away from a square.
In short, if you want to see even more amazing patterns, make a list of each cube and square and take the differences of them.
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99eff8 No.2910
I was waiting for you to show back up.
Fun timing considering "world events".
Give Chris our best ponies.
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99eff8 No.2911
>>2910
I can't unsee Rick portals as "rabbit holes" now…
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70de64 No.2912
>>2907
Here it is although this isn't exactly what I was looking for
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99eff8 No.2913
>>2912
gee… 0 sure seems to be an important thing there…
How does one get from 0 to n, as in (0.000…n), btw? When does the "1" show up?
cuz it's the same "n" that gets you from 0.999… to 1…
Right?
Amirite?
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70de64 No.2914
>>2913
Not sure what you're getting at. Here is the grid with all the multiples of any give number. Notice how there is a little box that is the same just repeated over and over again.
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70de64 No.2915
Here are the primes. Notice they have a little zig zag guy in every square.
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70de64 No.2916
>>2915
These are actually squares. I accidently saved over the prime image with the square image. I'll remake the prime one. With this square image you can see that there is one cell that kind of follows it's way through the charts. Sort of near the origin. I'd guess that is the entry for d*d.
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70de64 No.2917
Here are the prime entries
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70de64 No.2918
>>2917
This is the biggest version without the skew. You can really notice the patterns for this one. Notice there are squares for each thing and for odd bois you have a little zigzag guy going horizontal through the middle and for even bois you have a straight line going through the middle of a square.
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70de64 No.2919
Here is the grid for products of 2 primes. Notice that on each of these they have the same spicy boi zig zag grid that the odd entries had. These are also the only ones that matter for RSA.
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5167b5 No.2920
I've been looking into the patterns behind the 1, c rows.
So take c = 145. The row for a,b = 145 = (1, 61.0, 12, 11, 1, 145)
I've been wondering about the relationship between this N and the 5. So this doesn't let us find 'n', but it should give us some insight into how these are connected.
So we know for row (1, 5) we have the following:
{1:5:4:3:1:17}
{1:5:12:7:5:29}
{1:5:30:13:17:53}
{1:5:46:17:29:73}
{1:5:76:23:53:109}
{1:5:100:27:73:137}
{1:5:142:33:109:185}
{1:5:174:37:137:221}
{1:5:228:43:185:281}
{1:5:268:47:221:325}
{1:5:334:53:281:397}
{1:5:382:57:325:449}
{1:5:460:63:397:533}
If we split these up into two chains we get:
Chain #1
{1:5:4:3:1:17}
{1:5:30:13:17:53}
{1:5:76:23:53:109}
{1:5:142:33:109:185}
{1:5:228:43:185:281}
{1:5:334:53:281:397}
{1:5:460:63:397:533}
and
Chain #2
{1:5:12:7:5:29}
{1:5:46:17:29:73}
{1:5:100:27:73:137}
{1:5:174:37:137:221}
{1:5:268:47:221:325}
{1:5:382:57:325:449}
Now take the a, b's from chain #1 and generate the row for a=1, b=c. We get:
>>> rowForAB(1, 1 * 17)
(1, 5.0, 4, 3, 1, 17)
>>> rowForAB(1, 53 * 17)
(1, 421.0, 30, 29, 1, 901)
>>> rowForAB(1, 53 * 109)
(1, 2813.0, 76, 75, 1, 5777)
>>> rowForAB(1, 185 * 109)
(1, 9941.0, 142, 141, 1, 20165)
For chain #2 we get:
>>> rowForAB(1, 5 * 29)
(1, 61.0, 12, 11, 1, 145)
>>> rowForAB(1, 73 * 29)
(1, 1013.0, 46, 45, 1, 2117)
>>> rowForAB(1, 73 * 137)
(1, 4901.0, 100, 99, 1, 10001)
>>> rowForAB(1, 221 * 137)
(1, 14965.0, 174, 173, 1, 30277)
Note: I didn't compute every a=1, b=c for the different chains, but you should get the picture.
Today I've been looking into how the n grows and found two equations, one that explains it for the main chain (#1) and one for the second chain.
For chain #1 we can predict the n, d and x values without knowing the entries in (1, 5). Instead, we base it on the n (5) and d, x from (1, 1). I haven't found any good equation for (1, n) where n doesn't appear in (1, 1), but I'll look more into that later.
For even e:
n[t] * t^2 + d[t] * t + (x[t] / 2)
For odd e:
n[t] * t^2 + d[t] * t + ((x[t] + 1)/ 2)
This equation is only for the chain #1. For chain #2 we have instead:
n[t] * t^2 + d[t - 1] * t - x[t-1]
Again though, this doesn't solve for n. It doesn't solve much right now, but at least it explains the relationship between n and the big N's.
To give you a better understanding here is a few examples:
Let's try with (5, 1). We'll pick some values, we will start with finding the big N for values in (5, 27). So we have the row:
{5:1:34:7:27:43}
>>> n = 1
>>> 27 * n * n + 34 * n + (7 + 1)/2
65.0
>>> getAFromT(5, 65)
8323.0
>>> rowForAB(1, 69 * 245) # This is the 3rd entry in (5, 27), but the second entry in chain #1 after a=1,b=69
(5, 8323.0, 130, 129, 1, 16905)
>>> n = 2
>>> 27 * n * n + 34 * n + (7 + 1)/2
180.0
>>> getAFromT(5, 180)
64443.0
>>> rowForAB(1, 529 * 245)
(5, 64443.0, 360, 359, 1, 129605)
As for the d and x in those records, they are d = 2 * (27 * n * n + 34 * n + (7 + 1)/2) and x = 2 * (27 * n * n + 34 * n + (7 + 1)/2) - 1
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5167b5 No.2921
>>2920
I noticed my formulas might lead to some confusion.
For even e:
n[t] * t^2 + d[t] * t + (x[t] / 2)
For odd e:
n[t] * t^2 + d[t] * t + ((x[t] + 1)/ 2)
This equation is only for the chain #1. For chain #2 we have instead:
n[t] * t^2 + d[t - 1] * t - x[t-1]
Here [t] represents the t for n in (1, 1) while t is the t for (1, n[t]).
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1f670a No.2922
I'm already nearly finished with the white paper on just this subject.
http://www.kurzweilai.net/forums/topic/lets-talk-about-math-and-the-expression-of-energy-vs-surface-states?replies=8#post-826480
Also, the same software/algorithm, is essentially an infinite probability calculator, since there is a pattern in primes, you can use it as a quadratic equation to make a shortcut/s in diadic differential equations.
Ask me anything
What this furnishes is not only a way to break rsa, but a solution to that problem, furnishing truly unbreakable"The heat death of the multiverse would occur many times over" encryption, that is active in real time.
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1f670a No.2923
I have been busy, and haven't read all of the posts here. The goal, is creating a software that can be used on legacy machines, and allows them intercompatibility, and capability in mimd, and parallel computation/compression, decompression. The possibilities are endless.
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5167b5 No.2924
>>2923
MIMD as in Multiple Input Multiple Data (as opposed to SIMD)?
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1f670a No.2925
Let's have a look at some of my rudimentary, and crude design parameters
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1f670a No.2926
You are the input, but because it is a Turing complete idea within the Turing complete machine, it needs to make a decision. By it, I mean the software, so yes, mimd, is the way to go, it looks at multiple outcomes and therefore better uses the available hardware.
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5167b5 No.2927
I don't quite understand the pattern yet, but for (1, 17) the big N equation is:
17 * n * n - 4 * n - 4
17 * n * n + 4 * n - 4
And for (1, 29) it is:
29 * n * n - 12 * n - 6
29 * n * n + 12 * n - 6
The parts I don't understand yet is why 12 and 6 are important.
I'm assuming they refer to d-values and x-values.
For 17, we do have d = 4 at {1:5:4:3:1:17}
and for 29 we have {1:5:12:7:5:29}.
Both d's in this case match the d's in the equation used, but I haven't locked down why or if this is a coincidence. As for the x-part I'm still lost.
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1f670a No.2928
I tried to think in patterns of binary, rather than math, the iching, a 7,000 year old Chinese text, is very interesting in this. The pa kua. These are words, the computer, and math/vector analysis, don't start to become coherent until we have word lengths of 4 or more, from the human observable perspective.
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5167b5 No.2929
>>2927
For (1, 37) we have the following equations:
37 * n * n - 6 * n - 9
37 * n * n + 6 * n - 9
Again, {1:13:6:5:1:37}.
Here the d is correct, but I can't make heads and tails of the x value.
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99eff8 No.2930
>>2928
Heya! Just so you know, you can post multiple images at once (4-5 max). This thread is for figuring out the um…. genesis code? do we have an official name for it? lol
Anyway, a bit of stimulation and support is awesome, but if you wanna post a lot, please take a step into my oven: >>846
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1f670a No.2931
Understood, here is my prelim work for the white paper, for general public release.
In and Around 1945, mathematician, and physicist, John Von Neumann, and others, working on a project detailing the EDVAC, ENIAC, and it’s operation, set the paradigm for our modern digital world.
All modern hardware architecture from that point forward followed this basic design idea.
The fatal flaw, which is the very object of the solution that this paper covers, is the simple idea of the Von Neumann bottleneck, and in layman’s terms is the inability to have an instruction fetch, and perform data operation at the same time, or operate in a parallel capacity, based on the hardwares physical limitations.
The first problem, is to address whether a simple, elegant solution can be applied to all modern systems, and even the idea that legacy systems can not only be effected, but can be considered useful if the solution can be applied.
What is language? what, or who gives it meaning?
To answer this we look not to the computer, or machine that the user uses ex-corporis, but internally to the brain,that drives the genetic machine that we call the body.
The language that we humans use cannot be used effectively by a computer, because of binary principle.
What the casual reader of this paper may not know, is that this language, while designed before the writer of this papers time, constitutes a library of subjects, and objects.
When I use the word table perhaps a wooden object comes to mind, or perhaps an array? Whatever the idea, the object being referred to by the subject is stored, and completed in a parallel fashion faster than thought itself.
The real problem is not the Von Neumann bottleneck, but the speed, and efficacy of the user in inputting the information in a simple, and intuitive way that will translate into binary patterns that can be stored, written, retrieved, and re-arranged at a speed that the user finds comparable to actual thought, as well as the creation of a set of subject/object oriented patterns that the hardware can use to accomplish this task.
To create this illusion, and to allow it to interface in a way that is comfortable to the users liking, and in even a very basic form reflects the user, is based on 1: the hardware, and 2: the user.
This is the encryption itself.
Say the user has an older smartphone, the limitations of the hardware only allow the software to adjust to the user based on the physical specifications of the system that it’s installed on.
The other half of the encryption is the users own input, just as the idea of the bio-metric identification system makes identity possible, so too does the comparison between the user and the softwares interface with the hardware.
The users are human, and they change over time, so it makes sense that we would want a constantly rotating key that allows anonymity, and extra digital protection.
In order to break the encryption they would need the physical phone, and a way to physically replicate the user, and their personal usage patterns, more secure than a fingerprint.
To do that, you would need to know the user, and also know their location, which the software makes impossible with encryption.
Things like habits, location, family, and others will be taken into account, as this is the physical input to the device.
Let us say that the users device becomes stolen property, it will be easily traceable, and recoverable, as the design of the software is so that the imprint is made before the software can be fully used.
This makes theft worthless, and reuse of stolen devices impossible, except by the user assiged to them.
The idea is that the softwares evolution is the information that is recursively placed into the rotating encryption that is active, in real time, which is the very software actively managing all systems above it, from the hardware level.
The old patterns are replaced with new ones, as the device is used in new ways, and it receives input from the user.
The software cannot take over the world, and cannot perform tasks that require human action, nor can it even conceive that humans even exist.
The best way to describe it is that it is like a mirror, you get out what you put in, if the casual user put in the grocery list they will get back a grocery list, but if the user creates anything it will reflect the creative flair of the user in its every essence.
The limits are the imagination of the user, and clever users will take full advantage of their equipment by using their hardware to better lock their device, and unlock the full features of the user themself.
Let me be clear, in it’s infancy, just as our children must learn to become intelligent, the software itself is not super-intelligent, and cannot be so, it will take time, and various inputs to make it truly indistinguishable from human intelligence.
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5167b5 No.2932
Sorry guys, real world is catching up and I got work, but damn. Figuring out the relationship between n and big N seems like a big achievement.
I need a fresh mind for tomorrow. My brain is stuck on this pattern.
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1f670a No.2933
Make no mistake, we are working on the same thing, but from differing perspectives, I'm not here to play games, or post meme images.
The image posted is a small discovery I made entirely by accident. A Pattern in primes that is the shortcut I have been referring to. A si.ple, elegant quadratic equations that allows computational proof of unknown primes of any size, but is more than that. The interesting part is the link between this, and the golden mean, and pi, as they pertain to prime numbers.
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5167b5 No.2934
>>2933
Then you found the right spot.
If you don't mind sharing, what is the quadratic equation you found?
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1f670a No.2935
Let's just say that while I agree that this is important work, this is not the place for it. Fair enough? I have no grandiose plans to keep the info to myself, or accept all 7 millennium prizes for myself, along with beales money, at al. Or even desire the nobel prize, but it is sensitive because of its explicit use in the software I'm developing. There cannot be a link between it and me. I am #anon20204president.
I can assure you, this Is not a ruse, nor a lie.
Instead I'll use a form of encryption, that involves a hunt. I'll provide info, as clues so that only a clever poop slinging monkey can figure it out like I did.
A matrix of primes, as a spatial reference, or map……
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3ed644 No.2936
The c value at the record (0,2d) where a = d, is equal to 9*(c-1).
Example for c=145:
(0,2d) record:
(0,24,13) = {0:24:36:24:12:108} = 1296
(e,n) for c=145 record
(1,61,6) = {1:61:12:11:1:145} = 145
9*(145-1)=1296
I think generally, this can be written as:
c of (0,2d) at a = d
equals
9*(c-e) at (e,n) at a=1, b=c
I have looked at various records at e=1, 5, 6 and 8. Someone please confirm.
As the prime results share a relationship to 2d, perhaps there is a connection to those records as well?
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1f670a No.2937
Biggest clue I can give is that the universe isn't neat, it isn't clean. It exists as incommensurable and incomprehensible ideas
Stop trying to think in neat whole numbers
I started by finding what two algorithms that can be squared and get a prime, and continued to do the same…… viola!
Note: the primes exist between all real, whole numbers, and make a network behind the scenes, they connect to each other just as non primes do. Try multiplying pi and golden mean together and find the square of 5, and 79, and start peeking around. Notice in my diagram the primes listed in the margins? They make up a perfect cube of 5x5x5 of prime numbers that add sub mult and div like sudoku. There is a reason I expressed this in this way. And also why I chose the cone as a part of the diadic equation to represent 3d space of the cube these numbers are represented in.
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ff4dbe No.2938
>>2937
For someone who seems to know a lot of obscure information and have a "free mind", I'm surprised you'd use gmail.
>Try multiplying pi and golden mean together and find the square of 5, and 79, and start peaking around
3.1415926 * 1.618 = 5.083096827, but I have no idea what you mean about using that to find the square of 5 and 79
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c98aa4 No.2939
There is a reason I use gmail.
Great observation skills there. Maybe if you pay close attention you might have a chance at learning something other than arbitrary there anon.
Are you government? Or perhaps cicada3301, they would love this to be public release.
If you want it, earn it, like I did.
I didn't use apl, to see it in my head.
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99eff8 No.2940
>>2939
If you click "No.xxxx", you can reply directly.
Also, if ya stray toooooo far from the VQC training program, y'might need yer own oven.
There's a thread about when I "earned" mine.
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c98aa4 No.2941
The numbers that multiply together to get accurate whole primes are longer than on the pages I listed, I used smaller numbers for the simple sake of attempting to find a pattern, based on my educated guess. It turns out it was right, the clever reader, and observer will be able to ascertain what I did to get the trail of numbers I did. It's not complicated, and the quadratic equation can be done any number of ways, in fact, an infinite number of ways. I'm looking for the simplest, smallest most beautiful example of this equation, that Fermat "listed at the top of this thread" theorized about. Obscure….. hmm. I never believed that a millennial would be interested in such things there anon. That having been said if you don't like being wrong, or called out this is not the place for you, and I encourage you to find someone else to harass, I don't have the time to babysit. I will be the first to admit that I do not know everything, and before anyone tries anything stupid, I don't respond well to stupidity, this is the last generation we have to get this right, and I only have maybe 50-80 years left to provide my services to the world. Christopher langan is working on my other favorite subject of CTMU so this was what I started towards until I have a chance to go and visit him and have a serious heart to heart about the nature of the nature of the universe.
I will not entertain multiverse theory questions here, as we can clearly see from anons example mankind is not yet responsible enough for such subjects.
Love
#anon20204president
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99eff8 No.2942
GENERAL NOTICE!
I foo doggo pony showed Dnimeerf and I'm gonna set up his own side oven. Also suggested he lurk the previous threads. I'll run him through some chan ropes.
Thank you for your understanding!
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a876b4 No.2944
>>2940
Look, I'm not here to take over, I'm not here to stroke egoes, I see a lot of tail chasing.
Von Neumann was asked to write an essay for the layman describing what mathematics is, and produced a beautiful analysis. He explained that mathematics straddles the world between the empirical and logical, arguing that geometry was originally empirical, but Euclid constructed a logical, deductive theory. However, he argued, that there is always the danger of straying too far from the real world and becoming irrelevant sophistry.
I'm here to help, I have already made progress in this, as in a real world public release, I don't care about what you guys are doing, I think it's fantastic, and I want to genuinely help, but as I see it, and to the best I can figure, you are doing this the long way around.
There are also many solutions to te
The same problem. Im not a knoe it all, and im not hijacking here, just providing guidance, in an arena where the world has an opportunity at real nonlinear progress. I would be a fool to impede that, but I also must protect myself, my family, and my friends. Why on earth would I give contact info? Maybe because I am leaving myself open to real questions, and the opportunity for those interested to question me directly and get answers, just not here. Ideally to solve the issues you all have a real person brainstorming sesh would lead to UNBELIEVEABLE PROGRESS, but as I can understand the climate here that is also out of the question.
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99eff8 No.2945
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99eff8 No.2946
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f03c55 No.2947
>>2944
Freemind, please remove links to our workspace that you've placed on:
http:/ /www.kurzweilai.net/forums/topic/lets-talk-about-math-and-the-expression-of-energy-vs-surface-states?replies=8#post-826480
Would be great if you would delete your post there, thanks.
>>2945
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e8586e No.2948
>>2942
Thanks Topol… I was going to say something. Always glad to see new people but this practically seemed like a de-rail attempt. I love yall's progress here! Wonderful work by all the VQC nerds. There is a major Q happening going on right now if you guys have missed it. It sounds pretty HAPPENINGey and would be worth a look for everyone here. 8ch.net /greatawakening
Keep calm and do public math. :>) I love you guys!
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e8586e No.2949
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f03c55 No.2950
>>2948
Holee chit! BIG. Thanks for heads up Great Hobo!
>>2949 n/p
Great work everyone, following on the updates!
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e8586e No.2951
>>2950
Yeah, crazy shit. Not sure if anyone here follows the q boards.
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ab8fb8 No.2952
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ab8fb8 No.2953
>>2936
I think I'm seeing something in my head here.
WHY ISN'T C THE FIRST COORDINATE FOR THE FACTORIZATION?
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ff4dbe No.2954
>>2948
>THE PUPPET MASTERS HAVE BEEN REMOVED.
Well, if those aren't the most encouraging words Q has ever posted. Seems like great circumstances for the VQC to be figured out.
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ab8fb8 No.2955
I'll clean up code and share tomorrow. It's not nearly close to done, just cherrypicked the c's
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ab8fb8 No.2956
>>2955
Imagine valleys of perfect squares and each semiprime is inbetween two
289 - 144
361 - 169
256 - 121
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5167b5 No.2957
>>2955
How did you get the different numbers?
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ab8fb8 No.2958
>>2956
That's what I discovered.
The VQC is a sea of squares and remainders. Think about what happens if cells are identified with (n, e). Think about what happens if d^2+n is used instead of d^2 + e, or d^2 + f
>>2957
numbers above 1*c are little and big square for 1c,
numbers above ab are little and big square for ab
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ab8fb8 No.2960
>>2958
for any c, d^2 and (d+1)^2 are the 2 adjacent squares
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99eff8 No.2962
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3ed644 No.2963
>>2962
en.wikipedia.org/wiki/Thue–Morse_sequence
T3 = 01101001.
T4 = 0110100110010110.
0=R, 1=L
for any music anons, looks like a paradiddle.
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3ed644 No.2964
>>2936
duh. ab=c; a=d, b=9d, c=9(d^2).
continuing with (0,2d) analysis. Think I've found the boiling point of vinegar reference.
pic related shows movement of (0,2d) record to (0,1) using (n-1)*a transform, then to (1,1) using boiling point of vinegar transform.
Very funny. Not sure how useful in solving our prime problem.
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70de64 No.2965
>>2964
>using boiling point of vinegar transform
Ah yes, the classic boiling point of vinegar transform
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70de64 No.2966
>>2965
My previous pictures were fucked up. Let t = (b-a)/2. Then if we generate cells by (n-t,t) we have their entries on the rays which is the first pic.
If we generate cells by (t,n) (which corresponds to their parabolas) we get the second pic.
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3ff2ea No.2967
>>2952
I'm here working too lads. Taking a good look for answers hiding in plain sight.
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ab8fb8 No.2968
Nobody would like to know my 3rd factorization method? I may be able to finally pull it all together.
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ff4dbe No.2969
>>2968
Of course we would.
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3ff2ea No.2970
>>2968
Sure! Let's see it, Baker.
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70de64 No.2971
>>2967
Could plain sight be looking at this again?
n = (x^2+e)/2a
so then a new record would be
(e,n,d,x,a,b) -> (e,((x-1)^2+e)/2),x,x-1,1,1+2*(newX) + 2*(newN))
because that would generate another on the same vertical (same e) (which is the goal) but with a different N. Then maybe you could generate higher or lesser values in the cell until you can get to the next cell back. Then go back another cell, shift it to see if D = D, then go to another cell and repeat.
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3ff2ea No.2972
>>2971
Hey CA! Yeah, I'm back to analyzing all the "basic" diagrams and equations looking for something we missed. Can you make your idea work to generate new records? According to Baker, Chris says the answer is hiding in plain sight.
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70de64 No.2973
>>2972
Yeah but first I'd need to figure out the constant for Dc for a (e,n) cell such that D(t) = D(t-1) + 4tn + Dc
I haven't thoroughly investigated this equation but it might work. I'm working with my grids above right now
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3ed644 No.2974
Theory:
a) at (e,n) there exists a record where (b-a) = 2*d.
b) the n for that entry represents the a value in the prime solution we are looking for.
c) the d in the (b-a) = 2*d formula comes from the d in our initial c record.
Pic attached shows example at c=785.
In plain sight, perhaps in the c=145 example solution record where:
c: {1:61:12:11:1:145}
p: {1:5:12:7:5:29}
(29-5) = 2*12.
Perhaps there is a way to solve our existing equations based on this new relationship between a, b and 2d.
Also, in the attached picture, notice there is a second record that satisfies this equation that we can get to quite easily from c.
At (1,785, -378). This is the negative x for a record created at (e, c, t=c.t+c)
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ab8fb8 No.2975
>>2974
Now add the squares
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3ed644 No.2976
>>2975
Little square, confirmed. Not seeing it yet…
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ab8fb8 No.2977
>>2976
You're seeing what I saw. Try making a list of the values of (d+1)^2 - d^2 for (1,1).
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3ed644 No.2980
>>2977
Looks like (d+1)^2 - d^2 == f + 1.
It's a pity the small square formulas in terms of t only work at (e,1).
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51b282 No.2981
>>2974
Excellent work.
The manipulation of t I think will be part of solution.
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991b3a No.2982
There is a problem. Two men sat about idly talking about the constituency of the atom. It didn't take long for them to assume that not only you can seperate the two, but you can glean energy from their breakdown. They did this without considering first what they breakdown would mean for the future of energy, even knowing madam curies personal sacrifice. What we are proposing, here, on a public discourse, without knowing the purpose, and intent of the members, is to allow public access to a technology that will allow the cabal to decrypt global encryption, and at great personal danger the abolity of someone to masquerade as someone else, by using their personal account, with or without their permission. Those that continue here are very likely being used to solve this publically for someone with dubious means. I don't mean this to be a conspiracy, but if the wrong people get their hands on this, there is nothing left. I propose that we first work on a way to use this to secure personal privacy, and give back security, and liberty to the global populace, and those that have the solution to rsa, be EXTREMELY careful. This is not a toy. Infinite probability calculation allows for calculations for faster than light transmission of data, and pattern searching that transcends this universe….. once again, I'm not here to hijack the posts, but to inform, and prevent catastrophe.
I believe that this group has been infiltrated. I'm not joking. We, those of us this inspired by the beauty of the universe realize that the universe is impressive and beautiful, that all possibilities exist within all probability subsets. Vqc can be achieved without the use of cracking rsa. Why then talk about something that endangers the freedom, and liberty of every human being on earth? Are you all stupid? You are talking about vqc, meaning that you are among the most clever of us poop slinging monkeys. But no one has mentioned a probable solution that would keep out those that would misuse this technology from getting their cabbalistic greedy little hands on it.
Let's address the rats that have infiltrated this discussion. My e-mail is real, and I posted it to open myself up to questions, and answers. To say a few words, I am working g on interdimensional travel, using this tech. And in a way, that makes me a time traveller, but don't quote me on that. THIS friends, IS DISCLOSURE. I am working towards the second rennaisance for mankind. My goals are very likely not far from many of yours, unless you want to continue to divulge information that will allow those that are opposed to disclosure, and truth to have this tech, I strongly recommmend that we tread carefully. I have an encrypted chat where we discuss these things. Ask me about it, I have nothing to hide.
#anon20204president
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51b282 No.2983
>>2974
I get your t math now it based on potential not valid x
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8f59bc No.2984
>>2982
The only infiltration here appears to be you. Everyone here has been here from the beginning. You are the only outlier showing up and spraying your shit all over. Put it in the thread that was made for you since it is off topic. If you want to work on this project then contribute to it. If you have another then use your thread. The only infiltration here is done by you. And I verified you linking here from that (((kurzweil))) guys site that someone else posted. Bet you didnt delete that post as requested did you? :(
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991b3a No.2985
Then tell me, since you propose opposition, why you need rsa cracked to complete vqc, since it can be done without it.
Who are you, and who are you working for?
I'm working against those that vanguard vice. I'm not working so that others can financially gain worthless paper currenc6, and power over others, but you seem interested in shooting down anyone who opposes you. I'm not here to stop anyone, but as a way to complete vqc, I'm on topic, and im not diverting attention for my own ego. I question your identity, and motives. You are very likely the rat.
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991b3a No.2986
The reason I linked the forum, is to help, there are brilliant minds there, mr. Kurzweil is enlightened, and the forum contents are on topic, and consistent with the content discussed here, as well as a solution or workaround for rsa replacement. There will be no normies flooding in as anon would propose, and he seems to me to be a rather enigmatic fear monger here. I am so opposed, this work is the greatest achievement of mankind, and there is no room here for those that are hotdogging.
I welcome the "second renaissance" so that mankind has an opportunity to overcome the oppressive ideology of those that would prevent freedom, and gather and abuse power for themselves.
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fe8a56 No.2987
>>2984
Ray Kurzweil May be “enlightened” but he is chief engineer at Google, building the hive mind AI. He is in service to the (((enemy))). You lads seen the new church of AI over at Google? They’re seriously planning to worship the AI and merge with it. Yuck. So much for “freedom”. AJ has talked all about this. We must be getting close, lads. Read up on Kurzweil.
https://www.theguardian.
com/technology/2014/feb/22/robots-google-ray-kurzweil-terminator-singularity-artificial-intelligence
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5167b5 No.2988
He said earlier that he doesn't want to tell how how to solve what he is doing, but he is willing to give us hints so we can solve it ourselves. Literally, inside a thread about the exact same thing.
The guy has his own thread so just hide his posts and move on.
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fda808 No.2989
>>2982
>Those that continue here are very likely being used to solve this publically for someone with dubious means. I don't mean this to be a conspiracy, but if the wrong people get their hands on this, there is nothing left.
Everyone here has already considered that this is a ploy to "crowdsource" cracking RSA, if that is for disclosure reasons or dubious purposes is irrelevant as long as the results become public. Personally I'm fine with nothing left, time to build something new!
>…and those that have the solution to rsa, be EXTREMELY careful. This is not a toy.
This is on point, suggest you post it immediately. Strength in numbers! (30 UIDs, sigh, take precautions)
>probable solution that would keep out those that would misuse this technology
Everyone gets it or we all die trying to make it public. All tech can be misused, and those who do already have access to stuff way beyond this.
>Let's address the rats that have infiltrated this discussion
Not many who posted so far, but obviously this thread is monitored since day one (any intelligence agency not reading this are not really doing their job). Hi guys.
Your drawings are great and by all means contribute to this thread, just get rid of this email and scare propaganda shit. And create a new thread on interdimensional travel, I'm sure you'll find interested minds here.
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2fe6f6 No.2990
Afternoon.
As discussed previously.
One way to find a solution is to use the grid or virtual quantum computer in the following way:
Find the cell value at (e,1) where e is the remainder for c.
You are looking for a[t] = na
Remember
At that value, d[t] = na+x
Also
At that value, x[t] = x, the x value in the cell is equal to the x value at (e,n)
REMEMBER, the value of x at na in (e,1) is the SAME as x at (e,n)
REMEMBER, take d from all values of d[t] at (e,1) and there is a known patter of (n-1) as factor in these values of d[t]-d that is different (increasingly) from the pattern of factors of n in a[t]. It is THIS that gives the offset that is used to solve the problem and thus get the cell at (e,1) to do all the work for you.
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2fe6f6 No.2991
>>2989
RSA cracking is a consequence, not an aim.
This is a math problem, not a hackathon, though people are interested in this maths problem for many reasons.
A love of math is one big one.
The true aim is to show the utilisation of P = NP and build or grow from there.
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3ed644 No.2992
>>2991
welcome back. Any more crumbs?
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fda808 No.2993
>>2991
Well aware and looking forward to understanding how to use this for more advanced applications.
But RSA will be the first "proof" of P=NP, and I expect that to blow up no matter the intention of this thread.
Since you're here, would you mind dropping some more hints on sonoluminescence, pretty sure this is not worth printing yet
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5167b5 No.2994
>>2993
Actually, no. It isn't known where factoring is with regards to P or NP. Only that it is in BQP. However, being able to factorize fast will be a proof of the VQC. I think it's only a stepping stone in the right direction.
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fda808 No.2995
>>2994
I stand corrected. Always just assumed factoring primes was a P=NP problem.
Either way proof would give us an opportunity to recruit more minds for further progress, if we can do it without attracting shills. Alternatively its just a free-for-all on RSA keys, which is going to happen anyway… hmm
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3bb9d1 No.2996
>>2991
Good to see you VQC! Thanks for the crumbs. Back to work. When we find x for a given c value, it unlocks all the remaining variables including n.
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ab8fb8 No.2997
>>2980
Yes, and this is how it was meant to be. f is like an opposite remainder.
Imagine a ton of valleys of squares, and each semiprime we work on is inbetween two of them.
145 is inbetween 144 and 169.
e is the distance from c to the left valley, and f is the distance from c to the right valley.
Thus,
c - e = 144, or d^2
and
c+f = 169, or (d+1)^2
95 = 9^2 + 14
95 = 10^2 - 5
178 = 13^2 + 9
178 = 14^2 - 18
e is the remainder of the left, and f is the remainder of the right.
Therefore, you can represent c using f like so: c = (d+1)^2 - f
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ab8fb8 No.2998
>>2997
The mountains are the squares, and the valleys are the numbers who are not perfect squares.
d^2 + e
(d+1)^2 - f
n is the distance from d to the root of the big square.
j^2 or the little square is the distance from c to the big square.
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5167b5 No.2999
I've been looking into the latest, hot off the press, crumb from VQC. Here I subtracted 12 (d for c = 145) from the d's in (1, 1).
-10, -4, 6, 20, 38, 60, 86, 116, 150, 188, 230, 276, 326, 380, 438
5, , , 5, , 5, , , 5, , 5, , , 5,
, 4, , 4, , 4, , 4, , 4, , 4, , 4,
It shows that 5 (n) and 4 (n - 1) as factors in 20 and 60 (d=32, d=72 where a = 25, a = 61 respectively).
I tried some gcd-magic just to see how it handles and managed to find (n - 1) for some n = 17, but it doesn't appear to be a consistent solution as n's are all odd numbers in (1, 1) and (n - 1) will be even, thus contain at least 2 as a factor. However, this is very interesting.
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799348 No.3000
>>2988
Checked! And a quick idea of how to filter this guy would be amazing.
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ab8fb8 No.3001
>>2989
LOL! I agree. Their job would be to monitor threads like this. ;)
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1ccebc No.3003
>>2999
Checked!
>>2997
>>2998
Great explanation Baker! Thanks.
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84c232 No.3004
P=NP
https:/ /hooktube.com/watch?v=6ybd5rbQ5rU
…
Currently, many NP problems take a long time to solve. As such, certain problems like logistics scheduling and protein structure prediction are very difficult. Likewise, many cryptosystems, which are used to secure the world's data, rely on the assumption that they cannot be solved in polynomial time.
If someone were to show that NP problems were not difficult—that P and NP problems were the same—it would would have significant practical consequences. Advances in bioinformatics and theoretical chemistry could be made. Much of modern cryptography would be rendered inert. Financial systems would be exposed, leaving the entire Western economy vulnerable.
Proving that P = NP would have enormous ramifications that would be equally enlightening, devastating, and valuable…
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3ed644 No.3005
>>2990
Thank you. Didn't notice this post earlier.
>>2974
Just a quick update, the a[t]=na record also satisfies b-a = 2D.
So all that searching, and we had a valid starting position all along.
>>2997
>e is the remainder of the left, and f is the remainder of the right.
Much appreciate this clear explanation.
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ff4dbe No.3006
>>2995
Factoring known subprimes is O(n) currently, so it isn't a polynomial. You haven't posted much, which means I don't know how much you know about computer science, so I may as well explain it anyway (even if you already know, some lurkers probably don't).
Polynomial time means the number of calculations the computer has to make for a set of code can be calculated by n^k. When I mentioned O(n) earlier, I was referring to the same concept of the number of calculations a computer has to make. In O(n) time, there might be a big list of numbers or words or something n objects long, and you might iterate through each element of that list of objects to find the right one. That means it takes at most n calculations (if the list is 5 long, it's very quick, but if it's 10^39523 long, it'll take far too long for it to work). In polynomial time, there might be a list of objects n objects long, but the number of times you have to look through that list is n^k, where k is some constant number. This isn't a good example for the P/NP thing, but, for the sake of explaining polynomial time, generating the grid takes polynomial time, because the inner loop runs however many times as specified by the outer loop (meaning it's O(n^2), k = 2). If you had a loop within a loop within a loop and the outer loops determined the number of calculations done in the inner loops like with the grid, it would be O(n^3), k = 3. As n gets higher in any polynomial problem, the number of calculations necessary (and therefore the time taken to complete the code) goes up at an increasing rate (it isn't just a straight linear line).
NP (non-deterministic polynomial) is a name that refers to a group of problems that we can solve with code with some special properties. The solution to an NP problem can be verified in polynomial time. That means, for one of these problems, if you think you know the answer, you can figure out if it's correct or incorrect in polynomial time. The problem is that you can't actually necessarily figure out an NP problem's solution in polynomial time. This abbreviation, NP, does also cover all P problems (polynomial problems) that can be solved in polynomial time. NP covers any problem that can have a potential answer verified in polynomial time but it also covers problems that both can and can't be solved in polynomial time. Pic related.
Then there's this other class of problems called NP-complete problems. A problem is NP-complete if it's an NP problem and every other NP problem can be rephrased as a special case of it. An algorithm that solves an NP-complete problem can solve every other NP problem (those really time-consuming ones). That means, if we find a really quick way to solve an NP-complete problem, we can solve every other NP problem in a reasonable amount of time. There are a few NP-complete problems, such as the Hamiltonian Cycle problem. Say you had a map with a bunch of towns on it and you wanted to visit each town once in the shortest possible path. If you had a computer figure that out, it would need to go through each possible path starting from each possible city, which could take an extremely long time depending on how many cities there are. If you had an algorithm that could solve this problem quickly, you'd be able to solve all the others really quickly.
The P=NP problem refers to this picture. We know that all P problems fit the criteria of an NP problem. What we don't actually know (at least in mainstream mathematics) is whether all NP problems actually are all P problems. If there was a way to solve all NP problems in P time, that would mean P=NP. If there wasn't, and there were some problems we really couldn't solve in a reasonable amount of time, that would mean P!=NP.
If there's anything anyone wants me (or anyone other than me obviously) to explain any better, or if you want more information (or more context in relation to RSA), don't be afraid to ask. I only stop there because I have to go somewhere.
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ff4dbe No.3007
>>2985
>>2986
Hey there. You mentioned earlier that you hadn't read any of the previous posts. I don't know if you've caught up, but if you're planning to stick around and help out, even if it isn't necessarily with the RSA stuff, I'd highly recommend getting far more context than you seem to have. You can find links to archived threads (in chronological order) at the bottom of this post >>2556. Speaking of which, if you click the number at the end of each post header, it allows you to reply directly to someone, rather than just posting text (and then people have to figure out who it was directed towards).
Also, someone reported one of your posts. The last thing I want to do is ban you, but you've got to understand: this whole board is for discussing the entire scope of possibilities that arise from what we're all talking about (including what you're talking about), but this particular thread is for discussing RSA. We like to keep organized here, so if you want to discuss another topic (such as ways in which to use this kind of mathematics unrelated to RSA), you need to do it in another thread. One of the people here was kind enough to create one for you, which is right here >>2943. You can clearly help a great deal, which I'm sure we'll all be thankful for, but even if you don't like the idea of RSA-cracking technology being open source and available to anyone, that's what this particular thread is about. You're more than welcome to explain other things; we'd just ask that you do it in another thread is all.
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48f0b7 No.3010
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a2703a No.3011
I'm doing my best, here goes
How long will it take to calculate how long it will take to calulculate how long it will take to get from a-b?
That depends on the relationship of values that make a, and b subjective to objectification. This is also relative to lnaguagex and syntax used to express this. We cant use an equatio. Bigger than the number we are trying to solve for, that doesn't make sense either.
Where are We? I don't know about you, but I am thought swimming about in the infinite expanse of a poop slinging monkeys mind. I have no frame of reference. What i can do, is set recurseive arguements, so that i can create a map of subjects that can apply to objects, you would call this arguement a diadic equation. On one side are the empirical, and on the other side the logical. You would refer, and have reffered to this as p vs np.
You need a way to make this information transformative, so that it can make sense to the user. Can you read the fabric of the universe friend? I cannot, but as a clever poop slinging monkey, i know that this is exactly what computers do, and what they were created for. Human computers do this, but this is individualistic, and exists as a hardware state that cannot be effectively quantified within our finite lifespans.
Ok questions, or comments I will be avail, and I truly hope this helped.
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48f0b7 No.3012
>>2990
>at (e,1)
so n=1
>looking for a[t] = na = 1a = a
So a[t] = correct a?
>At that value, d[t] = na+x
So d[t] = 1a + x like always.
>x[t] = x
I'm confused this would just mean we're looking for the right x I'm not sure how much this other stuff helps unless I'm completly missing something
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a2703a No.3013
Rheomode, experimentation with language forms, develops into insight into the unbroken wholeness of existence, non-local causality.
This should help.
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ab8fb8 No.3015
>>3011
What kind of drugs are you taking?
>>3012
You are misunderstanding it. You are supposed to look for the real value of n in (e,1).
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3ed644 No.3016
>>3012
>>3015
Not sure if I'm understanding this hint either. I was under the impression we were looking for a matching x or matching t value going backwards from na.
Perhaps we are looking for matching calculations higher up.
See pic attached for c=145.
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ab8fb8 No.3017
>>3016
It's right there! x = 7 and a = na!
na for 145 is 25
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3ed644 No.3018
>>3017
>It's right there! x = 7 and a = na!
Sure. But without searching, how do we know that's the correct record?
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3bb9d1 No.3019
>>3016
>>3015
>>3012
Doing my best to follow you guys! What about the record PMA just showed at (1,1,6) where x=11 and a=61? It's a match for our (1,61,6) record. Thinking out loud here:
a[6]= 1*61 = na (from n=61 record up top)
d[6]= na + x = (n=61+11=x)
x[6]= 11 (same as original x record up top)
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3ed644 No.3020
>>3019
VA - that's our standard (n-1)*a transform from (e.n) to (e,1). Our starting position.
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ab8fb8 No.3021
>>3018
>>3019
>>3020
Try it on a number with a different e. Solution hiding in plain sight.
Also, ask yourself why numbers are indexed with distance from d^2 and distance from d to root of big square.
Why aren't they indexed with c or d? It's almost as if there is a similar property for all numbers with the same remainder (e). Dreamt of this
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3ed644 No.3022
Here's another example for c=901.
Only showing records where d[t]-d mod n-1 or a[t] mod n up to t=c.
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3ed644 No.3023
>>3021
I'm running for all my test cases. Not seeing anything obvious here.
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6b9bec No.3024
>>3023
Following along, would you please run for some smaller c's as well? Thanks.
87 (3x29)
115 (5x23)
143 (11x13)
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3ed644 No.3025
>>3024
No problem. See attached.
Also added a property at the end "dadiff". Represents diff between the d[t] calc and the a/n calc. Looking to either match the solution p.x, or the diff between c.x and p.x.
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6b9bec No.3026
>>3025
Ok, will review, was trying to understand the d[t] part, thanks!
87 (e=-13)
{ 6 : 35 : 9 : 8 : 1 : 87}
{-13 : 34 : 10 : 9 : 1 : 87}
{ 6 : 7 : 9 : 6 : 3 : 29}
{-13 : 6 : 10 : 7 : 3 : 29}
115 (e=-6)
{15 : 48:10 : 9 : 1 : 115}
{-6 : 47:11 : 10 : 1 : 115}
{15 : 4:10 : 5 : 5 : 23}
{-6 : 3:11 : 6 : 5 : 23}
143 (e=-1)
{22 : 61:11 : 10 : 1 : 143}
{-1 : 60:12 : 11 : 1 : 143}
{22 : 1:11 : 0 : 11 : 13}
{-1 : 0:12 : 11 : 11 : 13}
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3ed644 No.3027
Last one at c=785.
Interesting record at (1,1,379).
Diff calculation is 4, same as diff in x.
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520ee9 No.3028
>>3020
>>3017
Thanks for explaining the basics PMA! Is Baker correct that a=7 and na for 145 is 25? I'm working to understand over here, doing my best to follow along. Struggling a bit to keep up, but determined to learn.
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3ed644 No.3029
>>3028
Yes. Any (e,n) can be transformed to (e,1). The (1,1,4) record where x=7 and a=25 is the na transform of the prime solution for c=145.
However, we still don't have any calculation to navigate up the tree and determine if that is the solution record. We "know" it's correct. We just can't prove it.
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ab8fb8 No.3030
>>3029
Prime transform? Explain in more detail.
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3ed644 No.3031
>>3030
starting record at c=145
(1,61,6) = {1:61:12:11:1:145}
na transform to (1,1,6)
prime solution at a=5,b=29
(1,5,4) = {1:5:12:7:5:29}
na transform to (1,1,4)
Shortcut for the na transform is to take any record at (e,n,t) and create another record at (e,1,t). The a value for that record will be n*a from the original with n = 1.
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5167b5 No.3032
VQC mentioned that the pattern behind d[t] - d will give us the offset. The question is then, the offset to what?
And there is a pattern in the numbers of d[t] - d, for the numbers that contains (n - 1) as a factor. So we don't know which numbers, yet. Since we don't know which n to look for. But I have a feeling we aren't really going to look for n, but rather have n fall into our laps once we fully understand these patterns.
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ab8fb8 No.3033
Pictures are formatted like so:
d[t]-d{ } a[t]{ }——{difference between the two}——{difference between their squares}
odd and even.
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ab8fb8 No.3034
Hmm.. What to do with this information. Not sure why, but for every number that ends in 5, the element which contains the x value for the proper factorization is always in (e, 1, t-2) where t is the t value of the 1c factorization.
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ab8fb8 No.3035
It's easy to factorize any number that ends with 5 by dividing by 5.. But what if there is a formula just like that one for every other digit a number ends in?
Unsure. Need to look at it in other number systems.
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51b282 No.3038
Need some help!
I found two very interesting records in (e,n)
I was expecting a record at (2c,2c)
There isn't one.
But there are records at:
(2c, a,1) {2c: a: b: 0: b: b+2a}
(2c, b,1) {2c: b: a: 0: a: a+2b}
I [D]id [M]ention these to Chris a couple hours before the crumb yesterday.
Anyway was doing algebra on this mess and keep bouncing.
Did some geometry and found an even crazier record.
(2(c-aa), b-a, 1) {2(c-aa): b-a: a: 0: a: 2b-a}
I think this mess is solvable somehow.
Draw these suckers on graph paper! Very interesting.
Thoughts??
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ab8fb8 No.3039
>>3038
>>3038
One of the factors is 0? That's not valid.
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5167b5 No.3040
Okay, I've been looking more into the crumbs by Mr. VQC.
So we know if we do d[t] - d, we have some kind of pattern.
This relates to the equation to calculate an arbitrary d for e for n = 1. This we know how to do.
For even e it is:
(e / 2) + 4 * ( (t * (t - 1) ) / 2)
For odd e, it is:
(e + 1) / 2 + 2*t**2 - 1
Now, the fun thing is that the equation for a, is the opposite. To calculate an arbitrary a, for (e, 1) we do the opposite equations.
For even e, we do:
(e / 2) + 2*(t)**2
For odd e, we do:
e / 2 + 2 * (t * (t - 1)) + 1
Now I was looking into the crumb for c=145, we have d[0] - d = -10. If we use -10 as an offset for our calculations of d, we get:
-10 + 2 * t**2 - 2 (which is close to our a formula for even e).
When t = 4, we get 20. Now if we look at the calculations for n - a[t], we also have an offset.
So for n = 5 (a=5, b=29) we get 1 - 5 => -4. So we'll use -4 as our offset.
We'll do:
-4 + 2 * (t * (t - 1)).
For t = 4 we also get 20.
I'm thinking, maybe we are supposed to find the t, where the two equations intersect?
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5167b5 No.3041
>>3040
One downside is that we won't know the -4 offset, so this might be barking up the wrong tree. Since we only get -4 by doing 1 - 5 (and we are trying to find 5)
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f8a95b No.3042
>>3031
Great explanation, PMA! I get the concept now, thank you very much.
>>3032
Hey Isee! What about PMA's example right above your post?
starting record at c=145
(1,61,6) = {1:61:12:11:1:145}
na transform to (1,1,6)
(1,1,6) = {1:1:72:11:61:85}
prime solution at a=5,b=29
(1,5,4) = {1:5:12:7:5:29}
na transform to (1,1,4)
(1,1,4) = {1:1:32:7:25:41}
Thinking out loud here…
For c record (1,c)
[t] is the same in (e,1) and (e,n)
x[t] is the same in (e,1) and (e.n)
d[t] in (e,1) is original d (12) + (61-1) = transformed d = 72
a[t] = na = 1*61 = 61
For prime record:
[t] is the same in (e,1) and (e,n)
x[t] is the same in (e,1) and (e,n)
d[t] in (e,1) is original d (12) + (n-1)*a
transformed d is 12+ (5-1)*5 = 32
a[t] = na = 5*5 for transformed a
is there supposed to be a (n-1)*a connection for d[t]?
On a practical OpSec note, I'm very frustrated with my VPN. Any recommendations?
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f8a95b No.3044
Oops!
(n-1)*a works for the starting record d[t] as well. (61-1)*1= 60, 12+60 = transformed d = 72
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ab8fb8 No.3045
>>3044
>>3043
>>3042
If you know the t value that the factorization in (e,n) appears in you can solve it.
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51b282 No.3046
>>3039
x=0 isn't valid????????????????
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ab8fb8 No.3047
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51b282 No.3048
>>3047
Apologies for autistic response.
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769a5e No.3049
>>3045
Continuing with the d[t]-d) analysis.
Pic related is reconfigured test case for c=145 showing values where:
DN = (d[t] - d) mod (n-1) == 0
AN = a mod n == 0
Not seeing anything yet.
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5167b5 No.3050
I've been looking more into the patterns of (n - 1) for d, and I've found that there is a lot of reoccuring, but alternating patterns:
(1, 5, 1)
a[t] - n: 1, 2, 3, 4, 5, 6, .., 31
d[t] - d: 2, 4, 6, 8, 10, 12, 14, 16.., 30
(1, 5, 2)
d[t] - d: 2, 4, 6, 8, 10, 12, 14, 16.., 30
(1, 5, 3)
d[t] - d: 1, 3, 5, 7, 9, 11, 13, 15, .., 31
(1, 5, 4)
d[t] - d: 1, 3, 5, 7, 9, 11, 13, 15, .., 31
(1, 5, 5)
d[t] - d: 2, 4, 6, 8, 10, 12, 14, 16.., 30
(1, 5, 6)
d[t] - d: 2, 4, 6, 8, 10, 12, 14, 16.., 30
(1, 5, 7)
d[t] - d: 1, 3, 5, 7, 9, 11, 13, 15, .., 31
(1, 5, 8)
d[t] - d: 1, 3, 5, 7, 9, 11, 13, 15, .., 31
(1, 13, 1)
a[t] - n: 1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16, 18, 19, 21, 22, 24, 25, 27, 28, 30, 31
d[t] - d: 3, 9, 15, 21, 27
(1, 13, 2)
d[t] - d: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31
(1, 13, 3)
d[t] - d: 2, 4, 8, 10, 14, 16, 20, 22, 26, 28
(1, 13, 4)
d[t] - d: 6, 12, 18, 24, 30
(1, 13, 5)
d[t] - d: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31
(1, 13, 6)
d[t] - d: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31
(1, 13, 7)
d[t] - d: 6, 12, 18, 24, 30
(1, 13, 8)
d[t] - d: 2, 4, 8, 10, 14, 16, 20, 22, 26, 28
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5167b5 No.3051
>>3050
The code I've been using:
def generateGenesis(e=0, rows=10):
cells = []
genesis = genesis_block(e)
block = genesis
if e % 2 == 0:
offset = 2
else:
offset = 4
if e != 0:
cells.append(block)
for i in range(rows):
e, n, d, x, a, b = block
a = b
b = b + (offset + 4 * (i + 1))
c = math.fabs(a * b)
d = math.floor(math.sqrt(c))
x = math.floor(math.sqrt( (d + n)**2 -c ) - n)
block = (int(e), int(n), int(d), int(x), int(a), int(b))
cells.append(block)
return cells
def testing(e, d, n, rows=30):
cells = generateGenesis(e, rows)
print(e, d, n)
print('d[t] - d:')
for cell in cells:
ee, nn, dd, xx, aa, bb = cell
if (dd - d) % (n - 1) == 0:
print((xx + 1)/2, dd - d, (dd - d) % (n - 1))
# print()
# print('a - n:')
# for cell in cells:
# ee, nn, dd, xx, aa, bb = cell
# print((xx + 1)/2, aa - n, (aa - n) % (n - 1))
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5167b5 No.3052
>>3050
My notation is (e, n, t) and I only display a[t] once, since it's the same for the entire (e, n) row.
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5167b5 No.3053
>>3052
I haven't gotten any further with the patterns, though. Just wanted to share what I got. I still can't see the underlying patterns VQC is talking about. I'm sure some of you guys might though.
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68fb3c No.3054
>>3053
I can generalize the D's so you can solve for any D for any (e,1)
D values:
0 1 2 3 4 5 6 7 8 9 10
(0,1): 4, 12, 24, 40, 60, 84, 112, 144, 180, 220, 264
(+8, +12, +16, +20, +24, +28, +32, +36, +40, +44)
(+4, +4, +4, +4, +4, +4, +4, +4, +4, +4)
(1,1): 2, 8, 18, 32, 50, 72, 98, 128, 162, 200, 242, 288)
(+6, +10, +14, +18, +22, +26, +30, +34, +38, +42, +46)
(+4, +4, +4, +4, +4, +4, …)
(2,1): 1, 5, 13, 25, 41, 61, 85, 113, 145, 181, 221, 265
(+4, +8, +12, +16, +20, +24, +28, +32, +36, …
(3,1): 3, 9, 19, 33, 51, 73, 99, 129, 163, 201, 243
(+6, +10, +14, +18, +22, +26, +30, +34,
(4,1): 2, 6, 14, 26, 42, 62, 86,114,146
+4, +8, +12, +16, +20, +24, +28
(5,1): 4,10,20,34,52,74,100,130,164,202
+6, +10, +14, +18, +22, +26, +30, +34, +38
(6,1): 3,7,15,27,43,63,87,115
+4, +8, +12, +16, +20, +24
(7,1): 5,11,21,35,53,75,101,131,165
+6, +10, +14, +18, +22, etc
(8,1): 4,8,16,28,44,64
+4, +8, +12, +16, +20, etc
(9,1): 6,12,22,36,54,75,102
+6, +10, +14, +18
So for (e,1). The D value can be generated from T by
D(t) = D(t-1) + shift
where shift is added and incremented by 4 every time it is added. For odd e rows it starts at 6, for even e rows it starts at 4.
The starting point, D(0), as you can see above, goes like this
E = 1, 2, 3, 4, 5, 6, 7, 8, 9,
D0= 2, 1, 3, 2, 4, 3, 5, 4, 6,
So if you are at even E:
D0 = E/2
odd E:
D0 = 1 + (E+1)/2
Also for any (e,1) we know that for odd E we have:
x[0] = 1
even E:
x[0] = 0
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769a5e No.3055
See pics related for continued analysis of (e,1).
Examples are for c=65 and c=145.
Reminder for calculations we are comparing:
Calculation 1: (d[t]-d)/(n-1)
Calculation 2: a[t]/n
where d and n are the values from our entry record at (e,n). And d[t] and a[t] are from the (e,1) record being analyzed.
Only records where (n-1) and n are factors are shown.
Notes:
a) There are 2 sequences for each calculation.
b) One sequence increments by 1 and then is multiplied by a factor.
c) The other sequence increments by 2 and then is multiplied by a different factor.
d) Lines marked as "overlap" are where both calculations have (n-1) and n as factors.
Example:
c=145
the d[t] sequences are incremented by factors of 18 and 24.
the a[t] sequences are incremented by factors of 100 and 44.
No clue what to do with this information.
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ab8fb8 No.3056
>>3055
Is it computationally feasible to compute these factors?
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769a5e No.3057
>>3056
I really don't know.
First step is trying to understand what VQC was even talking about.
If this analysis is correct, then somewhere in here is an "offset" that gives us a connection to the prime solution.
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769a5e No.3058
>>3055
>>3056
>>3057
It's possible that the sequences for a/n add up to c-1.
c=65:
Sequence 1 factor = 36
Sequence 2 factor = 28
Total = 64 + 1 == c
c=145:
Sequence 1 factor = 100
Sequence 2 factor = 44
Total = 144 + 1 == c
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ab8fb8 No.3059
>>3058
Can you walk me through calculating the factors? Also, it is possible they are adding up to d^2 and not c-1. More testing required
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769a5e No.3060
>>3058
Also possible that for a/n, factor for sequence 1 comes from record (e,1,t+n) and factor for sequence 2 comes from record (e,1, n-t+1).
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769a5e No.3061
>>3059
Yes. Very possible it adds up to d^2. Was just my initial observation.
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921cd7 No.3062
(d+n)^2 - (x+n)^2 = dd + e
(d+1)^2 - (x+1)^2 = dd + e
dd + 2d + 2 - (xx + 2x + 2) = dd + e
dd + 2d + 2 - xx - 2x - 2 = dd + e
2d - xx - 2x = e
2d - e = x(x+2)
2d = x(x+2) + e
d = (x(x+2) + e)/2
?
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ab8fb8 No.3063
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921cd7 No.3064
>>3063
calculate any d value for any x in the first row (n=1). and x's go from x=e%2 and then you add +2 every time. Then from d-x you can get all As in the record
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ab8fb8 No.3065
Has anyone investigated my hypothesis that all numbers that end with the digit 5 have a solution in e,1 (x) whose t value is 2 minus the t value of the 1c element? It worked for all instances I tested it for.
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ab8fb8 No.3066
>>3065
You can always just divide it by 5, but at least it's a way to do it with the VQC.
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15a826 No.3067
Could the offset we are looking for be the difference between n and N for (a*a, b*b) and c*c records?
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ab8fb8 No.3068
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9f7aa6 No.3070
I found what he's talking about.
For (E,N,D,X,1,C) (our start record) in (E,1) there is always a record (e,1,d,x,a,b) where
D-d = N-1
and
a = N
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51b282 No.3071
>>3067
Wow good post!
How can you get anywhere if you don't know where you're going?
Goal record?
Goal records?
>>3038
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9f7aa6 No.3072
>>3070
(1, 61, 12, 11, 1, 145, 144)
X = 7
(1, 1, 32, 7, 25, 41, 16)
(1, 5, 12, 7, 5, 29, 24)
NA = 25
Goal:
(1, 1, 32, 7, 25, 41, 16)
1 1 72 11 61 85 24
-60 60 61 61
I meant D-d = -(N-1)
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9f7aa6 No.3073
>>3072
Also in this cell the X's are the same as the (a,b)=(1,c) cell. Then we can go straight to that cell through
(E,N,D,X,A,B)
to
(knowing (d-D = N-1 => N+D-1 = d)
(also since d-x=a we have: N+D-1-X = a
(E, 1, N+D-1, X, N+D-1-X, ?)
You could then just calculate B however you want.
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5167b5 No.3074
>>2990
I miss read him the first time. He states that d[t] - d is different from the pattern of factors of n in a[t]. Not the factors of (n - 1) in the difference of a[t].
The factors of 5, in a[t] at (1, 1) is:
2, 4, 7, 9, 12, 14, 17, 19, 22, 24, 27, 29
Which are two interleaving patterns, which we also know is the same as p*k + t (for k=1,2,…) and p*k + 1 - t. (In this case p = 5 and t = 2)
And as for the "different (increasingly)" I think he is referring to the difference between the numbers where (n - 1) is a factor of d[t] - d.
For example, take (1, 5, 2) where a = 5, b = 29 and c = 145.
Here d is 12, the (d[t] - d) that have 4 as a factor are the following:
-4, 20, 60, 116, 188, 276, 380, 500, 636
These increase with:
24, 40, 56, 72, 88, 104, 120, 136.
The difference between the increase is 16.
So could this 16 be what he refers to? Or could it be the 24? Or 24 + 16?
Nevertheless, we still don't know the (n - 1) we are looking for so I'm still a bit stumped.
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5167b5 No.3075
>>3074
I might have been a bit of a dumbass again.
So I checked the increasing values of (1, 13, 2) and the (d[t] - d) that contains (n - 1) as a factor is:
-36, 12, 60, 204, 300, 540, 684, 1020, 1212, 1644, 1884
This is, again, two interleaving patterns. Pattern one:
-36, 60, 300, 684, 1212, 1884
Starts with 96 in difference between -36 and 60, but then if you add 96 + 144, you'll get 300.
So for (1, 5, 2) the increasing number is 16, which is 4*4. For (1, 13, 2) it's 144, which is 12*12.
The other part of the interleaving pattern is:
12, 204, 540, 1020, 1644
Which starts with 192. Again, if you do 12 + 192 you'll get 204. If you do 204 + 192 + 144 you'll get 540.
So the increase is also 144 for the interleaved pattern.
However, for (1, 17, 1) and (1, 17, 2) it's increasing by 16 again. While (1, 17, 3) appears to be 64. Could be a special case since 17 isn't a part of (1, 1)?
For (1, 25, 2) we see again, two interleaved patterns: -72, -48, 48, 120, 312, 432, 720, 888, 1272, 1488
Pattern 1: -72, 48, 312, 720, 1272
Pattern 2: -48, 120, 432, 888, 1488
Both increase by 144 (12^2). However, this is for the record:
{1:25:80:43:37:173}
And 37 appears in (1, 13). So could be that we aren't looking for the 'n', but rather an 'n' where a exists?
I don't have much time to look for patterns right now, but once time opens up. I'll dig more.
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5167b5 No.3076
>>3075
As for the reason I dismissed the pattern before I checked (1, 13) was that (1, 5, 3), which has 30 as d appears to change with 2. So I don't know if the above is the silver bullet, but it's at least an interesting pattern.
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769a5e No.3077
Possible formula for the underlying factors to determine where a[t] % n == 0.
Pic related.
factor 1: 4*x
factor 2: (x-1)^2
Not sure if relevant.
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81b46d No.3078
Alright anons, I'm gonna swing for the fences like Babe Ruth.
(1,c)
(1,RSAc)
{e,n,d,x,1,RSAc}
solve for d and e.
n= (a+b)/2-d
x= d-a = d-1 = RSAx for 1,c
RSAt= (x+2)/2 for even, (x+1)/2 for odd.
Here's our starting position for any RSAc.
Now, use (n-1)*a to find the transform for the (e,1) position.
Then find the prime solution by finding the equivalent t value in (e,1).
I know it's a big swing, just looking for that home run. Thoughts?
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170bcb No.3079
So we can generate any cell in (e,1). Since we know that N = (XX+E)/2A, wouldn't that mean that 2NA = XX+E ?
Could we navigate to the cell where our new X is the same as our original D. Then we would have that that is divisible by N and A
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170bcb No.3080
>>3079
Problem with this is that (at least for c=145) the X values are odd and this would require them to be even (cuz our d is 12). In this case I believe we could go to our F record (basically d=d+1) to get the record where X is the same as the original D.
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ab8fb8 No.3087
>>3072
You have cells indexed with 7 variables in this post. What is the 7th??
>>3078
All RSA numbers are odd.
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51b282 No.3088
>>3080
>>3079
If e is even then you get {e, 1, a, 0, a, b}
If e is odd you get get {e, 1, a+1, 1, a, b}
You know e and n you can solve for a
Even a=e/2
odd a = (e+1) /2
So you can solve for the t=1 record.
Then you have to figure out x.
We know d(t)- d(t=1) partly due to n(t)-n(t=1) and partly due to x(t) - x(t=1)
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51b282 No.3089
>>3088
If we call original values E,N,T, D, X, A, B
Then a(E, 1, T) = NA
d(E, 1, T) = NA+X
For e even T=X/2+1 and e odd (X+1)/2
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5167b5 No.3090
Okay guys, I have one idea, no idea if it works. It doesn't sound like the thing VQC had in mind, but hear me out.
We know how to find the big N. And it appears as if every big N's exists in (e, 1). Now from what I've seen, the big N's D - d will contain (n - 1) as a factor.
If we can factorize the big N's (D - d), then we can figure out the (n - 1). Sounds silly, but what if we use a recursive algorithm. Essentially, we just keep factorizing the D - d value of the big N until we hit a prime number, in which we found one factor of our original (D - d).
To give you an example:
c = 145, this gives us:
>>> rowForAB(1, 145)
(1, 61.0, 12, 11, 1, 145)
>>> getTFromA(1, 61)
6.0
>>> generateNthRowFromGenesis(1, 6)
(1, 1, 72, 11, 61, 85)
>>> 72 - 12
60
>>> rowForAB(1, 60)
(11, 23.5, 7, 6, 1, 60)
>>> getTFromA(11, 23)
3.458039891549808
>>> generateNthRowFromGenesis(11, 4)
(11, 1, 37, 7, 30, 46)
>>> 60 / 30
2.0
>>> 37 - 7
30
>>> rowForAB(1, 30)
(5, 10.5, 5, 4, 1, 30)
>>> getTFromA(5, 10)
2.4364916731037085
>>> generateNthRowFromGenesis(5, 3)
(5, 1, 20, 5, 15, 27)
>>> 30 / 15
2.0
>>> rowForAB(1, 15)
(6, 5.0, 3, 2, 1, 15)
>>> getTFromA(6, 5)
1.0
>>> generateNthRowFromGenesis(6, 1)
(6, 1, 3, 0, 3, 5)
>>> 15 / 3
5.0
>>> 15 / 5
3.0
>>> # Our factors for 60 => 2, 2, 3, 5
Now I honestly don't know if this will work or not with bigger numbers. And I don't know if this is what VQC had in mind. But I think it's worth taking a look at.
As for those numbers where it turns into 10.5, I get the T for that value and round up.
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5167b5 No.3091
>>3090
Also interesting to note, d[t] - d for t = 2, 4, 6, 10, 12, 20, 30 and 60 all contains (n - 1) as a factor for c=145.
Those are all the result of different combinations of the factors of 60.
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ab8fb8 No.3092
>>3090
How about explaining what each method does. What is generateNthRowFromGenesis doing and what is getTFromA doing
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769a5e No.3093
Making a little progress.
Pics show various examples of the different factors for (d[t]-d) and a[t] formulas.
The formula (x^2 - f) can be used to find a valid factor of (d[t] - d). Sometimes it is the first valid factor below a record at (e,1, t+n).
How is this useful?
I believe we are searching for a formula for d.
For example, 901=17x53
Starting positions:
from c=901
(1,421,15) = {1:421:30:29:1:901}, f = 60
records at t = c.n + c.t
(1,17,436) = {1:17:23184:871:22313:24089}
(1,421,436) = {1:421:1772:871:901:3485}
Example formula that solves for the d value of (1,17,436) = 23184.
x^2-f = 29*29 - 60 = 781
(d)*(x^2 - f) - ( 2 * ( c - ( x^2 - f ) ) + 1 ) - 5
30*(781) - (2*(901 - 781) + 1) - 5
30*781 - (240+1) - 5
23430 - 241 - 5 = 23184
The 241 value is the next valid factor of d[t] - d after 781.
The 5 is just a value I added to the end because the numbers are really close. I was thinking of 2^2 + 1, or could be that one of the other formulas is wrong.
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ab8fb8 No.3094
>>3093
This guy's algorithm above you looks promising but I have no idea what the fuck getTFromA is doing.
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5167b5 No.3095
>>3094
It get's the t for a[t] in (e, 1).
For example in (1, 1) getTFromA(1, 5) will return 41, which is the a[t] where t = 5.
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ab8fb8 No.3096
>>3095
Based on what input variables? Just send it faggot
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5167b5 No.3097
>>3095
generateNthRowFromGenesis returns the nth row from genesis in (1, 1).
So generateNthRowFromGenesis(1, 5) will return {1:1:50:9:41:61}
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5167b5 No.3098
>>3097
Correction, from (e, 1)
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5167b5 No.3099
>>3096
getTFromA takes in e as the first parameter and a as the second. It returns t.
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ab8fb8 No.3100
>>3099
>>3098
>>3097
How. You are describing what it produces but not how it does
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5167b5 No.3101
>>3100
I've posted both code pieces before. Numerous times. If you're going to be an asshat, then you can look for it yourself.
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ab8fb8 No.3102
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9366ee No.3103
>>3101
You only posted that particular method once and the version you posted does not work the way you demonstrated. The result of 1 and 61 is 5, not 6
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191295 No.3105
Hello lads! Q is dropping bombs and we're going to break Twitter tonight! Don't forget to fire up your meme cannons and chip in! Trending #2 worldwide now. #ReleaseTheMemo.
See you all tomorrow for some more math fun!
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3ed644 No.3106
Still exploring d[t] formulas trying to find a link to any prime related record.
Trusty c=145 example in pic.
Records are filtered to show only valid factors of the d[t] and a[t] formulas.
I've added a column to test
( ( (d[t] - d) / (n-1) ) - e ) / d
to see if any records would stand out as more relevant. Interestingly, in all my test cases, (x^2 - f) - e is always a valid factor of d.
Still not sure if I'm heading down a valid path.
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28a942 No.3108
>>3105
Nuked a Japanese twitter account just for this.
Wew
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99eff8 No.3109
Neeeeerds?
BTC is not feeling so well.
What'd you doooooo?
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5167b5 No.3110
https://themarshallreport.wordpress.com/2017/11/24/the-mystery-q-solved-or-not-all-roads-lead-to-in-q-tel-bigger-than-big-brother/
Yeah, so I'm out guys. I believe this is just a LARP. We're not really getting anywhere, we're just running around in circles.
Kudos to you VQC, you had me going for a looong ass time, racking my brain back and forth. It's been fun.
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5167b5 No.3111
>>3103
Yeah, no. I've posted them in other threads along with the equation behind them.
def getDFromT(e, t):
if e == 0 or e % 2 == 0:
return (e / 2) + 2 * ( (t * (t - 1) ))
else:
return int( (e + 1) / 2) + 2*t**2 - 1
def getAFromT(e, t):
if e == 0:
return (e / 2) + 2*(t)**2
if e % 2 == 0:
return (e / 2) + 2*(t - 1)**2
else:
return int(e / 2) + 2*( (t * (t - 1)) ) + 1
def getTFromD(e, d):
if e % 2 == 0:
t = d - (e / 2)
t = t / 2
t = t * 4 + 1
t = t / 4
t = math.sqrt(t) + 1
return -1/2 + t
else:
t = d - (2 + int(e / 2))
t = t / 2
t = t + 1
t = math.sqrt(t)
t = t
return t
def getTFromA(e, a):
if e == 0:
t = a * 2 - e
t = math.sqrt(t) + 2
t = t / 2
return t - 1
if e % 2 == 0:
t = a * 2 - e
t = math.sqrt(t) + 2
t = t / 2
return t
else:
return math.sqrt((a - (e / 2))/2) + 0.5
I had some issues with e = 0, but anyways. There you have it. It should work fine for any (e, 1). As for every other function I've referenced they're either posted here or copied from College Anon's code from OP.
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769a5e No.3112
>>3111
Any further progress on d[t] hints?
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5167b5 No.3113
>>3112
Nah. Only thing I've observed is that the (d[t] - d) for record with a[t] = big N contains (n - 1) as a factor.
As for my recursive attempt, it doesn't work. It essentially divides by two. Feel free to try, it only worked on a few selected methods.
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769a5e No.3114
>>3113
I understand the recursion, but it's not going to be the solution.
If you get a chance, take a look at some of my screenshots. There are a number of paths to matching records that I've identified, but none that are consistent for different test cases.
If I'm reading the QVC hint correctly, the solution can be calculated from the na record.
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0bf39d No.3116
79*51
(60, 1890, 62, 61, 1, 3904, 3904, 3781, 1951.5)
(60, 1830, 61, 60, 1, 3781, 3781, 3660, 1890.0)
(60, 1770, 60, 59, 1, 3660, 3660, 3541, 1829.5)
(60, 1712, 59, 58, 1, 3541, 3541, 3424, 1770.0)
(60, 1654, 58, 57, 1, 3424, 3424, 3309, 1711.5)
(60, 1598, 57, 56, 1, 3309, 3309, 3196, 1654.0)
(60, 1542, 56, 55, 1, 3196, 3196, 3085, 1597.5)
(60, 1488, 55, 54, 1, 3085, 3085, 2976, 1542.0)
(60, 1434, 54, 53, 1, 2976, 2976, 2869, 1487.5)
(60, 1382, 53, 52, 1, 2869, 2869, 2764, 1434.0)
(60, 1330, 52, 51, 1, 2764, 2764, 2661, 1381.5)
(60, 1280, 51, 50, 1, 2661, 2661, 2560, 1330.0)
(60, 1230, 50, 49, 1, 2560, 2560, 2461, 1279.5)
(60, 1182, 49, 48, 1, 2461, 2461, 2364, 1230.0)
(60, 1134, 48, 47, 1, 2364, 2364, 2269, 1181.5)
(60, 1088, 47, 46, 1, 2269, 2269, 2176, 1134.0)
(60, 1042, 46, 45, 1, 2176, 2176, 2085, 1087.5)
(60, 998, 45, 44, 1, 2085, 2085, 1996, 1042.0)
(60, 954, 44, 43, 1, 1996, 1996, 1909, 997.5)
(60, 912, 43, 42, 1, 1909, 1909, 1824, 954.0)
(60, 870, 42, 41, 1, 1824, 1824, 1741, 911.5)
(60, 830, 41, 40, 1, 1741, 1741, 1660, 870.0)
(60, 790, 40, 39, 1, 1660, 1660, 1581, 829.5)
(60, 752, 39, 38, 1, 1581, 1581, 1504, 790.0)
(60, 714, 38, 37, 1, 1504, 1504, 1429, 751.5)
(60, 678, 37, 36, 1, 1429, 1429, 1356, 714.0)
(60, 642, 36, 35, 1, 1356, 1356, 1285, 677.5)
(60, 608, 35, 34, 1, 1285, 1285, 1216, 642.0)
(60, 574, 34, 33, 1, 1216, 1216, 1149, 607.5)
(60, 542, 33, 32, 1, 1149, 1149, 1084, 574.0)
(60, 510, 32, 31, 1, 1084, 1084, 1021, 541.5)
(60, 480, 31, 30, 1, 1021, 1021, 960, 510.0)
(60, 450, 30, 29, 1, 960, 960, 901, 479.5)
(1, 421, 30, 29, 1, 901, 901, 842, 450.0)
(1, 392, 29, 28, 1, 842, 842, 785, 420.5)
(1, 365, 28, 27, 1, 785, 785, 730, 392.0)
(1, 338, 27, 26, 1, 730, 730, 677, 364.5)
(1, 313, 26, 25, 1, 677, 677, 626, 338.0)
(1, 288, 25, 24, 1, 626, 626, 577, 312.5)
(1, 265, 24, 23, 1, 577, 577, 530, 288.0)
(1, 242, 23, 22, 1, 530, 530, 485, 264.5)
(1, 221, 22, 21, 1, 485, 485, 442, 242.0)
(1, 200, 21, 20, 1, 442, 442, 401, 220.5)
(1, 181, 20, 19, 1, 401, 401, 362, 200.0)
(1, 162, 19, 18, 1, 362, 362, 325, 180.5)
(1, 145, 18, 17, 1, 325, 325, 290, 162.0)
(1, 128, 17, 16, 1, 290, 290, 257, 144.5)
(1, 113, 16, 15, 1, 257, 257, 226, 128.0)
(1, 98, 15, 14, 1, 226, 226, 197, 112.5)
(1, 85, 14, 13, 1, 197, 197, 170, 98.0)
(1, 72, 13, 12, 1, 170, 170, 145, 84.5)
(1, 61, 12, 11, 1, 145, 145, 122, 72.0)
(1, 50, 11, 10, 1, 122, 122, 101, 60.5)
(1, 41, 10, 9, 1, 101, 101, 82, 50.0)
(1, 32, 9, 8, 1, 82, 82, 65, 40.5)
(1, 25, 8, 7, 1, 65, 65, 50, 32.0)
(1, 18, 7, 6, 1, 50, 50, 37, 24.5)
(1, 13, 6, 5, 1, 37, 37, 26, 18.0)
(1, 8, 5, 4, 1, 26, 26, 17, 12.5)
(1, 5, 4, 3, 1, 17, 17, 10, 8.0)
(1, 2, 3, 2, 1, 10, 10, 5, 4.5)
(1, 1, 2, 1, 1, 5, 5, 2, 2.0)
(1, 0, 1, 0, 1, 2, 2, 1, 0.5)
(0, 0, 1, 0, 1, 1, 1, 0, 0.0)
(0, 0, 0, -1, 1, 0, 0, 1, -0.5)
These cells are (E,N,D,X,A,B, DD+E, XX+E, (B-A)/2)
This is pretty neat. What I did was generate cells A,B = (1,C), then I generated another cell with C = XX+E. This is because the identity 2NA = XX+E, so I figured if A is a factor of C, then it must also be a factor of XX+E. So I decided to test out the XX+E cells. From the bottom up, if you look at N's you get (ignoring invalid cells)
0,1,2,5,8,13,18,25,32,41,50,61,72
0, (+1), (+1), (+3), (+3), (+5), (+5), (+7), (+7), (+9), (+9)
Also you will notice that the D and X values increment by one. If we can find a way to perhaps traverse UP the graph then we may be able to find de way
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769a5e No.3117
Found an interesting relationship between the c na record and the prime solution na record.
Pic is for c=145 example, but it applies everywhere.
Maybe this is obvious.
{1:1:72:11:61:85}
{1:1:32:7:25:41}
d = 72 - 32 = 40
x = 11 - 7 = 4
a = 61 - 25 = 36
b = 85 - 41 = 44
d-x = a
d+x = b
And if you look at the value of (x^2 - f) compared to the diff in d, the numbers are very close to a d[t] - d value in the list of valid factors of (d[t]-d) / (n-1).
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e7da8e No.3118
>>3117
Hey PMA! I’m still here working too. Nothing to report yet, but I’ve been working through all your examples. I noticed one thing: I think the formula is (n-1)*a for both the (1,c) record and the prime record. Looking at this (1,145) example over and over has been really helpful, thanks for the clean and clear output as always. I’m in till we finish this. Then on to more math fun!
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769a5e No.3119
>>3118
>I’m in till we finish this. Then on to more math fun!
VA - I agree completely! This problem has truly been an eye opener for me. Couldn't care less about RSA.
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ff4dbe No.3120
>>3109
Hey Topol, would you mind responding to this conversation starting here >>3008 in the Sonoluminescence thread? Someone's saying that VQC said some stuff about killing himself if the VQC wasn't public by Jan 1st and their only claim of evidence is that you probably saw it too (i.e. if you don't know what they're talking about it's probably bullshit).
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99eff8 No.3122
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769a5e No.3124
Just noticed the following for (e,1) records
b = d + x + 2
weird.
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ff4dbe No.3125
>>3124
Wild stab in the dark by someone who hasn't contributed for a week so maybe doesn't know what they're talking about: maybe 2 is that offset Chris mentioned and b for every e follows some pattern of d + x in some way with a similar offset.
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769a5e No.3126
>>3125
Very possible.
Checkout my observation of boiling point of vinegar.
>>2964
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ff4dbe No.3127
>>3126
Did you ever find the original post where he mentioned the vinegar thing? If you've found those numbers, maybe there was other important stuff in or around that post that you can use now.
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769a5e No.3128
>>3127
it was mentioned in RSA #2
>>2831
archive.fo/fyzAu
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81b46d No.3129
>>3124
Hey PMA! That formula works in n=1. For n>1 it is a+2x+2n. (e,1) is unique. As VQC said, (e,1) is the Row To Rule Them All.
Lads, sincere apologies for all my UID's. I keep getting kicked offline by my VPN.
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769a5e No.3130
>>3129
Thanks VA.
Seems like the d[t] hint is exposing some real power behind the scenes in the grid. Wish we knew more.
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81b46d No.3131
Hey PMA! can you look this over and give thoughts? I posted this a few days ago. Posting again for your analysis.
>>3042
"starting record at c=145
(1,61,6) = {1:61:12:11:1:145}
na transform to (1,1,6)
(1,1,6) = {1:1:72:11:61:85}
prime solution at a=5,b=29
(1,5,4) = {1:5:12:7:5:29}
na transform to (1,1,4)
(1,1,4) = {1:1:32:7:25:41}
Thinking out loud here…
For c record (1,c)
[t] is the same in (e,1) and (e,n)
x[t] is the same in (e,1) and (e.n)
d[t] in (e,1) is original d (12) + (61-1)*1 = transformed d = 72
a[t] = na = (1*61)*1 = 61
For prime record:
[t] is the same in (e,1) and (e,n)
x[t] is the same in (e,1) and (e,n)
d[t] in (e,1) is original d (12) + (n-1)*a
transformed d is 12+ (5-1)*5 = 32
a[t] = na = 5*5 for transformed a
is there supposed to be a (n-1)*a connection for d[t]?"
Thoughts, anons?
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769a5e No.3132
>>3131
VA - Excellent summary and understanding.
>is there supposed to be a (n-1)*a connection for d[t]?"
This is exactly what we are searching for. I still believe there is an elegant formula.
But notwithstanding the d[t] hint, I haven't found any combination of steps that is repeatable across different test cases.
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81b46d No.3133
>>3119
I just love math. RSA whatever. I also really love VQC's blend of spirituality, math, and truth. He attracted us all here with his crazy shit. I honestly have been super happy working with you all on this math challenge. This part of my self (math love) has been laying dormant for many years, and now it's wide awake.If all I get out of this is to commune with you fine Anons all around the world working on math, fuck it, I'm in 100%. Our small part in the #GreatAwakening will become huge when we unlock this. Remember, this has already been solved by other minds. We can solve it too. This is a crowdsourced effort to share this with the world. Don't lose sight, Anons! Persistence wins the day.
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769a5e No.3134
Latest test case of c=145 attached.
Still looking at the differences between c at na, and prime at na.
Added tests to compare the differences in d, a, and b against differences in x.
(d-d)/(x-x)
(a-a)/(x-x)
(b-b)/(x-x)
Another humorous references to the boiling point of vinegar? Coincidence?
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81b46d No.3135
Future generations will know that we did our duty for truth and justice. Grab your balls, faggots. For our kids and future generations, we need transparency and NO MORE LIES. That's really what we're working for here.
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ff4dbe No.3136
>>3135
Part of why I wish Chris gave us a bit more detail.
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81b46d No.3137
>>3134
Ok, PMA. Let's work. Can you give another (1,c) example that we can work on now? Pen, paper, and calculator handy over here. AlgebraAnons ready to work.
RATM for your listening pleasure:
https://youtu.be/fI677jYfKz0
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81b46d No.3138
take any (1,c)
derive d and e.
Use n=(a+b)/2-d to find n for (1,c)
Derive x,b
we now have {e,n,d,x,a,b} for our starting (1,c) position.
Now we transform.
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461fa6 No.3139
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81b46d No.3140
>>3139
Fuck yeah. "All your base are belong to us." I like hidden in plain sight. Let's find it, Anons.
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81b46d No.3141
>>3131
So the (e,1) record will always have equal t and x values as the (e,n) record? That makes it easy to find, right? We need to examine more (1,c) records to confirm this pattern.
So what's the connection between (1,c) and prime records? That's what we're working on now. Prime records will always have lower x and t values than (1,c). Can we make a pattern connection between the c and prime records?
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461fa6 No.3142
>>3140
>>3141
Try investigating the path to the solution for every every multiple of 5. If b is odd, then the solution in (e,1) is two cells above the (e,1) for the starting cell (setting n=1 and generating a cell.)
Particularly for multiples of 5, you can deduce that b is odd because it ends in the digit 5.
Example:
145 = {1:61:12:11:1:145}—(1, 61, 6)
(1, 1, 6) = {1:1:72:11:61:85}
(1, 1, 4) = {1:1:32:7:25:41}
7 = x for 5*29
(1, 5, 4) = {1:5:12:7:5:29}
Example 2:
95 = {14:39:9:8:1:95}—(14, 39, 5)
(14, 1, 5) = {14:1:47:8:39:57}
(14, 1, 3) = {14:1:19:4:15:25}
4 = x for 5*19
(14, 3, 3) = {14:3:9:4:5:19}
I don't know how many cells above the (e,1) solution is for even b, for multiples of 5. Maybe you can tell me. Have to work.
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81b46d No.3143
>>3142
Well that's the trick. How do we make (e,1) do the work for us?
For the examples shown, I'm seeing that the prime solution lies in c[t]-2 = prime[t]. However, that pattern may not hold as we move up.
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461fa6 No.3144
>>3143
Try investigating it.
Try 10, 15, 20, 25, 30 etc
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81b46d No.3145
>>3144
Baker, did you already figure it out? :)
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461fa6 No.3146
>>3145
It might look that way, but I haven't, I just found some things to investigate that are proving fruitful so far. Like the number systems crumbs. I'm starting to have some ideas on why he told me to look at it in binary. Binary is 1 and 0, the VQC is based on odd and even. And though not confirmed, every example I've tested has shown me that the solution cell for multiples of 5 is based on odd and even b.
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81b46d No.3147
>>3146
Chris said some of the patterns are easier to see in binary. You know this already, but posting it for everyone to see.
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99eff8 No.3148
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5167b5 No.3149
>>3119
Ugh, I agree. LARP or not, I can't help but looking more into this. Too many fascinating patterns from such a simple algorithm.
Yeah, yeah I'm out bla bla bla. I'm quitting tobacco, so I'm a bitch right now.
So I took a look at another angle of the (d[t] - d) and the pattern of n as factor of a. I haven't found any silver bullets or anything useful, but I did observe what appears to be another pattern.
If we look at the difference between (d[t] - d) and (a[t] - n) we can see there is a t where this difference is 0. At this t, we can find the n by d[t] - d - a. However, this t appears to be the result of the t from big N - (n + 1)/2. So it still requires knowledge of n.
The hint I'm going to look more into is this:
> I will post more on The Grid later but think about the distribution of n and (n-1) in cell (e,1) in the a[t] and d[t]
values respectively. This is the quick way to utilising f.
Also, has anyone looked into (n, 0) columns? VQC gave a hint about it:
> Tomorrow I will show the way in which na and (n-1)a relationships between a[t] and d[t] in row 1 cells (e,1) and the pattern in column zero (n,0), these can both be used to factor arbitrary c in < O(log l) where l is the length of c in bits.
Maybe something in (n, 0) will yield something interesting.
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5167b5 No.3150
>>3149
Take e = 1, n = 5, d = 30 as an example:
1.0 -28 -24 -4
3.0 -12 -20 8
5.0 20 -16 36
7.0 68 -12 80
9.0 132 -8 140
11.0 212 -4 216
13.0 308 0 308
15.0 420 4 416
17.0 548 8 540
19.0 692 12 680
21.0 852 16 836
23.0 1028 20 1008
25.0 1220 24 1196
27.0 1428 28 1400
29.0 1652 32 1620
31.0 1892 36 1856
Here we have:
t, (dd - d), (dd - d) - (aa - n), (dd - d) - ((dd - d) - (aa - n)).
We can se at some point t = 13, the (dd - d) - (aa - n) equals 0. Before that it was equal to -4, -8, -12 and after it is equal to 4, 8, 12, … So the change is 4.
We can get 13 by doing (30 / 2) - (n - 1)/2). However again, this requires us to know n.
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5167b5 No.3151
>>3150
For (1, 5, 46) we can predict the t:
(46 / 2) - (5 - 1)/2 = 23 - 2 = 21
And the following:
1.0 -44 -40 -4
3.0 -28 -36 8
5.0 4 -32 36
7.0 52 -28 80
9.0 116 -24 140
11.0 196 -20 216
13.0 292 -16 308
15.0 404 -12 416
17.0 532 -8 540
19.0 676 -4 680
21.0 836 0 836
23.0 1012 4 1008
25.0 1204 8 1196
27.0 1412 12 1400
29.0 1636 16 1620
31.0 1876 20 1856
If we could find another way of getting to t = 21, we can find n, since it's d[t] - d - a[t].
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8aa562 No.3152
>>3149
Yeah! Alright Isee. Glad you’re back.
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ab8fb8 No.3153
Wonder if this will drag on long enough for me to create a portable VQC APP.. I need one.
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ff4dbe No.3154
>>3153
For when you need to crack RSA numbers on the go huh?
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3ed644 No.3155
>>3149
>Maybe something in (n, 0) will yield something interesting.
See - welcome back. There was a discussion about this in the previous thread.
Not sure if the following is helpful.
>>2013
>>2016
I checked the original grid and the n=0 records do have some special properties:
e is negative and one of = 1^2, 2^2, 3^2, 4^2, etc.
x = sqrt(-e)
b-a = 2x
d-b = x
BigInteger newE = -(ter.d * ter.d);
BigInteger newX = ter.d;
BigInteger newA = ter.n;
TheEndRecord testNegative = TerFactory.CreateForEXA( newE, newX, newA );
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ab8fb8 No.3156
>>3155
A negative cell exists for every positive cell because f is literally just the opposite of e (if you think about my mountains and valleys analogy). That's YUUUGE part according to Chris.
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ab8fb8 No.3157
>>3155
They are the same thing. You can create the same e,n cell with f and (d+1) (which is also the ceiling square root of c)
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ab8fb8 No.3159
>>3158
It wasn't the Fibonacci number, but I've found a way to generate the a value for t=1 for all n=0 cells that exist
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ab8fb8 No.3160
>>3159
You can then generate any other element for n=0 and e<0 by adding (t-1) to the starting a value
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ab8fb8 No.3161
Completed the logic for generating n=0 cells
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e69cb4 No.3163
CA give input on whether >>1762 was a nothing burger or whether it is actually worth looking into as a solution
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51b282 No.3164
>>3161
>>3160
Found this yesterday thanks to your earlier posts and drawing a lot of records. e always -perfect squares for n =0
Goal to find N=0
More than one
N=0 is root of D
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5167b5 No.3165
I haven't had much time today, but I've also looked into the (n, 0) records. Haven't found any patterns, but I did just observe something "funny". I was curious about these records and decided to play with d-values from (1, 1).
Try this code:
def findOnes(en):
en = en ** 2
en = en * -1
for i in range(1, 1444 * 15):
ee, nn, dd, xx, aa, bb = createForEND(en, 0, i)
if aa < 0:
aa = aa * -1
if bb < 0:
bb = bb * -1
e, n, d, x, a, b = rowForAB(aa, bb)
if e == 1:
print(i, (e, n, d, x, a, b), (ee, nn, dd, xx, aa, bb))
It's a pretty rough brute force method, but it does show some interesting results.
So I checked with 2, 8, 18, 32 and 50 and it appears the "last" record that has a corresponding record in (1, 1) out of the lot is always equal to ((en ** 2) / 2 + 1). And it appears to give numbers related to the n's within the record from (1, 1) and the given d.
I don't understand it, but take a look:
>>> findOnes(2)
3 (1, 1, 2, 1, 1.0, 5.0) (-4, 0, 3, 2.0, 1.0, 5.0)
>>> findOnes(8)
9 (1, 5, 4, 3, 1.0, 17.0) (-64, 0, 9, 8.0, 1.0, 17.0)
33 (1, 1, 32, 7, 25.0, 41.0) (-64, 0, 33, 8.0, 25.0, 41.0)
>>> findOnes(18)
19 (1, 13, 6, 5, 1.0, 37.0) (-324, 0, 19, 18.0, 1.0, 37.0)
35 (1, 5, 30, 13, 17.0, 53.0) (-324, 0, 35, 18.0, 17.0, 53.0)
163 (1, 1, 162, 17, 145.0, 181.0) (-324, 0, 163, 18.0, 145.0, 181.0)
>>> findOnes(32)
33 (1, 25, 8, 7, 1.0, 65.0) (-1024, 0, 33, 32.0, 1.0, 65.0)
105 (1, 5, 100, 27, 73.0, 137.0) (-1024, 0, 105, 32.0, 73.0, 137.0)
513 (1, 1, 512, 31, 481.0, 545.0) (-1024, 0, 513, 32.0, 481.0, 545.0)
>>> findOnes(50)
51 (1, 41, 10, 9, 1.0, 101.0) (-2500, 0, 51, 50.0, 1.0, 101.0)
1251 (1, 1, 1250, 49, 1201.0, 1301.0) (-2500, 0, 1251, 50.0, 1201.0, 1301.0)
The first "record" also appears to be (d + 1) and the last appears to be (d**2 / 2) - 1. Not sure about the ones in between though.
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5167b5 No.3166
Another interesting this is this:
>>> findOnes(12) # d = 12 for c = 145
17 (1, 5, 12, 7, 5.0, 29.0) (-144, 0, 17, 12.0, 5.0, 29.0)
73 (1, 1, 72, 11, 61.0, 85.0) (-144, 0, 73, 12.0, 61.0, 85.0)
>>> findOnes(30) # d = 30 for c = 901
35 (1, 17, 18, 13, 5.0, 65.0) (-900, 0, 35, 30.0, 5.0, 65.0)
451 (1, 1, 450, 29, 421.0, 481.0) (-900, 0, 451, 30.0, 421.0, 481.0)
>>> findOnes(46) # d = 46 for c = 2117
51 (1, 29, 22, 17, 5.0, 97.0) (-2116, 0, 51, 46.0, 5.0, 97.0)
1059 (1, 1, 1058, 45, 1013.0, 1105.0) (-2116, 0, 1059, 46.0, 1013.0, 1105.0)
Here, every 'n' is an 'a' in (1, 5), but these records start off from d + n, so it won't solve anything. But maybe it sheds some light on these (n, 0) records?
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51b282 No.3167
Take a record. Keep adding one to d. Keep a and b constant. For each increase in d n goes down by one and x goes up by one. But e is decreased by 2*d+1 each time.
The formula for changing e by an integer m:
e(t+m) = e(t) - ( d(t)*m+ mm)
If you can find m=n(t) then adding n(t) to d then n=o which means d+x = b.
xx=-e
So you are looking for a negative e that is a perfect square. If the square root is x then you are done.
For example for 5,23
{15, 4, 10, 5, 5, 23}
{-6, 3, 11, 6, 5,23}
{-29, 2, 12, 7, 5, 23}
{-54, 1, 13, 8, 5, 23}
{-81, 0, 14, 9, 5, 23}
-e is perfect square equal to x
Trying to figure out how to predict when n=0 or -e = perfect square
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51b282 No.3168
>>3167
It looks like e+xx =2na fails when e is negative side. Have to look at the geometry.
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ab8fb8 No.3169
>>3168
>>3167
all positive elements exist as negative elements
e becomes -f
d becomes d+1
(-24, 4)
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ab8fb8 No.3170
>>3169
THIS IS HOW IT'S A MIRROR.
if you have (e,n,d) you have the cell.
if you have (f, n-1, d+1) you have the cell.
1 is subtracted from n because d is added to n and thus it requires one less integer to get to the large square.
Mirror. (e,n,d) = (f, n-1, d+1)
it's the same information
examples:
{1:5:12:7:5:29}
mirror=
{-24:4:13:8:5:29}
{20:1:22:4:18:28}
mirror=
{-25:0:23:5:18:28}
{-17:4:17:9:8:34}
mirror=
{16:5:16:8:8:34}
works every time.
f is a mirror of e
(d+1)^2 is the square after d^2
if n is the distance from d to (d+n)^2 (the large square) then
(n-1) is the distance from (d+1) to (d+n)^2
you see?
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3ed644 No.3171
>>3159
Can you share that formula for d?
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ab8fb8 No.3172
>>3171
I'm cleaning up my code and then I will share it. It's not the formula for d it is the formula for the first a value in an n=0 -e is a perfect square cell.
Then any other element in the cell can be generated by adding (t-1) to the first a value.
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3ed644 No.3173
>>3172
No problem. I've found a way to iterate values in (-e, 0).
Have you noticed that records exist matching the prime solutions?
c: (1,25,4) = {1:25:8:7:1:65} = 65
p: (1,1,2) = {1:1:8:3:5:13} = 65
(-16,0,3) = {-16:0:9:4:5:13} = 65
c: (1,61,6) = {1:61:12:11:1:145} = 145
p: (1,5,4) = {1:5:12:7:5:29} = 145
(-144,0,7) = {-144:0:17:12:5:29} = 145
c: (249,10065,71) = {249:10065:142:141:1:20413} = 20413
p: (249,1,3) = {249:1:142:5:137:149} = 20413
(-36,0,4) = {-36:0:143:6:137:149} = 20413
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51b282 No.3174
>>3167
>>3170
If you increase d(t) by m holding a and b constant then formula for e(t+m) :
e(t) - d(t)*m- mm
Finding m= n(t) solves RSA
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ab8fb8 No.3175
>>3173
Yes, that's the mirror! The formula for generating the first value of a for (-e, 0) is
take the value of sqrt(abs(e)) - 1
let's call that q
the first a value is ceil((q^2)/2)
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51b282 No.3176
>>3173
Notice square root of -e equals x for n=0
Which is the formula xx=2na-e
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ab8fb8 No.3177
Cleaned up the logic, just going to add in the generation of -e, n>0 cells (the left side)
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3ed644 No.3178
>>3175
Thank you.
Here's a very sloppy method for generating an (-e, 0) record for any other record. Haven't fully debugged as to why I can't create the record from (e,n,d,x,a,b) variables, but the e,x,a are certainly valid.
public static TheEndRecord CreateForZeroN( TheEndRecord ter ) {
BigInteger a = ter.a;
BigInteger b = ter.b;
BigInteger n = 0;
BigInteger x = (b - a) / 2;
BigInteger d = x + b;
BigInteger e = -(x * x);
return CreateForEXA( e, x, a );
}
Regarding the "mirror", it's a bit confusing.
We can get to the negative e side of the grid with e - ((2 *d) + 1).
Then there is the -x space at (e,n) where a and b values reverse.
And now there is the -e, 0 space where we can find matching prime results.
They all play a role.
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ab8fb8 No.3179
>>3178
I've made a method that generates it based on (e, n, t). The only thing I appear to be missing is a method that generates (0, n>0) cells.
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51b282 No.3180
Q posted this today.
Election fraud fries me!
Can't wait til they do something about it
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3ed644 No.3181
>>3180
(shocking)
>>3179
These are some methods I put together a while ago that mostly worked for any (e,n). I say mostly, because Teach found a case in (56,n) I believe that didn't quite fit.
Also, I think we also discovered the ability to iterate records by both (x+n) and (x-n) adjustments to d after this. But should be a reasonable starting position.
public static TheEndRecord GetNext( TheEndRecord ter ) {
BigInteger newd = GetNextD( ter );
return TerFactory.CreateForEND( ter.e, ter.n, newd );
}
private static BigInteger GetNextD( TheEndRecord ter ) {
if ( ter.n % 2 == 0 ) { // even
return ter.d + (ter.x + ter.n) + ter.n / 2;
} else { // odd
return ter.d + 2 * (ter.x + ter.n) + 2 * ter.n;
}
}
public static TheEndRecord GetPrevious( TheEndRecord ter ) {
BigInteger newd = GetPreviousD( ter );
return TerFactory.CreateForEND( ter.e, ter.n, newd );
}
private static BigInteger GetPreviousD( TheEndRecord ter ) {
if ( ter.n % 2 == 0 ) { // even
return ter.d - (ter.x + ter.n) + ter.n / 2;
} else { // odd
// to walk down the tree: new.d = orig.d - 2*(x+n) - 2n
// can also be written as ter.d - 2 * ter.x
return ter.d - 2 * (ter.x + ter.n) + 2 * ter.n;
}
}
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ab8fb8 No.3182
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6a1cc2 No.3183
>>3182
3rd record looks weird. Shouldn’t that be e=-144?
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ab8fb8 No.3184
>>3183
Nope, my method says that 5*29 does not exist in (-144, 0) unless you make t negative.
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ab8fb8 No.3185
Here's the code: pastebin.com/8Q6siWwj
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84fce1 No.3186
If anyone has the time, can I get a recap of everything I've missed this month? 95% done getting some crazy shit sorted out (for now)
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ab8fb8 No.3187
>>3184
Something interesting is that it shows 17, and 17^2-12^2 = 145
>>3186
Not much, just getting to know the VQC better, and integers in general. Read the previous threads
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6a1cc2 No.3188
>>3184
Just tested your {144:0:17:12:5:29} record.
It fails this check:
c == ( d * d ) + e
If you look at the original data in the grid, the t value in the (-e, 0) space is unreliable. See pic of records in (-16,0) with t = 2.
I don't believe you can generate these records using t as an input parameter.
>>3187
>Something interesting is that it shows 17, and 17^2-12^2 = 145
That is interesting. Will look into also.
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ab8fb8 No.3189
>>3188
You can't rely on d^2+e. Almost the entire left half of the VQC fails that test. May or may not be a real cell but it can still be generated.
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6a1cc2 No.3190
>>3189
>You can't rely on d^2+e
Not sure about that. I've been generating records all over the place (including the left half) without any failures just using that IsValid check I posted before. Happy to post it again.
The 17^2 - 12^2 = 145 comment is a good observation. Have confirmed on other records.
This makes sense, because normally (d+n)^2 - (x+n)^2 = c, and we are dealing with n = 0. So everything here is a perfect square.
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ab8fb8 No.3191
>>3190
That works because in the left cells the d value is incremented by one, and the e value is f. But d^2 + (-f) is not c.
Still, they are pretty much a mirror of each other.
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6a1cc2 No.3192
>>3186
MA - we're still searching.
Can't summarize everyone's work, but VQC did drop some hints again about (n-1)*a transforms from (e,n) into (e,1).
He mentioned that an offset can be found to solve the problem by analyzing (d[t] - d)/(n-1) factors and comparing them with a[t]/n.
>>2990
>>3134
I've posted an example for c=145 that shows my limited understanding of this.
There was also a recent reminder about records in the (-e, 0) space that also contain the prime results we are looking for.
Patterns and patterns all over the place. But nothing yet that links any c records to prime solutions.
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6a1cc2 No.3193
Just posting some additional tests of the (-e,0) space.
This includes records for both c and p (prime solution), and the (-d^2, 0) record they both share.
Interesting that the c value from (-d^2) less the original e equals the e value in the (-e,0) record.
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ab8fb8 No.3194
>>3193
How are you getting to the "c at (-e,0)" cell
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6a1cc2 No.3195
>>3194
Using the method I posted earlier with a=1, b=c.
>>3178
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6a1cc2 No.3196
>>3194
And corrected method is:
public static TheEndRecord CreateForZeroN( TheEndRecord ter ) {
BigInteger a = ter.a;
BigInteger b = ter.b;
BigInteger n = 0;
BigInteger x = (b - a) / 2;
BigInteger d = b - x; // this was posted incorrectly earlier.
BigInteger e = -(x * x);
return new TheEndRecord( e, n, d, x, a, b );
}
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ff4dbe No.3197
I've been going through Chris' old hints and found this
>I am going to show how to do it in a complexity of less than O(log t) where log t is the natural log of the length of c in bits.
>less than O(log t)
I wasted an entire week on that binary search wild goose chase when all I had to do is read the hints again properly. Nice. Anyway, that means that this algorithm's running time probably isn't something that will lead us to the solution, in case anyone still thought that might be the case.
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ab8fb8 No.3198
123 = {2:51:11:10:1:123} (2, 51, 6)
123 = {-21:50:12:11:1:123} (-21, 50, 6)
123 = 62^2 - 61^2
visit (0, 50, 61)
{0:50:264:120:144:484}
sqrt(484) = 22
22^2 - 123 = 19^2
22^2 - 19^2 = 123
22 + 19 = 41
22 - 19 = 3
123 = 3*41
numbers that aren't the diff of 2 squares are just as important
>>3197
>wasted
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ff4dbe No.3199
>>3198
Aside from it being something to do when I'm not busy and having the company of anonymous internet people, it was a bit of a waste of time in my opinion. I still can't tell if we're just waiting around for Chris to decide the time is right again or if we really are just going to keep going around in circles like this until it magically happens several months or years from now. No lurkers with malicious intent seem to have solved it, or we'd have heard about it by now. That means either there are no selfish people watching (that's a bit farfetched considering the recognition you'd get if you claimed it was you), Chris hasn't given us enough information to solve it (at this point I trust his judgement so I don't think it's that; plus, he figured all of it out himself, albeit, if memory serves, it took him years), or we just happen not to have figured it out yet (which Chris would be aware of and could solve by making it way more obvious than he has so far).
Here's some pointless c = c algebra I messed around with for a while. Maybe something magical will pop out if you stare at it for a really long time.
c = d^2 + e
c = (d+n)(d+n)-(x+n)(x+n)
Therefore
d^2 + e = (d+n)(d+n) - (x+n)(x+n)
d^2 + e = d^2 + n^2 + 2dn - x^2 - 2nx - n^2
d^2 + e = d^2 + 2dn - x^2 - 2nx
e = 2dn - x^2 - 2nx
x = d - a
Therefore
e = 2dn - (d-a)(d-a) - 2n(d-a)
e = 2dn - (d^2 - 2ad + a^2) - 2dn + 2an
e = 2an - d^2 + 2ad - a^2
n = ((a + b) / 2) - d
Therefore
e = 2a(((a+b)/2)-d) - d^2 + 2ad - a^2
e + d^2 = 2a(((a+b)/2)-d) + 2ad - a^2
c = 2a(((a+b)/2)-d) + 2ad - a^2
c = 2a(((a+(c/a))/2)-d) + 2ad - a^2
c = 2a(0.5(a+(c/a)) - d) + 2ad - a^2
c = 2a(0.5a + 0.5(c/a) - d) + 2ad - a^2
c = a^2 + c - 2ad + 2ad - a^2
c = c
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ab8fb8 No.3200
You know what we've neglected? Column 0.
What I did with 123 doesn't work with many others but I think you can get the point that it isn't a coincidence.
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ab8fb8 No.3201
>>3199
I don't believe anyone with malicious intent has the faith or intellect required to solve it.
We've defined so many terms and equations that you really are going to have a hard time understanding what we put out unless you interact with us, and I've spent so many hours on this because of faith.
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5167b5 No.3202
Remember chris giving a hint about negative x in (e, n) row's?
I remember I took a look at it, but I couldn't generate any. Instead, all I do is "jump" backwards. So trying to generate records with negative x for (1, 61) jumped me instead to (1, 41) at t = 1.
Maybe we should take another look, and see if either we can generate negative x records for (e, n) or if the jumping can reveal something?
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5167b5 No.3203
Okay, so I tested something and I can now generate negative x-values for (1, 61). There's two interleaved patterns here, one where it starts with 1, 101, and the other is 1, 145.
So to generate negative x records I started by generating the record where a = 1, b = 145 and one where a = 145 and b = 1.
This gives us:
(1, 61, 12, 11, 1, 145)
(1, 61, 12, -133, 145, 1)
Now I found this by trial and error, but the way to keep going in the "negative part" is by creating a record for E N X with this:
>>> rowForAB(1, 145)
(1, 61, 12, 11, 1, 145)
>>> rowForAB(145, 1)
(1, 61, 12, -133, 145, 1)
>>> createForENX(1, 61, -133 - 2*61)
(1, 61, 278.0, -255, 533.0, 145.0)
>>> createForENX(1, 61, -133 - 4*61)
(1, 61, 788.0, -377, 1165.0, 533.0)
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5167b5 No.3204
>>3203
Actually, createForENX(1, 61, -11 - 2 * 61) will also give you (1, 61, 12, -133, 145, 1).
And I just remember that maybe we were able to do this in the last thread?
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5ab110 No.3205
>>3202
I have spent the last weeks on the binary angle without really getting anywhere. Will post some thoughts and why I think the algebra is going in circles, starting with some crumbs
Couldn't find the direct quote, but think it went something like "odd or even is the first branch of the tree"
>At most O(log t), where log t is the natural logarithm of the length of c in bits
>Because a search exponential in the length of c is turned into a calculation which is the same complexity as finding a square root < O(log l) where l is the length of c in bits
Now, the only way I can think of is that length of c in _bits_ is relevant to the complexity of the problem is if they are a direct part of the calculation. The hint about odd and even makes sense since the LSB of a number is either 0 or 1. So the second bit (0 or 2) would be the second branch, and the third bit (0 or 4) the third… etc.
I have been trying all kinds of crap and still no closer to how this would map to a tree..
Per the latest hints, the binary patterns seem even more relevant. I generated the csv in binary and there are definitely some crazy shit going on but haven't quite figured it out (story of these threads..). I feel the difference between two squares is almost perfectly represented in base 2!
Now I haven't posted in a while due to lack of results but am here with you! We got this thing, don't go backing out now! And thank you all for keeping it going, I read every single post here
Basically I think the path we're trying to follow depends on making decisions on where to go next in the tree based on each cell we can jump to.
Also updating trip after the algo change, for when we need it in the future!
Remember the Ponies <3
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769a5e No.3206
>>3204
-x records seem to "line up better" than records in x. At least they give different target numbers to try and match to. See one of my sample pics.
You can create any corresponding -x record simply via an (e,n,t) formula.
for even e:
newT = 2 - t
for odd e:
newT = 1 - t
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769a5e No.3207
>>3193
Here are the same records in binary.
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5ab110 No.3208
>>3207
Suggest using fixed length {:b08} to {:b16} format strings etc. just so things line up - easier to find patterns that way
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5ab110 No.3209
>>3208
And obviously I meant {:08b} :)
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ab8fb8 No.3210
>>3207
Looks like I need to learn some binary.
You should also try the other bases, if you have the time. Chris said that there was one other base that helps you see the patterns, forgot what it was.
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769a5e No.3211
>>3208
Here you go. Padded right 24.
>>3210
>Looks like I need to learn some binary.
Likewise.
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5ab110 No.3212
>>3210
Nothing comes to mind relating to primes that would seem useful, without looking at them all. Binary is pretty easy!
I have some multithreaded bruteforce code to generate arbitrary formulas and functions ready to go, and access to enough cores to run it on, but am completely stuck at creating useful verification code that isn't exponential. If we could just jump from one record to the solution it would be easy, but there could be anywhere between 100 to 100000 jumps even for small numbers.
My thinking is just creating test cases where we already have the solution and using the known results to score each function. 100% hits for even small numbers is probably gonna be right on. Ideas on verification algos (ie. try formula, generate every possible record for the result using the previous variables, and generate the rest). This easily becomes a few thousand tries per formula, which is fine, but quickly becomes insane when you add jumps. Assuming an upper bound without any data to back it up will just cause false negatives and make the results useless.
>>3211
Checked! Thanks :)
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769a5e No.3213
>>3211
>Padded right 24.
Padded left.
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5ab110 No.3214
>>3212
By arbitrary formulas I mean every permutation of something like this (and more) def __init__(self, lit_min = -3, lit_max = 3):
self.var = list("endxabc")
self.lit = [i for i in range(lit_min, 0)] + [i for i in range(1, lit_max + 1)]
self.op = [ast.Add(), ast.Sub(), ast.Mult(), ast.Div(), ast.Mod(), ast.Pow()]
self.func = ["cos", "sin", "sqrt", "tan", "hyp", "gcd"]
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5167b5 No.3215
It seems like VQC makes an appearance and drops a hint, once a week. Maybe he'll stop by this week too with a hint.
I'm going to keep looking into the whole d[t] - d thing.
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769a5e No.3216
>>3215
Same. My next step is to try and iron down the patterns for d[t] - d and a[t].
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5167b5 No.3217
Since we started to talk about binary and VQC said some patterns are easier seen in binary I re-did patterns in >>3050 and looked at the d's in binary.
For (1, 5) (Note, since I struggle a bit with 0 and 1, - means 0 and k means 1). The code for bprint was found on stackoverflow.
>>> bprint(12)
------------kk--
>>> bprint(30)
-----------kkkk-
>>> bprint(46)
----------k-kkk-
>>> bprint(76)
---------k--kk--
>>> bprint(100)
---------kk--k--
>>> bprint(142)
--------k---kkk-
>>> bprint(174)
--------k-k-kkk-
>>> bprint(228)
--------kkk--k--
>>> bprint(268)
-------k----kk--
>>> bprint(334)
-------k-k--kkk-
>>> bprint(382)
-------k-kkkkkk-
12 gives d[t] - d: 2, 4, 6, 8, 10, 12, 14, 16.., 30 (Change of 2, or 4 if interleaved)
30 gives d[t] - d: 1, 3, 5, 7, 9, 11, 13, 15, .., 31
46 gives d[t] - d: 1, 3, 5, 7, 9, 11, 13, 15, .., 31
76 gives d[t] - d: 2, 4, 6, 8, 10, 12, 14, 16.., 30
100 gives d[t] - d: 2, 4, 6, 8, 10, 12, 14, 16.., 30
So we can see that the second LSB is either 1 or 0 and when it is, the alternating pattern is the pattern of d's where d[t] - d contains n-1 as a factor. Again, not sure if this is a fluke or a pattern.
For (1, 13)
>>> bprint(6)
-------------kk-
>>> bprint(38)
----------k--kk-
>>> bprint(68)
---------k---k--
>>> bprint(132)
--------k----k--
>>> bprint(182)
--------k-kk-kk-
>>> bprint(278)
-------k---k-kk-
>>> bprint(348)
-------k-k-kkk--
>>> bprint(476)
-------kkk-kkk--
6 gives 3, 9, 15, 21, 27 (Change of 6)
38 gives d[t] - d: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31
68 gives d[t] - d: 2, 4, 8, 10, 14, 16, 20, 22, 26, 28
132 gives d[t] - d: 6, 12, 18, 24, 30
182 gives d[t] - d: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31
278 d[t] - d: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31
348 d[t] - d: 6, 12, 18, 24, 30
476 d[t] - d: 2, 4, 8, 10, 14, 16, 20, 22, 26, 28
Here we have three (or four if you count (1, 13, 1)) distinct patterns of d[t] - d mod (n - 1).
The only bit 348 and 132 has in common is the 4th LSB, which is set to 1. While 38, 182 and 278 has the 4th and 2nd bit equal to 2 in common. 68 and 476 has 68 in common, but with more d's that give the same pattern, maybe there is more too it.
All in all though, I'm not sure if we are barking up the wrong tree here.
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5ab110 No.3218
>>3217
Welcome to my world! :)
There are patterns but I don't understand them, hence not posting here. Meant as inspiration iin case it triggers ideas, but could very well be the wrong tree (literally, binary numbers are a fractal tree on their own. if you want more on that just ask)!
With monospace fonts I prefer replacing 0 with space and 1 with X or | for readability, after you get over how binary works. But Chris seems to think its relevant, and if you can explain how the number or bits of a number maps to the O notation I would love to hear it. Binary search is close but not directly related to bits. Why be so explicit if its not part of our quantum computer?
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ab8fb8 No.3219
>>3217
I only got a few hours of sleep.. so hard to look at. I'm just going to analyze (0, n).
Chris said it was one column to rule them all!
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5ab110 No.3220
>>3218
>>3217
Fractal tree (can't be arsed to convert to horizontal, so use your scroll wheel)>>> for i in range(1, 50): print("{:08b}".format(i)[::-1].replace('0', ' ').replace('1', '|'))
...
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>>>
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ab8fb8 No.3221
>>3220
Oo.. I remember this. All numbers are a fractal!
wired.com/2011/01/partition-numbers-fractals/
researchgate.net/profile/Carlo_Cattani/publication/221433042_Fractal_Patterns_in_Prime_Numbers_Distribution/links/540499ef0cf23d9765a688ab/Fractal-Patterns-in-Prime-Numbers-Distribution.pdf
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5ab110 No.3222
>>3221
In my opinion fractals are the closest thing we have to explaining the universe with, well.. words and concepts.. Makes sense they apply to numbers too. Haven't seen any better ideas, anywhere. Then again we're in it ("the universe"), trying to understand so its not like its gonna work beyond our what our thougts tell it to anyway, like dividing by zero, except with magic :)
Will check out the links
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5167b5 No.3223
>>3220
Oh, wow. I had no idea that this beautiful pattern existed in something as simple as binary numbers.
Thank you! That's is amazing.
I played some with it:
Here we see (c, e, d, binary)
1 0 1 |
2 1 1 |
3 2 1 ||
4 0 2 |
5 1 2 | |
6 2 2 ||
7 3 2 |||
8 4 2 |
9 0 3 | |
10 1 3 | |
11 2 3 || |
12 3 3 ||
13 4 3 | ||
14 5 3 |||
15 6 3 ||||
16 0 4 |
17 1 4 | |
18 2 4 | |
19 3 4 || |
20 4 4 | |
21 5 4 | | |
22 6 4 || |
23 7 4 ||| |
24 8 4 ||
25 0 5 | ||
26 1 5 | ||
27 2 5 || ||
28 3 5 |||
29 4 5 | |||
30 5 5 ||||
31 6 5 |||||
32 7 5 |
33 8 5 | |
34 9 5 | |
35 10 5 || |
36 0 6 | |
37 1 6 | | |
38 2 6 || |
39 3 6 ||| |
40 4 6 | |
41 5 6 | | |
42 6 6 | | |
43 7 6 || | |
44 8 6 || |
45 9 6 | || |
46 10 6 ||| |
47 11 6 |||| |
48 12 6 ||
49 0 7 | ||
50 1 7 | ||
51 2 7 || ||
52 3 7 | ||
53 4 7 | | ||
54 5 7 || ||
55 6 7 ||| ||
56 7 7 |||
57 8 7 | |||
58 9 7 | |||
59 10 7 || |||
60 11 7 ||||
61 12 7 | ||||
62 13 7 |||||
63 14 7 ||||||
64 0 8 |
65 1 8 | |
66 2 8 | |
67 3 8 || |
68 4 8 | |
69 5 8 | | |
70 6 8 || |
71 7 8 ||| |
72 8 8 | |
73 9 8 | | |
74 10 8 | | |
75 11 8 || | |
76 12 8 || |
77 13 8 | || |
78 14 8 ||| |
79 15 8 |||| |
80 16 8 | |
81 0 9 | | |
82 1 9 | | |
83 2 9 || | |
84 3 9 | | |
85 4 9 | | | |
86 5 9 || | |
87 6 9 ||| | |
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5167b5 No.3224
>>3223
You can see how every time c is a perfect square, the d "resets" and starts counting up again.
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5167b5 No.3225
>>3224
Correction, the e resets.
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5167b5 No.3226
Here we have the first 15 records from (1, 1)
5 1 2 | |
65 1 8 | |
325 1 18 | | | |
1025 1 32 | |
2501 1 50 | | ||| |
5185 1 72 | | | |
9605 1 98 | | || | |
16385 1 128 | |
26245 1 162 | | | || ||
40001 1 200 | | ||| |
58565 1 242 | | || | |||
82945 1 288 | | | |
114245 1 338 | | | ||||| ||
153665 1 392 | | || | |
202501 1 450 | | ||| | ||
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5167b5 No.3227
>>3226
In that one the binary is a c in binary.
This one is d in binary for first 15 records of (1, 1):
5 1 2 |
65 1 8 |
325 1 18 | |
1025 1 32 |
2501 1 50 | ||
5185 1 72 | |
9605 1 98 | ||
16385 1 128 |
26245 1 162 | | |
40001 1 200 | ||
58565 1 242 | ||||
82945 1 288 | |
114245 1 338 | | | |
153665 1 392 | ||
202501 1 450 | |||
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5ab110 No.3228
>>3223
Glad you liked it. I still can't believe this is not common knowledge! Try telling people there are fractals in binary numbers and see what happens. Yet, one line of python here is enough :)
>>3226
>>3227
These are interesting but get back to me when we can do something with them. I've tried!
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769a5e No.3229
Attached are latest analysis of d[t]-d and a[t] for c=145, c=533 and c=6107. Just to give examples for various values of e.
These records are generated via a regular (e,1,t) function where t values begin from the original c at t=c.t + c.n, and then into negative t.
Included are differences for x and x/2, as well as those differences in a binary view where zeros are spaces, and 1's are #.
It looks like:
one of the sequences for a[t] is based on multiples of x.
one of the sequences for d[t]-d is based on multiples of (x+1).
There can be more than 2 interwoven sequences as the numbers get larger. See c=6107 as an example.
Not sure if this is adding any clarity.
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191295 No.3230
>>3201
Hear, hear. Well said, Baker.
>you really are going to have a hard time understanding what we put out unless you interact with us, and I've spent so many hours on this because of faith.
>>3205
>Remember the Ponies <3
>>3210
>Looks like I need to learn some binary.
Me too.
>>3186
MA! Good to see you.
>>3220
Wow, very cool binary pattern!
>>3229
PMA is a Math Machine. Are we stalled on finding a connection between c and p records? That jump between the two seemed like t-2 for the examples given.
Still working over here too, Anons. Nothing to report at this time. I'll keep plugging along!
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769a5e No.3231
>>3230
>Are we stalled on finding a connection between c and p records? That jump between the two seemed like t-2 for the examples given.
VA - Unfortunately, the more I dig into the data, the further away a solution seems. So much to learn.
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99eff8 No.3232
PathGawds!
I'm gonna fucker off for a bit.
Hit me if'n y'all need anything.
Don't lose your stamina! ^_^
<3 Ta +++
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5167b5 No.3233
>>3229
Regarding interwoven / interleaved patterns, you'll see theres 4 or 5 I think in (0, 8). 2 isn't a magic number, I suspect that you can find infinite interwoven patterns in any (e, n) if you extend the grid into infinity.
Maybe there is a limit, but I haven't spent any time looking into it. All we know is that for every an in (e, 1), the a will exist in (e, n). Thus assuming infinite long (e, 1) and given the p * i + t, p * i + 1 - t I conjecture (That's right guys, I'm a math guy now) that there's infinite of them.
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769a5e No.3234
>>3233
Thank you.
I've continued to look at the d[t] and a[t] differences and explored binary and factors of each.
Attached are latest tests for c=145, 533 and 6107.
I've added columns for t diff, t diff in binary, and t diff factors.
The factors are interesting, as they sometimes match the original x, x+1, x-1, d, d+1, or d-1 values indicated as [x], [x+], [x-], [d], [d+], [d-] respectively. I've also highlighted perfect square factors. Example is 4^(2) which indicates a 2x2 square.
Really interesting is the c=6107 example where a factor of 197 exists as a t diff factor in the d[t] - d analysis, and a factor of 31 exists as a t diff factor in the a[t] analysis.
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5167b5 No.3235
>>3234
I'm just thinking out loud here. VQC is very adamant about:
> Remember At that value, d[t] = na+x
> REMEMBER, the value of x at na in (e,1) is the SAME as x at (e,n)
> REMEMBER, take d from all values of d[t] at (e,1) and there is a known patter of (n-1) as factor in these values of d[t]-d that is different (increasingly) from the pattern of factors of n in a[t]. It is THIS that gives the offset that is used to solve the problem and thus get the cell at (e,1) to do all the work for you
Now, why is the x so important? Why do we need to care that x at na in (e, 1) is the SAME x as (e, n) if we are supposed to find na?
I mean, why would he stress that x is the SAME at (e, 1) as (e, n) if na already hit's the spot?
If we are supposed to find a pattern for (n - 1), then in the end, we'll now what n - 1 is, and then we know what n is and then we'll just do na / n to get a. So why is x important? Could it be that there is more to it than we think?
Also, why is d[t] = na + x so important? We know from that equation that d[t] - x = na, d[t] - na = x. Yet if we find the correct d[t] we have x and na, so again. Why are those three highlighted by VQC in the latest hint?
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5167b5 No.3236
>>3235
So if the x is important, maybe we aren't going to find n or (n - 1).
He also talks about an offset. That the pattern of (n - 1) in (e, 1) will give us an offset. So the offset will somehow give us x?
Maybe the offset will aid in generating numbers that will be potential x's? And at some point we have to use that to find (e, n) for a given c?
I'm probably talking about things everyone already knows. I just want to try and air my thoughts.
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769a5e No.3237
>>3236
I compared records at (e,1), (e,a), (e,b), (e,n), and (e,c) specifically because they share e, x and 2na values.
Figured that if we reduced the movement in variables, the solution would present itself. But I haven't been able to come up with a formula for d changes that accurately describes more than a few test cases.
I'm also leaning towards the offset being a difference in t or x. And will spend a bit more time on the (p * i + t, p * i + 1 - t) formulas you mentioned above.
Perhaps there is also something to be learned from looking at the difference between (-e,0) and the prime solution.
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ff4dbe No.3238
>>3235
>>3236
He did also say at one point that x marks the spot.
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94e2fe No.3239
>>3232
Thanks Topol! You are awesome. Your encouragement has always been appreciated.
>>3233
>(That's right guys, I'm a math guy now)
Fuck yeah, Isee. I'm learning to program over here. (Limited success so far, but working!)
>>3234
PMA hello! When we scale up in c and p, it looks like we have a prime (in a) and a multiple of 3 (in b). If we can figure out the a primes and tie them to a multiple of 3 pattern, that could work too.
>>3235
Hello Isee! If we have x, we can solve the entire element. X is the key. X marks the spot.
>>3236
If the offset can help us derive x, the problem is solved. We are on the right track, i think.
>>3237
PMA, t is a derivative of x. We find x, we solve this! But t in (e,1) is BIGLY important because it's our reference for (1,c) and prime records.
>>3238
X!!! Let's do this, faggots.
VQC's latest crumbs show that x[t] (and therefore the solution to all other vars) will be found in the (na)*1 and (na) for (e,1) vs (e,n). Prime and C records need a connection point.
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6a1cc2 No.3240
Here is another example of what doesn't quite work in different test cases, but works for c=901.
Might be considered a follow up to my post at >>3093 but using the (e,1) na records only to solve.
Starting na record for c=901
(1,1,15) = {1:1:450:29:421:481}
Destination na record for prime solution
(1,1,7) = {1:1:98:13:85:113}
New d formula:
newD = d - ( (d[t] diff) * (x+1) - (a[t] diff) * (x-1) )
the d[t] diff value can be calculated from (x^2 - f)
the a[t] diff value is d[t] diff - 4
In this example:
d[t] diff: 120
a[t] diff: 116
x+1 = 30
x-1 = 28
98 = 450 - (120 * 30 - 116 * 28)
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8aa562 No.3241
Hey lads. I'm working! Forgot my name/trip above. Feeling a bit defeated at the moment, not giving up tho. We've chased so many leads, and all on faith and hope to make our world better. We keep abandoning one line of search to chase another. I really thought c and p would yield a good result. Let's have a SOTU for VQC. What line of inquiry should we follow as we move forward?
VQC keeps saying there is a pattern in a[t] and d[t]. Can't see it yet tho! Fuggggg.
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ab8fb8 No.3242
>>3241
I can't work on it right now but I'll direct you to an important hint. Answer the question of, if c is a semiprime,
how many times does c^2 appear in column 0?
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ff4dbe No.3243
>>3242
It's 5, isn't it?
e.g. c = 91, a = 7, b = 13
91*91 = 8281
Possible sets of (a, b) are (1, 8281), (7, 1183), (13, 637), (49, 169) and (91, 91). That's 1 and c^2, c and c, a^2 and b^2, a and c^2/b, and b and c^2/a.
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Post last edited at
ab8fb8 No.3244
>>3243
And those all appear in column zero?
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ff4dbe No.3245
>>3244
Yeah, if c is a square then e = 0 regardless of a and b, right? That's why (a, b) = (2, 8) early on in e = 0.
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ab8fb8 No.3246
>>3245
That sounds pretty useful. If you squared it again it would probably appear in column 0 even more.
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ff4dbe No.3247
>>3246
If it was c^3 and c = 91
(1, 753571) (1, c^3)
(7, 107653) (a, a^2*b^3)
(13, 57967) (b, b^2*a^3)
(49, 15379) (a^2, b^3*a)
(91, 8281) (c, c^2)
(169, 4459) (b^2, a^3*b)
(343, 2197) (a^3, b^3)
(637, 1183) (c^2/b, c^2/a) or in other words c^4/c
Each factor is just every list of possible combinations of a, b and c that give c^k. I was trying to look for relationships between the cells I mentioned here >>3243 for a lower value of iMax than is necessary for c = 91 so it doesn't take hours to generate the grid, but for some reason it cuts off around n = 511 with setSize = 1 (I'm assuming it would be at t = 1 because VQC pointed out a while ago that in e=0's column, for subsequent t cells in a particular (e, n), a and b increment upwards in particular patterns. I'll find the post if you want. I read it in the pdf I posted in the last thread). With c^2 for c = 35, (a, b) = (1, 1225), i should equal 613, so that cell should show up at (e, n) = (0, 578). Can you think of why that isn't happening?
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ab8fb8 No.3248
>>3247
Set size is equal to the amount of t values that the VQC generator shows
{0:578:35:34:1:1225} (0, 578, 18)
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ff4dbe No.3249
>>3248
I know that. I think my code is wrong because even if I change setSize to 18 or more it doesn't show up. That's weird though. I thought it would have been at t = 1. Have you looked at the patterns as t increments always starting from lower numbers and occasionally a = one of the bs from before? What does the pattern of a and b look like before (0, 578, 18)?
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ab8fb8 No.3250
>>3249
Actually you are quite right that 1*1225 is the first valid element in (0, 578).
But the t required to generate is 18.
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ff4dbe No.3251
>>3250
Maybe something's wrong with your code. How could a = 0?
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ab8fb8 No.3252
>>3251
Nope, try calculating t for 1*1225. There are 17 more records before it
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ff4dbe No.3253
>>3252
What I meant was, why is a in each of the first of the cells in your picture 0? How do you make c if one of a or b is 0? I don't get it.
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ab8fb8 No.3254
>>3253
It makes absolutely no sense, yeah. But they're there. The 0*b records don't appear in the VQC, so I just assume they are to be ignored.
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ff4dbe No.3255
>>3254
You're using the Java version too, right? I'm trying to figure out how these cells are actually put into the grid. We know t = 18 for the cell we want based on math, but I'm looking at outputGrid and I'm not seeing how it would put it in at t = 18 instead of just the first instance of (e, n), which, for this cell, since those other ones aren't valid, would be t = 1.
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2b4079 No.3256
Regarding the d[t] - d hint and finding an offset.
We know the t we want is 4, and we know t=6 which is 61 (for c=145).
The pattern of dt - d is represented by the function 2*t^2 - 12. Similar to the normal equation for d. That function is negative for 1, 2 but for it's not for 3. What if that's the offset?
6 - 2 = 4.
Maybe the t-1 where d[t] - d is the first positive number represents (e, n)s t?
It appears to work for the numbers I checked
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2b4079 No.3257
>>3256
Ugh no, nevermind. It's not going to be the t, but we might explore it as the offset.
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51b282 No.3258
Think of the grid defining a space made up of perfect squares.
Then t equal d.
Think of n=0 records from baker. For an e perfect square there exists a record for every d positive integer where x = sqrt(abs(e))
Each increase of d by one is the next square.
Factor solution for c is n=0 then x tells you how far apart they are.
For each decrease in n by 1 d and x go up by 1 and a and b stay the same.
e(d+1) = e-(2d(d)+1)
Solution is -e=(n+x)^2
Solution is (-(n+x)^2, 0, d+n)
{-(n+x)^2: 0: d+n: x+n: a: b}
Notice d=long side x= short side
If you want to move multiple d at a time the formula is e+m = e-2dm-mm for increasing d
formula is e-m = e+2dm-mm for decreasing d
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5167b5 No.3259
>>3258
I know you've talked about this m, before but I coulnd't a description.
Could you explain m? And could you give an example of moving mutliple d's at a time?
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51b282 No.3260
>>3259
I just use m as integer variable for movement.
In this case the movement is between records describing the same a*b=c.
When you take a record (e, n, t) you can also describe it by changing d and e without changing a or b.
The relationship is d and x increase by m
n decrease by m
for e you use formula
>>3258
Its the same record just a different way of describing it. However when you move n steps then n=0 and you have your solution.
Amazingly and frustratingly if you use algebra on the formula you just get (n+x)^2.
Whenever you use algebra to solve in the grid you get something you already know.
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51b282 No.3261
Think t = d
Grid is infinite series of squares with side = d
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51b282 No.3262
>>3259
When you increase d by one subtract the difference between d and d +1 from remainder.
Its the same record.
You can do it multiple times it still describes a*b=c
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ab8fb8 No.3263
>>3260
I don't understand.
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d95295 No.3264
>>3263
You know how to increase d by one. Use f for e.
That is simply subtracting 2d +1 from e.
As d increased by one the remainder became negative because it's larger than c now.
A and b stayed the same.
X also goes up by 1 and n is reduced by 1. You can do this over and over. Remainder keeps getting more negative as d is increasingly larger than c. Each increase in d makes the remainder 2d+1 smaller than the previous d.
The formula allows you to move multiple Ds. When you have done that n times then e = -(x+n)^2
Then n=0 and x= x+n and that is the solution.
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d95295 No.3265
Just replaced my router cause it sucked. Now I have a new ID
>>3260
>>3264
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ff4dbe No.3266
>>3260
>>3264
>changing d and e without changing a or b
That doesn't make any sense. d = floor(sqrt(a*b)) and e = c - dd. They don't change if a and b don't change. What are you talking about? Are you using d and e to describe different variables? Because if you are it explains why I also don't understand this >>3258 stuff and you should pick different letters.
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d95295 No.3267
>>3266
>>3167
Look at the records in 3167 all have same a and b. They are different ways of expressing the same information as d grows larger the remainder the difference between d^2 and c becomes increasingly negative. Every move up 1 by d increases x by 1 and reduces n by 1. The goal is to get rid of n.
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d95295 No.3268
>>3266
Its really easy to understand if you draw it on graph paper. After you draw a few you realize that you are basically putting dots representing squares on the diagonal. d=t
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ff4dbe No.3269
>>3267
>>3268
How did you generate those records? There's only meant to be one value of negative e too, for a given static a and b.
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769a5e No.3270
>>3268
Just ran a quick test to confirm your records.
Here is some sample output for a=5, b=23. Also notice the -5 change in na.
(15,4,3) = {15:4:10:5:5:23} = 115; f=6; (x+n)=9; na=20; f+c=121; -x=-13
(-6,3,4) = {-6:3:11:6:5:23} = 115; f=29; (x+n)=9; na=15; f+c=144; -x=-12
(-29,2,4) = {-29:2:12:7:5:23} = 115; f=54; (x+n)=9; na=10; f+c=169; -x=-11
(-54,1,5) = {-54:1:13:8:5:23} = 115; f=81; (x+n)=9; na=5; f+c=196; -x=-10
(-81,0,5) = {-81:0:14:9:5:23} = 115; f=110; (x+n)=9; na=0; f+c=225; -x=-9
>>3269
>There's only meant to be one value of negative e too, for a given static a and b.
This doesn't appear to correct. There is only 1 valid at n=0.
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ff4dbe No.3271
>>3270
How do you generate more than one value of -e if a and b don't change then?
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ff4dbe No.3273
>>3272
But you haven't explained the logic by which you can create multiple records with the same a and b values but with changing e and d values when e and d are calculated based on a and b. You can't just change the formula to e-2*d-1. e = c - dd. That's what I don't understand.
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d95295 No.3274
>>3271
e = c -d^2
As d increases remainder becomes increasing negative.
>>3270
Your program is wrong. x = d - a it isn't negative
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ff4dbe No.3275
>>3274
d is also calculated based on a and b, so how are you meant to have changing values of d either? Do you not understand what I'm saying?
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d95295 No.3276
made slight typo corrected
>>3269
Test them!
I have given the rules for generating them over and over.
For any record you {E, N, D, X, A, B}
You can generate an equivalent record
{E-2*D-1, N-1, D+1, X+1, A, B}
If you do it again the the new record will be
{E-(2*d+1)-(2*(d+1)+1), N-2, D+2, X+2, A, B}
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769a5e No.3277
>>3274
Thanks. That -x calculation is used to jump into the -x space. It's calculated as -(x + 2 * n); I'll remove as it's confusing.
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7e0ce0 No.3278
I think if you check the GCD for d(in e,1) - D(from the start) and A the first entry where GCD != 1 is the X you want.
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7e0ce0 No.3279
>>3278
This works for every attempt I have done. I think this is the answer.
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ff4dbe No.3280
>>3276
You've also avoided answering my questions over and over. Jesus Christ anon. If you're right then all you have to do is explain your logic, and if you're wrong maybe I'll save you the time sink of a wild goose chase.
>you have specific values for a and b, say, a = 3, b = 5
>c = a*b (in this case 15), so if a and b don't change, c doesn't change
>d = floor(sqrt(c)) (in this case 3), so if c doesn't change, d doesn't change (also meaning if a and b don't change, d doesn't change)
>e = c - dd (in this case 6), so if c and d (and by extension a and b) don't change, e doesn't change
>if a and b don't change, d and e don't change
Why is it that whenever I ask where the logic in changing the formula is, all you say is "test them"?
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d95295 No.3281
>>3275
C=d^2 + e
C also equals (d+1)^2 + e - (2d+1)
f = e - (2d+1)
new record is {f, n-1, d+1, x+1, a, b}
Now do it again
new f = f - (2*(d + 1) + 1)
new record = {f - (2*(d + 1) + 1), n-2, d+2, x+2, a, b}
Draw them on graph paper
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d95295 No.3282
>>3280
Again c = d^2 plus e
Is it also possible for c to equal a bigger d^2 with a negative remainder? Is it possible that there are an infinite number of larger ds with increasingly larger negative remainders? All describing the SAME a and b
How about this pick a relatively small record like a =13 b = 37
{40, 4, 21, 8, 13, 37}
Now make e = -(x+n)^2
n=0
d=d+n long side
x = x+n short side
a=a b=b
{-144, 0, 25, 12, 13, 37} your solution record.
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769a5e No.3283
>>3282
Very interesting. Thank you for sharing this.
I've been able to reproduce the movement down to n=0.
Have you taken a look in the opposite direction?
newE = e + (2*d - 1)
newD = d - 1
newX = x -1
These entries literally go on for ever.
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ab8fb8 No.3284
>>3282
>>3283
Interesting. So he is saying you can take this path.
{40:220:21:20:1:481} (40, 220, 11)
{-144:0:25:12:13:37} (-144, 0, -62)
{40:4:21:8:13:37} (40, 4, 5)
For these purposes, I'm guessing I'm going to have to assume -t exists.
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ffc227 No.3285
Confirming operational status. Have other things to do now, taking render requests again to fill tomorrow.
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ab8fb8 No.3286
>>3282
I didn't understand it at the time, but Chris said that any number can be a remainder.
He said my "remainders in a sea of squares" was brilliant.
Must have hit the right note.
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ab8fb8 No.3287
>>3284
the code I wrote to generate e = negative perfect squares with n=0 matches the VQC.
pastebin.com/8Q6siWwj
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d95295 No.3288
>>3282
>>3258
{40, 4, 21, 8, 13, 37}
{-144, 0, 25, 12, 13, 37}
Use formula to change e from bottom of 3258
new e = e -2*d*m-mm
original e =40 d =21 n=4
so new e =40 -2*21*4-4*4=40-168-16=-144
e equals - (D+N)^2 or -(long side)^2
x=x+n or short side. Anyone rcognize the difference between two squares with n=0
Solution
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d95295 No.3289
>>3284
d=t
d=1 is a square 1^2
d=2 is a square 2^2
when you go to a negative remainder then d is getting bigger which means t is getting bigger
d=t
>>3283
You can go in the opposite direction but its the wrong way you want n to disappear. But you're right there are an infinite ways to describe a*b=c.
Also the formula for moving multiple Ds down is not the negative of moving up it is
e+2dm-mm
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769a5e No.3290
>>3289
If I am following you correctly, your new e formula enables a jump directly to the corresponding n=0 record using values of e, n and d.
But how is this a solution when the a and b values are still unknown?
What am I missing?
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d95295 No.3291
>>3284
>{40:220:21:20:1:481} (40, 220, 11)
I don't see where this record comes from.
The path I took goes directly from start record based on moving n steps into negative e space. d=t
There is a record for every t therefore d that can describe the original record.
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ab8fb8 No.3292
>>3291
That's just the 1*c record. What record do you start with?
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ab8fb8 No.3293
>>3288
This is actually amazing. Slowly starting to get it. d and x are equivalent to the big and small square because n=0
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d95295 No.3294
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d95295 No.3295
>>3293
Unfortunately we still don't have a way to figure n. But the grid is making more sense
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d95295 No.3296
>>3290
You aren't missing anything just better understanding the grid. We still don't know n.
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ff4dbe No.3297
>>3281
So let me get this straight. Is this what you're suggesting?
>there are infinite f values for a given (a, b)
>the first is e - 2d + 1
>if you use a value of f that isn't the original one, you're changing the formula to calculate f
>that means you also need to change the formula for every other variable
>this works because f is what you add to c to get a square and each successive f creates the next square higher than the larger square
>this means the changed formulae for the other variables relates to the smaller square
>as the squares increase or decrease (whatever's going on), they calculate c correctly all the same, just based on the changed formulae that come from how far into the infinite set of fs you are
This doesn't seem like what VQC had in mind, and I can't tell how this relates to finding n, but if that's what you're saying and it makes mathematical sense then I guess there might be something we can do with it maybe.
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d95295 No.3298
>>3297
Great find I don't remember that crumb.
Yes you can describe a*b=c with an infinite number of Ds and each D has a different f.
It may or may not help finding solution. I'm confident that as I learn more about the grid eventually the solution will appear.
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ff4dbe No.3299
>>3298
It seems like for each extension of these variables you'd need a separate dimension of the grid, though, right? Otherwise you'd have regular ds in the same side (either negative or non-negative) of the grid as any of the infinite other ds, and I can imagine a lot of them would overlap with other cells.
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d95295 No.3300
Anyone looked at the (f,1) records? They aren't just mirrors and they don't have negatives except for e. I have a formula but it uses nested if statements. Got a DM today referring to the (f,1) records. I've been playng with them for a few days.
f = 2d+1-e
When e = -f, a[t] = d[t]-d
Look for pattern of a[t] and n, and d[t]-d and (n-1)
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ab8fb8 No.3301
>>3297
>There are multiple patterns that would solve it.
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ff4dbe No.3302
>>3301
Firstly, what? Secondly, who are you quoting?
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ab8fb8 No.3303
>>3302
Chris. There are multiple patterns in the VQC that would solve this.
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ff4dbe No.3304
>>3303
Oh okay. I thought you were just using greentext wrong.
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6a1cc2 No.3305
>>3296
Thank you for clarifying.
>>3300
I looked into f records previously. But not with regards to the d[t]-d pattern. Will revisit.
Regarding the (-e,0) discussion earlier, attached pic are various test cases showing c and prime solution and their corresponding (-e,0) records.
In some cases, there is a defined relationship between the 2.
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d95295 No.3306
>>3299
I've been talking about negative e's. If you draw a record where e=-(x+n)^2 you will understand what I'm talking about. d=d+n longside. x=x+n shortside. a and b remain constant. To draw both the x and y axis contain equal values (squares) for a x d n b. E doesn't appear directly but is derived from other values.
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ab8fb8 No.3307
>>3306
e makes it all possible. It is the bridge from square to semiprime.
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d95295 No.3308
>>3305
The invalid e should be -(x+n)^2 which was 81 or -6561 that will work. I haven't looked at n,0 for (1,c) that a good idea thanks.
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4cf7a6 No.3309
>>3291
>{40:220:21:20:1:481} (40, 220, 11)
Ok, so transforming this to N=0, the new record is:
{-57600:0:241:240:1:481}
using:
new e = 40-2*21*220-220*220
new d = orig d + change in n, or 21+220=241
new x = orig x + change in n, or 20+220=240
So then, relation of:
{-57600:0:241:240:1:481}
to
{-144:0:25:12:13:37}
is… the next MisterE?
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ff4dbe No.3310
I've been looking into this >>3242 >>3246 stuff about finding c^2 in (0, n). I set up some grids and calculated it that way so I didn't have to wait a year for the grid to generate with a really high iMax. The idea with these grids is to see if there's a relationship between each possible c = c^2 record in the 0 column. For c^2, there are 5 cells in the 0 column: (1, c^2), (a, c^2/a), (b, c^2/b), (a^2, b^2) and (c, c). If c is a semiprime it won't be in column 0 unless it's the square of a prime (which we aren't looking for in RSA), but for all we know there might be a mathematical relationship in the c^2 cells that brings us to the answer.
The first image is (in the grid on the right) each possible combination of (a, b) on the x axis for a bunch of different c^2s on the y axis. I'll post the {e, n, d, x, a, b} and t values generated by each of these sets of (a, b) for a given c^2 in the next post.
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ff4dbe No.3311
>>3310
So these are each of the {e, n, d, x, a, b} and t values for a given c^2 based on each possible (a, b). We know c and we're looking for a and b (meaning we could also find n or x). Can we can use (a, b) = either (1, c^2) or (c, c) and mathematically get to one of the cells with one of a and b actually equaling a or b? Can anyone see any relationship?
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ab8fb8 No.3312
Your work wasn't in vain. Found something that might reduce the search space for n.
>>3308
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d95295 No.3313
>>3309
>>3305
I haven't looked at relations for between (n,0) for c,1 and a,b records it looks like the smaller e may be factor of larger (ABS). Question to check is difference between factor when a and b are close versus far apart. Big n small n. If there is a difference that's a big clue. There is a difference between the a of the (f,1) records when a and b are far apart. the opposite of what I would have guessed. So we finally have a way of reducing the time it takes to solve RSA
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d95295 No.3314
>>3312
All ears? eyes?
I don't feel any of this is in vain. Its just a huge puzzle that I keep seeing around the corner then when I get there I find another corner. Eventually there won't be anymore corners
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ab8fb8 No.3315
>>3314
When you keep doing the mirror jumps as you specified you always eventually get a perfect square. I hypothesize (but haven't thoroughly tested it) that it is always this perfect square where you will find the prime factorization of the number in (-e, 0).
It didn't work for any other numbers I tried, but with 145 and 287, the t value the factorization appears in is n-t, where n and t are the values from the 1*c element.
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d95295 No.3316
Just thought i would put up a few (f,1) records
E N D X A B
-48 16 31 20 11 83
-48 1 16 8 8 26
-17 8 34 17 17 67
-17 1 9 5 4 16
-23 77 58 45 13 257
-23 1 6 5 1 13
-84 2 59 16 43 79
-84 1 18 10 8 30
I think the relationship i saw yesterday with a in the f,1 records may have been a figment.
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ab8fb8 No.3317
>>3316
At first it seems pretty useless to need to find the t value where it appears because you've already found the small square when you find the perfect square, but if you could find the t value where the factorization appears maybe you could skip all of that, like how if you find the t value where the factorization appears in (e,n) you don't need to find n because the x value is in (e, 1, t).
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d95295 No.3318
>>3317
For movements in N d=t. Depends what variable is changing what t is
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d95295 No.3319
>>3317
As d increase you are subtracting 2d +1 from e then 2d+3 then 2d+5. I looked at patterns 500 deep. All of a sudden there is a perfect square. The grid is amazing at keeping its secrets.
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e135db No.3320
What if you could switch out e and n? e and n are both just a distance from a square.
n = i - d
n distance from d to root of large square
e = c - d^2
e distance from d^2 to c
>all numbers can be a remainder
(n,1)
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ab8fb8 No.3321
>>3320
That sounds interesting. What's d^2 + n?
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d95295 No.3322
>>3320
Excellent post. I think you may have to take a root of e to make it scalar. E isn't a distance its a 2 dimensional volume. So taking its root would be a distance.
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ab8fb8 No.3323
>>3322
It is a distance. It's the distance from c's adjacent square to c.
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ab8fb8 No.3324
I made a program that carries out this factorization.
It appears there is a straight path from semiprime to prime factorization. We just have to create a wormhole and go from start to end. :^)
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d95295 No.3325
>>3323
e=2na-xx both terms are two dimensional.
Also e= a*b -dd both terms again two dimensional.
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d95295 No.3326
>>3320
The idea of flipping e and n is excellent
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7e0ce0 No.3327
>>3278
Check this out guys I think this is it
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ab8fb8 No.3328
If you create a cell for 1*c
and then visit (t, N, t)
you find c
sometimes it's in (t, N-1, t) or (t, N-1, t+1), I'm trying to figure out the exact rule
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ab8fb8 No.3329
>>3328
{6:60:13:12:1:145} (6, 60, 7)
{8:127:17:16:1:287} (8, 127, 9)
{11:219:22:21:1:481} (11, 219, 11)
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ab8fb8 No.3330
>>3327
If that works, I'll program it and see how fast it is.
>>3325
I'm kind of lost now, but at least I have another factorization program, which is faster than -x jump bruteforce I made
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4d8060 No.3331
Look at how primitive Fermat's factorization is compared to VQC. Look at what an uncharted territory we're in. LOOK AT HOW CLOSE TO WINNING WE ARE!
>(0,n) one row to rule them all
>x marks the spot
>the square in 0,1 is the basis for all patterns in n=1
If the offset is discovered by subtracting d of d[t], why isn't anything subtracted from a[t]?
Reread Chris crumbs!
Non mathematical questions;
Why are people like you more qualified to solve this problem than mathematicians?
Why did mathematicians need to invent quantum computers just to find that which a semiprime divides into?
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4d8060 No.3332
Why do people like Q and Chris ask a bunch of questions and make you do the work?
Are you not happy to be pioneering a new mathematical device? Would you have learned anything if the solution was dumped onto /cbts/ with no challenge? Would the world be a better place? Would you Have a Message to Spread that Everyone can be a Genius?
One day you may not have any Hints.
One day the world may depend on YOU.
You're learning to think outside of the box. And per Q, you're learning to think.
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4d8060 No.3333
>>3332
>all for a LARP right?
Q—
Is elevating somebody curious enough to click on a typical larp thread to that of warrior for truth and justice not the most kindest act in written history?
VQC—
Is elevating somebody curious enough to click on a larp within a larp to somebody destined to upheave 50 years of number theory and cryptography not a kind act?
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4d8060 No.3334
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ff4dbe No.3335
>>3331
>Why are people like you more qualified to solve this problem than mathematicians?
We aren't all autistic by the way.
>(0,n) one row to rule them all
>x marks the spot
>the square in 0,1 is the basis for all patterns in n=1
So you're saying the solution has something to do with these posts?
>>3310
>>3311
Or are you saying something else (because nobody replied to those posts with anything)?
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ff4dbe No.3336
I'm looking at the 0 column and the relationship between c and c^2. I found a few weird things. For the square of a semiprime (probably every square but I'm only paying attention to that table of squared semiprimes), for (a, b) = (1, c^2), which will always be the first cell of the infinite set, t increases by the same number throughout the infinite set (e.g. 11, 21, 31, 41, 51, always +10, but that 10 changes set to set), a is the list of squares from 1 to infinity (the same 1, 4, 9, 16 each time in every set), b increases by a linearly increasing amount (e.g. 441, 484, 529, 576, 625, so it's +43, +45, +47, +49 etc), d also increases by a linearly increasing amount (e.g. 21, 44, 69, 96, 125, so it's +23, +25, +27, +29 etc), and x is linear with a gradient somehow related to t (e.g. 20, 40, 60, 80, 100, which is a gradient of t's increase * 2). Obviously e and n stay the same (0, 200) as you'll see in that table. The example I used for those numbers was c^2 = 441, c = 21.
Here's another example with c^2 = 225, c = 15. (e, n) = (0, 98). t is 8, 15, 22, 29 (+7 each time), a is 1, 4, 9, 16 etc like before, b is 225, 256, 289, 324, 361 (+31, +33, +35, +37), d is 15, 32, 51, 72, 95 (+17, +19, +21, +23), and x is 14, 28, 42, 56 etc (linear with a gradient of t's increase * 1 instead of *2 for some reason).
What that means is, if we use c^2 as our c, every variable follows a predictible pattern, although I don't know why each of the numbers starts where they are. Maybe it has something in common with (1, c). I looked at the (1, c) cells for c = 21 and c = 15 but the relationships are pretty weird and I can't figure out how they work or how they relate to their corresponding c^2 cell, but there's bound to be at least one if not several relationships there. It might just be how late early it is right now but I really hope someone else looks at this and finds something, because it definitely seems like an important relationship and Chris seems to have alluded to it.
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4cf7a6 No.3337
>>3299
>…for each extension of these variables you'd need a separate dimension of the grid, though,…
Have been seeing this as well. Appears the additional 'dimensions' are related to derivatives (rates of change, how the variables GROW), or second derivatives, sometimes interleaved. These are associated with the integer 'points' in the grid.
Gut and some research seems to be pointing toward a Clifford Algebra associated or overlayed on the grid. When we have those vectors for two points of interest, some inner product or other transformation would lead to a solution. Have been interested in learning about Clifford Algebra for a while as a foray into next level math understanding. Also, there were some suggestions earlier by Chris to dig into in the meantime, and the digging keeps turning up Clifford Algebra (will have to sauce that later).
Baker, we're over 700 posts. Can we consider including Cliff in the next bread, to complement Fermat in RSA#8? Some interesting stories, twists and turns in mathematics / physics, resurgence in his work decades later, and likely has a bearing on where we're headed. Touches on roots, modular number systems, quaternions, etc. It gets at the geometrization
of the real number system. Plus he wore an epic beard.
Some quotes from William Kingdon Clifford (they read like an anon credo!):
https:/ /libquotes.com/william-kingdon-clifford
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4cf7a6 No.3338
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12909e No.3339
>>3330
def factor(number):
C = number
A = 1
B = C
D = int(math.sqrt(C))
#print(D)
E = C - D*D
X = D - A
N = int((X*X+E)//2)
newX = E%2
newD = int((newX*(newX+2)+E)/2)
newA = newD - newX
while(gcd(newD-D,newA) == 1):
newX += 2
newD = int((newX*(newX+2)+E)/2)
newA = newD - newX
newA = D - newX
return newA
Here is my code for this. I've noticed that if you set D=1, then returned value is the negative factor of C (so with real D you get factor a, but with D=1 you get factor -a), so you don't even really need to calculate D, but it's here anyway. I can't factor RSA because (int((newX*(newX+2)+E)/2)) is too large for a float and it won't divide but I'm very confident that this works. I've tested it for some products of 2 primes (like 4 digits) and it has worked every time.
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12909e No.3340
>>3339
Here is output that semi-confirms it
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12909e No.3341
>>3340
>>3339
Also this was me I forgot my trip.
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5167b5 No.3342
Has anyone noticed this before?
(0, 5). Divide the a's by 5 and you'll get (1, 1) d's.
{0:5:20:10:10:40}
{0:5:60:20:40:90}
{0:5:120:30:90:160}
{0:5:200:40:160:250}
{0:5:300:50:250:360}
{0:5:420:60:360:490}
>>> 10/5
2.0
>>> 40/5
8.0
>>> 90/5
18.0
>>> 160/5
32.0
>>> 250/5
50.0
>>> 360/5
72.0
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5167b5 No.3343
>>3342
And the a's for (0, 4) is also equal to the d's of (1, 1)
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4cf7a6 No.3344
>>3288
>>3282
Thanks anon, this was an interesting dive, only part way through understanding (haven't dug into patterns at n=1).
>>3290
>But how is this a solution when the a and b values are still unknown?
>What am I missing?
Don't think you're missing anything, and not sure yet either, but useful for a quick jump to n=0 or n=1.
It also produces a new e that is a perfect square. The new t is always an integer that is the square root of the negative of the new e. This is expected because we know:
e=2an-x^2, and if n=0, e=-x^2
I think being able to transform to a perfect square column in the -e space may be useful once everything is worked out. There is a subsequent transform of 'a' needed, as illustrated below.
Attached images relevant.
First is the orig VQC, generated by Teach back in Dec and on pastebin. Showing only the -e perfect square columns, for n=0 and n=1. The gray bar is the first record, passing through {0:0:1:0:1:1}. Added x's starting at e=-9 above bar, as a sequence for 'a' could be extended upward with decreasing a values (moving t through zero to negative?).
Note: at n=0: d, a, and b increment by 1 for each t. x never changes. x^2=-e. There is a gap between a and b, which is 2*x, and this never changes for t sequences. Also, d-x=a for any n=0.
Note: observe how d increases as we move leftward decreasing e (each +1 increase in x). It's not linear.
seq:1,2,3,5,9,13,19,25,33
See: https:/ /oeis.org/A085913
A085913: Group the natural numbers such that the product of the terms of the n-th group is divisible by n!. (1),(2),(3,4),(5,6,7,8),(9,10,11,12),(13,14,15,16,17,18),(19,20,21,22,23,24),… Sequence contains the first term of every group.
1, 2, 3, 5, 9, 13, 19, 25, 33, 41, 51, 61, 73, 85, 99, 113, 129, 145, 163, 181, 201, 221, 243, 265, 289, 313, 339, 365, 393, 421, 451, 481, 513, 545, 579, 613, 649, 685, 723, 761, 801, 841, 883, 925, 969, 1013, 1059, 1105, 1153, 1201, 1251, 1301, 1353, 1405
There is a similar (same initially) sequence
See: https:/ /oeis.org/A099392
A099392: a(n) = floor((n^2-2*n+3)/2).
1, 1, 3, 5, 9, 13, 19, 25, 33, 41, 51, 61, 73, 85, 99, 113, 129, 145, 163, 181, 201, 221, 243, 265, 289, 313, 339, 365, 393, 421, 451, 481, 513, 545, 579, 613, 649, 685, 723, 761, 801, 841, 883, 925, 969, 1013, 1059, 1105, 1153, 1201, 1251, 1301, 1353, 1405
The second formula was easier in excel, but it breaks down at x=54. The d, a, b, and c get out of sync, and no longer check out (see highlighted region in screen cap). Maybe that's why the oeis.org sequence ends there? Maybe the A085913 sequence based on the factorial and groups would continue to work? Not sure, but this doesn't scale at this point for d, using this quick excel model.
- also highlighted in that image is e=-1849, the transform for 87 (1x87) to n=0: {-1849:0:44:43:1:87}, if anyone wants to find a link to the 87 (3x29) record transformed to n=0: {-169:0:16:13:3:29}
Ok, here are transformation examples with a few of the 'known' factors we've been using. Doesn't apply to getting from the 1*c to desired a*b factors yet, but interesting patterns. Refer to image with A085913 in top left, starting at e=0.
Take 85 (5x17)
{4:2:9:4:5:17}
transformed to n=0 is:
{-36:0:11:6:5:17}
Calc d for t=1 at x=6 is 19 (highlighted tan), and we want d=11, so transform 8 steps. a and b are 13 and 25, so taking away 8 from each is 5 and 17, 5x17=85, check.
87 (3x29)
{6:7:9:6:3:29}
to
{-169:0:16:13:3:29}
Calc d for t=1 at x=13 is 85 (highlighted tan), and we want d=16, so transform 69 steps. a and b are 72 and 98, so taking away 69 from each is 3 and 29, 3x29=87, check.
145 (5x29)
{1:5:12:7:5:29}
to
{-144:0:17:12:5:29}
Calc d for t=1 at d=12 is 73 (highlighted tan), and we want d=17, so transform 56 steps. a and b are 61 and 85, so taking away 56 from each is 5 and 29, 5x29=145, check.
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4cf7a6 No.3345
>>3339
>>3340
>>3341
Looks very interesting CA! Will have to dig into that when some time frees up, maybe you've got this!!!
>>3342
>>3343
Have been enjoying the patterns you point out Isee. Those are indeed interesting.
- also, some of the patterns PMA has pointed out with delta d's, and interleaved second differentials I see as well.
- for example, (1:5) attached pic.
- I'll throw a post together with some other patterns done visually.
Also, check out the interesting patterns when visualizing Fermat's last theorem, screen capped from a tube. I think this is what we have going on in a way.
>>3331
>… LOOK AT HOW CLOSE TO WINNING WE ARE!
Ok, but how do you KNOW we are close to winning?? Anything more to share?
> Reread Chris crumbs!
Yes, still working through them, picked this up a couple weeks ago to actually work through stuff (where Teach and all were at mid Dec). Each crumb leads to a deeper understanding of the VQC/GRID. One of the things that keeps me going on this, LARP or not, is that Chris has always been coherent in his crumbs. Granted, we could start dropping some crumbs with the grid now, even without an end solution, but it sides more with an 'informed faith' rather than a 'blind faith' at this point.
- we just need to convince Chris that a) we're ready for the next walk-through tutorial session, and b) it's Time for that Now to get ready for next phase of bigger picture! (hint hint Chris).
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4cf7a6 No.3346
Ok, here are a few playful images / exploration of patterns. Follows along with the fermat visualization above.
First, I took the orig grid output, e-64 to e64, t=1 to 10, n=0 to 63.
Zoomed out, you can see patterns right away.
Added darker shading to the cells with values. Some ranges only have a few t's, as the d, a or b capped out and stopped generation, but they would go on.
Wouldn't quite print right, so generated a png output for the left side and right side of the grid.
Could play in an image program, but chose to print these out, and play on a light table.
I see 2 parts to our goal: 1) be able to hop/transform to another cell that is valid (not empty), and b) able to go in proper direction(s) and transform appropriately along the way. Ala hopping the lilypads on the grid-pond.
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12909e No.3347
>>3346
I'm pretty sure these can be explained by looking at the D and X grids.
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769a5e No.3348
>>3300
Looked a bit further into (f,1) records, and analyzing the d[t] and a[t] differences.
See pics related for c=145, showing diffs at (e,1), (f,1) and (0,1).
In (f,1), there are more occurrences of the 5 factor in d[t] - d. But unfortunately, that pattern doesn't extend to other test cases. For c=6107, no records appear.
Similarly in (0,1) we get some matches at d[t]-d, but none at a[t].
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4cf7a6 No.3349
>>3347
That's cool CA, love your images!
Yes, that's exactly the pattern.
Also trying to look at the shifting / reflecting / twisting perspective.
Attached are the first set, looking at +e. Printed 2 copies and overlayed on light table. Then shifted. The darker cells are the ones where you would be able to jump and land on a cell (valid transform or not).
The perspective view makes the angled lines jump out. The strongest at n=0, e=0, which is probably the Root O D. These make sense with the math as well, including the slope (e can increase by 2 for each increase in one unit).
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4cf7a6 No.3350
>>3349
Ok, to finish off these playful patterns.
Looking at just the Negative e and shifting. These show the strongest lines, with a primary line slope of 2 running away from each perfect square at the n=0 origin. Other lines run different direction, opposite slope, but not 2.
Next reflecting along hortizontal axis (along an n) with the -e space. What is interesting about this, is that each perfect square e column has no gaps, meaning, you could jump from n to -n, back and forth. I think this would mean move along a perfect square column -e toward n=0, but, if you run into a gap, then skip over to the other side (plus n), and carry on toward n=0 until a gap, then jump back, and so on.
Finally a little rotation for fun - the VQC Vortex.
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5167b5 No.3351
Do we have a collection of Chris' crumbs?
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4cf7a6 No.3352
>>3348 Nice!
Interesting, the t diff binary pattern really jumps out visually.
Also reminds me of some patterns from a while back, think it was you?
Just that the a for an n, plus x, plus a for n+1 is the new a (at n+1), the a for n+1 is b for n. This looping pattern that cascades down.
Another pattern is the final digits for a, b, and c. For (2,1), c always ends in a 3, 7 or 1 digit.
For (1,1), it's always a 5 or a 1 for c. Similar patterns in a and b.
Reminds me to looking into modular forms for our a's, b's and c's for patterns and what they say about the rates of change, or about the factors / relationships.
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80a6b3 No.3353
>>3339
Just use the // operator for integer division instead and you won't be limited by then size of floats. Can even get rid ot the explicit int() conversions.
(newX*(newX+2)+E)//2
Alternatively since you're always dividing by two, its much faster to just shift right by one bit than performing an actual division
(newX*(newX+2)+E)>>1
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d7e5a7 No.3355
>>3353
Yeah I changed it to make it better. Now
newX = E%2
if(newX == 1):
dShift = 6
else:
dShift = 4
newD = int((newX*(newX+2)+E)//2)
newA = newD - newX
while(gcd(newD-D,newA) == 1):
newX += 2
newD += dShift
dShift += 4
newA = newD - newX
newA = D - newX
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d7e5a7 No.3356
>>3355
Also you can use any number for the D value. Also the D values are the same as the A values for the (e-1,1) cell.
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4cf7a6 No.3357
>>3356
> Also the D values are the same as the A values for the (e-1,1) cell.
-> Be careful, looks as if that only holds for initial e that are even. The D is same as the (e-1, 1, t+1) A if starting e is odd.
But, in the orig VQC output, this isn't always consistent in the negative e space, there are a few odds that are the same. Might be an artifact of where t=1 started and incorrect grid production (see attached image, e=-39 to e=-40 isn't same pattern as others).
And by extension, the D values are the same as the B values for the (e-1, 1, t-1) cell if e is even, and the same for the (e-1, 1, t) cell if e is odd.
Interestingly, this pattern holds in the negative e space for n=0, if jumping from perfect square to square (-9 to -16, -36 to -49), as shown in first image in previous post here >>3344. Need to validate further, but looks consistent.
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ff4dbe No.3358
>>3351
See >>1721
Hopefully you trust the people here enough to open a pdf. I made it. I couldn't think of a better filetype that could be uploaded to 8ch.
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ab40e5 No.3359
VQC, if you're reading, can I show this to my Number Theory professor? I don't know if I can trust him because he does work a liberal university, but it would be nice to get another set of eyes on it. Also is it dangerous to go public with this?
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ab8fb8 No.3360
>>3359
Hardly. It's just another conspiracy theory
.
.
.
Until it works.
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ab8fb8 No.3364
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ab8fb8 No.3365
Discovered something neat.
There's multiple solution records in (e,1), and there's more than one n that solves the number.
145
{1:61:12:11:1:145} (1, 61, 6)
{1:1:32:7:25:41} (1, 1, 4)
{1:5:12:7:5:29} (1, 5, 4)
{1:1:882:41:841:925} (1, 1, 21)
{1:-29:12:41:-29:-5} (1, -29, 21)
123
{2:51:11:10:1:123} (2, 51, 6)
{2:1:41:8:33:51} (2, 1, 5)
{2:11:11:8:3:41} (2, 11, 5)
{2:1:113:14:99:129} (2, 1, 8)
{2:-33:11:52:-41:-3} (2, -33, 27) [not sure why t changes here]
95
{14:39:9:8:1:95} (14, 39, 5)
{14:1:19:4:15:25} (14, 1, 3)
{14:3:9:4:5:19} (14, 3, 3)
{14:1:427:28:399:457} (14, 1, 15)
{14:-21:9:28:-19:-5} (14, -21, 15)
77
{13:31:8:7:1:77} (13, 31, 4)
{13:1:8:1:7:11} (13, 1, 1)
{13:1:8:1:7:11} (13, 1, 1)
{13:1:206:19:187:227} (13, 1, 10)
{13:-17:8:19:-11:-7} (13, -17, 10)
287
{31:128:16:15:1:287} (31, 128, 8)
{31:1:65:9:56:76} (31, 1, 5)
{31:8:16:9:7:41} (31, 8, 5)
{31:1:1697:57:1640:1756} (31, 1, 29)
{31:-40:16:57:-41:-7} (31, -40, 29)
7783
{39:3804:88:87:1:7783} (39, 3804, 44)
{39:1:36469:269:36200:36740} (39, 1, 135)
{39:-200:88:269:-181:-43} (39, -200, 135)
{39:1:1077:45:1032:1124} (39, 1, 23)
{39:24:88:45:43:181} (39, 24, 23)
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1b5a8c No.3366
>>3360
Obviously this is no indication of the speed of the algorithm, (I still can't get it for RSA) but it shows that it definitely works.
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ab8fb8 No.3367
>>3366
That the gcd factor algorithm? Upload the code so I can include it
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c5cd6b No.3368
>>3367
It's built into python. I'm assuming it's using Euclid's algorithm
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ab8fb8 No.3369
>>3368
Yeah, but your algorithm. I converted it to Java but made an error somewhere and it didn't work. So it'd be helpful if I could see the full code.
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c5cd6b No.3370
>>3369
def GCD(a, b):
if b == 0:
return a
else:
return GCD(b, a % b)
I think everything else is on here
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52c67f No.3371
Wow, Anons! I'm reading and absorbing all your posts. Excellent thoughts. Still can't solve for n, but we keep growing in understanding of the Grid. Also, the group we have here is excellent. Glad to be here, looking forward to contributing more when I have something to share. Deus Vult. Cras Es Noster. Carpe Diem. Let's keep going!!! Everyone here is doing their part. Feels good man!!
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52c67f No.3372
Reddit meme, lol! Just noticed. have a large stockpile. Here's a better one.
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4cf7a6 No.3373
>>3364
Thanks Baker, new bread looks superb, looking forward to the first slice.
>>3358
This is the pdf I use first for crumbs, nice compilation, have on phone. Probably everything needed is in there to unlock this with the (n)KEY.
Have been catching up on the crumbs, starting a couple weeks back, still a month behind. I need to dive into p next.
>>3371
>>3372
Hey VA, good to read you! Maybe a nice UK graphic will pull V in here.
>>3347
CA, I recall your other D grid here >>2890 that was interesting.
>>3233
>Regarding interwoven / interleaved patterns, you'll see theres 4 or 5 I think in (0, 8). 2 isn't a magic number, I suspect that you can find infinite interwoven patterns in any (e, n) if you extend the grid into infinity.
>Maybe there is a limit, but I haven't spent any time looking into it…
Hey Isee, seem to recall Chris specifically saying there are a limited number of patterns, and we should enumerate them (how they GROW). Can't find the crumb right now, but think this is critical. I see a 'pattern' as a 1st, 2nd… order rate of growth (derivatives, you know we're going to end up doing some sort of geometric calculus anons!) for each of our variables, and may include interleaved patterns. Believe we need to be able to extract/recognize/model those patterns from the relevant n=1 row elements, and apply the pattern to the relevant base values to get the jump(s)/shift(s) we need. We will benefit from using an appropriate CHOSEN prime number, multiplied by our c, to help/enable the transformation / ID of factors.
>>3159
>>3171
>>3172
Noted your D-sequence is the same I listed earlier here >>3344
Formula is: a(n) = floor((n^2-2*n+3)/2)
in Excel: =FLOOR.MATH(((E10+1)^2-2*(E10+1)+3)/2)
where E is an 'x' value, these increment by one. But as I mention, appears to break down later after the 53rd x. Perhaps the other method would work, or if you could share more details on how you're generating that would be appreciated.
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4cf7a6 No.3374
Last bit for today, is a look at n=0, and entering c-values into excel (doing manually still).
This gives a visual for how the "e space" between D values grows over time (+2 for each incremental D).
First column is the D value, sequence of integers. Second column indicates if D is prime (have a lookup table with first 100K primes that goes up to 1.2M). Then D^2, and then diff between squares of D floor and D ceiling. For each increase in D, +2 is added to the e value range.
So for our 145 example, between d of 12 and 13, the gap is 25 (max e, think one less than that actually, need to ignore perfect squares on either side where e=0), and actual e is 24 to reach 169. The 145 input, gap, and e are in the top row.
Another example shown, c=12,657,734,632, e=90,417. The e range up here is 225,013. Still small compared to RSA, but enough to play once all is figured out.
>>3078
Hey VA, if you want to input some numbers to hit that home run, here they are (from Chris), it would break my excel model though!
public static string Rsa100c = "15226050279225333605356183781326374297180681149613806886579084945801229632589528976540003506920061 39"; public static string Rsa100a = "37975227936943673922808872755445627854565536638199"; public static string Rsa100b = "40094690950920881030683735292761468389214899724061"; public static string Rsa100d = "39020571855401265512289573339484371018905006900194"; public static string Rsa100e = "61218444075812733697456051513875809617598014768503"; public static string Rsa100f = "16822699634989797327123095165092932420211999031886";//2d+1-e public static string Rsa100n = "14387588531011964456730684619177102985211280936"; public static string Rsa100x = "1045343918457591589480700584038743164339470261995"; public static string Rsa100x_plus_n = "1059731506988603553937431268657920267324681542931";
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c38ca6 No.3375
>>3374
>>3373
Mr E. My ExcelFag! Love you man. Those digits will break my Excel model too.
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ab8fb8 No.3376
>>3375
One of the things I tried was factorization of (e, n, t+na) instead of c itself because I thought that the smallest a for the c of t+na would also be a for c, however I discovered that this 145 is special, and that doesn't work for all numbers.
(t+na c for rsa617 has 17 as a, which obviously isn't a factor of rsa617.)
Perhaps there is a hidden pattern there. It is very easy to factorize the c from t+na, even for gigantic numbers, but it doesn't give you the right factorization.
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ab8fb8 No.3377
>>3376
Example:
t+na c for 145 = 145*533
the factorization of 145*533 (77285) = 5*15457
5 is a factor of 145, so we factored 145. But this doesn't work for the other numbers. Obviously c*any number is going to have the factors of c in it, but the smallest a isn't always the right a.
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ab8fb8 No.3378
But at least I know this is the right way to go, because I'm starting to see why knowing the pattern of a[t] and d - d[t] is useful - it gives you a way to go through e,1 faster, or without searching at all
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ab8fb8 No.3379
The bruteforce algorithm is the simplest fastest method of factorization that we've come up with so far. There is always one path to the factors of a semiprime - down.
So all it does is decrement t in (e,1) starting 1*c's t until it reaches the right x. There are only 2 steps to the factors of 145 this way.
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ff4dbe No.3380
>>3370
Are you sure it's a good idea to use recursion? If we're dealing with huge numbers, chances are you'll end up with stack overflow. We already had that problem with another method we tried.
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fbb317 No.3384
>>3380
def GCD(a,b):
while(b!=0):
temp = a
a = b
b = temp%a
return a
Would this work?
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ab8fb8 No.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?"
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5167b5 No.3386
>>3385
Is this something VQC has said?
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738c11 No.3387
>>3385
The first decision is hardly a decision. Anyone knows that if e is zero then you have it factored, so why include this. This is included because he wants to make the 2nd decision about division by 2. Remainder is odd or even is also the same as asking what is the remainder upon division by 2. Then I'd say the third step would have to do something with the remainder upon division by 3.
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ff4dbe No.3388
>>3385
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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ff4dbe No.3389
Last post so I can sticky the new one already
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