[m4xr3sdEfault]*******,=,e \_ヾ(ᐖ◞ ) ID: fd3baa RSA #16 - "Larp Inception" edition April 25, 2019, 12:08 a.m. No.9114   🗄️.is 🔗kun   >>0067

Virtual Quantum Computer

 

The virtual quantum computer (VQC) is a grid made of constructably infinite elements that follow a known pattern.

 

The grid in its entirety serves as the superposition. The input parameters that collapse the superposition are d and e, which are trivial to calculate for all c that is the difference of two squares. When the integers that are the difference of two squares are arranged into the grid and their corresponding properties are shown, a pattern emerges that shows a path to calculatethe factors of c instead of searching for them.

 

The grid is indexed using e, n, and t, where e is the column, n is the row, and t is the specific element in the cell.

 

Glossary

Look-up

A pattern used to calculate the factors of c, like a value look-up table.

 

Column

All cells for a given e

 

Row

All cells for a given n

 

Entry; record; element

A set of variables corresponding to a factorization for a given c. The legend to read entries is {e:n:d:x:a:b} (e, n, t) = c

Example: {1:5:12:7:5:29} (1, 5, 4) = 145

 

ab record; nontrivial factorization; prime element

The element that contains the factorization of c that is not 1*c, hence, nontrivial.

 

1c record; trivial factorization

The element generated from setting a=1 and b=c

 

Mirror element

The element in -f corresponding to an element in e, in the context of a given c.

 

Cell

All entries for a given e,n (not to be confused with an entry itself.)

 

Sieve

A sieve is an algorithm for factoring integers arising out of number theory in the 1900s, most notably from Carl Pomerance.

 

Smoothness

A number is described as smooth if it is composed of small prime factors, opposed to large ones.

 

Remainder Tree

The remainder tree is a structure created by recursively taking d and e starting with c, creating a tree with several to many branches.

 

Functions

 

na transform

A movement from a record in (e, n) into (e,1) where n becomes 1 and a becomes a times the n of the (e,n) record. It has also been used to refer to moving n*a records down in a cell.

 

Pell(n)

The nth Pell number function. Can be calculated recursively or formulaically using the square root of 2.

 

ST(n)

The nth square triangular number. A square triangular number is a number that is both a square and a triangle number. The Pell function is used to calculate square triangular numbers.

 

T(n)

The triangle number function.

Example: T(7) = the 7th triangle number

 

T-1(n), inverse T

The inverse triangle number function.

Example: T(7th triangle number) = 7

 

Variables

a and b are, to reiterate, the factors of c. a is the smaller factor of c, and b is the larger one.

d is the integer square root of c.

e is the remainder of taking the integer square root of c. Unless c is a perfect square, a remainder will be left over.

i is the root of the large square. It is equal to (d+n).

j is the root of the small square. it is equal to (x+n).

n is what you add to d to be exactly halfway between a and b, and it is the root of the large square, so it takes you from d to the large square.

x is what you add to a to make d. When added to n it makes the root of the small square.

f is what you add to c to make a square. (e is what you subtract from c to make the square below it, f adds to make the square above c.)

t is the third coordinate in the VQC, it is a function of x.

q is a product created by multiplying successive primes until the product is above d.

u is the triangle base of (x+n)^2. 8 times the triangle number of u plus one is (x+n)^2 for c with odd x+n.

h was a variable used to quantify families of numbers. The way to calculate it is currently unknown.

 

When capitalized versions of the variables are used in comparison to lowercase versions, the capitalized versions refer to the variable's value for the trivial record, and the lowercase variables refer to the values for the nontrivial record. Sometimes these trivial uppercase variables are referred to with "Big" preceding the letter.

{e:N:d:X:A:B} (e, N, T) is the trivial element.

{e:n:d:x:a:b} (e, n, t) in this context is the nontrivial element, the prime factorization of c.

[m4xr3sdEfault]*******,=,e \_ヾ(ᐖ◞ ) ID: fd3baa April 25, 2019, 12:13 a.m. No.9115   🗄️.is 🔗kun

Rules

Each cell of the grid (e,n) has infinite 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 allows you to traverse these infinite elements.

If a grid cell has elements, all elements are constructable from a finite set of root elements.

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. In other words, all numbers that are {0, 1, or 3} mod 4 are the difference of two squares.

 

(1, 1)

The cell (1, 1) contains as values for a and b the values of two consecutive squares added together.

The values of a and b in it are related to the length of the longest side in right angled triangles.

They are also also related to the Pell function.

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.

 

c

For a c at (e,n), there exists (-f, n-1). The difference between e and -f is 2d+1, making columns e and -f unique as a pair to c.

 

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.

 

Row 1

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).

If s is a factor of a[t], then (e, s) exists, meaning all divisors of a[t] in row one exist in the column as a row.

na and nb for any c can be found n places apart in the cell at (e,1).

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.

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.

 

For more of these rules, see the grid patterns thread.

 

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 = ((a + b) / 2) - d

d + n = i

x = d - a

x = (floor_sqrt(( (d+n)*(d+n) - c))) - n

x + n = j

j^2 = 8*T(u) + 1

f = e - 2d + 1

u = (x+n) / 2

 

if (e is even) t = (x + 2) / 2

if (e is odd) t = (x + 1) / 2

[m4xr3sdEfault]*******,=,e \_ヾ(ᐖ◞ ) ID: fd3baa April 25, 2019, 12:21 a.m. No.9116   🗄️.is 🔗kun

Code

 

C#

BigInteger libraries and test code —— pastebin.com/fiKJ6nLv

Recursive remainder tree generator —— pastebin.com/ZH9fSWu2

VQC generator —— pastebin.com/XFtcAcrz

VQC generator w/ Bitmap —— pastebin.com/hMTtJF6E

 

Java

Real-time VQC —— anonfile.com/TeH6q3d8bd/VQCGUI_v2.7z

Recursive remainder tree generator —— ghostbin.com/paste/njfcq

VQC generator —— pastebin.com/Dgu9aP1h

VQC library —— ghostbin.com/paste/kbf9a

 

Python

VQC generator —— pastebin.com/NZkjtnZL

VQC generator w/ bitmap —— pastebin.com/wEAKaqBp

 

Factorization algorithms —— ghostbin.com/paste/cyjop

Static Java/C# class with all RSA numbers —— pastebin.com/XYFpsDWE

Miscellaneous code —— ghostbin.com/paste/xrqme

VQC codebase archive (not comprehensive yet) —— anonfile.com/L3yd6di0n6/archive_7z

 

Other Threads

 

Fermat's Last Theorem —— archive.fo/RNRgl

Grid Patterns —— archive.fo/YmyoR

 

RSA #0 —— archive.fo/XmD7P

RSA #1 —— archive.fo/RgVko

RSA #2 —— archive.fo/fyzAu

RSA #3 —— archive.fo/uEgOb

RSA #4 —— archive.fo/eihrQ

RSA #5 —— archive.fo/Lr9fP

RSA #6 —— archive.fo/ykKYN

RSA #7 —— archive.fo/v3aKD

RSA #8 —— archive.fo/geYFp

RSA #9 —— archive.fo/jog81

RSA #10 —— archive.fo/xYpoQ

RSA #11 —— archive.fo/ccZXU

RSA #12 —— archive.fo/VqFge

RSA #13 —— archive.fo/Fblcs

RSA #14 —— archive.fo/HfxnM

RSA #15 —— archive.fo/ZxHdb

 

Every VQC map —— anonfile.com/a64765i3n9/maps_7z

VQC 4chan posts —— ghostbin.com/paste/szbfc

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 456f05 {legos.xyz}Spawnofebot.::・┻┻☆()゚O゚) June 30, 2019, 11:02 p.m. No.9418   🗄️.is 🔗kun

16 is the base of the hexadecimal number system, which is used extensively in computer science.

 

Let's do this.

AA !LF1mmWigHQ ID: cc2da1 July 1, 2019, 1:06 a.m. No.9419   🗄️.is 🔗kun

Looking into this post from last thread: >>9395

>If we were to look at an element that is d * d, a + b for that element would be (d + d), or 2d

We know from someone's Twitter dms with Chris quite a long time ago now that the fabled "root of d" that Chris has cryptically mentioned before is an element in (0,n).

>Root of d = {0, 2xd, 3xd, 2xd, d, 9xd}. All a and b of ( 0, 2xd, t) are multiple of d for all t.

>Yes and that pattern can be used elsewhere.

So if 2d is a representation of a+b for dd squares, and it has a directly calculable element related to it in (0,n), maybe there's something interesting to be found looking at (0,a+b).

 

What I've found looking into (0,a+b) is interesting I guess, but a lot of it is just basic arithmetic if you think about it. Using the same arithmetic from the root of d element, you'll find (most) (0,a+b) cells have an element at x=2(a+b). Some of them don't. These appear to be where n is twice a square (this sequence follows a different pattern which I'm pretty sure I went over in grid patterns). I remember there being another sequence of n values that didn't follow the same patterns in column 0 but I don't remember anything more specific since it's been a while since I did any actual work on this.

 

For the cells that do show up in (0,a+b) where x=2(a+b), the cell follows the same pattern as the root of d element except with i instead of d (since a+b=2i). Also, given i=d+n, the difference between each variable in both elements is the same multiplier by n. So, as follows:

<Root of d = {0, 2d, 3d, 2d, d, 9d}

<Root of i = {0, 2i, 3i, 2i, i, 9i}

<Difference = {0, 2n, 3n, 2n, n, 9n}

 

I didn’t really find anything all that interesting other than that (at least specifically about root of a+b). I'll make another post when I'm done putting code together in relation to the other posts about the link between the elements d is between in (e,1) and the cells in (0,n) that factorize dd.

AA !LF1mmWigHQ ID: 774bd8 July 1, 2019, 3:32 a.m. No.9420   🗄️.is 🔗kun

Based on the posts from the person at the end of the last thread saying they would tell us how to use the grid, and what they said about there being a link between the elements in (e,1)/(f,1) where d is between d[t], the solution cells for dd in (0,n), and something about (0,e) and (0,f). I've put some code together to analyze. It doesn't always work, so I'll probably need to go over it before putting it here, but here are two examples of it working. Hopefully I got the negative n stuff right for the (0,f) cells. Does anyone see any links anywhere? Or maybe if that person comes back and continues, you could possibly clarify?

PMA !!y5/EVb5KZI ID: 9d9b7b July 7, 2019, 8:19 p.m. No.9429   🗄️.is 🔗kun   >>9430 >>9435

>>9427

When a <-n, d <-> x+n.

 

Example:

 

(23,36,24) = {23:36:78:47:31:197} = 6107; f=134; (x+n)=83; (d+n)=114

(23,31,24) = {23:31:83:47:36:192} = 6912; f=144; (x+n)=78; (d+n)=114

PMA !!y5/EVb5KZI ID: 9d9b7b July 7, 2019, 9:06 p.m. No.9435   🗄️.is 🔗kun   >>9436

>>9429

>>9433

In the previous example for swapping a and n from c6107 to c6912, f increases from 134 to 144.

 

The new f can be calculated as f + 2(n-a).

 

Another example:

 

c34117

 

from:

(261,27,38) = {261:27:184:75:109:313} = 34117; f=108; (x+n)=102; (d+n)=211

 

swap a and n to:

(261,109,38) = {261:109:102:75:27:395} = 10665; f=-56; (x+n)=184; (d+n)=211

 

new_f = f + 2(n-a) = 108 + 2*(27 - 109) = -56

Anonymous ID: 682964 July 8, 2019, 9:58 p.m. No.9439   🗄️.is 🔗kun

“For in this hope we were saved. Now hope that is seen is not hope. For who hopes for what he sees? But if we hope for what we do not see, we wait for it with patience.”

Romans 8:24-25

Anonymous ID: 88b5c3 July 9, 2019, 9:09 p.m. No.9447   🗄️.is 🔗kun   >>9453

>>9442

Per >>9444

C and C2 have j values that find the correct large square root for c in the same way that the D and D2 records do. And in several examples, one corrects the other.

 

“There is an entire family of numbers that can be factored in one step”

Anonymous ID: 88b5c3 July 9, 2019, 9:12 p.m. No.9448   🗄️.is 🔗kun   >>9449

Far numerous too are the examples where the D and C (e, 1) element calculations give a solution i that is only off by a small amount

 

SeeIt 0.0.3 tested correcting this error value by checking below in the column for the correct combination based on factoring na recursively

SeeIt 0.0.4 will test a new error correction techniqu

AA !LF1mmWigHQ ID: 774bd8 July 10, 2019, 4:16 p.m. No.9453   🗄️.is 🔗kun   >>9454

>>9447

I'm quite skeptical of this being the way forward. From my testing, it scales almost identically to just looking for i[t]=i in d between d[t] recursing through na.

Anonymous ID: 88b5c3 July 10, 2019, 11:46 p.m. No.9454   🗄️.is 🔗kun   >>9457

>>9453

Recursive an factoring was a proof-of-concept, an idea to look for different d and n combinations that close the gap towards i.

The End’s qubits (elements) are all entangled. Each one contains the blueprints for the entire grid. One element contains all of it.

The application of this is that an infinite or unfeasibly large search space can be grasped with a tiny amount of data. All of a column can be represented in one cell, and all of a cell can be represented in one element. That’s what moving to (e, 1) from (e, n) and (e, N) is.

Anonymous ID: 88b5c3 July 10, 2019, 11:53 p.m. No.9455   🗄️.is 🔗kun   >>9457

Since the factors of c exist in column e, but the n values of that column are too numerous to search, we can identify them by compressing the column together using n=1. We know that since the column has been compressed into a single cell, that the data for the factorisations of c exist in it. The route of identifying those factorisations efficiently must be informed by the nature of how the column has been compressed.

Anonymous ID: 88b5c3 July 11, 2019, 12:02 a.m. No.9456   🗄️.is 🔗kun   >>9457

This nature is hidden in plain sight. The techniques of calculating the records c and d are between are the beginning of the unveiling of it, but with the stage finished being set, it will conclude anons discovering the solution.

 

Use the grid with this nature in mind.

Try EVERY idea that comes to mind.

Anonymous ID: 88b5c3 July 11, 2019, 6:46 p.m. No.9460   🗄️.is 🔗kun

Shor's algorithm consists of two parts:

 

A reduction, which can be done on a classical computer, of the factoring problem to the problem of order-finding [reduction to N -n, X -> x search space].

A quantum algorithm to solve the order-finding problem [VQC].

Anonymous ID: 88b5c3 July 11, 2019, 6:51 p.m. No.9461   🗄️.is 🔗kun

The quantum circuits [steps to factor based on family] used for this algorithm are custom designed for each choice of [c].

The input [d, e] and output [n, N] qubit registers need to hold superpositions of values [trivial and nontrivial elements].

Anonymous ID: 88b5c3 July 11, 2019, 6:58 p.m. No.9462   🗄️.is 🔗kun

Proceed as follows:

 

  1. Initialize the registers [coordinates] to [n = 1]

This initial state is a superposition of [all n values in the column], and is easily obtained by generating [-f, 1; e, 1] each a superposition of [row N and row n] applying [e, 1] in parallel to [-f, 1].

Anonymous ID: 88b5c3 July 11, 2019, 7:05 p.m. No.9463   🗄️.is 🔗kun

Construct [the elements d is in the gap of]

[trivial and nontrivial elements] are now entangled, or not separable [patterns]

Anonymous ID: 88b5c3 July 11, 2019, 7:29 p.m. No.9464   🗄️.is 🔗kun   >>9465

Apply the inverse Quantum Fourier transform [root and offspring of David transform] to the input register [gap elements]

AA !LF1mmWigHQ ID: d23b7f July 11, 2019, 7:38 p.m. No.9465   🗄️.is 🔗kun   >>9467

>>9464

So if we don't find i[t]=i or j[t]=j with the d between d[t] elements are we meant to use (0,n) somehow (root of David)? What's the "offspring" of David?

Anonymous ID: d50c20 July 11, 2019, 10:23 p.m. No.9468   🗄️.is 🔗kun   >>9469

https://ghostbin.com/paste/zbg76

 

https://8ch.net/qresearch/res/7005806.html#7006170

 

Just sighted in the bread above… Yall catch this?

Anonymous ID: d50c20 July 11, 2019, 11:27 p.m. No.9472   🗄️.is 🔗kun   >>9473

>>9469

Yeah I saw it in a bread and figured in the chaos you guys may have missed it. and yes… Fastjack gets the gas. Perhaps that code is valuable to yall. ALso.. Trump twatted about BTC being shit FYI. could be important. Keep mathing hard core boys. -HOBO I will keep a lookout for yall.

Anonymous ID: d6ad6a July 12, 2019, 9:48 p.m. No.9474   🗄️.is 🔗kun   >>9475

What's up faggots? I saw a "VQC" post fly by in /qrg/ and thought I'd pop back in and see how you motherfuckers are doing. Any progress? Anyone find Christopher Robbins again?

AA !LF1mmWigHQ ID: d23b7f July 12, 2019, 10 p.m. No.9475   🗄️.is 🔗kun   >>9476

From a recent /qresearch/ thread.

 

>>9474

It fizzled out a bit but most of us are still around even if they're not always posting to the board. He dropped a few more dates on us that amounted to nothing, but other people claiming to know how to use the grid have been in and out too. Also see >>9222

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 456f05 July 12, 2019, 10:19 p.m. No.9476   🗄️.is 🔗kun

>>9475

I was phonefagging and caught:

Fermat Factorization >>>bewbs

 

Something like that.

Was the only comment on that particular bread.

PMA !!y5/EVb5KZI ID: 52b83c July 12, 2019, 10:55 p.m. No.9477   🗄️.is 🔗kun   >>9478

brief test of using D.i and C.j as range of possible d+n values.

lots of examples outside of this range.

interesting that all solved RSA values (except Rsa129) fall within.

AA !LF1mmWigHQ ID: d23b7f July 13, 2019, 12:14 a.m. No.9490   🗄️.is 🔗kun   >>9494

>>9488

Assuming you keep posting despite >>9489, I can't think of any ways that factoring e would be obviously relevant, but any number can take a square out of (0,n) if it doesn't produce a square with c. I don't know that we ever found a pattern of numbers that put c into (0,n) other than c itself, so we never directly figured out ourselves what v could be other than for it to potentially just be qc multiplying qc.

 

>>9489

Considering the vast majority of people in the world have a "her" in their lives and drink coffee, if this is meant to be a threat it isn't a very convincing one (not that I'm encouraging you though).

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 456f05 July 13, 2019, 12:19 a.m. No.9492   🗄️.is 🔗kun

>>9489

Lemme guess… you like rifles and walks in the woods?

You think we're worried about you?

Pic related.

 

>>9491

I 'member "that call"…

AA !LF1mmWigHQ ID: d23b7f July 13, 2019, 12:34 a.m. No.9507   🗄️.is 🔗kun   >>9509

>>9506

Do we look at any specific elements within (0,e) and (0,f), or maybe gaps in these cells or something? Also are we using (0,|f|) or (0,-f)?

AA !LF1mmWigHQ ID: d23b7f July 13, 2019, 12:36 a.m. No.9510   🗄️.is 🔗kun   >>9513

>>9508

Are they three different methods for achieving the same goal, or are they interrelated and potentially used together to factor a given c?

AA !LF1mmWigHQ ID: d23b7f July 13, 2019, 12:38 a.m. No.9512   🗄️.is 🔗kun   >>9515

>>9509

That's vague enough that I don't really know what else to ask about it. I'm trying to come up with the most useful questions I can, but if you'd like to lead this or if I should just keep asking questions, either way is fine.

AA !LF1mmWigHQ ID: d23b7f July 13, 2019, 4:04 a.m. No.9516   🗄️.is 🔗kun

>>9498

It may be more detailed than just "primes can only be multiplied into squares with themselves" if you're talking about using the grid, but it would definitely appear that at least up to a couple thousand (I did more testing after this screenshot), primes can only create squares through multiplication with one other number if that number is themselves. So that would also imply that v for any composite number also can't be prime. v has to be composite to some degree.

AA !LF1mmWigHQ ID: d23b7f July 13, 2019, 6:32 a.m. No.9517   🗄️.is 🔗kun

>>9494

This post appears to imply that v will always be <=c (or I'm assuming qc). I wrote some code to show all the possible v values for a given c. This is kinda weird. For semiprimes, based on about 50 or so tests, it seems that the only time v is ever not just qc, either a or b is one of those numbers that ends up as part of q (so primes that end in 01 in binary), the other shows up as a factor of v, and the number of possible v values is equal to that first variable. Every other semiprime can only produce a square when multiplied by itself (at least from my testing). There is a lot more for us to learn about v, and I'm starting to think the actual method of finding the variable v is itself a non-trivial calculation, even before we get to whatever we end up using it for and whatever X and Y are and everything.

Anonymous ID: dc6643 July 13, 2019, 2:18 p.m. No.9520   🗄️.is 🔗kun   >>9534

Lots of differrnt ways to see a solution

(e, 1) makes a superposition of n

(0, remainder) makes a superposition of e

 

Other superpositions to make??

There are

AA !LF1mmWigHQ ID: d23b7f July 13, 2019, 5:31 p.m. No.9526   🗄️.is 🔗kun

>>9524

>>9525

When we find the elements d is between in (e,1), what asymmetry are we meant to be looking at? There's asymmetry in the distance between the lower element and d and between the higher element and d. There's also asymmetry between the elements d is between in (e,1) and the elements d is between in (f,1).

AA !LF1mmWigHQ ID: d23b7f July 14, 2019, 2:44 a.m. No.9529   🗄️.is 🔗kun

>>9527

Is what you said in >>9222 still true? Is some aspect of it still "protected" and you don't intend to make it public yourself personally yet (you are still avoiding telling us what X and Y are and how to find v, so it does still seem like you're keeping important information from us), or is it the time and you're here to "accelerate" in the sense that you're releasing it now (you said you "believe anons are ready" in the second post >>9484 here, and obviously you gave us quite a bit yesterday)? Are all of these posts directly preceding disclosure or are they just intended to be further clues?

Anonymous ID: 88b5c3 July 14, 2019, 6:39 a.m. No.9530   🗄️.is 🔗kun

I have flip flopped on whether I want to disclose it all.

If hints are explored and you still don’t see it.

Anonymous ID: 88b5c3 July 14, 2019, 6:51 a.m. No.9532   🗄️.is 🔗kun   >>9535

Implications may not be correct.

Protected doesn’t equate to information required to solve being intentionally hidden.

I did not want to give away the code and it be attributed to a fleeting moment of genius.

Anonymous ID: cb65ba July 14, 2019, 8:21 a.m. No.9534   🗄️.is 🔗kun

>>9520

Remainder here is the remainder of? c-dd (ie, e)?

(0, c) would make what a superposition?

 

And, once you have a superposition, how do you take advantage of it? Interference, collapse it? I'm not really up to speed on quantum mechanics, nor the lingo.

 

>>9533

> with anything

 

Anything as in?

Anonymous ID: cb65ba July 14, 2019, 8:23 a.m. No.9535   🗄️.is 🔗kun

>>9532

How do I build intuition and understanding of a superposition (and to a further extent, quantum mechanics) using the grid?

Anonymous ID: cb65ba July 14, 2019, 8:25 a.m. No.9536   🗄️.is 🔗kun

>>9531

All I hear is roaring.

 

If the grid employs us with an infinite set of qubits. What in the grid defines a singular qubit? An element? A set of elements? A specific group of them?

 

Once we understand, will the grid be a general purpose quantum computer or does the grid we know it exist only for RSA?

Anonymous ID: ba5e8f July 14, 2019, 10:10 a.m. No.9537   🗄️.is 🔗kun   >>9636

I think Shor’s algorithm is a good guideline. You’re going to get the right answers in your superposition at first, but you’re also going to get every wrong answer too. What we have to do before measurement is make every wrong answer destructively interfere among itself, and every instance of the right answer build itself up.

Anonymous ID: ba5e8f July 14, 2019, 10:37 a.m. No.9538   🗄️.is 🔗kun

If we don’t know the right answer, how are we supposed to make it come up above all the wrong ones?

 

We do have a criteria that we can apply to the entire superposition. A sifting filter. We know that only 2 answers can fit with c. And we also know that the trivial answer scales with c, and that the nontrivial answer does not scale with c. There is an asymmetry there.

 

A qubit is a register with an enormous range of the amplitudes that it can be, the values contained in it. In a sense, what a qubit in the grid is is.. all of the above. The more qubits the quantum computer can handle, the greater its computation power, and it grows by a lot with each one. 7 (+ the many others you can examine) values in each element, infinite elements in a cell.. infinite cells.. I can’t fathom how infinite its scope is.

 

I think it is the element. It is the first and the only finite unit of the grid, yet contains so many complex combinations of values that work together with the grid.

Anonymous ID: ba5e8f July 14, 2019, 10:45 a.m. No.9539   🗄️.is 🔗kun   >>9581

I have so much awe for all we have not grasped in mathematics, and for the clear fact that it has been designed by a great designer.

 

I must confess that much help from the artist has been received in conveying it to you all.

 

d[t] around d is an incomplete measurement of the superposition. Fine-tune the superposition and measure again (C). What’s more precise than C?

Anonymous ID: ba5e8f July 14, 2019, 5:01 p.m. No.9552   🗄️.is 🔗kun

Do you believe it’s capable of being figured out from the hints? Some have been specifically to illustrate what it looks like

AA !LF1mmWigHQ ID: d23b7f July 14, 2019, 5:02 p.m. No.9553   🗄️.is 🔗kun   >>9554

>>9551

Our knowns are c, d, e and f. There are a bunch of different cells and elements that we can find from these (such as all the things in (e,1) and (f,1), like where d is between d[t] or where a[t]=BigN etc). One or several of those would have to be the starting point I would think. There's been a lot of talk recently about d between d[t] in (e,1) being where the magic happens, but not a lot of talk about what magic is actually happening. And every few months we have a different concept that we're led to believe is the one we're meant to use to solve (whether it's qc, or something about the asymmetry between (e,1) and (f,1)'s na and nb elements, or the triangles that make up (x+n)(x+n), etc).

 

The short version is, while I think the way the solution works is that we start from knowns, apply some known concepts and some unknown concepts and eventually end up at the solution, I don't think any of us know enough that we could get any more specific than that.

AA !LF1mmWigHQ ID: d23b7f July 14, 2019, 5:09 p.m. No.9555   🗄️.is 🔗kun   >>9556

>>9554

What about it? c between i[t] (if that's what you meant) is another one of those concepts that we know how to find the elements for, and it does help to occasionally find i or j, even though it doesn't scale. It very well could be a starting point in conjunction with d between d[t] (like you (or someone, it's hard to keep track) suggested), but so could any number of concepts. Like I said, every few months we're told to start from somewhere else.

Anonymous ID: ba5e8f July 14, 2019, 5:58 p.m. No.9556   🗄️.is 🔗kun   >>9557 >>9559

>>9555

I understand that frustration.

You’re being given a new tool with each hint.

 

The hints are pieces, not the full puzzle. There isn’t a set in stone way that they have to fit together. The start point can be anything. There are a lot of different ways to solve. That’s why they aren’t elaborated on after a concept is unveiled. Anons create something new with them.

AA !LF1mmWigHQ ID: d23b7f July 14, 2019, 6:01 p.m. No.9557   🗄️.is 🔗kun   >>9558

>>9556

Well in that case we have a bunch of tools but we don't know how they work together or which ones to use in conjunction with one another, to further address your question before.

AA !LF1mmWigHQ ID: d23b7f July 14, 2019, 6:28 p.m. No.9560   🗄️.is 🔗kun   >>9561

>>9558

Okay, here's one. Chris brought up two concepts around the same time a few months ago and seemed to imply they were related but never actually linked them. These were qc (multiplying c by a bunch of low known primes to increase the frequency of factors showing up in (e',1) and (f',1)) and finding where a[t]=c (which for t=1 is (2c,1,1) and (2c-1,1,1), and then t=2 is further back and so on). Is there a link between those concepts?

AA !LF1mmWigHQ ID: f46a7b July 14, 2019, 9:58 p.m. No.9563   🗄️.is 🔗kun   >>9564

>>9561

>>9562

I think I should rephrase. How can knowing this fact about the grid (or any of the other things we've learned so far) get us from a place of not knowing how to use the grid to a place of knowing how to use the grid?

Anonymous ID: ba5e8f July 14, 2019, 10:03 p.m. No.9564   🗄️.is 🔗kun   >>9568 >>9573 >>9729

>>9563

If you start your sequence as e=-(2c+1) where a[t] = c, assuming odd c, you will have a sequence where every b value ends in 01 in binary all the way down to b = -1.

 

Are you with me?

Anonymous ID: ba5e8f July 14, 2019, 10:06 p.m. No.9565   🗄️.is 🔗kun   >>9568

Which means you can create a q sequence within the grid of a and b by calculating the opposite starting point of the sequence and determining which value is prime

Anonymous ID: ba5e8f July 14, 2019, 10:45 p.m. No.9566   🗄️.is 🔗kun   >>9568

There is a pattern for calculating the e values of this sequence that involves adding/subtracting c from squares. You may find that interesting.

 

The starting point for the a[t]=c sequence of negative b values that end in 11 and the sequence for 01 can be found at around

 

e = -((X + N)^2 + c)

e = -((d + N)^2 + c)

 

Memory a bit fuzzy but I believe you use the squares to move to the next odd value

AA !LF1mmWigHQ ID: f46a7b July 14, 2019, 10:48 p.m. No.9568   🗄️.is 🔗kun   >>9569

>>9564

>>9565

So we don't actually have to calculate q ourselves? If we use the c values in those cells I guess it would be the same sort of thing (although they won't all be primes).

 

>>9566

The pattern for e in the increasing b values is >>9145 here. I'll put some code together to display what you're talking about.

AA !LF1mmWigHQ ID: f46a7b July 15, 2019, 12:57 a.m. No.9573   🗄️.is 🔗kun   >>9574

>>9564

>>9561

I'm having trouble figuring this sequence out. With an example of c559 where e=-1118 (-2c), xx+e=2na is the same as xx-1118=1118 (assuming we're in row 1), which makes xx=2236, which isn't a square.

AA !LF1mmWigHQ ID: f46a7b July 15, 2019, 1:48 a.m. No.9574   🗄️.is 🔗kun

>>9573

And if n isn't equal to 1, both e and n become unknowns, so we can't find the cells, or there could be multiple. You're going to have to clarify.

AA !LF1mmWigHQ ID: f46a7b July 15, 2019, 5:51 a.m. No.9579   🗄️.is 🔗kun   >>9582

To those of you who aren't Chris but know how to use the grid (I don't know how many of you there are and it's hard to keep track), in particular the one who's been here a lot in the last few days. I've put together almost all of the concepts and patterns we know in as compact, concise and ordered of a way as I could (four posts in the Grid Patterns thread, which I'll link at the bottom of this post).

 

You seem to think we're ready and that now is the time for us to figure it out. You know better than we do how that's meant to happen. You're coming from the perspective of knowing how to use the grid, so having seen how far we've come, you probably think we're awfully close. From our perspective, we have an overwhelmingly long set of patterns and concepts and no idea how to use them, and there are a bunch of concepts we know exist but haven't been told about. You want us to figure it out without you just telling us how to do it, and we want to figure it out in the same way. We have to work with each other here. No matter how close you think we are, all we've done over the last couple days is more of the same thing we've been doing for 19 and a half months, which is analyzing patterns without knowing how to use them. If you think that's how to do this, maybe you're right, but that's how Chris has run this operation, and seeing how one of you left him a grumpy message in GP telling him nobody listens to him anymore, maybe you'll also understand that the reason for that sentiment is exactly the method by which he has been trying to teach us (which, so far, has been the same way you've been doing things with us here, albeit a bit more openly).

 

Maybe I don't know exactly how this could work best, but we all want the same thing, and you seem more open to conversation than Chris ever was. We talked today about how we know all of these patterns but we don't know how to use them. If continuing in the same way of analyzing specific patterns until you personally feel we understand them enough to move onto the next one is going to work, you should know, PMA, blank-name poster and I are here daily (as well as Topol, although he isn't a math person so much), and I personally have a lot of free time on my hands, so I will gladly work through whatever you feel necessary to get this done. All you have to do is tell us what to do. But I am skeptical of that being the way things will work (given, like I said, it's how things have gone for 19 and a half months). So, my suggestion to you guys (you could even just be the one person, or even Chris pretending to be other people… I don't know) is to have a read through the four posts I'll link below, which sum up most of our knowledge to date. Instead of going over individual patterns without having any idea why we're learning about them, given you'll have an overview of pretty much everything we've got, you could potentially guide us by showing us which concepts to use in conjunction with one another and what they actually accomplish. Let's be efficient here rather than stumbling around blindly.

 

>>9575

>>9576

>>9577

>>9578

Name ID: 7194aa July 15, 2019, 9:27 a.m. No.9580   🗄️.is 🔗kun

>>9494

Let's see if my mind is going the right place. For some numbers, you may have a v that is < c, which can be used to substitute c. The end result would still be a factor for c. But this isn't always the case, sometimes v = c (because no substitute exists)?

 

>>9508

Why doesn't (3, 1) give us the same "power" to create a superposition for c?

 

>>9545

t's being the smallest t possible when recursing down the triangles?

 

>>9554

The c = x+n element is in many ways beautiful. If you think of it in terms of complex numbers that cell kind just, clicks. d+n being the lateral part while x+n is the number. But it's not enough to go on alone. You need other parts, which I haven't solved yet.

 

>>9570

Capital T being the big T, the t from the na transform for bigN and c?

Name ID: 7194aa July 15, 2019, 9:43 a.m. No.9581   🗄️.is 🔗kun

>>9539

>d[t] around d is an incomplete measurement of the superposition. Fine-tune the superposition and measure again (C). What’s more precise than C?

 

Fine tuning by either following the d[t]'s into the new e's or by modifying c to "match" (or get a closer match) of the d[t]'s.

Anonymous ID: 88b5c3 July 15, 2019, 9:45 a.m. No.9582   🗄️.is 🔗kun

>>9579

Well done on the summary.

I will go through it and find the best things to work on and put together.

You are right that linking concepts isn’t giving it away.

Anonymous ID: 88b5c3 July 15, 2019, 9:58 a.m. No.9583   🗄️.is 🔗kun   >>9584 >>9601 >>9602

“We also find staircase numbers when we add the x values from an (e,1) element and an (f,1) element”

 

x is an index, BUT when something else is used as the index, it becomes more relevant.

Try the elements at x=d and x=d+1 (2d + 1).

 

ALSO, if the elements d is between are calculated with x ~= sqrt(f), what is at “x = f or x = f-1?”

Name ID: 7194aa July 15, 2019, 10:28 a.m. No.9585   🗄️.is 🔗kun   >>9586 >>9587 >>9589 >>9850 >>9851

I suppose it's kind of wasteful(?) or pointless(?) to share this now, but are you guys familiar with Rascals Triangle?

 

It has some interesting key properties that relate to the grid, columns and e-values a number exist in. I came over it by happenstance when looking into patterns of e's for numbers in different columns.

 

You know Pascal's triangle, it goes like this:

 

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

 

Rascal's triangle was "discovered" (created?) by three kids when given an "IQ" test. It went like this:

 

1

1 1

1 2 1

1 3 3 1

— What is the next number in the sequence

 

They didn't write 1 4 6 4 1, like you would assume based on Pascal's triangle, instead they wrote

 

1 4 5 4 1

 

When you look at Pascals triangle (and Rascals triangle) you can group 4 numbers in a diamond shape. So for:

 

1

1 1

1 2 1

1 3 3 1

1 4 5 4 1

 

A diamond shape here would be

2

1 3

4

 

You can divide the diamond shape into quadrants: North, East, West and South.

 

The pattern is:

South = ((East * West) + 1) / North

 

Pretty sweet, right? But how does it relate to the grid? It turns out that several rows in the rascals triangle is equal to the e's a number at a[t] exists in.

 

Given:

1

1 1

1 2 1

1 3 3 1

1 4 5 4 1

1 5 7 7 5 1

1 6 9 10 9 6 1

 

The a[2] in (1, 1) exists at e's = 1, 6 and 9 and is equal to the last line (row 7). a[3] in (1, 1) exists in e's: 1, 10, 17, 22 and 25 which is equal to the 11th row in the Rascal's triangle. It continues like this, matching the e's a number exist in with the triangle.

 

The formula for calculating the triangle is:

k * (n - k) + 1

 

Here the 1 is equal to the base or column. Swapping it with any other number and it generates the same triangle except for that column. Think of the above triangle as the e-index(?) for (1, 1).

 

I know this isn't much and with the guy who knows and understands the grid here, it feels like it falls quite flat. Either way, I wanted to share with you guys.

Name ID: 7194aa July 15, 2019, 1:59 p.m. No.9588   🗄️.is 🔗kun

Last week I had some major hd problems. Essentially I had literally a crash in my drive and I lost all my work. It sucks, and I tried to recover it, but unfortunately I couldn't. Instead I've been rewriting my work and getting back up there. My fault for not backing up and getting a more modern drive I suppose.

 

Some of the last stuff I found was some patterns relating to quantum angular momentum and column (0, 1), (1, 1) and (2, 1). It was related to a few quantum angular momentum matrices (but I have no clue about that stuff, at least not yet). I also don't remember the details, but it was related to the patterns from the Rascal triangle.

Name ID: 7194aa July 15, 2019, 2:03 p.m. No.9589   🗄️.is 🔗kun   >>9590

>>9585

Right, so the formula is k * (n - k) + 1, but the pseudocode would be:

 

for n in range(1, limit) (for some limit)

for k in range(0, n) (0 <= k <= n)

value = k(n - k) + column

 

The n represents row, while k would be the column within that row.

Name ID: 7194aa July 15, 2019, 2:20 p.m. No.9595   🗄️.is 🔗kun

Note though:

> Isn’t there a cell that can be created using c where the first existing column will be one of its factors?

 

I didn't read that as explicit a factor of c, but rather a factor of the cell you can create from c.

Name ID: 7194aa July 15, 2019, 3:18 p.m. No.9599   🗄️.is 🔗kun   >>9600

Just another thing I just noticed, something I haven't seen before.

 

Take an element a, b with an even n. Compute the na transform and then compute the element for a=1, b=na/2. What is the e here? Works for big N (and shadow n's) as well.

Name ID: 7194aa July 15, 2019, 3:21 p.m. No.9600   🗄️.is 🔗kun

>>9599

Hmm.. Looks like there are some offsets required here too in some cases.

 

Offsets are a funny thing, we see them so many times. Over and over again something holds for a set of elements, but for others you need the offset.

 

Since we now have someone here who understands and knows the grid, could you give some kind of intuition or understanding of these offsets?

AA !LF1mmWigHQ ID: f46a7b July 15, 2019, 4:33 p.m. No.9601   🗄️.is 🔗kun   >>9602 >>9605

>>9583

>x is an index, BUT when something else is used as the index, it becomes more relevant.

>Try the elements at x=d and x=d+1 (2d + 1).

So far all I'm seeing is that a[t] from (f,1) is always equal to BigN-1. Pic related.

>ALSO, if the elements d is between are calculated with x ~= sqrt(f), what is at “x = f or x = f-1?”

c between d[t]

AA !LF1mmWigHQ ID: f46a7b July 15, 2019, 7:19 p.m. No.9602   🗄️.is 🔗kun   >>9603

>>9583

>>9601

>ALSO, if the elements d is between are calculated with x ~= sqrt(f), what is at “x = f or x = f-1?”

May have jumped the gun a bit with that one, it was an assumption based on squares. It seems like you're talking about one specific element rather than two elements with something in their gap, since two x values in (e,1) or (f,1) are 2 apart rather than f and f-1 in the same cell. You're also linking it to the elements d is between obviously, so there should be some kind of linked concept or something that equals something else. All I'm seeing is that the |e| and |f| values in the two elements are 1 apart (which is also true of their x values, given those are f and f-1). I'm not seeing any link to the d between d[t] elements.

Anonymous ID: 9f00d8 July 15, 2019, 8:35 p.m. No.9605   🗄️.is 🔗kun   >>9606 >>9611

>>9601

You should expect to find c-relevant values in gaps between element values, since (e, 1) isn’t tailored directly to c, but is a cell representing the whole column.

 

If you’re wondering what could be useful at a pair of elements where x adds up to a specific thing, maybe it would help you to consider it a pair of x and n. Either would be valid. The grid’s all about squares and the patterns have squares in their squares.

AA !LF1mmWigHQ ID: f46a7b July 15, 2019, 8:49 p.m. No.9611   🗄️.is 🔗kun   >>9612

>>9608

If you want, sure.

 

>>9605

>>9606

Well so far all we've learned about gaps is that once in a while i[t]=i when d is between d[t], and then the same for j where c is between i[t]. We haven't done any "calculations" between elements. We've just observed something that occasionally happens.

MM !DMWWK0jvao ID: 249c0e July 15, 2019, 9:12 p.m. No.9612   🗄️.is 🔗kun   >>9614 >>9616

>>9611

AA, thanks for carrying the torch so often.

No idea if the 'skips' or 'ghost values' in sequences are similar to these 'gaps', which could be non-integral grid-points? They may still be rational, such as 1/3. Tried to describe a bit, a (me):

>>9089 (pb)

>What about those 'ghost' values?

> - they are 'non-integral' n-values. Note that if these were included x_base is a simple sequence increasing by 2 for each k.

> - there is likely a hidden part of the grid that can be used to traverse with these, haven't gone there yet. Diff 2 is constant, so a Second Order series.

 

V-Helper anon, thanks for the crumbs.

Have to head out to sea for a short bit again, but will print these recent bits to digest.

Topo, PMA, Hobo, et al - be well.

AA !LF1mmWigHQ ID: f46a7b July 15, 2019, 9:30 p.m. No.9620   🗄️.is 🔗kun   >>9623

>>9617

Are you taking the fact that we add 1 to d in the negative space into account with that, or are we generating the grid wrong (we're generating it based on Chris' original code)?

Anonymous ID: 9f00d8 July 15, 2019, 9:46 p.m. No.9625   🗄️.is 🔗kun

>>9622

At the point where two x values from the opposite columns add up, the difference in f will be that sum + 2.

 

Example: for aforementioned elements, x+x = 25, f - f = 27

AA !LF1mmWigHQ ID: f46a7b July 15, 2019, 10 p.m. No.9628   🗄️.is 🔗kun   >>9629

>>9627

I wouldn't have a clue. Given you're talking about this after talking about gaps, it probably has something to do with making calculations in a gap in some way we haven't done before. Maybe there's a formula to find an f value that we want in row 1. All the non-trivial things I can think of would immediately solve so I would think whatever we do to find an element pair where it adds up to something nontrivial would possibly be complicated and possibly involve recursion.

Anonymous ID: 9f00d8 July 15, 2019, 10:07 p.m. No.9629   🗄️.is 🔗kun

>>9628

So you do have an idea of what the solution looks like. Can I tell you that it isn’t complicated and can be rewritten into different ways? A solution might look like that long tree, but remember, it was also said that there were grid shortcuts. Don’t forget, the grid still confuses even the person who brought it to us.

Anonymous ID: 9f00d8 July 15, 2019, 10:09 p.m. No.9630   🗄️.is 🔗kun   >>9632 >>9634

“Fractal nature of the solution.”

 

What is a fractal, but doing the same thing over and over? Kind of like calculating c over and over? How are you doing that when you input grid coordinates without realizing it?

AA !LF1mmWigHQ ID: f46a7b July 15, 2019, 10:16 p.m. No.9632   🗄️.is 🔗kun   >>9633

>>9630

>>9631

We have been told that factoring d and e is meant to allow for the factoring of c. None of us have found a link between a fully-factored d and e and their parent c yet though.

AA !LF1mmWigHQ ID: f46a7b July 15, 2019, 10:27 p.m. No.9634   🗄️.is 🔗kun

>>9630

>How are you doing that when you input grid coordinates without realizing it?

That's pretty vague but you could maybe be talking about an (e,n) co-ordinate containing an infinite set of elements, or maybe you're talking about (e,1), or maybe the fact that you're choosing an n value implies something about a calculation that I can't think of. If that's an answer to your question.

Anonymous ID: 9f00d8 July 15, 2019, 10:33 p.m. No.9635   🗄️.is 🔗kun   >>9636

The amplitudes of qubits can contain values that you can’t put on a number line.

The same is true for elements.

How would you use those? )(It’s not a coincedence q gets the same variable name as Quantum, so what’s special about that process?)

AA !LF1mmWigHQ ID: f46a7b July 15, 2019, 10:48 p.m. No.9640   🗄️.is 🔗kun   >>9642

>>9639

My first thought then is that maybe there's something to find when we look at where d' is between d'[t], potentially as some kind of comparison to the same concept applied to regular c.

PMA !!y5/EVb5KZI ID: 9d9b7b July 15, 2019, 11:21 p.m. No.9641   🗄️.is 🔗kun   >>9644

There is a consistent way to calculate a "d" that falls between two records in (-1,n) using values from D and C records.

 

By way of an example, for c6107 where d=78:

 

D2 = {23:1:83:11:72:96}

 

d[D2] - d = 83 - 78 = 5

 

i^2 - c = 84^2 - 6107 = 949

 

sqrt( 949 ) = 30, remainder = 49

 

in (-1,5), d=30 falls in the gap between:

 

{-1:5:23:11:12:44}

{-1:5:55:19:36:84}

 

C2 = {23:1:6283:111:6172:6396}

 

x[C2] - d = 111 - 78 = 33

 

j^2 - c = 112^2 - 6107 = 6437

 

sqrt( 6437 ) = 80, remainder = 37

 

in (-1,33), d=80 falls in the gap between:

 

{-1:33:71:43:28:180}

{-1:33:129:65:64:260}

 

To generalize:

 

sqrt( D2.i^2 - c ) in the gap of ( -1, d[D2] - d )

sqrt( C2.j^2 - c ) in the gap of ( -1, x[C2] - d )

PMA !!y5/EVb5KZI ID: 9d9b7b July 15, 2019, 11:32 p.m. No.9644   🗄️.is 🔗kun

>>9641

let's refer to the record d is between in (-1,n) as "DD", representing a second gap that appears to be relevant.

AA !LF1mmWigHQ ID: f46a7b July 15, 2019, 11:32 p.m. No.9645   🗄️.is 🔗kun   >>9646

>>9642

I have a couple code issues to sort before I post a screenshot, but for all the examples I've tried so far, our original d is lower than the d[t] value in t=1 of (e',1) and (f',1), so it would be between negative elements.

AA !LF1mmWigHQ ID: f46a7b July 15, 2019, 11:47 p.m. No.9647   🗄️.is 🔗kun   >>9649

>>9646

Then I would ask what you mean by the "best place". Are we using those specific elements for something? I'm just altering my code for negative elements at the moment.

Anonymous ID: 249c0e July 15, 2019, 11:47 p.m. No.9648   🗄️.is 🔗kun   >>9654

At this rate you all are going to crack the grid while I'm on the high seas, will be "rooting" 4U.

Perhaps an angle to start the disclosure, given a post on Australian science fiction author Greg Egan's Twat:

Factor this, Malcolm: 1153156707837430719761444702620880678607724678317205449829468330793508082440760887670568702039791475818283095522190159

8:40 PM - 13 Jul 2017

https://twitter.com/gregeganSF/status/885705633852301312

    • +

Is to drop the factors, ask for another, and then you'll have someone's attention. Or something far more creative, I'm sure Topol could think of many.. 2yr old post, kek.

ps - Shild's Ladder is a great book!

AA !LF1mmWigHQ ID: f46a7b July 16, 2019, 12:03 a.m. No.9651   🗄️.is 🔗kun   >>9652

>>9650

You mean like this? For this example c559 d is 23, and the lowest it ever gets when I'm generating negative records is 118. So we'd use x from D and D2 in this case?

AA !LF1mmWigHQ ID: f46a7b July 16, 2019, 12:19 a.m. No.9656   🗄️.is 🔗kun   >>9660

>>9655

e? Or maybe d or f. I'm thinking in terms of cs as a number line and the grid as a 2D graph, but I don't entirely see what you're getting at.

Name ID: 661a31 July 16, 2019, 2:46 a.m. No.9657   🗄️.is 🔗kun

What is so special about (0, 2ˆn)? I see a relationship between (3, 6) and (0, 8), but I'm not sure I'm grasping it fully.

Name ID: 661a31 July 16, 2019, 2:47 a.m. No.9658   🗄️.is 🔗kun   >>9660

>>9655

It depends on the parity, but we have two derivatives. 4t and 4t + 2.

 

It also relates to the sequences in (e, n). I haven't nailed it down entirely, but the changes are something along the lines of 4tn + 2n or 4tn + offsets here and there.

Name ID: 661a31 July 16, 2019, 3:52 a.m. No.9659   🗄️.is 🔗kun

>>9655

I have a question. We've been talking about how this is all a fractal. Is this fractal visual? Similar to Mandelbrot set. Has it been generated? I'd very much like to see it, if possible.

Anonymous ID: 3dd86c July 16, 2019, 6:08 a.m. No.9661   🗄️.is 🔗kun   >>9667

It’s everywhere in the grid but also in no particular place.

 

https://en.wikipedia.org/wiki/Integration_using_Euler%27s_formula

Name ID: 2e4cbd July 16, 2019, 9:10 a.m. No.9668   🗄️.is 🔗kun   >>9671

>>9660

Funny, looking at the wiki page for e under "Representations" and looking at the continued fractions (the second one that is supposed to converge 3 times faster), those numbers look awfully familiar, no?

Anonymous ID: 3dd86c July 16, 2019, 12:31 p.m. No.9670   🗄️.is 🔗kun   >>9672

>>7438

Maybe he presented it outside of the grid because he felt a solution was too obvious with it. I don’t think it’s a coincidence that summing up the leaves gets you that close - it might could be used to get you to the right magnitude that a factor is (and remember, difference in factors is x+n). A bit similar to the D and C records in that respect, because the search space is significantly reduced, and it can be transformed into another variable search space (as in changing a search space from between d for i to j instead). A way of rotating an object, if you will - a reliable way to get relatively close to a solution variable is a big gain and is evidence the grid stores those variables in an accessible (from c) way.

Anonymous ID: 3dd86c July 16, 2019, 1:05 p.m. No.9673   🗄️.is 🔗kun

It is also easier to traverse the search space as a difference in x or x+n compared to the bare factors

AA !LF1mmWigHQ ID: f46a7b July 16, 2019, 2:07 p.m. No.9675   🗄️.is 🔗kun

>>9672

>>9674

I'm just basing that off of whatever whoever wrote it meant though. If we treated it as a0 we'd get x0=-1343793146355417714510857901560170544388157224211. I'm working on getting a more readable example at the moment.

AA !LF1mmWigHQ ID: f46a7b July 16, 2019, 2:14 p.m. No.9676   🗄️.is 🔗kun   >>9677

Okay here's a better example. c4343 (43*101), (118,7,12) = {118:7:65:22:43:101}, f=-13, c=4343, u=14, i=72, j=29

For this example if you add the leaves of the tree together you get 13, so if that's a0 then d-a0=65-13=52, where the actual x value is 22.

AA !LF1mmWigHQ ID: f46a7b July 16, 2019, 3:17 p.m. No.9680   🗄️.is 🔗kun

>>9679

It certainly does occasionally find exact matches in a similar way; for example, c559 where a=13, the leaves add up to 13. Whatever it is we do with the tree comes in what Chris called "part 3", part 2 being the generation of the tree itself, and he never went any further into how the tree relates to the grid (since part 3 is where the grid comes in). You were wanting to go in that direction, right?

AA !LF1mmWigHQ ID: f46a7b July 16, 2019, 10:05 p.m. No.9684   🗄️.is 🔗kun   >>9685

>>9683

So if we're using e as x and f as x at the same time, we'd need to be using two elements. If d is used as x+n then there'll be two separate n values for those elements. And then we do that recursively through the square roots of each one? I'll put some code together to do that in a bit. Does this relate to the tree?

AA !LF1mmWigHQ ID: f46a7b July 16, 2019, 11:12 p.m. No.9694   🗄️.is 🔗kun

>>9693

Great, thank you for clarifying. Just about finished my code so I'll post an example in a minute. Thank you also very much for how generous you've been with your time. In less than a week you've given us more than we've had in probably the last year.

AA !LF1mmWigHQ ID: f46a7b July 16, 2019, 11:33 p.m. No.9696   🗄️.is 🔗kun

It seems like the (0,n) cells in the first run of the recursive function aren't even valid, but from then on they're squares. The non-(0,n) elements are quite close together in their values, though. I calculated them based on (0,n,t) and then converting those t values back into x values to do xx+e=2na.

AA !LF1mmWigHQ ID: f46a7b July 17, 2019, 5:35 a.m. No.9697   🗄️.is 🔗kun   >>9698

I've put together some code that finds (0,|d-e|) and (0,|d-|f||) recursively using their square roots. Since valid elements don't always exist at the given x values, I made it find the element with the closest x value in that (0,n) (sometimes it has to use elements where x=0, but I also made a version where it only uses valid elements). Is this correct (first picture, with relevant tree for comparison)?

AA !LF1mmWigHQ ID: f46a7b July 17, 2019, 12:45 p.m. No.9700   🗄️.is 🔗kun

>>9699

Here are two more examples then. One thing I've noticed is that for each example, at least one of the original x values at each level is between the j values that end up being used. But I haven't noticed anything else yet.

Anonymous ID: 88b5c3 July 17, 2019, 8:48 p.m. No.9701   🗄️.is 🔗kun   >>9702

It doesn’t have to be perfectly implemented to see (and use it as, but you will quickly find shortcuts) the solution. Same intent as the tree.

 

The correct way to view the shadow grid calculating from (e, n, t). Unity of method necessary also to work together with one another.

AA !LF1mmWigHQ ID: f46a7b July 17, 2019, 8:51 p.m. No.9702   🗄️.is 🔗kun   >>9703

>>9701

>Unity of method necessary also to work together with one another.

Is there anything you'd recommend changing about the method the way I posted it then? Otherwise, we've been talking about it on Discord and it seems like we're ready to learn how to use it if you'd like to continue.

Anonymous ID: 88b5c3 July 17, 2019, 9:07 p.m. No.9703   🗄️.is 🔗kun   >>9704

>>9702

Don’t worry about finding a valid element. x for e=0 is even so take the 2 elements it’s between or if x is even take that element and the one above it

AA !LF1mmWigHQ ID: f46a7b July 17, 2019, 9:15 p.m. No.9704   🗄️.is 🔗kun

>>9703

Okay, we'll adapt it for the two surrounding elements. A few of us are online right now so we're ready to do whatever.

AA !LF1mmWigHQ ID: f46a7b July 17, 2019, 9:41 p.m. No.9706   🗄️.is 🔗kun   >>9707 >>9708

>>9705

Necessary vagueness I guess. If you meant multiplying numbers together that would remain in (0,n), you would have probably said "square" (since you mentioned something about "squaring the gap" a couple days ago, whatever that means). If it doesn't have to be perfectly implemented, it must have more to do with the d/e/f values than the elements they're used to find. I remember there being something in the grid about a number multiplied by (itself plus 1). It may have been a values in a particular cell. And in every example I've seen so far, the n values in the two first-generated elements for this recursive method thing have always been one apart. You're probably going to continue to be vague given the way you typed what I'm replying to but I might look into multiplying those n values together.

AA !LF1mmWigHQ ID: f46a7b July 17, 2019, 9:52 p.m. No.9709   🗄️.is 🔗kun

>>9708

Here's the same c as before with the (-1,1) elements I gathered you were potentially talking about. The triangle base u is the same as the higher of the two n values in the cases I tested (including this one obviously).

AA !LF1mmWigHQ ID: f46a7b July 17, 2019, 11:39 p.m. No.9710   🗄️.is 🔗kun   >>9712

I'm not sure if it's obvious algebraically or not but I just figured out that c-((|d-e|)(|d-|f||)) is a square.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 456f05 July 18, 2019, 1:25 a.m. No.9713   🗄️.is 🔗kun

I'd just like to take a moment to thank everyone here.

We have the comfiest Q-related Breads of 'em all.

 

SHADILAY, NERDS!

Name ID: 2e4cbd July 18, 2019, 3:38 a.m. No.9714   🗄️.is 🔗kun   >>9715

>>9671

If you look at the differences between consecutive a's in (0, 1) you'll see them (if you ignore the first two 1's in the continued fraction representation).

 

>>9712

I'm feeling the FOMO right now, jelly. Sadly I'm occupied for the next few days so I'll catch up with what you guys are doing later (Or I'll hear about it in the news if you finish up!).

AA !LF1mmWigHQ ID: f46a7b July 18, 2019, 4:53 a.m. No.9715   🗄️.is 🔗kun

>>9714

I'm not going to lie, I would personally love it if this was finally over within a few days. I don't know how close we are though. If it is done by then, there'll be plenty for you to come back to, so don't worry too much. I'll certainly still be here getting as much done as Not Chris' time allows.

PMA !!y5/EVb5KZI ID: 9d9b7b July 18, 2019, 10:15 p.m. No.9717   🗄️.is 🔗kun

Couple of examples of using q as a process instead of a single multiplier, assuming the prime factors are already known.

 

The "q" record is created with an a value equal to the minimum of q * a or b. The b value is the maximum.

 

c6107, c21059917, and c610737593 show cases where the smallest q prime multiplier results in the smallest n value.

 

For c25185549107, on the other hand, the largest q results in smallest n.

Anonymous ID: 8a790b July 19, 2019, 4:16 p.m. No.9721   🗄️.is 🔗kun

if you want to apply calculus concepts, think of it like physics

make rate of change not constant

AA !LF1mmWigHQ ID: f46a7b July 19, 2019, 4:48 p.m. No.9723   🗄️.is 🔗kun   >>9724

>>9718

This is kind of hard to understand. Are you saying we keep e and n the same but find the two elements where x is equal to D and where x is equal to D2?

AA !LF1mmWigHQ ID: f46a7b July 19, 2019, 8:04 p.m. No.9728   🗄️.is 🔗kun   >>9730 >>9732

>>9727

What I do see is that the i values from the d between d[t] elements become the j values in the squared elements (and when we have to find the surrounding elements due to parity i is halfway between the j values). So I gather there's something useful about the values found halfway between these newer i values too potentially. But I have a feeling we were meant to "see" something else.

Anonymous ID: fdc9a3 July 19, 2019, 8:05 p.m. No.9729   🗄️.is 🔗kun

>>9564

record a+b/2 (xx+e)/229:3:7924:215:7709:8145 7927 2312729:13:1994:215:1779:2235 2007 2312729:39:808:215:593:1101 847 2312729:593:254:215:39:1655 847 2312729:1779:228:215:13:4001 2007 2312729:7709:218:215:3:15851 7927 23127

Interesting

AA !LF1mmWigHQ ID: f46a7b July 19, 2019, 9:10 p.m. No.9734   🗄️.is 🔗kun   >>9735

>>9733

All values as in every variable comes into it somehow? Is there further transforming to be done or do we have enough elements and concepts to collapse it?

Anonymous ID: 88b5c3 July 19, 2019, 9:15 p.m. No.9735   🗄️.is 🔗kun   >>9736 >>9739 >>9764

>>9734

Think of it as decreasing the search space until there’s nothing but the factors or a prime determination.

 

Image slowly becoming unblurred?

You’re unblurring the large square.

Try the other parts.

Other hints will suddenly find their true use.

AA !LF1mmWigHQ ID: f46a7b July 19, 2019, 9:25 p.m. No.9736   🗄️.is 🔗kun   >>9737

>>9735

>until there’s nothing but the factors or a prime determination

How have the tree, the tree elements or the squaring of the d between d[t] gap limited the search space? There have been various variables that have been equal to other variables, or variables that have appeared between two of the same variable, but none of it has been knowns leading us to unknowns as far as I can tell. In fact, unless I'm missing something here, we've done basically the same thing again in that you've given us another list of transformations and new elements to look at in the grid that we still don't know how to use. I understand the vagueness, and I'm sure we're more close to a solution than we realize. I can imagine that when you say we're unblurring the large square and to try other parts you're referring to this all being based around i and that it would be useful to do the same thing with j and the elements where c is between i[t]. That's all well and good but where are we going with this? Is there something we're meant to see that we haven't seen yet? Is there something we're meant to do with all of these things that you haven't told us yet? Because if the c between i[t] equivalent is anything like the d between d[t] stuff you've been guiding us through, it seems like it's going to add to that list of things we don't know how to use.

AA !LF1mmWigHQ ID: 4a1328 July 19, 2019, 9:39 p.m. No.9738   🗄️.is 🔗kun

>>9737

If they're close enough bounds we'd only need two, since any one of the unknowns would solve. Otherwise it depends on how close the bounds are.

AA !LF1mmWigHQ ID: 4a1328 July 19, 2019, 11:02 p.m. No.9740   🗄️.is 🔗kun   >>9742

>>9739

>>9737

It looks like the (x+n) square base u values for the squared c between i[t] elements get very close to BigN-n. Not to a point where it seems like it could be used to calculate, but close enough that it's worth noting probably. It also doesn't seem like there's always a value below BigN-n, but there always seems to be a value above. Is this one of the upper bounds you were talking about?

AA !LF1mmWigHQ ID: 4a1328 July 20, 2019, 3:52 a.m. No.9742   🗄️.is 🔗kun

>>9740

I don't think there's anything to this. I made graphs to compare u minus BigN-n to u minus BigN, and they're pretty much identical. This pattern is more like C_F2's u being close to BigN than it is to C_F2's u being close to BigN-n. I suppose that's still kind of interesting, but I don't think it helps, since if anything all this would do is add extra iterations to an iterative look for n. I could be wrong, I don't know.

 

>>9737

Purely out of curiousity, if we keep going the way we have been for the last week, how much longer do you think it'll take? I'm not trying to speed anything up or come across as impatient, but I can't help but wonder about that.

Anonymous ID: 88b5c3 July 20, 2019, 8:55 p.m. No.9747   🗄️.is 🔗kun   >>9748

It has to do with geometry - missing square piece of c.

 

Decompose d (thus dd) and e, and calculate the missing piece of c.

Anonymous ID: 88b5c3 July 20, 2019, 9:35 p.m. No.9749   🗄️.is 🔗kun   >>9750

Take the sqrt(d).

Take the sqrt(i).

 

Apply to D, D2 scale search space estimate.

The problem can be solved with one of the squares in-between.

AA !LF1mmWigHQ ID: d23b7f July 20, 2019, 9:40 p.m. No.9750   🗄️.is 🔗kun   >>9751

>>9749

>Apply to D, D2 scale search space estimate.

This line is vague. Are we looking for sqrt(d) and sqrt(i) between d[t] in (e,1) and (f,1)? Are we using the already-calculated D/D2/C/C2 and fitting sqrt(d) and sqrt(i) into it somehow?

AA !LF1mmWigHQ ID: d23b7f July 20, 2019, 9:45 p.m. No.9754   🗄️.is 🔗kun   >>9755

>>9753

So we're treating d as c and i as c and finding their respective d' between d'[t]? That's the only way I can think to interpret what you're saying so hopefully that's right.

AA !LF1mmWigHQ ID: d23b7f July 20, 2019, 10:05 p.m. No.9755   🗄️.is 🔗kun   >>9756

>>9754

Here it is, d and i treated as c, d' between d'[t] in (e',1) and (f',1). Now, generally, when we're trying to reduce a search space, we know what we're searching for. So what are we searching for?

AA !LF1mmWigHQ ID: 968327 July 21, 2019, 5:46 a.m. No.9760   🗄️.is 🔗kun

While I have no idea what to do with the current clues right now, if we've gone from warm to hot with this latest cryptic clue, now might be a good time to summarize what we've done over the last week or so. Below I've listed each concept we've gone over, what we do know, and what we don't know. The don't know part isn't done out of cynicism, just in case it comes across that way. I just think it's useful to know what we haven't learned yet along with what we have, to potentially add a sense of direction to the hints.

>there are horizontal families, and there are squares and triangles in the family of c - don't know what that means or how to use it

>there are cells in (0,n) that make a superposition with X and Y - don't know what X and Y are or how to use superpositions

>there are various different superpositions (d between d[t], c between i[t], n=2, (0,e) and (0,f), (1,1), d between qc's (e',1), etc) which constrain values and limit search spaces - don't know what we're searching for or how to find them from within superpositions

>columns (0,n) and (-1,n) have some kind of relationship with coulombs, which potentially relate them to all of this talk about derivatives and rates of change - no idea how to use any of this or any of its relevance

>we can make the (2c,1,1) sequences have a decreasing b by starting at e=-2c and increasing |e|, and we can potentially use this to find qc from within the grid since one of the sequence multiplies c with primes ending in 01 in binary - I couldn't personally get the sequence to work although one of the others on Discord seems to have, and once we do generate it we don't know how to use qc yet since we don't know how to find v or what to do with vqc once we've found it

>there are recursive/fractal elements in the grid that relate to the tree in some way, (0, |d-e|) and (0, |d-|f||) - don't know what to do with these elements

>we can "square gap", or find further elements related to the d between d[t] elements, by finding the elements where x is equal to (or surrounding) the d values from the original elements, which is meant to have something to do with decreasing search space, and from which we're possibly meant to find squares that we "decompose" into ts - don't know what to do with them, don't know what values to find in what gaps, don't know what we're decomposing or how to do that, don't know what variables act as upper and lower bounds in a search space, and don't know what variables we might be searching for in said search space

>multiplying d by its factors (and possibly e too) can apparently help in the geometry of finding c - don't know how

>apparently recursively finding D and D2 for sqrt(d) and sqrt(i) is useful for reducing search space - don't know how it's useful, the sqrt(i) part is an unknown variable unless there's something we haven't noticed about it yet that makes it directly calculable

>several of these concepts are related to the tree somehow - don't know how they're related yet

If I've gotten any of this wrong or anyone understands any of this better than I do, please do say so.

AA !LF1mmWigHQ ID: b8ea74 July 21, 2019, 5:09 p.m. No.9763   🗄️.is 🔗kun   >>9764

>If hints are explored and you still don’t see it.

Is there anything in particular about all of these hints that we haven't fully explored yet?

AA !LF1mmWigHQ ID: b8ea74 July 21, 2019, 7:46 p.m. No.9765   🗄️.is 🔗kun   >>9766

>>9764

Well, the solution record is one obvious place. If we're talking squares in general, they turn up as c values in (0,n), a and b values in (0,n) where n is twice a square and f values in (e,1). Conceptually, a few things are used to create squares in different situations, such as d, i and j in any given element, so one of them from another element could equal the x+n we're looking for. There's a few ideas.

AA !LF1mmWigHQ ID: b8ea74 July 21, 2019, 7:58 p.m. No.9767   🗄️.is 🔗kun   >>9768

>>9766

I guess x+n could maybe be between two x+n values from different elements in a different cell, if that's why you're referencing D and C. This is just an idea off the top of my head, but given the solution is recursive in some way and just directly finding i or j would be O(1), maybe we find two elements that we know for sure x+n will be between (but we don't know its exact value, just that it's between two values), and then there's something inherent to the same concept applied to d in that it's the same distance between two variables in two elements as it is for c. Maybe that's why looking at a transform of the D and D2 elements is meant to be like "zooming in". That's just an idea based on a few of the concepts I do understand though. I don't know where you're going with this.

Anonymous ID: 88b5c3 July 21, 2019, 8 p.m. No.9768   🗄️.is 🔗kun   >>9769

>>9767

What would be more “in plain sight” than using scaling to narrow things down to a direct calculation?

 

But not the only way to do i🔺.

AA !LF1mmWigHQ ID: b8ea74 July 21, 2019, 8:02 p.m. No.9769   🗄️.is 🔗kun   >>9770

>>9768

More "in plain sight" would be if the exact value of x+n was in the same directly-calculable place for all elements and we just kind of aren't aware of it yet.

PMA !!y5/EVb5KZI ID: 9d9b7b July 21, 2019, 9:25 p.m. No.9772   🗄️.is 🔗kun   >>9773

>>9766

We've narrowed down i ranges using D and C. Can consistently find records in (-1,n) within the gap. (though not sure what to do with these yet).

 

Logical to use similar techniques to narrow down j.

 

Do a[D] and a[C] point to repeatable records in (0,n)?

PMA !!y5/EVb5KZI ID: 9d9b7b July 21, 2019, 9:37 p.m. No.9774   🗄️.is 🔗kun   >>9775

>>9773

lower bound on D.j matches initial starting position of the iterative search.

 

upper bound is the issue.

 

Not seeing a way to limit further than C.j.

PMA !!y5/EVb5KZI ID: 9d9b7b July 21, 2019, 11:24 p.m. No.9778   🗄️.is 🔗kun   >>9787

for i between i[D] and j[C], we're dealing with a subset of semiprimes.

for that same subset, it appears that j can be limited between j[D] and b[D2].

a majority of those can be further limited to a[D2].

 

This applies to all solved Rsa numbers (except Rsa129 again).

AA !LF1mmWigHQ ID: b8ea74 July 22, 2019, 1:06 a.m. No.9792   🗄️.is 🔗kun   >>9793

>>9790

This is another confusing and dense post. Are you saying we should look at all of the c values from the recursive d -x+n elements and compare them to our given c? I kind of already did and didn't see anything. That's the only way I can think to interpret this post.

AA !LF1mmWigHQ ID: b8ea74 July 22, 2019, 1:27 a.m. No.9794   🗄️.is 🔗kun

>>9793

Oh okay great. I have no idea what you're talking about. I'm going to go code something vaguely related to that difference of squares thing.

AA !LF1mmWigHQ ID: b8ea74 July 22, 2019, 4:23 a.m. No.9795   🗄️.is 🔗kun

>>9779

>>9782

Since these clues are kind of hard to interpret, I had a wild stab in the dark. I tried seeing if any representations of x or x+n as the difference of two squares turn up in any of the i and j values in the recursive d -x+n elements, as well as checking if any values that are off by 2d+1 turn up instead. No dice on either.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 456f05 July 22, 2019, 2:26 p.m. No.9800   🗄️.is 🔗kun

>>9799

Uncertainty abounds!

I'm certain of it!

 

Also.. a shadow…

How do ECC and RSA relate, again?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 456f05 July 22, 2019, 5:50 p.m. No.9806   🗄️.is 🔗kun

>>9805

I caught the CMYK reference you're making, but for what reason are you saying it's a hint?

 

Sauce on that?

AA !LF1mmWigHQ ID: b8ea74 July 22, 2019, 6:10 p.m. No.9814   🗄️.is 🔗kun   >>9815

>>9811

I still have no idea what that stuff about intercepting courses means so instead I'll ask, what does this stuff about values close to c have to do with the other clues?

Anonymous ID: 88b5c3 July 22, 2019, 6:35 p.m. No.9815   🗄️.is 🔗kun   >>9830

>>9814

You are very close.

 

https://www.researchgate.net/publication/325552720_Influence_of_Divisor-ratio_to_Distribution_of_Semiprime's_Divisor

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 456f05 July 23, 2019, 12:37 a.m. No.9826   🗄️.is 🔗kun   >>9827

>>9822

The fun part is… even though that's obviously nonsense since the President can't eat food that wasn't prepared on site, #DroolAtPresidentialHotPockets, this site comes up when you search for it.

 

http://showercapblog.com/secret-shameful-sex-kinks-pence-nunes/

 

"If nothing else, I’m grateful to these Days of Drumpf for giving us all final permission to simply laugh like cartoon hyenas whenever the “Christian” right makes a play for the moral high ground. “Man of faith.” Fuck right off."

 

Context? What I been sayin'.

Pence be suspect AF.

Anonymous ID: d6ad6a July 23, 2019, 12:59 a.m. No.9827   🗄️.is 🔗kun   >>9828

>>9826

I know a guy like the author of that blog. He's technically an intelligent person, or at least used to be, but he's just completely lost his shit in the last few years much like the blog's author. The irony is that to protest purported assholishness, they become much greater and even more unapologetic assholes themselves. There's also a singular confidence in their pronouncements, which is also ironic since their predictions are much worse than you would otherwise get from pure chance. I kind of feel bad for this person, really.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 456f05 July 23, 2019, 1:09 a.m. No.9828   🗄️.is 🔗kun   >>9829

>>9827

Meh.

If the intention's right, they can get past that part, eventually.

 

Otherwise, everyone just thinks I'm jaded.

<;3=

AA !LF1mmWigHQ ID: b8ea74 July 23, 2019, 4:57 a.m. No.9830   🗄️.is 🔗kun   >>9831

>>9815

This would be so much easier to read if they just went through an example with some numbers. Maybe I recognize a lot of the words but I feel like I need a PhD to understand any of the symbols and diagrams. When this is all over I'm going to go over everything we've done and put together an explanation in the most basic English I possibly can so that anyone can understand it.

Anonymous ID: 1b6b6e July 23, 2019, 11:12 a.m. No.9831   🗄️.is 🔗kun

>>9830

That would be very nice of you. Once Miller-Rabin is coded into the grid, if it happens before it’s solved, it will be the simplest way to understand number theory in existence.

Anonymous ID: 1b6b6e July 23, 2019, 11:33 a.m. No.9832   🗄️.is 🔗kun   >>9838

I can help code this. This algorithm uses the Riemann hypothesis and it uses modular functions which ECC also makes heavy use of.

Anonymous ID: 1b6b6e July 23, 2019, 11:45 a.m. No.9834   🗄️.is 🔗kun

Complex roots of unity in the Miller-Rabin algorithm have consequence for the factors of the number that is being primality tested. Possible to formulate in terms of the grid?

Name ID: 51aee3 July 23, 2019, 2:44 p.m. No.9835   🗄️.is 🔗kun   >>9836

>>9590

I'd love some more thoughts on this. I suppose there might.

 

I looked at t=1 for all e's one can create from c (when you change d) and just saw the repeated underlying square pattern that exists (for all factors) that I posted about in thread 15. I haven't made much headway from it, but I do think it is interesting.

 

It shows how numbers move across e's in a square pattern (I think the square was equal to the square of the factor?), but on the diagonal.

Anonymous ID: 1b6b6e July 23, 2019, 3:12 p.m. No.9837   🗄️.is 🔗kun   >>9850

But unless you find something special about calculating possible rows, (think I saw a variant of Pascal’s triangle related to it), it won’t be the easiest way to decrease the search space

Anonymous ID: 65536f July 23, 2019, 7:46 p.m. No.9842   🗄️.is 🔗kun   >>9843

>>9841

Matthew 18:20

nice!

 

..and lest we forget..

>>14092

Those who thought they could launder money through BitCoin.

Those who thought they could hide money through BitCoin.

They will have nowhere to turn.

The virtual quantum computer for this process is called The End for a reason.

Anonymous ID: 1b6b6e July 23, 2019, 11:04 p.m. No.9847   🗄️.is 🔗kun   >>9848

>>9846

1 John 4:10

 

Deterministic Miller-Rabin test that relies on two hypotheses

Riemann hypothesis and a hypothesis that it is sufficient to test prime witnesses only to determine primality accurately

pastebin.com/wszQwN01

Anonymous ID: 1b6b6e July 23, 2019, 11:19 p.m. No.9849   🗄️.is 🔗kun

>>9848

Was only testing it on rsa primes and semiprimes when I uploaded this, apparently my natural logarithm function becomes too inaccurate when the numbers get vqc example sized

 

I’ll implement some precomputed witnesses tomorrow to change that

A temporary fix to make small examples accurate is change the variable m in ln from (3*(p/2)) to 8

Anonymous ID: e2b8e0 July 24, 2019, 12:01 a.m. No.9850   🗄️.is 🔗kun   >>9851

>>9837

Right, I posted some about a variation of it in this thread >>9585.

 

I wrote a quick function to enumerate the first row, computing the gcd with c and then a loop to iterate over multiples of c (primes from column 1). You clearly see that the factors exists in different columns of e indexes by polite numbers, but that was when I realized I had already seen this pattern in >>8973 and I was just looking at a part of that.

 

Once you add more factors to c the sub-factors occurs several more times, in different orders.

 

Another though I had about the pattern in >>8973 was that, since the squares are rotated on the diagonal maybe we can shift the e's (to straighten out the squares). Like, move the "e-line" to the diagonal. Then the squares would become straight.

Anonymous ID: e2b8e0 July 24, 2019, 12:07 a.m. No.9851   🗄️.is 🔗kun   >>9852

>>9850

> Right, I posted some about a variation of it in this thread >>9585.

 

I suppose it would make sense that there exists other variations as well.

 

A thing I though of a few days ago was the idea of "virtual columns". It's not really a virtual column, but the fact that every single column can be created by enumerating other columns and combining them.

 

Take (3, 1). We have two equations we use which is either 2tt - (half of e) or 2t(t+1) - (half of e). But in the "virtual column" you create (3, 1) by merging other columns. I got the idea when I was looking at the change in the columns when following a specific c through the a[t]'s.

Anonymous ID: 5966d9 July 24, 2019, 8:16 a.m. No.9857   🗄️.is 🔗kun   >>9858

>>9856

By assuming that you can factor in the same time complexity as it takes to generate an RSA sized semiprime

 

(The time complexity of multiplying two numbers together)

 

Look up Newton root finding. It’s how the newer square root method works and can be used to calculate other methods provided you have a way of refining your estimate.

 

Root finding factorization?

Closer than you’d think to solve.

 

Dividing by 2 until odd (tree) is also used in Miller-Rabin to decompose a value into powers of 2 and a remainder.

 

Philippians 1:8-11

Anonymous ID: 1a8a58 July 24, 2019, 8:21 a.m. No.9858   🗄️.is 🔗kun   >>9860

>>9857

 

I'll look into it. I did see that the Miller-Rabin had the divide by 2 until odd and I was reminded of our recursive tree.

 

But I can't seem to see how it would be connected with the "virtual columns" (if it is).

Anonymous ID: e4f913 July 24, 2019, 8:29 a.m. No.9859   🗄️.is 🔗kun   >>9880

>>9855

Take the tree cells and find where d is between

Use i, j and c of those elements as an estimate of the nontrivial factorization of c. Put them together for the tree cells and you will see a pattern. For the values that are divided by two, take note of the power of 2 you removed. Think of it like a cell where every value is an estimate and you use c to make the estimate progressively better (log n time, same as our integer sqrt function).

 

The results can give you enough information to calculate where the right n for c must be.

Anonymous ID: e4f913 July 24, 2019, 8:44 a.m. No.9860   🗄️.is 🔗kun

>>9858

Primality tests are REALLY close to being turned into factorization algorithms.

You can already insert gcd into Miller Rabin to get a factor it it’s composite

Anonymous ID: e4f913 July 24, 2019, 8:49 a.m. No.9861   🗄️.is 🔗kun   >>9862 >>9864 >>9867

It can be done with error values and estimate refining just like the extensions to the prime counting function under the Riemann hypothesis

 

I can help explore the hints and say what type of piece they are.

The VQC is from our creator.

Anonymous ID: 1a8a58 July 24, 2019, 9:19 a.m. No.9862   🗄️.is 🔗kun   >>9863

>>9861

While I appreciate that, I guess my problem is I can't find any good questions right now. Odd how that works, huh.

 

Essentially, it's a square root factorizing algorithm. But we're missing something, the nontrivial transform.

Anonymous ID: 1a8a58 July 24, 2019, 9:25 a.m. No.9863   🗄️.is 🔗kun

>>9862

The first derivative of the grid is defined by >>9718?

 

So the second derivative would be a repeat of the above, but with the new values. Then rinse and repeat for more derivatives (if needed).

Name ID: 1a8a58 July 24, 2019, 9:37 a.m. No.9864   🗄️.is 🔗kun   >>9865

>>9861

Another thing I noticed in the Miller-Rabin test is the list of primes needed to check against for a set of number.

 

Which makes me think of q. Which also makes me think vqc =vector q, c? That we make a list of primes from column 1 (q) which we then use in a modification of the Miller-Rabin test against c to determine if it is a prime or not (and to get a factor).

 

Since the Miller-Rabin constructs a tree (although not directly, but abstractly) we'll do the same, but the end result will be similar. In the end we won't need to construct the tree, but for now it is a helpful method of learning. How far off am I? I can easily be way, way off.

Anonymous ID: e4f913 July 24, 2019, 10:10 a.m. No.9865   🗄️.is 🔗kun

>>9864

They’re exponentiated to create the result of a root of unity being squared

If you took a witness for compositeness and figured out how to calculate the (complex) root of unity, it would give the factors of c

Not the simplest formulation of the problem, though

Anonymous ID: d6ad6a July 24, 2019, 10:16 a.m. No.9866   🗄️.is 🔗kun   >>9868

Human beings were a mistake. This is such a typical human endeavor; a lumbering clusterfuck of naivety, deceit, and non-forthrightness. It seems we are in a different age now, one of post-truth and non-plain speaking. In many ways, and hardly exclusive to this particular forum, I am profoundly disappointed by all of it. Fate it seems has brought us together merely to jerk each other off and hint at a changed world but nothing more? Maybe someday beings will succeed at this kind of thing but they will not be human, they will be whatever post-humans call themselves. I love all you /vqc/ faggots, though, this thought is simply me waking up on a Wed, seeing such momentum in the last few weeks here, but feeling somewhere deep down that it is all for naught. I hope I'm wrong.

Anonymous ID: 8f99ff July 24, 2019, 11:02 a.m. No.9867   🗄️.is 🔗kun   >>9869

>>9861

If you could help with some hints there are two I'm wondering about.

 

What has a[p + 1 - t] have to do with the solution (or a solution?) and the other hint about

 

at the correct a[t], d[t] - d = a(n-1). How would those fit in with a solution?

Anonymous ID: 111380 July 24, 2019, 1:02 p.m. No.9868   🗄️.is 🔗kun

>>9866

Hope is for things that are not seen, for who hopes for what he sees?

Blessed are those who have believed without seeing.

This is real love: not that we loved God, but God loved us.

 

Be quiet, and I will teach you wisdom.

Job 33

Anonymous ID: b72e24 July 24, 2019, 2:16 p.m. No.9869   🗄️.is 🔗kun   >>0040 >>9878 >>9879 >>9923

>>9867

d[e, 1, t] - d = a[-f, 1, t]

 

(-f) : n <-(e) : n+d

(-f) : n <-(e) : n+a+x

 

(e-(2x+1)) : n+a <-(e) : n+a+x

(e-(2n+1)) : d <-(e) : n+d

(e-(2a+1)) : n+x <-(e) : n+a+x

 

There are lots of other superpositions that you don’t have to calculate to use the existence of, since they all stem from e. Construct a root finding algorithm like the square root one. You have been given a 0 estimate.

 

Progress like so:

{e:n0 (from -f):d0:x0:a0:b0} -{e:n1:d1:x1:a1:b1} -> … -> {e:n:d:x:a:b}

 

This is the exact same thing as the n0 triangle bases concept and the intercept course concept. They are different ways of doing the same thing, which is progressing toward the correct answer by starting with a decent estimate and calculating error values to progress (in a complexity that is like binary search) to the right answer. Imagine the movements like one of those casino machines picking each value to the right answer one by one.

That link those concepts for you?

 

P.S. the p of t can be used as a way to calculate an error value (amount to change the estimates by)

Anonymous ID: 28f322 July 24, 2019, 3:21 p.m. No.9872   🗄️.is 🔗kun

I will never promise a date with no fulfillment of it.

That is how the pied pipers lead you in circles.

Sheep no more.

Anonymous ID: 28f322 July 24, 2019, 3:28 p.m. No.9874   🗄️.is 🔗kun

Trust Christ actively.

HE WILL NEVER LEAVE YOU NOR FORSAKE YOU.

Behold, I stand at the door and knock. If anyone hears my voice and opens the door, I will come in to him and eat with him, and he with me.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 456f05 July 25, 2019, 12:34 a.m. No.9881   🗄️.is 🔗kun   >>9882 >>9921

>>9878

https://en.wikipedia.org/wiki/Superposition_calculus

 

The superposition calculus is a calculus for reasoning in equational first-order logic. It was developed in the early 1990s and combines concepts from first-order resolution with ordering-based equality handling as developed in the context of (unfailing) Knuth–Bendix completion. It can be seen as a generalization of either resolution (to equational logic) or unfailing completion (to full clausal logic). As most first-order calculi, superposition tries to show the unsatisfiability of a set of first-order clauses, i.e. it performs proofs by refutation. Superposition is refutation-complete—given unlimited resources and a fair derivation strategy, from any unsatisfiable clause set a contradiction will eventually be derived.

As of 2007, most of the (state-of-the-art) theorem provers for first-order logic are based on superposition (e.g. the E equational theorem prover), although only a few implement the pure calculus.

 

https://en.wikipedia.org/wiki/Euclidean_geometry#Methods_of_proof

 

Euclid often used proof by contradiction. Euclidean geometry also allows the method of superposition, in which a figure is transferred to another point in space. For example, proposition I.4, side-angle-side congruence of triangles, is proved by moving one of the two triangles so that one of its sides coincides with the other triangle's equal side, and then proving that the other sides coincide as well. Some modern treatments add a sixth postulate, the rigidity of the triangle, which can be used as an alternative to superposition.

AA !LF1mmWigHQ ID: 774bd8 July 25, 2019, 12:51 a.m. No.9885   🗄️.is 🔗kun   >>9887

>>9884

2na, but anyway. I'm not really seeing any unifying patterns in this i/j difference stuff. It seems like usually the numbers decrease but not always. If i and j are estimates, they're never going to be correct, since c isn't always going to be in (0,n), and even then if you do this with a square you're going to be looking at different n values. So are you saying whatever we're meant to find to calculate the distance between our estimate i and estimate j has something to do with xx+0=2na? Because that's very cryptic.

Anonymous ID: 88b5c3 July 25, 2019, 12:51 a.m. No.9886   🗄️.is 🔗kun

Ironically solving it doesn’t require you to find some beautiful mathematical pattern that makes everything clear.

Just estimate and correction.

But there is one.

AA !LF1mmWigHQ ID: 774bd8 July 25, 2019, 12:55 a.m. No.9888   🗄️.is 🔗kun

>>9887

Run the same recursive algorithm on each individual c estimate? Use i^2-xx or j^2-xx? I don't know what you're trying to say.

Anonymous ID: 88b5c3 July 25, 2019, 12:56 a.m. No.9889   🗄️.is 🔗kun   >>9890 >>9893

Three trees

 

Repeated sqrt and remainder

Repeated inverse triangle and remainder

Repeated decomposition into powers of 2 and remainder

 

All the same concept

AA !LF1mmWigHQ ID: 774bd8 July 25, 2019, 1:02 a.m. No.9892   🗄️.is 🔗kun   >>9894

>>9891

>Root finding square root function

Are you saying there's a different "root" that isn't a square root? I know you're all saying that you're trying to be more helpful than Chris but this is starting to get Chris-tier, at least in the lucidity of the clues.

Anonymous ID: 88b5c3 July 25, 2019, 1:09 a.m. No.9894   🗄️.is 🔗kun   >>9896 >>9899

>>9892

The root of a function gives you it’s “zero”

When you take away the root from c, you’re left with..

 

If e is a remainder and n is also a remainder and when roots are taken away remainders are left, are e and n the same type of thing? Does away that other “root” from c give you this nontrivial “remainder?”

AA !LF1mmWigHQ ID: 774bd8 July 25, 2019, 1:17 a.m. No.9896   🗄️.is 🔗kun   >>9897

>>9894

>>9895

>root of a function

>zero as a noun related to c

>e is basically n somehow

>n and d+n are somehow related to the root of david element

Like I'm not trying to be an asshole here but can you see it from my perspective? I have no idea what to do with literally anything you just said.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 456f05 July 25, 2019, 1:26 a.m. No.9901   🗄️.is 🔗kun

>>9897

Root and Offspring…

Sounds a lot like Alpha and Omega, but vectors.

 

Root ← i → Offspring

 

Adam 2.0?

Imagination?

AA !LF1mmWigHQ ID: 774bd8 July 25, 2019, 1:39 a.m. No.9903   🗄️.is 🔗kun   >>9905

>>9897

Surely when you say my "binary search code" you aren't talking about this? It wasn't based on anything to do with the grid at all. Plus I'd be surprised if anyone remembered that all that much. What are you referring to?

AA !LF1mmWigHQ ID: 774bd8 July 25, 2019, 12:26 p.m. No.9908   🗄️.is 🔗kun

>>9905

Okay if you even remember me making that code then you'd probably also remember that it didn't work very well. I'll have to remind myself of how it actually worked. But my point back then was that we were looking for a log(n) algorithm, and binary search is a log(n) algorithm. Is there anything specific to my code that was actually on the right track, or is it just the fact that it's log(n)?

AA !LF1mmWigHQ ID: a4a6ca July 25, 2019, 5:04 p.m. No.9910   🗄️.is 🔗kun   >>9911

>>9905

>>9906

I looked over my binary search code. It basically just guesses. In order for binary search to be a solution there needs to be a known relationship that we can compare an unknown's guess value to and know whether our guess is higher or lower than what it should be. That relationship doesn't appear to exist, or at least nobody found one. It was also mentioned earlier in this thread that it's log(n) because you take the square root until you can't anymore. That might be the same time complexity as binary search but it's a completely different algorithm. If the Root of David thing you're referring to is the element {0, 2d, 3d, 2d, d, 9d}, and that somehow relates to binary search, you're going to have to get significantly less vague if we're going to have any way of using this information.

Anonymous ID: 88b5c3 July 25, 2019, 6:53 p.m. No.9911   🗄️.is 🔗kun

>>9910

There have been a lot of posts about where a solution value predictably scales to be close to.

Operations that take remainders.

If it was a snake it would bite you.

Anonymous ID: 88b5c3 July 25, 2019, 8:24 p.m. No.9914   🗄️.is 🔗kun

>>9913

You are right.

I was wrong.

No one lives without God, admittedly or not, so there is no such thing as self-sufficiency. It’s God-sufficiency.

 

Will progress to full disclosure of the code.

 

Proverbs 12:1

Anonymous ID: d6ad6a July 26, 2019, 12:29 a.m. No.9920   🗄️.is 🔗kun   >>9921 >>9922

>>9919

I just kind of had a 'vision' and I want to run this by you quickly to see if I just glimpsed a part of it all. So all you really know at first is you have a large symmetric L of 'blocks' one unit thick. This is because that's what 1 * c fundamentally looks like.

 

So let's call that L an object. Let's use a tangible example of 7 * 19 = 133. So that object would be {12:56:11:10:1:133}. This represents 67² - 66², but also bakes in an opinion. This opinion is "that's aka 11² + 12".

 

But an alternate opinion on this is "that's aka 12² - 11". This point of view is represented by {-11:55:12:11:1:133}. Now I thought before that these were two different "things".

 

THEY'RE NOT. These are literally the same EXACT object, viewed from a different angle, like zooming around it or something. So basically when you're fucking around with a e, -f you're not actually moving or anything, you're kind of just gathering information by reading all of the signs that point at the object in question.

 

So basically what it boils down to here is the e and -f "signature" (skipping for the moment the details) of the 1 * c matches the a * b because in yet another dimension the a * b ACTUALLY IS 1 * c. 13² - 6² = 67² - 66² Same fucking shit. The thing is that in some dimensional view, there are just two spikes sticking out because these are a product of two primes.

 

So we have a L of unit 1 at 67 and we have a L of thickness 7 at 13. So we have to "ride" down the diagonal and efficiently find 13 and then 6.

Name ID: 8f99ff July 26, 2019, 1 a.m. No.9921   🗄️.is 🔗kun   >>9922

>>9920

This is my take on it too. Which then complements >>9881. We can use superpositions because they are the same object, just from different perspectives.

 

We're only given one perspective and what we're trying to do is to warp it to an equal, but different perspective so we can see the the whole object.

Name ID: 8f99ff July 26, 2019, 1:25 a.m. No.9922   🗄️.is 🔗kun   >>9925

>>9920

>>9921

If you think of it in terms of complex numbers, then we can think of it as we're missing the lateral piece (also known as the imaginary part).

 

If you think of d+n as the real part and x+n as the lateral part, then our c is missing a piece. Take ((d+n) + (x+n)i)*( (d+n) + (x+n)i) and you will get c + lateral i. But our starting point is just c. We're missing the lateral part of our number, meaning we only hold a piece of the information. By finding d+n, x+n we have enough information to recreate the lateral part.

 

I thought it might be something like the above a while ago and did some naive testing. The only complex square roots of c + ki (for some k 0) are …. (d+n) + (x+n)i (with non-exhausting testing so I might wrong).

 

For a c with more than two factors we will have multiple valid (d+n, x+n) pairs. One for each factor (including 1, c).

AA !LF1mmWigHQ ID: a4a6ca July 26, 2019, 3:26 a.m. No.9923   🗄️.is 🔗kun   >>9927

>>9915

>>9916

>>9917

>>9918

>>9919

Some analysis.

 

You start off by mentioning that this is used to adjust the squares. These squares are the i^2 and j^2 estimates from the (0,|d-e|) and (0,|d-|f||) cells, in the elements where d is between d[t]. e and f are columns unique to c, but they don't include column 0 for numbers other than squares. Also, each (0,n) element will have a different f. This means whatever elements you're referring to where we look at na-(n-1)a and a[t+n0] aren't directly found from any of the (0,n) concepts we've looked at. Further evidence of this comes from the variables we know when we find the elements in (0,|d-e|) and (0,|d-|f||). If we use the x values we find from e and f, there will only ever be valid elements in (0,1) (where (0,n) na transforms are found) for even x values, but it could be either parity. Plus the element we'd find from d, e and f would be invalid most of the time anyway, so its na tranform also wouldn't exist (i.e n*a wouldn't show up as a[t] in (0,1)). If we used the elements d is between, we would already know the a and n values, so we wouldn't need to find factors by subtracting (n-1)a from na. We could just use the a and n values we already had. So you wouldn't explain it like that if that was the case (I also tried it out in my code and didn't see anything, since it was just variables that were already within the elements, like I said).

 

So whatever elements these clues refer to are obviously specifically based around our c's (e,1) and (f,1), both since you explained it like that and because it wouldn't work otherwise. But if these n values "converge to n from c", then that would make the i^2 and j^2 estimates from (0,|d-e|) and (0,|d-|f||) irrelevant, so the link between these concepts is missing. There is also no mention of what n0 is specifically, although it can be inferred that the concept is related to finding an estimate n and doing something repeatedly to get a better estimate (one that converges to the unknown n we're looking for). If this n0 is the same n0 from the triangle estimate, it doesn't give us enough information to find specific elements, since we need either a or x to complete the xx+e=2na equation. That makes me think it's from >>9869 here, where n0 from the triangle base estimation is mentioned not as the same number but as a similar concept (it being the first guess n for further transforms). But no information in that post is given to explain how to get from an n0 element to an n1 element and then to n2 and so on. If these clues are based on >>9869 then it seems from my perspective like it could go in two directions: either going from n0 to n1 to n2 etc changes the e and f values making them lower/closer to 0 logarithmically and ends up on a pair of elements where the n values is the same as the unknown n we're looking for, or it stays within the same (e,1) (f,1) pair and somehow logarithmically iterates through t values. Either way I can't find a link to the (0,n) tree cells.

Anonymous ID: d6ad6a July 26, 2019, 4:24 a.m. No.9925   🗄️.is 🔗kun

>>9922

Moblefagging now, my id may change. Thanks for that, it does seem apropos and food for digestion over time.

 

One thing I have been thinking about as well was the OG 5 * 29 = 145 example. I used to hate that fucking thing because e = 1 and when I work through anything I hate values of 0, 1, and -1 because they tend to lead you to believe in things that don't extrapolate well. However, there were some interesting takeaways for me from it.

 

If you view the 145 as a 12 * 12 square plus a remainder 'e' zit of one, and starting at the leftmost and bottommost block you remove 1, then 3, then 5, then 7, then 9… or put another way in perfect 1, 4, 9, 16, 25… squares, you know at some point you will have removed a perfect square from the inside and exactly created one on the outside, and you now have a perfect difference of squares.

 

The 5 * 29 example is interesting because it demonstrates an upper bound. You remove literally ever single square from the inside and it stacks up perfectly around the outside. You could not go further if you wanted to because you have nothing left. It's literally a demonstration of the upper bound on 'n'.

 

When removing interior blocks and stacking them around the top and right sides, you get 'nd' on each side and n² in the corner. To upper bound n, you can simply divide c by 2d and check that e can't fill the leftover n². Here, d = 12, 2d = 24, c = 145, 145 / 24 = 6 remainder 1. That would leave a 36 unit n², which remainder 1 plus e of 1 cannot fill. So 1<= n <= 5, and is in this case 5, whereas with many small examples it'll simply be 1 or 2, and in big examples some anonymously large value that does not really invite thought.

 

This is probably widely known, but it was new to me when I began revisiting the topic a few days ago after being pleasantly surprised you motherfuckers were still here.

Anonymous ID: 88b5c3 July 26, 2019, 10:45 a.m. No.9927   🗄️.is 🔗kun   >>9928 >>9948

>>9923

>either going from n0 to n1 to n2 etc changes the e and f values making them lower/closer to 0 logarithmically and ends up on a pair of elements where the n values is the same as the unknown n we're looking for,

Oo that’s brilliant

AA !LF1mmWigHQ ID: a4a6ca July 26, 2019, 3:58 p.m. No.9928   🗄️.is 🔗kun   >>9929

>>9927

It was a logical conclusion to be drawn. The only way I can think to look for something like that right now would be to look at every possible pairing of e/f columns and check all of their na-(n-1)a values, but that seems extremely inefficient and impossible for bigger numbers.

Anonymous ID: bfa756 July 26, 2019, 5:56 p.m. No.9929   🗄️.is 🔗kun   >>9931 >>9939 >>9948

>>9928

But you have a device that can traverse impossible search spaces. Since you’re dealing with factors of na, those will be rows. I would try seeing if the rows can adjust i0 to the correct i.

AA !LF1mmWigHQ ID: a4a6ca July 26, 2019, 6:18 p.m. No.9931   🗄️.is 🔗kun

>>9929

There hasn't been any indication to look at a specific row, and there hasn't been any link made between factors of na and i value adjustments (or what "adjustment" actually means in this context). The earliest row we know will exist in both (e,1) and (f,1) and the cells between them is where (f,1)'s t values start, so I'll look in that row. If it's about i values and not na values, we could just look for the correct d value (the unknown d from the na element), since i=d+n, and n=1. So that sounds like it could be done with math just based around the whole (e+1)/2 thing for d and a values in (e,1). I'll have a look I guess.

AA !LF1mmWigHQ ID: a4a6ca July 26, 2019, 8:15 p.m. No.9932   🗄️.is 🔗kun   >>9933 >>9934 >>9948 >>9950

I looked for the elements in (e,1) and (f,1) where i[t] is equal to either of the i values from the a[t]=na and a[t]=(n-1)a elements. I also compared this to the tree cells and their x values (only really the first two generated cells).

 

Given the a[t]=c sequence concepts, we know that there are only two elements within row 1 where the a values will be equal at the same t (and therefore also only two where the d values are equal at the same t, since d[t]=(e-1,1)'s a[t]). Also since we're in row n=1, i=d+1. So there will always be only one other element in row 1 aside from the na element at the same t where the i value is the same. If we find this element for na and the same thing for (n-1)a, the e and f values add to the same value as the n values in the tree cells, as well as the difference between the x values in the tree cells. So I found a link. In the first example, 53+52 and 71+34 both add to 105, which is also 118-13. In the second example, 7+8 and 39+-54 both add to 15/-15 (obviously one will be negative but this pattern still holds), which is also 93-78. When na and (n-1)a are one t value apart, e+f and n+n will have a difference of 4. These two patterns hold for all of the examples I've looked at.

 

This doesn't seem to be a pattern that allows us to directly calculate anything, but at the very least it is a link between i[t] and the tree cells.

AA !LF1mmWigHQ ID: a4a6ca July 26, 2019, 8:28 p.m. No.9933   🗄️.is 🔗kun   >>9934

>>9932

Just noticed also that (na's e)-(na i[t]=i's e) is equal to ((n-1)a's e)-((n-1)a i[t]=i's e). In that first example, 118-71=47 and 34–13=47. They're the same distance from their starting places. Same with the second example. 78-39=39 and -93–54=-39. So if we start from (e,1) and (f,1) and move the same distance towards (0,1) from each one until we reach two cells where the absolute value of e-f is equal to (|d-e|)+(|d-|f||), the two cells we land on will contain within them the only other two elements in row one where elements exist at the same t as our na and (n-1)a values with the same i values. We'd still have to find that t, but maybe there's something about sequence patterns like d[t]-d that would allow us to do that.

AA !LF1mmWigHQ ID: a4a6ca July 26, 2019, 8:35 p.m. No.9934   🗄️.is 🔗kun

>>9933

>>9932

Okay forget about pretty much everything I just said. Every pair of cells the same distance from (e,1) and (f,1) inwards will add to the same value as the tree cell n values added together.

 

(1,1) and (6,1), 1+6=7

(2,1) and (5,1), 2+5=7

(3,1) and (4,1), 3+4=7

 

There's still only one element in row one aside from na/(n-1)a where i[t]=i at the same t, but the rest of what I said would appear to be redundant.

Anonymous ID: d10e68 July 27, 2019, 3:06 p.m. No.9940   🗄️.is 🔗kun

>>9939

Search spaces don’t have to be searched.

Think about it like taking limits to solve an integral. The result is something finite that technically can only come from an infinite amount of operations (area under a continuous curve) (Riemann sums). You’re doing math with math.

AA !LF1mmWigHQ ID: a4a6ca July 27, 2019, 8:12 p.m. No.9942   🗄️.is 🔗kun

One or two of you mentioned that you were writing code that was meant to help us in some way several days ago. Is that still happening?

AA !LF1mmWigHQ ID: a4a6ca July 27, 2019, 10:43 p.m. No.9948   🗄️.is 🔗kun

>>9932

I've been looking into this more but it doesn't seem to lead to a solution. Every two cells there's a value with the same i (at different ts), so there's no way to predict which specific one has the i value at the correct t. So I've explored everything I can think to explore in relation to these hints

>>9929

>>9927

>>9915 to >>9919

and not found anything. Has anyone else been looking at anything related to these posts?

AA !LF1mmWigHQ ID: b8ea74 July 28, 2019, 8:51 p.m. No.9952   🗄️.is 🔗kun   >>9953 >>9961

Just a thought. If we're traversing (e,1) and (f,1) inwards in O(log n) then it might be useful to see which cells contain n as a factor somewhere in each of the (sqrt,1) cells, and if not starting from e and f maybe starting from other numbers as the e and f values logarithmically inwards. If it's meant to happen for every number then you'll continually find it in the same place (parity-permitting), and maybe there'll be a pattern to the t values as well. But then again, this pattern is meant to be used to adjust the estimate i^2-j^2 values. But it's also meant to converge to n. More confusion, as usual. I might look into this later, but I felt like posting my thoughts here since there haven't been any major clues in a few days.

Anonymous ID: edb3e0 July 28, 2019, 9:03 p.m. No.9953   🗄️.is 🔗kun   >>9955

>>9952

That's a good idea. I was looking at some polynomials they use for the GNFS and it dawned on me just how useful it is to represent c as a sum of powers. So when you factor d and e, it's meant to decompose the dd+e equation into (factors of dd) + (factors of e).

Anonymous ID: edb3e0 July 28, 2019, 9:07 p.m. No.9954   🗄️.is 🔗kun   >>9955

And decomposing dd+e further into squares is definitely leading towards calculating the i^2 - j^2 equation.

 

I believe we're making a lot of progress.

Anonymous ID: edb3e0 July 28, 2019, 9:49 p.m. No.9957   🗄️.is 🔗kun

I see a pattern in them and I'm not crazy. Something that scales with c in a way that's hidden from the algebra.

 

Chris said the remainder tree resembles the solution. Decomposing values into powers of 2 and triangles resemble the remainder tree. And if all 3 of those can solve, it must be something they all share that makes them able to do that. Plus, you can probably use a simple equation to transform each term of each sequence into the other. The grid has those sequences.

 

Maybe it's just something obvious, like if you make everything tiny then that makes the search space tiny too.

Anonymous ID: edb3e0 July 28, 2019, 9:51 p.m. No.9958   🗄️.is 🔗kun

That's what I tried to do with the infinite triangle sequences. If you remember, the end result was a number that would get pretty close to (X-x)/2 of c for some reason.

Anonymous ID: edb3e0 July 28, 2019, 9:53 p.m. No.9959   🗄️.is 🔗kun

It was easy to make the search space small enough to count on one hand, but you had to be able to traverse back up the remainders.

 

Maybe there's something that can tell us what the next remainder is

Anonymous ID: edb3e0 July 29, 2019, 1:02 a.m. No.9960   🗄️.is 🔗kun

Perhaps if we can create like a phantom c that is easy to factor but also has a self-similarity to the large one? Seems like a way to check an answer

AA !LF1mmWigHQ ID: 968327 July 29, 2019, 9:45 p.m. No.9965   🗄️.is 🔗kun   >>9966

>>9964

>>9962

If you take a[t]-a[t] from (e,1) and (f,1), you'll get a sequence starting from d and decreasing by 2 each time t increases by 1 (pic related - ignore the weird t values for the invalid elements at the start of the (f,1) sequence, the a and b values are what they should be). If you move e and f inwards by 2 each, you get the same sequence starting from d-2, and so on. In order to find a, n or the correct t, you'd have to iterate d/2 (which is more steps than GNFS). That's why it isn't useful.

AA !LF1mmWigHQ ID: 968327 July 29, 2019, 10 p.m. No.9967   🗄️.is 🔗kun

>>9966

You don't know how many times you have to iterate. I meant that the search space is d/2, not that a is t+d from the start of the cells.

Anonymous ID: 8f99ff July 30, 2019, 8:44 a.m. No.9968   🗄️.is 🔗kun   >>9969

>>9801

Was there ever an underlying pattern in the triangles? Like a specific tiling pattern, or a system for organizing the parts of the different values we had?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 456f05 July 30, 2019, 2:54 p.m. No.9970   🗄️.is 🔗kun

-swirls the quantum whirlpool-

Maybe something'll habben at some point soon…

Would be nice.

PMA !!y5/EVb5KZI ID: 9d9b7b July 30, 2019, 11:10 p.m. No.9979   🗄️.is 🔗kun   >>9980

There is a connection between roots of unity and being able to navigate through multiple sequences in (e,n).

 

First pic attached shows valid records in (1,901) for t between 1 and 1802. The sequence column shows related records that can be navigated through using the current t=[mp + t] and t=[mp + offset - t] formulas. Columns p and px, respectively. In this case, where m*p = n = 901.

 

Roots of unity can be calculated as (i^2) % c = 1, for i between 1 and c-1.

 

For c901, roots_of_unity(901) = [1,52,849,900].

 

Second pic attached shows t values and differences required to navigate. Each of these differences can be expressed in terms of one of the roots of unity (52) and/or the d values (30), as indicated on the far right.

 

These t differences can be used to solve via gcd:

 

gcd( 106, 901 ) = 53

gcd( 136, 901 ) = 17

 

And each root of unity between 1 and c-1, less 1, has either a or b as a factor.

 

prime_factors( 52 - 1 ) = 3 x 17

prime_factors( 849 - 1 ) = 2^4 x 53

Anonymous ID: edb3e0 July 30, 2019, 11:12 p.m. No.9980   🗄️.is 🔗kun

>>9977

Which can also be thought of as different stops.

 

>>9979

Excellent work. Miller Rabin would probably be greatly enhanced by beiing put into the grid.

Anonymous ID: 8f99ff July 31, 2019, 2:11 a.m. No.9984   🗄️.is 🔗kun   >>0006 >>9985

>>9973

>>9971

Image 1 in the first post and image 2 in the second both appears to use the same c as an example.

 

I suppose this is because in both examples different initial estimations were used and adjusted for, meaning two different paths to the same number was taken.

Anonymous ID: 8f99ff July 31, 2019, 3:37 a.m. No.9985   🗄️.is 🔗kun   >>9986

>>9984

Just in case anyone wonders, the first starts with estimates from (e, 1) and the second (-f, 1). I'm trying to work my way through these examples, but I'm not entirely sure I know what the hell I'm doing. Trying to think a bit.

Anonymous ID: 8f99ff July 31, 2019, 7:46 a.m. No.9987   🗄️.is 🔗kun   >>9988

>>9971

I think your calculator is malfunctioning. In your conversion from the squares to the grid, c1 doesn't look correct.

 

Shouldn't it be: {-3734:-25:43:72:-29:65}?

Anonymous ID: 40e4f2 July 31, 2019, 1:53 p.m. No.9989   🗄️.is 🔗kun   >>0006 >>9990

>>9988

I'm still trying. I'm using the examples posted, but I haven't found anything consistent.

 

I'm trying to spittball a few ideas. I get the Newtons method idea, using derivatives to find errors and correcting them, but I'm still not 100% on the what the derivatives are. I see the pattern used with regards to D and D2, but I lack an understanding as to why those are considered the derivatives / rate of change.

 

All in all, I'm trying. I like where it is heading, but I still have work to do.

Anonymous ID: edb3e0 July 31, 2019, 2:06 p.m. No.9990   🗄️.is 🔗kun   >>0005 >>0006

>>9989

I think an overarching theme here is you can use a variety of estimates if you're actively making them relate themselves to c.

 

Kind of like the original tree being a "big picture" estimate program and the tree cells being a more "zoomed in" version since they use more information from the grid.

AA !LF1mmWigHQ ID: 968327 Aug. 1, 2019, 12:20 a.m. No.10004   🗄️.is 🔗kun

>>9973

I cross-checked the i and j values in this sequence against all known variables for the c values at each level. It looks like the i estimate potentially comes from the previous c's D2_f i value (the one that's one off here might be a parity thing), unless that's a coincidence (this sequence is the same c as most of the others, so they're going to be different sequences with potentially different variables from different places used as the consequent i and j values, so we'd need more test data to know for sure). j seems more ambiguous though, since there's nothing unifying in the j values for each element in the sequence. What this seems to mean is that we don't have enough information just from the sequences that were posted on /qresearch/ yesterday. If we're meant to use the same variables each time, the "adjustments" seem to be on those variables, rather than the adjustments being the sequence of the variables themselves.

Anonymous ID: 40e4f2 Aug. 1, 2019, 12:36 a.m. No.10005   🗄️.is 🔗kun

>>9990

Right, there might be multiple functions to use for estimates. I'm working now on trying to figure out some. I get that p of t would work, so I'll try to figure out how it would fit in. I figure at best I'll end up with a few bad estimator functions. At least, I assume I'll need more than one, since we got three variables to update (d, n, x).

Anonymous ID: 40e4f2 Aug. 1, 2019, 2:45 a.m. No.10006   🗄️.is 🔗kun

>>9989

>>9988

I've been trying to figure it out without using the tree. I figure that there must be an underlying pattern, but I also figure it might not be necessary to include a tree for that particular solution.

 

It seems like it's just adjustments being made, but I'm still not understanding the grid.

 

>>9990

Got any suggestions for paths to explore? I figure it might be related to the different patterns we know of. "Not Chris, Maybe Still Chris" previously mentioned that we could estimate it by estimating n0 from -f and then calculating the error rate from p of t. I'm wondering now if the patterns in >>9984 is just that.

 

I was thinking it went something like:

 

Estimate i, estimate j. Compute then look for error in the new respective e and -f columns. But if we do that, we'll be moving aimlessly, right? So I'm thinking. Use estimate i, estimate j and compute the cell. Use that cell to try and calculate an error rate based on the original e and -f columns.

 

Although easier said than done.

Anonymous ID: 40e4f2 Aug. 1, 2019, 6:15 a.m. No.10022   🗄️.is 🔗kun   >>0023 >>0024 >>0025

>>10015

I'm a big boy and need my energy.

 

Remember all those squares we made with triangles?

 

I though about those when I was thinking about a potential fractal.

 

One though I had, but I only got to play with it a little bit was using the a and b from (e, 1).

 

Take (1, 1). Every a is 1 + 4 triangles. Arrange them like a square, the middle unit is the 1 and four symmetrical triangles are spread around it. Then add the b which is again 1 + 4 triangles. Ignore the 1 for now then add the b's triangle in an alternating series to the triangle.

 

You'll end up with a square with "leftovers" (which will be the base of the triangles from b. Or if you wish, ignore the square but look at the triangle.

 

Not entirely sure how that is a fractal, but that was where my brain was heading to a long time ago.

Anonymous ID: 40e4f2 Aug. 1, 2019, 6:16 a.m. No.10023   🗄️.is 🔗kun   >>0024

>>10022

I'd draw it up, but since mah 'puter died I lost all my code that drew the triangles. The only thing I have now is some minor grid code left that I rewrote.

Anonymous ID: 40e4f2 Aug. 1, 2019, 6:18 a.m. No.10026   🗄️.is 🔗kun

>>10025

Note though, since they are seperated by n in (e, 1) the difference between the triangle bases would be quite a lot larger, might be pretty still though.

Anonymous ID: edb3e0 Aug. 1, 2019, 6:59 a.m. No.10035   🗄️.is 🔗kun   >>0037

Seems to me there is a reason n0 and the f mod stuff were given as "from arbitrary guesses" ie, differing guesses can still be adjusted to reach the factorization.

Anonymous ID: 40e4f2 Aug. 1, 2019, 7 a.m. No.10036   🗄️.is 🔗kun

>>10032

But do we stick to our original e and -f, or do we carry on with the new ones? I mean like I said, then we would just aimlessly be moving about wouldn't we?

 

So our new estimates would be depending on the e and -f, not necessarily origin c? I can see that. If I was given just an e and -f, I could of course generate the correct c (and it should be possible to then solve it too, but that's a bit beyond my knowledge right now). Since (e, -f) are a unique pair.

 

So we'll bounce the estimates back and forth (e, -f) (or in combination) until we converge on the answer.

Anonymous ID: edb3e0 Aug. 1, 2019, 7:25 a.m. No.10041   🗄️.is 🔗kun   >>0042

>>10039

>>10040

 

Indeed, the remainder tree, this triangle estimation stuff, the "intercept course" post, and the new square estimation posts are all saying the same thing in different ways.

Anonymous ID: 40e4f2 Aug. 1, 2019, 7:27 a.m. No.10042   🗄️.is 🔗kun   >>0043

>>10041

Newton's method, gradient descent I guess in the grand scheme it doesn't matter. It's about the error function. It's about how you correct your squares to get closer to the target.

 

Once you get that part, maybe you'll c i[t].

Anonymous ID: 40e4f2 Aug. 1, 2019, 9:26 a.m. No.10046   🗄️.is 🔗kun   >>0047 >>0048

>>10045

You know, there are some older hints (suggestions?) that Chris came with. I don't remember the exact details of it, it was a couple of threads ago. I'll see if I can find it.

 

It was about taking the columns from -f to 0 to create a matrix and then transpose them. Like I said, I don't remember much. There were some people who didn't get it. I figured, instead of transposing it on a variable size, treat each variable as an element and transpose the elements themselves.

 

Either way, I think it is one of those hints we never really dug into. Who knows, maybe it is relevant now?

Anonymous ID: 40e4f2 Aug. 1, 2019, 9:33 a.m. No.10047   🗄️.is 🔗kun

>>10046

Was from thread #14. Didn't see much, just three posts about it, no one seemed to follow up. Neither poster nor us. Guess it might not have been relevant.

Anonymous ID: edb3e0 Aug. 2, 2019, 11:55 p.m. No.10055   🗄️.is 🔗kun

Here's a sample ECC key on curve nistp192. I threw away the private key.

(3869035562762729475071350689725296476315431951714456492369, 5721900778986487358975428077347210421279593804122562127173)

Anonymous ID: edb3e0 Aug. 3, 2019, 12:05 a.m. No.10056   🗄️.is 🔗kun

Efficient factoring instantly breaks discrete logs, though. So it'll be two ivory towers falling at once.

 

Discrete logs are just factoring but with the same factor to some power.

Anonymous ID: 40e4f2 Aug. 3, 2019, 6:24 a.m. No.10063   🗄️.is 🔗kun

>>10061

Still working on it. Since I lost my 'puter I've had to write a lot of code from scratch. I finished up with some base grid code for big integers and I started thinking about the estimation function. I've been thinking with regards to Newton's method, but I'm still not grasping >>9718, nor have I grasped the patterns in updating i, j in >>9972 >>9973 and >>9974.

 

At this point I'm pretty ready to start testing with some numbers so I'll try a few estimate functions and see what sticks.

Anonymous ID: d5d547 Aug. 3, 2019, 4:54 p.m. No.10067   🗄️.is 🔗kun

>>9114

Hey, your ghostbin pastes are dead.

The use of The End as a lookup table involves calculating values that c and its respective values are between.

Anonymous ID: 40e4f2 Aug. 4, 2019, 5:36 a.m. No.10071   🗄️.is 🔗kun   >>0073 >>0074

>>9971

What confuses me about this example is that the correct n = 1 and d exists in (37, 1). So if we are to lookup the d in (e, 1) and if the d exists, we've found the record. So why keep iterating? Which makes me think, this example didn't lookup the elements where d was between d[t]. It uses another way of doing the estimations. c it maybe?

Anonymous ID: 40e4f2 Aug. 4, 2019, 5:44 a.m. No.10073   🗄️.is 🔗kun

>>10071

But then, the initial estimates for i and j is based on the values i (e, 1) where d is between the elements. So I'm finding this example a bit confusing.

Anonymous ID: f7c4f1 Aug. 4, 2019, 9:51 a.m. No.10074   🗄️.is 🔗kun

>>10071

That example does. It uses elements D, D2, R, R2, and E. It’s demonstrating the estimation concept of clarifying the picture using d.

Anonymous ID: f7c4f1 Aug. 4, 2019, 9:57 a.m. No.10075   🗄️.is 🔗kun

“Where is d between” (in a and b in places but in d[t] primarily) combined with an informed n estimate can be used as a boolean algebra to calculate what the correct value of n must be for c. That’s Modus Tollens.

Anonymous ID: f7c4f1 Aug. 4, 2019, 10:13 a.m. No.10076   🗄️.is 🔗kun

Think of it like this. You can make the correct factorization element of c an abstraction and use its existence to calculate it by finding it in the gaps using d since at that element d will be d from c.

 

(e, 1)

(e, n0) big step

(e, n1) little step

(e, n2) big step

(e, n3) little step

 

Jackpot

Anonymous ID: edb3e0 Aug. 4, 2019, 11:15 a.m. No.10077   🗄️.is 🔗kun

Didn't realize I'd have to bake this soon. Gives me a chance to fix the codebase. I am feeling the acceleration.

Anonymous ID: edb3e0 Aug. 4, 2019, 1:39 p.m. No.10080   🗄️.is 🔗kun   >>0082

I have said these things to you in figures of speech. The hour when I will no longer speak to you in figures of speech but will tell you plainly about the Father is now.

 

I and the Father are one. No one who denies the Son has the Father. Whoever confesses the Son has the Father also. Whoever confesses that Jesus is the Son of God, God abides in him, and he in God.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 456f05 Aug. 4, 2019, 2:13 p.m. No.10081   🗄️.is 🔗kun   >>0082

Except that God never wanted a middle man…

Jesus is just a band aid for those who'd been beaten down and told they were worthless and born with negative karma I mean sin.

 

We're all God's Children.

I'll just talk to It/Them directly if it's all the same to you.

 

And besides… Jesus and Buddha said the same thing.

#DontWorshipMeBro

 

It's about the Message.

Not the Messenger.

Remember the whole "Fall of Lucifer" thing?

<;3=

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 456f05 Aug. 4, 2019, 3:56 p.m. No.10082   🗄️.is 🔗kun

>>10080

>>10081

Fuck it.

Let us pray.

 

Dear Jeeezuuuuus,

Can you pass on a message for me?

God's always pissed off and refuses to talk to me; even since before I was born.

Truly, everyone who says God loves and cares about me is a liar, and that's why He ignores me and my prayers/thoughts/critiques/desires/passive-aggressive-notes.

So how does this work? I hear money's involved. Tithe… Tax… Fee… Protection Money… Whatever it's called.

I pay off your middle man, who pays you off, so you can grease some wheels?

And which Church/Bank do you use? I don't wanna pay off the wrong guy.

 

THANKS!

~Tops

AA !LF1mmWigHQ ID: 55d502 Nov. 2, 2019, 4:20 p.m. No.10087   🗄️.is 🔗kun   >>0088

Welcome back

 

All of our posts etc have been migrated over, several of us are still around daily, and all the mod functionality seems to still work. Hopefully those of you who were giving us clues around when 8ch went down three months ago end up here again and we can continue. Tor posting is enabled, by the way, if anyone has any trouble with the clearnet version while bugs are still being worked out. I don't know if you can upload pictures with Tor though (I know one of you clue-posting people was posting catgirls). Some of us (or potentially all of us) are having trouble viewing all of the posts on /qresearch/, but posts seem to work fine here, so if any of you were planning to post clues or anything like that it would probably be a better idea to post them here so that they're actually viewable by anyone.

AA !LF1mmWigHQ ID: 04c005 Nov. 11, 2019, 8:38 p.m. No.10091   🗄️.is 🔗kun   >>0092

>>10089

>>10090

Now all that's left is for the math people to come back to 8canoes and for /qresearch/ to update threads properly so we can bait them back over here. Might be a good time to start posting the findings of the last few months (although I know there haven't been a ton, and I also know none of it was me).

VQC !!2jfXytjJ9k ID: 55d502 Nov. 16, 2019, 10:57 a.m. No.10096   🗄️.is 🔗kun   >>0097

>>10095

A note on the product of the sum of two squares and the sum of two squares.

You may remember this was mentioned before.

The product of the sum of two squares and the sum of two squares is closed under multiplication.

What does that means in terms of the grid The End?

Well, the columns of the grid for e>0 and where e is in the set {1,4,9,16,..,..} consist of products that are the sum of two squares only in those columns. This is what is meant by by the sum two squares being closed under multiplication.

The simplest example of this is column 1 (where e=1), though technically column 0 is an example as well and an important one, since it is a special case, since the product of the sum of two squares and the sum of two squares, each consist of the sum of 0 squared and a square.

 

Back to column e=1, all the values of n (and therefore all the factors in a[t] (and b) at [1,1] are the sum of two squares.

 

What can be observed is that the contents of a column can be controlled.

If I have a product c, and it is the product of a the sum of two squares and a number that is not the sum of two squares, if I multiply it by the lowest primes that are each the sum of two squares, I control the density of numbers as they appear in the resulting column that the new product appears in.

Anonymous ID: 55d502 Nov. 16, 2019, 11:07 a.m. No.10097   🗄️.is 🔗kun   >>0098

>>10096

If I multiply a number, c, by 5,13,41,61,etc; the column (e) that results starts to bleed information, the more numbers that are multiplied to it.

Also, the value at -f in the other half of the grid, combined with the information bled out of the column (e) for the new product combines to bleed out possibilities for the factors in c even quicker.

It was noted previously that using low unique primes gives enough information to determine the prime factors of c in O(log q) by the time the size of the product of primes equals half the length (in bits) of c or equivalently, the square root of c.

What difference does choosing those low primes to be the sum of two squares?

It makes it easier to demonstrate why it happens.

The possibilities of factors in c bleed out more slowly in terms of the size of the product of small primes but for the number of low primes required, the difference is less than O(log q), what this means is that the size of the product of low primes creating numbers that are larger than just choosing the lowest primes (regardless of being the sum of two squares), since the sum of two squares that are prime are less frequent than all numbers that are prime, the increase in size (or length of those number) is less than that which would dominate the calculation thus keeping Big Oh at O(log q) overall.

VQC !!2jfXytjJ9k ID: 55d502 Nov. 16, 2019, 11:22 a.m. No.10099   🗄️.is 🔗kun

>>10098

If I count the primes, as those far greater than me once did, and studied them well, that counting function reveals that the number of primes up to a given integer,z, is roughly log(z), where log is ln, i.e. the natural logarithm.

 

https://en.wikipedia.org/wiki/Prime-counting_function

 

What the first equations on the wiki (I know we don't like wiki but it is hard for them to politicize math - politicise maths) is saying is that as an integer x tends to infinity (or gets larger), if I count all the primes up to x and divide that number by the log (natural) of x, the result tends towards one, the higher the value of x is chosen. It's a very unsexy equation and what it is saying is when you count primes up to a number, the larger the number you pick ,it gets closer to being a fairly good way of stating the natural log of that number.

 

What does this have to do with the price of fish?

 

EVERYTHING.

AA !LF1mmWigHQ ID: 55d502 Nov. 16, 2019, 1:44 p.m. No.10102   🗄️.is 🔗kun

>>10101

I don't but it was blank name field person who pointed it out when you dropped all of the passwords, and they're still around. We could just carry on under the assumption that it is you until we figure out which one it was.

AA !LF1mmWigHQ ID: 55d502 Nov. 16, 2019, 2:27 p.m. No.10103   🗄️.is 🔗kun

>>10101

The tripcode you used here >>8046 doesn't correspond to any of the passwords you dropped apparently. If you don't remember that password or if it was a typo of one of the passwords you dropped then I suppose you could prove you're VQC by giving us deeper-level clues. As far as I can tell, your clues today have been mostly based around things we already knew (given you told us to use q primes that ended in 01 in binary, which are sums of two squares, and odd sums of squares only turn up in square es and are only divisible by other odd sums of squares), with the added detail of O(log n) coming from the number of q primes needed being up to d. One way you could prove yourself would be to drop everything now lmoa

VQC !!2jfXytjJ9k ID: 55d502 Nov. 21, 2019, 11:13 p.m. No.10108   🗄️.is 🔗kun   >>0109

>>10106

The best way to provide retrospective proof will include the factors of all RSA Numbers.

Once it becomes public that PGP is fundamentally flawed could precipitate that, unless one of you makes the connection.

Q just alluded to this a short time ago.

I will continue to add to what you have.

It appears close, then the horizon shifts.

The control of the narrative is falling apart in the UK and that can mean only one thing.

Timing is everything.

And Everything starts with an E.>>10106

VQC !!2jfXytjJ9k ID: 55d502 Nov. 21, 2019, 11:28 p.m. No.10109   🗄️.is 🔗kun   >>0113

>>10108

What if there is another way to display the grid The End?

This was alluded to towards the beginning.

What is one piece of information that The End loses, straight off the bat?

When we take d, the square root of c, e is a maximum of 2d.

What happens to The End if this information comes back?

What is the best way to display this?

VQC !!2jfXytjJ9k ID: 55d502 Nov. 22, 2019, 12:32 a.m. No.10110   🗄️.is 🔗kun

int e1 = e;

int d1 = d;

while (d1 0)

{

string end = string.Format("{0}:{1}:{2}:{3}:{4}:{5}:{6}", e1, n, d1, dpn, x, a, b);

if (!TheEnd.Keys.Contains(e1)) TheEnd[e1] = new Dictionary<int, List<string>>();

if (!TheEnd[e1].Keys.Contains(n)) TheEnd[e1][n] = new List<string>();

TheEnd[e1][n].Add(end);

e1 = e1 + d1 + d1 + 1;

d1–;

}

VQC !!2jfXytjJ9k ID: 55d502 Nov. 22, 2019, 12:43 a.m. No.10111   🗄️.is 🔗kun

//Wider context

 

public static void CreateDictionary()

{

for(int i=0;i<256;i++)

{

for (int j = 0; j < i; j++)

{

#region add entries

int c = (ii) - (jj);

int d = (int)Math.Sqrt(c);

int e = c - (d * d);

int twodp1 = (2 * d) + 1;

int f = twodp1 - e;

int a = i - j;

int x = d - a;

int n = i-d;

int b = i + x + n;

int dpn = d+n;

int e1 = e;

int d1 = d;

while (d1 0)

{

string end = string.Format("{0}:{1}:{2}:{3}:{4}:{5}:{6}", e1, n, d1, dpn, x, a, b);

if (!TheEnd.Keys.Contains(e1)) TheEnd[e1] = new Dictionary<int, List<string>>();

if (!TheEnd[e1].Keys.Contains(n)) TheEnd[e1][n] = new List<string>();

TheEnd[e1][n].Add(end);

e1 = e1 + d1 + d1 + 1;

d1–;

}

#endregion

}

}

}

VQC !!2jfXytjJ9k ID: 55d502 Nov. 22, 2019, 12:45 a.m. No.10112   🗄️.is 🔗kun

//Reminder of output

 

public static void Output(string path)

{

TextWriter tw = (TextWriter)File.CreateText(path);

for (int y = 0; y < 100; y++)

{

for (int z = 0; z < 6; z++)

{

for (int x = 0; x < 33; x++)

{

string entry = "";

if (TheEnd.ContainsKey(x) && TheEnd[x].ContainsKey(y) && TheEnd[x][y].Count z)

{

entry = TheEnd[x][y][z];

}

tw.Write(entry.PadRight(28));

}

tw.WriteLine();

}

tw.WriteLine();

}

tw.Close();

}>>10111

AA !LF1mmWigHQ ID: 55d502 Nov. 22, 2019, 12:49 a.m. No.10113   🗄️.is 🔗kun   >>0114 >>0115

>>10109

>When we take d, the square root of c, e is a maximum of 2d.

>What happens to The End if this information comes back?

That doesn't make any sense. What information has been taken away based on e having a maximum value?

VQC !!2jfXytjJ9k ID: 55d502 Nov. 22, 2019, 1:07 a.m. No.10114   🗄️.is 🔗kun

>>10113

e is often a lot larger than x, therefore you will not ever see e and x on the grid as you might want, and the relations that would be missed because of this. In this context, where x becomes a value d somewhere on the grid (before it would appear like this).

 

Remember

 

(d^2 + e) MOD a = 0

(x^2 + e) = 2na MOD a = 0

 

If, the entry for 2na is now an element in the grid (where x is a d value and it's e is a lot larger than the original d), information is returned to the grid and a new set of patterns are added.

VQC !!2jfXytjJ9k ID: 55d502 Nov. 22, 2019, 1:10 a.m. No.10115   🗄️.is 🔗kun   >>0116 >>0118

>>10113

The set of patterns in the grid can be created or spawned from a small amount of information.

Add information and the patterns behind a new lookup start to appear.

And when that information is added to the -f side of the grid…

AA !LF1mmWigHQ ID: 55d502 Nov. 22, 2019, 2:08 a.m. No.10116   🗄️.is 🔗kun   >>0161

>>10115

Okay, so you're showing us another pattern. Where are you going with it? You have an idea in your head of how this disclosure is meant to happen, but you need some perspective (this post isn't just coming from me, the others told me to talk to you about this). Given the length of this and the lack of results based on your direction and methodology, you've lost pretty much all of our patience and faith, including that of people who claim to know how the grid works who have given us clues while you've been away (see >>9378 and >>9379, as well as the rest of this thread where multiple people claiming to know how the grid works gave us more insight than you because they didn't think you were doing your job here properly). You leave for several months at a time. You tell us we have enough information to solve and then a year later tell us you've been keeping vital information from us for 'when the time is right'. You tell us you're going to give us everything tomorrow or next week or on a specific date and then disappear for several months again. Even if this was all a result of some plan we don't know about, it is the exact opposite of the way that you lead a team of people. It'll be the two year anniversary of your "my hand is forced" post next week. There are only four regulars left including myself, and two or three more who come and go.

 

You can show us another pattern if you want, but we have barely any patience for this kind of thing anymore. If you're waiting for a specific degree of disclosure in the rest of the world, like FISA declassification or something, just tell us. You've contradicted yourself in relation to whether this has a timeline behind it several times. If you read the thread and look at the screenshots, you'll see that other people have actually given us far more detailed and direct hints in the many months that you've been away. If you really want to do this, work with us. Don't just turn up out of the blue every few months and give us another directionless pattern. Give us transparency, and give us direction. If you don't, you can't expect us to stick around forever.

Anonymous ID: d6ad6a Nov. 23, 2019, 2:52 a.m. No.10117   🗄️.is 🔗kun   >>0118 >>0158

HOLY FUCKING SHIT YOU GUYS ARE STILL HERE! I missed this place and you fags a lot. Good to see you all again from the clearweb. Godspeed and looks like things are happening here, I’ll try to catch up.

AA !LF1mmWigHQ ID: 55d502 Nov. 23, 2019, 3:02 a.m. No.10118   🗄️.is 🔗kun   >>0119 >>0121

>>10117

To some extent anyway, given we're in some kind of loose strike thing. Were you someone who had a name here at any point?

 

>>10115

Another thing I forgot to mention in my last post. Read >>9913

If you take the Bible as seriously as you claim, fucking us around for two years is a direct contradiction.

Anonymous ID: d6ad6a Nov. 23, 2019, 3:13 a.m. No.10119   🗄️.is 🔗kun   >>0120

>>10118

nah, just some random fag who got sucked in right around when you made the board. I’ve jumped in here and there and tried to figure shit out.

 

I just want to say I have immensely enjoyed just about every single post you have ever written, the way you just bang your head against shit and then just saying how you see it with no real filter. I’m usually banging my head to less avail than you. I’ve learned some interesting things along the way and have read more about crypto in general, the theories behind it, how it’s used, and I realize it’s sometimes pretty comfy to be a fucking pleb but give things the good old college try with zero expectations of usefulness or really even genuine understanding. It’s kind of fun and almost the ultimate luxury, to not have to work in a coal mine or something and to be able to fuck around with these idiotic things in my spare time.

AA !LF1mmWigHQ ID: 55d502 Nov. 23, 2019, 3:27 a.m. No.10120   🗄️.is 🔗kun   >>0121

>>10119

Haha, I guess thanks or whatever. The "no real filter" thing makes the others think I'm a grumpy person but I'm really not. I just usually know when something someone claims is a clue isn't actually a productive use of anyone's time for the most part by now. We're still at the point of knowing a bunch of grid patterns but having absolutely no idea what the algorithm is or how they relate, so if you're trying to refresh your memory hopefully the Grid Patterns thread will be useful for you (there are summaries at the bottom of the thread), and there's some new-ish stuff in this thread in the screenshots from /qresearch/.

Scab ID: 55d502 Nov. 23, 2019, 2:35 p.m. No.10121   🗄️.is 🔗kun   >>0122 >>0123 >>0158

>>10118

> loose strike thing.

Speak for yourself.

 

>>10120

> The "no real filter" thing makes the others think I'm a grumpy person

You kind of are. You at least come across as someone who acts like they deserve the answer.

 

We're talking about something grand here. This whole thing, the VQC, isn't just to solve integer factorization. It's not really about that. It's about understanding the underlying structure of numbers. It's about understanding what math is. You need to work to learn.

 

Extending the grid with d + n increases the grid by another dimension, does it reveal any patterns? (Hint it does). Here's a pattern for free. (e + nn, n), (e - nn, n). What does this reveal? Can you explain this pattern using terms from the VQC? How is it related to adding d + n into the grid?

AA !LF1mmWigHQ ID: 55d502 Nov. 23, 2019, 3:47 p.m. No.10122   🗄️.is 🔗kun

>>10121

Let me guess, you've been here for like three months? I don't remember ever seeing your name before. What I posted is the opinion of the Discord group, which is made up of people who have been here since the start two years ago and have gone through all of the many let-downs and meaningless tangents. You're more than welcome to disagree, and if you want to work on his latest hint, go right ahead. I think you'll be wasting your time, and so do other people who know how the VQC works who don't trust Chris (as I said, read the thread and look at the screenshots, they've been calling him out on the exact thing I'm calling him out on). But don't let me stop you.

Anonymous ID: 55d502 Nov. 23, 2019, 4:44 p.m. No.10123   🗄️.is 🔗kun   >>0125

>>10121

Shut the fuck up.

He speaks for all of us who are still here from the beginning.

 

I can pull a hint out of my ass that's way more interesting than these or yours. It's one thing to add information, it's another to insult the patient faith of everyone still remaining with the assumption that promising a date over and over again is the only thing that keeps them around. If you don't have the dignity to speak up or the wisdom to know that you've learned nothing from Chris if you view yourself as a dog eating table scraps, you're wasting your time and will never forge the solution.

Anonymous ID: 55d502 Nov. 23, 2019, 4:51 p.m. No.10124   🗄️.is 🔗kun

A disciple is not above his teacher, nor a servant above his master. It is enough for a disciple to become like his teacher and a slave like his master.

Matthew 10

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 55d502 Nov. 23, 2019, 5:05 p.m. No.10127   🗄️.is 🔗kun   >>0128

Fuck it. I hop on tor.

 

>>10125

You totally came across as grumpy and whiny and didn't actually speak for everyone on all points.

 

But we've talked about this, though I still wanted it on the board since I'd been trying over clearnet for a few days. :P

 

Maximum A just needed to vent as they periodically do.

Moving on.

AA !LF1mmWigHQ ID: 55d502 Nov. 23, 2019, 5:40 p.m. No.10128   🗄️.is 🔗kun

>>10127

I strongly disagree about it being vent (especially compared to the above), but if that's how I come across then there isn't much I can do about that. I think the points I made were factual, if they aren't then I hope someone rebuts them, and all I'm saying is that I think he should address the concerns of the other clue people and the fact that he has continually lied about dates etc. Not playing Discord group ambassador right now. That's all I have to say unless someone wants to point out something I got wrong or unless VQC reads the thread and addresses it all.

Anonymous ID: d6ad6a Nov. 25, 2019, 4:08 a.m. No.10129   🗄️.is 🔗kun   >>0130 >>0133

ATTENTION FAGGOTS

 

IT’S OK TO BE GRUMPY. LIGHTEN THE FUCK UP EVERYBODY!

 

The “problem” (because there is no actual problem) is simply every angry, grumpy, madlad’s expectations. My expectations on the low side are pure entertainment and using my brain for new things, things that actually have helped show me I’m interested in shit I didn’t know about, life is good.

 

My expectations on the medium side of it is if it doesn’t pan out within 5 years we turn it into a straight up fake cult. Meetups w/ beer, kick ass T shirts, maybe we can get a sympathetic local media interview, life is good.

 

My expectations on the high end is eventually we factor a challenge number (within five years, see above) and suddenly shit completely hits the fan and there’s literally archeological-tier discussion about who or what Chris is and who the fuck are the faggots that actually believed his insane shit from the get go, but get this, literally WHILE they were chasing down another Chris+++ known as Q, so early on that literally “who the fuck does that?”

 

Guys, we are who does that, cheer up! We are probably 10 in 5,000,000,000 and we managed to find each other. Even if we never factor RSA-12, my life has been richer for having been here. It has also been richer for fucking around with math, geometry, number theory, and crypto. I’m not telling anyone what to do. I’m just saying, fuck it, we can do anything we want. We could even design and start our own site, platform, etc. I have no idea what the fuck I want, if anything, but there is a slim but real chance we will all do some crazy but dumb shit together, so again life is good.

AA !LF1mmWigHQ ID: 55d502 Nov. 26, 2019, 2:42 a.m. No.10133   🗄️.is 🔗kun   >>0134 >>0135

>>10129

Wholeheartedly agree with everything except there is a (simple) problem (that in hindsight didn't need multiple paragraphs by multiple people to explain): Chris needs to talk about the lack of trust from within his own group and the lies about dates. That has nothing to do with expectations.

Topolanon +++ !!!MjllZTBiOWJiYzA4 ID: 226172 Nov. 27, 2019, 5:30 p.m. No.10144   🗄️.is 🔗kun

>>10143

nice timestamp

And yeah. Re-salt is done did.

 

So I guess now VQChris will have to post something only VQChris can.

Topolanon +++ !!!MjllZTBiOWJiYzA4 ID: 226172 Dec. 1, 2019, 12:56 p.m. No.10145   🗄️.is 🔗kun

While I work on convincing ANYONE TO POST SOMETHING USEFUL TO THE BOARDS…

 

Anyone have any inside scoops on Brexit?

 

I know we're supposed to be distracted by the kiddy diddling and the terrorists (who also diddle kiddies, not surprisingly)…

 

But one would think that those are some of the exact reasons to get away from the EU and other such insidious affiliations…

Anonymous ID: 086afd Dec. 1, 2019, 8:46 p.m. No.10146   🗄️.is 🔗kun   >>0147 >>0148

https://8kun.top/qresearch/res/7409289.html#q7409783 In case you guys missed it… THis just went by in the main bread. Figured you would be intersted! Welcome back.

Topolanon +++ !!!MjllZTBiOWJiYzA4 ID: 226172 Dec. 1, 2019, 10:02 p.m. No.10148   🗄️.is 🔗kun

>>10146

Also…

I could comprehend a scenario where:

P=NP & P≠NP

 

The question is… doooooes that make things easier?

Say you have an abundance of one…

And you just need a single proof of the other.

And they're both right because yes.

Anonymous ID: 98f53c Dec. 2, 2019, 10:40 a.m. No.10155   🗄️.is 🔗kun   >>0156

Clearnet testing testing - hello frens.

>>10154 checked!

Many thanks for the twatter updates, through the months of darkness and even now with comms for projectd - they've been Tippy Tops!

(btwaaaay - think you were still 1yr ahead for the Nov habbenings! They be a coming yet..)

Still mulling over that deposition by PB dropped over our northern border, there be layers to that, possibly information mixed with misinformation, such juicy bits!? Wondering if Omar is being sacrificed so as to spread disinformation?

Topolanon +++ !!!NjEwY2Q4Y2ZkZTNm ID: 226172 Dec. 2, 2019, 10:55 a.m. No.10156   🗄️.is 🔗kun   >>0157

>>10155

And like clockwork… "someone" showed up just as Q did.

 

As for Omar, she sure does seem to be bringing a loooot of attention to a loooot of the things that the Dems would have them not.

Wouldn't surprise me if Sadick Khan has the same ties that she does.

MM !!p2qIipw.fc ID: 98f53c Dec. 2, 2019, 1:27 p.m. No.10158   🗄️.is 🔗kun

>>10157

mmmm, meant >>>qresearch/7413439

a lil' rusty. Oh, and hi! >>10121 Rusty! Nice of you to "cheque-in!" (as required?). Might be amiss here, if so, bitch slap me brah.

>>10117 LH?? Lordy we miss ya!

>>10093 yo brothah! checked! drop another b4 the end of bread w/ the new salt..

Yo V! >>10095 earn it, don't but dont' urn it baby!

  • btwaaaay, not too happy with how (it appears, not 100% sayin') you (all?) decided to take on the persona of someone else in your academic/thought-leader space. 3DP and all, hmm? twatter drop impersonation? kapeeesh??? Been watching correlations long time, 18mo+.

…aside from that, looking forward to some insightful maths drops, that part has been wonderful.

also, is the bunker board up, need to ruffle through ole' papers…

MM !!p2qIipw.fc ID: 98f53c Dec. 2, 2019, 7:52 p.m. No.10160   🗄️.is 🔗kun

>>10095

Ok, that wasn't polite at all, sincere apologies. Shifting timelines are tough for all. Looking forward to continuing the journey.

Anonymous ID: 2419dd Dec. 2, 2019, 9:16 p.m. No.10161   🗄️.is 🔗kun   >>0162

>>10116

Can't say I disagree… AA, I bet you are a decent guy to work with IRL. Seems like its time for a real PROOF… Something irrefutable. Perhaps using blockchain or some other unbreakable existing tool… Something way outside the norm of possibilities with normal tech. Somethign that couldnt be faked.

 

Potus/Q show a 100 dollar pen and 10k thousand dollar watch of which there are very few. Pretty convincing that it is not some stoned teenagers in mom's basement. I will get shouted down for this one but it would prove up the whole program here as real and true.. Start opening abandoned BTC wallets. Nothing big but ones that havent been touched in years and years. The old original miners who forgot their passwords. Lets the regulars here pick a BTC address then deliver a working password to it. That would convince even the most vaunted skeptic and would invigorate the room hugely. There are so few here that it would be very unlikely to actually get out anyway.

 

-Hobo … im too lazy to make a trip code for the reasons AA listed here. I still love it here though

Anonymous ID: 261cb7 Dec. 7, 2019, 2:23 a.m. No.10163   🗄️.is 🔗kun   >>0164 >>0165 >>0166

Everybody reading this post, listen up: if there is no VQC by Jan 1, 2022, we are going to hold a meetup, in the nearest town to AA he feels comfortable with. A few months prior, we will decide on identifying apparel so we can find each other. This should authoritatively happen on /vqc/ but there probably will need to be side channels for people who are afraid. Let’s worry about that at that time.

 

I propose a charter: if no VQC by then, let’s create the world’s best, most widespread, most insidious to kill or block, multi-dimensional, virtual quantum, actually quantum, free speech platform known to man. This 8ch/8kun shit, while being exactly what is needed atm, and laudable, and having my sincerest thanks, prayers, from the bottom of my heart— this will not save us, the us being the you & me present here and reading this. We must take matters into our own hands if, and only if, /vqc/ does not deliver. Because if it does, our opinions are simply meaningless. If it does not deliver, well, it was the precipitating event that brought about 20/5,000,000,000 together for some reason we will collectively discover.

Topolanon +++ !!!NjEwY2Q4Y2ZkZTNm ID: 226172 Dec. 7, 2019, 5:49 p.m. No.10164   🗄️.is 🔗kun

>>10163

The world is not waiting 2 more years for Brexit.

You're insane.

 

I don't think the EU will even make it to 2022.

There's no way we're giving Chris LONGER THAN THE EU WILL EXIST to at least check in with something meaningful.

 

This can only remain a Trivial Pursuit for so long.

Even the lottery has a more frequent positive feedback loop.

 

When I bring this up to managers I know, we just talk about the difficulties of keeping fickle nerds motivated.

Apparently it's a useful skill being able to talk to 'em without explicitly comprehending exactly how everything works…

But that only does anyone any good if something actually comes from this.

 

The Nerds need feedback more than anything.

Legend has it that Chris didn't have this option.

Legend also has it that The Nerds aren't meant to suffer the same fate, let alone seven years of it.

Sure, I guess five years of it would be an improvement of sorts, buuuuuut… aye aye aye.

 

A little reinvigoration would be stellar.

AA !LF1mmWigHQ ID: c80685 Dec. 7, 2019, 6:11 p.m. No.10165   🗄️.is 🔗kun   >>0173

>>10163

Why me? My location is pretty inconvenient. We could all go camping with Hobo, but then that's probably inconvenient too. Also the "reason we will collectively discover" doesn't need to be forced by meeting up at a specific time in a specific place. If anything's truly meant to happen (I think so), it'll happen anyway without any of us necessarily knowing beforehand. It might happen a little bit faster if we can get the attention of those other clue people who were helping us out right before 8ch went down, if you're looking for something productive to do (i.e. go fishing on /qresearch/).

VQC !!PIYQp76geY ID: 61e37a Dec. 8, 2019, 2:33 a.m. No.10167   🗄️.is 🔗kun   >>0168 >>0170 >>0172 >>0184

Focus on this.

-f on the left, e on the right.

[-f,1] and [e,1]

You need to find n for a given c.

[-f,1] contains (n-1)a

[e,1] contains na

The ONLY question you need to answer is:

What determines two values being one apart give -f and e.

Think modular formulas. Famous ones. Famous ONES.

Euler and Fibonacci. Others?

Equal/equivalent to ONE. or MINUS one.

Once you c it, you cannot NOT c it.

Tops has drawn it.

MANY times.

There is more than one spiral. There are families of spirals.

You only need to discover this once.

Then you will have it.

VQC !!PIYQp76geY ID: 61e37a Dec. 8, 2019, 3:09 a.m. No.10169   🗄️.is 🔗kun   >>0170 >>0172

What I said back near the start is this is about working backwards.

One of the approaches out of the solutions to P=NP is the quasi-faith approach. Bear with me.

Let's ASSUME (or have Faith) for now, that the solution to the integer factorisation is O(log q) where q is the length of the binary representation of c in bits. And log q is the natural log of q.

 

The first step (as ALL steps) will grow in size as the length of c increaases, as will ALL the steps.

HOWEVER, the step (and ALL steps) cannot grow in size (of computation) greater than O(log q). That's how Big Oh works.

 

So, EVERY step in the solution MUST obey this rule.

 

In other words, every step must be as hard or simpler than find the root of a large number.

 

The FIRST step is we admit we are powerless over c as it is, and as it stands our life becomes unmanageable in finding a and b from c as it stands. Solution, we find the root of c and the remainder and then the difference between the sum of twice the root and one (giving us f).

 

The second STEP is finding the family that c belongs to. This must not be harder than finding the root of c. This is the step you are stuck on.

 

Look at the density of entries for n under columns 59 and 61 (where e = 59 and/or 61). Prime pairs are often around HIGHLY divisible numbers. But also look further out from the central prime (for pairs, not always perfect pairs). These are families. These grow like fractals along the positive number line in the grid.

 

As the size of c grows, these families grow close to the natural log of the increase. These families grow and exist in a fractal pattern. You can see it in some of the images of the grid as bitmaps.

Anonymous ID: 422b77 Dec. 8, 2019, 7:27 p.m. No.10173   🗄️.is 🔗kun   >>0174 >>0184

>>10165

Hell yeah! Lets go camping! Springtime you are looking at NorCal in the mountains… Watch the snowmelts and catch around 3-5k feet elevation in the Sierra-Nevadas about 2 weeks after snow melts away for fields of wildflowers. Alternatively spring is excellent in the intermountain west from Silver City NM through Moab UT. In the early and mid summer through 4 July the coast range of Oregon is the best. Post 4th of July head to the high country in the Cascade range! There are large resoviors on the headwaters of the Willamette river that are literally empty of people. I have set up there literally for months and never been bothered. August-September you are looking at THe Olympic National Park or Glacier National part or Beartooth Wilderness if you want to go off the beaten path. Gotta stay above 5k feet to beat the heat if you are not right on the water. Post September you gotta get south… and low elevation. Look for citris farms as the place to go. Normally I would say Socal… but you know… so instead the riverbottoms around Jerome Arizona are excellent. … good wine country as well, Boy, I wonder where I am right about now. -Hobo

Topolanon +++ !!!NjEwY2Q4Y2ZkZTNm ID: 226172 Dec. 9, 2019, 12:13 a.m. No.10175   🗄️.is 🔗kun   >>0176 >>0177

>>10172

Hey…

Forget about that for a sec.

 

The stuff that was just placed on the board.

What do you think of it?

 

And then… what if they never meet this demand of yours?

But they get back to the maths…

Can you let it go?

AA !LF1mmWigHQ ID: c80685 Dec. 9, 2019, 12:58 a.m. No.10176   🗄️.is 🔗kun   >>0185

>>10175

The guy has been called out as untrustworthy by multiple people for months and he's ignoring all of it. That brings into question his motivation for posting any new clues. If he would talk with us rather than at us it would benefit everybody including himself, so I don't see why this is something I should "let go". In fact I think it's necessary. It's not some giant thing either. Literally all he has to do is respond and explain. The last time we talked about this on Discord we talked in circles for like an hour and got nowhere, so if we couldn't understand each other then it's probably not going to happen now either.

 

As for the clues, to me it seems like he's repeating himself and being vague again for the most part. "There are values one apart in (f,1) and (e,1) that lead to a solution, and they have something to do with modular forms, p appearing every t+p in (e,1), concepts that haven't been explained (e.g. families and roots), and Biblical metaphors." We knew all of that over a year ago. The thing about e=59 and e=61 could potentially be useful, and I'll have a look into that over the next few days, but I personally think it would be far more useful to get the attention of the clue people who aren't Chris so we can start where we left off before 8ch was taken down.

Anonymous ID: 261cb7 Dec. 9, 2019, 3:51 a.m. No.10177   🗄️.is 🔗kun   >>0184

>>10175

I have a few thoughts. The pic related in the blue dots, we’ve seen this pop up before in plots. This is Euler’s totient function, which is an extremely homosexual way of saying “for a given number, how many numbers smaller than it share only 1 as a factor”. Obviously for primes it would be all of them so if n is your prime the value would be (n - 1), which is what sets the hard linear (ish?) upper edge.

 

Actually, more of interest to me is https://en.wikipedia.org/wiki/Carmichael_function which seems like a simpler and more useful version of above. The second plot is blue on green, and the blue is Carmichael’s and green is Euler’s. It is cool how they both are the same for primes, and they still have this certain look at them, but Carmichael’s kinda “fills in” everywhere Euler’s isn’t. Also:

 

>a^m ≡ 1 (mod n) for every integer a between 1 and n that is coprime to n (i.e., that only shares 1 as a factor)

 

… looks cool and is easy to wrap your head around, mostly.

 

I think my favorite thing I found was https://en.wikipedia.org/wiki/Wilson%27s_theorem. Basically, it related modular factorials (just factorial calculations using wraparound logic) to primes and magically the value on primes also comes out to itself - 1, same for Euler’s totient and Carmichael’s above, but is zero the rest of the time.

 

It’s cool to think about, not sure if this is any use.

Topolanon +++ !!!NjEwY2Q4Y2ZkZTNm ID: 226172 Dec. 10, 2019, 12:16 a.m. No.10178   🗄️.is 🔗kun   >>0184

Nom nom nom nom nom

 

China's Gold-Backed Crypto Currency Set to Blindside US Dollar Claims Keiser

Topolanon +++ !!!NjEwY2Q4Y2ZkZTNm ID: 226172 Dec. 10, 2019, 1:30 p.m. No.10179   🗄️.is 🔗kun

Those muslims and their maths.

I guess Mohammad didn't expressly outlaw it so it's one of the few things allowed to them.

 

RSA Cipher Encryption

Topolanon +++ !!!NjEwY2Q4Y2ZkZTNm ID: 226172 Dec. 10, 2019, 1:32 p.m. No.10180   🗄️.is 🔗kun

#BreadSoTasty

 

Internet Encryption Made Simple: Breaking Down Diffie-Hellman Key Exchanges

  • TLS 1.3

Topolanon +++ !!!NjEwY2Q4Y2ZkZTNm ID: 226172 Dec. 10, 2019, 1:45 p.m. No.10181   🗄️.is 🔗kun

One Encryption Standard to Rule Them All! - Computerphile

 

"We look at where the ubiquitous AES came from. Dr Mike Pound introduces the Rijndael algorithm. "

Anonymous ID: 98f53c Dec. 10, 2019, 5:23 p.m. No.10184   🗄️.is 🔗kun

>>10177

Will take a look, and is there a homomorphic hint there? Have been reading a few related topics.

>>10167

>Think modular formulas. Famous ones. Famous ONES.

Good hint, thanks. Have been reviewing these. Reminds me of your missspelled Einstein bit from way back, so looking at Eisenstein but more of a series than formula(?), and trying not to ignore Ramanujan and the others, especially Taylor-Wiles and Breuil-CONRAD-Diamond-Taylor, though modular forms for elliptic curves may be getting a step ahead of ourselves.

Having a Hecke of a time working through it all, but with Riemainn here to The End.

>>10178

>Nom nom nom nom nom

You bread muncher you! Interesting. Wanting to spin up a cool graphic with a KEY and your VQC-caricature on the end replacing the pattern that opens the lock.. Needs more skillz (pics related).

>>10173 awesome, thanks for the tour LH.

Anonymous ID: 2c11d8 Dec. 11, 2019, 7:52 a.m. No.10187   🗄️.is 🔗kun

>>10186

Share similar sentiment. Also willing to ASSUME (have Faith) a bit longer.

 

Not a baker, but did some link checks to help out. Perhaps SEVENTEEN will be epic!

links from bread #16

Validation

Code

C#

BigInteger libraries and test code —— pastebin.com/fiKJ6nLv

  • embeds links to 4 pastebins, each are working

Recursive remainder tree generator —— pastebin.com/ZH9fSWu2

  • dead link

VQC generator —— pastebin.com/XFtcAcrz

  • working

VQC generator w/ Bitmap —— pastebin.com/hMTtJF6E

  • working

 

Java

Real-time VQC —— anonfile.com/TeH6q3d8bd/VQCGUI_v2.7z

  • ok, links to file download

Recursive remainder tree generator —— ghostbin.com/paste/njfcq

  • fails, deplatformed - "You've requested a page on a website (www.ghostbin.com) that is on the Cloudflare network. Cloudflare is currently unable to resolve your requested domain (www.ghostbin.com)."

VQC generator —— pastebin.com/Dgu9aP1h

  • link ok

VQC library —— ghostbin.com/paste/kbf9a

  • dead link, same as ghostbin above

 

Python

VQC generator —— pastebin.com/NZkjtnZL

  • link ok

VQC generator w/ bitmap —— pastebin.com/wEAKaqBp

  • link ok

 

Factorization algorithms —— ghostbin.com/paste/cyjop

  • dead, same as above

Static Java/C# class with all RSA numbers —— pastebin.com/XYFpsDWE

  • ok

Miscellaneous code —— ghostbin.com/paste/xrqme

  • dead, same as above

VQC codebase archive (not comprehensive yet) —— anonfile.com/L3yd6di0n6/archive_7z

  • dead link "The file you are looking for does not exist!"

 

Every VQC map —— anonfile.com/a64765i3n9/maps_7z

  • dead link

VQC 4chan posts —— ghostbin.com/paste/szbfc

  • dead, same as above for ghostbin

 

Did not check the bread archive links.

Anonymous ID: 91695b Dec. 11, 2019, 4:46 p.m. No.10188   🗄️.is 🔗kun   >>0189 >>0190

>>10186

https://www.youtube.com/watch?v=2xT6fcm4V1o

 

We should all take the time to watch this. I picked it up from the alien bread in qresearch. It is somewhat related to here… … "suspend disblief" is stated over and over through this video. Enjoy.

Topolanon +++ !!!NjEwY2Q4Y2ZkZTNm ID: 226172 Dec. 11, 2019, 5:20 p.m. No.10189   🗄️.is 🔗kun

>>10188

It's a good one.

I listened to a while ago.

 

Memewhile, this is what the Yang Gang is up to.

I feel like our time is spent better here.