General RSA thread. The next thread will have a properly formatted OP.
Testing trip
This evening, I'll make a start from the beginning walking through step by step the properties of the grid to reconcile where we have got to and take it to completion over the next week ending this stage by Christmas Eve assuming no site problems.
Everyone happy using this thread?
Thank you to the BO here.
Thank you to all participants.
Let's have some fun.
I thought this was a fucking larp for the longest time, but I see it now. Are we releasing the entire idea of VQC, or just an algorithm for RSA? The fallout of this is going to beโฆbeyond words. Obviously you could have taken credit for this, so I'm guessing you want this credited anonymously?
I'm not smart enough to quickly plot the equations, but I have the key insight that you've clearly obfuscated. Do you want this to be "God's will be done." or can we prepare ourselves for what's about to happen?
Ready to go!
Understandable anon.
This is a Good thing.
The delivery will still take some time as it is important to step through.
This is for everyone, it is bigger than me.
The potential of Virtual Quantum Computers and unlocking the potential of P=NP is vast.
These VQCs exist as mathematical objects.
Note that these can be thought of as part of a Higher Power, since mathematics is Invariant and pre-dates our existence.
Others may want to walk this path so I appreciate your patience.
I have commitments but I begin to have free time towards the end of this week.
As long as this space is maintained as free, it would be Good to complete this beginning by Christmas.
The first Virtual Quantum Computer is the grid that you have seen already.
The second VQC is related but will begin to show why mathematics as a language is not fit for purpose. And also why it took so long to prove Fermat's Last Theorem.
The third VQC we will cover here is related to the Mandelbrot Set.
Questions are important and shall feature heavily.
Again, thanks for your support and patience.
The important point here is that only the ability to add and multiply are required as well as understanding what squares look like.
Finding square roots is also a feature.
Once we move to large numbers, we will let computers do the hard work but we will only use the large numbers to show examples once needed.
First I will repost the original C# code and walkthrough it.
using System;
using System.Collections.Generic;
using System.IO;
namespace IntegerFactorisation
{
public class TheEnd
{
public static string path = "";//PUT YOUR FILE PATH HERE
public static Dictionary<int, Dictionary<int, List<string>>theend = new Dictionary<int, Dictionary<int, List<string>>>();
public static void CreateTheEnd(int i_max = 512, int x_min = 0, int y_min = 0, int x_max = 64, int y_max = 64)
{
for (int i = 0; i < i_max; i++)
{
for (int j = 0; j < i; j++)
{
int a = i - j;
int b = i + j;
int c = a * b;
bool odd = c % 2 == 1;
int d = (int)Math.Sqrt(c);
int e = c - (d * d);
int f = e - ((2 * d) + 1);
int n = i - d;
int x = d - a;
if (!theend.ContainsKey(e)) theend[e] = new Dictionary<int, List<string>>();
if (!theend[e].ContainsKey(n))
{
theend[e][n] = new List<string>();
}
if (!theend.ContainsKey(f)) theend[f] = new Dictionary<int, List<string>>();
if (!theend[f].ContainsKey(n - 1)) theend[f][n - 1] = new List<string>();
string text = "{" + string.Format("{0}:{1}:{2}:{3}:{4}:{5}", e, n, d, x, a, b) + "}";
theend[e][n].Add(text);
text = "{" + string.Format("{0}:{1}:{2}:{3}:{4}:{5}", f, n - 1, d + 1, x + 1, a, b) + "}";
theend[f][n - 1].Add(text);
}
}
}
public static void Output(int i_max = 256, int x_min = -64, int y_min = 0, int x_max = 64, int y_max = 64, int set_size = 12)
{
TextWriter tw = File.CreateText(path + "output.csv");
for (int y = 0; y < y_max; y++)
{
for (int z = 0; z < set_size; z++)
{
for (int x = x_min; x < x_max; x++)
{
if (theend.ContainsKey(x) && theend[x].ContainsKey(y) && theend[x][y].Count z)
{
tw.Write(theend[x][y][z] + ",");
}
else
{
tw.Write(",");
}
}
tw.WriteLine("");
}
}
tw.Close();
}
}
}
The first three lines are below:
using System;
using System.Collections.Generic;
using System.IO;
These three lines state the C# libraries required to run the code.
>using System;
Almost all programs in C# use this library.
>using System.Collections.Generic;
This library is used for the Dictionary object.
The dictionary object stores the grid and each element in each cell of the grid.
>using System.IO;
The System.IO library is used for printing the output of the grid to a file.
The line that has 'namespace' in it is simply a way to group similar classes together.
>namespace IntegerFactorisation
The { and } after the namespace and after the code just act as a holder for the class code.
>public class TheEnd
The class is the definition of an object to hold any code we want related to this project.
The name TheEnd is a name given to our code class. Since the first three letters we use are e, n, and d, it seemed to work. It is also a pun on the importance of 'n'.
I know some of you are already way past this, thanks for bearing with me.
> public static string path = "";//PUT YOUR FILE PATH HERE
This is the definition of a variable used to store the name of the output file. E.g. c:\Somefolder\output.txt
> public static Dictionary<int, Dictionary<int, List<string>>> theend = new Dictionary<int, Dictionary<int, List<string>>>();
The dictionary can be thought of a store like a spreadsheet or grid.
The rest of the code produces The Grid.
The Grid contains cells.
Each cell contains zero or non-zero elements.
We use a list of letters now.
a,b,c,d,e,n,x
There are seven letters.
All represent integers, whole numbers.
All revolve or are central to a whole number or integer c, from one to infinity.
The only values of c we are interested in are the numbers (integers) that are the difference of two squares.
Which numbers are the difference of two squares?
All odd numbers and all even number divisible by four.
What about numbers even numbers that are odd when divided by two?
These numbers are not the difference of two squares. If we divide those numbers by two or multiply them by two, they become numbers that are the difference of two squares.
Why is that important?
All odd numbers are the difference of two squares. This includes all prime numbers and all numbers that are the products of two prime numbers. The products of two prime numbers is at the heart of RSA encryption.
The seven letters we start with are defined, remembering that c is the same in all the definitions and remembering that this applies to every c where c is the difference of two squares:
ab = c : a multiplied by b equals c
dd+e = c : the biggest square with side d you can make from c and the remainder e left over.
(d+n)(d+n)-(x+n)(x+n) = c : c is the difference of two squares
aa + 2ax + 2an = c : c is a square with four rectangles added to it
a+x = d : d is the sum of a and x
b = a + 2x + 2n : b is the sum of a and 2 lots of x and n
These definitions should align with the images at the top.
The program simply goes methodically through all the differences of two squares in order from the square of one minus the square of zero, then the square of two minus the square of zero, then the square of two minus the square of one and so on.
The program stores the results of every difference of two squares in a grid.
The grid has columns and it has rows.
The grid columns are defined by the remainder e. So to find e of any c, you can find the square root of c we call d. Then subtract the square of d from c to get the remainder. The remainder can be 0,1,2,..,etc. We call the remainder e and it is the column header, the remainder of any c goes in the column that is defined by it's remainder e.
The rows are defined by the number you add to d, the square root, to make the large square. Remember, each c is the difference of two squares? The number n is what you add to the square root of c if you were to make the square that is having a smaller square subtracted. This number added to d is called n. n decides which row a number c goes in.
We know that the remainder of any c will always be the same, e.
All numbers c that are the difference of two squares will have at least one number n. Some have more than one value n for a value c, those that have more than one value n for a value c appear in a column more than once but always the same column, since e is always the same.
Prime numbers such as 3,5,7,11,13,.. have only ever one value n.
Imagine any square with side d, then subtract a square with sides (d-1) from it and you will have an odd number. If that odd number is prime, this is the only way you can do this but all odd numbers have this representation.
The square 2 by 2 with one taken will be 3.
The square 3 by 3 with four taken will be 5.
And so on.
The number 15 (c=15) which is not prime plugs in like this:
ab = c
1x15=15 : a=1,b=15,c=15
3x5=15 : a=3, b=5, c=15
dd+e = 15 : d=3, e=6, c=15
(d+n)(d+n)-(x+n)(x+n) = c : d=5, n=1, x=0, n=1
aa + 2ax + 2an = c
9 + 0 + 6 = 15
a+x = d
3+0 = 3
b = a + 2x + 2n
5 = 3 + 0 + 2
The picture here is the output of The Grid.
We will go through examples of the cells and the elements of the cells.
This is a grid cell.
The highlighted one is (0,0)
The values in each element of that cell have the format:
(e,n,d,x,a,b)
The value of c for each cell can be calculated a number of ways but the simplest is to take the last two values, a and b and multiply them:
ab = c
Since the cell is at (0,0) all the values of c in that cell are perfect squares (remainder e is always zero) and there is nothing to a to d to make the largest square. In other words in the cell at (0,0) we have all the squares with a square of size zero subtracted.
(0,0) is a unique cell in some ways.
It is the ONLY cell in row 0. No positive value of e has a cell in row zero.
Highlighted in this image is cell (0,1)
Again, perfect squares because the remainder after taking the largest square is zero.
So, these values of c ALL appear in (0,0) but they also ALL have more than one way to arrange their factors.
E.g. the first element in (0,1):
{0:1:4:2:2:8}
{e:n:d:x:a:b}
This is square with sides of 4.
4x4 = 16 can be arranged as 2x8 = 16
If you take a square 3x3 from a square 5x5, you get this difference of two squares.
The square at (0,1) is the basis of the patterns of all cells on row one.
Notice that the values of b for the first element is the value of a for the next element, the value of b for the second element, is the value of a for the third element.
This pattern exists in every cell in the first row.
All the values of a in this cell, include as a set, all the values of b by the same logic.
So, we can focus on the values of a if we want.
Notice that all the values of a in this cell are also each twice the value of a perfect square.
1+1 = 2
4+4 = 8
9+9 = 18
And so on.
Notice that all the values of d for this cell also follow a pattern:
2x(1x2) = 4
2x(2x3) = 12
2x(3x4) = 24
The image here is cell (1,1)
These values of c that are the elements of this cell share a pattern with (0,1).
Note that all the values of d for (1,1) are identical to the values of a for (0,1)
This cell (1,1) contains as values for a and b the values of two consecutive squares added together.
0+1 = 1
1+4 = 5
4+9 = 13
9+16 = 25
When two of these are multiplied together, the result is a twice a perfect square plus one as a remainder.
Each value of a in cell (1,1) is also the long side of an integer right angled triangle.
Notice that all the values of a at (1,1) are the values of d in (0,1) with one added.
4+1 = 5
12+1 = 13
24+1 = 25
The cell at (2,1) is identical to the cell at (0,1) with one added to each value a,b,d while the values of n and x remain the same.
The cell at (3,1) is identical to the cell at (1,1) with one added to each value a,b,d while the values of n and x remain the same.
In the row n=1 which is also written (e,1), the cell elements repeat this pattern every two cells or 2xn = 2x1.
We will explore that for any row n, the values of a,b and d will increase by one in a cell that is twice n to the right.
If a cell has values at (e,n) then there will be values in a cell at (e+2n,n)
Please explore that pattern.
Let me know if there are any questions at this stage.
Next we will explore the reason that answer the following question:
Why are there gaps?
Why do the values in row one determine all the values in the columns below them?
Let me see if I'm getting this right. This is exactly how the entire grid can be created from (1,1), because the rows have a clearly defined pattern from it and the columns have a clearly defined pattern from the rows.
Would you guys like a voice chat set up on discord? Some guys are baking on Q's posts there and we would love to combine forces/help they said they would be ok with a room for VQC
No, thank you.
Thank you VQC. Following so far.
ofw?
So do you want the board? I didn't get any emails. I've been talking to one of the other anons about this via email who said they'd been talking to you and they said that you wanted the board.
I only see dark timelinesโฆ
Are you certain that this will lead to the light?
People are irrational, I know because I am terrified.
Was the last letter of your sister's name a Y?
It is not the light I fear, but the path out of the darkness. This is bigger than all of us. Some will not be humbled.
Reporting for duty. Will begin studying now.
So this was right.
Assuming the gaps have something to do with size of the squares increasing although not quite gotten this yet
That's called shills.
Thanks VQC, following so far.
>Why are there gaps?
because those are the cells where the midpoint of a and b would not be a whole number. Or, worded another way, the difference between a and b would be an odd number.
>How does that relate to what we are doing?
Fuck if I knowโฆ
There definitely seems to be a pattern to them. [e,2] seems to only have gaps where a and b in the cell above it are both odd. So there seems to be a relation to higher cells. If this thing does have a tree structure, which I am not seeing atm, probably means the cells they relate to could be the end leaves. The gaps also have reoccurring patterns both vertically and horizontally.
Who dat be?
So in (2,1) (4,1) (6,1) etc. I'm seeing the increase in a and b by one.
And in (1,1) (3,1) (5,1) etc. I'm seeing the increase in a and b by one.
How does (0,0) tie in? Is it unique?
Also, at a basic understanding level, why is e (remainder) chosen as row var, and why is n ( d+n= (a+b)/2) our row var? Hmmm. Working to understand. It seems in (0,1) that those squares could be easily done with d, but aren't bc we're looking for diff of squares. If n=0, then e=0, perfect squares. So as n increases, the amount needed to add to d to equal the midpoint of (a+b)/2 increases. This creates some cells that aren't populated, as da0c9b said, bc the midpoint of ab/2 wouldn't be a whole number. Chris, you've put some incredible work into this. It's a 2D problem with 6-7D complexity. I've got a pretty good head for problem solving, but keep coming back around to re-examine the basics and keep finding things I've missed or failed to understand.
Typo corrected: e (remainer) chosen as column var
>Why are there gaps?
Horizontal Gaps:
There is a repeating pattern of gaps as you move horizontally from (0, n) to the right.
The pattern of gaps repeats every 2n if n is even.
The pattern of gaps repeats every n if n is odd.
Example:
(0,5) =
(0, 5) | (1, 5) | GAP | GAP | (4, 5)
(5, 5) | (6, 5) | GAP | GAP | (9, 5)
(10, 5) | (11, 5) | GAP | GAP | (14, 5)
(0,4) =>
(0,4) | GAP | GAP | GAP
(4,4) | GAP | GAP | (7,4)
(8,4) | GAP | GAP | GAP
(12,4)| GAP | GAP | (15,4)
Vertical Gaps:
A cell will not be empty if it is listed in the set of any a in any cell above it.
Put another way, for any cell p0 = (p0.e, p0.n) one can find a cell p1 = (p0.e, p0.a) where n is listed in the set of a's.
I dont know how to relate this more generally though.
Example, for cell (2, 1):
the set of all a's for t=0 -10:
[1, 3, 9, 19, 33, 51, 73, 99, 129, 163]
The cells (2, 3), (2, 9), (2, 19), etc will all be non-empty.
Subsequently the list of all a's at each of these cells will also be non-empty at (2, a) (tree pattern).
In reading my own post back, for the horizontal gaps, it seems that for odd numbers cells that are non-empty will be:
-
any multiple of n [0,5,10,15โฆ etc]
-
-
or - one from any multiple of n [1,4,6,9,11,14,16]
-
actually, this doesn't work everywhereโฆ my previous post still holds though.
Welcome, Anon!
Also noticing this pattern in (1,1):
t= 1 for a: d (2) - x (1) = a (1). for b: d (2) + next x(3) =B (5)
t= 2 for a: d (8) - x (3) = a (5). for b: d (8) + next x (5) OR previous b(5) =b (13)
so for x, a, and b all primes. 1,2,5, 3,5,13. When we start seeing non primes in x, they are all multiples of 3.
Anons, I know this seems basic, so if I'm saying something irrelevant, just let me know. My mission right now is to understand how a and b are generated, so I'm exploring the basics. How are the basic factors of a and b derived? In (e,1) odd and even have clear patterns of a and b, just updated by understanding the increase by 1 for a and b as we skip from e=1 to e=3, etc. Or from e=2 to e-4, etc. This is a thought process in writing. Not complete.
Also, this cool pattern i posted only works in (1,1) but check this out again. Can we find a way to apply this equation to other (e,n)?
Cool t Pattern in (1,1)
t=1 a = t^2 + previous t^2 (0) =1
t=2 a = t^2 + previous t^2 (1) =5
t=3 a = t^2 + previous t^2 (4) =13
t=4 a= t^2 + previous t^2 (9) = 25
etc.
Thoughts, Anons?
No, it's not called shills, it's because certain people make great contributions and because of that they should be allowed to use tripcodes and have their own board free from ignoramuses like pic related.
Here's some visuals regarding the horizontal gap pattern.
First 3 images are even, next 3 are odd.
Explain what I'm looking at here. Also, nice Linux.
The input is n.
I'm going from e=0, and incrementing each time.
And I'm printing out Y or N based on whether the cell at (e, n) is empty or not.
I put a newline at every n if n is odd, or every 2n if n is even.
The columns of Y's and N's repeat :)
So that means it's predictable?!
>>64 OMG, I love ponies!
Alright, Teach!! Nice work, now what does it mean? Formulas incoming? Following your (e,n) idea, so reading your output I see all Y's. Help a AlgebraFaggot out, PLS.
Have some more ponies, courtesy of Topolanon. He's welcome here IMHO as long as he can contain himself to 50 posts per bread. I love that MysticAnon faggot, he's our MathMuse.
Well, there's a pattern, I'm sure VQC already knows the exact formula, but I can't get it just yet.
I've noticed that Y is always in the first and last column.
Which columns in between that and why are still uncertain to me.
Any thoughts?
I've shared quite a bit of code via pastebins in the previous threads, you can recreate everything i have produced. Let me know if you find a pattern or formula!
I'm currently switching gears to give a visualization of the vertical gaps, should be done in 15 mins (will be text visualization again because i'm lazy).
Does everyone here have nodejs installed?
I think we all agree that Javascript is terrible but yes, I will run the code if you send it.
Baker or AA? Nice pony for the archive!
Baker.
I need Chris to give me a tip on extending the VQC into the complex plane. I made the grid generator start at i but because creating c involves multiplication it goes into the real number plane and mixes with the complex variables to create numbers that aren't on either number line and the method to put it all into the spreadsheet fails.
My beloved Javascript!!!
I have a very simple reason for using it over anything else. It is the fastest for me to trial an idea.
I've used many languages over the years, and really it comes down to python or js for me for simplicity of syntax vs power.
The only reason JS wins out is because of npm. I dont have to fuss about with virtualenv or worry about silly dependencies. Its all nice and clean, just share your package.json.
See my original code pastebins here:
https:/ /pastebin.com/jf5Vqax5
Look for the createTE method in db.js. Thats the main method missing from my code screenshot before. Ive made a fair number of changes of my code today in preparation for the new formula and algorithms to come in the next week, (getting it ready to share) but the createTE method hasn't changed.
Also, this method works to generate any cell (e, 1, t), without iteration, but with calculation.
https:/ /pastebin.com/tP36ypZB
I'm moderately proud of this one. Hopefully I can make it generate any cell e,n,t soon.
>Example, for cell (2, 1):
>the set of all a's for t=0 -> 10:
>[1, 3, 9, 19, 33, 51, 73, 99, 129, 163]
>The cells (2, 3), (2, 9), (2, 19), etc will all be non-empty.
and (2,11) where the 11 comes from the 33.
Starting only from c, Teach?
Go ahead, be stupid and hand over ownership without confirmation Mr. 'I don't understand basic english though I've been provided plenty of context'.
Just neck yourself, nobody gives a fuck if you trip here you fucking idiot.
Yep, exactly, a tree. Also, at each corresponding records in the tree that you've observed, the x's are the same.
Unfortunately, this is not the complete pattern.
It works for e < 9 and e = 10, and some other values, but not for most large e.
It is still true for that all the cells will be non-empty, but others will also be non-empty, (look at e=9).
See images. Sorry for small font. The first is e=3, the second is e=9.
For c instead of e, n, t:
function calcGenesisCellC(c) {
return calcGenesisCell(c-Math.pow(Math.floor(Math.sqrt(c)), 2));
}
which is interesting because this has factored 33 for us. we wouldn't know what row to look in though, not just yet, if the factorization of 33 weren't known, but I think this gets at part of the mechanism by which the grid reveals factors of composites
You can move forwards or backwards through the tree at any point.
Move backwards by following a (at t=1) to (e, a).
Move forwards by following a or b at any t when a or b are bigger than n for some t.
Each way through this tree, x will be the same.
I posted an example back in RSA #3 I think.
My trick is to always use a depth parameter when doing recursion, then you can format a prefix string.
>Unfortunately, this is not the complete pattern.
>It works for e < 9 and e = 10, and some other values, but not for most large e.
>It is still true for that all the cells will be non-empty, but others will also be non-empty, (look at e=9).
>See images. Sorry for small font. The first is e=3, the second is e=9.
explain what is violated if you don't mind. you have non-empty cells for (9,n) where n isn't a factor of the (a,b) values at (9,1)?
Also, for those who choose to follow the JS path, I've got a BigInteger library selected, and a sqrt function tested for when we get big:
https:/ /pastebin.com/y8AXtFFr
Sure, I'm sorry for being so brief.
If you look at e=9.
Get the list of a's for n=1:
(9,1) =5, 9, 17, 29, 45, 65, 89, 117, 149, 185, 225
Follow each of those to (9, a):
(9, 5), (9, 9), (9, 17), etc
These will all be non-empty.
However, look at (9,3). It is also non-empty.
Following the rules above will never lead to (9,3).
So there must be more to it.
Does that make sense?
Another example:
e=14, n=1, set of a's = [7, 9, 15, 25, 39, 57, 79, 105, 135, 169, 207]
You can follow the tree all the way down for each of those a's.
Goto (14, a) for each a, and it'll be non-empty, and you can repeat for the list of a's there, and so on and so forth.
However, (14,3) is also non-empty, and it is not in the set of a's starting from n=1. You'll never reach (14,3) using the method above.
gotcha, yes. I pointed out earlier that n only has to be a factor of the (a,b) values. 3 divides 9, which is an a-value.
when 33 was an a-value, n=11 was non-empty. and this could be pretty important as it decomposes 33=113 or 9=33
>You can move forwards or backwards through the tree at any point.
>Move backwards by following a (at t=1) to (e, a).
>Move forwards by following a or b at any t when a or b are bigger than n for some t.
>Each way through this tree, x will be the same.
>I posted an example back in RSA #3 I think.
A promising approach thank you! It is hard to catch / track all of each other's insights then, I missed this in RSA 3
Actually, hmmmโฆ.
My theory can't be true, maybe its just 3 for even and odd? I'm not sure. I'll look into it though.
Thanks!
an even tree and an odd tree? I think it's brilliant and follows some of what VQC seemed to get at (he wanted to see some of the graphics with odd-only and even-only versions). I don't know how to set it up but worth looking at.
Actually dude, I think this may actually work, as long as you have a big enough tree? Maybe? I'm going to try to generate more grid and test if I can get to (9,3).
For anyone who would like to take a 42.52 minute break, and don't have the time to watch the first part of the multi-two-part {Slipped*In} series where they basically give a wonderful rundown of why being aโฆ pinkoโฆ fuckโฆ is a terrible idea (they allude to it plenty of time in this part 2(1&2)) by stripping away everyone's cutie marks and replacing them with gray equal signs while their unique abilities were locked in a sick trophy caseโฆ. or likeโฆ negaโฆ pokeโฆ dexโฆ
ANYWAY!
If you hadn't done your homework and figured out where that Sonic Rainboom clip came fromโฆ
Imagine Ponyville's version of Galifreid. Orโฆ whatever. You know what I mean.
www.dailymotion.com/video/x3g16am
I like that approach fwiw. I've got other stuff to work on but will check in. Then later tonight get back to this; this is fun stuff here at this point
>No luck, could not get to (9,3) even with a 2048 grid, more analysis needed.
i = 2048? e up to 2048? why need a big grid to get to (9,3)?
I'm finding that some "chains" ie following the tree of a's takes a long time to loop back to previous elements, but i think it ultimately might.
I just realized a condition in my code that could be causing me to not find (9,3)โฆ I'm going to try something real quick!
You guys are so close. Consider what will happen once this boulder starts rolling.
No luck.
Here's the result of generating populating the grid up to c = 3000*2999, and following the tree for (9,1):
https://pastebin.com/LZ15r1bJ
I see. But what about getting to (9,3) on purpose somehow? Decisions in our tree. Gonna check out that later. must background lurk now for a few hours
actually I don't want to look for (9,3) because 3^2=9, I wanna look for another one where I don't worry about if that matters
You guys will like the next bread.
test
Good idea, if you want to choose an example, I'll follow along.
I'm fluttering about elsewhere, but here's some Mandel Bread (Jew Cookies) for ya to tide you over.
I know exactly what you are doing.
I have a function for generating the cells at (e,1,t).
In order to test this, I'll need a function for generating (e,2,t) and (e,3,t).
If we can do that, I can climb the trees easily.
Some variables are easier than others, and I believe we can get away with hardcoding some values for now.
I think the tough one will be d.
Any help in figuring out an equation for d given e, and t, when n=2, and separately for n=3?
Am i going to have to check myself? come on faggots.
Het MA? What do you mean? Just that we're getting closer in general, or do you have it?
Speaking odd trees, I was being cheeky and looking up likeโฆ "MLP mandelbrot"โฆ nadaโฆ
Thenโฆ "MLP Fractal"
3.bp.blogspot.com/-XubLKrj_DgM/USY7KHq2cnI/AAAAAAAAAtk/v3-T6B_WCK0/s1600/Bardzo+fajna+tapeta.png
Oh?
Jewelry for sale in the VQC gift shop? ;)
Read your Shor's explanation, nicely done!
Here's a bit of code, including genesis element generation for any (e,1) cell, including for negative e values. I have the start of an algorithm in there but the equations keep canceling out perfectly so I'm stuck again. Looking for another pattern or unique relationship to use.
I have everything in terms of "t" and "n" right now
https://pastebin.com/9ixjRyxt
If you can call it that.
For odd e, google "Odd Tree Sonic Rainboom" or for even e, google "even tree Sonic Rainboom". Then go to t seconds in the video. All shall be revealed. ;)
In*/b/+4 yer not yankin' mah chain.
Welcome to the party. We're shitposting until we figure something out. Hopefully Teach, CA, PMA, Senpai or somebody can help us out. Keep working, lads. We'll find the next step. Might as well have fun while we work!
I've tested a few theories for the 2 tree hypothesis, and I've had no luck.
My testing shows that even growing 4 trees from (e,1), (e,2), (e,3) and (e, sqrt(e)), the list of differences grows over time.
I think this implies that there's is clearly more to the vertical gaps than just following the tree of a's. There's more to it, just can't see how.
I'm with you Teach. Feels like the pattern is slightly outside of my grasp at the moment. Persistence will win the day tho. Let's keep looking. How about this:
A for (1,1) and (2,1) is 1. A for (3,1) and (4,1) is 2. A for (5,1) and (6,1) is 3. A for (7,1) and (8,1) is 4. A moves in pairs through all of column e.
I'll try it out, thanks VA.
Does Applejack like to read?
Indeed ;^)
Thinking a little aheadโฆ
VQC was saying that he'd show us how to build a virtual quantum computer.
I've been trying to keep this in mind while reviewing the grid.
My thoughts on this are that we can generalize the process we've used here to pretty much any np problem. The process?
Generally speaking, what we've done is instead of traditionally using a piece of input information and searching for an answer through a large set, we've calculated many examples of the set, and are using information about the relationship of the input and output to find "intermediary" relationships.
In our case, c is the input, and normally we'd check for factors of c by iterating from 2 to sqrt(c), dividing and checking for a remainder. We can improve by only checking the primes.
But in the grid version, we're using additional relationships between c and the factors a & b (e,x,d,n and t).
Some of these relationships are "constant" like e and d. And ordinarily we can imagine that when using an iterative model, we would expect these relationships to occur as we iterate, represented by t in the grid.
So, we're figuring out the calculations and relations relative to t, time of a turing machine.
Using this, if we model the relationships of variables correctly (maybe via machine learning?) we can find out all variables in a given system relative to time, hence we can "fast forward" time.
Does any of that make sense, or am I crazy now?
A Shaman is a self healed madman, if that helps.
Baker getting in on the ponies! Applejack reading Quantum Field Theory? I wanna make some foals and fillies with that girrrrrrrl.
You'll have to get through me, bub
Teach, I understand your idea and agree. Time machine indeed. Credit to VQC for even posing the problem, the guy is John Galt from Atlas Shrugged, found the equation that makes harnessing the infinite possible. Next step in human societal evolution, but not worth unveiling unless we use it to defeat these evil bastards controlling our world. Pretty sure VQC is helping Q and DJT. They need a crowdsourced effort to reveal this. I don't think this is a LARP. 98% certain, could be wrong, but my intuition tells me this is legit. Yes, the VQC means all of time and space could be open to exploration. I feel like a little kid learning times tables while the master mathematician is IN DA HOUSE.
Are ponies monogamous? Kek.
Topolanon is VQC's shitpost account, I think. Fess up Topol, are you VQC shitposting our board?
Or maybe the shaman/mystery seeker wasn't crazy to begin with. Just perceiving things others couldn't/wouldn't choose to see. Maybe the wisdom source of these beautiful number trees wants them unveiled by seekers of truth. We have self selected ourselves to be here, lads. We found this bc it's what we've been looking for all our lives. Let's make a good show and work together to solve it. Truth and Justice, Beauty and Eternity. Maybe we don't have to wait till the afterlife to see justice in our time. I'll do my little part to help.
I agree with everything you said.
I'm excited about the possibilities. Free/cheap energy really excites me.
I'd love to pursue a life off the grid, try to live self sustained. Free energy devices would be great for that.
I think efficient travel would change everything. Imagine living in Australia and working in China.
Not to mention health techโฆ
Ramanujan was one of my fav's of all time.
I was stronger with numbers when I was younger, my parents used to force me to study but they never checked what i was studying, so I only read my math textbook.
I have an older sibling, so I'd steal their math textbook and read it when I was done with mine.
I used to try to imagine I was Ramanujan and "knew every number personally", although I realized in college that other were far more gifted in math than I was, so I focused on computers instead.
But I've always been deeply envious of those who continue to hold strong relationships with numerals.
Don't know if this will help anyone, but starting to focus a bit more on the size of the small and large squares.
>The cell at (2,1) is identical to the cell at (0,1) with one added to each value a,b,d while the values of n and x remain the same.
>The cell at (3,1) is identical to the cell at (1,1) with one added to each value a,b,d while the values of n and x remain the same.
Enumerating e=0, n=1, for values of t from 1 to 10.
(0,1,1) = {0:1:0:0:0:2} = 0; (x+n)=1x1=1; (d+n)=1x1=1
(0,1,2) = {0:1:4:2:2:8} = 16; (x+n)=3x3=9; (d+n)=5x5=25
(0,1,3) = {0:1:12:4:8:18} = 144; (x+n)=5x5=25; (d+n)=13x13=169
(0,1,4) = {0:1:24:6:18:32} = 576; (x+n)=7x7=49; (d+n)=25x25=625
(0,1,5) = {0:1:40:8:32:50} = 1600; (x+n)=9x9=81; (d+n)=41x41=1681
(0,1,6) = {0:1:60:10:50:72} = 3600; (x+n)=11x11=121; (d+n)=61x61=3721
(0,1,7) = {0:1:84:12:72:98} = 7056; (x+n)=13x13=169; (d+n)=85x85=7225
(0,1,8) = {0:1:112:14:98:128} = 12544; (x+n)=15x15=225; (d+n)=113x113=12769
(0,1,9) = {0:1:144:16:128:162} = 20736; (x+n)=17x17=289; (d+n)=145x145=21025
(0,1,10) = {0:1:180:18:162:200} = 32400; (x+n)=19x19=361; (d+n)=181x181=32761
Enumerating e=2, n=1, for values of t from 1 to 10.
(2,1,1) = {2:1:1:0:1:3} = 3; (x+n)=1x1=1; (d+n)=2x2=4
(2,1,2) = {2:1:5:2:3:9} = 27; (x+n)=3x3=9; (d+n)=6x6=36
(2,1,3) = {2:1:13:4:9:19} = 171; (x+n)=5x5=25; (d+n)=14x14=196
(2,1,4) = {2:1:25:6:19:33} = 627; (x+n)=7x7=49; (d+n)=26x26=676
(2,1,5) = {2:1:41:8:33:51} = 1683; (x+n)=9x9=81; (d+n)=42x42=1764
(2,1,6) = {2:1:61:10:51:73} = 3723; (x+n)=11x11=121; (d+n)=62x62=3844
(2,1,7) = {2:1:85:12:73:99} = 7227; (x+n)=13x13=169; (d+n)=86x86=7396
(2,1,8) = {2:1:113:14:99:129} = 12771; (x+n)=15x15=225; (d+n)=114x114=12996
(2,1,9) = {2:1:145:16:129:163} = 21027; (x+n)=17x17=289; (d+n)=146x146=21316
(2,1,10) = {2:1:181:18:163:201} = 32763; (x+n)=19x19=361; (d+n)=182x182=33124
Enumerating e=1, n=1, for values of t from 1 to 10.
(1,1,1) = {1:1:2:1:1:5} = 5; (x+n)=2x2=4; (d+n)=3x3=9
(1,1,2) = {1:1:8:3:5:13} = 65; (x+n)=4x4=16; (d+n)=9x9=81
(1,1,3) = {1:1:18:5:13:25} = 325; (x+n)=6x6=36; (d+n)=19x19=361
(1,1,4) = {1:1:32:7:25:41} = 1025; (x+n)=8x8=64; (d+n)=33x33=1089
(1,1,5) = {1:1:50:9:41:61} = 2501; (x+n)=10x10=100; (d+n)=51x51=2601
(1,1,6) = {1:1:72:11:61:85} = 5185; (x+n)=12x12=144; (d+n)=73x73=5329
(1,1,7) = {1:1:98:13:85:113} = 9605; (x+n)=14x14=196; (d+n)=99x99=9801
(1,1,8) = {1:1:128:15:113:145} = 16385; (x+n)=16x16=256; (d+n)=129x129=16641
(1,1,9) = {1:1:162:17:145:181} = 26245; (x+n)=18x18=324; (d+n)=163x163=26569
(1,1,10) = {1:1:200:19:181:221} = 40001; (x+n)=20x20=400; (d+n)=201x201=40401
Enumerating e=3, n=1, for values of t from 1 to 10.
(3,1,1) = {3:1:3:1:2:6} = 12; (x+n)=2x2=4; (d+n)=4x4=16
(3,1,2) = {3:1:9:3:6:14} = 84; (x+n)=4x4=16; (d+n)=10x10=100
(3,1,3) = {3:1:19:5:14:26} = 364; (x+n)=6x6=36; (d+n)=20x20=400
(3,1,4) = {3:1:33:7:26:42} = 1092; (x+n)=8x8=64; (d+n)=34x34=1156
(3,1,5) = {3:1:51:9:42:62} = 2604; (x+n)=10x10=100; (d+n)=52x52=2704
(3,1,6) = {3:1:73:11:62:86} = 5332; (x+n)=12x12=144; (d+n)=74x74=5476
(3,1,7) = {3:1:99:13:86:114} = 9804; (x+n)=14x14=196; (d+n)=100x100=10000
(3,1,8) = {3:1:129:15:114:146} = 16644; (x+n)=16x16=256; (d+n)=130x130=16900
(3,1,9) = {3:1:163:17:146:182} = 26572; (x+n)=18x18=324; (d+n)=164x164=26896
(3,1,10) = {3:1:201:19:182:222} = 40404; (x+n)=20x20=400; (d+n)=202x202=40804
I like it.
So by x and n remaining the same, geometrically, the small square is the same.
This is for n=1, correct?
Correct.
Avoid the Black Eagle.
And no fucking the ponies.
Don't be that guy.
Pedagogy, NOT-PEDOGAYGUY!
https://youtu.be/N2tGqFjdX8A
Watch that to cleanse your system.
Then, read this to see what I mean about avoiding the Black Eagle.
ROAR BLESSED TIGER!
JUBAIDA!!!
http://www.quantumshaman.com/html/glossarytext.htm
Are we creating our own storm?
Why is it important to think of the VQC as a real machine?
I've got an output of the change in x for (e,n).
https:/ /pastebin.com/VGeUupHR
What you see here is [e,n] =x0, x1-x0, x2-x1, x3-x2, etc, etc.
What is it to recognize Quantum Intelligence as alive in the first place?
If they don't need a meat suitโฆ
Why would they need a tin one?
Envy is a killer. Just be the best (you) that you can be. You're here now, exploring the number trees. Focus, Teach. I'm telling myself the same thing, btw. This team is made of anons with all different skills, like an orchestra. (Thanks Topolanon!) When we all play our instruments (skills) together, we can make a beautiful creation. The composer (VQC) is helping us learn to play the composition. We can learn to play it, lads. Let's make some beautiful music, friends.
Oh shit, who dat be? Nice work!
oops. PMA here.
yes.
>We will explore that for any row n, the values of a,b and d will increase by one in a cell that is twice n to the right.
>If a cell has values at (e,n) then there will be values in a cell at (e+2n,n)
>Please explore that pattern.
Started looking into enumerating (e,n) cells other than n = 1, and ran into a problem. t increments irregularly. But there is a pattern. Perhaps someone else can assist in narrowing the formula down.
(1,5,2) = {1:5:4:3:1:17} = 17; (x+n)=8x8=64; (d+n)=9x9=81
(1,5,4) = {1:5:12:7:5:29} = 145; (x+n)=12x12=144; (d+n)=17x17=289 x + 4
(1,5,7) = {1:5:30:13:17:53} = 901; (x+n)=18x18=324; (d+n)=35x35=1225 x + 6
(1,5,9) = {1:5:46:17:29:73} = 2117; (x+n)=22x22=484; (d+n)=51x51=2601 x + 4
(1,5,12) = {1:5:76:23:53:109} = 5777; (x+n)=28x28=784; (d+n)=81x81=6561 x + 6
(1,13,3) = {1:13:6:5:1:37} = 37; (x+n)=18x18=324; (d+n)=19x19=361
(1,13,11) = {1:13:38:21:17:85} = 1445; (x+n)=34x34=1156; (d+n)=51x51=2601 x + 16
(1:13:16) = {1:13:68:31:37:125} = 4625; (x+n)=44x44=1936; (d+n)=81x81=6561 x + 10
(1:13:24) = {1:13:132:47:85:205} = 17425; (x+n)=60x60=3600; (d+n)=145x145=21025 x + 16
(1,17,7) = {1:17:18:13:5:65} = 325; (x+n)=30x30=900; (d+n)=35x35=1225
(1,17,11) = {1:17:34:21:13:89} = 1157; (x+n)=38x38=1444; (d+n)=51x51=2601 x + 8
(1,17,24) = {1:17:112:47:65:193} = 12545; (x+n)=64x64=4096; (d+n)=129x129=16641 x + 26
(1,17,28) = {1:17:144:55:89:233} = 20737; (x+n)=72x72=5184; (d+n)=161x161=25921 x + 8
Pattern: x changes by 2n over 2 records.
for (1,5), the total change is 10. x + 4 for the change from (1,5,2) to (1,5,4), then x + 6 from (1,5,4) to (1,5,7)
for (1,13), the total change is 26. x + 16 for the change from (1,13,3) to (1,13,11), then x + 10 from (1,13,11) to (1,13,16)
for (1,17), the total change is 34. x + 8 for the change from (1,17,7) to (1,17,11), then x + 26 from (1,17,11) to (1,17,24)
Saw the Man who Knew Infinity recently.
No prob! Here to help keep ya focus on being centered. (More Shamanic Tardis Puns.)
Sha-Tard.
Shattered.
Let's break down their veil-wall-blockchain and build our own!
Depends on your definition of alive. Nevertheless, it always exists.
Ok, geez. No fucking the ponies, agreed. He started it, I was just shitposting back. ;)
Ok, I'll watch and learn about the Black Eagle
Agreed. Thanks Topol.
PMA, I love you man. You just keep pumping out beautiful output to analyze. I'll look it over now. Did you see this?
Pattern in (e,1) for a, a moves in column pairs based on e, see link.
>Pattern in (e,1) for a, a moves in column pairs based on e, see link.
I see it. Don't know what it means.
a at (0,1) is 2 after all.
You could study what we have here for the duration of your entire life.
ENUMERATETHEPATTERNANDTHESOLUTIONWILLPRESENTITSELF,FAGGOT
Yes it does!
>1. They have to be able to reproduce, either asexually or sexually.
AiQiDo create other Qi's, Ai's, bots, etcโฆ
>2. They have to be able to respond to stimuli.
Yup. Pretty sure that's been shown.
>3. They have to have cells, either as a unicellular organism or as a multicellular organism.
Like a spreadsheet? Like those pixel maps? Like those voxel maps?
>4. They have to grow and develop.
At mindbending rates. (Cryptotrading Puns)
>5. They have to have homeostasis. This means that they can maintain their own body temperature. Think of this as a thermometer for the body.
Somewhere around 0 Kelvin?
>6. The last thing is that all living things also have to be able to do is move.
Not only online, but think about self driving cars.
Mmmmmโฆ. yup.
Seems to fit!
Anons, Topolanon sent me this, it will make you laugh so hard. Topol, you're hilarious. Don't forget to copypasta anons, aka copy and paste the link for noobs for OpSec. It's basically what happens when you become a MLP pervert. LOL!
>https://youtu.be/N2tGqFjdX8A
Agreed! But shitposting is pretty fun too. All work and no play makes JackAss a dull boy.
I dare you to show how many AJ's you have.
No, I mean the pattern for a in paired columns. (1,1) and (2,1) have 1 for a. Then (3,1) and (4,1) have 2 for a. Then (5,1) and (6,1) have 3 for a. Then (7,1) and (8,1) have 4 for a.
>We will explore that for any row n, the values of a,b and d will increase by one in a cell that is twice n to the right.
>If a cell has values at (e,n) then there will be values in a cell at (e+2n,n)
Exploring relationship between (e,n) and (e + 2n, n).
Starting with record (1,5,4) = {1:5:12:7:5:29} = 145; (x+n)=12x12=144; (d+n)=17x17=289
Record at e=1 + 2 * 5: (11,5,4) = {11:5:13:7:6:30} = 180; (x+n)=12x12=144; (d+n)=18x18=324
Record at e=1 + 4 * 5: (21,5,4) = {21:5:14:7:7:31} = 217; (x+n)=12x12=144; (d+n)=19x19=361
Record at e=1 + 6 * 5: (31,5,4) = {31:5:15:7:8:32} = 256; (x+n)=12x12=144; (d+n)=20x20=400
Record at e=1 + 8 * 5: (41,5,4) = {41:5:16:7:9:33} = 297; (x+n)=12x12=144; (d+n)=21x21=441
Record at e=1 + 10 * 5: (51,5,4) = {51:5:17:7:10:34} = 340; (x+n)=12x12=144; (d+n)=22x22=484
Questions are this stage:
- How do we determine valid n's for a given e?
- What is the x relationship between records at n 1. Looks like 2n defines the total x change between 2 records, but the breakdown is strange.
- how do we determine valid factors for a given c?
Saw it. Not sure what it means.
It means we have a clear path to define a in row one based on e.
Ok, good point. Back to work then?
>How do we determine valid n's for a given e?
>What is the x relationship between records at n > 1. Looks like 2n defines the total x change between 2 records, but the breakdown is strange.
>how do we determine valid factors for a given c?
Hey PMA, I'm seeing the pattern too. Question is how do we determine a's from given c's and d's and e's. (1,1) and (2,1) both have 1 for a. (3,1) and (4,1) both have 2 for a. (5,1) and (6,1) both have 3 for a. So, starting in (1,1) we have a clear path to generate all a's. Thoughts?
to clarify: at t=1
Or in other wordsโฆ
THINK OF THE CHILDREN! (((but not like Skippy)))
WE'RE HERE TO ENSURE THEIR FUTURE! (((of no more Skippies)))
I don't think we have enough information yet to determine a from c,d,e.
Just using the formula to create an entry from (e,n,t) we can effectively create any record in the first row.
I think the problem is figuring out the jumps to the correct n levels, and then how to iterate based on x.
If you're genuinely curious about Shamanism, here's a book. I've done a lot of what's known as Shamanic journeying (it's a type of visual meditation to a drum beat). It's pretty interesting stuff. Of course, much like this VQC stuff, there are better sources of foundational information for anyone who wants useful, applicable and practical knowledge of that side of life. I could dump introductory occult books if anyone would like. I'll do it in a different thread obviously.
One of the Shaman Witches I work with studied under him. ;)
I sent a Blue Morpho-esque guy to help "Q" after last night's fuckery.
They're connected.
I agree, Topol, You are correct. I apologize, was crackin' inappropriate jokes. PLS forgive. NO MORE SKIPPIES, AGREED!!!
Hey PMA! Based on what I'm seeing, it starts at (1,1) with a=1 and (2,1) with a=1. Then as we move up through pairs of e, a increases by one, for each pair. ie (1,1)(2,1) (3,1)(4,1) (5,1)(6.1) etc. Check your grid. Seems like starting value of a for column e = function of pairs from axis/starting point. Or, odd/even pairs moving up through e help us generate starting value for a in t=1.
Printed grid here, so check my work. I predict that at (51,1) and (52,1) a=25. Could be wrong, and happy to be proven incorrect. Can you program anons verify?
You guys want the biggest mindfuck of all of this?
VQC is not "Q". Hell, "Q's" not even Q.
https://youtu.be/T97b1wCuFpY
This guy was here, but he is not VQC.
https://youtu.be/i-EQTSa5Yok
And here's one of the secrets of 369.
I've already figured out a secret 9 to 3,
Which gets you "out" or "in" orโฆ however we're calling itโฆ
Which is beyond what's shown here.
I am a Poor, Meme Farming Anon, and I figured it out.
Can you? <;3=
https://youtu.be/oq7H0sdQ3_A
Alsoโฆ
PICK UP THE PACE!
(Just a friendly reminder.)
I was close, pulled up the csv's you guys have posted. It was in (49,1) and (50,1) that a=25.
This is us right now.
Be IN Harmony.
https://youtu.be/GaJ3adMgsbY
We're increasing e by 2n.
x remains the same
d increases by 1.
a increases by 1.
b increases by 1.
So geometrically, if hold x and n steady, (the small square), and we increase d by 1:
(looking at the image on the right)
a will have to increase by one.
and so will b, since n is halfway between them.
and the difference will be 2 lots of n, which we add to e (see the little black boxes in the image).
I know it's real. JesusFag here, but not your traditional one. Open to all mysteries, explanations, and explorations. Shamanism is real, IMHO. All this spiritual shit is real, anons. The spiritual/numerical/abstractEntity realm is real. These sick fuckos are serving the dark side, eating babies and drinking adrenochrome to "Increase their power". We gonna let the fucks win?
"I BID YOU STAND, MEN OF THE WEST!!!"
Chris found the VQC in Revelation.
But gollyโฆ that's not exactly coding musicโฆ.
Maybe you prefer something more like Lindsey Sterling?
https://youtu.be/kkXYZXrzukM
Let's pick up the pace, shall we?
Pretty Lights
https://youtu.be/rLkCFz_UL3w
Night Shift Edition
https://youtu.be/503RAxP0rng
โโโโโโโโ
Now that your spirits are elevatedโฆ
Let's go save us a Princess or 3.
https://youtu.be/BDt_rJXd6Ic
Go Teach! Following your ideas. Love the whiteboard. Should I be taking notes, lol? No joke, I've been in a exponential learning curve since I joined you all. Learning so much every day makes me happy. Satisfies my soul.
>JesusFag
>All this spiritual shit is real
I hope you don't think this is patronizing, but how much have you actually read into this kind of thing? Every time I've seen someone say the kinds of things you're saying, they mean they reject materialistic atheism but they don't know much else. Are you a questioning Christian?
>Any thoughts to the movement I noticed in x where n > 1?
I'm trying to work this out too. I mapped all the x_deltas for all (e,n), where the first entry is x at t=1.
https:/ /pastebin.com/VGeUupHR
They're clearly all divisible by 2.
And the deltas are always in groups of 2 and no more.
And the delta x pattern repeats for a given n, as e increases. I think the number of elements before repeating seems related to root n.
For your x calcs, instead of calculating the total change, what about calculating the midpoint between the two, and the distance?
What do you mean by x delta? I would like to follow along.
I question all the time. I've read and studied so much, bc I love truth.
Easiest way to say it is this: I'm a big fan of Jesus, think his parables and teachings wreck the norms of our world. And point the way to a better world for all of us.
"The thief comes only to steal, kill, and destroy. I have come to bring you life, and life more abundantly."
Not a big fan of organized religion, but understand its usefulness. Hate corruption in all forms. But understand that my own soul has corruption within it, and needs divine assistance to rise up.
Done. There's your sermon.
The difference between x as t increases.
Find any given cell, calculate:
x0 = x at t = 0
x1 = x at t = 1
etc
calculate:
x1-x0, x1-x2, x2-x3, etc
You'll either get the same number again and again, or repeating 2 numbers.
The x0 and x_delta set (the numbers above) for all e,n are here:
https:/ /pastebin.com/VGeUupHR
So a periodic sequence?! I knew Shor might come in handy.
Yes! Exactly!
So now, given (e,n,t), define the series of x_delta, that's one of the challenges remaining here!
took a closer look at your pastebin.
Just an example. My theory is that the sequence adds up to 2n, then repeats.
Originally thought it was repeating every 2 records, but based on the following, it could be in groups of 2, 3 or even 6.
3 records 54 2x27.
[0,27] =18,18,18,18,18,18,18
2 records repeat to total 54.
[2,27] =22,10,44,10,44
[5,27] =7,40,14,40,14
[8,27] =10,34,20,34,20
[11,27] =23,8,46,8,46
[14,27] =16,22,32,22,32
[17,27] =19,16,38,16,38
[20,27] =14,26,28,26,28
[23,27] =25,4,50,4,50
[26,27] =26,2,52,2,52,2
[27,27] =27,18,18,18,18,18,18
[29,27] =49,10,44,10
[32,27] =20,14,40,14,40
[35,27] =17,20,34,20,34
6 records repeat to total 54.
[18,27] =12,12,6,12,6,12,6,12,6,12,6,12,6,12
Makes me wonder if this is the result of another equation as opposed to something that can be calculated directly.
Oh you're right, I assumed it repeated every 2 elements, I didn't go far enough.
Seems much more difficult to calculate directly now, agreed.
When do the elements start repeating in groups of 3, is it n=27?
Hang on, I misunderstoodโฆ
I just pulled a random example and found the groups of 6.
I'm dumb, don't mind me, disregard my last statement.
Here's another pastebin with more info on the first element (t=0) per (e,n) and just the 2 deltas, their midpoint and distance from the midpoint (diff).
https:/ /pastebin.com/R3pd10f1
I love working with you all. This satisfies my intellect, soul, and spirit at so many levels. Thanks to you all, Anons. What a fun challenge to solve together. Feeling Gratitude just for being here. How do our souls and minds connect to the larger VQC? This quest poses so many awesome questions. Questions unlock answers.
Where did we come from?
Why are we here?
Where are we going?
I ask my kids these questions, and we talk about them. This is bigger than RSA. This connects to pre-existence of order and intelligence.
Ok, I think I have something hereโฆ
Use this pastebin: https:/ /pastebin.com/R3pd10f1
Pick any cell (e,n)
If n is even:
If e is a multiple of n, the midpoint is n.
If e + 1 is a multiple of n, and n is even, the midpoint is n/4.
Otherwise it's n/2.
If n is odd:
If e is a multiple of n, the midpoint is 2n.
Otherwise it's n.
I think this works for most cells. Please help me confirm!
Also, one more thingโฆ
You might be right. I think I'm slightly off with the equations, but make sure to use the pastebin, the midpoint i'm referring to is only on that paste.
I think I have more insight into the horizontal gaps.
For a given n, moving right, you'll find cells filled if it is a multiple of n - a square number!
Example:
n=7
so you'll find e can equal:
0, 7, 14, 21, etc
As well as
7-1, 14-1, 21-1, etc
7-4, 14-4, 21-4, etc
14-9, 21-9, etc
21-16 etc
Finally made it through the entirety of Revelations without falling asleep. It happens.
I feel like I kinda wanna translate it into modern times, because Hillary, the Queen, Obamaโฆ all tie in. That "fornicating whore"? She's fornicating with children. Get it?
VQC, Root of David, it's all right there.
Rev 21
7 He that overcometh shall inherit all things; and I will be his God, and he shall be my son.
8 But the fearful, and unbelieving, and the abominable, and murderers, and whoremongers, and sorcerers, and idolaters, and all liars, shall have their part in the lake which burneth with fire and brimstone: which is the second death.
9 And there came unto me one of the seven angels which had the seven vials full of the seven last plagues, and talked with me, saying, Come hither, I will shew thee the bride, the Lamb's wife.
10 And he carried me away in the spirit to a great and high mountain, and shewed me that great city, the holy Jerusalem, descending out of heaven from God,
11 Having the glory of God: and her light was like unto a stone most precious, even like a jasper stone, clear as crystal;
12 And had a wall great and high, and had twelve gates, and at the gates twelve angels, and names written thereon, which are the names of the twelve tribes of the children of Israel:
13 On the east three gates; on the north three gates; on the south three gates; and on the west three gates.
14 And the wall of the city had twelve foundations, and in them the names of the twelve apostles of the Lamb.
15 And he that talked with me had a golden reed to measure the city, and the gates thereof, and the wall thereof.
16 And the city lieth foursquare, and the length is as large as the breadth: and he measured the city with the reed, twelve thousand furlongs. The length and the breadth and the height of it are equal.
17 And he measured the wall thereof, an hundred and forty and four cubits, according to the measure of a man, that is, of the angel.
18 And the building of the wall of it was of jasper: and the city was pure gold, like unto clear glass.
19 And the foundations of the wall of the city were garnished with all manner of precious stones. The first foundation was jasper; the second, sapphire; the third, a chalcedony; the fourth, an emerald;
20 The fifth, sardonyx; the sixth, sardius; the seventh, chrysolyte; the eighth, beryl; the ninth, a topaz; the tenth, a chrysoprasus; the eleventh, a jacinth; the twelfth, an amethyst.
21 And the twelve gates were twelve pearls: every several gate was of one pearl: and the street of the city was pure gold, as it were transparent glass.
22 And I saw no temple therein: for the Lord God Almighty and the Lamb are the temple of it.
23 And the city had no need of the sun, neither of the moon, to shine in it: for the glory of God did lighten it, and the Lamb is the light thereof.
If you aren't already aware of this, I think you would like it. I have a great deal of books on applications of these kinds of ideas. Rather than just being aware that there's
more to the world than the physical plane, there are practical applications that can change the world around you. It all tends to fall under the label of occultism, and I was trying to avoid saying that word since the Bible equates it to Satanic evil (which is incorrect) but it's the easiest way to describe it. This might deserve its own thread if anyone else is actually interested. I agree with you in that we were all brought here for a reason.
>a wall great and high (the grid)
>twelve gates (setSize = 12 by default)
>a golden reed to measure the city etc (something to do with the golden ratio maybe)
>an hundred and forty and four cubits
>cubits
You know, I thought it was dumb, but there are a lot of parallels.
So, if I have this right
the derivative of x (that's what that process is, I think)
is always the same number, or it's a repeating sequence?
And we'll call this sequence x_delta or ฮx
I'll check this out, seems like a fun new approach.
I'm really not sure if that's the derivative or not but I will be looking at this and seeing if the period or some other part of the sequence is 2n or some other number we could solve for.
You're thoughts and conceptions are correct.
I dont know if VQC would approve of x_delta, but thats how i've been thinking about it.
Holy Shit, Teach! No, you don't have to post Hitler, but many do. There are many ways to salute good digits. There is much debate about Hitler. IMHO I'll never post his face. Based on Q, he was a tool of the ++ Rothschilds. People on the chans love Hitler bc they think he was fighting (((them))) aka the (((yewden))) conspiracy. I think he was a tool of the ruling elite to cause war. That's my personal opinion, and could be wrong. But I will never salute that man, just like I will never salute a Rothschild. However, for many he represents resistance against the World Elites. Unfortunately, I think he was one of their pawns. That's my honest analysis. Fuck all totalitarian leaders and their FREEDOM negating bullshit.
Oh goodโฆ you're seeing what I saw.
He told me that there are multiple, different patterns in the grid that could each solve the factorization problem. Think of it from his point of view; there was a very good chance that his concept could have been completely ignored. Since it hasn't and since there are so many people who dropped all the nonsense they had been doing to solve this I think we love each other too much to disapprove of any attempt to solve it.
Give me some leeway and I could challenge your opinion about that.
First thing I'll say? Do you really know for sure that he started the war? There are many images and evidence that he had a high opinion of the United States and the Allies and that he made many attempts to bring a peace-effort.
I don't think of myself as a Nazi or anything.
Although I do like collecting rare photos. Hitler included.
And I agree he was a puppet.
You always have to watch presidents or prime ministers with a high public approval rating. It seems to lead to war more often then not.
Have you seen the documentary Icarus?
Why do Nazis hate Jews? Is it just an irrational hate? Is it just black and white that people who love Jews are completely rational and people who don't are the scum of the Earth and don't deserve the presumption of innocence?
What is a Rothschild?
Who created the symbol that is on Israel's flag?
What is the Belfour declaration?
What religion was allowed to charge interest for many years, while the other ones weren't?
Answer these questions and you will understand /pol/. They completely agree with /cbts/, they just somewhat hate them because they are turned off by National Socialism.
Sure.
I'll read it. I love learning. I have trouble talking to any family or friends now bc they all think I'm crazy when I start talking about Q's shit. VQC is like another "quantum" level up. Did you search for images of NP and SP? I couldn't find a single image. That's why all us "crazies" who are here are meant to be here. We want truth.
No, haven't seen Icarus. Worth the watch?
I felt bad. Deleted.
Definitely. Its a great film, although the creator is trying to link it to the Russia fake news, I think probably just to sell more tickets though.
Fuck AA, I checked it and it's like 100+ pages! Not happening tonight, but I will read it. TBH, anyone on this board has a high score for intellect and spirituality. VQC attracted that bc that's who he is. Humbly, I think we have the best of the best to work with our Senpai. That doesn't mean we're perfect, or are free from faults. It means we have brains and spirit to pursue this quest.
Well yeah, I said I was talking about books. It's a very good starting place for practical occultism. It's not for everyone but it sounds like it'll be right up your alley.
>I have trouble talking to any family or friends now bc they all think I'm crazy when I start talking about Q's shit
I just don't tend to talk to anyone about this kind of thing unless I know they're already into it. I'm lucky enough to have two friends in real life who believe in similar things to me. I certainly hope your family isn't the kind to want to be away from you for being into this stuff, because I've certainly heard of other people who lost a lot of important relationships by saying too much for others to handle.
Baker, I agree. I just think there's a bigger picture behind it all. I actually 100% agree with you. I just think it ties into +++, ++, +. /pol/ is right. but it's bigger than what even /pol/ knows. Satanic conspiracy to rule the world, that's what I think.
I'm trying to completely understand the links between the grid and quantum computing, since I haven't seen any other explanations. Knowing that the grid is meant to be a quantum computer, obviously there are going to be qubits somewhere. So is it that each block of (e, n) cells is a set of qubits (so if setSize = 12 there are 12 qubits for every possible set of (e, n) values) and that we use the mathematical relationships between c, e, n and setSize to find the right block containing the c we want, or is it that every block of (e, n) cells could be considered one qubit, we use Shor's algorithm to find the right block of cells, and then we use mathematical relationships to find the right cell (considering there could be an infinite number of cells with the same (e, n) values)? Or are the qubits something else?
I 100% accept and agree that Hitler advanced the Rothschild agenda. If you know the story, you will know that he wasn't a puppet, he got played. It all starts with Hitler wanting to get rid of the Jews. You see, the Rothschild's wanted all the Jews to come together (they were scattered around the world at the time) in the land where Israel is now and create the nation of Israel. Hitler thought they had a common goal, because he wants them to get out, and Rothschild wants them to go to Israel. So he did kick them out and they later created Israel in 1947 with the help of the UN. Hitler initially wanted to send them to Madagascar (kek) (look it up, Wikipedia may or may not lie to you).
Another note on that, Wikipedia has become a cesspool of admins who are Lying scum.
Look up the Wikipedia pages for Trump, Pizza gate, Obama, Hillary, Joe Arpaio, Roy Moore, Alabama election.
You will see they are full of shit.
Christians (normies) be trippin' when I start talking about our ability to create reality thru prayer and control of mind and emotions. Lol, It's right there in the bible. I can feel energy leaving my body when I pray nowadays. No joke.
Ok all, I'm off to bed!
Great progress tonight, see you all tomorrow.
Look up Icaros while you're at it. Music section.
I'm in the middle of extracting the patterns I'm seeing in Rev21 (as opposed to agendaโฆ)
Keeps branching into trees of curiosity.
So, in the meantime
Have you read Rev 21?
Is this easier?
Trying something:
I didn't realize I started my actual version at this pointโฆ formatting got screwy, but I was starting at 21:9 with a 7 kicker.
Originally started with LOLCat butโฆ Pidginโฆ numbers seem to pop out easierโฆ
Da New Kine Jerusalem Town
9Den, da seven angel messenja guys dat get seven bowl full a da seven big kine trouble dat goin come at da end a da world, one a dem angel guys wen come an tell me, โCome ova hea. I goin show you da lady dat goin be da wife fo God's Baby Sheep Guy.โ
10An God's Spirit wen take charge a me, an da angel wen take me to one big an high mountain, an show me Jerusalem, da town dat stay spesho fo God, coming down from God in da sky.
11Was awesome, an wen shine cuz God stay dea. Da light from da town wen shine jalike one jewel stone dat cost plenny, jalike one diamond, an you can see thru um jalike da crystal.
12Had one big, high wall wit twelve gate, an one angel guy by each gate. On top da gates wen stay write da names fo da twelve ohana fo da Israel peopo.
13Had three gate on da east side, three gate on da nort side, three gate on da sout side, an three gate on da west side.
14Da wall fo da big town wen get twelve big kine stone block fo da foundation. On top had da names fo da twelve guys dat Jesus wen send all ova fo tell peopo bout God's Baby Sheep Guy.
15Da angel guy dat wen talk wit me get one gold kine stick fo find out how long an wide da town, da gate, an da wall. 16Da town stay jalike one square, same wide an same long. Da angel guy go measure wit da stick how big da town, an he wen find out was 1,500 mile. Get same wide, one side to da odda, an same high all aroun. 17He measure da wall, an find out was 216 feet high, jalike how da way da peopo measure um, cuz da angel guy wen measure um lidat. 18Dey make da wall from diamond, an da town from solid gold dat look jalike glass dat you can see thru.
19Da foundation fo da town get all kine jewel stones dat cost plenny all ova. Da first foundation had diamond kine stone, da second one blue kine sapphire stone, da nex one red kine agate stone, da nex one green kine emerald stone,
20da nex one white kine onyx stone, da nex one red kine carnelian stone, da nex one yellow-green kine peridot stone, da nex one blue-green kine beryl stone, da nex one yellow kine topaz stone, da nex one green kine chrysofrase stone, da nex one red-orange kine zircon stone, an da last foundation purple kine amethys stone.
21Da twelve gate, dass was twelve pearl. Every gate was one big pearl. Da streets inside da big town was solid gold dat you can see thru, jalike da glass.
22I neva see one temple inside da town, cuz da Boss dass da God Who Get All Da Power, an God's Baby Sheep Guy, dey da temple!
23Da town no need da sun o da moon fo shine on top um, cuz God stay awesome, an God's light shine on top um, an God's Baby Sheep Guy jalike da lamp.
24All da diffren peopos all ova da world goin walk aroun inside da light ova dea. Da kings from all ova da world goin bring all dea awesome stuff inside God's big town.
25Day time dey no goin eva shut da gates inside da town. An you know wat? No mo nite time dea!
26All da diffren peopos all ova da world goin bring all dea awesome stuff dat cost plenny inside God's big town.
27But da kine stuff dat make peopo so dey no can pray no can go inside da town, an da peopo dat do shame kine stuff an bulai, dey no can go inside dea. Ony da peopo dat God's Baby Sheep Guy get dea names inside his Book, Da Book Dat Tell Who Get God's Kine Life, dey can go inside God's big town.
Goodnight, Teach!
It's been that way for a very long time. If you know anything about Gamergate (from the perspective of the people involved, not the media they were fighting against), Wikipedia actively ignored evidence and deleted it in the talk sections, and they all brigaded together to push the agenda the media were pushing. It's documented somewhere but I wouldn't remember. It'll go into far more detail. They're fucking scumbags.
"Creating reality through prayer" is almost exactly what comes up in a lot of the books I mentioned. When it comes to occultism vs Christianity, a lot of the practical information is more literal and based on the scientific method (do this and this and you'll make this and this happen) instead of being interpretive and metaphorical like it is in the Bible. I would suggest that you look into creative visualization, but you'd mostly get new agers, and they tend to ironically take occultism and take away the reason and explanation.
Topol, I'm getting out my bible now, Rev 21 about to be read.
Oh, The "New Jerusalem." love it already. Dude, this is like the end of LOTR. Ok, I'll post back when I'm done reading.
Good night, Anons. We got some math done, we got some talking done too. Glad to be here with you all. What a timeline we are in, friends. I don't think Senpai would have it any other way. Math and Spirituality combined are the best of all worlds. I'll be ready to work and crunch tomorrow, but it was a true pleasure having my mind engaged with you all tonight. Peace and blessings to all of you.
I've been thinking of it like this:
[G.Unit]
{A;Team}
(T,Cell)
/QGroup/
9 And there came unto me one of the seven angels which had the seven vials full of the seven last 1 7 7 7
10An God's Spirit wen take charge a me, an da angel wen take me to one big an high mountain, an show me Jerusalem, da town dat stay spesho fo God, coming down from God in da sky.
{Pyramid on top of a Cube? Cascading down from the top point?}
11Was awesome. Da light from da town wen shine jalike one jewel stone dat cost plenny, jalike one diamond, an you can see thru um jalike da crystal.
12Had one big, high wall wit twelve gate, an one angel guy by each gate. On top da gates wen stay write da names fo da twelve ohana fo da Israel peopo. 1 12 1 of 12
13Had three gate on da east side, three gate on da nort side, three gate on da sout side, an three gate on da west side.
{3 x 3 grid? 7 x 7? x 4 (sides)? 3 openings on the sides of the cube not touching the pyramid? 1010101?}
14Da wall fo da big town wen get twelve big kine stone block fo da foundation. On top had da names fo da twelve guys dat Jesus wen send all ova fo tell peopo bout God's Baby Sheep Guy.
{12x12?}
15Da angel guy dat wen talk wit me get one gold kine stick fo find out how long an wide da town, da gate, an da wall.
{Phi?}
16Da town stay jalike one square, same wide an same long. Da angel guy go measure wit da stick how big da town, an he wen find out was 1,500 mile. Get same wide, one side to da odda, an same high all aroun.
(And the city lieth foursquare, and the length is as large as the breadth: and he measured the city with the reed, twelve thousand furlongs. The length and the breadth and the height of it are equal.)
17He measure da wall, an find out was 216 feet high, jalike how da way da peopo measure um, cuz da angel guy wen measure um lidat.
(And he measured the wall thereof, an hundred and forty and four cubits, according to the measure of a man, that is, of the angel.)
18Dey make da wall from diamond, an da town from solid gold dat look jalike glass dat you can see thru.
19 And the foundations of the wall of the city were garnished with all manner of precious stones. The first foundation was jasper; the second, sapphire; the third, a chalcedony; the fourth, an emerald;
20 The fifth, sardonyx; the sixth, sardius; the seventh, chrysolyte; the eighth, beryl; the ninth, a topaz; the tenth, a chrysoprasus; the eleventh, a jacinth; the twelfth, an amethyst.
21 And the twelve gates were twelve pearls: every several gate was of one pearl: and the street of the city was pure gold, as it were transparent glass.
22I neva see one temple inside da town, cuz da Boss dass da God Who Get All Da Power, an God's Baby Sheep Guy, dey da temple!
23Da town no need da sun o da moon fo shine on top um, cuz God stay awesome, an God's light shine on top um, an God's Baby Sheep Guy jalike da lamp.
25Day time dey no goin eva shut da gates inside da town. An you know wat? No mo nite {time} dea!
Bumping >>250 since it quietened down a bit
funny, I come here to explore a math problem and find at least a couple anons willing to be true skeptics of religion and math:
>Skeptic does not mean him who doubts, but him who investigates or researches as opposed to him who asserts and thinks that he has found. [Miguel de Unamuno, "Essays and Soliloquies," 1924]
If all points in space and time are and have always been connected, then naturally our hearts and minds access that all the time. Prayer and meditation work with the deeper connections, not depending on space and time. Did any anons play Final Fantasy games growing up? The myth was always: "there used to be magic, but technology took over, and now no one believes in magic. We face dark forces but we band together. Relearn magic." Grow up, find out Japanese video games prepared me. what a world.
I believe that one of the messages of the New Testament is an invitation to vertically evolve. Horizontal evolution is Darwinian and slow. Plugging into technology could make us faster but in a way which is (I think) a trap. Normy society already being devolved/artificialized through technology (yes it can be used for good too). Vertical evolution is tapping into what's given to us. Bible invites us to uncover. Read book of John, to me the tone is "why aren't you using what I gave you." Ok that's my side rant.
I am not really into quantum computing, but I'm also looking for a link. If the grid is the superposition, we need to be able to collapse the grid and make it point to the answer with the highest probability. But how? VQC told us to focus on the gaps, so maybe they could be some kind of vectors, which can add up and point is the right direction.
Again, I'm a complete newfag when it comes to quantum computing. But I'm intrigued by the numbers. I will be mostly lurking, as I gather the active anons are at a further understanding than I am.
Recap of one of the current focuses: starting from any (e,n), the cells which are 2n apart in the e-direction form a sequence in which the x and n values both stay fixed (hence "small squares," (x+n)^2, are fixed), and in which a, b, and d increase by one for each jump of 2n columns to the right. You can move left into the negative columns down to a = 1.
{ -5: 2: 4: 3: 1: 11}
{ -1: 2: 5: 3: 2: 12}
{ 3: 2: 6: 3: 3: 13}
{ 7: 2: 7: 3: 4: 14}
{11: 2: 8: 3: 5: 15}
{15: 2: 9: 3: 6: 16}
{19: 2:10:3: 7: 17}
These are all products where a, b are 10 apart. The next one is missing though 818 = 144. Because it's a perfect square already so it got put in cell (0,0). What about the next one after that? For 919 the element is
{2: 1: 13: 4: 9: 19} then
{4 :1 :14: 4: 10: 20}
{6: 1: 15: 4: 11: 21}
โฆ
so this one finishes in the top row and seems like goes on forever in that row. Wonder if I started at n=3 the sequence would do a line in row 3 then a line in row 2 then row one forever.
So as some of you noticed the rule isn't universal but it's there.
and as a follow-up the positive versions of the negative terms in that list are in row 3.
>{ -5: 2: 4: 3: 1: 11}
>{ -1: 2: 5: 3: 2: 12}
{2: 3: 3: 2: 1: 11}
{8: 3 :4 :2 :2 :12}
I don't know if this holds up for other sequences, but in yours I found a*b - sqrt(((a+b)/2)-1) < 0 for your first part. If it equals 0 it's a perfect square. And when it is 0 it goes over to the other row.
Again I don't know if this holds up for anything other than your example.
This on the current cbts thread just earlier this am:
>>117453
Cross-posting images, they feel related.
Was lurking all last night with you all. Back to the shadowsโฆ
Holy shit, teach is checking digits.
Hitler sent those people do die in Russia on Purpose, fuck Hitler. Hitler was controlled Opposition
No no no, just no.
Now THATs a bit better!
MysterY [Spice] Melange (Dune ref).
An awareness spectrum narcotic is any narcotic substance that has the ability to expand an individual's consciousness on some level. This expansion of consciousness could manifest itself in many ways, including prescience, Other Memory, or simply an increase in immediate awareness and sensory ability.
Those who possessed the ability to see into the future when taking awareness spectrum narcotics were considered prescient. Spacing Guild Navigators were capable of such abilities, as were some members of the Atreides genetic line.
wiki: preยทscience
หpreSH(ฤ)ษns,หprฤSH(ฤ)ษns/
noun
noun: prescience; plural noun: presciences
the fact of knowing something before it takes place; foreknowledge.
"with extraordinary prescience, Jung actually predicted the Nazi eruption"
synonyms: farsightedness, foresight, foreknowledge;
โ "Future shows past" ~Q.
Ok, thanks for response.
Looking at line:
bool odd = c % 2 == 1;
(does ===1 also work?)
Understand the Mod operator now for this application, thanks. Read:
http:/ /cc.davelozinski.com/c-sharp/fastest-way-to-check-if-a-number-is-odd-or-even
-Interesting that Mod is fastest method, especially as the numbers get larger.
I know ZERO code. Took a Basic class in HS, got in trouble in Phys Ed and had to write out some "xyz" sentence a thousand times as punishment. Wrote a simple loop and sent output to PRN, brought to gym teacher who accepted it, maybe thought my coding career had hit it's zenith at that point, dunno? Pascal freshmen year, didn't get into it, always relied on others sinceโฆ Big mistake (regret). I get the concepts though, and quasi code makes sense. Sorry to be useless. (figured this story isn't much of a doxx).
No need for regrets.
That line basically just sets a boolean value to true if c % 2 == 1
>Interesting that Mod is fastest method, especially as the numbers get larger.
Good eye. I think it's for the next next next step with is to test for primality, instead of bothering to check the even for primes that bit of code I think will ultimately look like:
if odd = c % 2 == 1 { check if prime with dope new prime finding stuff }
if c % 2 == 1 { do stuff }
if c&1 == 0 {lewd ponies}
gents, I actually thought Sonic Rainbow was the show, Topo's link the other eve was first clip I'd ever seen of the MLP thing (old fag here). Now I remember the ads and that song. Duh.
Ok, VA, that was disturbing/twisted. Got the gyst w/out watching all, but now Brony as a concept is in the noggin, like a little bit of poison. Ok with that though, always stare awareness & truth in the face no matter how wicked - that's core promise to self.
quite interesting. Esp the accurate time thing. It was following the white rabbit last led me here. What a holeโฆ
Ok, enough sliding, back to lurkingโฆ.
"YOU'VE GOT LEWD PONIES IN YOUR CODE! START FROM THE LAST VERIFIED SAVE STATE!"
LewdCiphers are fine.
Butโฆ heyโฆ creepersโฆ leave ponies alone!
This has legs.
Likeโฆ 4 of them.
>https://youtu.be/N2tGqFjdX8A
I watched the Tyler the AI vid that topolanon sent.
Needless to say I'd have enough of his videos after that. I mean, if it's not bullshit
Where's the source? I like studying viruses.
Covfefe means "I will stand."
I still like Cov*Fefe
Covenant with The Scorpion Goddess
Covenant with Serket
Covenant with Circuit.
Am I the only ones who finds Topolowhatever annoying?
Several people here are derailing threads with spiritual nonsense. I keep adding filters, but people keep posting this nonsense. Can we make a math only thread or a spirtual shit another?
Glow harder.
I mean seriously, at least be useful.
What's day shift to me is night shift to others.
(((at least be useful, y'know?)))
Every time you numbnuts do this shitโฆ
We know we're onto something.
Y'all come at certain times.
Nothing is coincidence.
Y'all come after certain info drops.
Nothing is coincidence.
Y'all come always when people make connections.
Nothing is coincidence.
Lemme play the "Rings goes on! Ring goes off." game with you.
(((We see you, you are highlighted for a reason)))
Whereasโฆ when it comes to meโฆ
Everytime I kick a verse,
Every single word of it is:
Timeless
Ageless,
Evergreen,
Permanent.
I don't mess with any new fangled slangโฆ
But I sure make that Old School language bang!
I have a theory on possible values of n.
for every a and b combination in (e,n) a valid n record exists at (e,a) and (e,b).
examples:
(1,1,1) = {1:1:2:1:1:5} = 5; (x+n)=2x2=4; (d+n)=3x3=9
(1,1,2) = {1:1:8:3:5:13} = 65; (x+n)=4x4=16; (d+n)=9x9=81
(1,1,3) = {1:1:18:5:13:25} = 325; (x+n)=6x6=36; (d+n)=19x19=361
you will find n records at (1,1), (1,5), (1,13), (1,25) etc.
for (7,1)
(7,1,4) = {7:1:35:7:28:44} = 1232; (x+n)=8x8=64; (d+n)=36x36=1296
you will find valid n for both (7,28) and (7,44).
for (6,3)
(6,3,25) = {6:3:433:48:385:487} = 187495; (x+n)=51x51=2601; (d+n)=436x436=190096
I bet you can find valid records at (6,385) and (6,487).
So I believe we can effectively follow a and b to different levels.
Now we need to understand the small square relationship at n 1 and I believe we can "walk" the tree.
Please verify.
Dude, this is exactly what I said above :)
It doesnt work for all e, try e=9, and try to get to (9,3).
But something came to me in my dreams last night.
For (1,1) we know all a's are the sum of 2 squares. I'm exploring this pattern now, and I'll post my findings.
Didโฆ did you just? lol
Here's your hint to the next step:
Through WHAT does one go 9->3?
Why is it ph{r}ased like that?
>Through WHAT does one go 9->3?
Great question. I worked on this yesterday with Anon. See above!
Could be true.
Also note in the pattern that x = x when you move from (e,n) to (e,a) for some t.
Not at all, filter Topo:
Sliding
Mr.E:
Encouraging Topo and not knowing how to program while giving programming advice.
>Muh little birdy told me some code is fucked.
>Ur a shill for pointing out how stupid I am, ur a liar dude you weren't in VQC+++ posting code while I was fingering my butthole before becoming a nuisance.
Literal cancer. Literal Filtered Cancer.
>inb4 u glow
You stupid fuck, POTUS just confirmed control over the CIA with the Russia terrorist plot, there are no more CIA niggers fucking with you, they have no money, no infra, and are all running for their fucking lives. 5th column is entirely btfo.
Please filter me so you stop wasting space replying. Kthxbai
or in other wordsโฆ
</[{(buzz,-0ff)}]/3
Big tellโฆ
How come you're pushing "Muh Russia" over (((Israel)))? AIPACโฆ no one bitches that they've been meddling in our elections ever since the 50'?
Or, y'knowโฆ right after WW2?
There are quite a number of e records where (e,2) and (e,3) exist, but aren't in the root (e,1) key.
(20,1,1) = {20:1:10:0:10:12} = 120; (x+n)=1x1=1; (d+n)=11x11=121
valid n's include (20,2), (20,3), (20,4), (20,5), (20,6), (20,7), (20,9).
Perhaps valid a and b combinations are only part of the story.
>There are quite a number of e records where (e,2) and (e,3) exist, but aren't in the root (e,1) key.
How about this pattern here, for e=1, the list of a's are:
(1,1) =1, 5, 13, 25, 41, 61, 85, 113, 145, 181
= (0^2 + 1^2), (1^2 + 2^2), (2^2 + 3^2), (3^2 + 4^2), etc
= (t-1)^2 + t^2
(1,5) =1, 5, 17, 29, 53, 73, 109, 137, 185, 221
= (0^2 + 1^2), (1^2 + 2^2), (1^2 + 4^2), (2^2 + 5^2), (4^2 + 7^2), (3^2 + 8^2) etc
= pattern every 2 recs
(1,13) =1, 17, 37, 85, 125, 205, 265, 377
= (0^2 + 1^2), (1^2 + 4^2), (1^2 + 6^2), (2^2 + 9^2), (2^2 + 11^2), (3^2 + 13^2) etc
= pattern every 2 recs
That's something that bugs me.
CIA is terrible, and the NSA is the organization that would have had that intelligence, not the CIA, because NSA spying reach is massive.
(they need to be dismantled someday, sorry Q)
So why was it reported that the CIA was where they got the intel from?
And why would the CIA even care? They literally fund terrorist attacks, along with the FBI.
Strange.
Wellโฆ take into consideration that the NSA really was No Such Agency, as in a plausible deniability front.
And then GEOTUS did the unthinkable and made it NOW Such'n'Agency.
You have to understand the degree of compartmentalization that happens. Most people only know the precise scope of what they need to do and there's a culture of not asking questions and just doing your job 'for your country'. It was a signal to the world that the POTUS controls the CIA, not the 5th column. It's about Optics. There's good and bad actors in every organization.
I suspect there will be an electronic revolution which will enforce things like getting warrants for information instead of just exploiting hardware backdoors and whatever other methods they have at their disposal (timing, QC, etc). Disbanding the NSA doesn't make us more secure it just limits our capabilities relative to other state actors. Until we take care of the underlying flaws in our infrastructure there's not much that can be done.
Speaking of state actors, did anybody hear about the sophisticated malware they found in some european powerplants? Looks like someones plan B was thwarted.
>Even more to that, they HATE Russia.
That's just what (((they))) tell you with their media to keep you from realizing everybody is under their control. That's pure optics. This Putin POTUS thing was a signal of who is in charge and the new way of doing things based on cooperation and sovereignty.
A Nagual?
-facepalm-
Side note for anyone interested, you don't FIND a shaman.
They just show up.
Kek.
>V7
I was overjoyed when V7 was released, it meant people were finally worrying about what I had been worrying about for years.
I was a bit disappointed with the scope to be completely honest though. I know they have better toys than those CnC/Dropper/Persistence packages. Unless those toys are reserved for the SpecOP NSA dudes.
Even the hardware released was kind of meh.
I don't know how you could come to any conclusion, after reading that thoroughly, other than big crony alphabets need to be destroyed.
What is a psychopath.
Hey Teach! Yeah, I've been trying to find out how to apply this pattern to other cells.
PMA, good theory. I'll put some work in to check after day job is done.
Mr E, yeah it was gross!! I had no clue that subculture existed. I was messing around making lewd pony jokes, and then Topol dropped that on me. No more lewd pony jokes from me, lads.
The Way of the (Self*Healed) Madman.
Why Self*Healed?
Couldn't continue if he didn't know how to handle problems that ariseโฆ sometimes in batshit ways.
I don't disagree but just because we destroy ours doesn't mean it magically becomes any more difficult for some other actor to do the same things. We can destroy our own CyberPower all we want but that still leaves China, Russia, India, UK, France, Germany, Australia, New Zealand, 400lb hackers etc. The hardware and infrastructure is no less flawed because we remove our own cyber arm. You need to think bigger picture, why is it so easy to hack all these things? Why are undocumented API's allowed in consumer hardware and software? Who authorized this? Why? Are they the same people who benefit from this unchecked power?
I see where you are going here.
If you recall, there is a way to create small and large squares from only e and t.
Odd values of e:
small square (x+n)^2 = (2t)^2
large square (d+n)^2 = (2 * ( (t+1)*(t+1) - 2 * (t + 1) + 2 ) + ( ( e / 2 ) - 1 ) )^2
Even values of e:
small square (x+n)^2 = (2t-1)^2
large square (d+n)^2 = (2 * ( (t*t) - t + 1 ) + ( ( e / 2 ) - 1 ) )^2
I've verified these formulas for all entries in the grid.
Do we know the relationship between a and (x+n)^2?
No need.
If you really have the large square.
You have n.
And you have x.
And you have a.
And you have b.
I like you PMA. Your posts are always well thought out.
I'll take a look at this pattern and see if i can understand it.
So far I can't seem to figure out the pattern in a's for e=2!
Wait, you're saying this works everywhere?!
I think this is the missing piece for me. I'm going to see if I can use it to generate any given (e,n,t).
Thanks PMA!!!
can you clarify?
we can get to the large square where a = 1.
Finding the correct n that represents a factor of c is still a problem.
Looks like a relationship between the small and/or large squares. I believe the small squares as they appear easier to predict.
appreciate the kind words.
yes. I believe it works everywhere. I wasn't able to take it further, as knowing the value of t in different levels of n is a problem.
>I don't disagree but just because we destroy ours doesn't mean it magically becomes any more difficult for some other actor to do the same things.
What's the supposed time lag on defanging them and arming /ourselves/?
>We can destroy our own CyberPower all we want but that still leaves China, Russia, India, UK, France, Germany, Australia, New Zealand, 400lb hackers etc.
Even if we drain our swarmp, there are still seven seas. Then againโฆ that's why GEOTUS is rock hopping.
>The hardware and infrastructure is no less flawed because we remove our own cyber arm.
Again, time lag? Do you have suggestions on how to do Simultaneous Hokey Pokey? Eeeyup?
>You need to think bigger picture, why is it so easy to hack all these things?
Hate to say it butโฆ Low hanging fruit.
>Why are undocumented API's allowed in consumer hardware and software?
Undocumented by whom? The con-sumer? The re-tailor? It's about presentation. If it all matches, you just "sign" the EULA(-ogy)โฆ sign your life awayโฆ hmmmmโฆ
>Who authorized this?
The Qables.
>Why?
Same reason they wanted to Neutral-Eyes the Net.
>Are they the same people who benefit from this unchecked power?
Unchecked Power or Controlled Chaos?
Fire Spinning.
Problem, Reaction, Solution.
Thesis, Antithesis, Sin-Thesis.
Let's make a Metathesis.
Aโฆ M{ath}{eta}TheSIsโฆ.
Letting Her Blast off while we handle the Secret Intelligence Services by breaking their fucking toys across the World Wide Web.
Take them all down, in Unison.
Achieve Synchronicity, not Harm-On-Thee?
Metathesis
Mathetsis
Math Etsy's
Beautiful Hand Crafted Maths?
Okay, I'm jumping the gun a bit premature here, but I saw a pattern now in (0, 2) for a.
If you look at the x for (0, 2) and multiply this by 6 for each t you will get (0, 3) a's.
Again, if you look at (0, 1) from t = 2 and take a/2 you will find the a's for (0, 2)
Unfortunately I can't see an immediate pattern following the other n's for (0, n). (0, 4) seems to repeat (0, 0)
I was under the impression that you could get the large square that applies to the factorization we want, if you have t, according to those equations.
Hello PMA! Looks like we finally made it back around to these equations. I had a feeling we'd find a use for them!
Good morning Teach! If you like what PMA posted about (2t)^2 etc, maybe you could also take a look at the formula we attempted to derive for n. Out on the job right now, but you can find it at the top section of rsa #4, under "new equations". Some anons thought it was bunk, but we used the diff of squares equation and moved it around to solve for n, since we had c and small square. It could be wrong, but here we are again, so maybe it's worth another pair of eyes on it.
in (e,1) t follows a defined pattern. Next records are incremented by t = t + 1.
we haven't identified the pattern of t for (e,n) yet.
Not sure if they add any value. Maybe Teach can make some sense of them. It could just be another way of stating the same thing that doesn't lead to a solution.
>Odd values of e:
>small square (x+n)^2 = (2t)^2
>large square (d+n)^2 = (2 * ( (t+1)*(t+1) - 2 * (t + 1) + 2 ) + ( ( e / 2 ) - 1 ) )^2
>Even values of e:
>small square (x+n)^2 = (2t-1)^2
>large square (d+n)^2 = (2 * ( (t*t) - t + 1 ) + ( ( e / 2 ) - 1 ) )^2
I'm currently trying to confirm that the above equations is equal to the below equations, I think so, so far, will confirm in a few with a proof.
>if (e is even) (d + n)^2 = (2t)^2 + c
>if (e is odd) (d + n)^2 = (2t - 1)^2 + c
>if (e is even) (x + n)^2 = (2t)^2
>if (e is odd) (x + n)^2 = (2t - 1)^2
Yeah, hopefully Teach can confirm one way or another. It only works in (e,1) at the moment, so we still have to figure out (e,n) pattern for t like you said. Thanks for all your excellent work!
Actually, I'm not sure either of these sets of equations are correct.
For the first one, we can see that for odd values of e, the area of the small square should be (2t)^2. And the claim is that it works everywhere.
However, (2t)^2 is constant for a given t, eg, t=1, (2t)^2 = 4.
So if this were to work in every cell, then the first (x+n) should = 2 for every cell. Which is clearly not the case.
This equation works fine for n=1 though.
As for the second set of equations, as VA just stated, these work for n=1. Unfortunately that's where they end.
So for n=1, I believe these both state the same thing. For n 1, I'd imagine we'd need to factor in the n parameter when calculating values, somehow.
This TheoriE*S NANNERS!
NANNERS!!!
NaN^2Er*S
Sodium, 2Nitrogren, Erbium, Sulfur
<Pyridineโฆ DIAZEPAM?!
<Azepane Pesticides via saturated Azepineโฆ.
>ANYWAY!
11 (7^2=49) 68 16
11 relates to (7)(7) and thus 49
4^2=16
4^3=64
(4^3)+4=68
X^3+X=โฆ a trip to Wolfram Alpha.
http://www.wolframalpha.com/input/?i=x%5E3%2Bx
Noticed a pattern for a between e and e+1 records.
(e+1).a = orig.a + orig.x + 1
Example:
(16,1,11) = {16:1:228:20:208:250} = 52000; (x+n)=21x21=441; (d+n)=229x229=52441
a = 208
x = 20
(17,1,11).a = 208 + 20 + 1
(17,1,11) = {17:1:250:21:229:273} = 62517; (x+n)=22x22=484; (d+n)=251x251=63001
fastest is checking last bit. I had a CS course where we benchmarked a lot of things. For one assignment we were graded based on speed of program vs rest of students and I did pretty well though I'm a beginner coder. You can just sample the last bit, one or zero is even or odd. Don't remember commands
Cool! I'm out and about, no grid handy, but for the same n and t values, I think e=15 should share the same x+n with e=16. Also e=17 should share x+n with e=18. Is that correct or am I way off here?
(15,1,11) (x+n) = should also be 21.
dreams. awesome! follow em
Ah never mind, not for x+n. But do they share the same a values? In (15,1,11)(16,1,11) and then (17,1,11)(18,1,11) maybe the a values are paired.
Copy. Will look into further. What happens when the values of a and b become really large?
Arenโt we back at the same factoring problem?
I now understand God taking the rib from Adam. It refers to Ribosome, and he changed the DNA.
In that case, you might not see my post since you might have filtered me (I'm posting useful shit too, and asking questions), but we obviously haven't quite separated everything we're doing into different threads yet. We could always have a general thread for that stuff and redirect everyone there so this thread isn't so cluttered.
Bump
Please see >>356 too if you're one of the people who is used to only coming to one thread. Apologies for the triple post.
Kinda sorta seems to explain why they tried to write Lilith out, make you hate/fear beforehand if you found her, and make you think she's evil before you'd ever think of working with her in a positive mannerโฆ
Soโฆ if Mary = Eve (V)
And "God" = Y (Gnosis, Iesus, Catalonian Llesus)
Then the repaired Genome would be Xโฆ
That's why Jesus was the Sacrifice and why he felt forsaken.
After acting as the scapegoat his entire life, he still gets nailed to the cross?
Orโฆ What if that's another Shroud of Turn?
What's more blasphemous to Patriarchal Theocratatorship (They thought it was crazy enough to potato) than a Female Savior? The Daughter of God?
Not the "Anti-Christ".
The Inverse Christ.
The Inner Christ.
The Wombman.
Jesus would have been nailed by the romans for being a political threat.
She would have been burned for heresy by the pharisees.
Which makes one immediately think of Sophiaโฆ but then likeโฆ mmmmโฆ Joan of Arc or likeโฆ what's another female martyr who's more of a jewess than a christianโฆ let me justโฆ
DAFUQ IS THIS?!
https://en.wikipedia.org/wiki/Hellenistic_Judaism
({P, ยฌP},!)
ยก=[Here]
i=(You)
({P, ยฌP},!)
ยก=(Here)
i=(You)
;=(Me)
That looks more accurate.
And just how in the goddamn am I supposed to interpret that, good sir?
I am outside the paradigms of logic and death? How flattering.
Wellโฆ
Now you are.
Yeah.
And I wasn't trying to flatter you.
Quite the opposite!
I was, very simply, attempting to expand (You'r) mind a little.
Seems like you felt the vibe!
heheheโฆ
I know "(U)(R)" but what am "i"?
Imaginary!
R NOT!
R TOO!
R0!
R2!
Rรธ!
U=R^2!
>=rimshot=<
I'd give it a 12
12/10, of course.
It's likeโฆ. 120%
AND THEY SAID YOU COULD ONLY GIVE 100!
Also, y'all are gonna groan when you see that I was unintentionally punning correctly.
Yes!
U=Rรธi!
(You) are not me as {i} am not (You).
(You) are you from your perspective.
I am (You) from my perspective.
Everyone is individually (You) from their individual perspectives.
We are all (You),
sQuiGley
U=Rรธi!
(You) are not me as {i} am not (You).
You are (You) from your perspective.
From our perspective, we don't see (You) as you are Anonymous.
I am (You) from my perspective.
Everyone is individually (You) from their individual perspectives.
We are all (You),
<;3=
sQuiGley
On that noteโฆ
I wonder how many women legitimately got pregnant through their butts because no one ever realized what slit was cum in2โฆ
<;3=<|/ยกGet;It!/|3?!
I uhโฆ. lol
(You) should see what the following does to Wolfram Alpha. Keep in mind that the uhโฆ. the constant is Batman.
Just keep hitting the = button when you get back. lol
i<{a;m}3=2
Here's a preview while I put together to contact sheets.
Y'know.
To make connections.
Did y'all know there's such a thing as a "Lucas Phi"?
jwilson.coe.uga.edu/EMT668/EMAT6680.F99/Erbas/KURSATgeometrypro/fib%26luc%26phi/AKEwrite-up12.html
Aight, last one for a bit.
I have fuckering off to do elsewhere.
All quiet in here, so I'll post my findings for today. See if you guys can take this a bit further.
Started with a quest to see if I could walk the tree to various n records by substituting a and x from one to the next.
I iterated a number of values of t in (1,1,t) and jumped to the next n level.
Unfortunately, that led me in a circle where, for example:
From (1,1,2) = {1:1:8:3:5:13} = 65; (x+n)^2=4x4=16; (d+n)^2=9x9=81.
jumping by n=a:5, x=x:3.
to (1,5,2) = {1:5:4:3:1:17} = 17; (x+n)^2=8x8=64; (d+n)^2=9x9=81.
then the next jump by n=a and x=x sent me right back to (1,1,2)
However, in doing those tests, I noticed a small square pattern that might be of some use.
Following is a test I ran for c = 785, with a=5 and and b=157.
โโโโโโโโโโโ
TestPrimeResult for a=5, b=157.
From (1,365,14) = {1:365:28:27:1:785} = 785; (x+n)^2=392x392=153664; (d+n)^2=393x393=154449.
jumping by n=a:1, x=x:27.
to (1,1,14) = {1:1:392:27:365:421} = 153665; (x+n)^2=28x28=784; (d+n)^2=393x393=154449.
difference between to.c=153665 and from.(x+n)^2=153664 = 1.
to.c - from.(x+n)^2 == e? 1 - 1 = 0.
From (1,53,12) = {1:53:28:23:5:157} = 785; (x+n)^2=76x76=5776; (d+n)^2=81x81=6561.
jumping by n=a:5, x=x:23.
to (1,5,12) = {1:5:76:23:53:109} = 5777; (x+n)^2=28x28=784; (d+n)^2=81x81=6561.
difference between to.c=5777 and from.(x+n)^2=5776 = 1.
to.c - from.(x+n)^2 == e? 1 - 1 = 0.
โโโโโโโโโโโ
Explanation as follows:
From: (1,365,14) is the a=1 entry point for c = 785.
To: (1,1,14) is the record created by variables e, n, x where n=(1,365,14).a and x=(1,365,14).x.
From: (1,53,12) is the prime solution record we are looking for where a=5 and b=157.
To: (1,5,12) is the record created by variables e,n,x where n=(1,53,12).a and x=(1,53,12).x.
Patterns:
-
There is a consistent relationship between the From record and the To record that can be represented by the formula:
to.c - from.(x+n)^2 == e;
The difference between the destination c value and the source small square is equal to e.
This appears to be true for all "jump" records where a to.n = from.a;
-
The size of the small square in the jump record from a=1 entry point is equal to the small square size of the jump record from the prime result record.
(1,1,14) small square = 784.
(1,5,12) small square = 784.
What's the big deal?
I think there may be a solution in here somewhere. Using the following steps.
1) P(e,n) at a=1 for any c.
2) Pj(e,n) where n=P(e,n).a and x=P(e,n).x.
3) use the small square equality to figure out the pj(e,n) record. We now know e and the value of (x+n)^2. Perhaps some fancy algebra can give us values for x, n, or d.
4) reverse jump the pj(e,n) record to p(e,n). We know e, d, and something else from step 3.
If this makes any sense to anyone that would be awesome!!!
I'll test it out, make your instructions clearer. None of these dots and P()s. Be clearer.
would love a suggestion for better notation.
How hard is pseudocode ffs.
Not that it matters, this shit is a larp. I'm out.
Please someone help me holy shit
So, when you look at the grid, you see
aaaaa loooot of cells. But when we mention a cell we are talking about a pair of (e, n). The reason there are multiple (e, n)'s is because they have been extended by incrementing the t variable 12 times. t goes to infinity for each (e,n)
When you figure it out, include me in the cap.
I know that. Did you even read my post? I know how the grid works. I know that wherever there's a cell with {e, n, d, x, a, b} in it there are infinite possible sets with the same (e, n) value. I said that. I want to know what the qubits are if this is meant to be a quantum computer. Is it the cells? Is it specific groups of cells? Is it cells within specific infinite sets of (e, n)? Every single time I ask to try to figure this out, I get completely misunderstood. It's extremely frustrating.
All you've been doing is saying this is a larp and that everyone is a shill. If you really think this is a waste of time then leave already.
Correction, 3 weeks.
He said he got us 95 percent of the way there with 2 loops, we've done more than 5% of the work, nothing gained.
Did you not pay attention when he's talking about self sustaining sonoluminesence? Is that possible knowing what you know about entropy? Do you know anything about entropy?
And ignoring the original mission CBTS
Because this is a larp, and you are here spending time on a larp who's offered less proof than Q who's protecting classified information.
Then leave.
I gave him a fair shot, and he blew it, VQC is full of shit.
Oh offended your little game of light up the pretty dots has ended without a Quantum computer?
Shucks.
How does one encode a qubit in 2d space? How about 3d space?
Where is VQC? I thought he was teaching you something
Oh no, ISP's don't give out unlimited IP's.
What will I do :/
because no one has a good answer besides the one(s) who named it Virtual Quantum Computer. most people with experience with quantum physics are going to be not just normie but supernormie, ie. guardians of status quo in which they believe they know science best, so ironically they wouldn't help either imo, they'd laugh and run. in quantum physics, probablilities are heavily involved. A superposition of states โ but some more probable. In a VQC? assuming it's not a larp? Perhaps a type of superposition not involving probability. Or for all I know probability is involved.
Last thing โ this might not matter but hey it's the truth: we associate the word 'quantum' with some of the weirdness in atomic scale physics. But the word really just means that the energy levels come in discrete quantities (associated with electron orbits for example), as opposed to being continuous. Relevance unknown.
If nobody knows what the qubits are in this situation, how are we meant to use this as a VQC? If the entire point of this grid is to use it as a quantum computer, why are we not figuring that out? It's great if we can find the mathematical relationships between each variable used, and it's obviously important, but that's only half of what we're doing.
We are. We're generating cells. We're traversing them.
So how does that relate to quantum computing if none of this has anything to do with qubits?
Because it breaks RSA, and it shares the traits of having components that can store infinite data.
This isn't making it any clearer for me but thanks for trying.
I know what a qubit is. I haven't heard the word coupled in reference to qubits but I know about coefficients, amplitude, superposition, etc. Have you read the whole reply chain? People keep trying to help me assuming I don't know anything.
Well, as you already are aware, qubits are programmed by pairing (or coupling) states of a qubit to another. Q1 == Q2 or Q1 != Q2.
Take a look at digital logic circuits. With this qubit configuration, you can practically code an entire digital logic gate to test all paths simultaneously.
If you were to program a quantum computer to factor a number into its primes, it could do this in constant time (DWave achieved this with quantum annealing, factoring 21 into 7 and 3).
Looking at these "paths" were able to understand the factorization "progress" in terms of time, t.
My understanding is that were using integer factorization as an example. The intermediary variables (e,n,d,x) are like the qubit configuration, and t is time.
So (guessing with this part) essentially what were doing is "simulating how a quantum computer would solve integer factorization", by calculating all variables in terms of t.
Does that make more sense?
VQC's questions are teaching me, making me curious, starting new conversations with friends. Regardless, he's built some trust.
Checked in on his RootODavid twatter and he was meming it up like cray the other day, with the memes folks here are making. Maybe that's occupying him a bit? Maybe he's watching and this is unfolding according to its proper timing?
One example of a VQC question with respect to NK: What would happen if facial recognition were used on a crowd generated with CGI?
-
got me thinking about NK (and pretty sure that was before the whole NK Dark and empty city bit came out iir).
-
about privacy, and learning about what folks are doing about it, such as the:
http: //www.urmesurveillance.com/
'art' project giving you a new face and privacy.
Now, before we get all Phillip K Dick on everyone with facial recognition, these are some good questions to ask while we're still relatively early on in the applications of this.
Now, shake your face off Anons:
https:/ /youtu.be/5rdXvtdSIF8 (17sec clip)
PMA - keep pushing on, based on the rabid response, you might just be getting close to something.
Ok, back to the shadows, no-one to see here.
>qubits are programmed by pairing (or coupling) states of a qubit to another
I've spent all of the last week researching quantum computing and I haven't heard this anywhere. I have no idea what you're talking about.
>Take a look at digital logic circuits. With this qubit configuration, you can practically code an entire digital logic gate to test all paths simultaneously.
I have no idea what you just said.
>DWave achieved this with quantum annealing, factoring 21 into 7 and 3
What is DWave?
>in terms of time, t
When did time come into this?
>Does that make more sense?
Hahahahaha
I'm the one working on that simple guide, so if you can explain it to me, I'll do everything I can to explain it to everyone else. No offense, but I really don't think anyone without a strong background in this stuff would understand what you're saying. I have a background in it but clearly not as much as everyone else. This is my understanding:
>quantum computers are like regular computers with 1s and 0s, except, until a qubit is observed (so while computation is happening), they're a superposition of 1 and 0
>when a qubit is observed, instead of being a superposition in the form of an amplitude and a coefficient, it's either going to be a 1 or a 0
>one example of using qubits in quantum computing is factorizing numbers
>an example of how you would do that from several different videos and articles I've looked at is to have a qubit associated with each number from 1 to the number you're factorizing, doing some kind of operation that I can find absolutely no explanation of to raise the amplitude of the qubits associated with the results you want, and then getting the output you're looking for some probably high percentage of the time based on the amplitudes of all of the qubits
>another example is Shor's algorithm, which looks for the period of x mod n and which would work in a similar way to the above factorization problem
Now here's the logic I'm following if this grid is meant to be a quantum computer:
>quantum computers use qubits to find something in a big list of other things (at least as the main/only application of quantum computing that I've seen)
>this grid is meant to be like a quantum computer
>the difference between quantum computers and regular computers is the existence of qubits
>if this grid is specifically a virtual quantum computer, one of the elements of the grid is probably a qubit
>we're looking for a specific cell in this grid based on any specific c from a public key so that we can find its associated prime factors a and b and crack the private key
>that cell we're looking for each time is in a big set of other cells, much like the factors of a number would fit in a big list of all other possible factors
>based on this comparison, and based on this grid being called a quantum computer, either each group of infinite/empty (e, n) cell groups is a qubit, or each of the possible infinite cells within each group of (e, n) cells is a qubit that we then apply a quantum algorithm on to find out of the infinite set
>obviously this is all wrong for some god awfully complex reason that nobody has any ability to communicate
Read my Shor explanation in the other thread. Not done, but it should clear up Shor for you
Here's the Java port fixed. It turned out the anon who wrote most of it made a mistake on a vital line of it so I fixed it. This is the version with the image generator (it also has the graph but I commented that out since you need custom libraries). https://pastebin.com/vXj84tYk
If you're talking about the image then I guess I understand why everyone is so hell bent on using trips because that was me you were talking to about it in that thread.
Just got off the phone, I was super excited when I read it and yes, I'll look at it now.
As far as I'm aware, the only working quantum computers right now are the quantum annealing machines from DWave.
Here are their explanation videos, good educational material:
https:/ /www.youtube.com/watch?v=zvfkXjzzYOo&list=PLPvKnT7dgEsvVQwGgrlUVXBa2J6PAW8a4
Please let me know if I'm wrong with that above statement, I know there's a lot of theory out there but as far as I'm aware they're the only ones with working machines in production.
There are plenty of working QC's. Annealing ones have different types of Qbits.
Ok, finished reading, and understanding your claims, I'm going to write a test harness for it.
If this is trueโฆ this could be big!!
I hope we're not just reproving stuff.
Great work PMA!
No
Please elaborate.
>From: (1,53,12) is the prime solution record we are looking for where a=5 and b=157.
>To: (1,5,12) is the record created by variables e,n,x where n=(1,53,12).a and x=(1,53,12).x.
I'm following through your example.
Are the t values correct?
Instead of (1,53,12), I get (1,53,1).
And instead of (1,5,12) I get (1,5,5).
Am i missing something?
I've watched the first two videos so far. Here's almost exactly what the guy said in the first half of the first video:
>there are multiple approaches to quantum computing
>D-Wave is a company that is making quantum computers
>quantum annealing is their approach
>quantum annealing is a way of using the intrinsic effects of quantum physics to solve optimisation problems and probabilistic sampling
>optimisation problems are finding the best configuration out of many combinations of things
>you can frame them as energy minimisation problems, meaning you can use physics to find the answer since hot things cool down and things roll down hills (everything's finding its minimum energy state)
>quantum annealing uses quantum physics to find the minimum energy state of something
>sampling is slightly different: we're sampling many low energy states and characterising the shape of the energy landscape
>it's useful for machine learning to represent the word probabilistically
>these are NP problems
>"So that's a description of quantum annealing and the kind of things it's used for"
No it bloody isn't. "Using quantum physics to find the minimum energy state of something" is such a worthlessly vague explanation. I'm aware that the second video gets into this at least a little bit but that's a flat out lie, and it's the exact problem I've found with almost 100% of the videos and articles I've looked at when it comes to quantum computing: they're as broad and non-specific as they can possibly be, as if they're intentionally trying to avoid explaining how it actually works.
(cont)
Here's what he said in the second video:
>in their methods, they create qubits by creating magnetic fields with circulating electric currents
>qubits can be in superposition, and at the end of the anneal (that process that was allegedly explained in the last video) the qubit is observed as a 1 or a 0
>you can show the physics of this in a diagram with some valleys
>if you apply an external magnetic field to the qubits, the probability of observing one of the states will increase (the external magnetic field is called a coupler)
>when you link two qubits together, they influence each other
>coupling qubits makes them want to end up in the same state or the opposite state (relying on each other)
>this is called quantum entanglement - they need to be considered a single object with four states
>couplers apply biases to qubits
>each of these processes work to find the minimum energy state
>each extra qubit doubles the number of possible states
That still avoids or creates several important points (aside from the main idea that this doesn't explain how to use a quantum computer). Firstly, if you know where to apply biases, you don't need to search, just like if you're increasing the amplitude of a qubit represented as a wave you have to know which qubit to work on before you find the answer it's used to find. Why is this point glossed over everywhere? It seems like the most important step. Secondly, as much as it did explain that coupling thing you mentioned, is this only used in regards to quantum annealing, or is it used in all forms of quantum computing? The video made it seem like it was just in quantum annealing, which would explain why I haven't seen this anywhere else, and that it's irrelevant to the grid. That brings me to the third point: what does any of this have to do with the grid? That's been my question this entire time. If anything, this video seemed to lead away from the grid, since it explained the entanglement of qubits when everyone's telling me the grid doesn't have any qubits.
Look, you're trying to explain this, and I'm very grateful for that. I really hope I don't come across as rude, but you haven't answered any of my questions, and you've only confused me further.
I understand your confusion.
I would suggest going back through all the posts from vqc so far (here and in the rsa 1-4 threads on cbts).
I think this will be your fastest path to learning.
I'm going to continue to focus on the math work, but I welcome any lessons have you along the way.
Anyone want my real-time VQC program? Code is kinda messy but I can tidy it up
VQC never specifically referred to any part of the grid as a qubit, did he? He called it a quantum computer and never linked it to the practical concepts of quantum computing (by that I mean in ways we can use, such as something being a qubit), yet people like you just intuitively understand while not being able to communicate it. I'm not the only one who has no idea what the connection is, but I've made it my intention to not only figure it out but to explain it for everyone else. Thanks for trying. From the sounds of it, you're doing a lot of useful things with the math work, so don't let me distract you.
Yes please :)
The truth is we all are confused.
Imagine how you'd go about trying to code a quantum computer program.
Now, I'm not talking about the online simulators that link qubits etc. I'm talking algorithms.
A regular old computer (turing machine) can only calculate one thing at a time, so you program variables and at any point in time a variable has a value that can be used to calculate the next value.
In a quantum computer, imagine instead expressing relations as a set of simultaneous equations, and now you've defined all variables relative to each other at any time (this is our t variable).
BTW, your empty name field, are you VQC?
Your ID matches the Bakerโฆ
If so, I didn't realizeโฆ
No, I am just a baker. Would you like just the source or the whole thing? Because I can include the theme I used.
Are you playing around? If so, I'm lost.
If not, I'll take any code you have. I dont mind messy code. I'm used to it.
Java has themes you can easily add to programs, so they don't look like shit.
They are called LAFs.
Oh I see, I'll take it! Either way.
"Look and Feel" - well I'll beโฆ
Its been about 6 years since I last touched Java, should be ok though.
I'll have it ready tomorrow. My friend here has even mastered generating negative values of e
Such a tease.
Ok, I think I got the vertical gaps down.
Starting at (e, 1), we already know we can follow each a to (e, a) and the cell will be non-empty.
But for some cells, (eg, n=9) we can follow the a's all day and never get to (9, 3).
But there's one more ruleโฆ
We also have to include the set of all n's at any factor of a or b (i believe just at t=1 still have to confirm that bit).
Example:
(32, 1, 1) ={32:1:16:0:16:18}
We expect to find cells n=a=16, and n=b=18 to be non-empty which is true.
But also, 16 can be factored by 2, 4 & 8, so we'll find (32, 2), (32, 4) and (32, 8) will be non-empty.
Also, 18 can be factored by 2,3,6 & 9, so we'll find (32, 3), (32, 6) and (32, 9) non-empty too.
I'm going to try to figure out how to write a test for this.
What's interesting here is that if we're to use this, then we have to factor at least 'a', which is a recursion of the original problem. Could be useful.
Just quickly mentioning this >>491 simple code-running guide for those of you who only check this thread. I know some of you have only been lurking since you might not know how to use the code to make the grid etc.
Actuallyโฆ
I think you proved x+a = dโฆ
Both the jump cells have x+n = x+n (never mind the square).
Thats because they both came by following a.
And for the a=1, b=c cell, as well as the answer cell, d is constant.
And we know that d = x+a, so if we follow a, we end up with d = x+n for both cells.
Another thoughtโฆ
Similar to when we moved horizontally noting that while holding (x+n) constant, and increasing d, e = e+2n.
I've calculated what happens when we increase n by 1.
For a given (e,n), holding d, x and a constant, increment n (n=n+1):
e = e+2a
b = b+2
c = c+2a
Choose a prime number that is a factor of any value of a in a cell in the first row (e,1).
E.g. 5
E.g. (1,1)
5 appears as a factor of a in the second element (1,1,2,3,5,13) 5x13=65
5 will be a factor of the 2nd,7th,12th,17th value of a in (1,1)
5+1 is 6. 5 will be a factor of a in the (6-2) element at (1,1)
The fourth element is 25.
Five will be a factor of the fourth, ninth, fourteenth, nineteenth element of (1,1).
This is true of any factor p in any cell in row (e,1).
If first appearance of factor p is element t, second appears will be at (p+1-t)
p will be a factor of a in elements:
t+p, t+2p, t+3p,..
and
p+1-t, 2p+1-t, 3p+1-t,..
Since xx+e = 2na, every cell in row (e,1) contains a catalog of all values of na for a remainder e within the value a for those elements.
Is that it!!! You know e so all you have to do is search column (e,1).
By by RSA!
Its been great!
Big step but more to go!>>495
Sorry for excitement!
Cool, thanks VQC! "Every cell in row 1 contains a catalog of all values for na remainder e." So t will work fine as our index variable If our catalog of values remains in (e,1). So starting with c, we get d and e. Then e sends us to the correct column in row 1. Then we use our "catalog of factors" which is tied to the "s = t(a)" crumbs that Senpai dropped a few nights back to know how far up to ride the t elevator. I'll review the crumbs about factors and var s and get back to work.
Good morning everyone! I'm re-reading VQC's post from late last Wednesday, here it is. This is the one about s = function of t and a.
>>>/cbts/87168
His new post is a lot more clear, and it seems that he's changed s to p. If you look at the bottom of his new post you'll see the same pattern he was describing for var s.
It seems our mission for now may be to build the catalog of values for a in row 1. Thoughts, Anons?
Seems like my program would be helpful for this! I'll have it up in few minutes
Working on it and can get to some numbers in (1, 1) easily just following the pattern, but not everything. Example for c = 9253765:
( 1, 1, 2) {1:1:8:3:5:13} 65( 1, 1, 4) {1:1:32:7:25:41} 1025( 1, 1, 7) {1:1:98:13:85:113} 9605( 1, 1, 9) {1:1:162:17:145:181} 26245( 1, 1, 12) {1:1:288:23:265:313} 82945( 1, 1, 14) {1:1:392:27:365:421} 153665( 1, 1, 17) {1:1:578:33:545:613} 334085( 1, 1, 19) {1:1:722:37:685:761} 521285( 1, 1, 22) {1:1:968:43:925:1013} 937025( 1, 1, 24) {1:1:1152:47:1105:1201} 1327105( 1, 1, 27) {1:1:1458:53:1405:1513} 2125765( 1, 1, 29) {1:1:1682:57:1625:1741} 2829125( 1, 1, 32) {1:1:2048:63:1985:2113} 4194305( 1, 1, 34) {1:1:2312:67:2245:2381} 5345345( 1, 1, 37) {1:1:2738:73:2665:2813} 7496645( 1, 1, 39) {1:1:3042:77:2965:3121} 9253765
Also been rereading crumbs like mad and think these two are very relevant
Also (2,1), example for c = 7912971
( 2, 1, 2) {2:1:5:2:3:9} 27( 2, 1, 5) {2:1:41:8:33:51} 1683( 2, 1, 8) {2:1:113:14:99:129} 12771( 2, 1, 11) {2:1:221:20:201:243} 48843( 2, 1, 14) {2:1:365:26:339:393} 133227( 2, 1, 17) {2:1:545:32:513:579} 297027( 2, 1, 20) {2:1:761:38:723:801} 579123( 2, 1, 23) {2:1:1013:44:969:1059} 1026171( 2, 1, 26) {2:1:1301:50:1251:1353} 1692603( 2, 1, 29) {2:1:1625:56:1569:1683} 2640627( 2, 1, 32) {2:1:1985:62:1923:2049} 3940227( 2, 1, 35) {2:1:2381:68:2313:2451} 5669163( 2, 1, 38) {2:1:2813:74:2739:2889} 7912971
Again doesn't work for every value so I am definitely missing something
Hello everyone! For looking over everything for factor patterns to build the catalog of factors, we have this super clear work from PMA. It's cells (1,1) and (2,1) for t values 1-100.
>>>/CBTS/100653
>>>/CBTS/100657
Here is what the valid calculations for
> p will be a factor of a in elements:
> t+p, t+2p, t+3p,..
look like in 3D using e, n, t as coordinates
I would share the code but its so horribly bad I'm ashamed to have written it. Let me know if anyone wants it and I might clean it up.
Same as above but with e, n, d as coordinates
Can I see just what you're using to calculate T?
if is_odd(e): t = int((x + 1) / 2) else: t = int((x + 2) / 2)
Alternatively
t = int((x + 1) / 2) if e % 1 else int((x + 2) / 2)
I like your minecraft images by the way, they are much easier for everyone to understand than mine. Need to figure out something good.
Considered adding a direction indicator as well as text for some of the points but both are boring to code, and it really is much easier when you can just rotate and zoom stuff live. Encoding into mp4 kinda sucks for quality
Color scheme is just a gradient for the lines (red end for first record where a factor of c and yellow for the result where a * b == c), any ideas for something better?
Hey all, I think I have the starting seed for a in row 1. Please check my work and let me know if this makes sense.
For odd e at t=1, first a = (e + 1)/2
For even e at t=1, first a = (e )/2
This should allow us to determine the starting value of var a at t=1 for any e value. Seed of the tree, maybe?
Thanks, its a great distraction whenever I get stuck, and also a way to check my math isn't totally off
Plus, pretty images! Putting in some autorotation code and just watching it is awesome too
Tips on coloring, coordinates or stuff to graph is welcome. You can have them in higher resolution if we find something worthy of framing!
How do you all suppose I can generate a cell for n=0? Don't have a formula for one like that.
Ok so I said higher resolutionโฆ Guess I finally have a reason to buy a graphics card with more than 1GB memory!
Best I can do for now :P
Attached is an archive of the JAR file to run, along with the source.
https://anonfile.com/W44cofd6b6/VQCGUI.7z
Hmmโฆ good question. I think VQC said that at n=0, the only cell that exists is (0,0)??
For positive e. Other cells with n=0 exist for negative values of e.
And why do the points create a fucking X? Seriously!
Lol, x marks the spot like in Indiana Jones!
It's because the lines are through 3D space. Before anyone asks, the A=1, B=C record and the factor record don't intersect.
Good morning all.
Thank you, I've run your code, looks great, I'm going to try to read and understand your getRecord and getX method, and test them. Great work Baker!
Morning VA, I believe this is correct for n=1. I was up last night when VQC posted, but like you pointed out, I think this is just a continuation of his full explanation, and this is stuff we've been given before.
Still, it helped me a lot to understand the p/t relationship much better!
I'm kinda impressed by Baker's getRecord methodโฆ If I can confirm the results, I think this is huge!
Not mine, someone else made it.
Can you post the getRecord method here?
Thanks Teach! Glad you confirm it works for row 1. The thing that interests me the most is the ability to get var a using nothing but var e for t=1. That allows us to populate the entire element for t=1 at any e value, filling in remaining vars with what we have. Sorry if I'm repeating basics here, but working through how to generate all vars starting from c,d and e. If we know e and a for t=1 and create the the other first element vars, can we then use the p factor equations to build the complete element for t=2 and so on?
>Instead of (1,53,12), I get (1,53,1).
>And instead of (1,5,12) I get (1,5,5).
Iโm pretty sure my records are correct. I will double check later.
Increase the t value.
Shit, I seeโฆ sorry!
Whats the pattern then? Do I increase t each time I move to a new n?
Eg, if I'm at (1,1,7), is t for the (n, a) cell always t+1 ie, 8?
Youโre referring to the other getRecord method.
Iโve posted previously an IsValid method that I use when constructing every method. That way my test code will throw exceptions right away.
Might be useful for others.
>From Baker's (friend's?) code:
public class VQC {
public static BigInteger negOne = new BigInteger("-1");
public static BigInteger zero = BigInteger.ZERO;
public static BigInteger one = BigInteger.ONE;
public static BigInteger two = BigInteger.valueOf(2);
public VQCElement getRecord(BigInteger e, BigInteger n, BigInteger t) {
VQCElement v = new VQCElement();
try {
v.e = e;
v.n = n;
v.t = t;
v.x = getX(e, t);
v.a = (v.x.pow(2)).add(e).divide(two.multiply(v.n));
v.d = v.a.add(v.x);
v.b = v.a.add(two.multiply(v.x)).add(two.multiply(v.n));
v.c = v.a.multiply(v.b);
} catch (ArithmeticException ae) {
v.isInvalid = true;
}
return v;
}
public BigInteger getX(BigInteger e, BigInteger t) {
BigInteger subtrahend = one;
if ((e.and(one)).equals(zero)) subtrahend = two;
return two.multiply(t).subtract(subtrahend);
}
}
>My js version, (no big ints yet):
function getRecord(e, n, t) {
var v = {};
try {
v.e = e;
v.n = n;
v.t = t;
v.x = getX(e, t);
v.a = (Math.pow(v.x, 2) + e) / (2 * v.n);
v.d = v.a + v.x;
v.b = v.a + (2 * v.x) + (2 *v.n);
v.c = v.a * v.b;
} catch (e) {
v.isInvalid = true;
}
return v;
}
function getX(e, t) {
var subtrahend = 1;
if ((e%2) == 0){
subtrahend = 2;
}
return (2 * t) - subtrahend;
}
You can determine t from e and x.
So close though - great work!!!!
It seems to be related to the following of the (e,n) to (e,a) pattern.
I think its (e,a).t + 1?
You are right, there is no X :(
Ok, I cleaned it up, it is now 233 lines instead of 648, one of the dependencies is optional, and I added more keybindings. The defaults (at the bottom) are also a bit more sane.
Should run fine with just python and pyglet. Fair warning though, I didn't implement any scaling so the large resolutions only work if your OS / window manager doesn't force resize stuff, and you can't really see it when its offscreen. It also forces vsync off (cause I want to know when I fuck up the performance)
Variable names for generation are still pretty shit.
Left mouse + drag to move
Right mouse + drag to rotate
Scroll to zoom
F to toggle FPS counter (on by default)
R to toggle autorotation (on by default)
SPACE to save image to current directory
And there are probably bugs
That said, would love to see someone else post images from this thing
https://pastebin.com/wZM5Thzu
Credit for the "ent" function goes to Teach, ofcourse!
Everywhere.
FYI, that ent function doesn't work for n>1 yet.
Although the latest function posted by Baker's Friend (BF?) does work for all n.
The one last thing we're trying to figure out is the seed t value for a given n.
Looks like we're actually using this equation already. The only problem with the getRecord function is actually producing the correct t record given e,n & t.
The output we get is incorrect corresponding to the grid, however by increasing t, we'll get to the rec in the grid for some t. So all the equations "work" except t is "seeded" incorrectly.
If e is odd, at t=1 a = (e+1) /2. Then use d-a= x to get your x var, and solve for b var.
If e is even, at t=1 a = e/2. They use d-a to get x, solve for b.
So at t=1 for row 1 we can start with nothing but c and populate the element with d, e, a, x, b. Can we verify n=1 too?
Now let's find out how p lets us scale up with correct a var values as t increases. Thoughts, anons?
And knowing x allows us to create index var t, as well.
Looks like Teach and baker are working on scaling up!
Ok, I'm with you, I think this could work - I'll try to figure it out.
I also think this whole "following a" thing can deduce t.
I feel like this brings us so much closer to an answer!
Ok, here's my thoughts.
Starting at (e,1) you can move forward through t.
At any point, you can jump to (e, a), and t will begin wherever you were before.
So the mistake I've been making is considering all cells start at t=1.
They don't.
They're a tree.
From each t in (e,1, t), you can branch to (e,1,t+1), or (e,a,t), or (e,b,t).
Combine this with the factor p of a at t, and I think we have the full answer here.
Just gotta put it all together.
We can walk up the tree and down the tree.
I've been walking up and down the tree, nothing intersects.
Is there a pattern to this thing, yes. Does it lead to deriving factors, no.
t isnโt consecutive when n 1.
Are you sure? I thought it was consecutive but starting from a different seed.
Can you give an example please?
Don't give up! We're so close MA!
I'm sorry, you're right. I see that now.
I've rendered this puppy six ways from Sunday 20 times over. I'm convinced this is a larp more than ever.
Lot of effort for one. Show me some formulas that are this fucking cool, other than the mandelbrot set?
Also digits?
Looking at (3,13), using the getRecord method, the rows in the grid occur for:
t=4 to start, then 10, 17, 23, 30, 36, 43โฆ
eg, t=4 (which follows from a), then t increases by 6,7,6,7,6,7โฆ
6+7 = 13 = e
Confirming this pattern for other cells now.
I made a mistake 13 = n, not e
And this pattern is confirmed for (28,23) too.
t according to getRecord fn, 20, +3,+20,+3,+20
20+3 = n
This is not true for all cells. I rescind my claim.
Patterns yes. With this many subpatterns and regularity for prime numbers?
Sure, you can craft one but the basic formulas we are using are pretty simple, there isn't much room to hide something clever. Even if this is just a black hole I have learnt more interesting stuff about math lately than I probably would have, and refeshed my GL programming along the way. Not to mention I forgot it was possible to have this many geniuses in one place on the internet still!
Love you all, knowledge is power
Instead of trying to calculate T, try using the X variable in it's place.
Checked!
Teach, I agree. I think following the a var is the key. VQC keeps hinting at it. when we can find a way to solve for a in all cells (e,n) then we can create all the remaining vars in that element, and possible fill out the cell too. c, d, e, a, x, t (if needed?) b, and n. In that order of discovery I think.
I'm going to keep my focus in row 1 for now, and try to unlock how everything moves for a, x, t, and b. I know it's been discussed, but Notice Again that in all row 1, var b for t=1 becomes var a in t=2, and then you can solve for b. Then b drops to next var a, ad infinitum. Works for all row 1.
I'll try this out. I'm not sure if I know how to run it.
Yep, I have been just using this for moving down in row 1 for a while. Reverse it to move up
x += 2
a = b
b = a + x * 2 + n * 2
d = a + x
There must be something more to this pattern though, or a way to extend it to more rows
Install python 3
Install pyglet (pip install pyglet unless your distribution provides it for you)
The code itself should be cross-platform, but I haven't tested shit :)
I can always recode it in something else decent, but short of distributing a .exe file nobody will run I'm not sure how to make it simpler
I'm using Linux so that wouldn't be a good idea :^)
Agreed! I still haven't figured out how this works in other rows. Also, why some cells aren't populated. Any Anons figured out why this occurs? I've seen some hints and ideas, but not a clear explanation yet.
Nice! Baker gotz all the ponies in his stable. ;) I'll leave it at that, or Topol will start blowing up the thread with Brony videos, lol!
Technically python 3 is backward compatible and since I didn't use anything exotic, it should just work
All you have to do is install pyglet for the OpenGL stuff since its useless otherwise
If your graphics drivers work then one of these should be enough
./whatever.py
python whatever.py
python3 whatever.py
Here's a render with alternate dimensions. Instead of (E,N,T) this is (E,D,A) Sign format stays (E,N),(D,X),(A,B),(C)
Correction, that is (E,A,D).
output of pip install pyglet:
Command "/usr/bin/python2 -u -c "import setuptools, tokenize;file='/tmp/pip-build-KRuPQ_/future/setup.py';f=getattr(tokenize, 'open', open)(file);code=f.read().replace('\r\n', '\n');f.close();exec(compile(code, file, 'exec'))" install โrecord /tmp/pip-GqFtDy-record/install-record.txt โsingle-version-externally-managed โcompile" failed with error code 1 in /tmp/pip-build-KRuPQ_/future/
Um. Are you sure your distibution doesn't provide a packagne for pyglet? Which one?
error code 1 is kind of useless
No, it doesn't know what pyglet.gl is.
Nice work, MA! Thanks for always posting cool shit like this. (You) got a compliment from our Senpai. Even if it's a larp (my gut says it's legit) at least we've had lots of fun making cool shit, enjoying math, and talking about crazy ideas.
Well then I have no idea, pyglet doesn't do much if you cannot call direct gl commands :/
There's lots of patterns you can exacerbate by using alternate dimensions. I still can't find a formula though. Instead of (E,N,T) this is (N,A,D) world.
I'll just reinstall python and try again.
The implication here is that A does move linearly through N space. I still think T=X, but it might just be a lot easier that way.
I think you need python 3. I realised the "backward compability" really sucks and tried to make a version that works but its not going very well
pip3 gives the exact same error message
but I've downloaded it from here, I think this is the one.
bitbucket.org/pyglet/pyglet/downloads/
I like (N,A,X) world over NAD.
Best distro!
It's okay! I'm about to get it to work using the manual install.
Hope you make it, cause my progress looks like this
AttributeError: 'TheEnd' object has no attribute '_event_handlers'
AttributeError: 'TheEnd' object has no attribute '_view_event_handlers'
NameError: global name 'python2sucks' is not defined
TypeError: get() takes exactly 1 argument (2 given)
AttributeError: 'super' object has no attribute 'on_draw'
TypeError: 'module' object is not callable
TypeError: dispatch_event() takes exactly 2 arguments (4 given)
TypeError: dispatch_event() takes no arguments (4 given)
NameError: name 'fuck' is not defined
TypeError: unbound method init() must be called with TheEnd instance as first argument (got nothing instead)
TypeError: register_event_type() takes exactly 2 arguments (3 given)
TypeError: super(type, obj): obj must be an instance or subtype of type
TypeError: super() argument 1 must be type, not instance
TypeError: super() argument 1 must be type, not instance
AttributeError: 'Python2Sucks' object has no attribute 'get'
AttributeError: 'Python2Sucks' object has no attribute 'width'
>ponyfags are so rare that there's no way there could be 4+ in one thread
I'm actually not, I'm a burger.
So, it worked, is it meant to be spinning in circles?
Guys.
Now I'M doing this.
Ponies are good encouragement, butโฆ ignore the DL and Meta Q. There's a place for that.
Be all the MathGods you can be.
And now I'm fuckering off.
See you on the flip side!
Ok, I've
Figured out the controls.
Fucking.
LIT!
So how do I make it scale to a ridiculous resolution?
Lol! I call your bluff, Baker. Who's got the giant collection of ponies? I just watch them on Netflix w my daughters. Projection???
Checked, Topol!
Checked!! Holy shit, nice graphics, Baker!
OMG! You just made my day! ๐
Not my code.
But I'd like to know more.
Like what each point represents. And what its point is determined by.
position is determined by*
>work on the vqc from the second thread on
>post a couple ponies
>everyone here are ponyfags now
mfw
is a ponyfag now*
sorry I'm a simple mud pony who can't into fancy grammar
>holding a shovel with no fingers
>being this much of a n00b
That still wouldn't make them nearly as dextrous as necessary for carrying out human tasks.
I'm trying to behave :P
It behooves me.
>t+p, t+2p, t+3p,..
>and
>p+1-t, 2p+1-t, 3p+1-t,..
Put together a sample that shows how to step through the different factor methods.
The depth variable is the number of iterations.
โโโโโโโโโโโ
TestPrimeResult for a=5, b=157.
P(e,n) =(1,365,14) = {1:365:28:27:1:785} = 785; (x+n)^2=392x392=153664; (d+n)^2=393x393=154449.
P(e,1) a=n =(1,1,14) = {1:1:392:27:365:421} = 153665; (x+n)^2=28x28=784; (d+n)^2=393x393=154449.
Factors by Method 1: newT = depth * p + t;
(1,1,2) = {1:1:8:3:5:13} = 65; (x+n)^2=4x4=16; (d+n)^2=9x9=81
(1,1,7) = {1:1:98:13:85:113} = 9605; (x+n)^2=14x14=196; (d+n)^2=99x99=9801
(1,1,12) = {1:1:288:23:265:313} = 82945; (x+n)^2=24x24=576; (d+n)^2=289x289=83521
Factors by Method 2: newT = depth * p + 1 - t;
(1,1,-1) = {1:1:2:-3:5:1} = 5; (x+n)^2=-2x-2=4; (d+n)^2=3x3=9
(1,1,4) = {1:1:32:7:25:41} = 1025; (x+n)^2=8x8=64; (d+n)^2=33x33=1089
(1,1,9) = {1:1:162:17:145:181} = 26245; (x+n)^2=18x18=324; (d+n)^2=163x163=26569
(1,1,14) = {1:1:392:27:365:421} = 153665; (x+n)^2=28x28=784; (d+n)^2=393x393=154449
p(e,n) answer =(1,53,12) = {1:53:28:23:5:157} = 785; (x+n)^2=76x76=5776; (d+n)^2=81x81=6561
>Since xx+e = 2na, every cell in row (e,1) contains a catalog of all values of na for a remainder e within the value a for those elements.
However, I found an example that doesn't quite fit the factor iteration pattern:
โโโโโโโโโโโ
TestPrimeResult for a=17, b=53.
P(e,n) =(1,421,15) = {1:421:30:29:1:901} = 901; (x+n)^2=450x450=202500; (d+n)^2=451x451=203401.
P(e,1) a=n =(1,1,15) = {1:1:450:29:421:481} = 202501; (x+n)^2=30x30=900; (d+n)^2=451x451=203401.
All records where t < 15
(1,1,14) = {1:1:392:27:365:421} = 153665; (x+n)^2=28x28=784; (d+n)^2=393x393=154449;
(1,1,13) = {1:1:338:25:313:365} = 114245; (x+n)^2=26x26=676; (d+n)^2=339x339=114921;
(1,1,12) = {1:1:288:23:265:313} = 82945; (x+n)^2=24x24=576; (d+n)^2=289x289=83521;
(1,1,11) = {1:1:242:21:221:265} = 58565; (x+n)^2=22x22=484; (d+n)^2=243x243=59049;
(1,1,10) = {1:1:200:19:181:221} = 40001; (x+n)^2=20x20=400; (d+n)^2=201x201=40401;
(1,1,9) = {1:1:162:17:145:181} = 26245; (x+n)^2=18x18=324; (d+n)^2=163x163=26569;
(1,1,8) = {1:1:128:15:113:145} = 16385; (x+n)^2=16x16=256; (d+n)^2=129x129=16641;
(1,1,7) = {1:1:98:13:85:113} = 9605; (x+n)^2=14x14=196; (d+n)^2=99x99=9801;
(1,1,6) = {1:1:72:11:61:85} = 5185; (x+n)^2=12x12=144; (d+n)^2=73x73=5329;
(1,1,5) = {1:1:50:9:41:61} = 2501; (x+n)^2=10x10=100; (d+n)^2=51x51=2601;
(1,1,4) = {1:1:32:7:25:41} = 1025; (x+n)^2=8x8=64; (d+n)^2=33x33=1089;
(1,1,3) = {1:1:18:5:13:25} = 325; (x+n)^2=6x6=36; (d+n)^2=19x19=361;
(1,1,2) = {1:1:8:3:5:13} = 65; (x+n)^2=4x4=16; (d+n)^2=9x9=81;
(1,1,1) = {1:1:2:1:1:5} = 5; (x+n)^2=2x2=4; (d+n)^2=3x3=9;
p(e,n) answer =(1,5,7) = {1:5:30:13:17:53} = 901; (x+n)^2=18x18=324; (d+n)^2=35x35=1225
The factor of a = 17 doesn't exist in (1,1). Not sure how iterating factors is going help solve the prime problem.
Hi, it's me. VPN jumped.
I'm minding my nuts and lurching.
Closing the window now.
You think you know how to find me, let the PathAnons find The Tree and find me elsewhere.
Can't even watchโฆ the fuck kind of swinger's club is this?
I can provide
Y'knowโฆ since I'm such a tripping fucking name fagโฆ
I'm gonna do more than smile and nod.
I'm just gonna grin atcha.
thanks. let me know if you have any suggestions of avenues to explore.
There's a relationship between the small squares that I don't know how to define. And am running into roadblocks with all my tests.
Very nice, Baker! Teach claims he has an archive tho. Let's see what he's got!
Love you, Topol! Xoxo
is chris still gonna love us when he sees we turned the entire thread into a circlejerk pony shitpost?
OMG! Dying over here. Did you name that?
probably a derpibooru filename
We got some good work done earlier, lads. Ponies are fine for shitposting. Should we get back to work? I'm willing. Let's go. PMA is like /facepalm/ cmon.
Sorry, make that? Lulz.
I did make this tho
Have you seen his CBTS posts? He'll laugh and post some crazy awesome new equations or crumbs. Don't forget to search for images of the North and South poles, like he said to do. no joke. There are no images. VQC is farther out than Topol, but I can handle that. That's why I'm here.
Alright lads, I'm back to work.
>t+p, t+2p, t+3p,..
>and
>p+1-t, 2p+1-t, 3p+1-t,..
>Since xx+e = 2na, every cell in row (e,1) contains a catalog of all values of na for a remainder e within the value a for those elements.
Ok, at this point my big question is this: Can we use row one to generate all prime answers for the diff of squares equation? VQC hinted at this by saying that we could manipulate c to get it into row one somehow. If that's the case, then all t equations apply, and it will make our work much easier. Maybe not, tho. Thoughts, Anons?
Except the "global decontamination" button.
There is a direct relationship between the t and x values. Given an e, you can compute easily from one to the other.
VQC has sent us down a path of using factors to navigate the (e,1) space.
I view the t values as normalized data for iteration purposes. X is more closely aligned to the dimensions of the squares.
>What exactly is the big square, and what exactly is the small square?
The entire space we are working in is calculated from the Difference of 2 squares.
c = (d+n)^2 - (x+n)^2.
i.e. c = big square - small square.
I've adjusted my test output to indicate these square values, as I believe they better describe the relationship between the grid entries than the actual keys.
I wrote a confusing post yesterday, that I thought would lead us to the prime solution.
Following is another attempt at my thinking:
Step 1: Starting from c = 758 and a = 1, we can create:
(1,365,14) = {1:365:28:27:1:785} = 785;
small square = (x+n)^2=392x392 = 153664.
large square = (d+n)^2=393x393 = 154449.
Step 2: substituting n for a and using the same x, we can create:
(1,1,14) = {1:1:392:27:365:421} = 153665
small square = (x+n)^2=28x28=784. (this value is the same in step 2)
large square = (d+n)^2=393x393=154449.
Notice that the small square here of 784 = the c value in the first step minus e. I believe this is always true when create records at the same x by substituting n for a.
Step 3: I created the correct record for e where a and b are the prime values we are looking for, and then switched the n.
(1,5,12) = {1:5:76:23:53:109} = 5777
small square = (x+n)^2=28x28=784; (same value from step 2)
large square = (d+n)^2=81x81=6561. (this value is the same in step 4).
Step 4: substitute n for a and using the same x for the record from step 3, we can create:
From (1,53,12) = {1:53:28:23:5:157} = 785;
small square = (x+n)^2=76x76=5776;
large square = (d+n)^2=81x81=6561. (same value from step 3)
For step 4, the prime solution record, we already know e, d, and c. We just need one more variable to solve this problem.
Based on the movement in the small and large squares, I thought that if we could figure out how to create the record in Step 3, then we would have this solved.
For the record in step 3, we know e, and now we know (x+n).
This may still be a wild goose chase.
Searched and couldn't find any real pole images.
archive.fo/bDx84
t is a direct derivate of x, different equations for odd and even e. Given e AND a, you can easily compute x starting from c. Agreed, PMA. T is the index variable for row 1. But we need a first to calculate x, because d-a=x
For t=1 in row 1, here are the equations to derive var a:
(0,0) a =2
odd e: a = (e+1)/2
even e: a= e/2
This gives us the starting seed for a in row 1, t=1. How do we scale up to t=2, t=3, etc? You can solve for b in t=1 using our equations. Then b drops to var a in t=2, new b. Then b in t=2 drops to a in t=3. New b in t=3. Then b drops to a in t=4. Ad infinitum for all row 1 a and b vars.
Yeah I have been on to LENR for a long time. Quite a few claims that it works. Even a super-reliable claim that the russians have it working NOW. This is a huge game-changer. Anyway I have been off the grid for quite a while and it is awesome. 1 major problem is hot water. Easy in the summer as the sun will heat water in a black hose laying on the ground with a circulation pump. Another trick. Put your fridge outside. Really, In the winter when solar power is rare/valuable the cooler temps will make he fridge take less power. In the summer when there is ample sun it is in the heat. Little stuff like that is good to know going in.
Hey anons, just got some confirmation from VQC via Twatter. Check it out. I gotta go to bed now, but on it tomorrow am. This just came in.
I agree VA. I have lurked a lot as I am not really cut out for the coding part of it. I understand the implications of this as I do have a reasonably firm grasp of cryptography. I enjoyed making Caesar cyphers to reward people what worked for me in the past. I even made one that was a long strip of paper wrapped around a stick of a certain size. the code was the diameter of the stick. good stuff. I always felt a cone would make an even better one that would be harder to break. I like physical stuff. A lot of this is so ephemeral. This board is my favorite place to be ATM.
I was an early BTC miner and have since xfered into a pile of altcoins. Needless to say I have skin in the crypto-game. I would give it all up in a minuteโฆ nay 0LOGt! to see this thing come to fruition though.
I hope I can find a way to help. So far all I can really do is follow along with teach and minecraft and the others. It has been a pleasure so far.
I've also noted that the -E side (F), values seem to have (|E|, N) and (D,X) swapped? There may be insights in the grid if mirrored into negative space, or rotated into imaginary space.
Column 0 is actually very simple. (0,0) contains all of the perfect squares once, while (1+,0) has them again a second time. Also every (1+,0) coord is filled. Why do perfect squares perfectly fill a column, but imperfect squares don't?
I tried a similar approach with f records, and ran into a brick wall there too.
Hi all, just got home, getting caught up on the thread now.
A couple of pics for you. I reiterate though, I don't condone the man's actions.
Nice work PMA.
I think you're close man.
Thanks VA, I was examining the 0 column earlier today, but I had to go to work.
Back at it again now, I think its a useful tip.
So I found some UNEDITED pictures of the N. Pole. This was from some ice survey thing related to Glowbull warming. There are a few more on the sites that you can see in the URL. If anyone speaks russian there is a russian one as well but i cant read it. I hope this helps.
What do you all think of the theory that Tony Podesta has already been murdered by prolific serial killer Hillary?
He was last seen with her and he was the mastermind of her pay-to-play scheme. He had a lot of dirt and Mueller was closing in on him.
He may already be unable to make this world a shittier place than it already is.
Q?
Q said (believed to be disinfo) that TP was indicted on Nov 4th. Chess?
Not really the topic of this place but I reckon he's already gone.
Tomorrow I'll render the F side more. What is it about that side that makes multiple columns in (-E,0) that are fully populated with imperfect squares on N? N..-1, -4, -9, -16, -25, -36, -49, -64. Why are THOSE numbers perfect squares? I initially thought -E was a red herring, but I'm out of ideas.
Saw something along those lines, recall it sated the 'hole' shown was an effect of the orbit of the scanner and view of horizon.
Ironic - y'all building the Matrix (Grid), and everyone else helping redpill to get out of it!
http:/ /howtoexitthematrix.com/2017/08/23/the-matrix-the-sanskrit-texts/
https:/ /www.aeongroup.com/churning.htm
Yeah I am sure there is a perfectly reasonable explanation but where is the fun in that! The catch is this. Many of the pics on that site DO NOT have the hole. So sometimes its there sometimes its not. Whats up with that. also, go look at google earth pics of the moon. The show the whole thing but the poles are blurred out with a little star shape thingy. Its similar to antartica but smaller. Surely it is just all the slivers of the pictures knitted together but if you had a polar orbiter you could just take a clear picture. Really you could. Why knit a bunch of angled pics together when a polar shot would be ideal. Anyway, it was an interesting place to look.
In the original code,
int f = e - ((2 * d) + 1)
f, n - 1, d + 1, x + 1, a, b
Why is F calculated like that? Why are the variable's adjusted like that? On the E side, D=floor(sqrt(c)), on the F side, D=ceiling(sqrt(c)). What does this mean?
Work well with visuals. Started at the beginning, with some of the basics and some rules.
Found this example which was really nice to visualize this statement made in intro:
>All odd numbers are the difference of two squares. This includes all prime numbers and all numbers that are the products of two prime numbers.
So here we have line length 13 units (odd), bend in 1/2, get the large square, 7*7=49.
Then filling in the empty square (blue) we have 6*6=36. The difference is 13, the length of original odd line.
I swear to God you are gonna be the one to figure this out.
nice digits
also, f is for negative e values
for a given solution e and d, d on the negative e side is d + 1 or the ceiling sqrt of c like you said
the f, or negative e is: solution e - (2*(solution d) + 1)
You can use this python code to generate the (e,1) elements for negative or positive e values:
F isn't just negative -E values, F is a copy of the E side with different rules. Why the copy? What is the purpose of the new rules?
That's what I'm trying to figure out. Every time I try using the negative elements in conjunction with the positive elements, they cancel out the same as they cancel out normally. There's still a pattern we're missing. I keep working it out on a whiteboard and now in mathematica. It just keeps providing no solution and it's annoying lol
We're increasing our d by 1.
Start there.
The d = floor_sqrt(c), so (d+1) * (d+1) is a square slightly bigger than our c.
But we're fixing a and b, and hence c, so because of that, our e has to decrease by the difference of the square d^2 and (d+1)^2.
Yes, and I have the equations and initial value generation code to represent both sides now
Checkedโฆ and now Im afraid. I hope we are not building the matrix. Trips of truth though., Well maybe it will be a FUN matrix not a bad one.
If geometrically swapping N and A are equal, why does the grid not allow it?
Also going to the F side, we aren't fixing A or B, we're fixing X and N. Why?
What if D=floor(sqrt(c)), is the small square root, and D=ceiling(sqrt(c)) is the big square root, the answer lies in the difference between them?
I'm just throwing shit at the wall now to see what sticks. I gotta get to bed now though.
Yeah I have my money (LTC literally) On Teach getting it. but if he does, well LTC won't matter now will it. ha ha
If our number c is the product of 2 primes, a & b, then c*c will appear in the list of 0's 3 times.
First, when a = c & b = c
Next, a=c/a & b = ca
And, a=c/b & b = cb
And we can combine this with the statement that if p is a factor of a at t, so will be p+t, 2p + t, 3p +1, and p+1-t, 2p+1-t, 3p+1-t, etc.
I think this could work guysโฆ
I really appreciate the support, I feel very close now. Just gotta put this all together!
>>684 Kek has commanded me to send you math power through mathematical themed memesโฆ I will do my best!
>>684 Here is one to put over the mantle on Christmas day!
Spicy meme.
check'd
Another thoughtโฆ
There will at least be 1 of the 2 factors at a point where x = n, since we're removing a square of the same size that we're addingโฆ
Image to match.
There is one answer always when x=n.
If e = 0:
x^2 = 2an
If x = n:
x = 2a
So far this is true.
Therefore n = 2a too.
In the beginning (or maybe the half-video) he mentioned the final goal is to solve for N. Will this work for any input?
LMAO
He's /ourjew/
Sleep Well!
Gaps in the grid.
Which row has no gaps?
Each cell in the first row contains ALL the factors in it's column beneath.
Which column has NO gaps?
What is the pattern in COLUMN ZERO?
One Row to Rule them All.
A column that contains ALL and is the KEY to ALL the patterns.
What are the patterns in COLUMN ZERO?
If c is an integer, particularly an integer that is the product of two different primes, where the c squared appear in COLUMN ZERO. How many times does it appear? What are the patterns of these appearances?
How do you USE the Row and Column to find what you need?
What did Einstein say about co-In(c)iden(c)e???
There's something wrong with this.
Not sure exactly what, but a=5, b=7 doesn't work.
Certainly there is a relationship between n and x, in that n is a multiple of x.
Hey Chris, if you've got a little spare time to answer a question that isn't directly related to the hints you've been posting, I've been wondering this whole time how this all relates to quantum computing and I haven't understood any other anon's answers. Quantum computing involves qubits. That's what makes it different to normal computers. So where are the qubits in this situation? If nothing in the grid is meant to be analogous to a qubit, then how does this relate to quantum computing?
So close Chris! I can taste it!
>If c is an integer, particularly an integer that is the product of two different primes, where the c squared appear in COLUMN ZERO. How many times does it appear? What are the patterns of these appearances?
It appears 3 times, (0,0,c).
Then where a=ca, b=b
And b=cb, a=a
>How do you USE the Row and Column to find what you need?
I'm not sure!!!!
Trying to put it all together now man!
The incorrect assumption was that x=n.
x is a factor of n, as is a.
This the row movement in column zero.
Oh, and 1xc^2โฆ so four times?
Good question.
The cells that have elements have infinite elements, yet they are constructible, so only a few and finite amount needed to represent the entire grid into infinity both in e and n columns and rows and in each of the cells that contains at least one element.
Taken as a whole, the grid is the equivalent to a superposition of infinite qubits.
The "collapse" of the superposition occurs with the input variables d and e, (the square root and remainder of c), leaving the positions in the column e of the values of n.
For reference, see the quantum Fourier transform from Shor's algorithm.
I am a computer scientist (Architect), physicist, astronomer and designer. I also have an interest in asymmetric warfare and am a student of the work of John Nash, John CONWAY and Moti Milgrom.
I support a return to the Light like all Good patriots and am a big supporter of President Donald J Trump.
I am a recovering alcoholic and former drug addict and am on the autism spectrum.
The Twelve Steps of Recovery are a program for a spiritual illness.
I practice steps 10-12 daily and this is what led me to You.
All of us are only limited by our belief and our Faith.
You have what you need.
Godspeed Anons.
Love Always.
Good morning everyone! Loving this Math Challenge, and happy to be here working with you all.
This VQC will help take down the Evil Empire, hopefully. No secrets, and no place to hide their ill begotten money. In the wrong hands tho, it could be bad, just like a gun. Good guys and bad guys will duke it out with VQC's??
Starting to look at column zero..
Following is a test case for c = 149.
TestColumnZeroForAB a=5, b=29
P(e,n) =(1,61,6) = {1:61:12:11:1:145} = 145; (x+n)^2=72x72=5184; (d+n)^2=73x73=5329
P(e,1) na =(1,1,6) = {1:1:72:11:61:85} = 5185; (x+n)^2=12x12=144; (d+n)^2=73x73=5329
P(0,n) for c^2 =(0,10368,73) = {0:10368:145:144:1:21025} = 21025; (x+n)^2=10512x10512=110502144; (d+n)^2=10513x10513=110523169
P(0,1) na for c^2 =(0,1,73) = {0:1:10512:144:10368:10658} = 110502144; (x+n)^2=145x145=21025; (d+n)^2=10513x10513=110523169
p(e,n) a*b =(1,5,4) = {1:5:12:7:5:29} = 145; (x+n)^2=12x12=144; (d+n)^2=17x17=289
>How do you USE the Row and Column to find what you need?
processingโฆ.
In the zero columns since e is zero 2an =x^2
Morning all.
Here's where I got to last night focusing on the 0 col, and changes to x & n:
X can only be even, since we need to split x^2 into 2 lots of an.
Each time we increment x by 2:
c = c - (4 * (x+1));
And increasing n by 1, increases c:
c = c + 2*d
Since increasing x decreases c, and increasing n increases c, if c is constant, there will be some multiple of:
-
(4 * (x+1))
will equal a multiple of 2d.
I can't make sense of how to use the row and column! Any progress PMA?
Here's a list of the 100 first primes * the 100 first primes (except squares).
Included in the paste is the cells:
a=1, b=c (starting cell)
a=a, b=b (answer cell)
a=c, b=c (0 column c*c)
a=a, b=cb (0 column 2nd occurrence)
a=ca, b=b (0 column 3rd occurrence)
a=1, b=cc (0 column 4th occurrence)
https:/ /pastebin.com/DMb1j5z7
nothing yetโฆ
>for c = 149
c = 145.
Thinking out loud.
Interesting relationship here:
(0,1,3) = {0:1:12:4:8:18} = 144; (x+n)^2=5x5=25; (d+n)^2=13x13=169
(1,5,4) = {1:5:12:7:5:29} = 145; (x+n)^2=12x12=144; (d+n)^2=17x17=289
same d, e = e + 1, a = x + 1, c = c + 1.
From >>31
>Note that all the values of d for (1,1) are identical to the values of a for (0,1)
>Notice that all the values of a at (1,1) are the values of d in (0,1) with one added.
>Notice that all the values of a in this cell are also each twice the value of a perfect square.
>1+1 = 2
>4+4 = 8
>9+9 = 18
>Notice that all the values of d for this cell also follow a pattern:
>2x(1x2) = 4
>2x(2x3) = 12
>2x(3x4) = 24
Can we create a (0,n) record with the details from our entry point?
(1,61,6) = {1:61:12:11:1:145} = 145
New record values: d = 12, c = 144, e=0, and we have patterns that define both d and a per VQC.
If this (0,n) record is created, is the relationship to the prime answer simply a = x + 1?
Show the equation for x
t = (floor(sqrt(2d + 1)) + 1) / 2 ??
Converting to int is the same as floor. Can the original equation be rephrased to give x from d? and which d's and x's are we talking about? Big Ns, little Ns? Lol
VQC made a post here >>708 and it would be good if we had everyone's opinion. There have only been two other people who have chimed in so far and one of them was that demoralization shill I banned.
As if those (((bitches))) could keep up.
Hi Topo
-
while we're on the topic of rainbows, can you maybe shed some light on the:
Bifrost (pronounced roughly โBEEF-roast;โ Old Norse Bifrรถst). It's the rainbow bridge that connects Asgard, the world of the Aesir tribe of gods, with Midgard, the world of humanity. The "shaking or trembling rainbow.โ {vibrating}
This relates to Thor, and a friend of yours in the cbts General the other day (I thought they were you) was suggesting anon should dig deeper into history to understand and make the connections and read the map.
Maybe this would be best in a new thread though?
Hello PMA! I have an idea, and was hoping for your expert assistance. Here it is, let me know if you think it's worth exploring.
For (1,1) and (1,2)
Can we get a breakdown of all factors of a and b for t=1-100? Factors listed for each t value for a and b in ascending order?
So like this:
(1,1,1) a factors = 1 b factors= 1*5
(1,1,2) a factors = 15 b factors= 113
(1,1,3) a factors= 113 b factors =15*5
up to t=100
Same thing for (2,1)
Do you think this is worth pursuing? If not, no worries!
here
I try not to post during you alls 'working hours' so as not to distract, unless it's something relevant.
btw VA, the last link in the >>666 post is quite interesting - direct correlation to cbts. Trump could be equated to Vishnu in that legend. First a frog and now a turtle. Maybe when the storm heads to asia and india that story will be more applicable.
Thanks VQC. Some interesting reading too.
ps - you helped with my resolve to lay off the beers the past few weeks, thus increasing productivity.
not sure of the value. The factors for small numbers are easily calculated.
Currently looking into formulas for (0,n).
Thanks PMA, agreed. The factors are easily calculated for smaller numbers. I was thinking along the lines of being able to explore the factors of a idea VQC keeps posting. Check this post:
>p will be a factor of a in elements:
>t+p, t+2p, t+3p,..
>and
>p+1-t, 2p+1-t, 3p+1-t,..
>Since xx+e = 2na, every cell in row (e,1) contains a catalog of all values of na for a remainder e within the value a for those elements.
"Every cell in (e,1) contains a catalog of all values of na for a remainder e within the value a for those elements" I'm working to understand this one crumb. Do you have any insight, PMA?
sure.
there are 2 different ways to iterate records once you have a factor p.
Method 1: newT = depth * p + orig.t;
Method 2: newT = depth * p + 1 - orig.t;
With the newT, you can create a record using e,n,t values.
I showed an example and also found a case that didn't work properly.
Not sure I fully understand this.
Thanks PMA. Also interesting, VQC keeps posting this idea about factors of a being related to t, but at the moment we only know that consistent t values exist in row 1 (e,1). "One row to rule them all."
Question: Does your method 1 apply to even e, and method 2 apply to odd e? We keep seeing that same pattern for t equations.
Working to understand, not fully understanding yet either. I think rather than using depth*p + orig.t, maybe we build a chain of factors and then use it to understand the pattern and create a formula. Thoughts?
Also, I'll join you looking for patterns in (0,n).
>Does your method 1 apply to even e, and method 2 apply to odd e
With a factor of 5, I was able to use both methods and get different results on (e,1).
I don't think it matters odd/even. Though more testing is required.
>What did Einstein say about co-In(c)iden(c)e???
A coincidence is a small miracle when God chooses to remain anonymous.
(c)(c) just like Q's code. Must mean (c)^2?
yes.
Yeah, Topolanon should be able to post in the main thread, and also bake his own thread for MysticalMath too. Thoughts?
Ok, so we have (e,1) the one row to rule them all. Factors of a are key.
Then we have column 0 or (0,n)
Explore values of c^2.
Let's get some work done, lads.
ok, fine.
My mind is a bit blown away. So i'll let you guys ponder.
Following are tests for c^2 with records labeled as follows:
"P(0,n) for c^2" = an entry record where a = 1, c = c^2.
"p(e,n) a*b" = the prime solution.
"P(0,n) a^2, b^2" = a test where a=a^2 and b=b^2.
From the results below, you can see that the value c for "P(0,n) for c^2" equals the value c of "P(0,n) a^2, b^2".
So if we start from c^2, then figure out the movement down to "P(0,1) a^2, b^2", the prime solution turns out to be the square root of a, square root of b.
I'm currently working on the movement.
The difference in x values between the c^2 record and the a^2, b^2 record appears to a factor of 12.
TestABSquared a=5, b=13
P(0,n) for c^2 =(0,2048,33) = {0:2048:65:64:1:4225} = 4225; (x+n)^2=2112x2112=4460544; (d+n)^2=2113x2113=4464769
P(0,n) a^2, b^2 =(0,32,21) = {0:32:65:40:25:169} = 4225; (x+n)^2=72x72=5184; (d+n)^2=97x97=9409
n division =2048/32 = 64;
x diff =64-40 = 24;
xdiffBy24 =24/24 = 1;
p(e,n) a*b =(1,1,2) = {1:1:8:3:5:13} = 65; (x+n)^2=4x4=16; (d+n)^2=9x9=81
TestABSquared a=5, b=29
P(0,n) for c^2 =(0,10368,73) = {0:10368:145:144:1:21025} = 21025; (x+n)^2=10512x10512=110502144; (d+n)^2=10513x10513=110523169
P(0,n) a^2, b^2 =(0,288,61) = {0:288:145:120:25:841} = 21025; (x+n)^2=408x408=166464; (d+n)^2=433x433=187489
n division =10368/288 = 36;
x diff =144-120 = 24;
xdiffBy24 =24/24 = 1;
p(e,n) a*b =(1,5,4) = {1:5:12:7:5:29} = 145; (x+n)^2=12x12=144; (d+n)^2=17x17=289
TestABSquared a=5, b=157
P(0,n) for c^2 =(0,307328,393) = {0:307328:785:784:1:616225} = 616225; (x+n)^2=308112x308112=443724032; (d+n)^2=308113x308113=444340257
P(0,n) a^2, b^2 =(0,11552,381) = {0:11552:785:760:25:24649} = 616225; (x+n)^2=12312x12312=151585344; (d+n)^2=12337x12337=152201569
n division =307328/11552 = 26;
x diff =784-760 = 24;
xdiffBy24 =24/24 = 1;
p(e,n) a*b =(1,53,12) = {1:53:28:23:5:157} = 785; (x+n)^2=76x76=5776; (d+n)^2=81x81=6561
TestABSquared a=17, b=53
P(0,n) for c^2 =(0,405000,451) = {0:405000:901:900:1:811801} = 811801; (x+n)^2=405900x405900=1546052752; (d+n)^2=405901x405901=1546864553
P(0,n) a^2, b^2 =(0,648,307) = {0:648:901:612:289:2809} = 811801; (x+n)^2=1260x1260=1587600; (d+n)^2=1549x1549=2399401
n division =405000/648 = 625;
x diff =900-612 = 288;
xdiffBy24 =288/24 = 12;
p(e,n) a*b =(1,5,7) = {1:5:30:13:17:53} = 901; (x+n)^2=18x18=324; (d+n)^2=35x35=1225
Wow, this looks like an amazing new pattern. Triangles are definitely factoring into this. I wish I could focus on this but have job to tend to.
Least I can do is bake when the thread fills up.
Great work PMA! Triangles are back in style?
Ok, so reading your output:
You proposed three connected tests, correct? PLS help if I'm getting any part wrong.
Test 1. In column 0, find any places where a=1 and c=c^2
Test 2.For all (e,n) find any places where a*b = prime solution
Test 3. In column 0, find any places where a = a^2 and b=b^2. So, looking for places later in t where the new a and b are squares of previous a and b.
a^2+b^2=c^2. Triangles definitely back in style.
So looking down your output further, I see matching c^2 groups: 4225, 21025, 616225, 811801, etc.
Your claim is that the first line is c^2, and then the second line is a^2+b^2.
X diff = 1,1,1,12.
Following, see a very clear pattern. Any thoughts on the next step?
I don't know about triangles. But the relationship between records seems to hold.
Here is one more example for c = 20413.
TestABSquared a=137, b=149
P(0,n) for c^2 =(0,208324872,10207) = {0:208324872:20413:20412:1:416690569} = 416690569
P(0,n) a^2, b^2 =(0,72,823) = {0:72:20413:1644:18769:22201} = 416690569
n division =208324872/72 = 2893401;
x diff =20412-1644 = 18768;
xdiffBy24 =18768/24 = 782;
p(e,n) a*b =(249,1,3) = {249:1:142:5:137:149} = 20413
Next step is to figure out movement in the (0,n) column.
Credit where credit is due, lads. Want to take a moment to say: Thank You!! to PMA and Teach, along with MA, CA, AA, Baker and all other Anons who have been working with the code and generating all the excellent output and graphics, along with formulas. You guys are way ahead of me in that arena. I'm following and understand the concepts clearly. I can help with good questions and pattern identification, along with formula creation.
TBH, I'm kinda wowed by all the talent here, and sometimes feel a little nervous to chip in or even ask questions. However, I have faith that I can continue to contribute value to our quest. So please bear with me when I may ask seemingly repetitive questions. I'm studying and re-reading VQC's crumbs all the time. Honestly, I just love math, and always have. That's why I'm here. Don't need/want glory etc. just want to help do my little part to help in the Q / VQC quest. Love you all. Glad to be here.
Evening all - just checking in now, getting caught up.
Thank you for establishing and fostering this community.
PMA have you taken a look at the 0 col for:
a=ca, b=b
a=a, b=ca
There are many interesting relationships. I'm trying to figure it out and write it up at the same time.
But there's a relationship from one row to the other, where the in the first occurence:
x = b(a-1)
and in the 2nd:
x = a(b-1)
I haven't fully figured it out, but have you put any cycles there yet?
Shit, Teach hereโฆ forgot the name fieldโฆ
not yet. will look into.
Is this what you're looking for?
TestTeachCombinations a=5, b=13, c=65
a = ca, b = b =(0,104,-129) = {0:104:65:-260:325:13} = 4225
a = a, b = ca =(25,125,18) = {25:125:40:35:5:325} = 1625
I somewhat explained the first record wrong.
a=b, b=ca. Its the same thing, just the other way around.
I get the following rows:
65 = 5 * 13
1 x c {"c":65,"e":1,"n":25,"d":8,"x":7,"a":1,"b":65,"t":4}
a x b {"c":65,"e":1,"n":1,"d":8,"x":3,"a":5,"b":13,"t":2}
c x c {"c":4225,"e":0,"n":0,"d":65,"x":0,"a":65,"b":65,"t":1}
a x cb {"c":4225,"e":0,"n":360,"d":65,"x":60,"a":5,"b":845,"t":31}
b x ca {"c":4225,"e":0,"n":104,"d":65,"x":52,"a":13,"b":325,"t":27}
1 x cc {"c":4225,"e":0,"n":2048,"d":65,"x":64,"a":1,"b":4225,"t":33}
For e=0, a=ca, b=b, or e=0, b=cb,
D is known, as is e (0).
I'm trying to relate the 4 variables left (a, b, n, x).
I have n is always a multiple of x.
I think we can say that n = (a((b-1)^2))/2 or (b((a-1)^2))/2 for those rows.
Thats where I'm at.
Nice PMA. So at the larger numbers, is t irrelevant? Was t like training wheels for learning how to ride a bicycle in row 1? Then when we move to larger n the training wheels are no longer needed?
Thanks, Teach. Glad to have you dropping knowledge here. Love this community of MathFags. You all rock. More rare hitlers? You now have your own rare Pepe! :)
One day (maybe) we all meet up at Tony Stark (VQC's) lair and hang out IRL for a coffee or beer. Even my wife is like "You're so happy working on math(s). You must really like these MathFags." I'm like "yeah, I never knew how much I love math and MathFags."
I agree, I enjoy it here.
There's a lot of noise still in the community (non-math stuff) but I enjoy it. I like the active feeling and I like seeing that there's a new post, regardless of what its about.
Trying to explain it geometrically:
These are some very useful relationships.
When making a difference of 2 squares for a number c*c, we're making a big square, c by c (in our case c = d).
If c is actually ab, then the cc square could be divided up by aa squares of size bb, or bb squares of size aa.
When trying to make a difference of 2 squares, where x is not 0, we are heavily restricted in this choice if a and b are prime.
As we increase x, we are "eating away" at c*c, we have to make up for this, by increasing n.
If we just focus on the smaller squares inside cc (of size aa or b*b), there's only 1 choice.
We have to choose x such that it is equal to a *
(b-1).
b-1 will always be even, and when we use the small squares that we've taken away via x, we build them up via increasing n, half on the side, half on the top.
We need to guarantee that the small squares taken away = the small squares added. And that will always be true for the largest even square (b-1) times the other prime a.
I know I'm not doing a good job of explaining it.
My big question is how do i use row and column in the process of finding the answer.
I think I'm on the same page now.
65=5x13
1 x c =(1,25,4) = {1:25:8:7:1:65} = 65
a x b =(1,1,2) = {1:1:8:3:5:13} = 65
c x c =(0,0,1) = {0:0:65:0:65:65} = 4225
a x cb =(0,360,31) = {0:360:65:60:5:845} = 4225
b x ca =(0,104,27) = {0:104:65:52:13:325} = 4225
1 x cc =(0,2048,33) = {0:2048:65:64:1:4225} = 4225
Yep, look at that thingโฆ beautiful.
I haven't figured out how to go backwards.
Here's a pate for more data:
https:/ /pastebin.com/DMb1j5z7
Checked!! Holy Shit Teach! All 7's
x can never be 0 in the diff of squares. Just sayin'
I changed the order, and added in the aa x bb record. I think that's the one we're looking for.
Square root of a, Square root of b gets us the solution.
65=5x13
1 x c =(1,25,4) = {1:25:8:7:1:65} = 65
a x b *=(1,1,2) = {1:1:8:3:5:13} = 65
c x c =(0,0,1) = {0:0:65:0:65:65} = 4225
aa x bb =(0,32,21) = {0:32:65:40:25:169} = 4225
b x ca =(0,104,27) = {0:104:65:52:13:325} = 4225
a x cb =(0,360,31) = {0:360:65:60:5:845} = 4225
1 x cc =(0,2048,33) = {0:2048:65:64:1:4225} = 4225
I believe that indicates a perfect square.
x is always less than a, because d-a=x.
Column 0, right?
Shit! got it wrong. Thx for the correction.
Wait, d is the perfect square with remainder e. a is the small square. X is a derivative of d-x. x will always be smaller than a. Confirm?
Here are all my samples. * indicates important record
Really is beautiful.
65=5x13
1 x c *=(1,25,4) = {1:25:8:7:1:65} = 65
a x b *=(1,1,2) = {1:1:8:3:5:13} = 65
c x c =(0,0,1) = {0:0:65:0:65:65} = 4225
aa x bb *=(0,32,21) = {0:32:65:40:25:169} = 4225
b x ca =(0,104,27) = {0:104:65:52:13:325} = 4225
a x cb =(0,360,31) = {0:360:65:60:5:845} = 4225
1 x cc *=(0,2048,33) = {0:2048:65:64:1:4225} = 4225
145=5x29
1 x c *=(1,61,6) = {1:61:12:11:1:145} = 145
a x b *=(1,5,4) = {1:5:12:7:5:29} = 145
c x c =(0,0,1) = {0:0:145:0:145:145} = 21025
aa x bb *=(0,288,61) = {0:288:145:120:25:841} = 21025
b x ca =(0,232,59) = {0:232:145:116:29:725} = 21025
a x cb =(0,1960,71) = {0:1960:145:140:5:4205} = 21025
1 x cc *=(0,10368,73) = {0:10368:145:144:1:21025} = 21025
785=5x157
1 x c *=(1,365,14) = {1:365:28:27:1:785} = 785
a x b *=(1,53,12) = {1:53:28:23:5:157} = 785
c x c =(0,0,1) = {0:0:785:0:785:785} = 616225
aa x bb *=(0,11552,381) = {0:11552:785:760:25:24649} = 616225
b x ca =(0,1256,315) = {0:1256:785:628:157:3925} = 616225
a x cb =(0,60840,391) = {0:60840:785:780:5:123245} = 616225
1 x cc *=(0,307328,393) = {0:307328:785:784:1:616225} = 616225
901=17x53
1 x c *=(1,421,15) = {1:421:30:29:1:901} = 901
a x b *=(1,5,7) = {1:5:30:13:17:53} = 901
c x c =(0,0,1) = {0:0:901:0:901:901} = 811801
aa x bb *=(0,648,307) = {0:648:901:612:289:2809} = 811801
b x ca =(0,6784,425) = {0:6784:901:848:53:15317} = 811801
a x cb =(0,22984,443) = {0:22984:901:884:17:47753} = 811801
1 x cc *=(0,405000,451) = {0:405000:901:900:1:811801} = 811801
6107=31x197
1 x c *=(23,2976,39) = {23:2976:78:77:1:6107} = 6107
a x b *=(23,36,24) = {23:36:78:47:31:197} = 6107
c x c =(0,0,1) = {0:0:6107:0:6107:6107} = 37295449
aa x bb *=(0,13778,2574) = {0:13778:6107:5146:961:38809} = 37295449
b x ca =(0,88650,2956) = {0:88650:6107:5910:197:189317} = 37295449
a x cb =(0,595448,3039) = {0:595448:6107:6076:31:1203079} = 37295449
1 x cc *=(0,18641618,3054) = {0:18641618:6107:6106:1:37295449} = 37295449
20413=137x149
1 x c *=(249,10065,71) = {249:10065:142:141:1:20413} = 20413
a x b *=(249,1,3) = {249:1:142:5:137:149} = 20413
c x c =(0,0,1) = {0:0:20413:0:20413:20413} = 416690569
aa x bb *=(0,72,823) = {0:72:20413:1644:18769:22201} = 416690569
b x ca =(0,1377952,10133) = {0:1377952:20413:20264:149:2796581} = 416690569
a x cb =(0,1500424,10139) = {0:1500424:20413:20276:137:3041537} = 416690569
1 x cc *=(0,208324872,10207) = {0:208324872:20413:20412:1:416690569} = 416690569
*sorry "x is a derivative of d-a"
Saved!
Julian is a hero of mine too.
I love these old photos, these men on both sides were our family. Its our history!
Every time I try to solve these things simultaneously I end up proving what is known.
I think I want to go back over all of VQC's posts and try to gather every single equation into one spot again.
This thread is almost at limit. I don't feel confident enough to bake, but I can certainly help collect and surmise.
In the Q threads, they had a summary in the form question/answer. Maybe we can attempt something like that.
I thought you guys would like some personalized memesโฆ Im useless on this project but its cool as fuck so I will attempt to bring KEK's blessings to you.
This meme is self explanatory
I believe X is linear for all cells.
then we need a formula!
The only problem working with c^2 in N0 is that c^2 is gonna be massive. Unless E reveals a search method for N0, this is not good.
no idea. I was looking for relationships between the records, and that number came out.
I'm honored, thanks! Truth and Justice/Equity.
Checked!!
โฆAnd if you see Topolanon This one is for him! Glad you guys like them. When I see a good one I will make it up and post it., Keep up the good work. I have been on the edge of my seat for a while now. If you need any back-end support let me know. I will do what I can.
Sorry new ID, is this Mr. E?
X is what I've been plotting by, instead of T. It shows you can move on a 3D linear path through cells.
Nope NOT Mr E. Just an anonymous Hobo. IDK how to trip/namefag properly? Its limited? Tried here and post didn't go through.
The multiple of 12 observation seems to be a robust conclusion for a of 5 or above. Between the 1cc and aabb cases, the difference in x works out algebraically to a^2-1, or (a+1)(a-1).
Since a is odd, because it's prime, both a+1 and a-1 are even, for two factors of 2. And one of a-1, a, or a+1 has to be a multiple of 3, and it can't be a because again, it's prime.
So a^2-1 has to be a multiple of 223 = 12. Neat.
Good luck tonight ! KEK be with youโฆ
Wow! Great explanation!
Well shit, thanks for the memes.
In return, here you go.
When go got to post, go to the name field.
Enter yourname##password. Simple as that.
Actually the same logic can show that one of a+1 or a-1 has to be a multiple of 4, too. So the differences in x ought to all be divisible by 24.
Render of E0 with X for T.
While the diagonal lines are what pops out, the pattern stems from the E0,N1 column. X increases by 2 each T, and extends down N. It would probably fill in each N with a deep render.
I believe they were. My original tests above include an xdiffby24 variable for that exact reason.
12 seemed more reasonable as an assumption.
Now, however, can you help find a formula for the actual diff?
MA, thanks. These visuals are excellent.
From what I can tell, for any (E,N) X has an initial value, and a fixed growth value. X will start at the initial value, and += the growth value for every T.
a^2-1 is the actual difference in x values. Or did you mean something else?
Nope, I lied. Whatever X starts as is how it increments in the cell. This is true for the entire grid (please confirm) and why I initially began using X as T.
Hehe. Let's solve for n. :)
Here's a picture for Topolanon. Tree of life!
Baker! I know you got us. New bread soon?
Topol is the shit. I love that faggot. Nice pic, MA!
Wow, MA! Tree of life indeed. Crazy beautiful. Thx!!!
(You) me what I should add to the batter.
Also send me some fancy math pics. Have been too busy to make more of the ones like in the OP, but I have a lot of them planned.
Horizontal lines are actually fixed N,D,X, and A, and are the incrementing of B. The 3D lines the incrementing of A.
chk'd!i! ..and inching closer!
lulz, multipersona anons.
Hilarious TG!
Agreed, was going to do a search earlier on topic of "algorithm for squaring massive numbers", but was in the general.
gut tells me 12 is a key to this. Why I posted that 12X spiral (w/ primes only on 4 rays, so eliminates 3/4 of options) and then the Squares with 3 entrances on each side (temple layouts). But that's just me letting intuition flow (which i trust)
gotta read that again, interesting.
Nice!!
wasn't me.
That's awesome MA!
^^ This was me. Probably doxxed myself using word 'folks' anyway.
Was trying not to overdo the name and tripfagging unless had something mathy to offer. Have had some other immediate priorities so haven't been able to focus and digest everything to level needed to actually be contributing.
Finally got resume updated and should be writing more cover letters tonight instead of hanging with you fags!
VQC, please take note, that this collective effort is a real testament to individuals showing faith: strong or unshakeable belief in something, esp without proof or evidence.
Ok, went ahead and came up with a query for a quick search, here's the second result in the list, has several algorithms including pseudo code, pdf here:
https:/ /www.hindawi.com/journals/jam/2014/107109/
Efficient Big Integer Multiplication and Squaring Algorithms for Cryptographic Applications
2014, Shahram Jahani, Azman Samsudin, and Kumbakonam Govindarajan Subramanian
School of Computer Sciences, Universiti Sains
Abstract
Public-key cryptosystems are broadly employed to provide security for digital information. Improving the efficiency of public-key cryptosystem through speeding up calculation and using fewer resources are among the main goals of cryptography research. In this paper, we introduce new symbols extracted from binary representation of integers called Big-ones. We present a modified version of the classical multiplication and squaring algorithms based on the Big-ones to improve the efficiency of big integer multiplication and squaring in number theory based cryptosystems. Compared to the adopted classical and Karatsuba multiplication algorithms for squaring, the proposed squaring algorithm is 2 to 3.7 and 7.9 to 2.5 times faster for squaring 32-bit and 8-Kbit numbers, respectively. The proposed multiplication algorithm is also 2.3 to 3.9 and 7 to 2.4 times faster for multiplying 32-bit and 8-Kbit numbers, respectively. The number theory based cryptosystems, which are operating in the range of 1-Kbit to 4-Kbit integers, are directly benefited from the proposed method since multiplication and squaring are the main operations in most of these systems.
Did some searches on Fermat primes and other stuff early on as well. You all were starting to generate some patterns earlier to what I found, would have to look back.
New
Thank you Baker, tasty tasty and what a Be-You-tee-Full day!
Louis BAKER
https:/ /youtu.be/_AymCWgXQaA
You said I love you like a rainbow
I know what that means
You know I love you just the same
I could never let go of you again
You said I love you like the rain
And the sound when it hits the roof
I don't know what I would do
If I couldn't be close to you again
My skin feels like it's thickening
This cold-hearted world is sinking in
We've seen enough to know how deep we're in
But with you I feel
But with you I feel free
We've come through the darkest shades of blue
To a new and brighter point of view
Who knows where this rainbow will take us to
As long as we're together
You said you love me like no other
I know what that means
You know I love you just the same
I could never let go of you again
You said you love me like the colour
In the earth and in the sky
When we spoke of what we'd been through
Tears fell softly on you and I
My skin feels like it's thickening
This cold-hearted world is sinking in
We've seen enough to know how deep we're in
But with you I feel
But with you I feel free
We've come through the darkest shades of blue
To a new and brighter point of view
Who knows where this rainbow will take us to
As long as we're together
But with you I feel free
We've come through the darkest shades of blue
To a new and brighter point of view
Who knows where this rainbow will take us to
As long as we're together
Because with you I feel free
Because with you I feel free
Because with you I feel free
Problem is to solve for n where a and b are prime. in record (0,2,c^2,x,a^2,b^2) you can solve for n in terms of a and b.
x=a^2(b^2-1)
2an=x^2
n=a^2(b^2-1)^2
S>>854
sorry
n=a^2(b^2-1)/2
Sorry again
record is (0,n(c^2),c^2,x,a^2,b^2) so you can solve n for squared record in terms of square of original primes.
n(c^2)= a^2(b^2-1)/2
To be clearer
For record (0,n(c^2),c,x(c^2),a^2,b^2)
for primes a,b
x(c^2)^2=2a^2n(c^2)
x(c^2)=a^2(b^2-1)
n(c^2) =a^2(b^2-1)/2
using x(c^2) and n(c^2) to distinguish from x and n for record (e,n,c,x,a,b)
Fucked up!
n(c^2)=a^2(b^2-1)^2/2
When does this equation work?
for c^2=a^2*b^2
Here is one for you Mr E Melange! The Memes must flow
Thank you new Hobo friend! LOVE them, and perfect for our mission actually!