72180a No.7690 [View All]
The virtual quantum computer (VQC) is a grid made of infinite yet constructable sets that follow a known pattern. Like a quantum spreadsheet.
The grid is the superposition. The collapse of that superposition will be two input parameters, d and e which can be calculated easily for all integers, c, where c is the difference of two squares. Its purpose and our goal is to learn and show the TRUTH, one of them being P=NP. Cracking RSA will be a consequence.
When the integers that are the difference of two squares are arranged into the grid and their corresponding properties are shown, a pattern emerges that shows calculation instead of searching is possible.
Glossary
Column
All cells for a given e.
Row
All cells for a given n
The grid is indexed using e, n, and t, where e is the rows, n is the columns, and t is the specific element in the cell.
Entry, record, element
one set of variables that represents one factorization for a number.
an entry = {e:n:d:x:a:b} (e, n, t)
{1:5:12:7:5:29} (1, 5, 4) is a record AKA an element AKA an entry.
ab record, nontrivial factorization, prime record
the element that contains the factorization of c that is not 1*c, hence, nontrivial.
1c record, trivial factorization
the element generated from setting a=1 and b=c
Cell
All entries for a given e,n (not to be confused with an entry itself.)
Genesis cell
e,1
Remainder Tree
The remainder tree is the result of treating d and e as c's recursively until 1 is reached, creating a tree with several to many branches.
Functions
na transform
a movement from a record in (e, n) into (e,1) where n becomes 1 and a becomes a times the n of the (e,n) record. It has also been used to refer to moving n*a records down in a cell.
T
T of number or T(input) is the triangle number function. If our input is 7, T(7) returns the 7th triangle number
T-1, inverse T
the inverse function of the triangle number function that returns the index of a given triangle number. If our input is the 7th triangle number, the function returns 7.
Variables
The map's legend is {e:n:d:x:a:b}, where c is any number that is the difference of two squares, so odd numbers are included. It is the number you want to factor. It is the number that the a and b in an entry multiply to make.
a and b are, to reiterate, the factors of c. a is the smaller factor of c, and b is the larger one.
d is the integer square root of c
e is the remainder of taking the integer square root of c. Unless c is a perfect square, a remainder will be left over.
i is the root of the large square. it is the same thing as (d+n)
j is the root of the small square. it is the same thing as (x+n). i^2 - j^2, difference of squares.
n is what you add to d to be exactly halfway between a and b, and it is the root of the large square. So it takes you from d to the large square.
x is what you add to a to make d. When added to n it makes the root of the small square.
f is what you add to c to make a square. (e is what you subtract from c to make the square below it, f adds to make the square above c.)
g is the square root of c with decimals, opposed to d, which discards decimals.
t is the third coordinate in the VQC, it is a function of x.
u is the base of a triangle that helps us calculate (x+n) for certain c values. simply put, it is a representation of (x+n). 8 times the triangle number of u plus one is x+n.
s was a variable used to demonstrate patterns in (e, 1). See "(e, 1)."
When capitalized versions of the variables are used in comparison to lowercase versions, the capitalized versions refer to the variable's value for the trivial record, and the lowercase variables refer to the values for the nontrivial record.
{e:N:d:X:A:B} (e, N, T) is the trivial element.
{e:n:d:x:a:b} (e, n, t) in this context is the nontrivial element, the prime factorization of c.
____________________________
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b30a9a No.8536
which means sqrt(a) is our long side length of the triangle, with t and t + 1 being the other two sides.
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291f8b No.8538
Damn my intuition bells are going off right now Lads! What I’ve found in using my method is that when you begin building up the factors of the (x+n)^2 -1 area, you first arrive at a perfect square that is a potential match for Potential x+n or P(x+n). D, e, and f all determine what the first perfect square will be, and then
The big idea is finding the first perfect square built using the root of f.
And then using multiples of it.
F limits the possible correct values.
It’s a 2 step process. All larger x+n values can be built from the first perfect square.
Once you find the first perfect square, the other possible values are just multiples.
So PTB for x+n=6889 = 2(sqrt ((f-1)/8))-1 = 7
First match is 7 * 5 +1 = 36
Sqrt(36) = 6 = potential x+n
Here's the perfect square creation I'm talking about. This is building the x+n square using 4 rectangles (or 8 triangles) +1
Step 1: find the first perfect square.
PTB = 7
7 * 5 +1 = 36
potential triangle base value(u) is 6
potential triangle side value (u-1) is (6-1)=5
Step 2: correct x+n is a multiple of the correct value above ^^^
Formula is: u * (u-1) * 4 +1 = potential x+n
1. 6 * 5 *4 +1 = 121, sqrt(121)=11
2. 12 * 11 * 4 +1 = 529, sqrt(529)=23
3. 18 * 17 * 4 +1 = 1225, sqrt(1225)=35
4. 24 * 23 * 4 +1=2209, sqrt(2209)=47
5. 30 * 29 * 4 +1= 3481, sqrt(3481)=59
6. 36 * 35 * 4 +1= 5041, sqrt(5041)=71
7. 42 * 41 * 4 +1= 6889, sqrt(6889)=83 = correct x+n value
Step 3: use the quadratic to check/verify each iteration above ^^^ resulting in an integer answer (no remainder)
Sqrt(c + (x+n))-d = n
I'll run it on some more c values, I know this is c6107.
Just trying to find an idea that makes sense first, and then test it out.
Pretty cool to do this from memory with a calculator and pad. I love this problem. I think about it in the morning when I wake up, and it pops into my mind during the work day. I haven't posted much lately bc I had nothing new, but maybe this idea is worth exploring.
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0bee42 No.8539
Many are the plans in the mind of a man, but it is the purpose of the Lord that will stand.
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5ff613 No.8540
>>8536
But does this help with out problem at hand?
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5ff613 No.8541
>>8540
Ugh, with our problem at hand. It's still early in the morning.
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566843 No.8546
>>8526
I still lurk. I got sidetracked during the holidays. Still do work on this though. Noticed something recently. thanks to
>>8535
I was looking at (1,1).
In (1,1) the values for x and a are tied to a pythagorean triple,
(x, a-1, a) so that a*a = x*x + (a-1)*(a-1). Then I couldn't really follow what d was getting to, but I think this might be a way to tie together unrelated squares, because x*x, a*a, (a-1)*(a-1), d*d, are all related to eachother in some way in these equations. Granted there isn't much but forgive me.
Another thing I saw is that if you look at the records in row 1, all of their 'ones' place values repeat over time, no matter what base you are using. For instance, if we have a base equal to 45, then all of the values in column A % 45 will repeat over time. For instance, if we use base 60, then for every even e, the a's repeat every 30, then if we use an odd e, it repeats every 15. Hovever, if our base is odd, it will always repeat at that interval. Also within the repeating the same pattern occurs backwards so theres half as many options.
For base divisible by 4:
Even e:
A[t] % base = A[t+base/4] % base
Odd e:
A[t] % base = A[t+base/2] % base
Base divisible by 2:
A[t] % base = A[t+base/2] % base
Other base:
A[t] % base = A[t+base] % base
We could then iterate through the first t records in (e,1) because those would have the correct a%t values,
then maybe we could use modular arithmetic to solve it from there. Maybe mix in the -f entry to get two equations. Maybe deliberately use a base that is divisible by 4 so that d will be even and -f will be odd, and you'll have two different bases to use in the modular equation.
Lets use base 60 for c=145
We know that A[t] = na for some value in (1,1), and that the values A[t]%60 for this are pic related:
(1, 5, 13, 25, 41, 1, 25, 53, 25, 1, 41, 25,13,5,1,1,5,13,…)
We calculated all of these and there are a bunch of repeats so we filter them out
now our options are na == 1, 5, 13, 25, 41, 53 (mod 60)
Second pic related is -24.
If we look at (-24, 1) = (-f, 1), then there is an A[t] = n(a-1).
Now all the options are :
(6, 20, 38, 00, 26, 56, 30, 8, 50, 36, 26, 20, 18, 20, 26, 36, 50, 08, …
which can be reduced to n(a-1) == 6, 8, 18, 20, 26, 36, 38, 50, 56
We can check every permutation which would be 6*9 attempts = 54 iterations, and each time you'd do this.
Lets say I'm testing n(a-1) == 18, and na == 41 (mod 60)
na - n(a-1) == 41 - 18 (mod 60)
na - na + n == 23 (mod 60)
n == 23 (mod 60)
So I guess it would always turn out to be just getting the difference of all these numbers, which isn't too intense of a calculation. Of course once we get n (mod 60) it's not like we have n or anything. We might be able to narrow it down further with other equations and known variables but idk. Just thought this was neat and wasn't sure if anyone else had posted it.
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81cdd5 No.8547
YouTube embed. Click thumbnail to play.
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81cdd5 No.8550
YouTube embed. Click thumbnail to play.
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8a69d0 No.8552
>>7747
We've been trying to figure out the Parabola or perhaps, promulgation of rounding errors, for over a year now. Anyone figure it out?
Also, is Chris OK?
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89ec56 No.8553
It's all just one guy.
It's all just one guy.
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89ec56 No.8554
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27241f No.8555
>>8554
I like the way you higlighted specific cells but made the resolution of the picture low enough that you can't actually see the numbers. Nice touch, that.
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81cdd5 No.8556
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27241f No.8557
>>8556
I was being sarcastic. You could apply the colour thing to any set of numbers so it doesn't really matter. It looks like this is for 5x31. It doesn't really add any new information aside from highlighting (-1,n). We were already told to do that, but when I looked into (-1,n) I didn't find any unique patterns that changed anything, and it had the same patterns as the other negative square columns ((-4,n), (-9,n), etc). We were already told to do that colour diagram thing too, early last year. He said that the colours were significant. The only other time he used these same colours was in his triangle diagrams, but I don't remember being able to find any useful connections personally (plus nobody found anything originally (GA made some example diagrams but I don't remember which thread this was)).
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81cdd5 No.8558
>>8557
So you can explain 3 black segments, one gray segment, 2yellow highlights, the green segments, the purple segments, the 2 same-purple highlights separating green segments, and the 2 blue highlights?
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01587d No.8559
2d(n-1)+f-1= 2xn+ x^2
F and d limit possible n and (n-1) values.
Let’s work on this!
Is it a graph line where only certain integers work?
More about eliminating values than solving?
The algorithm is ancient.
According to VQC
How to combine multiple variable equations to run side by side.
a new form of algebraic solution
Underlying fractal patterns
Triangles and squares
And how they combine
Limiting possibilities
Factors contained within f.
(D+n)^2 - (x+n)^2=c
Simple and clean.
So much complexity and simplicity combined.
Mods and patterns
(Of mods)
Factoring each square
Multiple possibilities for perfect squares in (x+n)^2
Geometric patterns in the Grid
Geometric patterns in the 8Tu+1
Leading to a Grid shortcut based on understanding of the geometric patterns.
How do we factor the two squares (d+n)^2 and (x+n)^2 ?
Using only c d e f ?
I'm reworking equations over here.
I have too much time invested here to give up.
And I'll keep going till we solve this shit.
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81cdd5 No.8560
I bet these folks would SUPER appreciate it/be terrified if someone were to open that wallet up and give everyone their $190 million back…
https://www.marketwatch.com/story/crypto-exchange-customers-cant-access-190-million-after-ceo-dies-with-sole-password-2019-02-04
Crypto exchange customers can’t access $190 million after CEO dies with sole password
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566843 No.8561
>>8557
That was a tkinter app I made that would do it for any records. Never found much from it. Then again I never really used it because I saw the flaws and I made a better app. I think we could use the coloring to highlight these cells but then view them from a different perspective. I'm stil working on a program that will let us view the cells from any angle. AKA originally the axes are (e,n) but I'm trying to make something that will enable use to view from any axes and highlight any cells.
>>8559
>using only c d e f
I think that this could be the key. He probably chose those letters for a certain reason. Maybe theres a bunch of steps. c gives you d, d and c give you e, d and e give you f, f and e give you g, g and f give you h, …. up until 'n' I'd guess.
c d e f g h i j k l m n -> finally -> a or b
>VQC: Twelve steps are the key
c -> n is 12 steps if we follow this pattern. Now obviously I have no idea what these steps are in specific, but if this is the case then vqc has given us up to f. This is 1/3 of the way already.
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befb45 No.8562
>>8526
Still lurking here. Still my favorite board of all time. -Hobo
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81cdd5 No.8563
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2b6a32 No.8564
>>8561
We do have this one: >>8498 stating there are h families. So c gives d, c and d gives e. d and e gives f, f and e gives g? Then g and e gives h?
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4d8219 No.8566
Topol, here's one of yours!
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99b34d No.8570
Let's fill this one up, lads. When that's done, Fresh Bread is ready. Maybe we'll start kicking ass again??
>>8567
>>8567
>>8567
>>8567
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341dcd No.8571
I just noticed a cool pattern.
Starting at the (na transform) element,
a[t] (e) - a[t] (-f) = an increasing pattern.
128-127=1
100-97=3
76-71=5
56-49=7 which is (n-1)
So the difference between a[t] (e,1) compared to a[t] (-f,1) moves upward in an ascending pattern until the first element in a given e column.
for odd e, (e+1)/2 = a[1]
So this info greatly limits our search area.
Thinking out loud over here.
We limit the search using the (na transform) element and the a[1] element.
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341dcd No.8572
>>8571
To be clear, I’m not suggesting iterating by x.
I’m suggesting using every available piece of info to limit the search area.
BigN is also a limiting piece of info
Along with 2(sqrt((f-1)/8)-1
Set boundaries, then factor
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341dcd No.8573
More patterns.
This one has (-f,1) a[t]=c and its equivalent element in (e,1) is distance of (a) apart, 287 - 270= 7 = correct a value
Also, the a[t] = (bn) value = 328
in (e,1) 328 - 287 = 41 = b
so c, a, and b are available/calculable in a triangle pattern right next to each other in adjacent elements.
Here's a diagram.
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341dcd No.8574
Once you c it you can’t un-c it?? Why does a[t]=c appear in many of our cases?
for c287
2(sqrt(d))-1 = 7 = (n-1)
Must be a fluke, but still analyzing for patterns over here.
Lol, for c6107 2(sqrt(d))-1= 15 and correct (n-1)=35
Shared factor of 5. Interesting.
and 2(sqrt((f-1)/8)-1 = 7 which is another factor for 35
Hello PMA, Jan, and 5DAnon can I please request your assistance?
Could it be a characteristic of ONLY semiprime c values that we always have one a[t]=c element?
Makes sense that c always appears at a[t] since the prime factors have to show up again.
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341dcd No.8575
>>8574
And if that element exists for a given semiprime c, we can solve the problem with the adjacent elements. (maybe!)
You guys know how I work, calculator in hand with a pencil at the ready. If the idea sucks, that's fine. Working from small examples over here.
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344e49 No.8576
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349a57 No.8578
>>8576
So we have till 3rd of march to figure this out.
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344e49 No.8579
>>8578
I wouldn't hold my breath if I were you. How many times has he said "hey guys let's go through the solution right now/tomorrow/next week" and then acts like he never said that? And if he actually goes through with it this time, keep in mind, there are people who have been following along with this board for months if not more than a year who don't even understand it so if he legitimately goes through the solution (or "part 1", whatever that means) nobody will even know what the fuck he's talking about aside from maybe 5 or 6 of us who have put significant effort in already.
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344e49 No.8580
VQC, we know you've been posting on /qresearch/ with your name again. Why not come back here and say hi?
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74d41f No.8581
>>8580
VQC - I agree with AA, you should stop by and say hello.
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81cdd5 No.8582
>>8580
>>8581
Y'all are "special" if you think he wouldn't come back with:
"We never left.
It's time to return publicly.
VQC"
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258a51 No.8583
>>8580
Got any post ids for those? I know he also sometimes post hints there when people fish.
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81cdd5 No.8584
>>8583
I've heard that somewhere, too… Firsttimeposteranon47
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344e49 No.8585
In columns where e is a square, c is the sum of two squares. We already know this. What hasn't been pointed out is that in columns where e is a trianglular number, c is the sum of a square and a triangle. This could potentially be useful given some of the things that Chris has been saying on /qresearch/ recently (speaking of which, Topol, you should post all of those screenshots here rather than just putting them on Discord).
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026429 No.8586
>>8585
Yeah Topol, share the screenshots for those of us who don't hang out on discord.
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b69141 No.8587
aan(n-1) rings a bell with all these new posts.
Supposedly this type of integer is easy to spot.
Although we haven't discovered the identification process.
Lol.
product of triangles and squares, i think.
Best idea I can think of is to find the next largest perfect square, which should be (aann)
Subtract the difference of (aann) and aan(n-1)
For c145 it's 625-500 = 125
Then, aan(n-1) / 125 = 500 / 125 = 4 = (n-1)
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b69141 No.8588
>>8587
Above is a small example, but here is another key idea we have yet to solve.
aan(n-1) is supposed to be easy to spot when comparing (-f,1) and (e,1) a[t] values.
We worked for a week on this, and we need to finish it, as Saga correctly pointed out yesterday.
We fucking work our asses off on an idea, then bail on it when VQC distracts us with a new idea. That distracting faggot.
He's doing it on purpose to slow us down.
Let’s make a list of unsolved crumbs.
And then work on them independently, while sharing our results. That would work well for us as a group. We are all very independent souls but love to check out each other's ideas.
And we share a common purpose, to do our part for Q / VQC etc. and our fellow humans worldwide.
Thoughts, fellow Anons and Math Fam?
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b69141 No.8589
>>8567
Let's fill this bread up with some excellent posts about unsolved crumbs to work on. Then, we can move to our new bread.
>>8567
>>>8567
>>8567
>>>8567
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b69141 No.8590
The easiest method I can think of is this:
Get the (e na transform) and (-f na transform) elements.
Then create the list of a[t] values for both columns up to a[1].
Then simply move up subtracting e a[t] from -f a[t] and divide c by each result.
First whole integer result is prime a
It could be millions and millions of calcs, but it will run super fast, bc it’s so simple.
(e,1) gives us all factors.
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81cdd5 No.8591
>>8586
But that takes eeeeeeffooooooooort!
fiiiiiiiiiine
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2344ac No.8593
an interesting property, admitedly tested only on short integers
not directly connected to the problem presented, but nevertheless
sum of digits in base(b+1) is same as last digit in base(b)
e.g (base8):
0c1420=1+4+2+0=7(in base8) means its divisible by 8 and 7
0c3774=3+7+7+4=7(in base8) means its divisible by 7 and remainder when dividing by 8 is 4
0c7120=7+1+2+0=3(in base8) means its divisible by 8 and remainder when dividing by 7 is 3
0c6305=6+4+0+4=7(in base8) means its divisible by 7 and remainder when dividing by 8 is 5
0c6620=6+6+2+0=7(in base8) means its divisible by 8 and 7
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349a57 No.8594
>>8591
Hehe, the x^1 = d … x^(d-1) = d wasn't VQC, it was me. I was fishing for him and playing with the idea of roots of david being related to the roots of unity.
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81cdd5 No.8595
>>8594
-shrug-
The interaction got VQC's attention for the crumbs.
Fixed.
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81cdd5 No.8596
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344e49 No.8599
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344e49 No.8600
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344e49 No.8601
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