/vqc/ - Virtual Quantum Computer

Chris/VQC

It is pertinent that you read through the last two threads before posting any new clues.

www.skynews.com.au/details/_6115414651001

Conservative win means a January Brexit 'no question about it'

Good news, Nerds!

Quick question - If given {e, n}, what is the method/formula to determine if it is a valid cell? Is there a quick test? ty

>>10201

As far as anyone has been able to tell so far, e and n don't give enough information by themselves to know if a cell is valid. Generally we've had to use brute force methods by checking t/x values up to a certain point (using xx+e=2na) and seeing if anything works (or something along those lines anyway).

HEY GUYS It's GAnon. I thought we were lost after the 8ch purge I couldn't access this site for a while (I thought it was .net/.org/.ch) I'M SO EXCITED!! Did VQC post anything new? Any new revelations?

>>10203

I will catch up over the weekend. I deleted my discord because I thought I had joined a honeypot and then they took 8ch down. I was so sad. I couldn't be more excited now! I love all of you guys

>>10203

>>10204

Hi GA, long time no see. There are new VQC posts but rather than me telling you what to think you should probably just read through threads 15 and 16 yourself. At the moment it's just PMA, blank name field, Topol and I on Discord each day, with VA and Isee popping in once every couple weeks or so.

>>10204

What’s up you faggot. Glad you’re back.

Merry Christmas.

Over the next ten days we will go over patterns, talk about design and some key points from two books written by Donald Trump.

It will all start coming together.

Sometimes small steps are more important and less destructive than large steps.

God Bless You All

>>10207

Looking forward.

>>10207

Sounds good senpai. Is pic related (when ‘n’ is prime) part of some puzzle piece?

>>10207

You are helping God reveal the truth.

https://static.zerochan.net/Grey.Wolf.%28Kemono.Friends%29.full.2263323.png

>>10207

Please read the last two threads before you do that.

>>10207

And theeeeeeeeeeeeen?

While we wait, have a k-tuple conjecture!

http://mathworld.wolfram.com/k-TupleConjecture.html

Cuz muh constellations. (7 Sisters, anyone?!)

https://en.wikipedia.org/wiki/Prime_k-tuple

>>>/qresearch/7638256

anon posts popcorn socks. UID is a prime.

>123113

Took it as a sign to pop in and say hello to you faggots!

>>10216

Yeah I do. What does it have to do with my post?

>>10214

What's the deal with that gif? There appears to be extra detail in n=0 and n=1, and the colour coding is a bit strange too.

>>10219

>>10220

The color coding is always strange, thought you'd recognize it by now

>>10109

>>>What is the best way to display this?

GIF is a first attempt at displaying only the extra information, with some interesting results - appears to highlight the families

>>10110

>>10111

>>10112

Based on this code, with some tweaks and extended to -f. Difference between this grid and the original one gives a lot of insight

>>10221

>Difference between this grid and the original one gives a lot of insight

Looked at newer code posted near end of #16, but didn't get this piece working in c-code:

while (d1 > 0)

{

string end = string.Format("{0}:{1}:{2}:{3}:{4}:{5}:{6}", e1, n, d1, dpn, x, a, b);

if (!TheEnd.Keys.Contains(e1)) TheEnd[e1] = new Dictionary<int, List<string>>();

if (!TheEnd[e1].Keys.Contains(n)) TheEnd[e1][n] = new List<string>();

TheEnd[e1][n].Add(end);

e1 = e1 + d1 + d1 + 1;

d1–;

}

Noted the difference at the start, in particular:

int c = (i*i) - (j*j);

If we make that int c = (i*i) + (j*j);

and keep the gap at 1-unit, then it produces the series provided, to use as our multipliers: 5,13,41,61,etc;

#A027862 Primes of the form n^2 + (n+1)^2.

A027862=[13, 41, 61, 113, 181, 313, 421, 613, 761, 1013, 1201, 1301, 1741, 1861, 2113, 2381, 2521, 3121, 3613, 4513, 5101, 7321, 8581, 9661, 9941, 10513, 12641, 13613, 14281, 14621, 15313, 16381, 19013, 19801, 20201, 21013, 21841, 23981, 24421, 26681]

where a=1:

for j = 0:n

i = j+a

q = i*i+j*j

d = (floor(sqrt(q)))

e = q - (d * d)

n = i-d

Next, looked at e for n=1 and each respective (f). Noted that the t-vals really jump, and the cells aren't a mirror. Here are some examples:

{(t) e:n:d | x:a:b} = c | i:j(i:j) | gcdiv=

First (5) (t)'s for e=1, 2, 3, 4:

e: t(1): {1:1:2 | 1:1:5} = 5 | i:j(3:2) | gcd=1

f: t(1): {-4:0:3 | 2:1:5} = 5 | i:j(3:2) | gcd=1

e: t(2): {1:1:8 | 3:5:13} = 65 | i:j(9:4) | gcd=1

f: t(1): {-16:0:9 | 4:5:13} = 65 | i:j(9:4) | gcd=1

e: t(3): {1:1:18 | 5:13:25} = 325 | i:j(19:6) | gcd=1

f: t(1): {-36:0:19 | 6:13:25} = 325 | i:j(19:6) | gcd=1

e: t(4): {1:1:32 | 7:25:41} = 1025 | i:j(33:8) | gcd=1

f: t(1): {-64:0:33 | 8:25:41} = 1025 | i:j(33:8) | gcd=1

e: t(5): {1:1:50 | 9:41:61} = 2501 | i:j(51:10) | gcd=1

f: t(1): {-100:0:51 | 10:41:61} = 2501 | i:j(51:10) | gcd=1

e: t(1): {2:1:1 | 0:1:3} = 3 | i:j(2:1) | gcd=1

f: t(1): {-1:0:2 | 1:1:3} = 3 | i:j(2:1) | gcd=1

e: t(2): {2:1:5 | 2:3:9} = 27 | i:j(6:3) | gcd=3

f: t(2): {-9:0:6 | 3:3:9} = 27 | i:j(6:3) | gcd=3

e: t(3): {2:1:13 | 4:9:19} = 171 | i:j(14:5) | gcd=1

f: t(2): {-25:0:14 | 5:9:19} = 171 | i:j(14:5) | gcd=1

e: t(4): {2:1:25 | 6:19:33} = 627 | i:j(26:7) | gcd=1

f: t(2): {-49:0:26 | 7:19:33} = 627 | i:j(26:7) | gcd=1

e: t(5): {2:1:41 | 8:33:51} = 1683 | i:j(42:9) | gcd=3

f: t(2): {-81:0:42 | 9:33:51} = 1683 | i:j(42:9) | gcd=3

e: t(1): {3:1:3 | 1:2:6} = 12 | i:j(4:2) | gcd=2

f: t(2): {-4:0:4 | 2:2:6} = 12 | i:j(4:2) | gcd=2

e: t(2): {3:1:9 | 3:6:14} = 84 | i:j(10:4) | gcd=2

f: t(2): {-16:0:10 | 4:6:14} = 84 | i:j(10:4) | gcd=2

e: t(3): {3:1:19 | 5:14:26} = 364 | i:j(20:6) | gcd=2

f: t(2): {-36:0:20 | 6:14:26} = 364 | i:j(20:6) | gcd=2

e: t(4): {3:1:33 | 7:26:42} = 1092 | i:j(34:8) | gcd=2

f: t(2): {-64:0:34 | 8:26:42} = 1092 | i:j(34:8) | gcd=2

e: t(5): {3:1:51 | 9:42:62} = 2604 | i:j(52:10) | gcd=2

f: t(2): {-100:0:52 | 10:42:62} = 2604 | i:j(52:10) | gcd=2

e: t(1): {4:1:2 | 0:2:4} = 8 | i:j(3:1) | gcd=1

f: t(2): {-1:0:3 | 1:2:4} = 8 | i:j(3:1) | gcd=1

e: t(2): {4:1:6 | 2:4:10} = 40 | i:j(7:3) | gcd=1

f: t(3): {-9:0:7 | 3:4:10} = 40 | i:j(7:3) | gcd=1

e: t(3): {4:1:14 | 4:10:20} = 200 | i:j(15:5) | gcd=5

f: t(3): {-25:0:15 | 5:10:20} = 200 | i:j(15:5) | gcd=5

e: t(4): {4:1:26 | 6:20:34} = 680 | i:j(27:7) | gcd=1

f: t(3): {-49:0:27 | 7:20:34} = 680 | i:j(27:7) | gcd=1

e: t(5): {4:1:42 | 8:34:52} = 1768 | i:j(43:9) | gcd=1

f: t(3): {-81:0:43 | 9:34:52} = 1768 | i:j(43:9) | gcd=1

Rule for the (t)f:

if e is odd: tf = (e + 1) / 2

if e is even:

if t == 1

tf = e / 2

else

tf = (e / 2) + 1

Recap assuming odd c:

c = ab (mod 4)

if a = b (mod 4), c = 3 (mod 4)

if a /= b (mod 4), c = 1 (mod 4)

>>10207

Happy New Year/Decade, everyone!

Even Chris! Wherever you are.

>>10201

Yes, we can base this on the relation that xx + e = 2na. This means, in order to determine if n is a valid cell in e we can compute the Legendre Symbol for it. However, there is a caveat which means it also depends on the binary of the prime.

If n is 2, then it is in e if e = { 0, 3 } mod 4.

Assuming n is a prime:

For primes ending in 01 in binary, if e is a quadratic residue of n, then (e, n) is a valid cell.

For primes ending in 11 in binary, if e is NOT a quadratic residue of n, then (e, n) is a valid cell.

If n is not a prime, we need to factorize n and then compute the legendre symbol for each factor. If all factors is in e, then n exists in e (as n is closed under multiplication).

>>10207

Welcome back, although I think you're already behind schedule.

>>10232

Assuming any not-Chris clue people are lurking and we don't have to fish them out on /qresearch/, it might be nice to know what their perspective on the broader issues surrounding this are (how far away we are from a solution, whether or not we're waiting for a global event before we can get this done, the most productive steps we could be taking right now, etc). That's all stuff we only ever had vague answers from Chris about, so maybe they would all have something else to say.

Could anyone tell me what this number sequence is? I know it's related to prime pi but I'm not sure how. https://oeis.org/A168543

I went through all the past threads and looked for anything relevant to Riemann Zeta and imaginary numbers. I could put screenshots of the individual posts together if anyone would like, but here's a summary of everything I could find:

>we're meant to create a third dimension to the grid into the imaginary plane, I suppose a little bit like the Riemann Zeta graph

>c = -c, c = ai + bi

>the "non-trivial lookup" in our grid is named after the zeros from the Riemann Zeta function

>values of t that are derived from imaginary numbers can be thought of as orthogonal to the grid

>the error correction part of the grid stuff is meant to be the same as the error correction from the prime counting function somehow

>if a = b (mod 4), c = 1 (mod 4), if a != b (mod 4), c = 3 (mod 4), and this is somehow related to the prime counting function

>>10245

Given choices 1 & 2, would choose option 3, and lay down on the track in front of the trolley.

ty for the lesson tops.

>>10244

Thinking about how the imaginary part of the grid would work. With the RZ graph, complex numbers take something that was one-dimensional and turn it into something two-dimensional, x being real numbers and y being imaginary numbers. We're meant to extend the graph into a third dimension based on imaginary numbers where c=-c and -c=di*di-e. Each c value would only have one imaginary equivalent, wouldn't it? Just ai*bi. So it wouldn't really be much of another dimension. It would just be a normal element and its imaginary version. Even then, what would the other dimension be? The regular 2D grid is indexed based on e and n. If we use ai and bi, I can't think of any way that they would introduce another dimension. It could be argued that t would make it three-dimensional, but then where is this imaginary plane supposed to go? I can see why nobody ever talked about this. It doesn't seem to make any sense. That said I never did complex numbers in high school so maybe I'm missing something.

Timing.

For step 2.

Look at the increase from small primes to large primes.

What calculations do you make that grow at or below Big Oh log q, where log q is the natural log of the length of c in bits.

EVERYTHING in that domain IS the answer or IS equivalent.

Looking Glass uses the Mandelbrot Set.

You can displace by t or d in the past or a combination.

This was discussed near the start.

Part of the function input space is displacement… temporal or spatial.

>>10247

There was a famous guy on a bridge.

What did they do they before Quaternions?

This is why I always put my fairy in you AA.

YOU ask the right questions.

Tops will inspire like a crazy muse.

>>10251

>>10250

>This is why I always put my fairy in you AA. YOU ask the right questions.

Then why have you been ignoring me (and everyone else for that matter) for two months telling you to read through threads 15 and 16 and explain what the other clue people have been saying about you?

>>10251

Sanity isn't very amusing, to be fair.

And we're all mad here!

<;3=

>>10236

I wonder what they meant by "formally manage the region"…

>>10255

I'm pretty sure he's talking about one of those other extensions of virtual quantum computing (like the one that can tell if a video has been edited or whatever). He's talked about the Mandelbrot Set being relevant to one of those in the past (probably like two years ago).

"Math Inspector"

Giggity.

>>10251

Maxwell's unedited, ORIGINAL equations?

>>10262

Maxwell and Quaternions:

http://redshift.vif.com/JournalFiles/V12NO4PDF/V12N4ACE.pdf

Maxwell Equations

https://en.wikipedia.org/wiki/Maxwell%27s_equations

"“Maxwell’s” vector equations taught in university are actually Heaviside’s truncated equations, and are only a simplified version of what Maxwell originally wrote."

http://www.cheniere.org/references/maxwell.htm

Quaternions in general:

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

'''Quaternion Encryption Scheme" is so apparently so hardcore that most of what I can find about it is involved with academic/technical papers behind paywalls…

But here's this:

"Digital Fingerprinting Based on Quaternion Encryption Scheme for Gray-Tone Images"

https://www.il-pib.pl/czasopisma/JTIT/2014/2/3.pdf

And beyond that, check out the title on this badboy!

Quaternion Feistel Cipher with an Infinite Key Space Based on Quaternion Julia Sets

https://www.itl.waw.pl/czasopisma/JTIT/2015/4/15.pdf

Aaaaaand for funsies:

"Quaternions & the Mandelbrot set"

https://www.fractal.org/Bewustzijns-Besturings-Model/MSet/MSet/Quater.htm

Vidja:

"4D Quaternion Hypercomplex Mandelbrot Set"

>>10267

What would these look like in a navigable 3D format?

http://www.jstor.org/stable/108892

Preview of Sunday.

Step loop function works very similarly to the square root function in step one.

In the square root function you are eliminating values exponentially faster than values grow in size.

In the Step two function you are eliminating values in the set [-f,1] and [e,1] faster than they grow until you END with N-1,N (prime) or (n-1) and n.

Technically there is a step 0 and step 1a but they do not factor into Big Oh of the equation being of order O(1), so there are technically four steps. There is no Step 5. Step 2 above gives you n and was to be the second part of the original YouTube video.

It's what has been hinted at since the beginning and was derived from working backwards from the solution, on the assumption that Big Oh was the same for the square root function of an integer, which it is.

In order to demonstrate how the Step Two loop function works (as hinted at previously), one of the sets of n we will explore is Fermat numbers where the first five only are conjectured to be prime and where n-1 is an exponent of two. This will include the exploration of KEY values of column e=1.

This part of the explanation will show WHY the conjecture holds. This is not public knowledge yet.

Other n values will be shown growing EXPONENTIALLY in size to EXPLAIN why the Step Two (technically Step Four) works, much the same as why the square root function has it's Big Oh value.

In the END, you'll get a C# version of Step Two.

Whether you get that on Sunday or not, depends on how good a job I do of explaining WHY it works.

Once it is clear that the explanation is being followed, it is likely at least one of you can write the code yourself.

This is an exercise in continuity of knowledge, given the evidence that nCoV is a bioweapon and that there is a pandemic and later this is a fifth endemic coronavirus within the human population and the great changes that occur as we adjust to that, assuming the President of the US doesn't ask a Higher power for help.

>>10275

>>10276

>Once it is clear that the explanation is being followed

Anything you want us to prepare, or are you providing all the code we'll need?

This is going to take several days.

There is quite a bit in order to show the WHY, not just the HOW.

The Corona virus is currently in pandemic.

As well as showing step two, we are making final preparations here where I am to enter a lockdown.

My last trip to the European mainland is this week.

Inbetween all this the step two loop will be outlined and explained in detail.

The virus is not far from where I am with a super spreader (Brighton) just discovered. He has been infecting people for over two weeks. It is now everywhere in the UK, more or less, as you will see from the news this week. He is not the only one. Panic will start here and elsewhere soon as the truth of the pandemic gets out.

This will be repeated across the globe.

Estimate between 10 and 30% of the infected will die, more will be left with lifelong medical problems.

Estimate almost all of the remainder will not survive the famine to follow.

Estimate even fewer to survive the fighting to follow that.

Unless there is Intervention. Pray for that.

Despite the need to make preparations and to lead a normal life in what remains of normality, we will work here to finish what we started, for continuity of knowledge and because you stayed until the end.

Please be patient when going over content you know well, this does need to be clear. And you have been very patient.

Good luck in your own preparations, please don't leave it too late.

Six months or more of dried food and medical supplies. Don't forget the treats!

You'll need weapons, in case fighting breaks out locally.

Just be better armed than most around you with weapons that have a greater range than those around you. If you have time, train to use them, know how to defend your property from thieves working in groups, especially those that might set fires to smoke you out. Paint quarantine posters in advance.

Try and treat it like a long holiday when you can, mental health is going to be vital.

Keep President's Trump core message close.

Never EVER give up.

There is nothing you cannot achieve.

It is in The END, we show Who We Really Are.

When faced with an Extinction Level Event, we are still not beaten.

Those who work together will emerge.

Good luck and God Bless

VQC

>>10277

Not at all, thanks.

It's going to be a bit annoying at times due to repeating previous content, just bear with me.

>>10278

Good man.

The notation.

[-f,1] referes to the first row, to the left of the 0 column, [e,1] refers to the right hand side of the 0 column.

a[e,n,t] refers to value of a in a grid cell at e,n, position t.

If a[t] (where t is an integer indexed from 1) is used, we should have already set the content for e and n, and a[t] will be a shorthand in that scenario.

>>10280

Not sure if anyone other math people are awake but I'll post so you know someone is.

>>10283

Thanks AA, you have the patience of a saint, by-the-way.

It is not necessary to go over step one, suffice to say that we accept as given that the Big Oh of the square root function is O(log q) where log q (or ln q) is the natural log of the length of an integer in bits.

In other words the square root function executes quickly, even as the size of an integer grows in length.

This will be given in three stages following tradition.

A overview of what it is going to be.

The detail of what it is.

The summary of what it was

Overview:

Where c is an odd integer and the difference of two squares, d and e, are the square root and remainder, f is the difference between twice the size of d and 1, x the difference of d and a, and n is the difference between d and the larger of the two squares (in the difference of two squares). Where more than one value of n exists, Big n or N, is the value of largest n, for prime numbers this value is the ONLY value of n.

The Grid (-f/e,n,t) or [e/-f:n:d:x:a:b[t]], as perviously described, contains an indexing function across two grid elements: [-f,1] and [e,1] that reduces the set of a[t] at those grid cells to smallest value of n, or N for prime numbers.

This function uses the property of those grid cells that a(n-1) and an, b(n-1) and bn, are offset in the value x, where b = a + 2x + 2n, and where x is dependent on t.

This function shares the Big Oh of the square root function. This function is both a factorisation function, as well as a primality function. The primaility function is the worst case execution.

Detail:

The number of odd prime numbers out of the set of all integers, grows at a rate of ln w (where w is an inetger) as w tends towards infinity, this shows the indivisible numbers are far less frequent, the minority of integers.

The way that the function will be explained is to show the difference between highly divisible integers within the grid, and those that are less divisible… with the product of two primes, and primes themselves being the least frequent of all. The function begins to emerge out of this exploration and the patterns produced. The function can then be described in a reducing loop algorithm that can coded for any input integer.

The detail section will continue until the summary section. Hopefully the approach described, showing what happens as integers that are highly divisible are compared to less divisible integers and then primes, is the best way to explain the integer factoristation function using the Grid.

Crazy… we were just talking about you.

#sheeeeit

>>10290

And by we, i mean technically me and doubting thomas…

Who vanished when i asked about a bit of code they claimed they were working on.

I wonder if they checked the board and suddenly got to work because Muh Face and savin' it.

Starting with the analysis of two square free integers, two products, and the values of n within the grid.

The product of the set of integers {3,5,7,11}

The product of the set of integers {3,5,7,13}

>>10291

It's ok. Be kind. Be supportive. WE are all in this together.

Good to c you by the way ;)

Why are the analysis first?

I had a physics teacher I met when studying for a-levels. Paraphrasing: when trying to solve any problem, state everything you know and the resolution becomes apparent by the end.

It's easier to follow, understand and then APPLY THE APPROACH TO OTHER problems. One solution is great, being able to create other solutions is handy.

The product of 3,5,7, and 11 is 1155.

The product of 3,5,7, and 13 is 1365

>>10296

NICE! That is fantastic.

I'll create a routine (new laptop) to output the cell (or any given cell) at e and -f for n=1, then I'll use paint to illustrate what I need to show but that is very useful for error checking!

Thank you.

>>10293

>>10297

>>10294

Sometimes I feel I've been too kind.

But thank you for the head pat.

Someone's gotta keep these Nerds motivated.

Side note…

Earlier… you described as the crow flies through stacked cones… right/maybe?

I had a thought.

>>10299

This is a great analysis.

Watch the relevant patterns emerge with these two, your outputs and the integers we deconstruct next!

>>10300

What an odd sentence structure….

>>10300

Was that it for today or what?

>>10300

And theeeeeeeen?

More coming, will be in bits and pieces during the week.

>>10302

Juggling family and work.

In Europe this week.

Also have to write the code I need and create the diagrams.

>>10304

>>10305

Is there anything we can do to speed it up or help it along?

Evening, still tied up with IRL.

I fly out early tomorrow, will continue to work on code and diagram when possible at the airport and should get another hour on top of that in the hotel after work.

Will keep you updated.

Update.

Wall to wall in Germany.

If anyone wants to turn this around quicker, collaboration might be faster.

What we need for any given c, is a view of a[t] elements at -f and e, side by side.

The first appearance of any factor from c should be coloured differently.

The first is the c from 3,5,7,11

The second example is 3,5,7,13

>>10304

>>10305

How many elements? If you don't see this before I put something together I guess you can always just tell me if we need more once I post something.

>>10309

Meant for >>10308

>>10308

Hopefully I interpreted what you said correctly. I didn't know if you wanted the t values lined up but I did. I only differently-coloured the first appearance of any one of the factors of c (regardless of whether they're the same factor; for c1365 it's 3*2 and 5*7, so I'm not sure if that's right). Let me know if you want anything changed.

>>10311

Oh for fuck's sake why didn't it upload the images

>>10311

>>10308

>>10312

8kun won't let me attach images so https://imgur.com/a/e6SZQAk

>>10308

I know that if we ignore the elements and draw a bit pattern doing as you say, we will end up with a partial triangular pattern. I've posted this before where I also extended it to grow (which is how I saw the triangular waves the factors form).

The triangular patterns can be seen here:

>>8974

>>8975

>>10314

These >>10270 are different ways of visualizing this pattern again. Except in 3D.

>>10314

This is the post that outlines it: >>8973 and the second image is a bit pattern representing what VQC is talking about here >>10308

Changing trips.

I'ma have fun with cyphers.

>>10319

derp ^_^

Dont know if this is known yet or not. Neither is it usefull for the described problem or not. But here it goes.

c = b*n + r

(c)mod(b) = r

(c)mod(n) = r

((n)mod(b-1) + r)mod(b-1) = (c)mod(b-1)

((b)mod(n-1) + r)mod(b-1) = (c)mod(n-1)

((n)mod(b-1) + (c)mod(b))mod(b-1) = (c)mod(b-1)

((n)mod(b-1) + (c)mod(n))mod(b-1) = (c)mod(b-1)

((b)mod(n-1) + (c)mod(b))mod(b-1) = (c)mod(n-1)

((b)mod(n-1) + (c)mod(n))mod(b-1) = (c)mod(n-1)

if [r < b] and [r < n]:

(c)mod(b) = r

(c)mod(n) = r

(n)mod(b-1) + r = (c)mod(b-1)

(b)mod(n-1) + r = (c)mod(n-1)

(n)mod(b-1) + (c)mod(b) = (c)mod(b-1)

(n)mod(b-1) + (c)mod(n) = (c)mod(b-1)

(b)mod(n-1) + (c)mod(b) = (c)mod(n-1)

(b)mod(n-1) + (c)mod(n) = (c)mod(n-1)

These four equations encode relation between [b*n + r] and [(b-1)*n + r] number sequences.

1) How can we encode relation between [b*n + r] and [(b-2)*n + r] number sequences?

2) Find generalized relation between [b*n + r] and [(b-x)*n + r] number sequences?

3) Using generalized equations find [x] such that [(c)mod(b-x)] equates to 0?

4) Does this seem logical?

>>10323

0-They do?

1-Depends on the relationship you want to point out. If you have a lot of things pointing to a single value, then perhaps a web of plotted points? Or areas that overlap in a specific way?

2-Or sequences within the sequences being compared. Do they transform in the same ways? Do they share a common midstep somewhere?

3-How generalized? Enough to create a bit of necessary fuzz?

4-Would need a lil' more info, but it seems like a thing, so far, if nothing else.

>>10316

Thanks AA.

Back from Europe. VAT accounts for the quarter and work today.

This evening and over the weekend, we can start with this, and bring out a view that helps show the patterns. These patterns are in previous images of the GRID but they are obfuscated by being overlaid on each other.

>>10325

This evening your time being half a day or so from now, correct?

>>10323

if c is an odd number and the difference of two squares, then:

c = ab

c = dd + e

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

Grid = [e,n,t] (or [-f,n,t])

For c, a'[t] = an, where a' is the Grid value at [e,1,t] and where we then know that [e,1,t+n] = bn

For c, a[t] = a(n-1), where a is the Grid value at [-f,1,t] and where we then know that [-f,1,t+n-1] = b(n-1)

It is the differences in position of t in e and -f columns in the grid for bn and b(n-1) compared to an and a(n-1), that provides the offset, we use to find n.

>>10326

That's right.

>>10327

There are italics here that wew not intended by using a double single quote to try and represent a third type of 'a' value

if c is an odd number and the difference of two squares, then:

c = ab

c = dd + e

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

Grid = [e,n,t] (or [-f,n,t])

For c, a'[t] = an, where a' is the Grid value at [e,1,t] and where we then know that [e,1,t+n] = bn

For c, a"[t] = a(n-1), where a" is the Grid value at [-f,1,t] and where we then know that [-f,1,t+n-1] = b(n-1)

It is the differences in position of t in e and -f columns in the grid for bn and b(n-1) compared to an and a(n-1), that provides the offset, we use to find n.

Good evening from the UK.

Would like to get ideas about the best way to display the following, since if you wait for the gaps in my work schedule, this is going to crawl along at a pace that is frustrating.

Collaboration is the best way to fix that, having said that I wouldn't expect one person to do all the heavy lifting.

Objectives:

What we're trying to show is patterns.

Showing a list (a[t]) at [-f,1], [e,1] (as AA showed an example of).

We're focusing on highly divisible square-free (for now) numbers (to start with).

We want to show the position of potential values of "an" in the list for [e,1] . The more divisible the value of c, the more values of "an" there will be in the list at [e,1]. We want to compare that to the list of values at [-f,1] for "a(n-1)".

REMEMBER column -f, HAS to have ALL the value (n-1), not just in [-f,1] but obviously in the COLUMN ITSELF, since every factor in [-f,1] is a value of 'n' for the column at -f. REMEMBER THIS. It allows you to force a column to do what you want (we won't need this for the integer factorisation but it shows one way (another way) how to use the grid as a virtual quantum computer) by multiplying c by a value.

By understanding how [-f,1] and [e,1] behave for highly divisible numbers, our analysis will start to show a set of patterns (four) for all odd numbers that are the difference of two squares.

For each of those families, there becomes a pattern to reduce the set of possible factors down again, and so on until there is a factor multiplied by a value of 'n' or only Big N (c is prime). As we know by now an O(log q) algorithm for integer factorisation whose worst case is finding prime is pretty striking.

>>10330

What do you mean by "potential" values of an? I could put diagrams together that show all the ans and a(n-1)s for both c values.

>>10330

I didn't write all of this code for nothing, just tell me what you want a diagram of and I'll put it together.

>>10332

Is 8kun not letting me upload images again?

>>10330

>>10333

https://imgur.com/a/tRwbDkt

Here are both c values and all of their an and a(n-1) elements.

>>10330

You're doing the same shit you always do, telling us you'll do something on a specific day and never even showing up. If you don't do it now you never will. I don't care if I have to do all of the heavy lifting. From what I've gathered, I'm the only one from the original group who is still willing to give you the time of day. Just tell me what needs to be done in order for you to stop dicking around in this context, rip the bandaid off and explain everything, and I will do all of it immediately. I don't have anything better to do. Let's go.

>>10329

>>10330

Now that I’m divorced I finally have time to do shit I like again, like work on math(s)!!

Looks like we’re back to (-f,1) and (e,1).

Howdy faggots. Hope you’re all well.

>>10341

I have no expectations of success. I still, and will for as long as it exists, read posts here because I enjoy thinking about it. I am constantly impressed with the shit you all come up with, it amazes me.

> Realistically, what is something you think VQC would answer?

Personally I would like to know if Wilson’s theorem has anything to do with it, since factorials were mentioned, and astoundingly, the knowledge behind that goes back to the year 1000ish. It appears to be useless, due to the size of factorials, but why otherwise was there a modular, modular, modular … factorial, factorial, factorial push unless to lead to this apparent cul de sac?

When do we start designing /vqc/ clothing, I need a new hoodie.

>>10342

I meeeeean… I have a few designs rolling around in my head and some that I've whipped up but I never really did anything with 'em.

A GHIDRA thingy with a Muncher head… could be fun.

Any particular design requests?

I'm re-evaluating at the moment.

I was in Hanau stopped in a Taxi at the petrol station moments before the first shooting began next to the petrol station.

I'll attach a screenshot of the Whatsapp conversation the next morning from the Taxi driver.

Not sure if huge coincidence or near miss.

>>10345

>>10344

So you must have seen the /qr/ posts then

>>10344

I say this with no emotional connection to that situation of yours, but thinking about it logically, if that wasn't a coincidence, and you barely escaped that just for suggesting that you would explain the VQC, then the understanding of it would have died with you, and potentially will if it's tried again given it didn't work out. If I were in your situation this would convince me to dump everything immediately.

>>10347

By dump I mean post here by the way

>>10345

>>10344

Spooky! Keep safe!

>>10344

>was in Hanau … moments before the first shooting began

You always say, it's all about time, so good timing on your part. Close enough to a lightning strike to get your attention, but not enough to get zapped. Mebbe a Higher Power is sending a message?

>>10346

>So you must have seen the /qr/ posts then

Think your fishing expedition caught a few in the net..

>>10341

Great questions, pondered the other day. Good to read you.

>>10350

>Mebbe a Higher Power is sending a message?

If he was given the VQC by a higher power, it seems like that higher power is giving a bit of a nudge to get on with it, in my view. He said himself about the continuity of knowledge and the dangers of nCov. He, like any of us, could die at any moment. He almost did the other day, and if he had, a significant task bestowed upon him would have been uncompleted.

Go to 5:25…

Seems like a way to go from 2D to 3D.

Square, circle, sphere, Cube.

>>10352

Indeed.

We will make progress. I will try to be here once a day at least.

You guys are amazing.

I would have thrown the tables up in the air out of frustration long ago.

There is a HUGE incentive to go get Bitcoins, to hack this or that but all extrapolated roads for me never end well.

This has always been about you. YOU.

My AA Higher Power (God Chris) will lead the way.

I have to be at the airport in 4 hours to fly to Germany again.

Thank you for sticking with this.

It's been a burden in some ways and I feel like I let you down. But it does have to be done the right way. With the right timing.

This is just the beginning. I know it doesnt feel like that.

You will honestly not care about this within a day of realising what is next, perhaps fond(ish) memories of frustration. I can empathise. I was completely alone during this, except for the Higher Power, kept throwing bones. Just when I gave up… inspiration. I thought… again and again… this is it! Then the Offset.

What every person (and they all knows maths better than) looks for is somehow to find equations so that the number of variables is covered.

Two variables, I need non-equivalent equations for two minimum.

You can see where I'm going…

Re: dead man switch. No need. I am sure your Higher Power will find a way. I think I might be someone else's dead man's switch ;)

>>10341

Somewhat this.

Thank God for Tops.

And every(o)ne else.

>>10355

Firstly, what's with the change in trip? Secondly, what needs to be done in order for you to post the code? I will do all of it right now. Just tell me what to do.

>>10357

Sorry about the trip. One of the characters was wrong. Cut n paste (RIP)

We'll create the code together.

It will take the input (d,e) and reduce it to n or N if n does not exist (prime).

>>10358

I'm talking about diagrams and stuff. You said collaboration would be the best way forward and then talked about specific diagrams you'd need (e.g. >>10335 ). Give me a list and I'll get them all done right now.

>>10357

It's the short cut. THE short cut.

Being able to apply the GCD algorithm across a SET.

Can you imagine?

It's the equivalent of that.

Sounds good.

Do you know how much trouble you can get into on day one of having that?

Wikileaks insurance files. Open.

Anthony Weiners insurance file. Open (theres a copy at an onion address)

Clinton's server (and what she gave the Chinese). Open

>>10359

It's almost 3am. I'll connect from Heathrow.

Diagrams.

For a highly divisible number we want to show the different "n" values in [e,1]

The more factors a number has, the more values of n I [e,1]

What's the best way to show this?

>>10361

Hey! Missed you pal. Thanks for hanging in there. And for everything you do.

Feels like we're on the home straight!

Its 2.45, gotta drive to the airport soon. Will chat from there.

>>10362

I did that for your two examples >>10335 here. Would you like me to expand on those diagrams or do them in some other way maybe?

>>10360

I think we're all well aware of the ramifications. If it's the right time for disclosure then all of it is irrelevant. There's a long and complicated story behind this but the short-ish version is that I've been messing with these binaural beat meditation things the Monroe Institute developed (they also helped the CIA to develop the training program for the Project Stargate remote viewers) and there's a way you're meant to be able to use them to ask questions about various things. Every time I've asked a question related to VQC and gotten a usable answer, it turns out to be true. During one of them, I was told that I had VQC-related work to do on February 7th, and that was the day I convinced you to start posting here again. There was another one that said Sunday, and that was the day I fished on /qr/ and you came back here about posted about Hanau. There have been a whole bunch of those throughout the last year or two. The main point is that back in October I deduced through this thing that full VQC disclosure would happen in February. Whether anything I just typed sounds ridiculous or stupid to anyone here or not, the thing clearly works (I also used it to get someone I hadn't talked to in months to text me, and he did within two minutes of the end of the audio tape about something completely irrelevant to me that had happened a week prior). Whether it actually happens before the end of the month or not, given what happened in Hanau, there's clearly some crazy shit going on, and the ramifications of this going public are irrelevant to the importance of getting it done.

AA, we'll approach this methodically and analytically. You will follow it with ease.

Tops will no doubt produce some excellent and inspiring art along the way.

Coincidence we're all here?

I think this was always the to complete this.

The Three Musketeers.

Perfect.

>>10335

Are you able to show [-f,1] and [e,1] for 3x5x7x11 and 3x5x7x13?

The value of -f and e should not change.

This is normal, we'll get to the diagram we need.

>>10364

Hairs standing on end.

>binaural beat meditation

My wife just introduced me to this.

>>10364

One of the best posts I've ever read.

>>10367

Those are (f,1) and (e,1) for 3x5x7x11 and 3x5x7x13, what do you mean? Is there an error I didn't spot? I'll put a screenshot of the actual grid entries together if that'll help.

>>10368

>>10369

The tapes are all available to torrent if you want a link, or if you want more information about binaural beats etc.

>>10355

>My AA Higher Power (God Chris) will lead the way.

As in oversoul/christ consciousness?

>I have to be at the airport in 4 hours to fly to Germany again. (from Heathrow)…

I wonder if this is disinfo or just letting people know so there's a trail… interdasting…

>I think I might be someone else's dead man's switch ;)

I wonder how your appearances line up with certain timings, such as the hearing today in that place you supposedly were/are…

>>10356

<;3=

>>10360

But what about that Great Fire Wall of China?

How would that go/work?

Lock the censors out?

Unlock everyone else?

I feel like that'd be part of "what she gave the Chinese"…

>>10363

Feels like?

Even if it's God's Timetable, aren't you the one charged with managing it in this scenario?

Or… do you have a higher-but-not-highest up?

I will be here as witness. I extend any spiritual protection within my power.

While we wait for Senpai to return, I present a video to which you shall react…

"Meh. Pretty much."

>>10374

Thanks AA, I'm having connections issues in the German office including images, will try from hotel.

>>10380

Anything else you want me to put together in the meantime? I'd imagine you'll want the same diagrams for cs that have a different number of factors at some point for example, although I don't know if you want to do that after you've explained whatever you're going to explain about 1155 and 1365.

>>10380

Doesn't look like the economic impact of this will matter since the market appears to be on a significant decline because of nCov. More evidence that now is the time for you to explain the solution as soon as humanly possible perhaps?

https:// www.marketwatch.com/story/the-dow-is-off-by-more-than-a-1000-points-and-heading-for-its-second-worst-point-drop-in-its-history-heres-how-the-stock-market-tends-to-perform-afterward-2020-02-24

http:// archive.is/VoFqV

Trying to post. Middle of the night in Germany. Had connection problems all day.

>>10382

Agreed.

>>10381

Notice anything interesting about these two numbers. Just by looking at them?

The first pattern we will be showing is key.

There are two types of odd integer.

The sum of two squares and not the sum of two squares.

This is further broken into two.

The product of two numbers where the factors are both the sum of two squares or both are not. Or one is and the other is not (with the fourth sub type being made of one being the smaller or larger factor).

Prime numbers are included, since 1 is the sum of the square of 0 and 1, making prime numbers all belong to one family, this will be interesting/important later.

This is all about information carrying. Integers carry all the information you need about them to tell you everything. We just do not recognise it because we are not trained to see it.

In the special case where two factors are the same (squares), integers still carry the information to determine this faster than using a root function.

You may not be surprised by this next piece of information.

We will be switching to binary.

Makes everything a [bit] MORE black and white.

When displaying the a[t] elements of [e,n] in binary, what do we see? What repeats? And what does not? This is the carrier information and the determinant, since the properties of [e,1] are a wave function of sorts.

>>10384

>Notice anything interesting about these two numbers. Just by looking at them?

Nothing that isn't already covered in the grid patterns thread, like one of them having an and a(n-1) at the same t and one of them having them at t and t+1, for example. Whatever you're implying you'll probably have to point out yourself.

>>10251

Are you saying doing something like a graphic where you can have a coordinate (x, y, i) and then a vector for direction you're looking, then on that plane you have (j, k, l) for the RGB values or something like that?

>>10387

Similar.

>>10386

The integer, if a highly divisible number will shows it's constituents. Information is hard to hide in very divisible (smooth) numbers. I.e. 1365 1300 65 same with the smaller number divisible by 11.

The algorithm we will produce is based on information density. The factors of an integer are impossible to hide. They have a SIGNATURE. Signatures are IMPORTANT.

Remember. If a factor is in column 'e' for any e, it is present exactly twice in [e,1] within two positions that together, they ALWAYS equal that factor plus 1. And the factor is always a value of n in column e. If even one factor did not behave this way, none would. Just like the construction of the Mandelbrot Set. Just like the non-trivial zeros of the analytic continuation of Riemann Zeta function, just like the UNIVERSTALITY of the Riemann function.

ALL integers are CONNECTED.

NOTHING is random.

Quantum versus Analog.

How can you calculate the Nth hexidecimal place of PI without calculating the predecessors?

Spigot function?

The biggest discovery in history towards using virtual quantum computer functions.

THE BINARY REPRESENTATION OF AN INTEGER CONTAINS THE TRAVERSAL CODE FOR INTEGER FACTORISATION.

It just has to be performed in the correct order at the CORRECT SCALE.

>>10389

Do we need any more specific diagrams for binary numbers?

>>10389

I appreciate how much time you've spent here lately a great deal, but the way you immediately leave after posting one of these clue posts is a little odd. Are you stalling or are you just unreasonably busy?

Quick note to say I will be here tomorrow, been travelling most of the day. Hope your week is going well and your preparations are done well in advance, just in case supplies do get harder in your area.

>>10360

Why are you acting like linear-time integer factorization will open the Wikileaks insurance files? It would take a significant time chunk studying the algorithm and dissecting it to reverse it even if the problems had similar solutions. As it stands, what you are saying has little to do with reversing AES.

>>10355

Hacking Bitcoins not only includes solving the Discrete Logarithm problem over Elliptic Curves, but reversing SHA1. Why are you teasing these things so deceptively as if we have even solved THIS problem?

>>10394

Is it tomorrow, yet?

Summary of VQC posts from February so far

>there's a loop function that works similarly to the square root function in that it eliminates values exponentially faster than they grow

>the algorithm eliminates an/a(n-1) element pairs in (e,1) and (f,1) until it outputs either both N-1 and N (if it's prime) or the smallest valid n-1 and n

>values are eliminated based on patterns in (e,1) and (f,1) that are easier to see in binary

>there are four patterns for all odd numbers

>there are two important types of odd number

<sum of two squares (which appear in columns where e is a square)

<not sum of two squares (which appear at any other e column)

>there are four important configurations of a and b values for every c

<a and b are both sums of two squares

<neither a nor b are sums of two squares

<a is a sum of two squares and b is not

<b is a sum of two squares and a is not

Other potentially relevant grid patterns

>if an integer p is a factor of a[t], then p will be a factor of a[p+1-t] for ALL cells in row n=1

>if p is a factor in a[t] then there exists (e,p)

>in (e,1) we have elements where a[t]=an and a[t]=bn. In (f,1) we have elements where a[t]=a(n-1) and a[t]=b(n-1). These elements are n and n-1 apart in terms of t respectively

>in (e,1) where a[t] = bn, d[t-1]-d = b(n-1)

>the a[t] values in (1,1) are all odd sums of two squares (although not every odd sum of two squares appears as an a[t] value). The a[t] values in (1,1) are also only divisible by numbers which are also odd sums of two squares

>all products (integers) c that are the sum of two squares appear (only) in columns where e=0,1,4,9,16,25…

>>10385

>>10389 ty

Helluva fishing expedition in the open waters today AA, duly noted.

Also well done on the trips audits.

And interesting on the binaural beats and meditation experience.

>>10395

>>10396

Hope you didn't spook the fish with those posts mate. Indeed, we're still on step 1/2/3 or whatever the steps are for this program.

Apologies for the delay.

The switch to binary is to show the order in which the reduction algorithm runs across the set in a[t] for any given c at [-f,1] and [e,1] between the value of a[1] and a[t] = Nc, where all values of n are found, where there is the empty of values, c is prime and the algorithm terminates, hence why finding a prime is a worst case, though given this is O(log q) where q is the length in bits, this is still lightning fast.

Remember, [-f,1] MUST contain a value of (n-1) for each (even if only one) value of 'n' in [e,1].

Is it possible to highlight this for the previous examples?

What are the values of 'n' and (n-1) for those?

>>10391

I'm following a pattern.

I'm not determining that pattern.

>>10390

100%

>>10392

Love it

>>10402

It's been almost a month since you said you'd give us the solution. You're clearly too busy to do this during the week if you can't even manage one post per day after saying you would. Europe, Asia, the US and the Middle East are all in full-on pandemic empty-supermarket-shelves doomsday prep mode. The economy and stock market are down the shitter. You almost died in a fucking mass shooting. Anyone else would have already done it by now. Is there any reason why we can't finish this right now?

>>10395

Answered above

>(What is a) spigot function (algorithm)?

It's a funny old world.

>>10401

I could calculate each of the n and n-1 values and put them in the spreadsheet cells beside their row in the table. Do you want this for the binary version or the base 10 version?

>>10403

You already have what you need.

Start writing the algorithm.

Each step in the loop should remove AT LEAST half of the set between a[0] and the value for BigN (or BigN-1) multiplied by c.

Write the algorithm, start with the skeleton and I will correct it.

We can cut to the HOW, and if you'd like, we can come back to the diagrams of WHY.

Start with how you think the algorithm will begin, then there is a reducing loop based on a condition.

>>10407

I'll be doing this in Java then, it's the language I'm best with. Based on your specifications, here's a beginning skeleton. I forget how to do code tags here but I don't want you to fucking leave for half a week again so I'm doing this in a hurry.

public BigInteger factor(BigInteger d, BigInteger e){

ArrayList<BigCell> e1t0tocBigN = makee1list(d, e); //assuming this function is irrelevant since we can do it and c*BigN appears in (e,1) where x=c-d

for(int i=0; i<e1t0tocBigN.size(); i++){

//something

}

}

>>10408

>>10407

public BigInteger factor(BigInteger d, BigInteger e){
    ArrayList<BigCell> e1list = makee1list(d, e);
    ArrayList<BigCell> f1list = makef1list(d, e);
    BigInteger c = (d.multiply(d)).add(e);
    for(int i=0; i<e1list.size(); i++){
        if(isOddType1(c)){
            //something
        } else if(isOddType1(c)){
            //something
        } else if(isOddType1(c)){
            //something
        } else if(isOddType1(c)){
            //something
        }
    }
}

The four isOddType things are for those four patterns you talked about.

>>10409

Come to think of it, making the lists for (e,1) and (f,1) would take a very long time and take up a lot of space in memory for big numbers. Is there a way around that?

>>10410

>making the lists for (e,1) and (f,1)

So… why make full lists then?

What correlations are you looking for?

Can you filter the results that way?

>>10410

>>10407

Sounds like he is describing a binary search. Meaning we wouldn't need to generate all the elements from a[0] to a[t] = Nc, but instead the midpoints. Ie, first element to look at would be halfway between a[0] and a[t] = Nc, then we discard one half and look at the next.

Although when it comes to develop the algorithm such a method would probably be useful.

>>10409

The gist from my understanding

1. Create element for a=1, b=c

2.1. Transform element to row 1 (NA transformation)

2.2. Transform element to negative column (NA transformation for -f)

3. Convert x to index t

4. Loop (initial values: low = 0, high = t from step above)

4.1. Generate element between t=low, t=high (mid point? a[(low + high)/2])

4.2. Evaluate cell

4.2.1. If correct => return ?

4.2.2. Else if tooHigh => high = (low + high)/2 - 1

4.2.3. Else if tooLow => low = (low + high)/2 + 1

4.2.4. Else => return is prime?

The place we're stuck at is evaluation, how do we correctly evaluate the element?

>>10413

If it's a type of binary search we wouldn't need a=1 b=c, we'd just need t=0 and t for x=c-d for (e,1) as the top and bottom points, since he said that was the range. He also said "at least" half, which would imply there's something different about the midpoint, or that it's a different type of algorithm.

>>10414

You're right and I agree with the "at least" half. I was thinking a bit about

>>10407

> Each step in the loop should remove AT LEAST half of the set between a[0] and the value for BigN (or BigN-1) multiplied by c

The naive interpretation of set would be elements between a[0] and Nc transform, unless he is being coy about the phrasing.

public BigInteger factor(BigInteger d, BigInteger e){

    BigInteger c = (d.multiply(d)).add(e);
    BigInteger cBigNx = c.subtract(d);
    BigInteger cBigNt = BigInteger.valueOf(0);
    if((e.mod(BigInteger.valueOf(2))).equals(BigInteger.valueOf(0))){
        cBigNt = (cBigNx.add(BigInteger.valueOf(2))).divide(BigInteger.valueOf(2));
    } else {
        cBigNt = (cBigNx.add(BigInteger.valueOf(1))).divide(BigInteger.valueOf(2));
    }

    Element aLow = e1Element(e, 0) //t=0
    Element aHigh = e1Element(e, cBigNt) //a[t]=c*BigN

    while(!((aLow.a()).equals(aHigh.a()))){
        if(isOddType1(c)){
            //something
        } else if(isOddType2(c)){
            //something
        } else if(isOddType3(c)){
            //something
        } else if(isOddType4(c)){
            //something
        }
    }

}

Element would be an object representing an element obviously, and e1Element would be a function that just returns an element from (e,1). I would think since you haven't said anything about needing (f,1) that we'd find the element at the same t when we go to make calculations maybe.

>>10416

Why don't we define some constants for 0, 1 and 2, or at least assume we have them. Then we can remove the need for "BigInteger.valueOf(..)". Should make the code slightly more readable. Something like ZERO, ONE, TWO as big integer constants.

>>10416

Also I'm unsure of our while condition. The basis for this is only in column e, ignoring column -f.

>>10417

public BigInteger factor(BigInteger d, BigInteger e){

    BigInteger zero = BigInteger.valueOf(0);
    BigInteger one = BigInteger.valueOf(1);
    BigInteger two = BigInteger.valueOf(2);
    BigInteger c = (d.multiply(d)).add(e);
    BigInteger cBigNx = c.subtract(d);
    BigInteger cBigNt = zero;
    if((e.mod(two)).equals(zero)){
        cBigNt = (cBigNx.add(two)).divide(two);
    } else {
        cBigNt = (cBigNx.add(one)).divide(two);
    }

    Element aLow = e1Element(e, zero) //t=0
    Element aHigh = e1Element(e, cBigNt) //a[t]=c*BigN

    while(!((aLow.a()).equals(aHigh.a()))){
        if(isOddType1(c)){
            //something
        } else if(isOddType2(c)){
            //something
        } else if(isOddType3(c)){
            //something
        } else if(isOddType4(c)){
            //something
        }
    }

}

I forget if saying "x = zero" assigns the object pointer or the value for BigIntegers so a couple of those might need to be changed.

>>10418

He said he would correct it so when he's back from whatever the fuck prevents him from being able to post here for reasonable lengths of time he can elaborate, but since he usually frames this from the perspective of (e,1) with comparisons to (f,1) as opposed to using both and comparing them to each other (e.g. he mentioned c*BigN in (e,1) without any mention of c*BigN-1 in (f,1)) I thought that would likely be the way to do it.

Here's the whole thing in a class, with the Element class too. I didn't really change much about the function in >>10419 so refer to that. All we need are the conditions in the loop function.

vqc.java: https://pastebin.com/DGTDrpxN

Element.java: https://pastebin.com/bJyynTSQ

Obligatory "where the fuck is Chris"

I would imagine that those four odd types might have something to do with 4 columns containing all primes in mod12.

Going off the rails a bit:

c = a * b; where a and b are primes and b > a gives:

0 = c (mod b)

a = c (mod (b - 1))

a = c (mod phi(b))

Going thru some properties of: https://en.wikipedia.org/wiki/Modular_arithmetic

Modular multiplicative inverse can be calculated in O(log n). And we have two unknowns and two equations involving a and b. Can we frame this question in such a way to be solvable by Extended Euclidean Algorithm?

>>10421

>I would imagine that those four odd types might have something to do with 4 columns containing all primes in mod12.

I thought it would have been based on sums of two squares, since he mentioned that there are four relevant types of odd c based on sums of squares >>10385 here. I've never heard of that mod 12 thing. I mean it could be that, but it would make more sense if it was something Chris already said I would think. If any number is the sum of two squares it'll end with 01 in binary, and if it isn't it'll end in 11 (aside from 57 in my tests for some reason). There might be a better way that also tells you about a and b that Chris can tell us.

Summary for Chris if you turn up today so you can maximize time spent giving us code

There's a function here >>10419 that takes in d and e, finds the (e,1) elements for t=0 as a minimum and a[t]=c*BigN as a maximum, and then initiates a loop that changes the maximum and minimum until they're the same (at which point you'd return n or whatever).

All we need to know is

-what the four criteria are for odd numbers and how to figure out where they apply (I figured it's probably the four sums of two squares things but >>10421 thinks it's something else)

-how to get rid of at least half of the elements each time the loop runs like you said

-any other corrections we aren't aware of

That's only three small things left to do. Surely you can do the all of it today.

The 17-Klein Bottle

All you need is one extra equation.

One offset.

The difference between -f and e, the first cell.

Why do the numbers fit?

The first thing you do if you become very rich is worry about protecting those close to you.

Better not to play.

>>10430

Are they protected? You did agree several times that the time is right.

>>10429

Are you going to give us the equation? If clues were the way forward we would have figured it out ourselves.

>>10429

Has the equation been posted in the past?

>>10337

I love you man… Every time I come check in you crack me up. What ever happened to Teach? That dude was pretty cool.

>>10436

I can't tell if you're implying that you're laughing at me or with me but sure. Teach hasn't been here since like 2018.

>>10429

>>10430

The longer you stall the more you contradict everything you've ever said about higher powers and timing. I swear to god you're going to give me a fucking aneurysm.

>>10440

Have you tried what he has been suggesting? Look at the grid in binary, look for patterns that would allow you to remove elements not worth looking at?

>>10441

If I'd found anything useful I would have said. The only stuff I found was related to 00/01 and sums of two squares in those diagrams I posted. How about you?

>>10440

Not yet, still looking at it, trying to see any pattern he talked about.

>>10442

The guy doesn't stick to the timeline, but that's his thing. We shouldn't let it control our emotions. In the end it's just a math problem. It's okay to take a few days off and just get some fresh air. Might even help to gain some perspective on everything.

>>10443

I agree with you and everything but it's clearly a bit more than "just a math problem"

>>10444

Not really. It boils down to a math problem, but the consequences of the problem are big.

We're not here to deal with the consequences, nor to ponder upon them. We're here to learn math in a new way, one that allows us to solve problems the old way can't.

The loop section executes once for each possible length of 'a' in bits, where 'a' is less than 'd'.

The loop section executes a maximum of length of 'd' in bits (half of 'c' in bits).

>>10446

Now to play the "is this the only thing he's going to post today" game

>>10446

Nothing you've posted for the last week and a half has been productive. It seems to me like all you're doing it making it clear that you haven't abandoned us and that you're waiting for something. If a global pandemic with entire countries going under lockdown and an economic and stock market crash that's probably going to be worse than 2008 isn't enough for you then what the hell could you possibly be waiting for? Or are you just too scared of what comes next? Why not be transparent about it and have a conversation instead of stringing us along like you always do?

>>10450

a is going to be at least 3 (which is two bits long) and at most d (the square root is the highest possible a value). The range of possible lengths of a in bits is two to however long d is in bits, or d's length minus one. We already know whether a is going to be odd or even since if c is even the lowest a value (which he said is what the algorithm finds) is 2. That determines the first bit (1 if a is odd, 0 if a is even). If you include that as a step, that makes it the length of d in bits, and seems to imply that in the worst case (primes) we have to do a calculation for each bit one by one, like you said. In an O(log n) algorithm, you halve the search space every time you run through the loop, and d is half the length of c in bits, so assuming you have a preliminary O(1) step I suppose technically it's O(log n) where n is the length of c in bits.

We also know based on this stuff >>10423 and what he said here >>10385 that we can probably figure out the second bit (second from the right) based on sums of two squares (although we don't know about when a and b aren't both sums of two squares or not sums of two squares yet, just when they both are or both aren't, and there are also a few unaddressed anomalies like 57 which doesn't fit the bit pattern of the rest of them from what I remember). That reinforces the idea that we're doing a calculation for each bit. If it's a loop algorithm, that means we're going to be doing a calculation or a set of calculations over and over again. He did say here >>10429 that we need one more equation. From what I gather, we do some unknown calculation to some as yet unidentified variable and then repeat the sums of two squares thing based on a different bit, and we do it over and over. Aside from not knowing what that equation is or where to apply it, how to figure out which of a or b is a sum of two squares and which isn't if they aren't the same, and the anomalies in those bit patterns, the other main confusion is what role an and a(n-1) in (e,1) and (f,1) from t=1 to x=c-d play and how you can halve the number of elements you're looking at in any way that remotely relates to any of what I just explained.

Long story short, metaphorically, he's drawn out a set of directions on the map, but the directions start two towns over. Hence what I said above about him obviously not making these posts to actually get us anywhere but to stall for time for whatever reason.

>>10452

Sooooo…. why aren't we 2 towns over?

>>10453

Because he hasn't given us directions to where the directions actually start

>>10454

Still no Genesis Point?

Where does the gap lead?

What even defines "the gap"?

https://oeis.org/A000010

Euler totient function phi(n): count numbers <= n and prime to n.

-comes back days later to see something sitting in the quick reply box-

Apparently I was doing something and fucked off.

I wonder where Chris went…