VQC !!Om5byg3jAU ID: b02913 April 3, 2018, 2:10 a.m. No.5368   🗄️.is 🔗kun   >>5370

>>5284

That was me asking a question about whether it was worth continuing the fight against alcoholism, staying sober, a massive personal struggle, and also whether I had messed up by participating in Discord, I saw some good in them at the time.

I think the response from Q was general to all anons who get down and/or struggle.

VQC !!Om5byg3jAU ID: b02913 April 3, 2018, 2:28 a.m. No.5377   🗄️.is 🔗kun   >>5378 >>5406 >>5533

>>5376

There's a reason I'm back now.

I've got to go through a fair bit of setup as I've wiped and replaced my primary devices and restore to this temporary one.

I believe in the plan of Q, POTUS and WW support.

This is truly BIGger than imagined.

Anonymous ID: b02913 April 3, 2018, 2:49 a.m. No.5382   🗄️.is 🔗kun   >>5390 >>5406

>>5380

That's a good question.

I will stick here to the process of you all discovering how the VQC works and the subsequent consequences of P = NP and tech, etc.

We've come a long way together.

Through the hard times and the good.

We're still in the foothills.

The view at the top will be worth it.

VQC !!Om5byg3jAU ID: b02913 April 4, 2018, 1:06 a.m. No.5413   🗄️.is 🔗kun   >>5414 >>5416 >>5420 >>5428 >>5430 >>5450 >>5467 >>5508 >>5594

>>5408

You are closer than you think.

It is efficiencies.

From O(EXP) to O(log t).

From exponential complexity class (well, sub exponential with the general number field sieve) to the natural log of the length of the integer c in bits.

Breaking the problem down.

First by roots.

Then by triangular numbers.

I'll clean up my set up, then walk through a clean version.

Perhaps one day I'll pastebin the whole dump of the last seven years and show you the tortured/enlightened path through all this.

The probability of it all happening by accident seems remote. Just needed someone like Trump/Q et al to come along.

 

NOTE: You don't need to know the general number field sieve. It is a horrible approach in terms of understanding. Just know that it is based on making the Fermat factori(z)ation method more efficient.

 

FUN and IMPORTANT later: any coincidences in the Mandelbrot Set? E.g. one of them relates to Pi.

VQC !!Om5byg3jAU ID: b02913 April 9, 2018, 5:11 a.m. No.5481   🗄️.is 🔗kun   >>5486 >>5501 >>5534 >>5671 >>5739

>>5430

Hi VA, and everyone else, you make a good point.

I think the release and impact on integer factorisation has been anticipated.

I also see nothing that prevents anyone from progress in mathematics.

This is simply the solution of a mathematical problem after all, at heart, but in reality it is so MUCH more.

Thanks for your patience. I am waiting on new equipment at the moment.

To answer other questions, after the next big step, the RSA numbers will be solvable on the laptop I owned ten years ago, before even.

Going from a search to a calculation is an almost inconceivable step.

Think of a Rubik's cube.

Trying to solve it without a plan.

Then remembers those books on how to solve it?

You have to know which side to start with and which steps to take.

All we will end up doing with the next step is figuring out a pretty pattern with odd squares with a single unit hole in the middle.

The symmetry is beautiful in the solution.

When that symmetry cannot exist in more than one way, you have a neat way of spotting prime numbers.

VQC !!Om5byg3jAU ID: b02913 April 9, 2018, 5:14 a.m. No.5483   🗄️.is 🔗kun

>>5438

The improvements on this method became the sieves, ending up with the general number field sieves.

The start of the general number field sieve talks about selecting a polynomial and that there is no known way to do this. This is critical. You don't have to know about all that, but in essence, what we are doing is bypassing that random step.

VQC !!Om5byg3jAU ID: b02913 April 9, 2018, 5:23 a.m. No.5485   🗄️.is 🔗kun   >>5498 >>5501

>>5456

>>5456

Most of this is spot on.

The main point to make about O(log t) is that the amount of effort in the calculation doesn't increase by ten if an extra decimal digit is added, it increase by the natural log.

The natural log of 100,000 is only two more than 10,000 as an example.

1000,000 is only two more than that (approx 13 : floor)

In reality, the process will be optimised to be AT MOST O(log t). It's basically dependent on how efficient your square root function is. Anyone got the Big Oh for the square root function?

VQC !!Om5byg3jAU ID: b02913 April 9, 2018, 5:29 a.m. No.5491   🗄️.is 🔗kun   >>5500 >>5517 >>5521

>>5486

On exactly the right lines.

The 'heart' of the problem is that breaking the problem down involves two sets of triangle solving (otherwise it is prime), with one set of triangles (say half of the eight) working on a triangular problem that is one unit longer than the other set. This is one reason why no current approach I have seen works.

Then we'll show how the grid (The End) solves/shows the solution.

VQC !!Om5byg3jAU ID: b02913 April 9, 2018, 5:30 a.m. No.5492   🗄️.is 🔗kun   >>5533

I should have new equipment by the end of the week. If not, I'll get admin access to install what I need on the current borrowed equipment.

Then there is a week before my time comes under pressure, so hopefully next week is going to be big!

VQC !!Om5byg3jAU ID: b02913 April 9, 2018, 8:55 a.m. No.5499   🗄️.is 🔗kun

>>5498

I posted some code for sqrt of big integers earlier on.

My understanding is that the complexity of finding square roots is at most O(log t) where t is the length or size of the number whose root we are finding (remember we don't mind what the remainder is as a fraction or decimal, just the integer part, once the integer part is found, the remainder is simple to find).

A note on Big Oh; what you're looking for in any process is the part that grows the fastest.

 

E.g.

 

x squared plus a billion x + a trillion

 

x^2 + 10^9x + 10^12 would still be dominated by x^2 at scale. From x=1 increasing to x=one million and up, the total of the equation begins to be dominated by x^2 and as x becomes a quadrillion, the other parts become less significant and so on.

 

One way to look at complexity or Big Oh is to think what if the size of operand (in our case right now, it is c) in the operations of the process tends towards infinity. What dominates? That's your Big Oh. For the process we are looking at, the square root function would dominate. I.e. there is nothing that would dominate MORE than that.

VQC !!Om5byg3jAU ID: b02913 April 13, 2018, 3:49 p.m. No.5545   🗄️.is 🔗kun

Thanks for waiting.

Pick up new equipment tomorrow after 3pm UK time.

Will be here everyday for seven days after that as well as tomorrow.

Thanks again for waiting.

VQC !!Om5byg3jAU ID: b02913 April 13, 2018, 3:57 p.m. No.5546   🗄️.is 🔗kun   >>5547 >>5549 >>5550

>>5533

Nice digits.

What we will move onto quickly later is using the Mandelbrot Set on both spatial and temporal transforms for signal input.

That will make your head spin.

Someone takes an illegal picture? Rotate the view.

Want to see what happened in previous 'frame'? Etc.

A perfect resonator like the application of the Mandelbrot Set on signal data is remarkable in a (growing) array of solutions. I'm surprised it is not in the public domain yet.

VQC !!Om5byg3jAU ID: b02913 April 14, 2018, 9:43 a.m. No.5572   🗄️.is 🔗kun   >>5573 >>5574

Anons, I was emailed a death threat this morning at 10:12am to an email address I haven't given out.

The threat was to myself and my family.

Please stand by.

VQC !!Om5byg3jAU ID: b02913 April 15, 2018, 3:20 a.m. No.5591   🗄️.is 🔗kun

>>5574

Not sure, it was quite subtle.

Life Insurance offer that could have been spam but knew my details to a point too far and talked specifically about my family in a pointed way.

Let me put some more measures in place.

There is plenty to work on to come up to speed in the exercise as it is VERY relevant for all the subsequent pieces.

VQC !!Om5byg3jAU ID: b02913 April 15, 2018, 3:24 a.m. No.5592   🗄️.is 🔗kun   >>5593 >>5599 >>5603 >>5604 >>5662

>>5575

>>5576

>>5582

>>5583

Brilliant.

If you colo(u)r the values of f in the grids that show the triangles, patterns will be added, ESPECIALLY if you divide f up into 8. The remainder of f when divided by 8 is a key to what values of d and (n-1) can be used.

Playing with these patterns will organically lead you exactly where we will be going.