Thanks for the bread, Baker! Very tasty. Thread Jesus says "Well done, my good and faithful servant."
>We have a method to calculate what n would be if we used our blue base of ((f-2) div 40). This n will call n0, to make it different from the value of n that will be our solution.
>We will add a method to calculate what would be missing from the triangles if we could only fill them with n0 squared and multiples of 2d.
>That gap will allow us to use more geometry of triangular numbers to establish what multiple of 2d could be added to make our eight triangles.
>That geometry with multiples of 2d will give us the solution, if it exists (which in our example we know it does). If doubling the width of the chunks of (f-2) didn't give us a solution, the number c would be prime, the worst case (doubling the width of (f-2) chunks very quickly determines this as the number of times it doubles is logarithmic in the length of (f-2)).
>So far, it should become clearer that increasing the length of c adds to the number of calculations in the logarithmic of half the length c in bits. Hence why the overall complexity is < O(log m) where m is the length of c in bits.
>Bear with me as we walk through the rest in stages.
So since f is a derivative of c, d, and e we can use it's value to find n0. Then we use multiples of of n0 + 2d to fill the triangle. It looks like VQC is saying to double the width of (f-2) chunks until we get a match. If no match, c is prime. Thoughts, Anons?
>Recursively grow until we find a match?
Theory: Basically, there are only certain values of n that can exist, and they're multiples of n0 or n0 + 2d. This is what we've been looking for, lads. A way to narrow down the n search.
n0 is the triangle seed for the c var tree!
Sure, AA! But I'm still working to understand. VQC is pretty quick to respond about questions to his twatter: @ChrisRootODavid. Ask him directly, he's pretty quick to respond. We are all Confused at this point. Moving forward tho.
Twatter is fun, just gotta be secure.
I think a big key to understanding n0 is that we're working with Grid examples that have very small numbers relative to RSA examples. Maybe we should work through RSA 100? We have the answers, so it makes a perfect walkthrough. Thoughts, Anons?
So maybe let's pick a bigger number from the grid then. In the 10,000 range for c.
>If the important thing about the base is having a number between n and x+n, then there has to be something in the original c, d, e or f values that tells us what to use.
PMA, I agree. Our starting value of c contains all we need to find the prime factors. Looks like triangles will save the world, lol. So here's my current thoughts.
n0^2 + multiples of 2d can fill the remaining part of one triangle. Pic to follow, trying to work it out now.
Ok, so I'm working on the new diagrams. Have lots of questions, and will post new diagrams later tonight. Don't want to clutter up the board without working through some real examples.
One important question worth asking at this point: in the small square (x+n)^2, why is x^2 (red) the same color as 2d(n-1) in the big square? Accident or correlation?
Hello lads! Working over here, re-reading all crumbs. Working on the algebra side to understand the (x+n)^2 formulas. Have fam stuff all day today, but here thinking and and checking in. Here's what I have so far, and it goes with the small red square in >4678.
(x+n)^2 = n^2 -1 + 2d(n-1) + 2d + 1 - e
Also, you can sub in 2xn + xx for 2d(n-1) + f - 1. Trying to understand the (x+n) square and how it relates to all the formulas. Thoughts?
TBH, the diagrams produced are already excellent, so I felt like I was doing pointless work trying to build another. Great job MM Isee, and PMA on all the new ones!
Hey Baker! I think you're right. VQC said the grid provides a shortcut.
Thanks Topol!
Hey MM! I was just using that element to verify that the formulas worked. Using PMA's n0 method >>4879 he was able to get a couple of exact matches for (x+n) using multiples of n0. Check out his second sheet of calcs.
Big Idea: If we can find (x+n) using triangle geometry, then we don't even need to know the individual values of x and n. We use the quadratic to solve for n, then plug in n and solve for x.
Actual Order of Operations Looks like this:
Start with c
Get d,e,f
Get n0: (f-2)/8 (for smaller f) n0=(f-2)/40 (for larger f)
Multiply n0 by increasing factors until a match for (x+n) is found. This was my triangle diagram idea. PMA tested it in spreadsheet form and got a match.
Verify correct match by crosschecking using quadratic formulas to see if answer matches the prime answer we're looking for.
Solve for n first.
Then plug in n to solve for x,a, and b.
Still a work in progress, but we are very close to
Thanks PMA! Studying now. Also, I agree with you here:
>I think you should come up with a better name for these methods!
I propose "CAprime.one and CAprime.two :)
Lost my trip. Baker, you there? I'm looking forward to some excellent new bread! New loaves? Here's some loaves for you lads.