Here are some Mersenne trees for you. These are without trimming bits / dividing by two.
True, and the resulting tree makes more sense when trimming. Still, >>3653 gets the right answer without completely trimming the 40. So did we just get lucky with small numbers or are there rules for how many bits to trim? Maybe this is too simplistic anyway, given the hints about larger numbers being more complex, and triangular numbers.
I tried just coloring triangular numbers instead of primes but there are to few of them to really do anything on their own… Can't figure out how to relate the recursive tree to our grid so I'm a bit stumped at the moment
Yes, we need to split into branches (return two or more results) for each recursion, otherwise there is no tree. The rules for how to manipulate numbers (trim all bits, only trim bits if number is not triangular, ect.) for each step are up to us. But VQC was pretty explicit on trimming all zeros, not trimming some zeros. So assuming the tree we can generate now is right, step 3 has to be walking it and summing numbers in some way?
Wish I was better at recursive programming. I got pretty stuck trying to add dimensions to a binary tree in addition to branching…
Just summing up the leafs for a completely trimmed RSA100 tree and dividing by two gets us to
40364365001756683226800431241044541563293164124405
which is pretty close to our factor
40094690950920881030683735292761468389214899724061
but not close enough. Maybe we can use the grid on this.
This is what I meant by "walking" the tree. I've been considering it but the only thing I can come up with is starting with an untrimmed tree and trimming one more bit for every possibility, and regenerating the lower nodes. Then checking the leafs again for a factor.. this is when I figured I could just generate a complete multidimensional tree in the first place. Still doable computation wise, not nearly log n but if we get a solution tnat may give more hints.
Then I realized you have to try all permutations of the multidimensional tree, not just walk it once, so we're back to exponential…
So maybe something like walk up the tree until you find a prime, triangular number, or some grid variable, and choose your next direction based on that (up or next branch down)… thinking out loud as well
Trying something new. RSA100 with 1xcc for each number. Let me know if anyone has ideas for what functions to use
I may have posted the wrong image, that was 1xc, not 1xcc. Or I misunderstood everything completely, whatever. So here they are with bonus 2xcc. I am not sure how to add cats yet.
And apparently bioinformaticsfags don't like having dark backgrounds in their programs so I had to hack the shit out of it to get this.
Checked, nice trips
Good work both of you, so far any correlations I've found appear to be just luck and work for one number only.
In other news I wrote a variant of the tree-generation algo in C and ran some tests:
RSA100 0.07ms
RSA2048 0.4ms
RSA4096 0.7ms (self generated key)
RSA8192 1.5ms (self generated key)
RSA16384 3ms (self generated key)
And just for fun, M74207281 * M77232917 (two largest known primes multiplied by each other) takes… ~42 seconds. Somehow I don't think anyone will be using RSA keys that large. More surpised my computer didn't blow up trying that!
Screwed that up when converting code to post formula. Supposed to be x+n, not x*n. So if you have x_plus_n:
n = sqrt(x_plus_n**2 + c) - dx = x_plus_n - na = d - x
I have been doing something similar on the RSA100 tree trying to find close matches for x or x+n using various formulas for every combination of the numbers in the tree, up to ~10 different numbers. Summing nodes in any combination isn't any good. Neither is summing triangular numbers, or trimmed versions of triangular numbers. Initial runs on multiplication aren't looking too hot.
for i in range(1, 42): print(i) #for comb in combinations(traverse(tree), i): for comb in combinations(map(lambda n: (n*(n+1))//2, set(traverse(tree))), i): result = sum(comb) ...check how close result is to targets...
I think the C version counts as instant though, just need to figure out what to checkclang -O3 -march=native -mtune=native -lgmp tree.c -o tree0.000493 seconds - create_tree0.000018 seconds - traverse(tree_count)2587 nodes
Those timings are for RSA2048, not RSA100. Sure beats python.
I have to agree, we're not really solving much on our own, just collectively understanding the crumbs. Seems almost designed that way.
The fractal part has been obvious since day one imo, and is one of the main reasons I'm here. They just got way more obvious in tree form than looking at a flat grid. I think the tree is only neccesary for factoring numbers though, and not related to navigating an infinite fractal grid, although recursion still applies. Think of it as a tool, not a solution.
The worst part is I'm not really interested in solving RSA, I want to see what we can come up with using this. A software quantum computer ffs, the possibilities are ridiculous. And given what we've already seen its pretty fucking fast too.
I'm not sure what the strategy behind this slow drip of info is since Chris (or the "Chris entity") is obviously smart enough to have one, and its kinda working well (we're still here).. Trying to drop this knowledge any other way would probably be suicide and it'll get wiped from the internet, so that may be part of it.
A darker twist would be that some group already has this math (the elite hoards knowledge) and are already using it in secret. We know the NWO plan depends on chaos and the financial system is already set up to fail, so if (((someone))) didn't want to get the blame, what better way than having a bunch of channers release it in a timed way that blows shit up good…
^^ The above may be a bit out there and I'd like to think my intuition is good enough to avoid such traps, but knowing what we're up against I'm not so sure. Think its worth considering at least.
Now lets get back to math!
Sooo.. didn't mean to kill the thread.. too weird? too real? too crazy?
Everyone working on math? Or upgrading their VPNs?
Already know where I stand on the spoiler part, and will continue anyway but assumed it'd start a discussion at least. Post some trees or something!