Hey all - I'm around too. No progress to report either.
Welcome back, and thanks for the tips.
A few possible answers to the questions…
1+8T for odd squares:
Draw odd square, remove middle square, divide remaining into 4 equal rectangles, by drawing out from the middle square.
Each rectangle's sides are only 1 different from each other, which is the same as 2 equal triangles.
See image.
Why is (n-1) important?
I think it has to do with the same property.
A square with 1 row or 1 column removed is 2 equal triangles of (n-1) each.
C'mon CA share! :) hopes are up.
Thanks PMA. So have you come to any conclusion about the e%4 parity of x+n yet?
Also a few more comments, sorry if these are obvious.
A square with even side length can be divided into 4 equal smaller squares, so we can reduce a big square of even parity into 4 smaller squares repeatedly until they're odd.
A while ago vqc posted about base. I think I've kinda figured out how that plays into it…
Example c=145, in base 2 = 10010001
c = 2^7 + 2^4 + 1.
c = 2^3(2^4 + 2) + 1
c = 2^3(2(2^3 + 1)) + 1
c = 8(2(8(△1) + 1)) + 1
Typed out my previous comment before I saw this.
See my latest post :)
But you do have an interesting idea here…
Now we can break down either even or odd squares into triangles, and you're saying we can reverse the entire process and move up instead of down?
I like that thought.
Hey guys!
Thanks VQC for the guidance!
This is pretty overwhelming all this new info - love it though!
First off, Topol! Great to put a face to the ideas! I've actually watched some of your videos before not knowing it was our very own!
VA I'll keep that tab open too, and hopefully I can help out a little.
This is cool CA! Multidimensional Triangles!
So I've been reading and trying to understand all the crumbs. It seems that this one in particular might be useful to focus on:
I've been playing with a simple example, with small numbers just to try to understand:
a = 3
b = 29
c = 87
d = 9
e = 6
f = 13
(f-2)/8 = 1
(f-2)%8 = 3
x = 6
n = 7
xpn = 12
half_xpn = 6
test_correct_base = 7
(f-2)/8 factors = 1,1
So if we go back to our original calculation of f in the grid, the associated values to f were calculated by using d=d+1, x=x+1, and n=n-1.
And this crumb is saying that we can move not in increments of 1, but in increments where the last d we add to f has the same "left over" as (f-2)%8.
The one part I'm most confused about is the division by 5:
I can see that we can test for small divisors, such as 2, 3, or 5, but I don't get why we do that specifically.
Comments on my thinking guys?
Also, I just want to add, I'm very open to helping create the single html page. I have all the pieces ready to go, and I'm totally down to collaborate!!
I'd like to share all my code, I'm just concerned about linking it back to my personal identity somehow.
Also, I stopped posting JS snippets because I thought I was the only one using JS!
Well, I was thinking about using a git provider, to share the code, because i'm updating it constantly, and that would be the easiest way to share and collab.
I could host the git server on:
1) my home server - linked to my ip
2) my amazon server - linked to my cc
3) use another cloud provider - trying to find a secure one that doesn't track IP (also, concerned about keys now but that's a separate issue)
Any suggestions?
Perhaps he simply chose 5 "arbitrarily" because it was easy to see.
So I have another question, is x or n larger? Is it possible to tell?
How do we go about aiming for this n<HERE<x+n?
The second I have something small thats worth sharing I will. Otherwise, I'll upload my work to dropbox or something - thanks.
Thanks, I'm going to head to bed to think about this more, and hopefully dream of triangles.