Dumb answers to dumb questions

[sniffnoy]

Why do I even have that recursion in the first place? It's to account for if beta gives me an output that's of the same value, but structurally different... but beta can't do that! Except in the trivial case. Which raises the big question: Why the hell am I calling it in the trivial case? Skip that one! Get on with the searching actual cases! And eliminate the dumb recursion to handle an unnecessary case!

...this will probably simplify things a lot, but it won't be enough to fix the function fully. Oh well. It's a good start.

Comments

[sniffnoy.livejournal.com]

Well, this entry is pretty obviously just muttering to myself - dealing with dumb program internals - but as for what I'm actually writing... let me pull up the entry where I described the problem... here it is.

Stuff that's happened since then: On the upper bound thing, Josh has pushed things down to 2.7, and maybe 2.65? His lemmas used to not be good enough for the latter. He's also shown, if we define F(n)=max over k=1 to n of f(k), that F(n)-3log₃n goes to infinity. Which is pretty far from actually showing they're not asymptotic, but that's what we have. He also has some idea about putting upper bounds on the average of f(k) over k=1 to n, but he hasn't done much with that and actually I want to check that out too once I get this thing working.

As for me, after much tedious hand computation, I pushed things up from a≤13 to a≤16 ( :P ), but finally realized how to automate my computations. My method is ridiculously inefficient, but I do want to see how far I can push it. What the program actually computes is, well, I'll define g_r(k) to be the rth-highest number you can make using exactly k ones using addition and multiplication. The point of this program is to compute g_r(k), except, it's not just to compute it for specific r and k, because that wouldn't really prove anything. Rather, it takes as input, r, and a congruence class mod 3 for k, and (when it finally works) will return a number N and an affine function h, such that for k≥N and of the given congruence class, g_r(k)=h(g_1(k)). (It's been pointed out to me, that if I'm so sure the function is affine, why not just determine it at two points and be done with it? My answer is: A. the numbers involved would be really large, g_1(k) is about 3^(k/3), and B. there's the problem of that lower bound, I'd need to plug in k≥N - without doing these computations, I don't see any good way to get an a priori lower bound; I think best I thought of was, like, N=6r or something, which is a huge overestimate, and makes the numbers involved that much larger; admittedly, I haven't really thought about that problem much at all, because I'm pretty sure writing this thing (which just does the same things I was doing by hand) is the right way to go...)