Chronicles of Harry

Things happening inbetween my returning to NJ and my going up to Boston this Friday.

Kongai: Well, I *haven't* started playing Kongai again. I basically stopped playing sometime during winter quarter, and never really majorly started again. Then during Spring quarter I started having more free-time-with-nothing-else-to-do but instead of getting back to Kongai I decided to reread ASoIaF[0]. The funny thing is, I've still been playing the challenges on Kongregate and getting the cards. The only card I have 0 of is Deadly Poison, and it's been that way for several months. I'm sure if I were actually playing, even with the low drop rate, I would have had it long ago. I have all the characters, and I keep a file listing how many I have of each item (since in order to be able to make any deck, you need 3 of each item[3]); I've got 5 3-ofs, 11 2-ofs, 9 1-ofs, and Deadly Poison as my sole 0-of. And, of course, now I don't have much time to start playing again before I'll be up at PROMYS with no time to.

I wonder if the metagame's changed since I stopped? The reason I decided to stop using Marquis was because so many people were using guys that were really good against him, but if the metagame's shifted maybe my old deck would be good. Otherwise, well, with not having played in so long, it's not like relearning my Anex deck will be so much harder than just relearning the game...

Addition and multiplication: See this entry on Josh's blog. Read the comments too.

Josh has indeed come up with a better upper bound, namely, (59/20)log₂n. Where the 59/20 is not the best possible his proof allows, but an approximation for an irrational he didn't bother to solve for exactly[4]. He might have improved it since then? He sent me a copy of the proof for 59/20, I read through it but it had a lot of errors I had to correct for as I read (not real errors, as it all worked, though I didn't check everything - but stuff would be written down wrong and you'd have to figure out what he actually meant), he says he has a cleaned up version but he hasn't sent it to me yet. From numerical data Josh thinks you should still be able to do much better, closer to 2log₂n, but if he has improved it I think he still has it above 5/2.

He also sent an email to Richard Guy asking about the current state of the problem - apparently f(p)=f(p-1)+1 is known to be false, counterexample presumably found by massive computation, namely 353942783.

What do I have to do with any of this? Well there's also the problem of proving that f(2^a*3^b)=2a+3b (a,b not both zero). I have zero idea how to, say, tackle the case of general powers of 2, but I have managed to prove it true for low a, namely, a≤13. The proof for a≤10 is very straightforward, for a≤13 somewhat less so; but trying to extend this proof method much further looks quite icky. And certainly hard to generalize. I haven't really been spending a lot of time on this like Josh has - his upper bound improvements are way harder than the easy stuff I've done! It didn't even take me 2 days to figure out a≤13. I would expect it's not new, though Guy doesn't mention any such result in his email, in relation to (2^a)(3^b) he provides only numerical data. So, hey, it could be new just because this isn't something anyone's bothered to think about before. I have no idea.

-Harry

[0]Man, I'd really forgotten so much of what happened during book 4. May have something to do with the fact that when I got into the series, it was the first 3 that were out, so I read them all at once, and 4 later. Or it may have to do with the fact that 4 is kind of boring by comparison. :P Here's hoping book 5 comes out this year.
[3]Well, 5 if you play 5-card, but A. who cares about 5-card and B. It's not even possible to have 5 of each currently and C. Even if it were possible, that would still take really, really long.
[4]For some reason I keep writing "59/60" instead of "59/20"... initially he wrote 23/8, but apparently that was in error. Also, technically the proof he sent me is for 1+(59/20)log₂n, but A. that's asymptotically the same and B. Josh tells me you can get rid of the 1 (for n>1), it's just annoying to prove.