Chronicles of Harry

OK. My initial writeup of this is *finally* done. I hope it's clear enough. Well, I'll show it to Josh before anything else. Interesting thing: When compiling it, TeX gave me several "Underfull \vbox" messages - but not all the badnesses were 10000! Take a look:

Underfull \vbox (badness 10000) has occurred while \output is active [6]
Underfull \vbox (badness 10000) has occurred while \output is active [7]
Underfull \vbox (badness 2213) has occurred while \output is active [8]
Underfull \vbox (badness 5316) has occurred while \output is active [9]
Underfull \vbox (badness 1546) has occurred while \output is active [10]
Underfull \vbox (badness 10000) has occurred while \output is active [11]

Is 10000 just what it caps badness at, or something?

(Oh, and that modification to the code I made? Made absolutely no difference in the output at all.)

-Harry

Still writing up my result. First I wrote up the easy parts; then I spent a bunch of time rearranging the easy parts to make more sense; then I spent a bunch of time editing that so that it flows better. Now I'm writing the hard parts. I'd say there are three hard parts, of which I am almost done with the second (describing how I've computed g_r(k) without the abstraction).

It occurs to me that I really haven't checked the articles in Guy's email to Josh about this - since Guy didn't mention anything that seemed relevant, I assumed there wasn't, but I should really check... but now I've changed my point of view on what I've actually proved, maybe there was. It would be pretty annoying if someone else proved the same thing, but never used it to show that f(2^18 3^m)=36+3m, and therefore wasn't mentioned... I should also check the references on Sloane, it has some sequences related to f in it. (And you can be sure once this is done, I'm going to submit this sequence of numbers. By which I mean, numbers n such that f(m)<f(n)⇒m<n, which are given by the two forms I've listed earlier. Because that sequence is not in there. Which I guess does suggest nobody has done this before. :) )

Lie algebras is cool, alpha[0] topology is boring so far, teaching 105 is hard; but you probably guessed all that.

Regarding topology: Yeah, I have to take alpha topology. When it was QR signup time, I figured of course the algebra, I'll try my hand at the topology, and I won't even bother with the analysis. Well, algebra was no trouble. Topology... it's Saturday morning, I sit down with the first half of the test, realize I don't know half this stuff, and decide "screw it, I'm going back to sleep".

Regarding 105: My second lesson was a real disaster as I didn't prepare properly, but that hasn't happened since. My boardwork still really needs work. You know what else I'm terrible at? All the non-teaching parts of the job, or that require explaining things other than mathematics (course policies, etc). To them it's crucial, but to me it's quite secondary and I tend to forget about it when not explicitly reminded. Now on Fridays they have me down in the MathLab helping kids individually, and there, of course, I feel much more at home. None of my students has yet to show up to my office hours, though, and I'm inclined to think that's a bad thing.

...enough math for now.

So Aubrey[3] left for Australia for some solar car competition this Monday, leading to a game night on Sunday. Played Shadows over Camelot. If you haven't heard of it, it's a co-op game, with the possibility that one player is a traitor. The way that's determined is, there are 8 loyalty cards, and each player gets one; one of the loyalty cards reads "traitor". So with the maximum of 7 people, which we had, there's probably a traitor. Well, guess what. We had no traitor, and we had 7 people, and we still only barely pulled off a win... I guess I see why I'd seen people saying on BGG[4] that they often play without one. I guess co-op games have to be hard by nature, huh? In the same way that one-player games have to be hard. It's not really something I'm used to outside of video games. Oh, and I should note, we also didn't realize that the rulebook said that you shouldn't discuss the specific contents of cards in your hand/secret cards that you'd seen, so we had another advantage that way. I think that's a bit silly; what's the point of limiting player communication? I suppose the point is to actually make it a co-op game, with different roles. Basically because people can communicate so much, in a board game, that's pretty hard, and I'm inclined to think it's not really something very well-suited to the boardgame format at all. Video games can accomplish this pretty easily, by contrast. In any case, I would think better than telling the players not to communicate is to, you know, give them some incentive not to, such as by having a traitor... or multiple, if one is not enough... (mind you, I have no objection to rules against people revealing their stuff to prove what they say, just to rules limiting what people can say.)

Still on the whole Truth House is kind of boring so far compared to Tufts House, i.e. people don't seem to be around as much. I'm guessing the fact that the lounge isn't the center of the house like in Tufts House contributes to that... we also have very few videogames in the lounge, which is kind of sensible, as there have been some break-ins... but right now we just have the house Wii (which is bolted to this frame which is... yeah) with pretty much just Brawl for it, and my Dreamcast (as who's going to steal a Dreamcast?). Still, I've seen other games being played, just nobody is willing to leave them down there.

So I found the third game store, the one I saw on the way in; it too isn't too far away. And it's pretty cool - it has old used video games for lots of old systems; there were Atari 2600 games there. No Dreamcast though. :-/ It's called "Get Your Game On". And, get this, they're hosting a Zendikar prerelease. Maybe I should go? Maybe this is a time for me to get back into Magic? I'll admit, though, that I haven't actually computed how much money I'll have - this first month I've been really just living off of what I have stashed away without too much thought; soon I'll have to actually start figuring out just how much disposable income I actually have. But hell, maybe it's worth going to this anyway. Of course, knowing me, there's a good chance the correct answer is "No, you shouldn't go buying cards, because that would require finding people to play against, which would probably largely mean going to organized events, and you're too lazy for that."

I swear, there was somehing else I wanted to mention, the same thing I forgot about last time - I know this because I remembered it briefly; but I've forgotten it again.

I haven't started playing Kongai again. The internet connection here is really pretty inconsistent. I guess Kyle[5] has appointed a network steward now, no? Not that he'd likely know immediately just what's going on, as I think he's a first-year here too, but oh well. It's certainly the sort of thing he should be trying to fix...

Of course, why do I think I'll have so much time? I'm teaching 105, and largely I've had time so far because only now have I had to begin actually grading things, not to mention my own homework which has basically just begun...

-Harry

[0]Alpha sequences: At Michigan there are basic sequences in algebra, topology, analysis, for grad students. Unlike at Chicago, these really are pretty basic, and certainly not mandatory... but you do have to take them if you don't pass the appropriate qualifying review. Well, or you could just take the QR again later, but if you didn't pass the QR, you probably need to take them... or at least when you do as badly as I did... see above.
[3]Mentioned him briefly before - another math PhD student, who lived in Truth House, and plays a lot of boardgames, and does parkour, and is generally pretty awesome, but will not be here this semester.
[4]I really rarely visit BGG anymore...
[5]The work manager, responsible for assigning people's jobs. This is the same Kyle who actually plays in Brawl tournaments. There is another Kyle, who is, IINM, the network steward I was just speaking of.

...barring any errors in the program, anyway. And some of its output was a bit suspicious. Regardless, I'm pretty certain of it by this point, and failing all else I can always check it all by hand, which would be a bunch of case checking, but not nearly as much as it would have been previously. I really doubt any mistakes it may have made are ones that would affect the final result.

So, here we go: Let f be the function Josh and I have been working on; let k∈N. Then if k is not of either of the following forms:
A.) 2^m3ⁿ for m≤10
B.) 2^a(2^b3ⁿ+1)3^m for a+b≤2
Then the following holds:
1.) If ⌈3log₃k+(5-6log₃2)⌉≡0 (mod 3), then f(k) is at least this quantity.
2.) If ⌈3log₃k+3(1-log₃2)⌉≡0 or 1 (mod 3), then f(k) is at least this quantity.
3.) f(k)≥3log₃k+1.

(Note that 1, which gives the largest bound, is only a tiny improvement over 3, with a difference of only about 0.21 - and of course this whole exercise is such a tiny improvement on the standard bound in the first place! - but it can matter. For instance, say we want to prove f(2¹⁸3ⁿ)=36+3n; the last inequality isn't going to cut it. Note that as far as 2^a3^b goes, these bounds are only good enough for a≤18, not 19; 19 was a mistake.)

EDIT 9/12: There was a typo in the lists below; Instead of "4*3^n+1" I wrote "4*3^n+4".

Furthermore, not all the exceptions listed above necessarily are exceptions; the completely reduced lists of numbers for which 1, 2, and 3 fail (where the condition is in fact met, obviously) are:
1.) 16*3^m, 128*3^m, 1024*3^m, (4*3^n+4)*3^m for n≥1, (4*3^n+2)*3^m for n≥1, (4*3^n+1)*3^m.
2.) 2*3^m, 8*3^m, 16*3^m, 64*3^m, 128*3^m, 512*3^m, 1024*3^m, (2*3^n+2)*3^m for n≥1, 40*3^m, (2*3^n+1)*3^m for n≥1, 14*3^m, 38*3^m, 5*3^m, 13*3^m.
3.) 2^n*3^m for n≤9 (except when k=1), (3^n+1)*3^m, 20*3^m, 40*3^m, 7*3^m, 19*3^m, 14*3^m, 38*3^m, 5*3^m, 13*3^m.

And that's all, folks!

...OK, OK, at some point I'm going to have to start writing up a proof. But really! There it is!

Yaaaaaay!

-Harry

...did I say my proof worked for a≤19? I meant 18, apparently. Did I say 3log₃(3/2)? I meant 3log₃(4/3) (minus the exceptions - though the better bound is still often true). Oops. (Still have yet to prove *this*, mind you, but I'm pretty certain.)

-Harry

So I've been thinking about what I've doing as being working on the 2^a3^b problem, but in fact powers of 2 don't seem to enter much into it. I think I should be able to take this stuff (once I've properly proved the regularities in g_r(k), I mostly know how but I haven't done all of it - and actually I still haven't checked that they're good enough) and prove something of the form "For all numbers n not of the following forms, [list of forms], f(n)≥[something bigger than 3log_3(n)]". Depends how good the numbers are, though.

-Harry

And, according to the program, for 3 not dividing k, k≥16, g₅(k)=... 7/9g(k) !

OK, I haven't technically verified that absolutely everything is working properly, but, it is giving me actual results for r>2, and that's new. And I have checked a lot. And all its low results match up with my hand computations.

Man, each of these computations took me, like, hours before. Computers are so great. Now to start making a table of results! Later I'll see what these actually allow me to prove.

(Basically, so far, there's never been a constant term; and what I want is for that to remain the case, and for the size of the linear term to get small faster than the number of 2s in its numerator gets large. I don't think the second part should be a problem, but I suspect that eventually a constant term will appear, and then I won't be able to make use of it (or at least not in the same way).)

-Harry

...yeah, somehow I don't think I'm going to ever write down most of the things I said I was going to. Perhaps if I'm bored (which I assuredly will be) I'll get around to some of it. (Today I had a headache and spent much of the day napping.) I can't even transcribe the T-shirt for you, because I don't have one. I forgot to take one at first, and later, whoops, order was screwed up, they're out of mediums. So, they'll be sending one here, to the house.

I guess now I finally have time to get back to work on the (2^a)(3^b) problem. By which I mean, getting that program to work so I can see how far I can push the method I've been using. Today's progress: Removing the "FIX" tags from things I've already fixed, and writing a summary in the comments of what the *current* problem is! Whee. I think I may want to do some reorganization...

-Harry

I know there was more stuff I intended to write, but I'm forgetting lots of it right now. So this is pretty random stuff, but I guess that's pretty usual.

I said I thought the program was pretty close to done, but I'm not so sure. A lot of the "debugging" I've done is not so much debugging but rather rewriting things so as to not be totally wrong. It's like, imagine I wrote a list length function, and defined it by len [] = 0, len (x:xs)=len xs - and not because I forgot to add one, but because I truly misthought things and concluded that was how you write length. That's the sort of errors I've been finding. Right now the program can't compute anything but the trivial r=1 case; earlier I made a fix that allowed it to compute r=2, 3|k, but then I made another fix and it broke that. :P At any rate I'm a lot closer than I was before, though whether or not that's actually "close" remains to be seen. Unfortunately I've not had much time to work on it each day.

Lucas pointed out to me that the axiom of infinity is over-specific: All you need is the existence of a Dedekind-infinite set. Then, take on it a non-surjective injection, pick an element that's not hit, and take the intersection of all subsets containing it and closed under this function, and you've got a perfectly good model of ω. And then, using axiom of replacement, you can show the existence of our ordinary ω. I wonder why set theorists do it that way? I guess it's a convenience thing - you can represent the whole numbers any way you want, but that's an especially convenient way. Much like Church numerals in lambda calculus.

Funny thing - I had been trying to prove the infinite version of Arrow's Theorem - in the infinite case it doesn't need to come from a dictator, just from an ultrafilter - since seeing it as an exercise somewhere, but failing horribly. Then Henry Cohn decided to do a combinatorics problem set that was just a 4-part proof of the finite version; parts 2 through 4 only applied to the finite case, but part 1 turned out to be just the hint I needed. Yay!

Didn't nap today; guess that's fixed.

Have not written anything about reunion. Perhaps will at some point.

-Harry

Well, initial coding is done. First test: r=1, k=0 mod 3. Simplest possible case, should hardly have to do any computation at all. Actual result: The program uses up a bunch of processor and doesn't return an answer. Yay. Now I'm going to have to learn how to debug Haskell.

-Harry

After much tedious computation, much of which accomplished little and the latest of which finally extended the result from a≤14 to a≤16, I think I finally see how to automate these computations. It's going to be a pain to program, though. I think I'm going to want to remember how to use programming languages other than C...

-Harry