Comments on: The abc game http://bit-player.org/2012/the-abc-game An amateur's outlook on computation and mathematics Tue, 07 May 2013 12:54:10 +0000 hourly 1 http://wordpress.org/?v=3.4.2 By: Andrew Au http://bit-player.org/2012/the-abc-game/comment-page-1#comment-5279 Andrew Au Wed, 19 Sep 2012 00:39:03 +0000 http://bit-player.org/?p=1270#comment-5279 The explanation of the problem is awesome! The explanation of the problem is awesome!

]]> By: Eli http://bit-player.org/2012/the-abc-game/comment-page-1#comment-5219 Eli Wed, 12 Sep 2012 01:35:45 +0000 http://bit-player.org/?p=1270#comment-5219 THANKS for writing the clearest, most enjoyable explanation of the abc problem I've read. With the news of the rumored proof there are so many math-geeks now interested on the subject but I'm afraid most of us got completely disillusioned at the ridiculous level of jargon involved in most other explanations. THANKS for writing the clearest, most enjoyable explanation of the abc problem I’ve read. With the news of the rumored proof there are so many math-geeks now interested on the subject but I’m afraid most of us got completely disillusioned at the ridiculous level of jargon involved in most other explanations.

]]> By: brian http://bit-player.org/2012/the-abc-game/comment-page-1#comment-5217 brian Tue, 11 Sep 2012 18:36:01 +0000 http://bit-player.org/?p=1270#comment-5217 @Craig: The ABC@Home project offers plenty of data for curve-fitting. The file records just a and c for each hit, so you need to compute b and R and the ratio of logs. I'm curious too, and I'll try to do a simple histogram for a subset of the data in a day or two. @Craig: The ABC@Home project offers plenty of data for curve-fitting. The file records just a and c for each hit, so you need to compute b and R and the ratio of logs. I’m curious too, and I’ll try to do a simple histogram for a subset of the data in a day or two.

]]> By: Jim Lyon http://bit-player.org/2012/the-abc-game/comment-page-1#comment-5216 Jim Lyon Tue, 11 Sep 2012 18:27:10 +0000 http://bit-player.org/?p=1270#comment-5216 If N(h) is the number of hits for value h, it's not that strange that N(h) is finite when h > 1, but infinite when h = 1. Now it would be mighty strange if the limit of N(h) as h approaches 1 were finite. Is it? If N(h) is the number of hits for value h, it’s not that strange that N(h) is finite when h > 1, but infinite when h = 1.

Now it would be mighty strange if the limit of N(h) as h approaches 1 were finite.

Is it?

]]> By: brian http://bit-player.org/2012/the-abc-game/comment-page-1#comment-5215 brian Tue, 11 Sep 2012 17:53:26 +0000 http://bit-player.org/?p=1270#comment-5215 <em>R</em> ≠ <em>c</em>. Thank you all. Amazing how much computation you can avoid with a little thinking! R ≠ c. Thank you all. Amazing how much computation you can avoid with a little thinking!

]]> By: Craig http://bit-player.org/2012/the-abc-game/comment-page-1#comment-5214 Craig Tue, 11 Sep 2012 17:07:31 +0000 http://bit-player.org/?p=1270#comment-5214 Is anything known, even empirically, about the asymptotic behavior of the number of solutions as h approaches 1? (Aside from the fact that it goes to infinity.) Like, does it go as 1/(h-1)? 1/(h-1)^2? exp[1/(h-1)]? Is anything known, even empirically, about the asymptotic behavior of the number of solutions as h approaches 1? (Aside from the fact that it goes to infinity.) Like, does it go as 1/(h-1)? 1/(h-1)^2? exp[1/(h-1)]?

]]> By: Sean Palmer http://bit-player.org/2012/the-abc-game/comment-page-1#comment-5213 Sean Palmer Tue, 11 Sep 2012 15:29:59 +0000 http://bit-player.org/?p=1270#comment-5213 From jgeralnik on HN https://news.ycombinator.com/item?id=4505264 "I may have misunderstood something, but I believe it is simple to prove that R cannot equal c. Because c is relatively prime to both a and b, there exists some prime p which b is divisible by and c is not. Because b is divisible by p, R=rad(a,b,c) is also divisible by p. Since R is divisible by p and c is not, R!=c." From jgeralnik on HN https://news.ycombinator.com/item?id=4505264 “I may have misunderstood something, but I believe it is simple to prove that R cannot equal c.

Because c is relatively prime to both a and b, there exists some prime p which b is divisible by and c is not. Because b is divisible by p, R=rad(a,b,c) is also divisible by p. Since R is divisible by p and c is not, R!=c.”

]]> By: lmm http://bit-player.org/2012/the-abc-game/comment-page-1#comment-5211 lmm Tue, 11 Sep 2012 14:24:05 +0000 http://bit-player.org/?p=1270#comment-5211 Surely R=c is trivially impossible. R = c => R | c => R(a) | c => a and c have a common factor. Surely R=c is trivially impossible. R = c => R | c => R(a) | c => a and c have a common factor.

]]> By: Carlos Fuentes http://bit-player.org/2012/the-abc-game/comment-page-1#comment-5210 Carlos Fuentes Tue, 11 Sep 2012 14:22:48 +0000 http://bit-player.org/?p=1270#comment-5210 Great article. I was going to point out the bit about c=R but I see I've been beaten to it. Still great way to make the topic accessible. Thank you. Great article.

I was going to point out the bit about c=R but I see I’ve been beaten to it. Still great way to make the topic accessible. Thank you.

]]> By: Matt http://bit-player.org/2012/the-abc-game/comment-page-1#comment-5209 Matt Tue, 11 Sep 2012 13:00:07 +0000 http://bit-player.org/?p=1270#comment-5209 Assuming c=R, log c=log R (log c)/(log R)=1. Because this is never the case, c and R are never equal. Other than this small oversight, very nice article :) Assuming c=R,
log c=log R
(log c)/(log R)=1.
Because this is never the case, c and R are never equal.

Other than this small oversight, very nice article :)

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