Comments on: Gruenberger’s prime path

By: Craig

Craig — Fri, 19 Feb 2010 03:16:07 +0000

Sorry, that’s vector. C++’s STL includes special-case code that represents that type using a packed bit array (a clever trick).

By: Craig

Craig — Fri, 19 Feb 2010 01:34:56 +0000

Brian, my sieve is nothing special. It’s about a dozen lines of C++. The only things worth noting are that I use a vector, for which the STL cleverly packs the bits, and that I don’t store the primality of even numbers (particularly apropos here). I can send you my code if you want.

I know there are fancier sieves, but I’m not enough of a number theorist to drop in anything more clever than what I’ve got. It seems like you’d want some sort of windowed approach, in which you deduce the primality of a billion numbers at a time as you head out on your walk.

For the record, no return to (0,0) for N <= 4e9. The walk ends at 115220,257903, and made 94983231 left turns and 94978580 turns.

By: brian

brian — Thu, 18 Feb 2010 18:47:02 +0000

@Craig: 2 x 10^9 is exactly twice as far as I got. (I’m curious about your seive. Indeed, I’m curious in general about the best practical primality test in a problem like this. When I tested a couple of simple-to-code alternatives, the results were not quite what I expected.)

@Nate: Thanks for asking about the artwork. The short answer is that it’s all home-baked. The longer answer is that I should write a separate post on this subject, since a couple of people have asked recently. I’ll try to do that in the next few days.

By: Stephan Mertens

Stephan Mertens — Thu, 18 Feb 2010 17:36:57 +0000

My guess is that the RMS distance of the prime based, non-reversing random walk is proportional to \sqrt{n \log n}.

By: Nate

Nate — Thu, 18 Feb 2010 16:14:29 +0000

I like the visual style of your graphics in this post (and others). Is there a particular graphics package that you use? Thanks. Nate.

By: Craig

Craig — Thu, 18 Feb 2010 15:12:45 +0000

Cute question. For the record, I’ve verified that the path doesn’t return to (0,0) for N up to 2,000,000,000 (which is far enough for now; my sieve will start to fail after 31 bits). Upon reaching N=2,000,000,000, we find that the path has taken 49,112,581 left turns and 49,109,703 right turns, and finds itself at position (113,156, 248,033). I want to say “it doesn’t look like it’s coming back home”, but that’s the usual naivete with respect to random walks.

I’d like to compute which location in the plane is visited most often for N up to 2,000,000,000, but that’s a bit more of a hassle (it demands much more memory).

By: Cody

Cody — Wed, 17 Feb 2010 16:20:57 +0000

An interesting problem to lift my spirits this morning. Thanks!

By: Russ Abbott

Russ Abbott — Wed, 17 Feb 2010 04:04:14 +0000

I joined the CS Department at Cal State Northridge while Fred was there. One of the things that stands out in my mind was his advice to beginning programming students. When writing a program, first make it work; then make it pretty. Elegance and abstraction weren’t among his primary considerations when writing software.

Those were the days of the key punch and batch processing. Fred thought it was good discipline to have to wait an hour or more for each compile/execute run. That way one would look more carefully at one’s code before submitting it.

The good old days!