of interest to educational math geeks

[jrtom]

(Yes, I mean you, hypgnosis--as if there were any doubt. And probably amnesiadust, too.)

+plus magazine

Plus is an internet magazine published five times a year which aims to introduce readers to the beauty and the practical applications of mathematics. Whether you want to know how to build a sundial, how to keep your messages safe or what shape the universe is, it's all here.

(No, I'm not getting paid by these folks, I just think it looks interesting.)

Comments

Re: fascinating, Captain

[jrtom.livejournal.com]

Actually, same Hamilton (http://scienceworld.wolfram.com/biography/HamiltonWilliamRowan.html) (although it's not obvious from that bio, oddly enough, although I got there from the Wolfram article on Hamiltonian circuits (http://mathworld.wolfram.com/HamiltonianCircuit.html)).

Basically, a Hamiltonian path is a graph traversal that visits each vertex exactly once; a Hamiltonian circuit is a H. path that ends up at the vertex that it started at. (Not to be confused with an Eulerian circuit, which passes over each edge exactly once.) The problem of finding a Hamiltonian circuit in a graph is more commonly known as the Travelling Salesman Problem, one of the more well-known NP-hard graph theory problems.

minor oops

[jrtom.livejournal.com]

(I should have said, for H. path: "visits each vertex exactly once, and passes over each edge no more than once".)

Re: minor oops

[(Anonymous)]

Dude. Is it possible to have a path that visits each vertex once but visits some edge more than once?

Re: minor oops

[jrtom.livejournal.com]

I emphasized the wrong aspect, apparently. What I meant to emphasize is that a Hamiltonian path need not visit every edge (unlike an Euler tour, in which each edge must be traversed). You are, of course, correct in suggesting that you can't have a path that visits each vertex once but some edge > once.

"I've got egg on my face from both sides now..."