Re: [math-fun] Games of Ramsey and van der Waerden
Gareth wrote: << On Monday 30 October 2006 00:57, Daniel Asimov wrote:
A square is defined here as four *distinct* points of GxG -- P,Q,R,S -- such that the segments PQ, QR, RS, ST are of equal length, and the angles PQR, QRS, RSP, SQP are right. A square is determined by its set of vertices.
Oh -- and the metric used on GxG is just the cartesian product of that on G with itself, where G is a regular octagonal curve.
OK. Allow me to suggest that that's less natural than the different definition I proposed, because it seems freaky to consider {a1,c1,e1,g1} a square. But it's your game, so you get to choose the rules :-).
If you mean (0,0),(0,2),(0,4),(0,6), that's *not* a square -- all angles are straight, not right. A regular octagonal curve has the same metric as a circle (with 8 equally spaced points on it). The 64 vertices derive their metric from the metric on S^1 x S^1 = R^2 / Z^2 (as metric spaces). The distance between two points P,Q of R^2 / Z^2 is the length of the shortest line segment in R^2 whose endpoints project to P,Q. PQR is a right angle when the shortest segments PQ,QR make a right angle, or equivalently when they have lifts PQ^, QR^ in perpendicular directions R^2. Sorry to have been unclear. (Or am I missing something?) --Dan
Oh -- and the metric used on GxG is just the cartesian product of that on G with itself, where G is a regular octagonal curve.
OK. Allow me to suggest that that's less natural than the different definition I proposed, because it seems freaky to consider {a1,c1,e1,g1} a square. But it's your game, so you get to choose the rules :-).
If you mean (0,0),(0,2),(0,4),(0,6),
(Yup.)
that's *not* a square -- all angles are straight, not right. A regular octagonal curve has the same metric as a circle (with 8 equally spaced points on it).
Ohhh, I see. I entirely misunderstood. I thought you meant the metric induced by the obvious embedding of G in R^2. With the clarification you've now provided, I agree that that's a fine definition, not least because it's equivalent to the one I prefer :-). -- g
participants (2)
-
Daniel Asimov -
Gareth McCaughan