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Another backgammon puzzle: more serious effort
Posted By: Jason Lee In Response To: Another backgammon puzzle: more serious effort (David Levy)
Date: Friday, 1 March 2013, at 10:10 p.m.
Let G bet the set of all possible backgammon games. Assume it is finite--this will lead to a contradiction, negating the assumption.
G contains games G1...Gn where n is the (finite) number of games in the set. For each Gx, define a function f(Gx) = the number of times the opening position occurs in game Gx. [Aside: we know that the opening position can recur, see Kershaw's Backgammon Funfair]. Over the set G, f(Gx) can take on at most n different values, so there is some number, say N, between 1 and n+1 that did not occur.
It is easy to construct a backgammon game where the opening position recurs N times. That game could not have been in set G, the set of all backgammon games, a contradiction. Therefore the assumption that G is finite fails.
If Jason disagrees, I'm cheating with my set theory somehow...
I think your proof is correct but overkill in a sense. When I teach how to write proofs, I always encourage students to avoid, where possible, a proof by contradiction. It's a perfectly valid method of proof, but it can obscure the real reason why something is true.
Here, you have correctly reached the contradiction, but it hides the real reason, which you articulate in the course of your proof. The heart of your proof is here:
It is easy to construct a backgammon game where the opening position recurs N times.
That's all you really need. By being able to construct a game where the opening position repeats exactly N times, you have demonstrated the existence of an infinite set of games.
JLee
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