You could play this using some of Jim's candy bars instead, if there are any left. (But you only get to eat them if you beat Richard's score ...) WFL On 11/13/14, Richard Guy <rkg@ucalgary.ca> wrote:
Dear all, It's not exactly a game, but funsters may be interested in (or already know about) the following: Start with a pile (tower) of n building blocks (cubes). Dismantle the tower by splitting into two towers of heights a and n-a, giving yourself a score of a(n-a). Continue with one of the resulting towers, and repeat until there are n towers of height 1. How to maximize your score ? E.g., 7 --> 4,3 --> 1,3,3 --> 1,1,2,3 --> 1,1,2,2,1 --> 1,1,2,1,1,1 --> 1,1,1,1,1,1,1 gives a score of 12+3+2+2+1+1 = 21. Can you do better? R.
________________________________________ From: math-fun [math-fun-bounces@mailman.xmission.com] on behalf of Dan Asimov [dasimov@earthlink.net] Sent: Thursday, November 13, 2014 1:51 PM To: math-fun Subject: Re: [math-fun] Games of no strategy
Although I doubt this is what Jim had in mind . . .
. . . It is well known that for countable games where each of two players A, B alternates in creating a binary string of order-type omega (== {0, 1, 2, 3, ...}) with a subset W(A) in {0,1}^omega and its complement W(B) determining who won according as which of W(A) or W(B) the string ended up in . . .
. . . then the statement:
"There is always a strategy for A or B regardless of the dichomy W(A)+ W(B)"
(aka "the Axiom of Determinacy") is not consistent with the Axiom of Choice.
So according to AC, there is some W(A) + W(B) for which there exists no strategy for either player.
I suspect this means that the outcome does not depend on what either player does. But I'm not absolutely sure of that.
--Dan
On Nov 13, 2014, at 8:00 AM, James Propp <jamespropp@gmail.com> wrote:
What are fun examples of combinatorial games that (like Conway and Paterson's game of Brussels Sprouts) appear to be games of strategy but whose outcome doesn't depend on what either player does? . . .
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