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Topic: Odd victory condition (Read 273 times) 

rmsgrey
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Odd victory condition
« on: Feb 23^{rd}, 2018, 10:04am » 
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(shamelessly appropriated from BBC Radio 4's daily puzzle a bit over a week ago): Two people are playing a takeaway game with a twist: On each player's turn, they take 1, 2 or 3 tokens from the pile, and play continues until all tokens are taken. So far, so standard, but the winner is determined not by who takes the last turn, but by how many tokens each ends up with. The player with an odd number of tokens wins. The game starts with 23 tokens in the pile (and none in either player's possession). Assuming perfect play, who wins, and what's their strategy?


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Grimbal
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Re: Odd victory condition
« Reply #1 on: Feb 23^{rd}, 2018, 1:07pm » 
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Interesting! It looks like a symmetrical poset game, but it is more tricky than that. If you consider A and B to have symmetrical roles where each wants an odd sum, then their goals are not necessarily opposed in the analysis, i.e. if you start with an even number, they both can win. If you consider that A tries to get an odd sum and B tries to prevent that to happen, then their goals are opposed but their roles are not symmetrical. B does not change the parity of A's sum. So even though the game is actually symmetrical, you cannot evaluate positions in terms of "from this position the next player wins". Or not easily.

« Last Edit: Feb 23^{rd}, 2018, 1:15pm by Grimbal » 
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towr
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Re: Odd victory condition
« Reply #2 on: Feb 24^{th}, 2018, 4:21am » 
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hidden:  I think you can just split the odd and even cases to build the win/lose table If A has an odd number of tokens, and there are N tokens remaining, then B must have an odd number of tokens iff N is odd. (Given we started with 23.) Similarly if A has an even number of tokens, B must have an even number iff N is odd So look in the appropriate column to see if there's a move where (s)he's in the losing position. even odd 0 Lose Win 1 Win Lose 2 Win Win 3 Win Win 4 Win Lose 5 Lose Win 6 Win Win 7 Win Win 8 Lose Win 9 Win Lose 10 Win Win 11 Win Win 12 Win Lose 13 Lose Win 14 Win Win 15 Win Win 16 Lose Win 17 Win Lose 18 Win Win 19 Win Win 20 Win Lose 21 Lose Win 22 Win Win 23 Win Win 

« Last Edit: Feb 24^{th}, 2018, 4:24am by towr » 
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rmsgrey
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Re: Odd victory condition
« Reply #3 on: Feb 24^{th}, 2018, 7:46am » 
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on Feb 23^{rd}, 2018, 1:07pm, Grimbal wrote:if you start with an even number, they both can win. 
 There are a couple of options for dealing with this. One is to say that you can, indeed, play for a collaborative victory (which is a much less interesting game to analyse since the only plays that matter are the ones that could leave just 1 coin remaining). Another is to say that you can only start with an odd number in the pile. A third is to say that you can start with any number in the pile, but, if it's even, one player starts with an extra token (or, equivalently, one player wins on even; the other on odd). Follow up question: What if, rather than 13, you take 1n coins (where n is fixed for an entire game). How does that change things?


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towr
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Re: Odd victory condition
« Reply #4 on: Feb 24^{th}, 2018, 12:23pm » 
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on Feb 24^{th}, 2018, 7:46am, rmsgrey wrote:Follow up question: What if, rather than 13, you take 1n coins (where n is fixed for an entire game). How does that change things? 
 I don't think that qualitatively changes the solution, though the outcome may be different. You'll just have to look at more rows in the table to see if there's a winning move. And if you change the goal to 1 mod 3, or 1 mod M, you just add more columns, but it also doesn't really change anything, I think. Multiple players would make things more complicated.


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