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July 1st, 2017, 02:40 AM   #1
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The game of pebbles

Hi,

We consider the game at 2, which consists of a set of N stones, from which one can remove a pebble, or c> 2 pebble, the winner being the one who does not take the last stone. Is there a winning strategy?

Cordially.
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July 1st, 2017, 09:47 AM   #2
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Math Focus: सामान्य गणित
What are the other rules of the game?
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July 1st, 2017, 01:15 PM   #3
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The game of pebbles is a two-player game. The game starts with N stones, and a number c>2 is fixed. On a player's turn, he or she must remove either 1 or c stones. The player who does not take the last stone is the winner.

Is there a winning strategy?
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July 1st, 2017, 01:32 PM   #4
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If you take c stones you lose, so the only strategy is to take 1 and hope.
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July 1st, 2017, 02:10 PM   #5
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Suppose N = 6 ,c = 3

The player who goes first cannot lose

If he takes 3 the other player losses

If he takes 1 and the other player takes 1 then he takes 3 , first player wins again

If he takes 1 and the other player takes 3 then he takes 1 , first player wins again

Suppose N = 5 , c = 3

The player who goes first cannot win

If he takes 1 the other player takes 3

If he takes 3 the other player takes 1

Looks like it depends on the choices of N and c so there is no winning strategy for all N and all c


Last edited by agentredlum; July 1st, 2017 at 02:22 PM.
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July 1st, 2017, 05:09 PM   #6
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This would be a better way to illustrate:
13 pebbles; players take 1,2 or 3

last to pick loses : so winner's strategy is to leave 5 pebbles

whoever goes first picks 2: then sure to win
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July 2nd, 2017, 01:19 AM   #7
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There are a loosing position,

the solution here :

But another difficult, the case, when the choice is 1,2 or 4 stones.
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