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Timeline for A Nim game variant

Current License: CC BY-SA 3.0

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Apr 18, 2011 at 13:37 comment added Jean-Pierre Thank you for your comments. Quantumelixir said:"the first player will always win ". Are you sure? For example: How do you play and win if you are the first player and have to play the position (2,2,1)?
Apr 18, 2011 at 7:31 history edited quantumelixir CC BY-SA 3.0
gap in proof
Apr 18, 2011 at 4:30 history edited quantumelixir CC BY-SA 3.0
updated answer to reflect changed question
Apr 18, 2011 at 4:24 comment added quantumelixir Well, in that case too, the first player will always win but for a different reason. See the updated answer; hope this helps answer your question.
Apr 17, 2011 at 22:30 comment added Jean-Pierre In this problem, only one in the game, a player may pass his turn. There is only ONE pass available for BOTH players .Once one player has passed, no other player can pass. For example, if I give my opponent the position (1,1), if my opponent -uses the PASS, I let him with (1) and I win -plays (1), I use the PASS and I win Other example, if I give my opponent the position (2,2,1), if my opponent: -uses his pass, I let him with (2,2) and I win -plays (2,2), I pass and I win -plays (1,2), I let him with 1 and I win -plays (1,2,1), I give my opponent the position (1,1) and I win
Apr 17, 2011 at 17:54 comment added quantumelixir I've divided the set of positions into winning and losing ones and from that the conclusions regarding the variants with a pass move follow. I don't understand why my conclusion is not true for the reason I give. (You can check my previous edit for this question where I explain the strategy, which is the same as the one pointed out by @thomas-andrew above later)
Apr 17, 2011 at 17:44 comment added whuber "Zero-sum" is not a relevant adjective for this game. We're really working with Conway/Berlekamp/Guy's partizan games. A pass gives a move to one side, to be sure, but this guarantees a win only when the value of the game for that person was already greater than -1. Because the values of Nim are extremely close to 0, your conclusion is indeed true, but not for the reason you give.
Apr 17, 2011 at 16:22 history edited quantumelixir CC BY-SA 3.0
correct definition
Apr 17, 2011 at 16:02 history answered quantumelixir CC BY-SA 3.0