A move consists in removing an end coin, whereby the coin's immediate (and the only) neighbor becomes naturally an end coin available for the removal on one of the successive moves.
The players take turns, 1st defined randomly.
When all the coins have been removed, the game ends, and the players count their bounties. The player with the larger amount wins.
Devise a winning strategy for one of the players.
| No Solution Yet | Submitted by Ady TZIDON |
| Rating: 5.0000 (1 votes) |
An even number of coins
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Yes, if there are an even number of coins, the 1st player seems to be in a controlling position. If the sum of the even coins is greater than the sum of the odd ones, pick the last coin (which is even) and keep picking even coins (there is always one available). If the sum of the even coins is less than the sum of the odd ones, pick the first coin (which is odd) and keep picking odd coins (there is always one available). As we have already seen, this does not guarantee the greatest bounty, but it does guarantee a win. For instance, consider our previous example of 6-5-3-5.
If the sum of the even coins equals the sum of the odd ones, the first player can always guarantee at least a tie. A win may be available to the 1st player, but a simple strategy is not clear to me.
For instance, if the layout is 2-4-4-1-2-3, the first player wins by taking 3, the second player must take a 2, and now the total of the odd coins does not equal the total of the even ones and the number of coins is even. Being ahead by 1 already, a win is easy.
On the other hand, 2-4-4-3-2-1 looks to me like a tie against best play.
Edited on November 29, 2015, 11:15 am
| Posted by Steve Herman on 2015-11-28 19:42:34 |
