Variants on the game of Nim

Résumé du document

Long before rock and roll, television, and computer generated dinosaurs, a group of stones was all that was needed to provide hours of entertainment and learning for the Chinese. Thousands of years ago in China, the fascinating game of Nim was created and continues to challenge minds today. Nim is a strategic game with a secret order hidden within it by mathematics. Without applying mathematical logic, it is difficult to win. But, if you look for a pattern, you just might outsmart your opponent. A limited number of objects are arranged in rows, and a move consists in removing objects in one row. One wins by removing the last object. The solution of this game is known, and we can always tell who is going to win, assuming the two players are playing optimally. In this paper, we will first describe the solution for the Chinese Nim, and find the winning strategy for two variants of the Chinese Nim, where the players can remove matches from more than one row, or see what happens when the number of matches one can remove is limited.

Extraits

[...] If it is a winning move, then we are done. Otherwise, it is a losing move. But in the latter case, this means that the second player has a good counter-move. It is immediate that any position reachable from the new position (with M N 1 squares) is also reachable from the initial position (with M N squares). Hence, the first player could have gotten to that position right away. This famous proof can be used in practice as follows. [...]

[...] Math. Monthly 81 (1974), 876–879. M. Artin, Algebra, Englewood Cliffs, N.J.: Prentice Hall, c1991. G. Hardy and E. Wright, An introduction to the theory of numbers, 5-th ed, Oxford University Press D. Mazzoni, Three player Nim and more, Senior Thesis, Harvey Mudd College Berlekamp, Elwyn R., Conway, John H., Guy, Richard K. Winning Ways for your mathematical plays. New York: Academic Press, 1982: 75-6. [...]

[...] As before, the game is even if S = otherwise the game is odd. Let us consider a simple example for n = 2. The initial game is there are four rows of six, seven, twelve, and fourteen objects, respectively. We can remove objects from two rows, so we calculate the Nim-sum modulo 0 + 0 + 1 + 1 0 0 1 As the Nim-sum is not zero, the game is odd. To make the Nim-sum zero, the first player simply removes seven objects from the third row and eleven from the fourth one: (6,7,12,14) Notation 2. [...]

[...] The player who picks up the last object wins. There exist positions such that no matter how one plays, one cannot possibly win the game, provided the opponent plays optimally. Such positions are called (for the player whose turn it is to move next). The positions which are not losing are called “winning”. Which positions are losing for the player whose turn it is to move? The answer is determined by the of the rows. Definition 1. The Nim-sum of r rows with mi elements in the ith row i is obtained as follows: 1. [...]

[...] Let i be the number of digits in the binary representation of S. Choose a row with p objects such that the ith digit of the binary representation of p is a 1. Such a row exists, because at least one row must have contributed to the 1 in the ith digit of S. Let q be the Nim-sum of all the remaining rows. Now, we simply take away objects from the chosen row until it has exactly q objects. [...]