# Removing pawns - the game

Here is a simple game I've invented (if the idea is not fresh, then please let me know):

The game is played on a board. The board has some (finite) number of lines drawn on it. A pawn is placed on each intersection point of (two or more) lines. Two players take alternate turns removing pawns. On each turn, a player removes one or more pawns. All pawns removed in a single turn have to be taken from the same line. The player who cannot make a move loses (alternatively: the player who takes the last pawn wins).

Here is my question: For what values of m and n does the player who begin have a winning strategy when the game is played on an $n\times m$ rectangular grid?

• The game on a rectangular or triangular board is called "Two-Dimensional Nim" by Aviezri Fraenkel in "The Lighter Side of Mathematics: Proceedings of the Eugène Strens Memorial Conference." Fraenkel organized combinatorial game displays at the conference, and suggested the $3 \times 5$ board as a good one to play. It is not the same game as the one called "Two-Dimensional Nim" in Winning Ways. Some results are given without proof by Jim Ferry at the discussion groups.google.com/forum/#!topic/sci.math/-EHfgnl74_0 – Zack Wolske Nov 5 '14 at 21:49
• In particular, in response to your specific question, one of the messages in the link says that the first player has a winning strategy in the $3\times 5$ and $3\times 7$ cases. – Gabriel C. Drummond-Cole Nov 6 '14 at 4:50

Here is a partial answer. I'll assume $m$ and $n$ are the number of intersections rather than the number of squares.
Obviously if $m$ or $n$ is $1$ then the first player can win.
I expect the general odd by odd case to be much harder because $3\times 3$ is a second player win by case analysis and the strategy appears to have no symmetry. Also the Sprague-Grundy values of the subsets of the $3\times 3$ square do not seem to have any easy pattern.