Wednesday, April 15, 2009

Making Intersections

This is a game for any plural number of players.

The game consists of a number of rounds, that number being a multiple of the number of players.
Each player takes turns being the offense player.

Before each round, draw an array of n-by-n dots (which correspond to the vertices of a grid of (n-1)-by-(n-1) squares).

On each round, all players take turns drawing straight vertical or horizontal line-segments from any dot to any other in the same row or column such that no line-segments cross any previously drawn segments and such that no line-segment coincides with any previously drawn line-segment. (But any line-segment may connect with a previously-drawn segment at a dot.)

The round ends when the players have drawn m line-segments, where m is a predetermined multiple of the number of players, and is picked by agreement among the players -- and m is the same for every round.
Note: m should be < n^2 - 2n + 4 = (n-1)^2 +3, which is the minimum number of segments needed to connect every dot to each of its orthogonal neighbors. (That this is the minimum number of segments is simply proved; and proving this may be a fun exercise for your amusement, if you don't want too hard a puzzle.)

After m line-segments are drawn, the offense player for the round receives a point for every dot connected to by 3 or 4 line-segments. (For the purpose of counting points, a single line-segment drawn by a player on a move and passing through -- and not terminating at-- a dot counts as 2 "line-segments" meeting at that dot.)

Players add up the points they received on the rounds that each was offense, and the highest total score wins.

Thanks,
Leroy Quet

PS: If this game is played on a rectangular array of j by k dots, the minimum number of straight horizontal and vertical line-segments (each of any length) needed to connect every dot to its orthogonal neighbors, without any line-segments crossing, is:
j*k - k - j + 4, or so I think.