The game Slice is played using a m×n rectangular piece of paper as a board. Players alternate turns, on each turn they choose a rectangle and cut it into two rectangles, each with integer side lengths. The last player who is able to cut a rectangle is the winner. If 1≤m≤20 and 1≤n≤20, for how many of the 400 different starting games does the first player have a winning strategy, no matter how the second player plays?
Details and assumptions
For a 1×1 board, the second player is a winner.
4 answers
20
is 20 correct?
I have a winning strategy, but perhaps there are others.
Draw a line of symmetry to split the game into two identical games. Whatever the next player does, you would do the mirror image. Therefore you will end-up splitting the last rectangle.
Based on this, there are only two ways to draw the first line to make sure you win the game.
Draw a line of symmetry to split the game into two identical games. Whatever the next player does, you would do the mirror image. Therefore you will end-up splitting the last rectangle.
Based on this, there are only two ways to draw the first line to make sure you win the game.
Since the player not able to make a move loses, the first player wins if and only if mn is even. Go test it out on small mn!
So the answer is 300. (actually we first look at odd products - there are 10*10 = 100.There is a total of 20*20 = 400 configurations. So the answer is 400 - 100 = 300.)
So the answer is 300. (actually we first look at odd products - there are 10*10 = 100.There is a total of 20*20 = 400 configurations. So the answer is 400 - 100 = 300.)