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.
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.
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.)
To determine the number of starting games in which the first player has a winning strategy, we need to consider each possible starting game and analyze the winning strategy for the first player.
We can use a systematic approach to count the number of winning strategies for the first player. Let's start by analyzing the smallest possible boards.
For a 1x1 board, the second player is guaranteed to be the winner according to the given assumption.
For a 1x2 board, the first player can choose to make a cut that splits the board into two 1x1 boards. Since we know the second player wins in a 1x1 board, any move by the second player will result in a win for the first player. Therefore, all starting games on a 1x2 board result in a win for the first player.
Similarly, for a 2x1 board, the first player can make a cut that splits the board into two 1x1 boards. Again, this guarantees a win for the first player regardless of the second player's moves.
For larger boards, we can analyze the winning strategy by breaking down the board into smaller components. Consider a 2x2 board:
1 1
1 1
The first player can make a cut horizontally, vertically, or diagonally to divide the board into two smaller boards. Let's consider each case:
1. Horizontal cut:
The board becomes two 2x1 boards:
1 1
and
1 1
As we analyzed earlier, the first player can guarantee a win on a 2x1 board, regardless of the second player's moves.
2. Vertical cut:
The board becomes two 1x2 boards:
1 | 1
1 | 1
Similarly, the first player can guarantee a win on a 1x2 board.
3. Diagonal cut:
The board becomes two 1x1 boards:
1 | 1
- - -
1 | 1
Again, as we know from the assumption, the second player wins on a 1x1 board.
In all cases, the first player can guarantee a win on a 2x2 board. This pattern continues for larger boards as well.
Using this analysis, we can conclude that for any rectangular board with dimensions m x n, if either m or n is even, the first player will always have a winning strategy.
Now, let's analyze the possible combinations of m and n values to determine the total number of starting games where the first player has a winning strategy:
1x1: 0
1x2: 1
2x1: 1
2x2: 1
2x3: 2
3x2: 2
3x3: 4
...
Continuing this approach, we can calculate the number of starting games for larger board dimensions. By summing up the counts for all 400 different starting games, we will have the answer to the original question.
Note: Since the board dimensions are limited to 1 ≤ m ≤ 20 and 1 ≤ n ≤ 20, this systematic approach to analyze each combination is feasible. However, for larger board dimensions, alternative approaches such as dynamic programming might be more efficient.