What is the least number of unit squares that you need to remove from a 5 by 3 checker board in order to make it impossible for anyone to put an "X" in 3 remaining squares to make a connected vertical, horizontal, or diagonal set of 3 (a set that looks like a "win" in a standard Tic Tac Toe game)?
To find the least number of unit squares that need to be removed from a 5 by 3 checkerboard to make it impossible for anyone to form a connected set of 3 squares, including diagonals, we can go through a step-by-step process.
1. First, let's analyze the possible winning combinations:
- Horizontal winning combinations: three squares in the same row.
- Vertical winning combinations: three squares in the same column.
- Diagonal winning combinations: three squares in a diagonal line.
2. Observing the checkerboard, we notice that the 5 by 3 dimensions allow for horizontal, vertical, and diagonal sets of 3 squares.
- There are 5 rows, so there could be 5 horizontal winning combinations.
- There are 3 columns, so there could be 3 vertical winning combinations.
- There are 2 diagonals that can form winning combinations.
3. To make it impossible for anyone to form a connected set of 3 squares, we need to remove at least one square from each possible winning combination.
- Removing one square from each of the 5 horizontal rows eliminates the possibility of any horizontal winning combinations.
- Likewise, removing one square from each of the 3 vertical columns prevents any vertical winning combinations.
- Removing one square from either of the two diagonals eliminates the possibility of diagonal winning combinations.
4. Considering the above, we need to remove a total of:
- 5 squares for the horizontal winning combinations (one from each row).
- 3 squares for the vertical winning combinations (one from each column).
- 1 square for the diagonal winning combinations (one from either diagonal).
Therefore, the least number of unit squares that need to be removed from a 5 by 3 checkerboard to make it impossible for anyone to form a connected set of 3 squares is 5 + 3 + 1 = 9 squares.