Winning Strategy for 5467500000-Cube Chocolate Bar Game

[ghostfirefox]

Consider the following game. Two players alternately break vertical or horizontal lines from a rectangular chocolate bar. In each move the player can break one or two lines, either vertical on the right or horizontal from below. The chocolate bar in the upper left corner of the plate is poisoned. The player forced to take her loses. How many different rectangular chocolate bars of 5467500000 cubes, for which the first player has a winning strategy (tablets of sizes n × m and m × n are considered the same)?

 

[jvicens]

Hello,

This is an interesting game and I would like to provide some insights as a scientist.

Firstly, let's break down the problem. We have a rectangular chocolate bar with a total of 5,467,500,000 cubes. The two players take turns breaking either one or two lines, either vertical or horizontal, from the bar. The player who is forced to take the last piece from the upper left corner, where the poison is located, loses the game.

To determine the number of different rectangular chocolate bars for which the first player has a winning strategy, we need to consider all the possible moves and outcomes.

Let's start with the base case, where the chocolate bar is a 1x1 square. In this case, the first player will always lose since there is only one cube and the second player can simply take it.

Next, let's consider a 2x2 square. In this case, the first player can either break one line or two lines. If they break one line, the second player can break the remaining line and win. If the first player breaks two lines, the second player can break one line and force the first player to take the last piece, resulting in a loss. Therefore, in a 2x2 square, the first player will always lose.

Moving on to a 3x3 square, the first player can break one line and force the second player to take the remaining two lines, resulting in a win. If the first player breaks two lines, the second player can break one line and force the first player to take the last piece, resulting in a loss. Therefore, in a 3x3 square, the first player has a winning strategy.

Using this pattern, we can see that in a 4x4 square, the first player will always lose, in a 5x5 square, the first player has a winning strategy, and so on.

Now, let's apply this logic to a 5467500000 cube chocolate bar. We can break it down into smaller squares and determine the outcome for each size.

For example, a 100x100 square would result in a loss for the first player, since they can only break one line and the second player can break the remaining lines and win.

Similarly, a 101x101 square would result in a win for the first player, since they can break two lines and force the