Hexagonal Checkers 2

To start, we place the number $r^{n-1}$ into each cell, where n is the row that the cell is in and $r=\frac{\sqrt5-1}{2}$ . We then define the characteristic value of each state of the game to be the sum of all the cells that have a checker on them. Notice that for each move, the characteristic value either stays the same or decreases (this is true because $r^{n+2}+r^{n+1}=r^n$ .) In other words the characteristic value at the beginning of a game will be greater than or equal to the characteristic value at the end of the game. The characteristic value of the desired state (a checker in the top cell) is $1$ . We notice by the formula for the arithmetico-geometric sequence that the characteristic value for a checkerboard with all the cells filled would be $\frac{7+3\sqrt5}{2} \approx 6.854$ If the line were below the seventh row, the maximum possible characteristic value for the beginning state would be $\frac{-81+31\sqrt5}{2} \approx .867$ Therefore it is impossible to get a checker into the top cell when $n=7$ . The $n=6$ case can be shown by example. See also a variation on this problem: Hexagonal Checkers .

Note: Unfortunately, when I found the $n=6$ case by hand, it took me too many moves to be short enough to share here. If you would like to verify the $n=6$ case for yourself please go ahead and do so. I would recommend finding the solution backwards (i.e. starting with the top cell filled and replacing one checker with two checkers in the appropriate manner until you get all checkers below the sixth row). Also, if you find it easier, you can notice that the game on a triangular board is equivalent to the game on one quadrant of a square board.

Edit: The question has been rephrased from checkers on a triangular checkerboard to pegs in a triangular pegboard, so that is why my solution uses the original terminology.

This is the Pablito's army variant of Conway's soldiers . The maximum number of empty rows for this variant is $\boxed{6}$ .

This is not a full solution, and is designed to expand on John Ross's solution. We want to put an number in each position such that the sum of all positions is finite, and yet, when you take the sum of all populated positions, making any move strictly increases, or strictly decreases the sum. This sum will then take on the role of a characteristic value, which we can use to prove impossibility.

The easiest way to populate the cells with these numbers is to put 1 at the top, and then for each subsequent row, put the value of any individual position on the previous row, multiplied by some x, where $|x| < 1$ . Now, suppose that we make a move that takes a peg from two rows and places it on the row above. We can define our numbers such that doing this does not alter the characteristic value. In this case, it follows that $x^2+x=1$ . Solving with the quadratic equation gives us two candidate solutions, $x = \frac{\sqrt{5}-1}{2}$ and $x = \frac{-\sqrt{5}-1}{2}$ . Looking back to our restriction, $|x| < 1$ we can discard the second possibility, leaving just one possible $x$ value, $\frac{\sqrt{5}-1}{2}$ . This value may look familiar, as it is related to the golden ratio - in fact, it is the reciprocal of phi.

To check that the characteristic value is valid, we consider all possible moves. Moving up does not change the characteristic value, but moving sideways or down will decrease the characteristic value, which means that it can be used to prove impossibility.