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Here is one way you could begin to prove that the board can be cleared for all values of $m$ and $n$. Proof by induction. [incomplete] Case 1: $k = 1$ Here we take k = 1 to mean the smallest possible board with the given conditions $(m \ge n \ge 2)$ which would be a $m = n = 2$ or $2m \times 2n$ = $4\times4$ board. Theorem: For an Othello board of size $...


Yes, I found this video on YouTube that has this "perfect game". Terrible music by the way


The upper bound limit is likely to be quite a bit lower than 60. For the purposes of this I will assume that you are WHITE. In order for you to have a valid move in Othello, you need a line formed by the empty space and at least one of each colour. The largest number of valid moves made possible by the presence of a single WHITE counter is 8 - one for each ...


if the question still stands, in the Wthor database of 100.000 real games, the maximum number of legal moves is 21. And theaverage number is about 9.5 Olivier


The diagonal symmetry with opposing colours on opposing sides of the axis isn't really possible because there are pieces on the symmetry axis too. Since OP agreed that this observation presents a problem, here's the "most symmetrical" finished position with 4 empty spots I could find. The checkered spots can be any colour you want (as long as they are all ...


Remark regarding the tie-breaker rule: Solution:


I wrote a hill-climbing algorithm that searches for configurations with the maximum number of legal moves. I've made sure that the initial 4 cells are filled, the number of white and black tokens is the same, and all the tokens form a single connected component.


In 20,000 random games, the longest list of legal moves for any position was 24. After 10,000 the longest list was 22. You could make an estimate by playing many random games and recording the average and standard deviation of the move list lengths at every ply. Or you could make a pair of players that try to maximize the number of legal moves for either ...

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