#### 14 Jun 2003 Bram»(Master)

Rubik and Megaminx pun

A came up with a unified solution which applies both to Rubik's cube and Megaminx. It's a nice solution to either one even without that property.

Hexagon Game

The hexagon game I mentioned earlier is easier to win than necessary. When designing a game, one wants there to be no draws, but within that restriction you want it to be as difficult as possible to win, so as not to truncate interesting game play.

The new winning condition is that a player wins if they make a move such that after that if all the empty cells are filled in for the opponent then the opponent does not have any of the following:

• A single group which touches four edges

• Three separate groups each of which touch three edges

• Two separate groups each of which touch three edges and one of which touches three edges no two of which are adjacent

I'll now go over each of the specific winning configurations in detail and explain the reasoning behind them. My notation will be that the edges of the hexagon are numbered 1-6 going clockwise and that (a, b, c) describes a single group touching edges 1, 2, and 3. Multiple groups are grouped together to describe a player's position, so ((1, 3), (3, 5)) means that a single player has two groups, one connecting sides 1 and 3 and the other connecting sides three and five. To describe a whole board position I'll give two sets of groupings in a row, the first for player A and the second for player B.

Here are the winning configurations, not including rotations and reflections. Some entries will include two whose reasons for being winning are related. You may wish to get out two different colored pens and draw the actual diagrams.

((1, 2, 3, 4))

Consider the complete board ((1, 2, 3, 4)) ((1, 4, 5, 6)). This is rotationally symmetric, so it must be a winning configuration.

((1, 2, 3), (1, 4, 6))

Consider the complete board ((1, 2, 3), (1, 4, 6)) ((4, 5, 6), (1, 3, 4)). This is rotationally symmetric, so it must be a winning configuration.

((1, 2, 3), (1, 4, 5))

Consider the complete board ((1, 2, 3), (1, 4, 5)) ((1, 5, 6), (1, 3, 4)). This is symmetric about a line through the middle of sides 1 and 4, so it must be a winning configuration.

((1, 2, 3), (1, 3, 4), (4, 6))

Consider the complete board ((1, 2, 3), (1, 3, 4), (4, 6)) ((4, 5, 6), (1, 4, 6), (1, 3)). This is rotationally symmetric, so it must be a winning configuration.

((1, 2, 3), (1, 2, 4), (1, 5))

Consider the complete board ((1, 2, 3), (1, 2, 4), (1, 5)) ((1, 5, 6), (1, 4, 5), (1, 3)). This is symmetric about a line through the middle of sides 1 and 4, so it must be a winning configuration.

((1, 3, 4, 6)) and ((1, 2, 3), (1, 4), (4, 5, 6))

Compare the completed positions ((1, 3, 4, 6)) ((1, 2, 3), (4, 5, 6)) and ((1, 3, 4), (1, 4, 6)) ((1, 2, 3), (1, 4), (4, 5, 6)). Player A is strictly better in the first case, therefore A should win in the first case and B should win in the second case.

((1, 3, 4, 5)) and ((1, 2, 3), (1, 4), (1, 5, 6))

Compare the completed positions ((1, 3, 4, 5)) ((1, 2, 3), (1, 5, 6)) and ((1, 3, 4), (1, 4, 5)) ((1, 2, 3), (1, 4), (1, 5, 6)). Player A is strictly better in the first case, therefore A should win in the first case and B should win in the second case.

((1, 2, 3), (3, 4, 5), (1, 5, 6)) and ((1, 3, 5), (1, 5, 6))

Compare the completed positions ((1, 2, 3), (3, 4, 5), (1, 5, 6)) ((1, 3, 5)) and ((1, 2, 3), (3, 4, 5), (1, 5)) ((1, 3, 5), (1, 5, 6)). Player A is strictly better in the first case, therefore A should win in the first case and B should win in the second case.

I'm very happy with this game. There is a clear canonical winner in every position, and making any of the won positions happen is very subtle. It's missing the property that even if you kept playing after the game ended it would still be clear from the board position who the winner was, but that only happens with completely symmetric final positions, and having initiative play a strategic role may be a good thing.