"Cut-the-Knot" is quite entertaining! I like to invent abstract games, and a number of my game ideas are available on my webpage at http://www.rpi.edu/~vanpag/. One of these in particular, described below, has a very mathematical flavor which might make it appropriate for Cut-the-Knot.

The following is a permutation game I invented, called "Odds vs. Evens".

This game uses 2N cards numbered 1 through 2N, where N is at least 4. Player-I owns the odd numbered cards, while player-II owns the even numbered cards.

The game begins with the cards arranged in increasing order, from left to right.

On his turn, a player must switch the positions of two cards, i and j, (a player may switch ANY two cards, whether he owns either of them or not), such that, before the switch, i < j and i is to the left of j. Passing is not allowed.

The goal for each player is to arrange his cards in decreasing order from left to right, but not necessarily consecutively. A player wins if his goal is achieved without his opponent's goal being simultaneously achieved. Note that the switch which creates the winning arrangement may have to be made by the
losing player. If both players' goals are achieved simultaneously, then the game ends in a tie.

Here is a sample game, N=4 (Switched cards indicated by bars):

1 2 3 4 5 6 7 8 <---
_ _
1 2 5 4 3 6 7 8 I 1
_ _
1 2 5 7 3 6 4 8 II 2
_ _
4 2 5 7 3 6 1 8 I 3
_ _
4 2 8 7 3 6 1 5 II 4
_ _
4 7 8 2 3 6 1 5 I 5
_ _
4 7 8 6 3 2 1 5 II 6
_ _
7 4 8 6 3 2 1 5 I 7
_ _
8 4 7 6 3 2 1 5 II 8
_ _
8 7 4 6 3 2 1 5 I 9
_ _
8 7 6 4 3 2 1 5 II 10
^ ^ ^ ^

Player II wins the game.

Note that after move 1, player-I could win by switching 1 and 7, so player-II defends by switching 4 and 7 in move 2. But then player-II could win by switching 2 and 8, so player-I defends against this by switching 1 and 4 in move 3. And so on.

There are many variations on this theme, but this seems the simplest. I would like to know if there is an easy winning strategy for this game.

Gregory K. Van Patten