- a player throws N coins; he then puts one or more aside and rethrows the rest
- this throwing is repeated until he no longer has any to throw
- highest score (dice) or maximum number of heads (coins) wins

Z = number of zeros (tails)

- if Z = 0, quit
- if Z = 1, throw the zero
- if 1 < Z < N, save one one, throw the other N-1 coins
- if Z = N, save a zero, throw the other N-1 coins

Two players alternate placing X's on a rectangular grid. No two X's may appear adjacent along a side or across the diagonal at a corner. The last X wins.

Some theory: The "indicator" for a position is: make all possible moves from the given position.

Evaluate the indicator of each of these successor positions.

The indicator of the first position is the smallest number which is not the indicator of a successor position. The indicator of the null position is 0. The second player wins iff the indicator is 0. Example of calculating an indicator for the 3 x 3 board: There are 3 distinct moves possible -- corner, side, center. Playing in the center leaves the null position, indicator 0. Playing on the side leaves a 1 x 3 line, indicator 2. Playing in the corner leaves a 3 x 3 L, indicator 3. The smallest number not appearing in our list is 1, so the indicator of a 3 x 3 square is 1. For two boards (not touching) played simultaneously, the indicator is the XOR of the indicators for the separate boards. For any position, the indicator is <= the maximum game length.

PROBLEM: Find some non-exponential way to compute the indicator of a given position. For lines, a period of 34 is entered after the line is about 80 long. For Ls: if one leg is held fixed, the indicator (as a function of the other leg) seems to become periodic with period 34. The time to enter the period becomes greater as the fixed leg increases.

- On an odd x odd board, the 1st player wins.
- On a 4 x N board, the 2nd player wins.
- On a 6 x 6 board, the 1st player wins by playing at the center of one quarter.

white: pawns at QN3 and KN7, knight at QN4, bishop at KB7, king at QB2;

black: pawn at QN3, king at QR6. White mates in three moves.

- The player starts with each of nine windows open, showing the digits 1 - 9.
- Roll two dice.
- Cover up any digits whose sum is the sum on the dice.
- Iterate throwing and closing windows until the equality of sums is impossible.
- Your score is the total of closed windows (highest wins).

PROBLEM: 6 dots is minimum to ensure no stalemate with 2 players; how many dots are required with 3 players?

A B C D E F G H I J K L M N P Q . S T U V W X Y Z 1 2 3 4 5 6 7 8

. B . D E . . . . . . . . . P . . . T U V W . . . . . . 4 . . 7 .

A B C D E F G H . . . L . N . . . . . U V W . . . 1 2 3 . 5 6 7 8

Another useful observation is that the pegs and their original hole
positions fall into four equivalence classes in which they stay
throughout the game. Thus the four pegs which can reach the center on
the first move are the only ones that ever can. Similarly, the peg
jumped over on the last move must be in one of the two classes of
eight members which get reduced on the first move. The program's main
heuristic was to reduce the larger classes first.

With its heuristics disabled, the program simply scanned lexicographically (left to right in the inner loop, then top to bottom) for a peg which could move. At one point, there is a peg which can move two ways; it chose west. Twelve moves from the end it stopped and went into an exhaustive tree search, in which it found two basically different wins. (Try it yourself.)a b a c d c a b a b a b a c d c . c d c a b a b a b a c d c a b a

. . . . . . . . . . K . . . . Q . . . . . . X Y Z 1 2 3 4 5 6 7 8

REMOVE CAN END WITH ONLY THE PEG 1 1, 7 = 10, 13 2 2, 6, 11, 14 4 3 = 12, 4, 9, 15 5 13