Saturday, March 14, 2009

Permutations Of Divisors

This is a game for any number of players. (Gosh darn, no grids this time.)

This game consists of a number of rounds, where the total number of rounds is predetermined and is a multiple of the number of players.

Players take turns choosing integers, one integer per player per round.
On a round, the player whose turn it is to chose picks any positive integer that has not yet been chosen in the game to be placed at the end of a growing list of integers.
(So, after the player picks a number during round n, there are then exactly n integers in the list.)

Say this list (the "divisor list") is (d(1),d(2),...,d(n)).
On the nth round, after the nth term is appended to the divisor list, each player (by themselves and in secret) then tries to come up with a positive integer m such that, if (d'(1),d'(2),...,d'(n)) is a player's permutation of the divisor list, then
d'(j) divides (m+j-1) for all j where 1 <= j <= n, if such a permutation exists.
In any case, each player tries to find a permutation of the divisor list where as few terms as possible are not in their same position as they are in the original divisor list, and where as few of the members of the divisor list as possible do not divide the numbers in the "multiple list" they are paired with (where the "multiple list" is the list of consecutive integers from m to m+n-1).

So, in other words, a player's score on a round begins as the number of j's where d'(j) = d(j), where {d'(j)} is the player's own permutation of the divisor list, and where d'(j) divides m+j-1. Then, to get the player's score for that round, subtract the number of j's where d'(j) does not divide (m+j-1) (1 <= j <= n).

A player grand score is the sum of his/her scores from each round.

The player with the highest grand score wins.

For example, let us say that n = 6. And the divisor list looks like this:
1, 2, 4, 3, 17, 5.

A player then chooses this multiple list:
13, 14, 15, 16, 17, 18.

That same player then choses this permutation of the divisor list:
1, 2, 3, 4, 17, 5.

Only the 3 and the 4 are out of place. And only the 5 does not divide its respective integer in the multiple list.

So this player for this round gets
3 - 1 = 2 points, because only 3 integers in his/her divisor list permutation (Those 3 integers are 1,2,17) both divide their respective integers in the multiple list and are not out of order from their positions in the original divisor list.

Leroy Quet

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