Here is another game that uses an n-by-n grid drawn on paper.

This game is for 2 players.

Players take turns placing 1,2,3,... n^2 into empty squares of the grid, one number each turn.

(So, player 1 places the odd numbers into the grid, and player 2 places the evens.)

Let H represent the collection of variables h, where each h is the sum of a HORIZONTALLY-adjacent pair of integers in the grid.

Let V represent the collection of variables v, where each v is the sum of a VERTICALLY-adjacent pair of integers in the grid.

(So, H and V each consist of n*(n-1) variables.)

Player 1 then chooses a variable d which divides at least one of the h's.

Player 1 gets as a score:

d*(number of h's that are divisible by d)

Player 2 then chooses a variable d' which divides at least one of the v's.

Player 2 gets as a score:

d'*(number of v's that are divisible by d')

The player with the largest score wins.

For example, let us say two poorly-playing players have this n=4 grid:

08 09 01 10

03 11 16 07

02 14 15 05

04 12 06 13

H =

(17,10,11, 14,27,23,

16,29,20, 16,18,19)

Let us say that player 1 picks 4, then he gets 4*3 = 12 points.

(Because h's = 16, 20, and 16 are divisible by 4.)

Had he picked 29, he could have gotten 29, however.

V =

(11,5,6, 20,25,26,

17,31,21, 17,12,18)

Player 2 picks 3 and gets 3*4 = 12 points.

She should have picked 31.

Yes, you can always score at least n*(n-1) by picking d or d' = 1.

As implied in the example, a player is responsible for finding his/her own score.

Thanks,

Leroy Quet

## Sunday, June 12, 2011

## Saturday, June 4, 2011

### Palindromic Rows/Columns

This game is for 2 players.

Start by drawing an n-by-n grid on paper.

(Who would have guessed??)

The players take turns filling in empty squares of the grid, one square per move. Each player fills in floor(n^2/4) squares, so that there are a total of 2*floor(n^2/4) squares (about 1/2 of grid) filled in at game's end.

Player 1 gets as a score the number of grid-squares whose states need to be changed (from filled to unfilled, or vice versa) in order to make each ROW a palindrome.

Player 2 gets as a score the number of grid-squares whose states need to be changed in order to make each COLUMN a palindrome.

The player with the SMALLEST score wins.

{I know that the rules could have had Player 1 score with the number of squares needed to make the *columns* into palindromes, and Player 2 could have had the rows, then the *highest* score wins; but if each player attempts to determine their own score, then with my way each player has an incentive to be as efficient as possible in determining how many squares need to be changed to achieve the palindromes.}

Thanks,

Leroy Quet

Start by drawing an n-by-n grid on paper.

(Who would have guessed??)

The players take turns filling in empty squares of the grid, one square per move. Each player fills in floor(n^2/4) squares, so that there are a total of 2*floor(n^2/4) squares (about 1/2 of grid) filled in at game's end.

Player 1 gets as a score the number of grid-squares whose states need to be changed (from filled to unfilled, or vice versa) in order to make each ROW a palindrome.

Player 2 gets as a score the number of grid-squares whose states need to be changed in order to make each COLUMN a palindrome.

The player with the SMALLEST score wins.

{I know that the rules could have had Player 1 score with the number of squares needed to make the *columns* into palindromes, and Player 2 could have had the rows, then the *highest* score wins; but if each player attempts to determine their own score, then with my way each player has an incentive to be as efficient as possible in determining how many squares need to be changed to achieve the palindromes.}

Thanks,

Leroy Quet

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