(Is this game original? I myself may have come up with something similar earlier.)

This is a game for 2 players.

You have an n-by-n grid drawn on paper (as almost always).

n should be >= 8, I suggest.

In the first part of the game, the players take turns placing a total of n x's into the grid, where one x is placed in an empty square of the grid each move.

In the second part of the game, players take turns. On a turn, a player draws a line (through the centers of the intervening squares) either up, right, down, or left from the last x drawn to by the other player. The (maybe bending) line may take at most one right-angle turn. And it must end at an x not drawn to/from yet. (Player 1 draws from any of the x's on her/his first move.) (What I mean by "the line may take at most one right angle turn" is that the x's will either be connected by a single straight line-segment {if both x's are in the same row or column} or they will be connected by two perpendicular, connected straight line-segments {if the x's are in both a different row and a different column}).

Lines can't pass through a line already drawn in the second part of the game. And the line cannot pass through an x on its way between two other x's during a move. The line may, though, share a corner with, or coincide partially with (both line-segments horizontal or both line-segments vertical), a line drawn previously during the second part of the game.

A player must move if it is possible.

The last player able to move LOSES.

(Note: Unlike some other games I have posted recently, more than one x or no x's at all may be in any row or column.)

Thanks,

Leroy Quet

## Saturday, November 27, 2010

## Tuesday, November 9, 2010

### Multiplications Within The Permutation

This is a game for any plural number of players. Let the number of players be m.

Start with an n-by-n grid drawn on paper, where n = k*m + 1, k is some integer >= 3.

In the first part of the game, players take turns placing x's in empty squares of the grid, one x per turn, such that no more than one x is in each row and in each column of the grid.

After exactly n x's are placed in the grid, the first part of the game is over.

The second part of the game starts with a 1 being placed in the leftmost square with an x in it. Players then take turns.

On the jth move (starting at move # 1) of the second part of the game, the moving player places a (j+1) in any square with an x and without a number already in it.

He/she gets added to her/his score:

|x(j)-x(j+1)| * |y(j)-y(j+1)|,

where x(j) is the number of squares from the bottom of the grid where the square with the j in it is located, and y(j) is the number of squares from the left side of the grid where the square with the j in it is located.

So, what we are adding to the moving player's score (the score of the player writing a j+1 in a square) is the product of {the change in horizontal distance between the squares with j and j+1 in them} and {the change of vertical distance between the squares with j and j+1 in them}.

When the nth x is numbered with a n= m*k+1, the game is over.

Largest score wins.

Here is an example:

n=7. m =2.

. 3 . . . . .

. . 6 . . . .

. . . . . . 5

. . . . 7 . .

1 . . . . . .

. . . 4 . . .

. . . . . 2 .

Squares 1 to 2: 5*2 = 10

Squares 2 to 3: 4*6 = 24

Squares 3 to 4: 2*5 = 10

Squares 4 to 5: 3*3 = 9

Squares 5 to 6: 4*1 = 4

Squares 6 to 7: 2*2 = 4

Player 1 gets: 10+10+4 = 24 points.

Player 2 gets: 24 + 9 + 4 = 37 points.

Player 2 wins.

Thanks,

Leroy Quet

Start with an n-by-n grid drawn on paper, where n = k*m + 1, k is some integer >= 3.

In the first part of the game, players take turns placing x's in empty squares of the grid, one x per turn, such that no more than one x is in each row and in each column of the grid.

After exactly n x's are placed in the grid, the first part of the game is over.

The second part of the game starts with a 1 being placed in the leftmost square with an x in it. Players then take turns.

On the jth move (starting at move # 1) of the second part of the game, the moving player places a (j+1) in any square with an x and without a number already in it.

He/she gets added to her/his score:

|x(j)-x(j+1)| * |y(j)-y(j+1)|,

where x(j) is the number of squares from the bottom of the grid where the square with the j in it is located, and y(j) is the number of squares from the left side of the grid where the square with the j in it is located.

So, what we are adding to the moving player's score (the score of the player writing a j+1 in a square) is the product of {the change in horizontal distance between the squares with j and j+1 in them} and {the change of vertical distance between the squares with j and j+1 in them}.

When the nth x is numbered with a n= m*k+1, the game is over.

Largest score wins.

Here is an example:

n=7. m =2.

. 3 . . . . .

. . 6 . . . .

. . . . . . 5

. . . . 7 . .

1 . . . . . .

. . . 4 . . .

. . . . . 2 .

Squares 1 to 2: 5*2 = 10

Squares 2 to 3: 4*6 = 24

Squares 3 to 4: 2*5 = 10

Squares 4 to 5: 3*3 = 9

Squares 5 to 6: 4*1 = 4

Squares 6 to 7: 2*2 = 4

Player 1 gets: 10+10+4 = 24 points.

Player 2 gets: 24 + 9 + 4 = 37 points.

Player 2 wins.

Thanks,

Leroy Quet

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