Game. Any plural number of players. n-by-n grid. (I suggest for a 2-player game an n of about 4 to 6 for beginners.)

The first player to move puts an x in any square.

The players take turns. (The player who last drew an x is player A. The player who is now choosing where to put an x is player B.)

The players take turns who is player A and who is player B.

After player A draws an x, she then tells player B how many squares from player A's recent x that player B can put his x. (This distance is k squares.)

Player B then chooses a direction (up, down, right, or left-- whatever is possible, given the edges of the grid) from player A's x that he shall put the new x.

Player B can only put an x in an empty square. And this x must be the number (k) of squares dictated by player A from player A's last x (as I already said).

And, if possible, player A MUST dictate the distance (k) to an empty square from her x.

Last player able to move is the winner.

So, to be the last player, you want to move into an empty square that is the last empty square in that column and row.

I actually played this game with someone else. (I admit, I hardly ever play my games with other people before publishing them. I have a hard time finding people willing to play.)

For novices, I found, the game goes along without too much use of strategy until the last few moves.

I would guess that if the players are more advanced, then strategy would come in sooner. But maybe not. Maybe, maybe, it doesn't really matter how good you are until the last few moves.

Thanks,

Leroy Quet

## Thursday, July 29, 2010

## Monday, July 26, 2010

### Subdivide And Acquire Game

Here is a game for any plural number of players. Start with an n-by-n grid drawn on paper, where n is larger if there are more players.

Each player has a colored pencil of unique color.

In the first part of the game, players take turns drawing straight line-segments -- one line-segment each turn. A player can draw a line-segment from either the edge of the grid at a vertex, or from the end of another line-segment. (Multiple line-segments can join at one point.) The line-segments are drawn to any vertex of the grid (either empty or already occupied by a line segment), such that no line-segment is drawn through another segment or along another segment or through a vertex occupied by line-segments. (Although, as I said, a line-segment may end at a vertex already occupied by another segment.)

After the grid is subdivided into n*(number of players) sections, the second part of the game begins.

Players take turns filling in sections of the grid with their colored pencils. After each player fills in n sections, the score is determined.

Players add up the total area of all the sections in each player's color, with the area of a grid-square being 1. (This may be tricky because some sections will most probably have non-integer areas.)

Let a player's total area be m; then the winner is the player where

number of divisors of floor(m) is the SMALLEST.

Thanks,

Leroy Quet

Each player has a colored pencil of unique color.

In the first part of the game, players take turns drawing straight line-segments -- one line-segment each turn. A player can draw a line-segment from either the edge of the grid at a vertex, or from the end of another line-segment. (Multiple line-segments can join at one point.) The line-segments are drawn to any vertex of the grid (either empty or already occupied by a line segment), such that no line-segment is drawn through another segment or along another segment or through a vertex occupied by line-segments. (Although, as I said, a line-segment may end at a vertex already occupied by another segment.)

After the grid is subdivided into n*(number of players) sections, the second part of the game begins.

Players take turns filling in sections of the grid with their colored pencils. After each player fills in n sections, the score is determined.

Players add up the total area of all the sections in each player's color, with the area of a grid-square being 1. (This may be tricky because some sections will most probably have non-integer areas.)

Let a player's total area be m; then the winner is the player where

number of divisors of floor(m) is the SMALLEST.

Thanks,

Leroy Quet

## Friday, July 23, 2010

### Transform -- The Game

This is a game for any plural number of players.

(No grids this time, sorry.)

There are a number of rounds in this game. The number of rounds is a multiple of the number of players. The players take turns being "the permutator", where each player is the permutator the same number of rounds.

Before starting any of the rounds, the players agree on a positive integer n. (n is the same value for all rounds in the game.)

On a round, all of the players (including the permutator) start the round by each coming up with an ordered list of n integers (positive, negative, or 0). Each player's numbers must be distinct, in that no integer occurs more than once in a particular player's list. The players each keep their lists secret from the other players for now.

Let player p's list for any particular round be {b(p,k)}, k = 1,2,3,...n.

Next, all players who are not the permutator take turns choosing terms of a list of n distinct integers(positive, negative, or zero; no integer more than once). (So, if the number of players is m, I guess to be fair, n should be a multiple of (m-1).)

Let this list be {a'(k)}, k = 1,2,3,..n.

Then, after the list is complete, the permutator forms any permutation {a(k)} of {a'(k)}.

Then everyone reveals their b-lists.

Each player p forms a sequence of n integers in this manner:

c(p,k) = sum{j=1 to n} a(n+1-j) b(p,j). k = 1,2,3...,n.

Player p's score for this round is the number of primes in {c(p,k)}.

After all the rounds are played, the players add up their scores for all the rounds. The player with the largest grand score wins.

In a variation, instead of the number of primes being the criterion for scoring, the players decide amongst themselves before each round what will be the criterion for a number in the c-list to score. Fibonacci numbers? Squares? Where each c(p,k) is coprime to c(p,k-1)?

Or maybe the choice of criterion should be totally up to the permutator (with veto power from the other players), and expressed at the beginning of each round before the b-lists are constructed.

Thanks,

Leroy Quet

(No grids this time, sorry.)

There are a number of rounds in this game. The number of rounds is a multiple of the number of players. The players take turns being "the permutator", where each player is the permutator the same number of rounds.

Before starting any of the rounds, the players agree on a positive integer n. (n is the same value for all rounds in the game.)

On a round, all of the players (including the permutator) start the round by each coming up with an ordered list of n integers (positive, negative, or 0). Each player's numbers must be distinct, in that no integer occurs more than once in a particular player's list. The players each keep their lists secret from the other players for now.

Let player p's list for any particular round be {b(p,k)}, k = 1,2,3,...n.

Next, all players who are not the permutator take turns choosing terms of a list of n distinct integers(positive, negative, or zero; no integer more than once). (So, if the number of players is m, I guess to be fair, n should be a multiple of (m-1).)

Let this list be {a'(k)}, k = 1,2,3,..n.

Then, after the list is complete, the permutator forms any permutation {a(k)} of {a'(k)}.

Then everyone reveals their b-lists.

Each player p forms a sequence of n integers in this manner:

c(p,k) = sum{j=1 to n} a(n+1-j) b(p,j). k = 1,2,3...,n.

Player p's score for this round is the number of primes in {c(p,k)}.

After all the rounds are played, the players add up their scores for all the rounds. The player with the largest grand score wins.

In a variation, instead of the number of primes being the criterion for scoring, the players decide amongst themselves before each round what will be the criterion for a number in the c-list to score. Fibonacci numbers? Squares? Where each c(p,k) is coprime to c(p,k-1)?

Or maybe the choice of criterion should be totally up to the permutator (with veto power from the other players), and expressed at the beginning of each round before the b-lists are constructed.

Thanks,

Leroy Quet

## Friday, July 16, 2010

### TraverX

This is a game for any plural number of players. Start with a grid of n-by-n squares ((n+1)-by-(n+1) lines) drawn on paper, where n is even, and where I suggest that n is >= 12.

At the beginning of the game, a small x is drawn at the intersection of the middle horizontal line and middle vertical line.

Players take turns moving. On the k-th move (the kth move considering all the players' moves together) the player "traverses" j(k) = (k-1)(mod(n-1))+2 intersections from where the last player last put an x.

(So, for k = 1,2,3,4,...,n-1,n,n+1,n+2..., the number of intersections traversed is 2,3,4,5,...,n,2,3,4,..., repeating 2 through n.)

(On the first move, the first player traverses 2 positions from the central x.)

The player can "traverse" j(k) intersections in the direction of either right, left, up, or down, and then may change direction at any time at most once, and traverse perpendicularly to their initial direction for the remainder of the j(k) intersections traversed. The player then places an x at the intersection they land upon. It is only acceptable for players to land upon (at the j(k)th intersection traversed) an intersection without an x already drawn upon it. Players may, though, traverse over intersections with x's already on them, or not.

After a player writes down an x, he/she gets a point (points are bad in this game) for every other x already written on the same vertical line and same horizontal line as their x.

The game continues until any player cannot move anywhere (given j(k) and the lack of available intersections).

The player with the SMALLEST score wins.

Thanks,

Leroy Quet

At the beginning of the game, a small x is drawn at the intersection of the middle horizontal line and middle vertical line.

Players take turns moving. On the k-th move (the kth move considering all the players' moves together) the player "traverses" j(k) = (k-1)(mod(n-1))+2 intersections from where the last player last put an x.

(So, for k = 1,2,3,4,...,n-1,n,n+1,n+2..., the number of intersections traversed is 2,3,4,5,...,n,2,3,4,..., repeating 2 through n.)

(On the first move, the first player traverses 2 positions from the central x.)

The player can "traverse" j(k) intersections in the direction of either right, left, up, or down, and then may change direction at any time at most once, and traverse perpendicularly to their initial direction for the remainder of the j(k) intersections traversed. The player then places an x at the intersection they land upon. It is only acceptable for players to land upon (at the j(k)th intersection traversed) an intersection without an x already drawn upon it. Players may, though, traverse over intersections with x's already on them, or not.

After a player writes down an x, he/she gets a point (points are bad in this game) for every other x already written on the same vertical line and same horizontal line as their x.

The game continues until any player cannot move anywhere (given j(k) and the lack of available intersections).

The player with the SMALLEST score wins.

Thanks,

Leroy Quet

## Monday, July 5, 2010

### Vertical/Horizontal Guessing Game

Here is a game for two players, and, you guessed it, it uses an n-by-n grid drawn on paper. (I suggest an n of at least 8.)

First, the players each secretly guess how many squares will be filled in before the game terminates. Each player writes their guess down and hides the guess from their opponent.

Next, fill in with a pen/pencil the center square (if n is odd) or one of the 4 center squares (if n is even).

(*)The players then both secretly choose a direction. Player 1 chooses either up, vertically-steady, or down. Player 2 chooses either left, horizontally-steady, or right. The players each write down their choices.

Then the players both reveal their choices simultaneously.

The next square filled in has the vertical direction chosen by player 1 and the horizontal direction chosen by player 2 from the last square filled in; and this next square filled in is adjacent (touching on a side or on a corner) to the last square filled in. (There are 9 possible combinations of directions, including not changing the square at all {when both players pick steady}.)

Note: If the last filled in square is on the edge of the grid, then this limits what directions can be stated by one or both of the players.

If that adjacent square is already filled in, then the game is over.

But if that square is empty before being filled in on the current move, then the game continues. (Go to *.)

When the game is over, the winner is the player whose guess for the number of squares filled in is closest to (either greater than or lesser than or equal to) the actual number of squares filled in.

(By the way, if both players choose the direction 'steady' at the same time, the game ends then, of course.)

Thanks,

Leroy Quet

First, the players each secretly guess how many squares will be filled in before the game terminates. Each player writes their guess down and hides the guess from their opponent.

Next, fill in with a pen/pencil the center square (if n is odd) or one of the 4 center squares (if n is even).

(*)The players then both secretly choose a direction. Player 1 chooses either up, vertically-steady, or down. Player 2 chooses either left, horizontally-steady, or right. The players each write down their choices.

Then the players both reveal their choices simultaneously.

The next square filled in has the vertical direction chosen by player 1 and the horizontal direction chosen by player 2 from the last square filled in; and this next square filled in is adjacent (touching on a side or on a corner) to the last square filled in. (There are 9 possible combinations of directions, including not changing the square at all {when both players pick steady}.)

Note: If the last filled in square is on the edge of the grid, then this limits what directions can be stated by one or both of the players.

If that adjacent square is already filled in, then the game is over.

But if that square is empty before being filled in on the current move, then the game continues. (Go to *.)

When the game is over, the winner is the player whose guess for the number of squares filled in is closest to (either greater than or lesser than or equal to) the actual number of squares filled in.

(By the way, if both players choose the direction 'steady' at the same time, the game ends then, of course.)

Thanks,

Leroy Quet

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