Here is a game for any plural number of players. Start with an m-by-m grid drawn on paper. (I suggest that m be about 8 to 10 for beginners.) Draw the grid large enough so that two integers can be written in each square.
In the first phase of the game, players take turns writing the positive integers 1 to m^2 in order into the squares of the grid. One number is placed in any empty square of the grid on each move. (So, if there are 2 players, one player writes in the odd numbers, and the other player writes in the even numbers.)
Let the variable d (d for 'divisor') start the second phase of the game with a value of 1.
At the start of the second phase of the game, player 1 then writes the value of d, which is 1, alongside any number in the grid (in the same square as the number).
The players thereafter continue to take turns. On a move, a player chooses any square of the grid that has not yet had a second number written in it, but is adjacent to (in the direction of above, below, right of, or left of) any square that has had a second number written in it.
He/she then writes down in the square (with one number) any* positive divisor of the number in that square.
The variable d then becomes that divisor.
* The value of d, however, must change each move. The same divisor number cannot be written in two squares on two consecutive moves.
The absolute value of the difference between the older recent value of d (the divisor written by the previous player to move) and the new value of d (the divisor written by the current player moving) is then added to the currently moving player's score.
Note: The goal of the game is to get the LOWEST score. So, it is advantageous to change the value of d by as little as possible on a move. (Changing the value of d by 1 is the best a player can hope for on a move.)
The game continues until each square has exactly two numbers in it.
(So, there are a total of m^2 moves in the first phase of the game, and m^2 moves in the second phase of the game.)
As I said before, the player with the lowest score wins.
I would suggest that the divisor numbers (the values of d) be written smaller than the numbers written during phase 1 of the game, or be written in another color than the first numbers placed in the squares.
PS: The only problem I can see with this game is if the last square to get a second number has a 1 in it, and the previous (next to last) player to move placed a 1 as the second (divisor) number in some square. (This is a problem because d must change each move.)
Then, in that case, the second phase of the game ends after m^2 - 1 moves.
Thanks,
Leroy Quet
Tuesday, March 17, 2009
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.
Thanks,
Leroy Quet
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.
Thanks,
Leroy Quet
Saturday, March 7, 2009
Connecting Graphs
This is a game for any number of players.
Start with an n-by-n grid lightly drawn on paper. Or perhaps a square array of dots (representing a grid's vertices) would be better.
Players take turns connecting any ADJACENT pair of dots/vertices with a straight line-segment, one segment each move. Only pairs of dots that have not been previously connected may be connected on any move. Although any particular dot/vertex can have multiple line-segments drawn to it.
Line-segments may be vertical or horizontal. In a variation of the game, diagonally adjacent vertices/dots may be connected as well, as long as no line-segments cross. (In the variation, each line-segment is drawn either N, NE, E, SE, S, SW, W, or NW.)
Borrowing a term from graph-theory, a "graph" of connected line-segments (each line-segment drawn during the game's play) is a "connected graph" if one can trace along the line-segments from any vertex of the graph to any other vertex (possibly, but not necessarily, connecting each vertex to any other in multiple ways).
Whenever 2 distinct connected graphs are combined into 1 connected graph by a line-segment, the player drawing that line-segment gets a point.
No points are obtained for starting a new connected graph, for extending a single connected graph (in a way that does not connect to another connected graph), or for connecting a connected graph back to itself.
The game is over as soon as there is exactly one connected graph, no more, of line-segments drawn on the grid. The initial single connected graph, formed when the first move of the game is made, does not end the game. (Otherwise, what a dumb game this would be!)
Highest score wins.
How does allowing diagonally drawn line-segments alter the game, I wonder?
This game sounds familiar too. Where have I stolen the idea from?
Thanks,
Leroy Quet
Start with an n-by-n grid lightly drawn on paper. Or perhaps a square array of dots (representing a grid's vertices) would be better.
Players take turns connecting any ADJACENT pair of dots/vertices with a straight line-segment, one segment each move. Only pairs of dots that have not been previously connected may be connected on any move. Although any particular dot/vertex can have multiple line-segments drawn to it.
Line-segments may be vertical or horizontal. In a variation of the game, diagonally adjacent vertices/dots may be connected as well, as long as no line-segments cross. (In the variation, each line-segment is drawn either N, NE, E, SE, S, SW, W, or NW.)
Borrowing a term from graph-theory, a "graph" of connected line-segments (each line-segment drawn during the game's play) is a "connected graph" if one can trace along the line-segments from any vertex of the graph to any other vertex (possibly, but not necessarily, connecting each vertex to any other in multiple ways).
Whenever 2 distinct connected graphs are combined into 1 connected graph by a line-segment, the player drawing that line-segment gets a point.
No points are obtained for starting a new connected graph, for extending a single connected graph (in a way that does not connect to another connected graph), or for connecting a connected graph back to itself.
The game is over as soon as there is exactly one connected graph, no more, of line-segments drawn on the grid. The initial single connected graph, formed when the first move of the game is made, does not end the game. (Otherwise, what a dumb game this would be!)
Highest score wins.
How does allowing diagonally drawn line-segments alter the game, I wonder?
This game sounds familiar too. Where have I stolen the idea from?
Thanks,
Leroy Quet
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