Wednesday, June 23, 2010

Predicate -- A Game of Numbers And Creativity

This is a simple game that has the potential to go horribly wrong...

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

Each player has a different colored pen/pencil/crayon.

Make a number line with the positions immediately beneath it labeled in order with 1 through m*n, where n is some positive integer decided ahead of time by the players.

The players take turns. On a PLAYER'S k_th move, he/she writes (with the pen/pencil/crayon of his own color) the number k just above any one of the empty positions along the number line.
After every player has written n -- after a total of m*n moves, and the number line is filled up -- the next part of the game begins.
(When the first part of the game is complete, every integer k occurs exactly m times on the top of the number line.)

But before writing down the numbers, each player comes up with a rule for scoring points. The players all write down their rules, and only reveal them after the number line has been filled with numbers.

Each rule completes this sentence:

A point is scored for a player for every integer in the player's color where _______.

The rule must be based on the position number (below the line) of the integer (above the line) being tested , and/or on the neighboring integers written during play (above the line).

A rule must NOT be based on the colors of the integers or on any external variables.

The rules may use any mathematics the players personally choose.

All the rules apply to all the players' numbers fairly.
In other words, the players EACH come up with a rule, and all the rules are used to test all the players' numbers, and the points obtained (in respect to all the rules) by each player are summed.


An example of some rules:

A point is scored for a player for every integer k in the player's color where _______.
* k is next to exactly one integer of opposite parity.
* k = the number of divisors of its position-number.
* k divides the sum of its immediate neighbors.
* k is coprime to the sum of all numbers to its left.

(My examples use basic number theory, but you can involve other branches of mathematics.)

Largest score wins.

Variation:
Play on a grid instead of number line.
Involve the number of the column and/or the number of the row of each number being tested, as well as neighboring numbers, possibly.

Any unforeseen (by me) problems with this game?

Update: (6/30/10) I changed this game so that the rules the players choose are written down BEFORE the numbers are placed along the number line. The rules are kept secret until after the number line is filled with numbers.

Thanks,
Leroy Quet

Thursday, June 10, 2010

Upward-Rightward Game

Here is a game for 2 players, using an n-by-n grid drawn on paper.


Each player has a pen/pencil of a color different than their opponent's color.

First, the players take turns filling in the squares, one square per move in this part of the game.
The first player fills in the lower left square with his/her color. Thereafter, each player fills in the square (with the player's own color) either immediately to the right of or immediately above the last square filled in by the previous player.

This crooked "line" of squares continues until it reaches the upper right square. (So the last few squares filled in this way may be forced to be in the top row or most rightward column.)


Then, in the second part of the game, the players take turns filling in a number of squares. On a single move, a player fills in any empty square (of any color) above or right of any square filled in before in the game. Then, in that same move, the player may fill in any number of empty squares where each square filled in that move is above or right of the square filled in previously by the same player during that move. This crooked line of squares may terminate at any time, but must contain at least one square.


When all n^2 squares of the grid are filled in, the game is over.

Player 1 gets as a score:
sum{k=1 to n} (product of lengths of runs of squares in row k)

Player 2 gets as a score:
sum{k=1 to n} (product of lengths of runs of squares in column k)

A "run" contains consecutive squares (in a specific row or column) all of the same color (either color), bounded on each side by squares of the opposite color (or bounded by the end of the row/column).

Largest score wins.

Here is a completed sample 6-by-6 game:
x x x o X O
x x x o O x
o o o O X o
o X O X o o
o O o o o x
O X o o o o

(Capital X and O are drawn during first part of game. Lower-case letters are drawn during second part of game. Sample game played without strategy.)

Player 1 gets (sum over rows):

3*1*1*1 + 3*2*1 + 4*1*1 + 1*1*1*1*2 + 5*1 + 1*1*4 = 24 points.

Player 2 gets (sum over columns):

2*4 + 2*1*1*1*1 + 2*4 + 3*1*2 + 1*1*1*3 + 1*1*2*1*1 = 29 points.

Player 2 wins.

And of course, players can play with one pen, and "fill" the squares each with a different symbol.

Thanks,
Leroy Quet

Wednesday, June 9, 2010

Deja Vu Divisors

Here is a game for two players.

The players take turns picking any integer from 1,2,3,...., r that has not been picked (by either player) previously in the game, where r is some large integer (such as 1000 or 10000 or more).

Let this picked number be m. (m was picked by the player temporarily called the "provider".)
The same player then picks any integer k where 1 <= k <= m. (k may have been picked earlier any number of times in the game.)

The other player (the "finder") then tries to find any positive integer n not equal to m (n can be arbitrarily large and either picked previously during the game or not) such that:

Both d(m) = d(n) and d(m+k) = d(n+k), where d(j) is the number of divisors of j.

The finder may also provide a proof that no such n (not equal to m) exists.

If the finder either finds an n or proves there is no such n fitting the conditions, then the finder gets a point.

Players then switch who is the provider and who is the finder.

Players play an even predetermined number of moves, and the player with the largest score wins.

Thanks,
Leroy Quet