## Monday, November 24, 2008

## Tuesday, November 18, 2008

### f(g(x)) = g(f(x)) (a game)

This game is for 2 players.

First, the players each secretly come up with two mathematical functions (two functions per player), f(x) and g(x).

The functions, f(x) and g(x), are defined and real for all real x, and they are continuous.

Next, the players take turns each being the "proposer" and the "solver".

First in a round the proposer reveals his/her functions.

The solver tries to determine (within a given time period) whether

f(g(x)) = g(f(x))

either for (1 of 3 choices) no real x's, a finite number of real x's, or an infinite number of real x's.

If the solver can't do this in a specific finite period of time, then the proposer gets a point. Otherwise the solver gets a point.

The players then switch who is the proposer and who is the solver.

(So, with only the two rounds, this is a low-scoring game.)

Example: f(x) = sin(x). g(x) = e^(x+1).

So, f(g(x)) = sin(e^(x+1)). g(f(x)) = e^(sin(x)+1).

The number of x's where sin(e^(x+1)) = e^(sin(x)+1) is precisely the number of pairs of integers (m,n) such that:

e^(1- pi/2 +2*n*pi) = pi/2 + 2*pi*m (I think).

I don't personally know if this equation is solvable at all for any integers m and n, let alone if there are a finite number of pairs (m,n) or an infinite number of pairs of integers m and n. So, if I was the solver and if the proposer proposed f(x) = sin(x) and g(x) = e^(x+1), then the proposer would get a point that round, and I would get no points that round.

First, the players each secretly come up with two mathematical functions (two functions per player), f(x) and g(x).

The functions, f(x) and g(x), are defined and real for all real x, and they are continuous.

Next, the players take turns each being the "proposer" and the "solver".

First in a round the proposer reveals his/her functions.

The solver tries to determine (within a given time period) whether

f(g(x)) = g(f(x))

either for (1 of 3 choices) no real x's, a finite number of real x's, or an infinite number of real x's.

If the solver can't do this in a specific finite period of time, then the proposer gets a point. Otherwise the solver gets a point.

The players then switch who is the proposer and who is the solver.

(So, with only the two rounds, this is a low-scoring game.)

Example: f(x) = sin(x). g(x) = e^(x+1).

So, f(g(x)) = sin(e^(x+1)). g(f(x)) = e^(sin(x)+1).

The number of x's where sin(e^(x+1)) = e^(sin(x)+1) is precisely the number of pairs of integers (m,n) such that:

e^(1- pi/2 +2*n*pi) = pi/2 + 2*pi*m (I think).

I don't personally know if this equation is solvable at all for any integers m and n, let alone if there are a finite number of pairs (m,n) or an infinite number of pairs of integers m and n. So, if I was the solver and if the proposer proposed f(x) = sin(x) and g(x) = e^(x+1), then the proposer would get a point that round, and I would get no points that round.

## Tuesday, November 11, 2008

### Grid-Squiggly Game

This game is played on an n-by-n grid (n by n squares, {n+1} by {n+1} lines) that is lightly drawn on paper. The game is for 2 or more players.

First, players each secretly pick a positive integer m between 1 and n^2. (See below.)

The first player to move draws a line segment, one grid-square in length, along any vertical or horizontal line of the grid. (But don't draw along the border of the grid.)

Players then take turns drawing a line-segment each turn, where the line-segment is one grid-square in length, and goes from any vertex with a line-segment drawn to it (by any player) to any adjacent vertex that does not yet have a line segment drawn to it. (The drawn-to vertex is immediately above, below, right of, left of the drawn-from vertex.)

No line segments go along the border of the grid, but line segments can connect to vertices along the border of the grid.

Players continue to draw segments until there is no place they can draw them. (A total of n^2 + 2n - 4 segments will be drawn.)

Next, with a pencil of a color different that their opponents' pencil colors, each player takes turns (completely) filling in sections of the grid, one section each move. Each "section" is bounded by the lines the players drew and by the border of the grid.

When the whole grid has been colored in, count the number of squares filled in by each player.

Players then reveal the numbers (m) they picked at the game's beginning.

The player whose number of squares filled in is closest to the number they picked at the game's beginning (m) wins.

Thanks,

Leroy Quet

First, players each secretly pick a positive integer m between 1 and n^2. (See below.)

The first player to move draws a line segment, one grid-square in length, along any vertical or horizontal line of the grid. (But don't draw along the border of the grid.)

Players then take turns drawing a line-segment each turn, where the line-segment is one grid-square in length, and goes from any vertex with a line-segment drawn to it (by any player) to any adjacent vertex that does not yet have a line segment drawn to it. (The drawn-to vertex is immediately above, below, right of, left of the drawn-from vertex.)

No line segments go along the border of the grid, but line segments can connect to vertices along the border of the grid.

Players continue to draw segments until there is no place they can draw them. (A total of n^2 + 2n - 4 segments will be drawn.)

Next, with a pencil of a color different that their opponents' pencil colors, each player takes turns (completely) filling in sections of the grid, one section each move. Each "section" is bounded by the lines the players drew and by the border of the grid.

When the whole grid has been colored in, count the number of squares filled in by each player.

Players then reveal the numbers (m) they picked at the game's beginning.

The player whose number of squares filled in is closest to the number they picked at the game's beginning (m) wins.

Thanks,

Leroy Quet

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