Crystal Bennett
Joshua Chukwuka
Advisor: Dr. K. Berg
Chomp.










Game Setup
Rules of gameplay
Positions
Theorem 1
Proof of Theorem 1
Approach to 3x5
Approach to 3xn
Theorem 1
Theorem 2
Theorem 3
Game Setup
Standard
game
consists of a grid
of size mxn.
m
= # of rows, n
= # of columns.
Position
on grid
denoted by (i, j)
Game Setup
(1, 1) square marked by “P” for
poison is the least desired square.
 The
 You
lose the game if you take the (1, 1)
piece.
Rules of Gameplay
• Players 1 and 2 alternate removing
squares from the grid.
• Removing an (m, n) piece means taking
(m, n) and (𝑚, 𝑛) where 𝑚 ≥ 𝑚, 𝑛 ≥ 𝑛.
Rules of Game play.
P
P
P
Positions

Let’s consider the position [u, v, w]
where:
 u = # of blocks in row 1.
 v = # of blocks in row 2.
 w = # of blocks in row 3.
If [u, v, w] is a position in Chomp.
 Then 0≤w≤v≤u.
Plan

We will analyze Chomp positions for w =
0, 1, 2.
Theorem 1
 Theorem: The complete list of all
losing positions [u, v, w], where w ∈
0, 1, 2 is:
○ 𝐿 = { [1, 0, 0], [ 2, 2, 1], [3, 1, 1], [u,
u-2, 2] (u ≥ 4), [u, u-1, 0] (u ≥ 1)}.
Proof
 Let
𝐿 be the list from the Theorem
𝑊 consist of all [u, v, w] with w ≤
2 which are not in 𝐿.
 Let
Proof
will show that every position in 𝑊
can be moved to a position in 𝐿.
 We
 We
will show that any move from any
position in 𝐿 moves it into 𝑊.
Proof
• 𝑊 𝑖𝑠 𝑎 𝑠𝑒𝑡 𝑜𝑓 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑠.
• 𝑊 = { [𝑢, 𝑣, 𝑤] |𝑤 ≤ 2, [𝑢, 𝑣, 𝑤] is not
in 𝐿}
• [0, 0, 0] ∈ 𝑊.
Proof
• Notice:
• We will only make moves from a
position [u, v, w] where u ≥ v ≥ 2 to
a position [𝑢, 𝑣, 𝑤] where 𝑢 ≥ 𝑣 ≥ 𝑤.
Proof
• When 𝑤 = 0, the following positions
are in the set 𝐿:
• [1, 0, 0]
• [u, u-1, 0]
Proof
• For [u, v, w], w = 0; the set of
positions in 𝑊 can be described as:
• [u, u, 0]
• [u, u-k, 0], where k > 1.
Proof
• For the position [u, u, 0],
[u, u, 0]→[u, u-1, 0] by taking piece
(2, u).
• For the position [u, v, 0], where v ≤
u – 2,
[u, v, 0]→[v+1, v, 0] by taking the
piece (1, v+2).
Proof
• When 𝑤 = 1, the following positions
are in the set 𝐿:
• [2, 2, 1]
• [3, 1, 1]
Proof
• For [u, v, w], w = 1; the set of positions
in 𝑊 can be described as:
• [u, u, 1] with u ≠ 2.
• [2, 1, 1]
• [u, u-1, 1], u ≥ 3
• [u, 1, 1], u ≥ 4
• [u, u-k, 1] where 2 ≤ k < u - 1, u > 3
Proof
• For the position [u, u, 1],
[u, u, 1]→[2, 2, 1] by taking the (1, 3)
piece.
• For the position [u, u - 1, 1],
[u, u, 1]→[u, u - 1, 0] by taking the
(3, 1) piece.
Proof
• For the position [u, u - k, 1],
[u, u - k, 1]→[2, 2, 1] by taking the
(1, 3) piece.
Proof
• For [u, v, w], w = 2; the set of
positions in 𝑊 can be described as:
• [u, v, 2], where v ≠ u -2.
Proof
• When w = 2, the following positions
are in the set 𝐿:
• [u, u-2, 2]
Proof
• For the position [u, v, 2], where v = u
or v = u-1; [u, v, 2]→[u, u-2, 2] by
taking the (2, u-1) piece.
• For the position [u, v, 2](where v <
u-2), [u, v, 2]→[v+2, v, 2] by taking
the (1, v+3) piece.
Proof
So for [u, v, w], w ≤ 2
• We have shown that every [u, v, w]
in 𝑊can be taken to a position in 𝐿
in one move.
Proof
• Then every winning position for
𝑢, 𝑣, 𝑤 , 𝑤 ≤ 2 can be taken to a
losing position in one move.
Proof
• Now, we will show that for any
position [u, v, w] in 𝐿, we can move
to a position in 𝑊.
Proof
 [1, 0, 0] → [0, 0, 0]
Proof
• Because the position [1, 0, 0] is in 𝐿,
the only move from [1, 0, 0] is the
move:
• [1, 0, 0]→[0, 0, 0]
• And since [0, 0, 0] is in W. The
game is now over.
Proof
 [2, 2, 1] → [2, 2, 0]
 [2, 2, 1] →[1, 1, 2]
 [2, 2, 1] →[2, 0, 0]
 [2, 2, 1] →[1, 1, 1]
Proof
 [u, u - 1, 0] →[u - 1, u - 1, 0]
 [u, u - 1, 0] →[u, u - k, 0]
 [u , u - 1, 0] → [u – k , u - k, 0]
Proof
 [u, u - 2, 1] →[u, u - 2, 0]
 [u, u - 2, 1] →[u, u - k, 1]
 [u, u - 2, 1] →[u-1,u – 2, 0]
 [u, u - 2, 1] →[u - s, u - k, 1]
Proof
 [3, 1, 1] → [3, 1, 0]
 [3, 1, 1] →[3, 0, 0]
 [3, 1, 1] →[2, 1, 1]
Proof
• The first player can force the game
to go from 𝑊 position to 𝐿 position
and back to 𝑊 position.
Theorem 2
For any complete chomp grid of size m by n, there is a
winning strategy. In other words, any one player before the
starting of the game can always be sure of winning.
PROOF:
Because it is a complete m by n grid, it will have a square at
its upper-right hand corner . Either taking that square is a
winning move or it is not.
Case 1: If it is a winning move, player 1 makes it and wins the
game.
Case 2: If it is a losing move then player 2 has a winning
move in response to the first move which he makes and
then wins the game.
However, player 1 could have made player 2 move because it
would require the removal of the square at the upper right
hand corner and he would win the game.

Theorem 2.
P
Theorem 3
For any chomp grid of size m by n (where m=n), that is a m by
m grid, there is a winning strategy and the winning move is to
take the (2,2) square.
If the first player take the (2,2) square we would have 1 row and
1 column. Excluding the poisoned square,, there is (m-1)
columns and rows.
Then, the second player can only chomp from the bottom
squares or the left squares. Since both are equal, it does not
matter which squares player 2 takes from. His move always
makes them unequal and the winning strategy for player 1 is
to make his next move such that the 1nth row and
column(without poison) have equal number of squares.
This way, player 1 takes the last square from either the row or the
column and then player 2 is left with the poisoned square
and then loses.

Theorem 3
P
Theorem 4
For all stair case kind of chomp grid of step size 1, that
is any row has one square more than the row above it
there is a winning strategy.(except the first staircase)
Case 1: For the staircase with three rows, the winning
move is to chomp the (2,2) square.
Case 2: For any staircase, the winning move is to chomp
any square that makes the staircase a 2 by n with a
winning move already made.
PROOF:
Case 1 is true by theorem 3.
Case 2 is true by theorem 1.
 Therefore, there is always a winning strategy for all
staircase(excluding the (1,1))

Theorem 4 contd.
Approach to 3x5



Start at the 2x2 Square finding the winning positions.
Add one square at a time to the grid solving for the
winning moves up until the 3x5 grid.
Analyze different types of staircase grids where the
bottom, middle, and top row all have different
lengths.
P
Approach to 3x5


The position (3,4) is a winning position for the 3x5
grid.
How do I prove this.
 Analyze the winning moves for all grids up to the
3x5 .
 Identify any patterns, or relations that may exist.
𝑊1
P
Approach to 3x5
𝑊
1
P
*
*
*
*
Approach to 3x5
𝑊
1
P
*
*
*
Approach to 3x5
𝑳
1
P
*
*
Approach to 3x5
𝑊1
P
*
3x6 Chomp Grid.
Losing Position.
W1
P
3x6 Chomp Grid.
Possible Cases.
CASE 10 CASE 11 CASE 12
CASE 7
CASE 8
CASE 9
CASE 1
CASE 2
CASE 3
CASE 4
CASE 5
CASE 6
3x6 Chomp Grid.
Case 1
• Case 1 is true by the Rules of Game play
3x6 Chomp Grid.
Case 2
P
• Case 2 is true by Theorem 2
3x6 Chomp Grid.
Case 3
P
• Case 3 is true by Theorem
2.
3x6 Chomp Grid.
Case 4
P
Case 4 is true by Theorem 2.
3x6 Chomp Grid.
Case 5
P
• Case 5 is true by
Theorem 1.
3x6 Chomp Grid.
Case 6
P
Case 6 is true by Theorem 1.
3x6 Chomp Grid.
Case 7
P
Case 7 is true by Theorem 1.
3x6 Chomp Grid.
Case 8.
P
Case 8 is true by the square chomp
Theorem 3.
3x6 Chomp grid.
Case 9
P
3x6 Chomp Grid.
Case 10
P
Case 10 is true by Theorem 3.(2xn)
3x6 Chomp Grid.
Case 11
P
Case 11 is true by Theorem C.
3x6 Chomp Grid
Case 12
P
Case 12 is true by Theorem c
Approach to 3xn




Use 2x2 up to 3x5 findings to solve larger grids at
finding the winning positions.
Add one square at a time to the grid solving for the
winning moves.
Analyze different types of staircase grids where the
bottom, middle, and top row all have different
lengths.
Record winning positions to see if we can find a
relationship between the winning positions.
3x7 Chomp Grid.
3x7 Chomp Grid.
CASE 15
CASE 16
CASE 17
CASE 18
CASE 8
CASE 9
CASE 10
CASE 11
CASE 12
CASE 13
CASE 14
CASE 1
CASE 2
CASE 3
CASE 4
CASE 5
CASE 6
CASE 7
Q U E S T I O N S
COMMENTS
CONCERNS