Mathematics and GamesMathematics and Games Parker Glynn-Adey University of Toronto...

Mathematics and Games

Parker Glynn-Adey

University of Toronto

parker.glynn.adey@utoronto.ca

March 23, 2014

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 1 / 28

Outline

1 What are games?

2 The 15 Game

3 Nim

4 A Safari in the Land of Games

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 2 / 28

This is not a game.(Although, there is a lot of mathematics in Angry Birds.)

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 3 / 28

This is not a game.

(Although, there is a lot of mathematics in Angry Birds.)

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 3 / 28

This is not a game.(Although, there is a lot of mathematics in Angry Birds.)

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 3 / 28

Backgammon (∼1100AD) and Poker (∼1850AD).

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 4 / 28

Backgammon (∼1100AD) and Poker (∼1850AD).

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 4 / 28

Chess (∼1400AD) and Go (∼2000BC)

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 5 / 28

Chess (∼1400AD) and Go (∼2000BC)

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 5 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.

Goal: To collect up numbers adding up to 15.Setup: Write the numbers one through nine on a piece of paper.Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:2 3 4 5 6 8 9 (L = ,R = )

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.Goal: To collect up numbers adding up to 15.

Setup: Write the numbers one through nine on a piece of paper.Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:2 3 4 5 6 8 9 (L = ,R = )

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.Goal: To collect up numbers adding up to 15.Setup: Write the numbers one through nine on a piece of paper.

Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:2 3 4 5 6 8 9 (L = ,R = )

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.Goal: To collect up numbers adding up to 15.Setup: Write the numbers one through nine on a piece of paper.Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:2 3 4 5 6 8 9 (L = ,R = )

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.Goal: To collect up numbers adding up to 15.Setup: Write the numbers one through nine on a piece of paper.Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:2 3 4 5 6 8 9 (L = ,R = )

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.Goal: To collect up numbers adding up to 15.Setup: Write the numbers one through nine on a piece of paper.Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:2 3 4 5 6 8 9 (L = ,R = )

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.Goal: To collect up numbers adding up to 15.Setup: Write the numbers one through nine on a piece of paper.Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:2 3 4 5 6 8 9 (L = ,R = )

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.Goal: To collect up numbers adding up to 15.Setup: Write the numbers one through nine on a piece of paper.Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:2 3 4 5 6 8 9 (L = ,R = )

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.Goal: To collect up numbers adding up to 15.Setup: Write the numbers one through nine on a piece of paper.Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:1 2 3 4 5 6 7 8 9 (L = 0,R = 0)

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.Goal: To collect up numbers adding up to 15.Setup: Write the numbers one through nine on a piece of paper.Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:1 2 3 4 5 6 7 8 9 (L = 0,R = 0)

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.Goal: To collect up numbers adding up to 15.Setup: Write the numbers one through nine on a piece of paper.Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:1 2 3 4 5 6 7 8 9 (L = 0,R = 0)

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.Goal: To collect up numbers adding up to 15.Setup: Write the numbers one through nine on a piece of paper.Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:

�A1 2 3 4 5 6 7 8 9 (L = 0,R = 1)

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.Goal: To collect up numbers adding up to 15.Setup: Write the numbers one through nine on a piece of paper.Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:

�A1 2 3 4 5 6 �A7 8 9 (L = 7,R = 1)

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Rules of The 15 Game (A Toy Game)

Players: Two.Goal: To collect up numbers adding up to 15.Setup: Write the numbers one through nine on a piece of paper.Play:

Players alternate taking turns.

Both players start with total zero.

On your turn: cross out a number, and add it to your total.

If your total is 15, you win.

Example:

�A1 2 3 4 5 6 �A7 8 9 (L = 7,R = 1)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 6 / 28

Observations:

Sometimes no one wins.

5 is a good opening move.

The Fifteen Game feels oddly familiar.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 7 / 28

Observations:

Sometimes no one wins.

5 is a good opening move.

The Fifteen Game feels oddly familiar.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 7 / 28

Observations:

Sometimes no one wins.

5 is a good opening move.

The Fifteen Game feels oddly familiar.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 7 / 28

Observations:

Sometimes no one wins.

5 is a good opening move.

The Fifteen Game feels oddly familiar.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 7 / 28

8 1 63 5 74 9 2

The 15 Game is Tic-Tac-Toe.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 8 / 28

8 1

6

3 5 74 9 2

The 15 Game is Tic-Tac-Toe.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 8 / 28

8 1

6

3

5

74 9 2

The 15 Game is Tic-Tac-Toe.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 8 / 28

8 1

6

3

5

74 9

2

The 15 Game is Tic-Tac-Toe.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 8 / 28

8 1

6

3

5 7

4 9

2

The 15 Game is Tic-Tac-Toe.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 8 / 28

8 1

63 5 7

4 9

2

The 15 Game is Tic-Tac-Toe.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 8 / 28

8 1

63 5 7

4

9 2

The 15 Game is Tic-Tac-Toe.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 8 / 28

8

1 63 5 7

4

9 2

The 15 Game is Tic-Tac-Toe.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 8 / 28

8 1 63 5 7

4

9 2

The 15 Game is Tic-Tac-Toe.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 8 / 28

8 1 63 5 74 9 2

The 15 Game is Tic-Tac-Toe.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 8 / 28

8 1 63 5 74 9 2

The 15 Game is Tic-Tac-Toe.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 8 / 28

Nim

Nim was the first game to be exhaustively analyzed mathematically.

In 1901, Charles Bouton published a solution to the game.This result started the field of mathematical game theory.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 9 / 28

Nim

Nim was the first game to be exhaustively analyzed mathematically.In 1901, Charles Bouton published a solution to the game.

This result started the field of mathematical game theory.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 9 / 28

Nim

Nim was the first game to be exhaustively analyzed mathematically.In 1901, Charles Bouton published a solution to the game.This result started the field of mathematical game theory.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 9 / 28

Rules of Nim

Players: Two.

Goal: To be the last player to pick up a stone.Setup: Lay out heaps of n1, n2, . . . , nk stones. (ni ≥ 0)Play:

Players alternate taking turns.

A move consists of removing at least one stone from a heap.

If you can’t move, you lose.

Example:

(1, 2)R (1, 1)

L (0, 1)

R (0, 0) (Left loses.)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 10 / 28

Rules of Nim

Players: Two.Goal: To be the last player to pick up a stone.

Setup: Lay out heaps of n1, n2, . . . , nk stones. (ni ≥ 0)Play:

Players alternate taking turns.

A move consists of removing at least one stone from a heap.

If you can’t move, you lose.

Example:

(1, 2)R (1, 1)

L (0, 1)

R (0, 0) (Left loses.)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 10 / 28

Rules of Nim

Players: Two.Goal: To be the last player to pick up a stone.Setup: Lay out heaps of n1, n2, . . . , nk stones. (ni ≥ 0)

Play:

Players alternate taking turns.

A move consists of removing at least one stone from a heap.

If you can’t move, you lose.

Example:

(1, 2)R (1, 1)

L (0, 1)

R (0, 0) (Left loses.)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 10 / 28

Rules of Nim

Players: Two.Goal: To be the last player to pick up a stone.Setup: Lay out heaps of n1, n2, . . . , nk stones. (ni ≥ 0)Play:

Players alternate taking turns.

A move consists of removing at least one stone from a heap.

If you can’t move, you lose.

Example:

(1, 2)R (1, 1)

L (0, 1)

R (0, 0) (Left loses.)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 10 / 28

Rules of Nim

Players: Two.Goal: To be the last player to pick up a stone.Setup: Lay out heaps of n1, n2, . . . , nk stones. (ni ≥ 0)Play:

Players alternate taking turns.

A move consists of removing at least one stone from a heap.

If you can’t move, you lose.

Example:

(1, 2)R (1, 1)

L (0, 1)

R (0, 0) (Left loses.)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 10 / 28

Rules of Nim

Players: Two.Goal: To be the last player to pick up a stone.Setup: Lay out heaps of n1, n2, . . . , nk stones. (ni ≥ 0)Play:

Players alternate taking turns.

A move consists of removing at least one stone from a heap.

If you can’t move, you lose.

Example:

(1, 2)R (1, 1)

L (0, 1)

R (0, 0) (Left loses.)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 10 / 28

Rules of Nim

Players: Two.Goal: To be the last player to pick up a stone.Setup: Lay out heaps of n1, n2, . . . , nk stones. (ni ≥ 0)Play:

Players alternate taking turns.

A move consists of removing at least one stone from a heap.

If you can’t move, you lose.

Example:

(1, 2)R (1, 1)

L (0, 1)

R (0, 0) (Left loses.)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 10 / 28

Rules of Nim

Players: Two.Goal: To be the last player to pick up a stone.Setup: Lay out heaps of n1, n2, . . . , nk stones. (ni ≥ 0)Play:

Players alternate taking turns.

A move consists of removing at least one stone from a heap.

If you can’t move, you lose.

Example:

(1, 2)

R (1, 1)

L (0, 1)

R (0, 0) (Left loses.)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 10 / 28

Rules of Nim

Players: Two.Goal: To be the last player to pick up a stone.Setup: Lay out heaps of n1, n2, . . . , nk stones. (ni ≥ 0)Play:

Players alternate taking turns.

A move consists of removing at least one stone from a heap.

If you can’t move, you lose.

Example:

(1, 2)R (1, 1)

L (0, 1)

R (0, 0) (Left loses.)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 10 / 28

Rules of Nim

Players: Two.Goal: To be the last player to pick up a stone.Setup: Lay out heaps of n1, n2, . . . , nk stones. (ni ≥ 0)Play:

Players alternate taking turns.

A move consists of removing at least one stone from a heap.

If you can’t move, you lose.

Example:

(1, 2)R (1, 1)

L (0, 1)

R (0, 0) (Left loses.)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 10 / 28

Rules of Nim

Players: Two.Goal: To be the last player to pick up a stone.Setup: Lay out heaps of n1, n2, . . . , nk stones. (ni ≥ 0)Play:

Players alternate taking turns.

A move consists of removing at least one stone from a heap.

If you can’t move, you lose.

Example:

(1, 2)R (1, 1)

L (0, 1)

R (0, 0)

(Left loses.)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 10 / 28

Rules of Nim

Players: Two.Goal: To be the last player to pick up a stone.Setup: Lay out heaps of n1, n2, . . . , nk stones. (ni ≥ 0)Play:

Players alternate taking turns.

A move consists of removing at least one stone from a heap.

If you can’t move, you lose.

Example:

(1, 2)R (1, 1)

L (0, 1)

R (0, 0) (Left loses.)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 10 / 28

Rules of Nim

Players: Two.Goal: To be the last player to pick up a stone.Setup: Lay out heaps of n1, n2, . . . , nk stones. (ni ≥ 0)Play:

Players alternate taking turns.

A move consists of removing at least one stone from a heap.

If you can’t move, you lose.

Example:

(1, 2)R (1, 1)

L (0, 1)

R (0, 0) (Left loses.)

Let’s play!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 10 / 28

Quiz!

Play and win: (10, 11).

Is (17, 0) a good position?Is (2, 2, 3, 3, 5) a good position?

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 11 / 28

Quiz!

Play and win: (10, 11).Is (17, 0) a good position?

Is (2, 2, 3, 3, 5) a good position?

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 11 / 28

Quiz!

Play and win: (10, 11).Is (17, 0) a good position?Is (2, 2, 3, 3, 5) a good position?

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 11 / 28

Observations:

Nim is tricky!

There are lots of options.

Empty heaps don’t matter.

A single heap is a good position.

Equal heaps ‘cancel out’.

Let’s solve Nim!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 12 / 28

Observations:

Nim is tricky!

There are lots of options.

Empty heaps don’t matter.

A single heap is a good position.

Equal heaps ‘cancel out’.

Let’s solve Nim!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 12 / 28

Observations:

Nim is tricky!

There are lots of options.

Empty heaps don’t matter.

A single heap is a good position.

Equal heaps ‘cancel out’.

Let’s solve Nim!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 12 / 28

Observations:

Nim is tricky!

There are lots of options.

Empty heaps don’t matter.

A single heap is a good position.

Equal heaps ‘cancel out’.

Let’s solve Nim!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 12 / 28

Observations:

Nim is tricky!

There are lots of options.

Empty heaps don’t matter.

A single heap is a good position.

Equal heaps ‘cancel out’.

Let’s solve Nim!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 12 / 28

Observations:

Nim is tricky!

There are lots of options.

Empty heaps don’t matter.

A single heap is a good position.

Equal heaps ‘cancel out’.

Let’s solve Nim!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 12 / 28

Observations:

Nim is tricky!

There are lots of options.

Empty heaps don’t matter.

A single heap is a good position.

Equal heaps ‘cancel out’.

Let’s solve Nim!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 12 / 28

A quick review of numbers

2014 = 2 · 1000 +

0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 +

1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 +

4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1

= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 +

0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 +

1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 +

4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

=

(2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

=

1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 +

512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 +

256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 +

128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 +

64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 +

16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 +

8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 +

4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 +

2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

=

210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 +

29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 +

28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 +

27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 +

26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 +

24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 +

23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 +

22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 +

21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

A quick review of numbers

2014 = 2 · 1000 + 0 · 100 + 1 · 10 + 4 · 1= 2 · 103 + 0 · 102 + 1 · 101 + 4 · 100

= (2, 0, 1, 4)10

= 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2

= 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22 + 21

= (1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0)2

Theorem

Every number n can be written as a sum of powers of two in a unique way.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 13 / 28

Nim addition

We define the following operation on binary digits:

⊕ 0 1

0 0 11 1 0

Definition

For a pair of numbers n and m, write:

n = (nk , nk−1, . . . , n0)2 m = (mk ,mk−1, . . . ,m0)2

We define their Nim-sum:

n ⊕m = (nk ⊕mk , nk−1 ⊕mk−1, . . . , n0 ⊕m0)2

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 14 / 28

Nim addition

We define the following operation on binary digits:

⊕ 0 1

0 0 11 1 0

Definition

For a pair of numbers n and m, write:

n = (nk , nk−1, . . . , n0)2 m = (mk ,mk−1, . . . ,m0)2

We define their Nim-sum:

n ⊕m = (nk ⊕mk , nk−1 ⊕mk−1, . . . , n0 ⊕m0)2

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 14 / 28

Nim addition

We define the following operation on binary digits:

⊕ 0 1

0 0 11 1 0

Definition

For a pair of numbers n and m, write:

n = (nk , nk−1, . . . , n0)2 m = (mk ,mk−1, . . . ,m0)2

We define their Nim-sum:

n ⊕m = (nk ⊕mk , nk−1 ⊕mk−1, . . . , n0 ⊕m0)2

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 14 / 28

Nim addition

We define the following operation on binary digits:

⊕ 0 1

0 0 11 1 0

Definition

For a pair of numbers n and m, write:

n = (nk , nk−1, . . . , n0)2 m = (mk ,mk−1, . . . ,m0)2

We define their Nim-sum:

n ⊕m = (nk ⊕mk , nk−1 ⊕mk−1, . . . , n0 ⊕m0)2

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 14 / 28

Nim addition

We define the following operation on binary digits:

⊕ 0 1

0 0 11 1 0

Definition

For a pair of numbers n and m, write:

n = (nk , nk−1, . . . , n0)2 m = (mk ,mk−1, . . . ,m0)2

We define their Nim-sum:

n ⊕m = (nk ⊕mk , nk−1 ⊕mk−1, . . . , n0 ⊕m0)2

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 14 / 28

Nim addition

We define the following operation on binary digits:

⊕ 0 1

0 0 11 1 0

Definition

For a pair of numbers n and m, write:

n = (nk , nk−1, . . . , n0)2 m = (mk ,mk−1, . . . ,m0)2

We define their Nim-sum:

n ⊕m = (nk ⊕mk , nk−1 ⊕mk−1, . . . , n0 ⊕m0)2

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 14 / 28

Examples of Nim Addition

3⊕ 2 =

(2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

=

(1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1,

1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 =

(4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1,

0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0,

1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

Examples of Nim Addition

3⊕ 2 = (2 + 1)⊕ (2)

= (1, 1)2 ⊕ (1, 0)2

= (1⊕ 1, 1⊕ 0)2

= (0, 1)2

= 1

5⊕ 5 = (4 + 1)⊕ (4 + 1)

= (1, 0, 1)2 ⊕ (1, 0, 1)2

= (1⊕ 1, 0⊕ 0, 1⊕ 1)2

= (0, 0, 0)2

= 0

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 15 / 28

5⊕ 7 =

(1 + 4)⊕ (1 + 2 + 4)

= (1⊕ 1)⊕ (0⊕ 2)

⊕(4⊕ 4)

= 0⊕ 2⊕ 0 = 2

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 16 / 28

5⊕ 7 = (1 + 4)⊕ (1 + 2 + 4)

= (1⊕ 1)⊕ (0⊕ 2)

⊕(4⊕ 4)

= 0⊕ 2⊕ 0 = 2

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 16 / 28

5⊕ 7 = (1 + 4)⊕ (1 + 2 + 4)

=

(1⊕ 1)⊕ (0⊕ 2)

⊕(4⊕ 4)

= 0⊕ 2⊕ 0 = 2

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 16 / 28

5⊕ 7 = (1 + 4)⊕ (1 + 2 + 4)

= (1⊕ 1)⊕ (0⊕ 2)

⊕(4⊕ 4)

= 0⊕ 2⊕ 0 = 2

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 16 / 28

5⊕ 7 = (1 + 4)⊕ (1 + 2 + 4)

= (1⊕ 1)⊕ (0⊕ 2)

⊕(4⊕ 4)

=

0⊕ 2⊕ 0 = 2

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 16 / 28

5⊕ 7 = (1 + 4)⊕ (1 + 2 + 4)

= (1⊕ 1)⊕ (0⊕ 2)

⊕(4⊕ 4)

= 0⊕ 2⊕ 0 = 2

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 16 / 28

The Winning Nim Strategy

For a Nim position (n1, . . . , nk) let:

V (n1, . . . , nk) = n1 ⊕ n2 ⊕ · · · ⊕ nk

Examples:

V (1, 1) = 1⊕ 1 = 0.

V (1, 2) = 1⊕ 2 = 3.

V (4, 5) = 4⊕ 5 = 1.

V (1, 2, 3) = 1⊕ 2⊕ 3 = 1⊕ 2⊕ (1 + 2) = 0.

Definition

A position is good if V 6= 0. A position is bad if V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 17 / 28

The Winning Nim Strategy

For a Nim position (n1, . . . , nk) let:

V (n1, . . . , nk) = n1 ⊕ n2 ⊕ · · · ⊕ nk

Examples:

V (1, 1) = 1⊕ 1 = 0.

V (1, 2) = 1⊕ 2 = 3.

V (4, 5) = 4⊕ 5 = 1.

V (1, 2, 3) = 1⊕ 2⊕ 3 = 1⊕ 2⊕ (1 + 2) = 0.

Definition

A position is good if V 6= 0. A position is bad if V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 17 / 28

The Winning Nim Strategy

For a Nim position (n1, . . . , nk) let:

V (n1, . . . , nk) = n1 ⊕ n2 ⊕ · · · ⊕ nk

Examples:

V (1, 1) = 1⊕ 1 = 0.

V (1, 2) = 1⊕ 2 = 3.

V (4, 5) = 4⊕ 5 = 1.

V (1, 2, 3) = 1⊕ 2⊕ 3 = 1⊕ 2⊕ (1 + 2) = 0.

Definition

A position is good if V 6= 0. A position is bad if V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 17 / 28

The Winning Nim Strategy

For a Nim position (n1, . . . , nk) let:

V (n1, . . . , nk) = n1 ⊕ n2 ⊕ · · · ⊕ nk

Examples:

V (1, 1) = 1⊕ 1 = 0.

V (1, 2) = 1⊕ 2 = 3.

V (4, 5) = 4⊕ 5 = 1.

V (1, 2, 3) = 1⊕ 2⊕ 3 = 1⊕ 2⊕ (1 + 2) = 0.

Definition

A position is good if V 6= 0. A position is bad if V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 17 / 28

The Winning Nim Strategy

For a Nim position (n1, . . . , nk) let:

V (n1, . . . , nk) = n1 ⊕ n2 ⊕ · · · ⊕ nk

Examples:

V (1, 1) = 1⊕ 1 = 0.

V (1, 2) = 1⊕ 2

= 3.

V (4, 5) = 4⊕ 5 = 1.

V (1, 2, 3) = 1⊕ 2⊕ 3 = 1⊕ 2⊕ (1 + 2) = 0.

Definition

A position is good if V 6= 0. A position is bad if V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 17 / 28

The Winning Nim Strategy

For a Nim position (n1, . . . , nk) let:

V (n1, . . . , nk) = n1 ⊕ n2 ⊕ · · · ⊕ nk

Examples:

V (1, 1) = 1⊕ 1 = 0.

V (1, 2) = 1⊕ 2 = 3.

V (4, 5) = 4⊕ 5 = 1.

V (1, 2, 3) = 1⊕ 2⊕ 3 = 1⊕ 2⊕ (1 + 2) = 0.

Definition

A position is good if V 6= 0. A position is bad if V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 17 / 28

The Winning Nim Strategy

For a Nim position (n1, . . . , nk) let:

V (n1, . . . , nk) = n1 ⊕ n2 ⊕ · · · ⊕ nk

Examples:

V (1, 1) = 1⊕ 1 = 0.

V (1, 2) = 1⊕ 2 = 3.

V (4, 5) = 4⊕ 5

= 1.

V (1, 2, 3) = 1⊕ 2⊕ 3 = 1⊕ 2⊕ (1 + 2) = 0.

Definition

A position is good if V 6= 0. A position is bad if V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 17 / 28

The Winning Nim Strategy

For a Nim position (n1, . . . , nk) let:

V (n1, . . . , nk) = n1 ⊕ n2 ⊕ · · · ⊕ nk

Examples:

V (1, 1) = 1⊕ 1 = 0.

V (1, 2) = 1⊕ 2 = 3.

V (4, 5) = 4⊕ 5 = 1.

V (1, 2, 3) = 1⊕ 2⊕ 3 = 1⊕ 2⊕ (1 + 2) = 0.

Definition

A position is good if V 6= 0. A position is bad if V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 17 / 28

The Winning Nim Strategy

For a Nim position (n1, . . . , nk) let:

V (n1, . . . , nk) = n1 ⊕ n2 ⊕ · · · ⊕ nk

Examples:

V (1, 1) = 1⊕ 1 = 0.

V (1, 2) = 1⊕ 2 = 3.

V (4, 5) = 4⊕ 5 = 1.

V (1, 2, 3) = 1⊕ 2⊕ 3

= 1⊕ 2⊕ (1 + 2) = 0.

Definition

A position is good if V 6= 0. A position is bad if V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 17 / 28

The Winning Nim Strategy

For a Nim position (n1, . . . , nk) let:

V (n1, . . . , nk) = n1 ⊕ n2 ⊕ · · · ⊕ nk

Examples:

V (1, 1) = 1⊕ 1 = 0.

V (1, 2) = 1⊕ 2 = 3.

V (4, 5) = 4⊕ 5 = 1.

V (1, 2, 3) = 1⊕ 2⊕ 3 = 1⊕ 2⊕ (1 + 2)

= 0.

Definition

A position is good if V 6= 0. A position is bad if V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 17 / 28

The Winning Nim Strategy

For a Nim position (n1, . . . , nk) let:

V (n1, . . . , nk) = n1 ⊕ n2 ⊕ · · · ⊕ nk

Examples:

V (1, 1) = 1⊕ 1 = 0.

V (1, 2) = 1⊕ 2 = 3.

V (4, 5) = 4⊕ 5 = 1.

V (1, 2, 3) = 1⊕ 2⊕ 3 = 1⊕ 2⊕ (1 + 2) = 0.

Definition

A position is good if V 6= 0. A position is bad if V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 17 / 28

The Winning Nim Strategy

For a Nim position (n1, . . . , nk) let:

V (n1, . . . , nk) = n1 ⊕ n2 ⊕ · · · ⊕ nk

Examples:

V (1, 1) = 1⊕ 1 = 0.

V (1, 2) = 1⊕ 2 = 3.

V (4, 5) = 4⊕ 5 = 1.

V (1, 2, 3) = 1⊕ 2⊕ 3 = 1⊕ 2⊕ (1 + 2) = 0.

Definition

A position is good if V 6= 0.

A position is bad if V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 17 / 28

The Winning Nim Strategy

For a Nim position (n1, . . . , nk) let:

V (n1, . . . , nk) = n1 ⊕ n2 ⊕ · · · ⊕ nk

Examples:

V (1, 1) = 1⊕ 1 = 0.

V (1, 2) = 1⊕ 2 = 3.

V (4, 5) = 4⊕ 5 = 1.

V (1, 2, 3) = 1⊕ 2⊕ 3 = 1⊕ 2⊕ (1 + 2) = 0.

Definition

A position is good if V 6= 0. A position is bad if V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 17 / 28

The Winning Nim Strategy

If V (n1, . . . , nk) = 0, and it is your turn, you will lose.

If V (n1, . . . , nk) 6= 0 then there is a move, ni n′i , so thatV (n1, . . . , n

′i , . . . , nk) = 0, and your opponent will lose.

That is: Bad positions lose. Good positions win.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 18 / 28

The Winning Nim Strategy

If V (n1, . . . , nk) = 0, and it is your turn, you will lose.

If V (n1, . . . , nk) 6= 0 then there is a move, ni n′i , so thatV (n1, . . . , n

′i , . . . , nk) = 0, and your opponent will lose.

That is: Bad positions lose. Good positions win.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 18 / 28

The Winning Nim Strategy

If V (n1, . . . , nk) = 0, and it is your turn, you will lose.

If V (n1, . . . , nk) 6= 0 then there is a move, ni n′i , so thatV (n1, . . . , n

′i , . . . , nk) = 0, and your opponent will lose.

That is: Bad positions lose. Good positions win.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 18 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ =

V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′

⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕

0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

=

V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕

(V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

=

(V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

=

(n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V .

Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni .

Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni .

V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

If V = 0, and there is a legal move, then any ni n′i will makeV ′ 6= 0.

If V 6= 0 then there is a legal move, ni n′i which will make V ′ = 0.

We write V = V (n1, . . . , nk) and V ′ = V (n1, . . . , n′i , . . . , nk).

We compute:

V ′ = V ′ ⊕ 0

= V ′ ⊕ (V ⊕ V )

= (V ′ ⊕ V )⊕ V

= (n′i ⊕ ni )⊕ V

We then get our two claims:

If V = 0 and ni 6= n′i then V ′ 6= 0.

If V 6= 0 then take ni to have the same largest digit as V . Letn′i = V ⊕ ni . Note: n′i ≤ ni . V ′ = ((V ⊕ ni )⊕ ni )⊕ V = 0.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 19 / 28

(1, 2)V=3

R (1, 1)V=0

L (0, 1)V=1

R (0, 0)V=0 (Left loses.)

What about (15, 7, 9, 11) ?

15 = 1 + 2 + 4 + 87 = 1 + 2 + 49 = 1 + 8

⊕ 11 = 1 + 2 + 8

10 = 2 + 8

We must remove: 2 + 8 = 10 from either: 15, 10, 11.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 20 / 28

(1, 2)V=3R (1, 1)V=0

L (0, 1)V=1

R (0, 0)V=0 (Left loses.)

What about (15, 7, 9, 11) ?

15 = 1 + 2 + 4 + 87 = 1 + 2 + 49 = 1 + 8

⊕ 11 = 1 + 2 + 8

10 = 2 + 8

We must remove: 2 + 8 = 10 from either: 15, 10, 11.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 20 / 28

(1, 2)V=3R (1, 1)V=0

L (0, 1)V=1

R (0, 0)V=0 (Left loses.)

What about (15, 7, 9, 11) ?

15 = 1 + 2 + 4 + 87 = 1 + 2 + 49 = 1 + 8

⊕ 11 = 1 + 2 + 8

10 = 2 + 8

We must remove: 2 + 8 = 10 from either: 15, 10, 11.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 20 / 28

(1, 2)V=3R (1, 1)V=0

L (0, 1)V=1

R (0, 0)V=0

(Left loses.)

What about (15, 7, 9, 11) ?

15 = 1 + 2 + 4 + 87 = 1 + 2 + 49 = 1 + 8

⊕ 11 = 1 + 2 + 8

10 = 2 + 8

We must remove: 2 + 8 = 10 from either: 15, 10, 11.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 20 / 28

(1, 2)V=3R (1, 1)V=0

L (0, 1)V=1

R (0, 0)V=0 (Left loses.)

What about (15, 7, 9, 11) ?

15 = 1 + 2 + 4 + 87 = 1 + 2 + 49 = 1 + 8

⊕ 11 = 1 + 2 + 8

10 = 2 + 8

We must remove: 2 + 8 = 10 from either: 15, 10, 11.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 20 / 28

(1, 2)V=3R (1, 1)V=0

L (0, 1)V=1

R (0, 0)V=0 (Left loses.)

What about (15, 7, 9, 11) ?

15 = 1 + 2 + 4 + 87 = 1 + 2 + 49 = 1 + 8

⊕ 11 = 1 + 2 + 8

10 = 2 + 8

We must remove: 2 + 8 = 10 from either: 15, 10, 11.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 20 / 28

(1, 2)V=3R (1, 1)V=0

L (0, 1)V=1

R (0, 0)V=0 (Left loses.)

What about (15, 7, 9, 11) ?

15 = 1 + 2 + 4 + 87 = 1 + 2 + 49 = 1 + 8

⊕ 11 = 1 + 2 + 8

10 = 2 + 8

We must remove: 2 + 8 = 10 from either: 15, 10, 11.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 20 / 28

(1, 2)V=3R (1, 1)V=0

L (0, 1)V=1

R (0, 0)V=0 (Left loses.)

What about (15, 7, 9, 11) ?

15 = 1 + 2 + 4 + 8

7 = 1 + 2 + 49 = 1 + 8

⊕ 11 = 1 + 2 + 8

10 = 2 + 8

We must remove: 2 + 8 = 10 from either: 15, 10, 11.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 20 / 28

(1, 2)V=3R (1, 1)V=0

L (0, 1)V=1

R (0, 0)V=0 (Left loses.)

What about (15, 7, 9, 11) ?

15 = 1 + 2 + 4 + 87 = 1 + 2 + 4

9 = 1 + 8⊕ 11 = 1 + 2 + 8

10 = 2 + 8

We must remove: 2 + 8 = 10 from either: 15, 10, 11.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 20 / 28

(1, 2)V=3R (1, 1)V=0

L (0, 1)V=1

R (0, 0)V=0 (Left loses.)

What about (15, 7, 9, 11) ?

15 = 1 + 2 + 4 + 87 = 1 + 2 + 49 = 1 + 8

⊕ 11 = 1 + 2 + 8

10 = 2 + 8

We must remove: 2 + 8 = 10 from either: 15, 10, 11.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 20 / 28

(1, 2)V=3R (1, 1)V=0

L (0, 1)V=1

R (0, 0)V=0 (Left loses.)

What about (15, 7, 9, 11) ?

15 = 1 + 2 + 4 + 87 = 1 + 2 + 49 = 1 + 8

⊕ 11 = 1 + 2 + 8

10 = 2 + 8

We must remove: 2 + 8 = 10 from either: 15, 10, 11.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 20 / 28

(1, 2)V=3R (1, 1)V=0

L (0, 1)V=1

R (0, 0)V=0 (Left loses.)

What about (15, 7, 9, 11) ?

15 = 1 + 2 + 4 + 87 = 1 + 2 + 49 = 1 + 8

⊕ 11 = 1 + 2 + 8

10 = 2 + 8

We must remove: 2 + 8 = 10 from either: 15, 10, 11.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 20 / 28

(1, 2)V=3R (1, 1)V=0

L (0, 1)V=1

R (0, 0)V=0 (Left loses.)

What about (15, 7, 9, 11) ?

15 = 1 + 2 + 4 + 87 = 1 + 2 + 49 = 1 + 8

⊕ 11 = 1 + 2 + 8

10 = 2 + 8

We must remove: 2 + 8 = 10 from either: 15, 10, 11.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 20 / 28

Winning Ways for Your Mathematical Plays

Winning Ways for Your Mathematical Plays byGuy, Conway, and Berlekamp. (Left to Right)

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 21 / 28

Peg solitaire and the Rubik’s Cube.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 22 / 28

Peg solitaire and the Rubik’s Cube.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 22 / 28

Sprouts and Hackenbush.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 23 / 28

Sprouts and Hackenbush.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 23 / 28

Dots and Boxes

Dots and Boxes by Berlekamp, and a tournament game.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 24 / 28

Your Move

Nim like games.

Chess problems.

Go problems.

Lateral thinking.

Tic-Tac-Toeproblems !

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 25 / 28

Your Move

Nim like games.

Chess problems.

Go problems.

Lateral thinking.

Tic-Tac-Toeproblems !

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 25 / 28

Your Move

Nim like games.

Chess problems.

Go problems.

Lateral thinking.

Tic-Tac-Toeproblems !

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 25 / 28

Your Move

Nim like games.

Chess problems.

Go problems.

Lateral thinking.

Tic-Tac-Toeproblems !

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 25 / 28

Your Move

Nim like games.

Chess problems.

Go problems.

Lateral thinking.

Tic-Tac-Toeproblems !

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 25 / 28

Your Move

Nim like games.

Chess problems.

Go problems.

Lateral thinking.

Tic-Tac-Toeproblems

!

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 25 / 28

Your Move

Nim like games.

Chess problems.

Go problems.

Lateral thinking.

Tic-Tac-Toeproblems !

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 25 / 28

The Golden Age of Board Games

Amazons, Catchup, Slither.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 26 / 28

References

Berlekamp, Elwyn R., John H. Conway, and Richard K. Guy

Winning Ways for Your Mathematical Plays, Volume 1-4.

AMC 10 (2004): 12.

Berlekamp, Elwyn R.

The Dots-and-Boxes Game: Sophisticated Child’s Play.

AK Peters, Ltd., 2000.

Silverman, David L.

Your Move: Logic, Math and Word Puzzles for Enthusiasts.

Dover Publications, 1991.

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 27 / 28

Thank you!

parker.glynn.adey@utoronto.ca

pgadey.wordpress.com

Parker Glynn-Adey (UoT) Math & Games March 23, 2014 28 / 28