games, hats, and codes mira bernstein wellesley college sums 2005

Games, Hats, and Codes

Mira Bernstein

Wellesley College

SUMS 2005

A game with hats (N=3)

Each player is randomly assigned a red or blue hat.


Each player can see the color of his teammates’ hats but not his own.


BLUE REDPASS

Players simultaneously guess the colors of their own hats. Passing is allowed.


BLUE REDPASS

At least one correct guessNo incorrect guesses

WIN!

Some observations

• A player gets NO information about his own hat from looking at his teammates’ hats

• NO strategy can guarantee victory

• Easy strategy: one player guesses, everyone else passes.

• Can the players do better than 50%?

A better strategy for N=3

Instructions for all players:

If you see two hats of the same color, guess the other color.

If you see two different colors, pass.

Possible hatconfigurations

Guesses#1 #2 #3


Guesses

RED

#1 #2 #3


Guesses

RED

BLUE

#1 #2 #3


Guesses

RED

BLUE

RED

#1 #2 #3


Guesses

RED

BLUE

RED

BLUE

#1 #2 #3


Guesses

RED RED

RED

BLUE

RED

BLUE

BLUE BLUE

#1 #2 #3


Guesses

RED RED RED

RED

RED

BLUE

RED

BLUE

BLUE

BLUE BLUE BLUE

#1 #2 #3

√√√√√√

X

X


√√√√√√

X

X

Probability of winning:

75%

•How is this possible?

•Can we do better?

•What about N >3?


Guesses

RED RED RED

RED

RED

BLUE

RED

BLUE

BLUE

BLUE BLUE BLUE

#1 #2 #3

6 correct guesses

6 incorrect guesses

# correct = # incorrect

RED RED RED

RED

RED

BLUE

RED

BLUE

BLUE

BLUE BLUE BLUE

#1 #2 #3

Guesses

Why?

The same instructions that lead to a correctguess in one situation…

… will lead to an incorrect guess in another.

BLUE

Player #3

BLUE

Player #3

In general…

If the game is played once with each possible configuration of hats, then

# correct guesses = # incorrect guesses

True for any number of people N

True for any deterministic strategy S

Why does the N=3 strategy work?

It takes only one correct guess to win!

Strategy:

Spread out the good guesses.

Concentrate the bad guesses.

#1 #2 #3

In general…• When played over all hat combinations

with N players, any strategy produces k correct guesses and k incorrect guesses. (The value of k depends on the strategy.)

• Best possible guess arrangement:– 1 correct guess per winning combination– N incorrect guesses per losing combination

• A strategy which achieves this is called a perfect strategy.

Do perfect strategies actually exist?

N = 3: Yes!

Other N?

#1 #2 #3

Some terminology

H: set of all possible hat configurations

(sequences of 0s and 1s)

Distance in H: number of places in which two elements of H differ.

1 0 0 1 0 1 1 0 1

1 0 1 1 0 1 0 0 1

Distance: 2

Some terminology

Ball of radius r around h H: the set of allconfigurations whose distance from h is at most r.

h: 1 0 0 1

B1(h): 0 0 0 1 1 1 0 1 1 0 1 1

1 0 0 0 | B1(h)| = N+1 center 1 0 0 1

Some terminology

• S: a (deterministic) strategy

• L: set of all hat configurations where a team playing according to S loses

• W: set of all hat configurations where a team playing according to S wins

L W = H

An important fact

Suppose h and h’ are elements of H that differ only in the i th entry.

If, according to strategy S, Player i guesses correctly in h, then he guesses incorrectly in h’, and vice versa.

4th player’s guess

h: 0 1 0 0 1 0 √

h’: 0 1 0 1 1 0 X

Corollaries

Theorem 1: Every element h W is within distance 1 of some element of h’ L.

Proof: Suppose Player j guesses correctly in h. Let h’ be the hat configuration that differs from h only in the j th entry. Then Player j must guess incorrectly in h’, so h’ is in L.

Corollaries

Theorem 2: In a perfect strategy S, every element h W is within distance 1 of exactly one element of L.

Proof: Suppose h differs from h1 L in the i th entry and from h2 L in the j th entry. Since S is a perfect strategy, all players guess incorrectly in h1 and h2. But then Players i and j must both guess correctly in h, which cannot happen in a perfect strategy.

Corollaries

In other words….

Theorem 1: Every element h H is contained in a ball of radius 1 around some element of L.

Theorem 2: In a perfect strategy S, the balls of radius 1 around elements of L do not overlap.

Codes

A perfect code of length N is a subset L in H such that the balls of radius 1 around points of L

• include all of H

• do not overlap

A perfect strategy yields a perfect code!

H

A perfect code

H

= points of L

A perfect code

Perfect code perfect strategy

Instructions for Player i:

– If the hat configuration might be in L:

Guess so that if it’s in L, you’ll be wrong.

– If you can tell that the hat configuration is not in L: Pass

Results (for hat configuration h):– If h is in L: Every player guesses wrong.

– If the hat configuration h is not in L:

There is a unique element h’ of L which differs from h in one place -- say the i th place.The i th player can’t tell if the

configuration is h or h’, so he guesses away from h’, correctly. All others pass.

Perfect code perfect strategy

000

111

100

010

001 101

110

011

Example: N=3

L = {000,111}

Example: N=3

L = {000,111}

000

111

100

010

001 101

110

011


– If the hat configuration might be in L, guess so that if it’s in L you’ll be wrong.

Translation: If you see two 0’s or two 1’s, guess the opposite number.

Example: N=3

L = {000,111}


– If you can tell that the hat configuration is not in L, pass.

Translation: If you see two different numbers, pass.

Example: N=3

L = {000,111}

How good are perfect strategies?

Theorem: If S is a perfect strategy for N players then the probability of winning is N/N+1.

Proof: In every ball, N out of the N+1 points correspond to winning configurations.

Example: If N=3, the probability of winning with a perfect strategy is ¾. There can be no better strategy than the one we found.

Do perfect codes exist for N>3?

Theorem: A perfect code of length N can exist only if N=2m-1 for some integer m.

Proof: A perfect code splits H into disjoint balls of radius 1. Each of the balls has N+1 points and H has 2N points, so 2N is divisible by N+1. Thus N+1 is a power of 2, so N=2m-1.

Example: We know a perfect code of length 3 = 22-1. But what about 7, 15, 31,…?

Error-correcting codes

I love you!

11010…

11010…

I love you too!


I love you!

11010…

10010…

You what?


I love you!

1001100


I love you!

1001100

1000100


I love you!

I love you too!

1001100

1001100

A different game: Nim

RulesTake any number of stones

from any one pile

A B


RulesWhoever takes the last stone wins the game

A B


A B


A B

Two equal piles: bad news for whoever is next to move.


A B

From now on, B simply imitates A’s moves…


A B

WINLOSE

… so B is guaranteed to take the last stone.

Nim is…

• Combinatorial: no element of chance

• Impartial: same moves available to each player

• Finite: must end after a finite number of moves

In any Nim position, exactly one of the players has a winning strategy.

Who wins?

P-position: Previous player wins

N-position: Next player wins

e.g. the empty game is a P-position

e.g. two equal piles is a P-position

e.g. two unequal piles is an N-position

Nim strategy: Figure out which positions are P-positions and move to them!

More on P and N

P-position: cannot move to a P-position; move only to an N-position (or not at all)

N-position: can move to at least one P-position

If you combine two Nim games into one:• P + P = P: cannot move to another P+P• N + P = N: can move to P+P• N + N = ? (depends on the games)

Position vectors

A Nim position with all heaps of size N can be described by a vector of length N.

(1,0,0,2,3)1 heap of size 52 heaps of size 23 heaps of size 1

Position vectors

Two equal heaps are a P-position, so we can ignore them (P+P=P, N+P=N).

(1,0,0,2,3)1 heap of size 52 heaps of size 23 heaps of size 1

Position vectors

Thus we can replace all the entries in the position vector with 0’s and 1’s.

Binary vector

1 heap of size 52 heaps of size 23 heaps of size 1

(1,0,0,0,1)

Legal moves

Let X, Y be binary position vectors for Nim.

One can move from X to Y if• X >Y (as binary numbers)• Distance from X to Y is 1 or 2

Examples:

1 0 1 0 1 1

Legal moves


One can move from X to Y if• X >Y (as binary numbers)• Distance from X to Y is 1 or 2

Examples:

1 0 1 0 0 1

Legal moves


One can move from X to Y if and only if• X >Y (as binary numbers)• The distance from X to Y is 1 or 2

Examples:

1 0 1 0 0 0

P and N: review

P-position: cannot move to a P-position; move only to an N-position (or not at all)

N-position: can move to at least one P-position

P and N revisited

Theorem: The distance between two P-position vectors is ≥ 3.

Proof: If the distance were 1 or 2, you could move from the larger vector to the smaller one. However, it is impossible to move from a P-position to a P-position.

Corollary: Balls of radius 1 around P-positions do not overlap.

P and N revisited

Theorem: Any vector X with distance ≥ 3 from all smaller P-positions is a P-position.

Proof: Since you cannot move from X to any P-position, X itself must be a P-position.

Corollary: We can look for P-positions inductively.

How to look for P-positions

• Start with (…,0,0)

• At each step, look for the next smallest sequence that has distance ≥ 3 from every previously-found P-position.

• The preceding theorems guarantee that this procedure will give you all and only P-positions!

Why look for P-positions?

• To win at Nim.

• The set of all P-positions with heaps of size N will give us non-overlapping balls of radius 1. If N=2m-1,we may get a perfect code.

• To find perfect strategies for the hat game.

P-positions for heaps of size 7

0 0 0 0 0 0 0 Start with 0

0 0 0 0 1 1 1 Smallest v with three 1’s

0 0 0 1 ? ? ? Impossible!


0 0 0 0 0 0 0 Start with 0




0 0 0 0 0 0 0 Start with 0


0 0 1 1 ? ? ?


0 0 0 0 0 0 0 Start with 0


0 0 1 1 0 0 1 Next smallest

0 0 1 1 ? ? ?


0 0 0 0 0 0 0 0

0 0 0 0 1 1 1 A

0 0 1 1 0 0 1 B

0 0 1 1 ? ? ?


0 0 0 0 0 0 0 0

0 0 0 0 1 1 1 A

0 0 1 1 0 0 1 B

0 0 1 1 1 1 0 A+B



0 0 0 0 0 0 0 0

0 0 0 0 1 1 1 A

0 0 1 1 0 0 1 B

0 0 1 1 1 1 0 A+B

0 1 0 1 ? ? ?


0 0 0 0 0 0 0 0

0 0 0 0 1 1 1 A

0 0 1 1 0 0 1 B

0 0 1 1 1 1 0 A+B

0 1 0 1 0 1 0 C

0 1 ? ? ? ? ?


0 0 0 0 0 0 0 0

0 0 0 0 1 1 1 A

0 0 1 1 0 0 1 B

0 0 1 1 1 1 0 A+B

0 1 0 1 0 1 0 C

0 1 0 1 1 0 1 C+A

0 1 1 0 0 1 1 C+B

0 1 1 0 1 0 0 C+A+B


Can we have any other P-positions that begin

0 1 X X X X X ? D

Then we would have D+A, D+B, etc: 16 total

Can we have 16 P-positions with heaps 6?

(16 balls of radius 1) (7 points per ball) > 26

Impossible!


0 0 0 0 0 0 0 0

0 0 0 0 1 1 1 A

0 0 1 1 0 0 1 B

0 0 1 1 1 1 0 A+B

0 1 0 1 0 1 0 C

0 1 0 1 1 0 1 C+A

0 1 1 0 0 1 1 C+B

0 1 1 0 1 0 0 C+A+B


0 0 0 0 0 0 0 0 1 0 0 0 ? ? ?No!

0 0 0 0 1 1 1 A

0 0 1 1 0 0 1 B

0 0 1 1 1 1 0 A+B

0 1 0 1 0 1 0 C

0 1 0 1 1 0 1 C+A

0 1 1 0 0 1 1 C+B

0 1 1 0 1 0 0 C+A+B


0 0 0 0 0 0 0 0 1 0 0 1 ? ? ?

0 0 0 0 1 1 1 A

0 0 1 1 0 0 1 B

0 0 1 1 1 1 0 A+B

0 1 0 1 0 1 0 C

0 1 0 1 1 0 1 C+A

0 1 1 0 0 1 1 C+B

0 1 1 0 1 0 0 C+A+B


0 0 0 0 0 0 0 0 1 0 0 1 0 1 1 D

0 0 0 0 1 1 1 A

0 0 1 1 0 0 1 B

0 0 1 1 1 1 0 A+B

0 1 0 1 0 1 0 C

0 1 0 1 1 0 1 C+A

0 1 1 0 0 1 1 C+B

0 1 1 0 1 0 0 C+A+B


0 0 0 0 0 0 0 0 1 0 0 1 0 1 1 D

0 0 0 0 1 1 1 A 1 0 0 1 1 0 0 D+A

0 0 1 1 0 0 1 B 1 0 1 0 0 1 0 D+B

0 0 1 1 1 1 0 A+B etc.

0 1 0 1 0 1 0 C

0 1 0 1 1 0 1 C+A

0 1 1 0 0 1 1 C+B

0 1 1 0 1 0 0 C+A+B


16 P-positions

8 points per ball A perfect code!

128 = 27 points

A vector space over 2 with basis

0 0 0 0 1 1 1 A

0 0 1 1 0 0 1 B

0 1 0 1 0 1 0 C

1 0 0 1 0 1 1 D

The Hamming Code for N=7

Our code is the kernel of the matrix

1 1 1 1 0 0 0

A = 1 1 0 0 1 1 0

1 0 1 0 1 0 1 over 2

The columns are the numbers 1-7 in binary!

This code is called the Hamming Code.


A quick way to check if v ker(A):• Record the numbers corresponding to

the positions where v has a 1• Write these numbers in binary, with

leading zeros if necessary• Add the numbers as binary vectors:

1+1 = 0, no carry!• v ker(A) iff you get 0


Example: v = ( 1 0 1 0 0 0 1 )

v has 1’s in positions 7,5,1

7 = 1 1 1

5 = 1 0 1

1 = 0 0 1 padding

0 1 1 addition with no carry

The answer is not 0, so v is not in the code.

Finding the nearest code vector

v = ( 1 0 1 0 0 0 1 )

Which coordinate of v should we change?

• When we “added” 7,5, and 1, we got 0 1 1.

• If we change the “3” coordinate of v from 0 to 1, we’ll have to “add” another 0 1 1.

• 0 1 1 + 0 1 1 = 0 0 0 !

v’ = ( 1 0 1 0 1 0 1 ) is in the code.

The Hamming Code for N=2m-1

Let A be the mN matrix whose columns are the numbers 1 through N, written in binary.

For N=15:

1 1 1 1 1 1 1 1 0 0 0 0 0 0 0

A = 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0

1 1 0 0 1 1 0 0 1 1 0 0 1 1 0

1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

ker(A) is called the Hamming code of length N.

The Hamming Code for N=2m-1

Every N-vector v is within distance 1 of exactly one code vector v’.

A(v), read as a binary number, gives the coordinate in which v and v’ differ!

The Hamming code is a perfect code, which makes error-correction especially simple!

Homework

• Prove that the Hamming Code really is perfect and that A(v) gives the coordinate of the error.

• Using the Hamming code, find a practical strategy for Nim.

• Using the Hamming code, find a practical strategy for the hat game for 7 players.

A bit more about codes

Not all codes are perfect: – In a general error-correcting code, the balls

around the code points are disjoint, but do not necessarily include every point in the space.

– In a general covering code, the balls cover the whole space, but may overlap.

– A perfect code is both an error-correcting code and a covering code.

A bit more about codes

• In a general error-correcting code, the balls around the code points may have radius r>1.

• The Hamming codes are (essentially) the only perfect codes with radius 1.

• The is only one other perfect code: the Golay code for N=23, with radius 3.

Lexicodes

We used the P-positions of a game (Nim)

to construct a code (the Hamming code).

In general, the P-positions of many impartial games correspond to well-known codes!

These can be constructed in increasing lexicographic order, starting at 0 -- hence the name lexicodes!

References

• Lexicographic Codes: Error-Correcting Codes from Game Theory, John H. Conway and N.J.A. Sloane, IEEE Trans. Inform. Theory, 1986

http://www.research.att.com/~njas/doc/lex.pdf

• On Hats and Other Covers, H. Lenstra and G. Seroussi, 2002 (preprint)

www.hpl.hp.com/infotheory/hats_extsum.pdf

games, hats, and codes mira bernstein wellesley college sums 2005

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