Games, Hats, and Codes

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Games, Hats, and Codes

• Mira Bernstein
• Wellesley College
• SUMS 2005

2
A game with hats (N3)
Each player is randomly assigned a red or blue
hat.
3
A game with hats (N3)
Each player can see the color of his teammates
hats but not his own.
4
A game with hats (N3)
BLUE
RED
PASS
Players simultaneously guess the colors of their
own hats. Passing is allowed.
5
A game with hats (N3)
BLUE
RED
PASS
At least one correct guess No incorrect guesses
WIN!
6
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?

7
A better strategy for N3

• Instructions for all players
• If you see two hats of the same color, guess
  the other color.
• If you see two different colors, pass.

8
Possible hat configurations
Guesses
1
2
3








9
Possible hat configurations
Guesses
1
2
3
RED







10
Possible hat configurations
Guesses
1
2
3
RED


BLUE




11
Possible hat configurations
Guesses
1
2
3
RED


BLUE
RED



12
Possible hat configurations
Guesses
1
2
3
RED


BLUE
RED


BLUE
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Possible hat configurations
Guesses
1
2
3
RED RED

RED
BLUE
RED
BLUE

BLUE BLUE
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Possible hat configurations
Guesses
1
2
3
X
RED RED RED
RED
RED
BLUE
RED
BLUE
BLUE
BLUE BLUE BLUE
v
v
v
v
v
v
X
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Possible hat configurations
Probability of winning 75
X
v
v
v

• How is this possible?
• Can we do better?
• What about N gt3?

v
v
v
X
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Possible hat configurations
Guesses
1
2
3
RED RED RED
RED
RED
BLUE
RED
BLUE
BLUE
BLUE BLUE BLUE
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Guesses
1
2
3
RED RED RED
RED
RED
BLUE
RED
BLUE
BLUE
BLUE BLUE BLUE

• 6 correct guesses
• 6 incorrect guesses
• correct incorrect

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Why?
The same instructions that lead to a
correct guess in one situation
BLUE
Player 3
will lead to an incorrect guess in another.
BLUE
Player 3
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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

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Why does the N3 strategy work?
1
2
3

• It takes only one correct guess to win!
• Strategy
• Spread out the good guesses.
• Concentrate the bad guesses.

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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.

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Do perfect strategies actually exist?
1
2
3

• N 3 Yes!
• Other N?

23
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

24
Some terminology

• Ball of radius r around h ? H the set of all
• configurations 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) N1
• center 1 0 0 1

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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

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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 players guess
• h 0 1 0 0 1 0 v
• h 0 1 0 1 1 0 X

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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.

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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. ??

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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.

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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!

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A perfect code
H
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A perfect code
H
points of L
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A perfect code
H
points of L
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Perfect code perfect strategy

• Instructions for Player i
• If the hat configuration might be in L
• Guess so that if its in L, youll be wrong.
• If you can tell that the hat configuration is not
  in L Pass

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Perfect code perfect strategy

• 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 cant tell if the
  configuration is h or h, so he guesses away
  from h, correctly. All others pass.

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Example N3 L 000,111
011
111
001
101
010
110
000
100
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Example N3 L 000,111
011
111
001
101
010
110
000
100
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Example N3 L 000,111

• Instructions for Player i
• If the hat configuration might be in L, guess so
  that if its in L youll be wrong.
• Translation If you see two 0s or two 1s,
  guess the opposite number.

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Example N3 L 000,111

• Instructions for Player i
• If you can tell that the hat configuration is not
  in L, pass.
• Translation If you see two different numbers,
  pass.

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How good are perfect strategies?

• Theorem If S is a perfect strategy for N players
  then the probability of winning is N/N1.
• Proof In every ball, N out of the N1 points
  correspond to winning configurations.
• Example If N3, the probability of winning with
  a perfect strategy is ¾. There can be no better
  strategy than the one we found.

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Do perfect codes exist for Ngt3?

• Theorem A perfect code of length N can exist
  only if N2m-1 for some integer m.
• Proof A perfect code splits H into disjoint
  balls of radius 1.
• Each of the balls has N1 points and H has 2N
  points, so 2N is divisible by N1.
• Thus N1 is a power of 2, so N2m-1.
• Example We know a perfect code of length
• 3 22-1. But what about 7, 15, 31,?

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Error-correcting codes
I love you!
11010
I love you too!
11010
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Error-correcting codes
I love you!
11010
You what?
10010
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Error-correcting codes
I love you!
1001100
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Error-correcting codes
I love you!
1001100
1000100
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Error-correcting codes
I love you!
1001100
1000100
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Error-correcting codes
I love you!
1001100
I love you too!
1001100
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A different game Nim
Rules Take any number of stones from any one pile
B
A
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A different game Nim
Rules Whoever takes the last stone wins the game
B
A
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A different game Nim
B
A
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A different game Nim
B
A
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A different game Nim
B
A
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A different game Nim
B
A
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A different game Nim
Two equal piles bad news for whoever is next to
move.
B
A
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A different game Nim
Two equal piles bad news for whoever is next to
move.
B
A
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A different game Nim
From now on, B simply imitates As moves
B
A
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A different game Nim
From now on, B simply imitates As moves
B
A
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A different game Nim
From now on, B simply imitates As moves
B
A
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A different game Nim
From now on, B simply imitates As moves
B
A
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A different game Nim
so B is guaranteed to take the last stone.
WIN
LOSE
B
A
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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.

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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!

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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 PP
• N P N can move to PP
• N N ? (depends on the games)

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Position vectors

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

1 heap of size 5 2 heaps of size 2 3 heaps of
size 1
(1,0,0,2,3)
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Position vectors

• Two equal heaps are a P-position, so we can
  ignore them (PPP, NPN).

1 heap of size 5 2 heaps of size 2 3 heaps of
size 1
(1,0,0,2,3)
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Position vectors

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

1 heap of size 5 2 heaps of size 2 3 heaps of
size 1
(1,0,0,0,1)
Binary vector
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Legal moves

• Let X, Y be binary position vectors for Nim.
• One can move from X to Y if
• X gtY (as binary numbers)
• Distance from X to Y is 1 or 2
• Examples

1 0 1
0 1 1
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Legal moves

• Let X, Y be binary position vectors for Nim.
• One can move from X to Y if
• X gtY (as binary numbers)
• Distance from X to Y is 1 or 2
• Examples

1 0 1
0 0 1
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Legal moves

• Let X, Y be binary position vectors for Nim.
• One can move from X to Y if and only if
• X gtY (as binary numbers)
• The distance from X to Y is 1 or 2
• Examples

1 0 1
0 0 0
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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

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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.

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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.

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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!

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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 N2m-1,we may get a perfect code.
• To find perfect strategies for the hat game.

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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 1s
• 0 0 0 1 ? ? ? Impossible!

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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 1s
• 0 0 1 0 ? ? ? Impossible!

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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 1s
• 0 0 1 1 ? ? ?

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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 1s
• 0 0 1 1 0 0 1 Next smallest
• 0 0 1 1 ? ? ?

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P-positions for heaps of size ? 7

• 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 ? ? ?

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P-positions for heaps of size ? 7

• 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 AB
• 0 1 0 0 ? ? ? Impossible!

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P-positions for heaps of size ? 7

• 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 AB
• 0 1 0 1 ? ? ?

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P-positions for heaps of size ? 7

• 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 AB
• 0 1 0 1 0 1 0 C
• 0 1 ? ? ? ? ?

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P-positions for heaps of size ? 7

• 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 AB
• 0 1 0 1 0 1 0 C
• 0 1 0 1 1 0 1 CA
• 0 1 1 0 0 1 1 CB
• 0 1 1 0 1 0 0 CAB

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P-positions for heaps of size ? 7

• Can we have any other P-positions that begin
• 0 1 X X X X X ? D
• Then we would have DA, DB, etc 16 total
• Can we have 16 P-positions with heaps ? 6?
• (16 balls of radius 1) (7 points per ball) gt 26
• Impossible!

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P-positions for heaps of size ? 7

• 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 AB
• 0 1 0 1 0 1 0 C
• 0 1 0 1 1 0 1 CA
• 0 1 1 0 0 1 1 CB
• 0 1 1 0 1 0 0 CAB

86
P-positions for heaps of size ? 7

• 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 AB
• 0 1 0 1 0 1 0 C
• 0 1 0 1 1 0 1 CA
• 0 1 1 0 0 1 1 CB
• 0 1 1 0 1 0 0 CAB

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P-positions for heaps of size ? 7

• 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 AB
• 0 1 0 1 0 1 0 C
• 0 1 0 1 1 0 1 CA
• 0 1 1 0 0 1 1 CB
• 0 1 1 0 1 0 0 CAB

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P-positions for heaps of size ? 7

• 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 AB
• 0 1 0 1 0 1 0 C
• 0 1 0 1 1 0 1 CA
• 0 1 1 0 0 1 1 CB
• 0 1 1 0 1 0 0 CAB

89
P-positions for heaps of size ? 7

• 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 DA
• 0 0 1 1 0 0 1 B 1 0 1 0 0 1 0 DB
• 0 0 1 1 1 1 0 AB etc.
• 0 1 0 1 0 1 0 C
• 0 1 0 1 1 0 1 CA
• 0 1 1 0 0 1 1 CB
• 0 1 1 0 1 0 0 CAB

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P-positions for heaps of size ? 7

• 16 P-positions
• 8 points per ball A perfect code!
• 128 27 points
• A vector space over Z2 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

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The Hamming Code for N7

• 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 Z2
• The columns are the numbers 1-7 in binary!
• This code is called the Hamming Code.

92
The Hamming Code for N7

• 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
• 11 0, no carry!
• v ? ker(A) iff you get 0

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The Hamming Code for N7

• Example v ( 1 0 1 0 0 0 1 )
• v has 1s 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.

94
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,
  well 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.

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The Hamming Code for N2m-1

• Let A be the m?N matrix whose columns are the
  numbers 1 through N, written in binary.
• For N15
• 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.

96
The Hamming Code for N2m-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!

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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.

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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.

99
A bit more about codes

• In a general error-correcting code, the balls
  around the code points may have radius rgt1.
• The Hamming codes are (essentially) the only
  perfect codes with radius 1.
• The is only one other perfect code the Golay
  code for N23, with radius 3.

100
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!

101
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

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