Hat Guessing Games

1
Hat Guessing Games

• Ang Wei Zou
• Peter Tigges

2
Introduction

• There are 2 people in the room
• A hat of either red or blue is placed on both
  peoples heads
• Both hats can be of the same color
• They can see each others hat colors, but not
  themselves
• Is there a guessing strategy that can maximize
  the number of right answers collectively?

3
Red Or Blue?
4
Considerations

• Both players guess simultaneously, so the guess
  of the first player cannot transmit any
  information to the other
• No form of communication is allowed between those
  2 people in the room
• A strategy can be formed between the two people
  prior to entering the room

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Guessing Strategy
1st Actual Hat 2nd Actual Hat
B R
R B
B B
R R
1st Guess 2nd Guess
R R
B B
B R
R B
1st Person Guesses the Opposite of 2nd Persons
hat color 2nd Person Guesses the Same as 1st
Persons hat color
6
Theory 1

• If there are n players and k different colors,
  then there exists a strategy guaranteeing at
  least n/k correct guesses

7
Proof

• Let there be n players and k color hats
• Number the players 1 to n and the colors 1 to k.
• Every ith player guesses as if the sum of all
  hats are equals to i Mod k.

1st Actual Hat 2nd Actual Hat
0 1
0 1
1 1
0 0
1st Guess 2nd Guess
0 0
1 1
1 0
0 1
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Undirected Vision

• In the Undirected Case, we assume that if A can
  see B, then B can see A as well.
• The obvious strategy is to pair up players as
  best as possible and implement the strategy as
  mentioned before. Hence the next theorem
• Let G be an undirected graph with M a maximum
  matching of G and lets define H(G) as the maximum
  number of correct guesses. Then H(G) M
• G has people as vertices and their sight as
  directed edges

9
Example
E
D
A
B
C
F
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Restricted Directed Vision

• In the Directed Case, we do not assume that A can
  see B if B can see A
• In this case there are no specific values for
  H(G). However, there are simple upper bounds and
  lower bounds.

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Theorem

• Given a directed graph G let c(G) denote the
  maximal number of vertex disjoint cycles in G,
  and let F(G) denote the minimum number of
  vertices whose removal from G makes the graph
  acyclic. Then c(G) H(G) F(G).
• The lower bound is formed by realizing that we
  can guarantee at least one correct answer for
  every cycle

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Vertex Disjoint Cycle

• A directed cycle.
• Sample

13
Acyclic

• A directed graph with no directed cycles.
• Sample

14
Upper Bound

• Lets assume the graph G has N vertices and that
  the set of vertices whose removal would make G
  acyclic is called K
• We arrange the graph such that K set of vertices
  are one the left side and everything else is on
  the right side
• After the hats are set, we can choose each of the
  last n-k hat colors as to force the corresponding
  player to guess incorrectly given the colors of
  the preceding players

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Example
16
Bounds

• These upper and lower bounds are not sharp.
• To prove that the upper bound is not sharp,
  consider G as triangle. F(G) is 2, but we know
  that H(G) is 1.

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Usage of Hamming Codes

• Developed by Richard Hamming in World War 2 that
  contributed to the fields of information theory,
  coding theory and cryptology.
• The Hamming weight of a string is the number of
  symbols that are different from the zero-symbol
  of the alphabet used.
• Example

Alphabet String Hamming weight
0,1 11101 4
0,1 11101000 4
0,1 00000000 0
' ',a-z hello world 10
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Hamming Weight Example

• Consider the sight graph below. How can we
  implement Hamming Weights to come up with an
  optimal guessing strategy?
• This is a directed graph with hat colors 0 or
  1
• G(a) B E G(b) C E G(c) D E G(d)
  A E 1 with mod 2 arithmetic
• G(e) 1 if(A B,B C,C D,A D 1) has
  Hamming weight 1
• 0 if(A B,B C,C D,A D
  1) has Hamming weight 3.

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Hamming Weight Example

• Since a, b, c, and d are all making their guesses
  based on es hat color, they will make either 1
  or 3 correct guesses depending on what e is
  wearing.
• e can then guess correctly and get at least one
  right answer, his own hat color, or guess
  incorrectly and get at least 3 correct answers
  from the rest

20
Using hypercubes to approach the game

• An n dimensional hypercube is called an n-cube
• Vertices - all 2n binary arrangements of length n
• Edges - join vertices that differ by one letter
• Ex. In the 4-cube there is an edge between 1101
  and 1100. This can be written as 110 where the
  indicates the spot not known. Applied to the hat
  problem, this would mean the fourth player can
  see a 1, 1, 0 for the first, second, and third
  players. Here the fourth player must guess a 1
  or 0 so the direction of the edge between 1101
  and 1100 can represent the chosen guess.
  (1101?1100 if the player guesses 1)

21
Hypercubes in balanced stategies

• Theorem If there are n players and k different
  hat colors, then there exists a strategy which is
  balanced. That is, if xi of the players are
  wearing hats of color i (1 ? i ? k), then at
  least xi/k of the people wearing color i guess
  correctly for each value of i.
• Constructing the hypercube involves grouping the
  vertices of the hypercube in i levels. Each
  vertex is in level i if the binary arrangements
  Hamming weight is i. The up-degree (down-degree)
  at a vertex in level i is the number of edges
  between that vertex and vertices in level i 1
  (i - 1).
• Proof First proven where k 2 then extended to
  all values of k. Shows that at each vertex every
  edge coming in from below (above) is paired, if
  possible, with an edge going down (up)

22
Hypercubes and optimal stategies

• In the 2-color game when n is even, each player
  is as likely to guess one hat color as to guess
  the other under many optimal strategies. (Optimal
  strategies are unbiased)
• Proposition Suppose the set of hat colors is 0,
  1 and there are n players playing an optimal
  strategy, where n is even. For any fixed player
  i, looking over all possible hat placements, the
  number of times that player i guesses 0 is the
  same as the number of times that player guesses
  1.
• Proof When n is even an optimal stategy
  corresponds to an n-cube with the in-degree equal
  to the out-degree at each vertex (the strategy
  corresponds to a Eulerian walk on the n-cube).
  The edges that point up represent guesses of 1
  and the edges pointing down represent guesses of
  0. Since it is a Eulerian walk the up and down
  edges are equal.
• This proposition is extended to games with more
  than 2 colors.

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The limited hats game

• Consider the game with only a limited supply of
  hats. Let H(n a1, a2, , ak) denote the max
  number of correct guesses that we can gaurantee
  when there are n players and a1 hats of the first
  color, a2 hats of the second color, and so on up
  to ak of the kth color.
• Ex. Three players, two blue hats and two red
  hats. Using the strategy of player a guessing
  the opposite of what player b is wearing, player
  b guessing the opposite of what player c is
  wearing, and player c guessing the opposite of
  what player a is wearing will guarantee 2 correct
  answers or H(3 2,2) 2.

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Bibliography

• G. Aggarwal, A. Fiat, A. V. Goldberg, J. D.
  Harline, N. Immorlica, and M. Sudan,
  Derandomization of auctions, Proceedings of the
  37th Annual ACM Symposium on Theory of Computing,
  2005, pp. 619 625.
• Steve Butler, Mohammad T. Hajiaghayi, Roberty D.
  Kleinberg, Tom Leighton, Hat Guessing Games, SIAM
  J. Discrete Mathematics, Vol. 22, No. 2, pp.
  592-605.
• Fred S. Roberts, Barry Tesman, Applied
  Combinatorics. Pearson Prentice Hall, New
  Jersery, 2005, pp. 633.