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Hat Guessing Games Ang Wei Zou Peter Tigges Introduction There are 2 people in the room A hat of either red or blue is placed on both people’s heads Both hats can be of the same color They can see each other’s hat colors, but not themselves

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hat guessing games

Hat Guessing Games

Ang Wei Zou

Peter Tigges

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

1st Person Guesses the Opposite of 2nd Person’s hat color

2nd Person Guesses the Same as 1st Person’s hat color

theory 1
Theory 1

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

proof
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.
undirected vision
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
example
Example

E

D

A

B

C

F

restricted directed vision
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.
theorem
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
vertex disjoint cycle
Vertex Disjoint Cycle
  • A directed cycle.
  • Sample:
acyclic
Acyclic
  • A directed graph with no directed cycles.
  • Sample:
upper bound
Upper Bound
  • Let’s 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
bounds
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.
usage of hamming codes
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:
hamming weight example
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.

hamming weight example19
Hamming Weight Example
  • Since a, b, c, and d are all making their guesses based on e’s 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
using hypercubes to approach the game

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)

hypercubes in balanced stategies
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 arrangement’s 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)

hypercubes and optimal stategies
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.

the limited hats game
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.

bibliography
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.