The Hat GameThe Hat Game

11/19/0411/19/04

James FiedlerJames Fiedler

ReferencesReferences Hendrik W. Lenstra, Jr. and Gadiel Hendrik W. Lenstra, Jr. and Gadiel

Seroussi, Seroussi, On Hats and Other CoversOn Hats and Other Covers, , preprint, 2002, preprint, 2002, http://www.hpl.hp.com/research/info_theohttp://www.hpl.hp.com/research/info_theory/hats_extsum.pdf .ry/hats_extsum.pdf .

J.P. Buhler, J.P. Buhler, Hat TricksHat Tricks, The Mathematical , The Mathematical Intelligencer Intelligencer 24 24 (2002), no. 4, 44 – 49.(2002), no. 4, 44 – 49.

Sarah Robinson, Sarah Robinson, Why Mathematicians Why Mathematicians Now Care About Their Hat ColorNow Care About Their Hat Color, New , New York Times, Science Times, p. D5, April York Times, Science Times, p. D5, April 10, 2001.10, 2001.

A Famous PuzzleA Famous Puzzle

The SetupThe Setup A group of n players enter a room A group of n players enter a room

whereupon they each receive a hat. Each whereupon they each receive a hat. Each player can see everyone else’s hat but player can see everyone else’s hat but not his own. not his own.

The players must each simultaneously The players must each simultaneously guess a hat color, or pass.guess a hat color, or pass.

The group loses if any player guesses the The group loses if any player guesses the wrong hat color or if every player passes.wrong hat color or if every player passes.

Players are not necessarily anonymous, Players are not necessarily anonymous, they can be numbered.they can be numbered.

The SetupThe Setup

Assignment of hats is assumed to be Assignment of hats is assumed to be random.random.

The players can meet beforehand to The players can meet beforehand to devise a strategy, but no devise a strategy, but no communication is allowed inside the communication is allowed inside the room.room.

The goal is to devise the strategy The goal is to devise the strategy that gives the highest probability of that gives the highest probability of winning.winning.

The 2 color, 3 player caseThe 2 color, 3 player case

Of course, designating one player to Of course, designating one player to guess randomly while every other player guess randomly while every other player passes gives probability 50% in the passes gives probability 50% in the binary case.binary case.

The best strategy turns out to have 75% The best strategy turns out to have 75% winning probability.winning probability.

The strategy: If any player sees that the The strategy: If any player sees that the other two players have the same hat other two players have the same hat color, he guesses the opposite color. color, he guesses the opposite color. Otherwise the player passes.Otherwise the player passes.

The 2 color, 3 player caseThe 2 color, 3 player case

The group wins whenever exactly The group wins whenever exactly two hats have the same color, which two hats have the same color, which will happen ¾ of the time.will happen ¾ of the time.

Everyone guesses incorrectly when Everyone guesses incorrectly when all three hats have the same color, all three hats have the same color, which will happen ¼ of the time.which will happen ¼ of the time.

Thus 75% chance of winning.Thus 75% chance of winning.

Perfect StrategyPerfect Strategy

This is an example of what we will This is an example of what we will call a perfect strategy (for any call a perfect strategy (for any number players and hat colors): number players and hat colors): Every winning configuration will Every winning configuration will have one person guessing correctly have one person guessing correctly and the others passing, every losing and the others passing, every losing configuration will have all the configuration will have all the players guessing incorrectly.players guessing incorrectly.

Perfect StrategyPerfect Strategy

This is the best that can be done.This is the best that can be done. From the New York Times article:From the New York Times article:

Binary GameBinary Game

Let the two hat colors be 0, 1.Let the two hat colors be 0, 1. For n numbered players, each player i For n numbered players, each player i

sees n-1 hats, which can be thought of as sees n-1 hats, which can be thought of as a vector of length n-1 over Fa vector of length n-1 over F22. .

The strategy will tell him whether to The strategy will tell him whether to guess 0, 1, or pass. Think of this as a guess 0, 1, or pass. Think of this as a function function

where Vwhere Vn-1n-1 is the is the vector space of dimension n-1 over Fvector space of dimension n-1 over F22. .

BinaryBinary

We assume the strategy is We assume the strategy is deterministic, that it tells each deterministic, that it tells each player exactly which option to take player exactly which option to take for any hat configuration.for any hat configuration.

The vector The vector is a is a complete description of the strategy. complete description of the strategy. We’ll call the vector a We’ll call the vector a deterministic deterministic n-player strategyn-player strategy..

BinaryBinary

LetLet be all the hat be all the hat configurations for which our strategy configurations for which our strategy wins.wins.

By the rules, we haveBy the rules, we have

..

Also, Also, ..

1-Coverings and 1-Coverings and StrategiesStrategies

This means that the set of all losing This means that the set of all losing strategies are a 1-covering of the strategies are a 1-covering of the linear space Vlinear space Vnn. .

Conversely every 1-covering Conversely every 1-covering determines a strategy for which the determines a strategy for which the winning configurations are the winning configurations are the complement of the 1-covering.complement of the 1-covering.

Strategy from a 1-Strategy from a 1-CoveringCovering

Let C be a 1-covering of the space Let C be a 1-covering of the space VVnn. Let C determine a strategy as . Let C determine a strategy as follows: if player i sees the vector follows: if player i sees the vector (w(w11, …, w, …, wi-1i-1, w, wi+1i+1, …, w, …, wnn) and if ) and if exactly one x=0,1 puts exactly one x=0,1 puts (w(w11, …, w, …, wi-1i-1, x, , x,

wwi+1i+1, …, w, …, wnn)) outside of C, player i outside of C, player i guesses x, otherwise passes.guesses x, otherwise passes.

VVnn – C are winning configurations. – C are winning configurations.

Strategy from a 1-Strategy from a 1-CoveringCovering

Proof: Let C be a 1-covering and suppose Proof: Let C be a 1-covering and suppose the configuration (wthe configuration (w11, …, w, …, wnn) lies outside ) lies outside C. There is an i such that C. There is an i such that (w(w11, …, w, …, wi-1i-1, w, wii+1, +1,

wwi+1i+1, …, w, …, wnn)) lies inside C. Then i correctly lies inside C. Then i correctly guesses his hat color and for every other guesses his hat color and for every other player j there is at least one value (wplayer j there is at least one value (w jj) ) that puts the vector outside of C. Thus that puts the vector outside of C. Thus every other player guesses correctly or every other player guesses correctly or passes and Vpasses and Vnn – C – C {winning config.s}. {winning config.s}.

Strategy from a 1-Strategy from a 1-CoveringCovering

Converse: If (cConverse: If (c11, …, c, …, cnn) is in C then ) is in C then for this configuration everyone will for this configuration everyone will pass or someone will guess pass or someone will guess incorrectly. If there’s a choice x for i incorrectly. If there’s a choice x for i for which (cfor which (c11, …, c, …, ci-1i-1, x, c, x, ci+1i+1, …, c, …, cnn) is ) is outside of C, player i will choose that outside of C, player i will choose that color, incorrectly. If no such choice color, incorrectly. If no such choice occurs, every player passes. Thus occurs, every player passes. Thus {winning config.s} {winning config.s} V Vnn – C. – C.

Hamming ConnectionHamming Connection

Thus (the complements of) 1-coverings Thus (the complements of) 1-coverings are synonymous with winning are synonymous with winning configurations, and the best strategy for configurations, and the best strategy for any n will be given by the smallest 1-any n will be given by the smallest 1-covering code for that n.covering code for that n.

Thus perfect 1-coverings give optimal Thus perfect 1-coverings give optimal solutions when they exist and since solutions when they exist and since Hamming Codes are perfect 1-coverings, Hamming Codes are perfect 1-coverings, they give the optimal strategy for n=2they give the optimal strategy for n=2rr-1, -1, any positive integer r.any positive integer r.

Perfect = Perfect Perfect = Perfect

Perfect 1-coverings C correspond to Perfect 1-coverings C correspond to perfect strategies.perfect strategies. Proof: If C is a perfect 1-covering then no Proof: If C is a perfect 1-covering then no

two codewords are within distance 1. Then two codewords are within distance 1. Then if a given configuration is within C, every if a given configuration is within C, every player will see one option for their hat color player will see one option for their hat color that puts the configuration outside of C and that puts the configuration outside of C and every player will guess wrong. If a given every player will guess wrong. If a given configuration is outside of C then for one configuration is outside of C then for one player only will there be one option that player only will there be one option that puts his hat color outside of C, the rest will puts his hat color outside of C, the rest will pass.pass.

Perfect = PerfectPerfect = Perfect Conversely, if we have a perfect strategy, call Conversely, if we have a perfect strategy, call

the set of losing configurations C, let the set of losing configurations C, let cc=(c=(c11, …, , …, ccnn) be in C. We know already that C is a 1-) be in C. We know already that C is a 1-covering so letcovering so let

(c(c11, …, c, …, ci-1i-1, x, c, x, ci+1i+1, …, c, …, cnn) be outside of C. Given ) be outside of C. Given this configuration i will see (cthis configuration i will see (c11, …, c, …, ci-1i-1, c, ci+1i+1, …, , …, ccnn) and correctly guess hat color x. Now, ) and correctly guess hat color x. Now, suppose suppose

(c(c11, …, c, …, ci-1i-1, x, c, x, ci+1i+1, …, c, …, cnn) is within distance 1 of ) is within distance 1 of another element of C (spheres of radius 1 another element of C (spheres of radius 1 around elements of C are not distinct). around elements of C are not distinct).

Perfect = PerfectPerfect = Perfect

Say switching cSay switching cjj gets us to this other gets us to this other element. Then players i and j would element. Then players i and j would both guess a (correct) hat color in the both guess a (correct) hat color in the configuration configuration

(c(c11, …, c, …, ci-1i-1, x, c, x, ci+1i+1, …, c, …, cnn). Then the ). Then the strategy would not be perfect. Thus the strategy would not be perfect. Thus the sphere of radius 1 around the elements sphere of radius 1 around the elements of the 1-covering C are distinct, which of the 1-covering C are distinct, which means C is a perfect 1-covering.means C is a perfect 1-covering.

Hamming AgainHamming Again

The only perfect 1-coverings in the The only perfect 1-coverings in the binary case are Hamming Codes, so binary case are Hamming Codes, so all perfect strategies in the binary all perfect strategies in the binary case occur for n=2case occur for n=2rr-1.-1.

The n=3 case corresponds to the The n=3 case corresponds to the code (losing configurations) {000, code (losing configurations) {000, 111}.111}.

Probability of LosingProbability of Losing

For Hamming Codes of length n we For Hamming Codes of length n we have |C| = 2have |C| = 2nn/(n+1), reaching the /(n+1), reaching the sphere packing bound.sphere packing bound.

The probability of losing with a The probability of losing with a Hamming Code strategy is then Hamming Code strategy is then

PPLL =|C|/2 =|C|/2n n = (2= (2nn/(n+1))/2/(n+1))/2nn = 1/(n+1). = 1/(n+1).

Other Hamming-based Other Hamming-based StrategiesStrategies

For 2For 2rr-1 < n < 2-1 < n < 2r+1r+1-1, we can -1, we can construct a strategy based on the construct a strategy based on the Hamming Code as follows.Hamming Code as follows. The first 2The first 2rr-1 players ignore the players -1 players ignore the players

22rr, …, n and play according to the 2, …, n and play according to the 2rr-1 -1 game, the players 2game, the players 2rr, …, n always pass., …, n always pass.

It is known that these strategies are It is known that these strategies are optimal for n=2optimal for n=2rr, but not for larger n., but not for larger n.

Probability of LosingProbability of Losing

Probability of losing for these Probability of losing for these strategies isstrategies is

.. The lower bound is from the sphere The lower bound is from the sphere

packing bound, attained when n=2packing bound, attained when n=2rr-1.-1. The upper bound comes from the The upper bound comes from the

worst case Hamming-based strategy. worst case Hamming-based strategy.

Best Known Linear and Best Known Linear and Nonlinear StrategiesNonlinear Strategies

For n > 8 the optimal solution is For n > 8 the optimal solution is unknown except when n=2unknown except when n=2rr or n=2 or n=2rr--1.1.

It is known that there are nonlinear It is known that there are nonlinear 1-coverings that approach the 1-coverings that approach the sphere packing bound as n goes to sphere packing bound as n goes to infinite. Thus there are strategies infinite. Thus there are strategies that approach the winning that approach the winning probability of 1.probability of 1.

The q-ary GameThe q-ary Game

Same rules, now the set of hat colors Same rules, now the set of hat colors is Q={0, …, q-1}. is Q={0, …, q-1}.

Valid strategies are now synonymous Valid strategies are now synonymous with strong coverings:with strong coverings: A strong covering C A strong covering C Q Qnn is such that is such that

for all (wfor all (w11, …, w, …, wnn) in Q) in Qnn – C there is – C there is an i such that for all x in Q – {wan i such that for all x in Q – {wii} (w} (w11, , …, w…, wi-1i-1, x, w, x, wi+1i+1, …, w) is in C., …, w) is in C.

Strong Covering from a Strong Covering from a StrategyStrategy

Proof: If w = (wProof: If w = (w11, …, w, …, wnn) is a winning ) is a winning configuration then some player i configuration then some player i guesses his hat color wguesses his hat color wii correctly from correctly from the information (wthe information (w11, …, w, …, wi-1i-1, w, wi+1i+1, , …, w…, wnn). Thus for x ≠ w). Thus for x ≠ wii, the , the configuration (wconfiguration (w11, …, w, …, wi-1i-1, x, w, x, wi+1i+1, …, w, …, wnn) ) will cause player i to guess incorrectly will cause player i to guess incorrectly and lose. Thus the losing configurations and lose. Thus the losing configurations form a strong cover.form a strong cover.

Strategy from a Strong Strategy from a Strong CoverCover

Converse: Let C be a strong Converse: Let C be a strong covering. Determine a strategy as covering. Determine a strategy as follows: given the information (wfollows: given the information (w11, , …, w…, wi-1i-1, w, wi+1i+1, …, w, …, wnn), if there is ), if there is exactly one choice x such that (wexactly one choice x such that (w11, , …, w…, wi-1i-1, x, w, x, wi+1i+1, …, w, …, wnn) is outside C ) is outside C then player i guesses hat color x, then player i guesses hat color x, otherwise passes. otherwise passes.

Strategy from a Strong Strategy from a Strong CoverCover

This gives a valid strategy: If (wThis gives a valid strategy: If (w11, …, w, …, wnn) is ) is outside C then there is at least one player outside C then there is at least one player for whom there is only one choice (the for whom there is only one choice (the correct one) for which (wcorrect one) for which (w11, …, w, …, wi-1i-1, w, wi+1i+1, …, , …, wwnn) is outside C. Everyone else has at least ) is outside C. Everyone else has at least one choice so guesses correctly or passes. one choice so guesses correctly or passes. If (cIf (c11, …, c, …, cnn) is inside C then any player ) is inside C then any player that sees exactly one option that puts the that sees exactly one option that puts the configuration outside C then he guesses configuration outside C then he guesses incorrectly. Otherwise everyone passes.incorrectly. Otherwise everyone passes.

Perfect Strong CoveringPerfect Strong Covering

If C is a strong covering, then If C is a strong covering, then

If equality holds C is called perfect, If equality holds C is called perfect, and perfect strong coverings and perfect strong coverings correspond to perfect strategies.correspond to perfect strategies.

Unfortunately, for q > 2 and n > 1 Unfortunately, for q > 2 and n > 1 perfect strong coverings do not perfect strong coverings do not exist. exist.

Analog of Sphere-Analog of Sphere-Packing BoundPacking Bound

For a strong covering C, For a strong covering C,

Proof: Consider ordered pairs (x,y) Proof: Consider ordered pairs (x,y) where x is a winning configuration, y where x is a winning configuration, y a losing configuration and they differ a losing configuration and they differ in one coordinate. Let S be the set of in one coordinate. Let S be the set of all these ordered pairs. If we fix x all these ordered pairs. If we fix x then there are at least q – 1 choices then there are at least q – 1 choices for y since C is a strong covering.for y since C is a strong covering.

Analog of Sphere-Analog of Sphere-Packing BoundPacking Bound

Thus (q – 1)|VThus (q – 1)|Vnn - C| - C||S| or |S| or

(q – 1)(q(q – 1)(qnn - |C|) - |C|) |S|. If we fix y then |S|. If we fix y then there are at most n choices for x, so |there are at most n choices for x, so |S|S| n|C| and rearranging n|C| and rearranging

(q – 1)(q(q – 1)(qnn -|C|) -|C|) |S| |S| n|C| we get n|C| we get

Perfect = PerfectPerfect = Perfect Perfect strong coverings correspond to perfect Perfect strong coverings correspond to perfect

strategies.strategies. Proof: Let C be a perfect strong covering. Then Proof: Let C be a perfect strong covering. Then

we get equality (q – 1)(qwe get equality (q – 1)(qnn -|C|) = |S|= n|C| from -|C|) = |S|= n|C| from last slide. If last slide. If ww = (w = (w11, …, w, …, wnn) is not in C then for ) is not in C then for one player i there will be exactly one hat color one player i there will be exactly one hat color x=wx=wii which puts (w which puts (w11, …, w, …, wi-1i-1, x, w, x, wi+1i+1, …, , …, wwnn) outside C. The left side of the equality above ) outside C. The left side of the equality above means that there are exactly q-1 ways to change means that there are exactly q-1 ways to change ww to put the configuration inside C. All of these to put the configuration inside C. All of these changes are already used by the ichanges are already used by the ithth coordinate. coordinate.

Perfect=PerfectPerfect=Perfect Thus we can change the others freely, which Thus we can change the others freely, which

means any other player j will have more than means any other player j will have more than one choice that puts the configuration outside of one choice that puts the configuration outside of C and will pass. Let C and will pass. Let cc = (c = (c11, …, c, …, cnn) lie inside C. ) lie inside C. Then there are exactly n changes we can make Then there are exactly n changes we can make to to c c to push it outside C, based on the right side to push it outside C, based on the right side of the equality above. If any more than 1 of of the equality above. If any more than 1 of these changes can be made in any coordinate, C these changes can be made in any coordinate, C would not be a strong covering. Thus each would not be a strong covering. Thus each player will see 1 hat color that puts the player will see 1 hat color that puts the configuration (that he sees) outside of C and configuration (that he sees) outside of C and every player will guess incorrectly. Thus we every player will guess incorrectly. Thus we have a perfect strategy.have a perfect strategy.

Perfect = PerfectPerfect = Perfect

Converse: Just as the above Converse: Just as the above direction this is similar to the binary direction this is similar to the binary case. I’ll skip the converse for now.case. I’ll skip the converse for now.

A Strong Covering for Q-A Strong Covering for Q-ary Caseary Case

Let n=2Let n=2rr-1, and let -1, and let MM be a check be a check matrix for the binary Hamming Code matrix for the binary Hamming Code of length n. Let v be in Qof length n. Let v be in Qnn, , φφ::QQnn→Q→Qrr,, φφ((vv) = ) = MvMvTT. In terminology from . In terminology from Wednesday Wednesday φφ finds the syndrome of finds the syndrome of vv. Let . Let

Then C is a strong covering.Then C is a strong covering.

A Strong Covering for Q-A Strong Covering for Q-ary Caseary Case

Let Let

Then C is a strong covering.Then C is a strong covering. Proof: If Proof: If ww = (w = (w11, …, w, …, wnn) is in Q) is in Qnn – C, then – C, then

some coordinates of some coordinates of φφ((ww) are zero. There ) are zero. There is a column is a column mmjj of of MM whose coordinates whose coordinates are 1 exactly when the coordinates of are 1 exactly when the coordinates of φφ((ww) are zero. Then for all ) are zero. Then for all αα in Q in Q**, , φφ((ww) + ) + α α mmj j is in C, since this latter vector no is in C, since this latter vector no longer has any zero coordinates.longer has any zero coordinates.

Probabiltiy of LosingProbabiltiy of Losing

This strategy loses with probabilityThis strategy loses with probability

which goes to zero as nwhich goes to zero as n..

A More General A More General ConstructionConstruction

Slightly smaller strong coverings can Slightly smaller strong coverings can be achieved with the following be achieved with the following generalizationgeneralization

where wt is the Hamming weight, where wt is the Hamming weight, the number of nonzero coordinates.the number of nonzero coordinates.

BoundsBounds

It is known that there are strong It is known that there are strong coverings that can do better than coverings that can do better than either of these q-ary constructions, either of these q-ary constructions, with a losing probability ofwith a losing probability of

Constructions have not been found Constructions have not been found to reach this limit.to reach this limit.

VariationsVariations

In the full version of their paper, In the full version of their paper, Lenstra and Seroussi consider the Lenstra and Seroussi consider the following variations.following variations. Non-uniform distributionsNon-uniform distributions Randomized playing strategiesRandomized playing strategies Symmetric strategiesSymmetric strategies Zero-information strategiesZero-information strategies

ReferencesReferences Hendrik W. Lenstra, Jr. and Gadiel Hendrik W. Lenstra, Jr. and Gadiel

Seroussi, Seroussi, On Hats and Other CoversOn Hats and Other Covers, , preprint, 2002, preprint, 2002, http://www.hpl.hp.com/research/info_theohttp://www.hpl.hp.com/research/info_theory/hats_extsum.pdf .ry/hats_extsum.pdf .

J.P. Buhler, J.P. Buhler, Hat TricksHat Tricks, The Mathematical , The Mathematical Intelligencer Intelligencer 24 24 (2002), no. 4, 44 – 49.(2002), no. 4, 44 – 49.

Sarah Robinson, Sarah Robinson, Why Mathematicians Why Mathematicians Now Care About Their Hat ColorNow Care About Their Hat Color, New , New York Times, Science Times, p. D5, April York Times, Science Times, p. D5, April 10, 2001.10, 2001.