How to make the most of a shared meal: plan the last bite first
HHOW TO MAKE THE MOST OF A SHARED MEAL:PLAN THE LAST BITE FIRST.
LIONEL LEVINE AND KATHERINE E. STANGE
Abstract.
If you are sharing a meal with a companion, how best tomake sure you get your favourite mouthfuls? Ethiopian Dinner is a gamein which two players take turns eating morsels from a common plate.Each morsel comes with a pair of utility values measuring its tastiness tothe two players. Kohler and Chandrasekaran discovered a good strategy— a subgame perfect equilibrium, to be exact — for this game. We givea new visual proof of their result. The players arrive at the equilibriumby figuring out their last move first and working backward. We concludethat it’s never too early to start thinking about dessert.
Introduction
Consider two friendly but famished acquaintances sitting down to dinnerat an Ethiopian restaurant. The food arrives on a common platter, and eachfriend has his own favourite and not-so-favourite dishes among the spread.Hunger is a cruel master, and each of our otherwise considerate companionsfinds himself racing to swallow his favourites before his comrade can scoopthem up. Each is determined to maximize his own gastronomic pleasures,and could not care less about the consequences for his companion.An
Ethiopian Dinner is a finite set D = { m , . . . , m n } , with m i = ( a i , b i )whose elements are called morsels . Each morsel m i is an ordered pair ofreal numbers ( a i , b i ). Two players, Alice and Bob, take turns removing onemorsel from D and eating it. The morsels are discrete and indivisible: eachmorsel can be eaten exactly once, and the game ends when all morsels havebeen eaten. The larger the value of a i , the tastier the morsel for Alice; thelarger b i , the tastier for Bob. We assume that the players know one another’spreferences , and that the preferences are totally ordered, that is, a i (cid:54) = a j and Date : September 23, 2011.2010
Mathematics Subject Classification.
Key words and phrases. efficiently computable equilibrium, nonzero-sum selectiongame, subgame perfect equilibrium. a r X i v : . [ c s . G T ] S e p LIONEL LEVINE AND KATHERINE E. STANGE b i (cid:54) = b j for i (cid:54) = j . Alice’s score is the sum of the a i for the morsels m i sheeats, while Bob’s score is the sum of the b i for the morsels m i he eats.In such a game, the players are not adversaries; in fact, the game mayend quite peaceably and successfully for both players if they have dissimilartastes. The question we are interested in is this: if a player acts rationally tomaximize her own score, and assumes that her meal partner does the same,what should be her strategy?Eating your favourite morsel on the first move of an Ethiopian Dinner isnot necessarily a good strategy. For example, if the dinner is D = { (1 , , (2 , , (3 , } , then Alice’s favourite morsel is (3 , , ,
2) for a total score of 4. Instead Aliceshould snag (2 ,
3) on the first move; after Bob takes (1 , ,
1) for dessert and a total score of 5.If deciding on the first move in an Ethiopian Dinner appears complicated,the last move is a different matter. The subject of this paper is a strategydiscovered by Kohler and Chandrasekaran [12], which we call the crossoutstrategy . Its mantra is:“
Eat your opponent’s least favourite morsel on your own lastmove. ”To arrive at this strategy, each player reasons informally as follows. Myopponent will never choose her least favourite morsel, unless it is the only oneleft; therefore, unless this is my last move, I can safely save my opponent’sleast favourite morsel for later.This reasoning predicts that if, say, Bob has the last move of the game,then Bob’s last move will be to eat Alice’s least favourite morsel. Becausethis is a game of perfect information, both players can use this reasoningto predict with certainty the game’s last move. We now cross out Alice’sleast favourite morsel from the dinner D to arrive at a smaller dinner D (cid:48) in which Alice has the last move. The same reasoning now implies that onher last move, Alice will eat Bob’s least favourite morsel in D (cid:48) . We thencross out Bob’s least favourite morsel from D (cid:48) and proceed inductively, al-ternately crossing out Alice’s least favourite and Bob’s least favourite amongthe remaining morsels until all morsels have been crossed out. The crossoutstrategy is to eat the last morsel to be crossed out. Strategies and equilibria.
To convert the informal reasoning above intoa proof that crossout is a “good strategy,” we need to define both words!A strategy is any function s from nonempty dinners to morsels, such that s ( D ) ∈ D for all nonempty dinners D . Playing strategy s means that if it OW TO MAKE THE MOST OF A SHARED MEAL 3 is your turn and the remaining set of morsels is D , then you eat the morsel s ( D ). For example, the Alice greedy strategy chooses the morsel m i ∈ D suchthat a i is largest. Either player could play this strategy. (It would be prettyspiteful of Bob to do so!) Other examples are the “competitive” strategythat chooses the morsel maximizing a i + b i , the “Alice cooperative” strategychoosing the morsel maximizing a i − b i , and the “Alice masochistic” strategychoosing the morsel minimizing a i . A strategy can be as complicated as youlike: perhaps if the number of morsels is prime, you greedily choose yourown favourite morsel, and otherwise you spitefully choose your opponent’sfavourite.The appropriate notion of good strategy depends on the class of gamesone is considering. Ethiopian Dinner is a nonzero-sum game: one player’sgain may not be the other’s loss. In such a game, the basic requirement ofany pair of good strategies (one for Alice, one for Bob) is that they forma Nash equilibrium , which means that neither player can benefit himself bychanging strategies unilaterally.A Nash equilibrium represents a stable, predictable outcome: Alice candeclare, “I am playing my equilibrium strategy, and you’d do best to playyours.” If Bob responds rationally by playing his own equilibrium strategy,then both players know how the game will turn out.
Subgame perfect equilibria.
A game may have many equilibrium strat-egy pairs, some with better outcomes than others, so one tends to look forequilibria with further desirable properties. Which properties again dependson the class of games being considered. In game theory lingo, Ethiopian Din-ner is a perfect-information non-cooperative game in extensive form . Thatis, both players know the values a , . . . , a n and b , . . . , b n (perfect informa-tion); the players may not bargain or make side deals (non-cooperative); andthe players alternate making moves (extensive form).Non-cooperative games model situations in which the players have noway of communicating (our dinner guests may be too busy stuffing theirfaces to bother with conversation) or are forbidden to collude. For instance,airlines are forbidden by law from colluding to fix prices. Colluding to fix theoutcome of a meal is still legal in most countries, but Alice might neverthelessbe dissuaded by cultural taboo from making propositions like “If you payme fifty cents I promise not to eat any more spinach.”An Ethiopian Dinner with n morsels is certain to end in n moves. A widelyaccepted notion of a good strategy for games of this type (perfect informa-tion, non-cooperative, extensive form, finite length) is the subgame perfectequilibrium . This is a refinement of the Nash equilibrium which requires thatthe strategies remain in equilibrium when restricted to any subgame. In our LIONEL LEVINE AND KATHERINE E. STANGE B o b ’ ss c o r e − → Alice’s score −→ Figure 1.
Plot of the score pairs for all possible outcomes ofa permutation dinner D of size 14. The large dot • at upperright represents the score Alice and Bob receive if they bothplay the crossout strategy. The light dots • represent scores forstrategy pairs of the form ( s, c ) for s arbitrary: these are all ofthe outcomes Alice can obtain playing against Bob’s crossoutstrategy. According to Theorem 1, among these outcomesshe does best when she herself plays crossout. Dark dots • represent the outcomes of all other strategy pairs. Producedusing Sage Mathematics Software [17].case, a subgame is just a subdinner consisting of a subset of the morsels,with the same player moving last. A subgame perfect equilibrium is robustin the sense that even if one player, say Bob, makes a “mistake” on a partic-ular move by deviating from his equilibrium strategy, Alice can confidentlycontinue playing her equilibrium strategy because the same strategy pairis still an equilibrium of the resulting subgame. See, e.g., [15] and [16] forbackground on these concepts.Let c be the crossout strategy described above for Ethiopian Dinner. Wewill give a new proof of the following theorem, which is due to Kohler andChandrasekaran [12]. Theorem 1.
The pair ( c, c ) is a subgame perfect equilibrium. OW TO MAKE THE MOST OF A SHARED MEAL 5
In other words, if Alice plays crossout, then Bob cannot benefit himself byplaying a different strategy, and vice versa.Figure 1 illustrates Theorem 1 in the case of a particular permutationdinner , that is, a dinner of the form D = { (1 , b ) , (2 , b ) , . . . , ( n, b n ) } where b , . . . , b n is a permutation of the numbers 1 , . . . , n . Each dot in thefigure represents the outcome of a strategy pair, with Alice’s score plottedon the horizontal axis and Bob’s score on the vertical axis, for the followingpermutation dinner of size 14: D = { (1 , , (2 , , (3 , , (4 , , (5 , , (6 , , (7 , , (8 , , (9 , , (10 , , (11 , , (12 , , (13 , , (14 , } . To visualize Theorem 1, note that the large dot • in Figure 1, which repre-sents the outcome when both players play crossout, is rightmost among allpossible outcomes achievable by Alice given that Bob plays crossout (suchoutcomes are indicated by light dots • ).Recall that any function s from dinners to morsels such that s ( D ) ∈ D for all dinners D counts as a strategy. In particular, let k be the number ofmoves that Alice makes in D . For any subset A ⊂ D of size k , there is astrategy pair ( s, s (cid:48) ) such that if Alice plays s and Bob plays s (cid:48) , then Alicewill eat precisely the morsels in A and Bob will eat the rest. This is how wecomputed the set of all dots in Figure 1.Part of the difficulty of proving that a pair of strategies is in equilibrium– and also part of the power of the result once it is proved – stems fromthe vast number of possible strategies. How are we to rule out that anyconceivable strategy s , which might be arbitrarily subtle, clever and com-plicated, performs better against c than does c itself? For all we know,Alice has access to unlimited computational resources, and her strategy s could involve factoring enormous integers or solving large instances of thetraveling salesman problem. Nevertheless, Theorem 1 ensures that if Bobplays c , then Alice would do at least as well playing c as s . In a sense, atheorem about equilibrium is a theorem about the limits of intelligence: Anequilibrium enables you to hold the line against a smarter opponent! Crossout is an efficiently computable equilibrium.
In games arisingin the real world, for instance in evolutionary dynamics and in economics,the appeal of the Nash equilibrium concept is twofold. First, it can explainwhy we observe certain strategies and not others. Second, even in the case ofa game that has multiple equilibria and lacks a well-defined “best” outcome,knowing an explicit equilibrium provides certainty. Alice simply announces
LIONEL LEVINE AND KATHERINE E. STANGE her intention to play crossout, refers Bob to the proof of Theorem 1 andtrusts that his own best interest compels him to follow suit. What mighthave been a tense evening with an unpredictable outcome becomes a morerelaxed affair in which each player can predict in advance which morsels shewill be gobbling up.To reap these benefits, the players must be able to compute an equilibriumpair, not just know that one exists! A recent strand of research, popularizedby the slogan “ if your laptop can’t find it, then, probably, neither can themarket ,” has explored the tendency for equilibria to be extremely difficultto compute [9]. The general existence proof for subgame perfect equilibria[15, VIII.2.10] uses a backward induction from the last move: if convertednaively into an algorithm, it would seem to require searching through expo-nentially many move sequences in order to find an equilibrium. This kind ofbrute force search is typically out of the question even for games of moderatesize (for example, an Ethiopian Dinner of n morsels has n ! possible movesequences). For this reason, it is always interesting to identify special classesof games that have efficiently computable equilibria. The crossout equilib-rium for Ethiopian Dinner is an example: if both players play the crossoutstrategy, then they eat the morsels in reverse order of the crossouts. In thiscase, the entire move sequence of the dinner can be worked out in the order n log n time it takes to sort the two lists a , . . . , a n and b , . . . , b n . Proof of equilibrium
Dinners and strategies. A dinner is a finite set of morsels D = { m , . . . , m n } . Each morsel m ∈ D comes with a pair of real numbers u A ( m ) , u B ( m ) repre-senting its utility to Alice and Bob. We often write m as an ordered pair, m = ( u A ( m ) , u B ( m )) . We adopt the convention that
Bob has the final move by default. Sincemoves alternate, the first move is determined by the parity of n . Alice hasthe first move if n is even, and Bob has the first move if n is odd.A strategy is a map assigning to any non-empty dinner D a morsel s ( D ) ∈ D to be eaten by the first player. Suppose that P ∈ { Alice, Bob } is a player, D is a dinner, and that P plays strategy s . If it is P ’s turn to move, heselects morsel s ( D ) and receives payoff u P ( s ( D )). The remaining dinner is D − s ( D ), with his opponent to move. Suppose his opponent plays strategy t . OW TO MAKE THE MOST OF A SHARED MEAL 7
The score v DP ( s, t ) of player P is defined by the recurrence v DP ( s, t ) = v D − s ( D ) P ( s, t ) + u P ( s ( D )) if P plays first in D , v D − t ( D ) P ( s, t ) if P plays second in D ,0 if D = ∅ . (1)where for m ∈ D , the dinner D − m denotes D with morsel m removed.Since D has finitely many morsels, equation (1) defines v DP ( s, t ) uniquely.Our convention in denoting a player’s score is that his own strategy isalways the first listed in the ordered pair.Formally, we can regard Ethiopian Dinner as a single game whose positionscomprise all finite dinners. A pair of strategies ( s, t ) is a subgame perfectequilibrium for this game if v DA ( s (cid:48) , t ) ≤ v DA ( s, t ) and v DB ( t (cid:48) , s ) ≤ v DB ( t, s )for all strategies s (cid:48) and t (cid:48) and all finite dinners D . The crossout strategy.
After giving the formal definition of the crossoutstrategy described in the introduction, we explain how to visualize it using a“crossout board” and prove the lemma that lies at the heart of our argument,the Crossout Board Lemma (Lemma 2).Let D be a set of n morsels. Write (cid:96) A ( D ) for Alice’s least favourite morselin D , and (cid:96) B ( D ) for Bob’s least favourite morsel in D . Let D = D , and D i +1 = D i − m i , i = 1 , . . . , n − m i = (cid:40) (cid:96) A ( D i ) , i odd (cid:96) B ( D i ) , i even . The sequence of morsels m , m , . . . , m n is called the crossout sequence of D . Note that m is Alice’s least favourite morsel in Dm is Bob’s least favourite morsel in D − m m is Alice’s least favourite morsel in D − m − m m is Bob’s least favourite morsel in D − m − m − m ...Now suppose D is a dinner (i.e., a set of n morsels with Bob distinguishedto move last). The crossout strategy c is defined by c ( D ) = m n . Note that if
LIONEL LEVINE AND KATHERINE E. STANGE •• ••• •• • − −− − − − − − B o b ( b ) − → Alice ( a ) −→ D •• •• •• • − −−− − − − B o b ( b ) − → Alice ( a ) −→ D − m Figure 2.
Left: Example of a crossout board for a dinner D with 8 morsels. Labels on the axes indicate the crossoutsequence. Right: The crossout board for the dinner D − m , inwhich a morsel m has been removed. By the Crossout BoardLemma, each label on the right is at least as far from theorigin as the corresponding label on the left.both players play the crossout strategy, then they eat the morsels in reverseorder of the crossout sequence: m n = c ( D ) m n − = c ( D − m n ) m n − = c ( D − m n − m n − )... m = c ( D − m n − · · · − m ) . Thus m , which is Alice’s least favourite morsel in D , is eaten by Bob onthe last turn. Crossout boards.
To prepare for the proof of Theorem 1, it is convenientto illustrate the crossout sequence with a crossout board , as in Figure 2. Wedisplay the dinner on a Cartesian coordinate plane. Each morsel m = ( a, b )is graphed as a dot at coordinate ( a, b ). Since we assume that the players’preferences are totally ordered, each vertical or horizontal line passes throughat most one morsel. The crossout sequence itself is indicated by writing the OW TO MAKE THE MOST OF A SHARED MEAL 9 number (or label ) i on the a -axis below m i if i is odd, and on the b -axis tothe left of m i if i is even.Figure 2 shows the crossout board of the dinner D = { (1 , , (2 , , (3 , , (4 , , (5 , , (6 , , (7 , , (8 , } and of D − m , where m is the morsel (6 , Lemma 2 (Crossout Board Lemma) . Let D be a dinner, and (cid:98) D ⊂ D asubdinner. For each k = 1 , . . . , | (cid:98) D | the location of label k in the crossoutboard of (cid:98) D is at least as far from the origin as the location of label k in thecrossout board of D .Proof. Let B be the crossout board for D , with crossout sequence m , m , . . . , m | D | . Let (cid:98) B be the crossout board for (cid:98) D , with crossout sequence (cid:98) m , (cid:98) m , . . . , (cid:98) m | (cid:98) D | . For morsels p and q of D , we write p < D q to mean that p appears before q in the crossout sequence for D . If p and q are also morsels of (cid:98) D , then wewrite p < (cid:98) D q to mean that p appears before q in the crossout sequence for (cid:98) D . In particular, for any 1 ≤ j, k ≤ | (cid:98) D | we have m j < D m k ⇐⇒ j < k ⇐⇒ (cid:98) m j < (cid:98) D (cid:98) m k . (2)Given 1 ≤ k ≤ | (cid:98) D | , let us say k is jumpy if the label k is strictly closer tothe origin in (cid:98) B than in B . We will show that there are no jumpy labels.Let P be the player who places the label k (so P is Alice if k is odd, Bobif k is even). When P places the label k on board B next to the morsel m k , this morsel is the closest available to the origin along P ’s axis. If k is jumpy, then the morsel (cid:98) m k is closer to the origin along P ’s axis, which (cid:31)(cid:31)(cid:31)(cid:31)(cid:31)(cid:31)(cid:31) (cid:31)(cid:31)(cid:31)(cid:31)(cid:31)(cid:31)(cid:31) k (in (cid:98) B ) k (in B ) • • (cid:98) m k m k • origin axis Figure 3.
If label k is closer to the origin in (cid:98) B than in B ,then (cid:98) m k must have been labeled already in B by some j < k ,i.e. (cid:98) m k < D m k .means that (cid:98) m k is unavailable , that is, it was already labeled in B by some j < k (Figure 3). Hence k is jumpy = ⇒ (cid:98) m k < D m k . (3)Note also that if (cid:98) m k < (cid:98) D m k , then in the crossout sequence for (cid:98) B , both (cid:98) m k and m k are available at step k and (cid:98) m k is chosen. Therefore, the label k isplaced closer to the origin in (cid:98) B than in B . Hence (cid:98) m k < (cid:98) D m k = ⇒ k is jumpy . (4)Now suppose for a contradiction that one of the labels 1 , . . . , | (cid:98) D | is jumpy,and let k be the smallest jumpy label. Since (cid:98) m k ∈ (cid:98) D and (cid:98) D ⊂ D , the morsel (cid:98) m k also belongs to D . Let j be its label on the crossout board of D ; that is, m j = (cid:98) m k . (5)Then k is jumpy = ⇒ (cid:98) m k < D m k by (3)= ⇒ m j < D m k by (5)= ⇒ j < k by (2)= ⇒ (cid:98) m j < (cid:98) D (cid:98) m k by (2)= ⇒ (cid:98) m j < (cid:98) D m j by (5)= ⇒ j is jumpy by (4) OW TO MAKE THE MOST OF A SHARED MEAL 11 •• ••• •• • B B B B A A A A B o b ( b ) − → Alice ( a ) −→ Figure 4.
A crossout board showing the sequence of play:Alice eats the morsel above the label A i on turn i , and Bobeats the morsel to the right of the label B j on turn j .That is, j < k and j is jumpy. But k was the smallest jumpy label. Thiscontradiction shows that there are no jumpy labels, completing the proof. (cid:3) The crossout scores χ A ( D ) and χ B ( D ) are the scores for Alice and Bobwhen both play the crossout strategy: χ A ( D ) = v DA ( c, c ) = m + m + · · · + m (cid:98) n/ (cid:99) ,χ B ( D ) = v DB ( c, c ) = m + m + · · · + m (cid:100) n/ (cid:101)− . These scores are easy to read off from the crossout board. The unlabeled morsel locations on a player’s axis are precisely the utilities of the morselshe eats if both players follow the crossout strategy. Therefore, the crossoutscores χ A ( D ) and χ B ( D ) are obtained by summing the unlabeled locations(marked with dashes in Figure 2) on the a - and b -axes respectively. Forinstance, for the board D pictured in Figure 2, we have χ A ( D ) = 4+5+6+8and χ B ( D ) = 3 + 5 + 6 + 8.If we also wish to show the order of play, then we can label the a -coordinateof the morsel eaten by Alice in turn i with the symbol A i , and the b -coordinate of the morsel eaten by Bob in turn j with the symbol B j asshown in Figure 4. Alice’s score is the sum of the a -coordinates labeled with A ’s, and Bob’s score is the sum of the b -coordinates labeled with B ’s.In D of Figure 4 we see that Alice, who plays first, eats her favouritemorsel (8 ,
7) on her first turn. In the remaining game D − (8 , first but does not eat his favourite morsel (1 ,
8) until his last move (for suchis Alice’s loathing for it that he can safely ignore it until the end). Aninteresting property of the crossout strategy, which we leave as an exerciseto the reader since it is not needed for the proof of the main theorem, is thatif both players follow it, then the first player eventually eats her favouritemorsel.
The main lemma.
The next lemma shows that neither player can improvehis crossout score by choosing a different first morsel.
Lemma 3 (Main Lemma) . Let D be a dinner, and let m be a morsel of D .Let P be the player to move first in D . Then u P ( m ) + χ P ( D − m ) ≤ χ P ( D ) . Proof.
We compare the crossout boards for D and D − m (Figure 2). In thelatter, a morsel has been removed. Player P is the second player to move in D − m , so he swallows one fewer morsel in D − m than in D . This meansthe number of labels on P ’s axis is the same in the crossout boards of D and D − m . By the Crossout Board Lemma 2, each label on the board for D − m is no closer to the origin than the corresponding label on the boardfor D . Therefore the sum of the labeled positions on P ’s axis is at least aslarge in D − m as in D . Hence the sum of the unlabeled positions on P ’saxis is no larger in D − m than in D . For D , this sum is the crossout score χ P ( D ). For the board D − m , this sum consists of the score χ P ( D − m ) plusthe utility u P ( m ) of the removed morsel m . (cid:3) Proof of Theorem 1.
Let D be a dinner of n morsels, and let c be thecrossout strategy. We induct on n to show that for any player P ∈ { A, B } and any strategy s , v DP ( s, c ) ≤ v DP ( c, c ) . The base case n = 1 is trivial because c is the only strategy: In a game withone morsel, the only thing you can do is eat it!On to the inductive step. Suppose first that P is the first player to movein D . Let m = s ( D ). Then v DP ( s, c ) = u P ( m ) + v D − mP ( s, c ) by (1) ≤ u P ( m ) + v D − mP ( c, c ) by the inductive hypothesis ≤ v DP ( c, c ) by Main Lemma 3.It remains to consider the case that P is the second player to move in D .Letting m = c ( D ), we have by the inductive hypothesis and (1), v DP ( s, c ) = v D − mP ( s, c ) ≤ v D − mP ( c, c ) = v DP ( c, c ) OW TO MAKE THE MOST OF A SHARED MEAL 13 which completes the proof.
Concluding Remarks
We have analyzed Ethiopian Dinner as a non-cooperative game, and foundan efficiently computable subgame perfect equilibrium, the crossout strategy.Here we discuss its efficiency, and mention some variants and generalizations.
Pareto efficiency and inefficiency. An outcome of a dinner D is a par-tition of its morsels between Alice and Bob. An outcome is called Paretoinefficient if there exists another outcome that is at least as good for bothplayers and is strictly better for one of them. Equilibrium strategies may re-sult in an outcome that is Pareto inefficient, as demonstrated by the famousPrisoner’s Dilemma, in which both players do better by mutual coopera-tion than by mutual defection even though mutual defection is the uniqueequilibrium [15].For the permutation dinner shown in Figure 1, we see that the crossoutoutcome ( c, c ) is Pareto efficient because there are no dots lying (weakly)both above and to the right of the crossout score ( • ). In fact, Brams andStraffin [6, Theorem 1] prove that the crossout outcome is always Pareto ef-ficient with respect to the “pairwise comparison” partial order on outcomes.On the other hand, the crossout outcome is not always Pareto efficient for thescore function we have been considering, the sum of the utilities of morselseaten. Among permutation dinners, the smallest Pareto inefficient examplesoccur for dinners of size 6, for which there are two: D = { (1 , , (2 , , (3 , , (4 , , (5 , , (6 , } ,D = { (1 , , (2 , , (3 , , (4 , , (5 , , (6 , } . If each player employs his greedy strategy in dinner D , then Alice’s score of6 + 4 + 3 is equal to her crossout score, while Bob’s score of 6 + 5 + 1 is onepoint better than his crossout score. The outcome that improves on crossoutfor D is rather interesting: Bob can do one point better than his crossoutscore of 11 without hurting Alice’s score, but only if Alice magnanimouslyleaves him the most delicious morsel (6 ,
6) instead of gobbling it up on thefirst bite! Bob must then return the favor by refraining from eating (3 , c, c ) outcome wasPareto efficient, and in all but 241 it was weakly Pareto efficient (that is,no other outcome resulted in strict improvements for both players). The improvement in scores achieved by alternate outcomes was small: in anexhaustive check of all possible game outcomes in all 10 000 dinners, thelargest improvement for any player was less than 8%. These findings providesome evidence that the crossout strategy is reasonably efficient. Generalized payoffs.
The outcome of an Ethiopian Dinner is a partitionof the index set { , . . . , n } into a set A = { i , . . . , i (cid:98) n/ (cid:99) } of (cid:98) n/ (cid:99) morselseaten by Alice and a set B = { j , . . . , j (cid:100) n/ (cid:101) } of (cid:100) n/ (cid:101) morsels eaten by Bob.We have assumed that the final scores (payoffs) for Alice and Bob take theform p A = (cid:88) i ∈A a i and p B = (cid:88) j ∈B b j . This particular payoff function is not essential for the argument, however.Let f A : R (cid:98) n/ (cid:99) → R and f B : R (cid:100) n/ (cid:101) → R be functions that are strictly increasing in each coordinate, and symmetricwith respect to permutations of the coordinates. Then the Ethiopian Dinnergame with payoffs p A = f A ( a i , . . . , a i (cid:98) n/ (cid:99) ) and p B = f B ( b j , . . . , b j (cid:100) n/ (cid:101) )has crossout as its optimal strategy. Indeed, the proof we have given usesonly the relative order of the a i and the b j , and not their actual values.One could also generalize the payoff function so that Alice’s payoff dependsnot only on the morsels she ate but also on the morsels Bob ate, and viceversa. A natural choice is p A = α A (cid:88) i ∈A a i + β A (cid:88) j ∈B b j , and p B = α B (cid:88) i ∈A a i + β B (cid:88) j ∈B b j . That is, Alice’s payoff is α A times the sum of her own utilities of the morselsshe ate, plus β A times the sum of the utilities to Bob of the morsels Bobate. Bob’s payoff is defined similarly. The ratios β A /α A and α B /β B measurethe degree of altruism of the two players. The scenario of friends eating inan Ethiopian restaurant might correspond to values of these ratios strictlybetween 0 and 1. One can also imagine scenarios where β A /α A >
1: perhapsAlice is Bob’s mother and the morsels in question are brussels sprouts.All of these games turn out to be equivalent to Ethiopian Dinner. Supposewe are considering the above payoffs p A and p B on the dinner D consistingof morsels m i = ( a i , b i ) for i = 1 , . . . , n . Translating all of a player’s utilities OW TO MAKE THE MOST OF A SHARED MEAL 15 • • • • • • • •− − − − − − − − B o b ( b ) − → Alice ( a ) −→ competitive • • • • • • • • −−−− − − − − B o b ( b ) − → Alice ( a ) −→ cooperative Figure 5.
Crossout boards for a fully competitive and fullycooperative dinner.by an additive constant has no effect on strategy, so we may assume that (cid:80) ni =1 a i = (cid:80) ni =1 b i = 0. Then (cid:88) j ∈B b j = − (cid:88) i ∈A b i and (cid:88) i ∈A a i = − (cid:88) j ∈B a j . Now let D (cid:48) be the dinner consisting of morsels m (cid:48) i = ( α A a i − β A b i , β B b i − α B a i ) , for i = 1 , . . . , n. Any strategy s on D has a corresponding strategy s (cid:48) on D (cid:48) (which chooses m (cid:48) i whenever s chooses m i ), and p DP ( s, t ) = v D (cid:48) P ( s (cid:48) , t (cid:48) )for both players P ∈ { A, B } . In other words, the modified payoff in D equals the usual Ethiopian Dinner payoff in D (cid:48) . Now let s be the strategyon D for which s (cid:48) = c (that is, s (cid:48) is the crossout strategy on D (cid:48) ). Then thepair ( s, s ) is an equilibrium for the modified payoff dinner D .We distinguish two extreme cases.If α A = β B = 1 and α B = β A = −
1, then the game is zero-sum. Inthe terminology of combinatorial game theory, each morsel m = ( a, b ) is a switch { a | − b } , so the full game is a sum of switches. The morsel ( a, b )has temperature a + b , and optimal play proceeds in order of decreasingtemperature (see [2] for background). The equivalent Ethiopian Dinner D (cid:48) has morsels ( a i + b i , a i + b i ) of equal appeal to both players, and crossout on D (cid:48) gives the same decreasing-temperature play. This dinner could result ina couple of burned tongues!If α A = α B = β A = β B = 1, then the game is fully cooperative. Bothplayers have the same goal of maximizing their joint welfare. Since the gamerules constrain them to alternate moves, the optimal play is the following.Order the morsels m , . . . , m n so that a i − b i is a decreasing function of i .Alice takes morsels m , . . . , m (cid:98) n/ (cid:99) , and Bob takes morsels m (cid:98) n/ (cid:99) +1 , . . . , m n .In this case, the equivalent Ethiopian Dinner D (cid:48) has morsels ( a i − b i , b i − a i )and crossout on D (cid:48) gives the optimal strategy just described. Figure 5shows examples of crossout boards for a zero-sum (competitive) dinner anda cooperative dinner. Combinatorics at the dinner table.
We can measure the “cooperative-ness” of a permutation dinner by its inversions. Let π = ( π , . . . , π n ) be apermutation of 1 , . . . , n . For each pair of indices i < j such that π i > π j , wecall the pair ( i, j ) a left inversion of π and the pair ( π i , π j ) a right inversion of π . Both players should be pleased with a permutation dinner if it has alot of inversions, because each inversion represents a pair of morsels m i , m j such that Alice prefers m j while Bob prefers m i . Hopkins and Jones [11]show that if the left inversions of π are a subset of the left inversions of π (cid:48) ,then Alice’s crossout score for the permutation dinner π (cid:48) is at least as goodas for π . In fact they show more: there is a bijection between the set ofmorsels Alice eats in π and the set she eats in π (cid:48) such that each morsel eatenin π (cid:48) is at least as tasty to Alice as the corresponding one in π . (Alice prefersa prime piece of pie to an ordinary one: after all, who wouldn’t?) Likewise,Bob prefers dinners with a lot of right inversions. (Curiously, although rightinversions are in bijection with left inversions, set inclusion of right inversionsinduces a different partial ordering on permutations than does set inclusionof left inversions, as the reader can verify for permutations of 3 elements!)Consider permutation dinners having an even number of morsels, n = 2 k .If we define an outcome for Alice as the set of utilities of the morsels sheeats, then the number of conceivable outcomes (over all permutation dinnersof size 2 k ) is the binomial coefficient (cid:0) kk (cid:1) . But not all of these outcomes areachievable if both players play the crossout strategy. For instance, Alicenever has to eat her least favourite morsel (of utility 1), and she eats atmost one of her three least favourite morsels (of utilities 1, 2 and 3). Ingeneral, for each j ≥ j of her 2 j + 1 least favouritemorsels. Hopkins and Jones [10, 11] show that if both players play crossout,then the number of attainable outcomes for Alice is the famous Catalannumber C k = k +1 (cid:0) kk (cid:1) , and the number of attainable outcomes for Bob is C k +1 . OW TO MAKE THE MOST OF A SHARED MEAL 17
Cake cutting and envy-free division.
There is a large literature on cake-cutting [7] in which a cake (identified with the interval [0 , envy-free partition of the cake, whichmeans that each player prefers the piece assigned to him over the piecesassigned to the other players. When the cake is comprised of indivisibleslices (as, for example, in a game of Ethiopian Dinner), this criterion becomesimpossible to achieve in general, and finding an envy-minimizing allocation isa hard computational problem [14]. The outcome of the crossout strategy isreasonably close to envy-free: the first player is not envious, and the secondplayer’s envy is bounded by the utility of his favourite morsel. Anotherway of achieving an approximately envy-free allocation is described in [14,Theorem 2.1]. The algorithm described there is even faster than crossout,because it does not require sorting the lists of utilities. In the case when anenvy-free allocation exists, the undercut procedure of [5] gives an algorithmfor finding one. First move advantage.
Who takes the first bite at dinner might seem oflittle consequence, but imagine instead that Alice and Bob are taking turnsdividing up the assets of their great uncle’s estate. Alice, choosing first,claims the mansion, leaving Bob to console himself with the sports car. Toreduce the advantage of moving first, Brams and Taylor [8, Ch. 3] proposewhat they call “balanced alternation,” which uses the Thue-Morse sequence
A, B, B, A, B, A, A, B, B, A, A, B, A, B, B, A, . . . as the order of turns. Alice chooses the first item, then Bob chooses thesecond and third, Alice the fourth, and so on. The Thue-Morse sequenceis the unique sequence starting with A that is fixed under the operation ofreplacing each term A by A, B and each term B by B, A . Here is anotherdefinition that the reader might enjoy proving is equivalent: the terms areindexed starting from n = 0, and the n -th term equals A or B according towhether the sum of the binary digits of n is even or odd. Many amazingproperties of this sequence can be found in [1].A further innovation of [8] aimed at reducing the first move advantage issimultaneous choices. On each turn both players choose a morsel, and if theychoose distinct morsels then each eats his choice. If they choose the samemorsel, then that morsel is marked as “contested” and simultaneous playcontinues with the remaining uncontested morsels. When only contestedmorsels remain, the players revert to the Thue-Morse order of play.It would be interesting to quantify the intuition that the Thue-Morse ordertends to produce a fair outcome. Some assumptions on the utilities would be needed: In the example of the estate, if the value of the mansion exceedsthe combined value of all the other assets, then the first player retains anadvantage no matter what. Inducing sincerity.
Mechanism design asks how the rules of a game shouldbe structured in order to induce the players to act in a certain way. Acommon goal is to induce the players to make sincere choices – that is,choices in line with their actual preferences. For example, the voter whodeserts his preferred candidate for one he deems has a better chance ofwinning is making an in sincere choice. Making sincere choices in EthiopianDinner means playing the greedy strategy of always eating your favouriteavailable morsel. Because the greedy strategy pair is not an equilibrium,the players have an incentive to make insincere choices by playing the morecomplicated crossout strategy. How could the rules of Ethiopian Dinner bemodified so as to make the greedy strategy pair an equilibrium? Brams andKaplan [4] (see also [3, ch. 9]) design a mechanism that allows the playersof an Ethiopian Dinner to offer one another trades, and show that its effectis to make the greedy strategy pair an equilibrium. Open questions
We conclude by describing a few natural variants that we do not knowhow to analyze.
Delayed gratification.
Suppose that the utilities u A ( m ) and u B ( m ) de-pend not only on the morsel m , but also on when it is eaten. A naturalchoice is to value a morsel eaten on turn i with λ i times its usual value,for a parameter λ <
1. Thus, a morsel declines in value the longer it re-mains on the plate (perhaps the delicate flavours are fading). The choice ofexponential decay λ i corresponds to the common assumption in economicsthat a payoff received in the future should be discounted to its net presentvalue according to the prevailing interest rate. If the interest rate is α , then λ = 1 / (1 + α ). In the resulting game, each player feels an urgency to eather favourites early on. Because time-sensitive payoffs break the symme-try assumption, our proof of equilibrium does not apply. Can the crossoutstrategy be modified to produce an equilibrium? Inaccessible morsels.
Ethiopian food is served atop injera , a layer ofspongy bread that can only be eaten once it is revealed. If the game isplayed with the requirement that the order of consumption must respect afixed partial ordering on the morsels, the crossout strategy may not be anallowable strategy. What should take its place?
OW TO MAKE THE MOST OF A SHARED MEAL 19
Imperfect information.
How should your dinner strategy change if youdon’t know your opponent’s preferences? Perhaps the morsels come in dif-ferent types and there are many identical morsels of each type. Then youmight start by playing greedily, while trying to infer your opponent’s pref-erences from his play. Once you start to form a picture of his preferences,you can start playing a crossout-style strategy. Bob seems to be stayingaway from the spinach, so you can probably leave it for later. But was it anhonest signal when he ate all those chickpeas? Can deceit pay in a dinnerof imperfect information?
Three’s a crowd.
Ethiopian Dinner resembles the process of draft picksin sports. Each team participating in the draft has its own belief abouthow much each player is worth, and the teams draft players one at a timeaccording to some predetermined order of play. Typically, many teams (morethan two!) participate in the draft. Brams and Straffin [6] point out anumber of pathologies in the case when the number of teams is greater thantwo. For example, it may be to a team’s advantage to choose later in thedraft. For players with certain specific preferences (called “gallant knights”in [13]) the crossout strategy applies to games with any number of players.But the crossout strategy does not seem to apply to games of three or moreself-interested players, which leads us to end with a question. Is there anefficiently computable equilibrium for Ethiopian Dinner with three or moreplayers?
Acknowledgements.
The first author’s research was partially supported byan NSF Postdoctoral Research Fellowship. The second author’s researchwas supported by NSERC PDF-373333 and NSF MSPRF 0802915, and waspartially performed while the author was at Simon Fraser University and thePacific Insitute for the Mathematical Sciences at the University of BritishColumbia. We thank the Centre for Experimental and Constructive Mathe-matics at Simon Fraser University for providing computer resources, and weare grateful to Elwyn Berlekamp, Kevin Doerksen, Eric LeGresley, JamesPropp, Scott Sheffield and Jonathan Wise for useful suggestions. Thanks toRenato Paes Leme for bringing reference [12] to our attention, and to anony-mous referees whose astute suggestions significantly improved the paper.We especially thank Asmara Restaurant in Cambridge, MA for the sharedmeal that inspired this paper. (The clear winner on that occasion was thefirst author’s wife, whose strategy was to ignore all talk of game theory andeat all of her favourite morsels while her dinner companions were distractedby analyzing the game!)
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E-mail address : [email protected] Department of Mathematics, Stanford University, 450 Serra Mall, Build-ing 380, Stanford, CA 94305. http://math.katestange.net
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