fair cake cutting

8/8/2019 Fair Cake Cutting

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An Envy-Free Cake Division ProtocolAuthor(s): Steven J. Brams and Alan D. TaylorSource: The American Mathematical Monthly, Vol. 102, No. 1 (Jan., 1995), pp. 9-18Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2974850

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Steven J. Brams and Alan D. TaylorOur startingpoint is the well-knownparentalsolution to the problemof dividingacake between two children so that each child thinks he or she has been treatedfairly:The parent instructsone of the childrento cut the cake into two pieces inany way he desires. The other child is then instructed o choose whicheverpieceshe prefers. This two-step sequence of instructions,known as "cut-and-choose,"providesa simple exampleof the kind of game-theoreticalgorithm hat Even andPaz [11] call a "protocol."Associatedwith the cut-and-chooseprotocol s a naturalstrategy or each child:The first child cuts the cake into two pieces that he considers o be equal, and thesecond child chooses a piece that she considers o be at least as large as the otherpiece. Notice that each child's strategy guarantees him or her "satisfaction,"regardlessof what the other child does.The general version of this problem involves n people ("players"),each ofwhom has his or her preferences over subsets of the cake given by a probabilitymeasure.1An allocationof the cake amongthe players s said to be proportional ifeach playerreceives a piece of size at least 1/n (in his or her own measure),and itis said to be envy-free if each playerreceives a piece he or she would not swap forthat received by any other player. It is easy to see that an envy-freeallocation sproportional,but the conversefails unless n = 2. Thus, for example,every one ofthree playersmay think his or her piece is at least 1/3, but a playermay think thatone of the other playershas a largerpiece.The results on proportionaland envy-freeallocationsobtained over the past 50yearstend to fall into one of four classes: i) ExistenceTheorems; ii) Moving-KnifeSolutions; (iii) Algorithms;and (iv) Protocols. We say something about each inturn.

Existence theorems, dating back to the 1940s, are often based on some versionof Liapounoff'sConvexityTheorem [20]. Typically, hey establishthe existence ofan ordered partition of the cake corresponding o an envy-free allocation, oftenwith some additionalproperty uch as: all the measuresof all the pieces are exactly1/n [21]; or the pieces are connected sets [27 and 31]; or the allocation is alsoPareto-optimal 30 and 4]. In the words of Rebman [24, p. 33], however, theseresults provide "no clue as to how to accomplish uch a wonderfulpartition."There are two well-knownmoving-knife olutions. The first is due to Dubinsand Spanier [10] and is a moving-knifeversion of the Banach-Knaster ast-1If one wantsto abandon he cake metaphor,and literallyworkwith arbitrary robabilitymeasureson some set C, then even cut-and-choose an fail. For example, f both childrenhave their preferencesgiven by the same 0-1 valued measure, .e., the same ultrafilter, hen both will want the same pieceregardless f how the cake is divided.For everythingwe do in the presentpaper, t suffices o assume hat our measuresare all definedonthe same algebrav of subsetsof C and satisfy he following wo properties: i) for everyset P E a?and every finite k, P can be partitioned nto k sets of equal measure,and (ii) for every P, Q E a?,either P can be trimmed o yield a subset the same size as Q, or vice-versa.

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An Envy-FreeCake Division Protocol


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diminisher chemefor n people that we shallpresent later. (The Dubins-Spanierschemeis easy to describe:a knife is slowlymoved along the top of the cake sothat all the slices made are parallel. Each player calls "cut" when he or she iswilling to take the resultingpiece as his or her allocation.) The second is amoving-knife cheme due to Stromquist 27] that yields an envy-freeallocationamongthree people. (This one is not so easy to describe,because envy-freeness sconsiderablyharderto obtainthanproportionality.)Moving-knife chemes, however,are not the step-by-stepprocesses one usuallyassociateswith the term "algorithm."A good example of what one would call analgorithm n this context is Woodall's scheme [32] for producing an allocationwherebyeach participantgets strictlymore than 1/n of the cake (according o hisownmeasure).This algorithm equires,as partof the input, a piece P of cake andtwodistinctnumberscgand p such that Player1 thinksthe measureof P is cg,andPlayer2 thinksthe measureof P is ,l3.An envy-freeversionof this algorithm s in[5].Finally,there is what Even and Paz [11] call a "protocol,"and this is the onlykind of result we are going to analyze in the present paper. Since we will bepresentingexamplesof protocols,as opposed to provingtheir non-existence,wecan afford the same level of informality n the descriptionof what is meant by aprotocol as Even and Paz used.A protocol s a computer-programmablenteractiveprocedure that can issuequeriesto the participantswhose answersmayaffect future decisions.It may issueinstructions o the participants uch as: "Choose k pieces from amongthese mpieces"or "Partition his piece into k subpieces."The protocolhas no informationon the measure of the variouspieces as seen by the differentparticipants this isprivateinformation.Moreover, if the participantsobey the protocol, then eachparticipantwill end up with a piece afterfinitely many steps.Still following[11], we define a strategyor a participant o be an adaptivesequence of moves consistent with the protocol, which the participantschoosesequentiallywhen called upon by the protocol.A protocol is proportionalf eachof the n participants as a strategy hat willguaranteehim at least 1/n of the cake(by his own measure), ndependentlyof the other participants' trategies. Purelyfor convenience,we will henceforthuse only the masculinepronoun.)Departingfrom [11], we will call a protocol envy-freef each of the n participantshas astrategy hat will guaranteehim a piece that is, according o his own measure, atleast tied for largest.A constructiveproof of the existence of, say, a proportionalprotocol involvesproducing hree separate things:the rules of the protocol,a strategy or each ofthe players,and an argument hat the strategiesdo, in fact, guaranteeeach playerhisproportional hare.We distinguish ulesand strategiesby demanding hat rulesbe enforceableby a referee implementing he protocol.Thisfmeansthat a statement ike "Player 1 cuts the cake into n pieces" is anacceptablerule,whereas a statement ike "Player1 cuts the cake into n pieces thathe considers to be equal" is not. This is because the latter statement cannot beenforced by the referee, who has no knowledge of Player l's measure and socannot tell if the rule has been followed or not.In presentingprotocols, we will separate rules from strategies by placing allstrategicaspects in parentheses.This providesone with the option of readingtherules alone in a reasonably moothway. All arguments hat the strategiesperformas advertisedare placed between steps and labeled as "Aside." For example, inour method of presentation,cut-and-choosebecomes:

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Cut-and-ChooseStep 1. Player 1 cuts the cake into 2 pieces (that he considersto be the samesize).Step 2. Player 2 chooses a piece (that she considers to be at least tied for

largest).Aside. Clearly,Player l's strategyguaranteeshim a piece of size exactly 1/2in his measure,while Player2's strategyguaranteesher a piece of sizeat least 1/2 in her measure.The modern era of cake cutting began with Steinhaus'observation"during hewar [World War II]" [25, p. 102] that the cut-and-choose protocol could beextended to yield a proportionalprotocol for three players (see [18]). He thenasked if it could be extended to yield a proportionalprotocol for the case n > 3.(Steinhaus,however,never used the word "protocol.")His questionwas answeredin the affirmativeby Banach and Knaster and reported in [25] and [26]. TheSteinhaus and Banach-Knaster protocols introduced two key ideas that wouldresurface n the envy-freesolutions 15 and 50 years later.The first idea was that of having an initial sequence of steps resulting n onlypart of the cake's being allocated to one player n this case). The sequence is thenrepeated a finite numberof times, after which the entire cake has been allocated.The second idea and perhaps he more importantof the two was that of havinga player trim a piece to a smallersize.Explicit mention of the lack of a constructive procedure for producing an

envy-free allocation among more than two people dates back at least to Gamowand Stern [14]. The first breakthroughs n this problemoccurred n the late 1950sand early 1960s, when the protocol solution to the envy-freeproblem for n = 3was found by John L. Selfridge, and rediscovered independently by John H.Conway.These solutions also involved trimmingand an initial allocation of onlypart of the cake; they were widely disseminatedby R. K. Guy and others, andeventuallyreportedby Gardner 15],Woodall [32], Stromquist 27], and Austin [1].The moving-knife olutionof Stromquist 27]was found two decades later, as was ascheme due to Levmore and Cook [19], which can be recast as quite a differentmoving-knife olution to the envy-freeproblemwhen n = 3. Still other envy-freemoving-knife chemes for three people [7] and, more recently, our people [8] havebeen discoveredand are summarized n [6].The extensionof the Selfridge-Conway rotocol to the case of even four peoplehas remained an open, and much-commentedupon, problem. See, for example,Gardner[15], Rebman [24], Stromquist 27], Woodall [31], [32], Bennett et al [3],Webb [29],Hill [16],[17], and Olivastro 22]. In what follows,we solve this problemby producingan envy-freeprotocol for arbitrary .We have chosen a uniform presentationof four protocols that highlightstheevolution of two important deas namely,trimming,and the use of sequences ofpartialallocations.Historically, hese four protocolsarose over a period of 50 yearsand nicely illustrate how ideas in mathematicsare built, one upon another. Theprotocolswe present are:

1. The proportionalprotocol for n = 3 (Steinhaus).2. The proportionalprotocol for arbitrary (Banach-Knaster).3. The envy-freeprotocol for n = 3 (Selfridge,Conway).4. The envy-freeprotocol for arbitrary .

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Before turningto the protocolsthemselves,we must acknowledge he help ofseveralpeople. Ourinterest n fair divisionwassparkedby Olivastro 22].Valuablemathematicalcontributionswere made by William Zwicker and Fred Galvin.Indeed, the presentversion of our envy-freeprotocol owes much to the reworkingof an earlierversionby Galvin.Specificobservations ndcommentsby DavidGale, SergiuHart,TheodoreHill,WalterStromquist,William Webb, and DouglasWoodallalso provedhelpful. Inaddition, we have benefited from conversationsand correspondencewith thefollowingpeople:Ethan Akin, JuliusBarbanel,JohnConway,MortonDavis, KarlDunz, Shimon Even,A. M. Fink, Peter Fishburn,MartinGardner,Richard Guy,D. Marc Kilgour, Peter Landweber, Jerzy Legut, Herve Moulin, DominicOlivastro,BarryO'Neill, PhilipReynolds,WilliamThomson,Hal Varian, C:harlesWilson, and H. Peyton Young.The first protocolwe present is a generalization f cut-and-choose o a propor-tionalprotocol for three people. This is the one found by SteinhausduringWorldWar II.

TheProportionalProtocol or n = 3(Steinhaus,circa 1943)Step 1. Player 1 cuts the cake into 3 pieces (that he considersto be the samesize).Step 2. Player2 is given the choice of either passing, .e., doing nothing(whichhe does if he thinks2 or moreof the pieces are of size at least 1/3), ornot passing and labeling 2 of the pieces (that he thinks are of size

strictly ess than 1/3) as "bad."Step 3. If Player2 passed in step 2, then Players 3, 2, and 1, in that order,choose a piece (that they considerto be of size at least 1/3).Aside. In thiscase, each playerreceivesa piece of size at least 1/3 in his ownmeasure.This is true of: Player 3, because he chooses first;Player 2,because he thinkseither 2 or 3 pieces are that large, and so at leastone of them will still be availableafterPlayer3 chooses his piece; andPlayer 1, becausehe made all 3 pieces of size 1/3.Step 4. If Player2 did not pass at Step 2, then Player3 is given the same tmrooptionsthat Player2 had at Step 2. He ignores Player2's labels.Step 5. If Player 3 passed in Step 4, then Players 2, 3, and 1, in that order,choose a piece (that theyconsiderto be of size at least 1/3).Aside. In this case, as before, eachplayerreceives a piece of size at least 1/3* , Mln nls own measure.Step 6. If Player3 did not pass at Step 4, then Player 1 is required o take apiece that bothPlayer2 and Player3 labelled as "bad."Aside. Note first that there certainly must be such a piece. At this point,Player 1 has received a piece that he thinks is of size exactly 1/3,whichboth Player 1 and Player 2 think is "bad," i.e., of size strictlyless than 1/3.Step 7. The other two pieces are reassembled,and Player2 cuts the resultingpiece into two pieces (that he considers o be the same size).Step 8. Player3 chooses one of the two pieces (that he considers o be at leasttied for largest).Step 9. Player2 is given the remainingpiece.Aside. This is just cut-and-choosebetween Players 2 and 3, which ends theprotocol.



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The second protocol we present followedquicklyon the heels of the first. It isthe Banach-Knaster rotocol,offered in response o Steinhaus'questionof whetherhis result could be extendedfrom 3 to n people.Note here the introduction f theidea of trimming,whichwill be furtherexploited n both of the upcomingenvy-freeprotocols.ProportionalProtocol or Arbitraryn(Banach-Knaster,irca 1944)

Step 1. Player 1 cuts a piece P1 (of size l/n) from the cake.Step 2. Player 2 is given the choice of either passing (which he does if hethinks P1 is of size less than l/n), or trimminga piece from P1 tocreate a smallerpiece (that he thinks s of size exactly /n). The pieceP1, nowperhaps rimmed, s renamed P2. The trimmings re set aside.Step 3. For 3 < i < n, Player i takes the piece Pi_l and proceeds exactly asPlayer 2 did in Step 2, with the resultingpiece now called Pi.Aside. For 1 < i < n, Player i thinks that Pi is of size less than or equal tol/n. We also have that P1 z *** 2 Pn. Thus, every playerthinksPnis of size at most l/n.Step 4. The last player to trim the piece, or Player 1 if no one trimmed t, isgiven Pn.Aside. The playerreceivingPn hinks it is of size exactly l/n.Step 5. The trimmingsare reassembled,and Steps 1-4 are repeatedfor theremainderof the cake, and with the remainingn - 1 players n placeof the originaln players.Aside. The playerwho gets a piece at this second stage is getting exactlyl/(n - 1) of the remainderof the cake; he, and everyoneelse, thinksthis remainders of size at least(n - l)/n. Hence, he thinkshis pieceis of size at least l/n.Step 6. Step 5 is iterated until there are only 2 players eft. The last 2 playersuse cut-and-choose.Aside As before,each playerreceivesa piece that he thinks s of size at leastl/n.This ends the protocol.

The next protocolwe present is the Selfridge-Conwaynvy-freeprotocolfor thecase n = 3. (There are slight differences in the presentationsof Selfridge andConway;we follow the latter.)This protocol nvolvesan elegant combination f thetrimming dea introducedby Banach-Knaster nd the basic framework hat Stein-haus used. It also introduces the important notion of one player'shaving an"irrevocable dvantage" ver anotherplayer,followinga partialallocation.Envy-FreeProtocol or n = 3(Selfridge,Conway, irca 1960)

Step 1. Player 1 cuts the cake into 3 pieces (that he considersto be the samesize).Step 2. Player 2 is given the choice of either passing (which he does if hethinks two or more pieces are tied for largest), or trimminga piecefrom (the largest)one of the three pieces (to create a tie for largest).IfPlayer 2 trimmed a piece, then the trimmings are named L, for"leftover,"and set aside.

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Step 3. Players3, 2, and 1, in that order, choose a piece (that they considertobe at least tied for largest) rom amongthe 3 pieces, one of which mayhave been trimmed n Step 2. If Player 2 did not pass in Step 2, thenhe is required o choose the piece he trimmed f Player3 did not.Aside. Notice that only part of the cake has been allocated. This yields apartition {X1, X2, X3, L} of the cake such that {X1,X2, X3} iS anenvy-free partial allocation. The lack of envy is true of: Player 3,because he chooses first;Player2, because he made at least two piecestied for largest,and so at least one of them will still be availableafterPlayer 3 chooses his piece; and Player 1, because he made all threepieces of size 1/3, and the trimmedone has definitelybeen taken byeither Player3 or Player2.Step 4. If Player2 passed at Step 2, we are done. Otherwise,either Player2 orPlayer 3 received the trimmed piece, and the other received anuntrimmedpiece. Whichever player received the untrimmed piecenow divides L into 3 pieces (that he considersto be the same size).Call this player the "cutter"and the other the "non-cutter."Aside. We will refer to Player 1 as having an irrevocabledvantagever thenon-cutter.The point is that, since the non-cutter eceived he trimmedpiece, Player 1 will not envy the non-cutter,regardlessf how L islater divided among the three.Step 5. The three pieces into which L is divided are now chosen by theplayers in the order: non-cutter first; Player 1 second; cutter third.(Each chooses a piece at least tied for largestamongthose available ohim when it is his turn to choose.)Aside. At this point, the entire cake has been allocated.Since the non-cutterchooses his piece of L first, he experiencesno envy. Player 1 does notenvy the non-cutter,since he had an irrevocableadvantageover him,and Player 1 does not envy the cutter, because he is choosinghis pieceof L before the cutter does. Finally, the cutter experiencesno envysince he divided L into three equal pieces.This ends the protocol.

The final protocolwe present is our envy-freeprotocolfor an arbitrary umberof players.This result was announced n [9, 12, 13, 23]. A brief discussionof someimportant differences between this protocol and the three earlier ones, and acouple of importantopen questions,follow.The centralfeature of our envy-freeprotocol, ike that for the n = 3 protocol, sthat players trim pieces of the cake to create ties, rendering them indifferentamong these pieces. When n > 3, however,one needs to start the trimmingandchoosing process leading to an envy-free partial allocation with more piecesthan there are players.As an informal illustrationof how to achieve an envy-free partial allocation,suppose there are four people. Have Player 1 cut the cake into 5 equal pieces.Player2 then trims 2 pieces, creatinga 3-waytie for largest.Player3 then trims 1piece, creatinga 2-waytie for largest.The playersnow choose in the order:Player4, Player3, Player2, Player 1, with the middle two playersrequired o take a piecethey trimmed f one is available.Clearly,each player hinkshis piece is at least tiedfor largest.The burdenof our demonstration f the n-personenvy-freeprotocol sto show that a full allocation of the entire cake can be accomplished n a finitenumberof steps.14 [JanuaryN ENVY-FREE CAKE DIVISION PROTOCOL


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For simplicity,we will presentonlythe n = 4 versionof the envy-freeprotocol.The extension to arbitraryn is fairlystraightforwardnd left to the reader. Inoutlineform,the protocolgoes as follows:One player(chosenhere to be Player 2 for laternotationalsimplicity), uts thecake into4 equalpieces, handsthese out, and asksif anyoneobjects.If, say,Player1 objects,then Players1 and 2 (alone)go throughseveralstepswhich yieldsixsets(the Ys and Zs in Step 7 below)to be used as a startingpartition in place of thefive equalpieces) for the kind of trim-and-chooseequenceamongall fourplayersthat we illustratedtwo paragraphsearlier. This trim-and-choosesequence isrepeatedagainand againuntil we arriveat a partialallocation n whichPlayer1has an irrevocable dvantageoverPlayer2 (the "aside"afterStep 15below).Fromthispoint on, we neverhave to worryaboutPlayer 's objectingbecause of envyforPlayer 2. Repeating this at most once for each pair of players results in anenvy-freeallocationof the entirecake after finitelymany steps.

Envy-FreeProtocol orArbitralyn(the n = 4 version)(1992)Step 1. Player2 cuts the cake into 4 pieces (thathe considers o be the samesize), keeps one piece, and handsone piece to each player.Step 2. Eachof the otherthreeplayers s askedwhetheror not he objectstothis allocation. A playerobjects iff he enviessome other player.)Step 3. If no one objects,then each keeps the piece he was givenin Step 1,and we are done.Step 4. Otherwise,we choose the smallest i so that Player i objected.Fornotational simplicity,assume i= 1. Player 1 now chooses a pieceoriginallygiven to some otherpIayer whom he envied)and calls thatpiece A. The piece originallygiven to Player 1 is called B.Aside. Once we have A and B, the other two pieces in the allocation romStep 1 are reassembled.Thatpart of the cakewill be allocated ater.Note that Player1 thinks A is largerthan B. Player2 thinks A andB are the same size.Step 5. Player1 nownames a positive ntegerr 2 10(chosen so that,for anypartition of A into r sets, Player 1 will prefer A, even with the 7smallest accordingto Player 1 pieces in the partitionof A re-moved,to B).Aside. Player 1 can easily choose such an r. That is, the union of the 7smallestpieces is certainlyno largerthan 7 times the averagesize ofall r pieces.,Hence, Player 1 simplychooses r large enough so that7,u(A)/r < ,u(A) - ,u(B),where ,u is his measure.Step 6. Player2 nowpartitionsA into exactlyr sets (thathe considers o bethe same size), and does the sameto B.Step 7. Player 1 chooses (the smallest) 3 sets from the partitionof B and

names these Z1,Z21 Z3 He also chooses either (the largest) 3 setsfromthe partitionof A (if he thinksthese are all strictly argerthanall the Zs), and trims at most 2 of these (to the size of the smallestamongthe three),or he partitions the largest)one of the sets in thepartitionof A into 3 pieces(thathe considers o be the samesize). Ineithercase, he namesthese Y1,Y2,Y3.Aside. Player l's strategy n Step 7 guarantees hat he will think all three Ysare the same size, and each strictly argerthan all three Zs. This is

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true even if he chooses the secondoption.2 Player 2 thinks all threeZs are the same size, and each is at least as large as all three Ys.Step 8. Player 3 takes the collection of 6 pieces, and either passes (if hethinks there already s at least a 2-waytie for largest),or trims(thelargest)one of these (to the size of the next largest),thuscreatingatleast a 2-waytie for largest).Step 9. Players4, 3, 2, and 1, in that order,choose a piece from amongthe 6Ys and Zs as modifiedin Step 8 (that they consider to be largest ortied for largest),with Player3 required o take the piece he trimmedif it is available.Player2 must choose Z1, Z21 or Z3. Player 1 mustchoose Y1,Y2, or Y3.Aside. This yields a partition {Xl, X2, X3, X4, L1} of the cake such that{X1,X2,X3,X4} is an envy-freepartial allocation, and L1 is theleftoverpiece. Moreover,Player1 thinkshispiece X1 is strictlyargersay by E than Player 2's piece X2.Step 10. Player 1 namesa positive integer s (chosen so that [4,ul(L1)/S]s< ,where ,u1 s Player l's measure).Aside. The integer s specifies how manytimes the playerswill iterate thebasic trim-and-choose equence to follow. Notice that if the ruleswere instead to allow the iterations to continue until Player 1 said"stop" (which he could strategicallydo at the point at which hethinks the leftovercrumb s smallerthan the advantagehe has overPlayer 2), then there is no guaranteethat a strategicallymisguidedPlayer1 wouldnot keep the gamegoingforever.Step 11. Player1 cuts L1 into 5 pieces (that he considers o be the same size).Step 12. Player2 takesthe collection of 5 pieces, selects (the largest)3 pieces,andtrims(the largest)2 or fewer of these (to the size of the smallest,therebycreatingat least a 3-waytie for largest).Step 13. Player3 takesthe collection of 5 pieces, perhapstrimmed n step 12,selects(the largest) wo, andtrims, f he wantsto, (the largest)one ofthese (to the size of the smallest, huscreatingat least a 2-way ie forlargest).Step 14. Players4, 3, 2, and 1, in that order,choose a piece (that theyconsider

2The proof runs as follows: We are assuming hat both A and B have been partitioned nto rpieces, and that B is not onlysmaller han A but smallereventhan A with the smallest7 piecesof A'spartitionremoved.Arrange he sets in both partitions romlargestto smallestas A1,A2,. . ., Ar andB1,B2,. . ., Br. Let ,u denote Player 1's measure, and suppose, for contradiction, hat both of thefollowinghold:1. ,u(Br_2) 2 ,u(A3),whichholds if A1, A2, and A3 are not all strictly argerthan Br_2, Br_l,and Br.2. ,u(Br_2)2 ,u(Al)/3, whichholds if Al cannotbe partitioned nto 3 sets all largerthan Br_2,Br_1,and Br-It followsfrom 1 that:3. ,u(B7U *** U Br_3)2 ,u(A3U *** U Ar_7) since there are r - 9 sets in each union, and thesmallestone of the Bs is at least as large as the largestone of the As.It followsfrom 2 that:4- y([B1 U B2 U B3] U [B4 U Bs U B6]) 2 ,u(Al u A2),sinceeach of the blocksof 3 Bs is larger haneach of the As.But 3 and 4 clearlydemonstratehat:5.y(B)2y(AlU *-- UAr-7)Thisis the desiredcontradictionince the set on the right s A with the smallest7 pieces of its partitionremoved.



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to be largestor tied for largest),with Players3 and 2 required o takea piece they trimmed f one is available.Step 15. Steps 11-14 are repeated s - 1 more times, with each applicationofthese four steps applied to the leftover piece from the precedingapplication.Aside. This yields a partition {X1,X2, X3,'X4, L2} of the cake such that{X1, X2, X3, X4} iS an envy-free partial allocation, and such thatPlayer 1 thinks that X1 is larger than X2 U L2. We now declarethat Player 1 has an irrevocabledvantage ver Player 2, and webegin creation of a subset of {1,2,3,4} X (1,2,3,4}, which we call">v" for "irrevocable dvantage," y putting(1, 2) E AS.Step 16. Player 2 cuts L2 into 12 pieces (that he considers to be the samesize).Step 17. Each of the other players declares himself to be of type A (if heagrees all the pieces are the same size), or type D (if he disagrees).

Player2 is declared to be of type A.Step 18. If D x A c - S, then we give the 12 pieces to the players n A, witheach of thetn receivingthe same numberof pieces. In this case, weare done.Step 19. Otherwise, we choose the lexicographically east pair (i, j) fromD x A that is not in AS, and we return to Step 4 with Player i inthe role of Player 1, Player in the role of Player2, and L2 in placeof the cake.Step 20. Steps5-18 are repeated.Aside. Eachtime we pass throughStep 15, we add an orderedpair to AS.Notice that since D x A c {1, 2, 3, 4} X {1, 2, 3, 4}, and - a?c

iterations.At this point, we conclude at Step 18 with an envy-freedivisionof the entire cake.This ends the protocol.There is an important way (pointed out to us by several people) that theenvy-freeprotocolfor even n = 4 differs from the envy-freeprotocol for n = 3:For n = 3, the number of cuts needed is at most 5, regardlessf what the

measu-s are. For n = 4, the numberof cuts needed can be made arbitrarilyargeby a suitable choice of the four measures(althoughthe moving-knife olution [8]for the four-personproblemgives a boundednumberof cuts). This raises:Question 1. Is there a boundednvy-freeprotocol for n = 4 or n > 4?There is another slightly more subtle (and perhaps related) way in which theenvy-freeprotocol differs from the others:The three earlier ones also work in thecoXntextf what are called "CD preferencerelations" in [2]. (A CD preferencerelation s a complete, reflexive, ransitivebinaryrelationthat satisfies a partition-ing postulate,a trimmingpostulate,and a weakadditivitypostulate.)The envy-freeprotocol,on the other hand, requireswhat is call an "ArchimedianCD preferencerelation" n [2]. The main result in [2] is the fact that a CD preference relation isinduced by a finitely additive measure in the obvious way iff it is Archimedian.This raises:Question 2. Is there an envy-freeprotocol for n = 4 or n > 4 that works in thecontext of non-Archimedianreferencerelations?It turns out that techniques similar to those used in the n-person envy-freeprotocol can also be used to solve the "chores"problem[15],wherein each player1995] AN ENVY-FREE CAKE DIVISION PROTOCOL 17


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wants to minimize the amount of cake he or she receives. This and relatedquestions e.g., the Pareto-optimality f allocations)are discussed n [6].REFERENCES1. A. K. Austin,Sharinga cake,MathematicalGazette66, no. 437(1982),212-215.2. J. Barbaneland A. Taylor,Preferencerelationsand measures in the contextof fair division,Proceedingsf theAmericanMathematicalociety to appear).3. S. Bennett, et al, FairDtvisions:GettingYourFairShare,HIMAP HighSchoolMathematics ndIts Applications]Module9 (Teachers'Maunal),1987.4. M. Berliant,K. Dunz, and W. Thomson,On the fair divisionof a heterogeneouscommodity,Journal f Mathematicalconomics21 (1992),235-240.5. S. J. Bramsand A. D. Taylor,A note on envy-free akedivision,Journal f CombinatorialheoryA (to appear).6. S. J. Bramsand A. D. Taylor,Fair Division:Proceduresor AllocatingDiuasible nd IndivasibleGoods(to appear).7. S. J. Brams,A. D. Taylor,andW. S. Zwicker,Oldand newmoving-knifechemes,Mathematical

Intelligencerto appear).8. S. J. Brams,A. D. Taylor, and W. S. Zwicker,A moving-knife olution to the four-personenvy-free akedivisionproblem,preprint.9. P. J. Campbell,Mathematics, 995Encyclopedia ritannicaYearbook f Scienceand the Future(1994),379-383.10. L. E. Dubins and E. H. Spanier,How to cut a cake fairly,AmericanMathematicalMonthly68(1961),1-17.11. S. Even andA. Paz, A note on cake-cutting,DiscreteAppliedMathematics (1984),285-296.12. D. Gale,Mathematicalntertainments,Mathematicalntelligencer5, no. 1 (1993),48-52.13. For AUPracticalPurposes:ntroductiono Contemporary athematics, d ed., W. H. Freeman,New York,1994,ch. 13.14. G. Gamowand M. Stern,Puzzle-Math, iking,New York,1958.15. M. Gardner,aha! Insight,W. H. FreemanandCompany,New York(1978), 123-124.16. T. P. Hill, Stochasticnequalities, MSLectureNotes 22 (1993),116-132.17. T. P. Hill, Fair-division roblems,preprint.18. B. Knaster,Sur le problemedu partagepragmatiquede H. Steinhaus,Annales de la SocietePolonaisede Mathematique9 (1946),228-230.19. S. X. Levmoreand E. E. Cook,SuperStrategiesor Puzzlesand Games,Doubleday,GardenCity,NY (1981),47-53.20. A. A. Liapounoff,On completelyadditivevector functions, Izv. Akad. Nauk SSSR (1940),465-478.21. J. Neyman,Un theoremed'existence,C. R. Acad. Sci. Paris222 (1946),843-845.22. D. Olivastro,Preferred hares,TheSciences,March/April 1992),52-54.23. D. Olivastro,Object Lessons" Solutionsand Sequelae),TheSciences, uly/August(1992),55.24. K. Rebman, How to get (at least) a fair share of the cake, in Mathematical lums, RossHonsberger, ditor,MathematicalAssociationof America 1979),22-37.25. H. Steinhaus,The problemof fair division,Econometrica6(1948),101-104.26. H. Steinhaus,Sur la divisionpragmatique,Econometricasupplement)17 (1949),315-319.27. W. Stromquist,How to cut a cake fairly, AmericanMathematicalMonthly87, no. 8 (1980),640-644. Addendum, ol. 88, no. 8 (1981),613-614.28. W. Stromquist nd D. R. Woodall,Sets on whichseveralmeasuresagree,J. Math.Anal. Appl.108(1985),241-248.29. W. A. Webb,An algorithm or a stronger airdivisionproblem,preprint.30. D. Weller, Fair divisionof a measurable pace, Journalof Mathematical conomics14, no. 1(1985),5-17.31. D. R. Woodall,Dividinga cake fairly,J. Math.Anal. Appl. 78, no. 1 (1980),233-247.32. D. R. Woodall,A note on the cake-division roblem,Journalof Comb.Theory A) 42 (1986),300-301.DepartmentfPolitics DepartmentfMathematicsNewYorkUniversity UnionCollegeNewYork,NY10003 Schenectady,Y1230818 [JanuaryN ENVY-FREE CAKE DIVISION PROTOCOL

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