algorithmic solutions for envy-free cake cutting · ﬁnding an envy-free cake cutting is the same...

OPERATIONS RESEARCHVol. 60, No. 6, November–December 2012, pp. 1461–1476ISSN 0030-364X (print) � ISSN 1526-5463 (online) http://dx.doi.org/10.1287/opre.1120.1116

© 2012 INFORMS

Algorithmic Solutions for Envy-Free Cake Cutting

Xiaotie DengDepartment of Computer Science, University of Liverpool, Liverpool L69 3BX, United Kingdom; and

Department of Computer Science, City University of Hong Kong, Hong Kong{[email protected]; [email protected]}

Qi QiDepartment of Industrial Engineering and Logistics Management, Hong Kong University of Science and Technology, Hong Kong,

[email protected]

Amin SaberiDepartment of Management Science and Engineering, Stanford University, Stanford, California 94305,

[email protected]

We study the problem of finding an envy-free allocation of a cake to d+1 players using d cuts. Two models are considered,namely, the oracle-function model and the polynomial-time function model. In the oracle-function model, we are interestedin the number of times an algorithm has to query the players about their preferences to find an allocation with the envy lessthan �. We derive a matching lower and upper bound of �41/�5d−1 for players with Lipschitz utilities and any d > 1. In thepolynomial-time function model, where the utility functions are given explicitly by polynomial-time algorithms, we showthat the envy-free cake-cutting problem has the same complexity as finding a Brouwer’s fixed point, or, more formally, it isPPAD-complete. On the flip side, for monotone utility functions, we propose a fully polynomial-time algorithm (FPTAS)to find an approximate envy-free allocation of a cake among three people using two cuts.

Subject classifications : fair division; cake cutting; envy-free; FPTAS; fixed point; PPAD.Area of review : Optimization.History : Received April 2010; revisions received March 2011, September 2011; accepted January 2012.

1. IntroductionSuppose you have a cake and you would like to divide itamong n persons. Each person may have a different opinionas to which part is more valuable. Is there a way of cuttinga cake into n pieces with n−1 cuts and allocating one pieceto each person so that everyone values his or her piece noless than any other piece? Is there an efficient algorithmthat finds such a way, or commonly what is known as anenvy-free solution?

Since the beautiful works of Banach and Knaster (seeBrams and Taylor 1995) and then Steinhaus (1948, 1949),the envy-free cake-cutting problem has been a favoriteamong mathematicians for its properties and algorithmicsolutions (see, e.g., Even and Paz 1984, Brams and Tay-lor 1995, Woeginger and Sgall 2007). Its core applica-tions in operations research and related fields, including fairscheduling, resource allocation, and conflict resolution havebeen studied extensively, where every participant desires asolution that optimizes its own utility.

1.1. Applications in Operations Researchand Economics

Many interesting operations research applications withinthe cake-cutting framework are presented by Thomson(2007). Some of the examples include allocating staffto time-intensive tasks such as scheduling police patroloperations and allocation of cleaning tasks to maintenance

crews. Thomson also studies territory-splitting applications,for example, for shop placement, and proposes a cake-cutting protocol.

Similar territory-splitting operations are also discussedin Sherstyuk (1998) for understanding partisan gerryman-dering schemes. Again, the author uses fair cake-cuttingapproaches to develop methods that maintain electoral fair-ness. It is also worth noting that the issue of connectednessof the districts, the property also enforced in our problem,is of vital importance for this application.

In the same spirit, Moulin (2007) targets the issue of fair-ness in scheduling and queuing decisions. Meunier (2008)studies the more abstract problem of fairly allocating adiscrete object represented as a necklace. Other discretevariations of fair allocation have also attracted a lot ofattention, both in algorithms (Lipton et al. 2004, Asadpourand Saberi 2007, Bansal and Sviridenko 2006) and commu-nication networks (Kelly et al. 1998, Kumar and Kleinberg2006, Kleinberg et al. 1999).

There is also an extensive literature in the broader con-text of envy-free allocation. Halland and Zeckhauser (1979)develop an implicit market procedure to elicit honest prefer-ences from individuals’ private preferences by assigning theplayers efficiently to goods or positions such that their dif-ferent objectives are maximized within the choices. A truestory example, reported in Pratt and Zeckhauser (1990),deals with the allocation of two trunks of silver left behind

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Deng, Qi, and Saberi: Algorithmic Solutions for Envy-Free Cake Cutting1462 Operations Research 60(6), pp. 1461–1476, © 2012 INFORMS

to eight grandchildren. The designed procedure is in accor-dance with a market equilibrium approach, divided in threesteps: eliciting utility functions, solving for the equilibrium,and implementing a randomization step for the final integersolution. The story shows a real success of using the equi-librium procedure to solve the envy-free allocation prob-lem. The utility function here is not an oracle, but explicitlygiven. That makes it more efficient than using the oraclemodel to query the players from time to time.

In the context of bankruptcy management, Abdelghaniet al. (Elimam et al. 1996) find a solution for fair andequitable distribution of debtors’ mixed assets to theirrespective creditors. In a different direction of firm acquisi-tion and merging, Hirsch (1986) studies the transition fromhostile ambushes to smooth transition of the power, provid-ing incentives to sustain order within the business. Whilethe study focuses on the social aspect of the changes, itreveals the underlying principles that have made the socialchange: an amiable model that tries to maximize the utilityof everyone involved, a situation advocated in the envy-freecake-cutting paradigm.

1.2. Related Work

Various notions have been proposed to capture fairness.A related fairness criteria that is less demanding than envy-freeness is proportionality. A solution is proportional ifeach person gets a piece that he or she values more than1/n of total.

For two players, finding a proportional solution is quitesimple: cut and choose. One gets to cut the cake, andthe other gets to choose which piece. The cut and chooseprotocol results in an envy-free, and hence also a propor-tional, solution. A generalization of this scheme for find-ing a proportional solution for n > 2 players is discoveredby Banach and Knaster (see Brams and Taylor 1995) andSteinhaus (1948, 1949). Later, Even and Paz (1984) givean elegant divide-and-conquer algorithm for this problemthat runs in O4n logn5. Very recently, Edmonds and Pruhsprove that this algorithm is essentially optimum for deter-ministic algorithms (Edmonds and Pruhs 2006a) and givea randomized algorithm with running time O4n5 (Edmondsand Pruhs 2006b).

The problem can also be relaxed to allow more thann− 1 cuts. In that case, there is an envy-free solution forthree players found around 1960 by Selfridge and is redis-covered independently by Conway (see Brams and Taylor1995 for details). The solution for the n-player envy-freeallocation is not known until the result of Brams and Taylorin 1995 (Brams and Taylor 1995).

For the problem under our consideration, that is then-person envy-free cut problem with exactly n− 1 cuts,progress in complexity analysis has been limited. The exis-tence of such a solution is proven by Stromquist (1980)with a fixed-point argument. His proof implies that an�-approximation can be found in time exponential in inputsize O4log 1/�5, where � > 0 can be considered as an

insignificant extra amount in the value of their utilityfunctions that the players would not care. Later, Simmonsand Su (Su 1999) prove a similar result by Sperner’s lemma.

Stromquist (1980) also proposes a moving knife proce-dure for cutting a cake for three players, that results inexactly two cuts. We will explain this procedure in detailsin §4. It is easy to see that Stromquist’s solution requiresfrequent recalculations of the knife positions, and it maytake an exponential time. We will show a stronger result:the particular fixed point found by Stromquist needs anexponential number of queries about the values of theplayers.

1.3. Our Contributions and Methodology

The main question studied in this paper is whether it ispossible to find an envy-free allocation by an efficientprocedure. Because utilities can be arbitrary continuousfunctions, it may be impossible to describe the envy-freesolution with a finite number of digits. Instead, we wouldbe interested in solutions with � > 0 relaxation in envy.That is, we are interested in a partition of the cake amongthe participants such that each would not envy another’spiece more than an �.

For simplicity, let N = 1/� throughout our discussion.We are interested in whether there are algorithms withthe time complexity polynomial in O4logN5, i.e., whetherthere is an algorithm in time polynomial in the bit size ofthe approximation parameter.

The approximation approach allows us to consider thediscrete version of the problem. A triangulated simplex is avery useful tool in the study of computational complexity.In a triangulation, a d-dimensional simplex is partitionedinto a collections of smaller d-dimensional simplices suchthat any two of them have either no common element or aface of them. Such a d-dimensional simplex in the trian-gulation will be called a base simplex or a base cell in thefollowing discussion.

1.3.1. The Oracle-Function Model vs. the Polyno-mial-Time Function Model. We are interested in twotypes of functions for computing the utilities or the pref-erences of the players. In the polynomial-time functionmodel, the agents have to represent their utilities fully bya polynomial-time algorithm. In the oracle model, on theother hand, the algorithm gets a partial picture of the util-ities of the agents after every question. Because of this,our problem is harder under the oracle-function model thanunder the polynomial-time function model. This is simplybecause the oracle-function model can admit many morefunctions than the polynomial-time function model.

Both models are quite natural. Oracle-function modelsare suitable in situations when the agents are unable orunwilling to communicate their full utility functions. Thiscan happen because the full utility function is too long(e.g., when N is large in our context) or when it is costlyfor the agent to evaluate his or her utility function for

Deng, Qi, and Saberi: Algorithmic Solutions for Envy-Free Cake CuttingOperations Research 60(6), pp. 1461–1476, © 2012 INFORMS 1463

every possible scenario. Oracle-function models are alsoused in the context of learning theory and combinatorialauctions (Blum et al. 2004, Blumrosen and Nisan 2007,Dobzinski et al. 2010).

The polynomial utility functions are suitable when theutilities can be described concisely, for example, when theyare linear or well-behaved concave functions. In this model,it is assumed that the agent can represent his or her utilityby a simple program or a polynomial-time function.

For both models, we prove that the time complexity isequivalent to that of finding a fixed point for general utilityfunctions.

1.4. Main Results

We establish three main results.1. When the preferences of the players are provided as

polynomial-time algorithms, we prove that our cake-cuttingproblem is PPAD-complete (as defined in §2.4). By provingPPAD-completeness, we establish that the complexity offinding an envy-free cake cutting is the same as finding aBrouwer’s fixed point, or finding a Nash equilibrium or amarket equilibrium.

2. We also study the case in which the preferences aregiven by an oracle function. In other words, we considerthe setting in which the algorithm can query the preferencesof the players by showing them a particular cake cuttingand asking for their preferred piece (or pieces) in the cut.In this setting, we are interested in the minimum number ofqueries needed for finding an �-approximate cake cuttingfor d + 1 players. We derive a matching lower and upperbound of �4N d−15 for the query complexity (see §2.4 formore details).

Despite a strong connection, mathematical and com-plexity-wise, between equilibrium computation and fixed-point computation, this is the first matching query complex-ity result for an equilibrium computation problem. Forexample, we know of no such result for Nash equilibrium,which is also PPAD-complete with a strong tie to fixed-point computation.

3. For the special case of monotone utility functions,we prove there is a fully polynomial-time approximationscheme (see Definition 7 in §2.2) for finding an approx-imate envy-free allocation of a cake among three peopleusing only two cuts. In other words, we are able to find anapproximate solution in time polynomial of the number ofplayers and O4log41/�55.

1.4.1. PPAD Completeness. Our proof starts with theproblem of finding a fully colored base simplex in a trian-gulated d-dimensional simplex with a valid Sperner color-ing, which is a known PPAD-complete problem.

We define a discrete problem in which the goal is tofind an approximate envy-free cut. Let N be a large inte-ger. Consider all the d-cuts represented by barycentriccoordinates Á = 4�01�11 0 0 0 1�d5 with �0 = X0/N , �1 =

X1/N 1 0 0 0, �d = Xd/N , where Xis are integers between 0

and N and∑

Xi =N . Note that 4�01�11 0 0 0 1�d5 is a pointon the standard d-dimensional simplex. We may use X =

4X01X11 0 0 0 1Xd5 as the integer representation of a d-cut.To prove that our problem is in PPAD, we use Kuhn’s

triangulation (Kuhn 1960) to triangulate the simplex so thatthe corner points of the base cells represent the discretecuts defined above for some N . Then, the reduction is doneby a two-stage process: labeling and coloring.

The usefulness of Kuhn’s triangulation in efficientimplementation has also been noted by Su and used inhis implementation, the Fair Division Calculator (at Su’swebpage http://www.math.hmc.edu/su/fairdivision/calc/),a very interesting applet that does fair division of ahomotopy algorithm.

Let V be the set of vertices of the triangulation. We finda labeling L2 V → 80111 0 0 0 1 d9 such that ∀X1Y ∈ V onthe same base cell, L4X5 6= L4Y5. Then for any labeledvertex X, we define a coloring C2 V → 80111 0 0 0 1 d9 suchthat C4X5 = i if player L4X5 prefers the ith piece of thecut X. By Sperner’s lemma, the resulting triangulation has afully colored base that corresponds to an approximate envy-free solution. This reduces the problem to finding a fullycolored base in a Sperner simplex and therefore establishesthat the problem is in PPAD.

On the other hand, we design a reduction based on theBROUWER problem (Brouwer 1910; Papadimitriou 1994;Chen and Deng 2008, 2009a, b) for its proof of PPAD-hardness.

1.4.2. Matching Lower and Upper Bound in theOracle-Function Model. We derive a �4N d−15 timematching bound for the query complexity of cake cuttingfor d + 1 players with Lipschitz utilities. The tight upperbound requires a binary search method that finds a balancedcut of the simplex. It is made possible by Kuhn’s triangu-lation and our labeling rule that allows an efficient paritychecking of the index of the boundary simplices of 4d−15-dimension. Take the case of three players as an example:because two cuts are possible to obtain an envy-free cut-ting, we have two variables. Therefore, the problem can bedecided on a two-dimensional space. We need to find a tri-angle of all three colors on its vertices on the Kuhn’s trian-gulation over a two-dimensional space. To find one, we firstdivide the triangulated Sperner simplex (with a valid color-ing of 8011129) into two polygons P1 and P2. By the stan-dard degree theory, the number of fully colored edges by80119 on the boundary has the same parity as the numberof fully colored triangles by 8011129 in the polygon. Wethen count the numbers of edges on the boundary that arecolored by both 0 and 1 for both polygons and denote themas N1 and N2, respectively. This can be done in time O4N5.By Sperner’s lemma, the number of triangles with all threecolors in the original simplex is odd. Therefore, either N1

or N2 is odd. Without loss of generality, we assume N1 isodd. Then there must be a fully colored triangle in P1. Thesize of the problem is reduced by 1/2. By doing the above


steps recursively, we can find one fully colored triangle intime O4N5 + O4N/25 + · · · + O415 = O4N5. The abovemethod can be generalized to high-dimensional problems.By using the same argument, we only need to examine theedges on the boundary, which reduces the complexity onthe exponent from d to d− 1.

For the lower bound, the results are obtained by a reduc-tion from BROUWER (Deng et al. 2011).

1.4.3. FPTAS for Three Players with Monotone Util-ity Functions. For the special case of monotone utilityfunctions, we are able to utilize the monotonicity to con-struct an �-approximate envy-free solution in time polyno-mial in log4N 5 for three players.

We rely on the general approach of binary search on theparity but exploit the monotonicity of the best choice alongcertain lines in the simplex. We will sketch the main ideaof the approach in the next two paragraphs and leave thedetails to the last section.

Any player with a monotone utility function would preferA to B for B ⊆ A. Therefore, when one cut is fixed, oneof the three pieces is fixed. Any player’s preference onthe other two pieces will change monotonically as anothercut changes from left to right. We use the coordinate X =

4X01X11X25, along the line of fixed X0, and let X1 increasefrom 0 to N − X0. The preference of a player’s choicewill start with the last piece, to the first piece, and to themiddle piece (one or two of them may be missing). Similarmonotonicity holds when X2 is fixed. For each player, wecan find the break point of the choices by binary search inO4logN5 time.

Because of the monotonicity along those two direc-tions, we can calculate the indices of cut edges efficiently.Therefore, the indices of the two regions split by the cutcan be decided quickly. We will stay on the region withan odd index so that we should terminate with one that isa diamond-shaped polygon consisting of at most two basecells. By parity, one of the two cells is a fully colored basetriangle. The overall query complexity and running timewill be of O4log2 N5.

1.5. Organization

The rest of the paper is organized as follows. We first intro-duce the relevant concepts, the mathematical and computa-tional models in §2.

In §3, we present the complexity results for finding anenvy-free cake-cutting solution, with a matching algorith-mic bound under the oracle-function model, as well asa PPAD-completeness proof for the polynomial functionmodel.

In §4, we show that for three players there is an algo-rithm for finding an �-approximate envy-free solution intime polynomial in log4N 5 under certain conditions. Incomparison, we prove that Stromquist’s solution demandsan exponential number of queries.

Finally, we conclude our work with discussion and openproblems in §5.

2. Definitions and Basic ConceptsIn this section, we will briefly review the basic definitionsand results in combinatorial topology and computationalcomplexity that will be used in the proof.

2.1. Sperner’s Lemma

In general, a d-simplex ãd4X01X11 0 0 0 1Xd5 is the con-vex hull of d + 1 independent vectors X01X11 0 0 0 1Xd,i.e., ãd = 8v � v =

∑di=0 �iXi1�i ¾ 01

∑di=0 �i = 19, where

4�01�11 0 0 0 1�d5 are the barycentric coordinates of v. Forv ∈ãd, define �4v5= 8i � �i > 09.

A face of a simplex ãd4X01X11 0 0 0 1Xd5 is any simplexãk4Xi1

1Xi21 0 0 0 1Xik

5 where 0 ¶ i1 < i2 < · · ·< ik ¶ d.Triangulation of a simplex is a procedure that partitions

the original d-simplex ãd into several small d-simplicesã11 0 0 0 1ãm such that for ∀ i1 j , ãi ∩ãj is either empty or acommon face.

Let V be the union of the vertices of all ãis. Sperner’slemma is defined on V , the vertices of a triangulated sim-plex with Sperner coloring.

Definition 1 (Sperner Coloring). For a triangulatedsimplex ã with vertex set V , a coloring

�2 v → 80111 0 0 0 1 d9 is called Sperner coloring

if ∀v ∈ V 2 �4v5 ∈ �4v50

A Sperner colored triangulated simplex has a specialproperty that is best explained by the concept of index.

Definition 2. For a d-dimensional simplex with verticesassigned all colors 80111 0 0 0 1 d9, we define its index as 1.Otherwise, it is defined to be zero. For a triangulated poly-gon in d-space, its index is defined to be the sum of theindices of its base simplices modulo two.

For a polygon P , denote by ¡P the boundaries of P .Note that the triangulation of P induces a triangulation of¡P into 4d−15-dimensional simplices. Naturally, we defineindexd−14¡P1�5 =

∑

�d−1∈¡Pindexd−14�d−11�5. Note that a

4d − 15 dimension cell has index 1 if its vertices haveall colors in 80111 0 0 0 1 d − 19, 0 otherwise. The followingresult is useful (see, e.g., Todd 1976).

Proposition 1. index4P1�5≡ indexd−14¡P1�5mod 2.

With the restriction of Sperner coloring, we immediatelyobtain the following:

Theorem 1 (Sperner’s Lemma, 1928 (Sperner 1928)).For a simplex that is triangulated and colored by Spernercoloring, the number of simplices in the subdivision withindex 1, i.e., its vertices all colored differently, is odd.

Sperner’s lemma states that a triangulated simplex witha Sperner coloring has a base cell such that all its verticeshave different colors. Such a simplex is commonly referredto as a Sperner Simplex.


2.2. The Model and Envy-Free Solutions

Following the cake-cutting literature, we represent the cakeas a line segment L= 60117. To model the preferences, weassign a utility function ui defined over each line segmenton L to each player i ∈ I . We may more generally definethe utility function over the Borel space of the line segmentL= 60117. However, as we are considering a d-cut solutionfor d + 1 players, a function defined on each subsegmentof L would be enough for our study.

Our utility functions are required to satisfy the followingtwo conditions:

• Nonnegativity condition: ui4�5= 0 and ui4 6= �5 > 0.• Lipschitz condition: For any pair of intervals A,

B ⊆ L, where A = 6a11 a27 and B = 6b11 b27, �ui4A5 −

ui4B5�¶K×�4A�B5 for a fixed constant K, where �4 · 5is the Borel measure and A�B = 4A\B5∪ 4B\A5.

The first condition, which is also called the “hungry con-dition,” is the main requirement in determining a continu-ous solution. The second condition opens up a possibilityto discretize the function for an approximate solution.

A cut in the middle will be along the point 1/2, whichcan be denoted by 41/211/25, meaning that the first part isthe left half of the whole, and the second part is the righthalf of the whole. In general for a set I = 80111 0 0 0 1 d9 ofd+1 players, a cut into d+1 pieces through a d-cut can berepresented by a 4d+15-component vector 4�01�11 0 0 0 1�d5with the property that

∑di=0 �i = 1. Here again, the d-cut

splits the cake, from left to right, pieces in fractions �j ,j = 0111 0 0 0 1 d, of the whole.

By the nonnegativity condition of utility functions, everyplayer will strictly prefer the nonzero segments to the zerosegments. More formally, consider any boundary point prepresented as 4�01�11 0 0 0 1�d5 such that �j = 0 but �i > 0.For the d-cut defined by p, every player strictly prefers theith segment to the jth.

It is easy to see that the set of all possible d-cuts forma d-dimensional simplex ãd. As a side note, consider anytriangulation of such a simplex and color the vertices ofthe triangulation by the preferences of one of the players.Because of the nonnegativity property of the utility func-tions, the resulting coloring would be a Sperner coloring.

We may define an envy-free solution for the cake-cuttingproblem as follows:

Definition 3 (An Envy-Free Cut). A partition or a d-cutS is envy-free if different players prefer different segmentsof S, i.e., there is a permutation �2 I → S such that for allj 6= i,

ui4�4i55¾ ui4�4j550

That is, for each player, its assigned piece of the cake isthe best among all the d + 1 pieces, according to its ownutility function.

Theorem 2 (Stromquist 1980, Su 1999). There is anenvy-free cake-cutting solution for d + 1 players that usesonly d cuts.

2.3. Simmons-Su’s Method for FindingApproximate Solutions

Su (1999) explicitly described the first computational pro-cedure for finding an approximate envy-free cake-cuttingsolution, by a labeling process on a triangulated simplex,citing Simmons as the one who first outlined these ideasusing Sperner’s Lemma (Simmons 1980).

Simmons-Su’s method is based on a discretization andtriangulation of the simplex corresponding to a d-cut. Itfirst labels all vertices of the triangulated d-simplex withd+1 numbers such that each base d-simplex is labeled onits vertices with all d+ 1 numbers. We will refer it to as a4d+ 15-labeling of the triangulated simplex.

Definition 4 (Simmons-Su Labeling). Given a triangu-lated simplex, a labeling of the vertices is a Simmons-Sulabeling if, on each base simplex, every vertex is labeleddifferently.

Recall that the Sperner’s lemma says for any triangulatedsimplex with Sperner coloring, the number of the fully col-ored base cells is odd. If a triangulated simplex is labeledby a Simmons-Su labeling, then all base cells are labeleddifferently. Note that any triangulated simplex can be col-ored by Sperner coloring, but not every triangulation admitsa Su’s labeling. See Figure 1 as an example.

Definition 5 (Simmons-Su Simplex). Given a triangula-tion with a Sperner coloring and a Su’s labeling, a basesimplex is called a Su’s simplex if it is fully labeled andcolored, i.e., all its vertices are labeled differently and col-ored differently.

Clearly, a Simmons-Su labeling together with a Spernercoloring admit a Simmons-Su simplex by Sperner’s lemma.

Theorem 3. For a triangulated simplex that is labeledby Simmons-Su labeling and colored by Sperner coloring,there exists at least one Simmons-Su simplex.

Figure 1. An example of triangulation not admittingany Simmons-Su labeling scheme.


This immediately leads to an approximate solution con-cept for cake cutting. For a vertex u of label i, it is coloredby j if player i prefers the jth piece on cut u. The nonneg-ative property of the utility functions guarantees the colorsare Sperner colors. Therefore, there exists a Su’s simplexin the triangulation.

On a Simmons-Su simplex, the d + 1 labels with d + 1different colors indicate that the optimal choices of d+ 1players are different. Even though those preferences areexpressed for different d-cuts, the corresponding d-cuts areclose enough to form an approximate solution when thebase simplices are sufficiently small.

By refining the triangulation, one can use the above argu-ment to find a continuous fixed point and therefore, anenvy-free solution.

Su (1999) used the barycentric triangulation of a sim-plex. At each level, a simplex is triangulated into 4d + 15!simplices recursively in dimensions as follows. For dimen-sion d = 1, a simplex is a line segment, and its barycentrictriangulation is a partition of the line segment into two byinserting the center point of the line segment. A d > 1dimensional simplex has d + 1 faces. Each face is par-titioned into a 4d − 15-dimensional barycentric triangula-tion (with a total of d! cells of 4d − 15-dimension). Eachcell forms a d-dimensional simplex with the center of theoriginal simplex of d-dimension. With a total of d + 1faces, there are a total of 4d+ 15d! = 4d+ 15! simplices ofd-dimension in this triangulation.

It continues in levels after levels until the desired gran-ularity is reached. The barycentric triangulation can belabeled by a Simmons-Su labeling and therefore it canprove the existence of the envy-free solution. However,from a complexity perspective, barycentric triangulationsare problematic because they create simplices with largeaspect-ratios that make the process converge rather slowly.The distance between two points in the same base simplexin the barycentric triangulation will remain relatively longwith respect to the level of triangulation.

In §3, we will adapt the Simmons-Su approach forKuhn’s triangulations. By giving polynomial-time algo-rithms for producing the triangulation, and showing that aSimmons-Su labeling is possible for Kuhn’s triangulation,we come up with hardness proofs both for the polynomial-function model and the oracle model. We will define thesecomplexity models in the next subsection.

2.4. Complexity Models and Classes

We will study the complexity of the envy-free cake-cuttingproblem for two types of models: the oracle-function modeland the polynomial-time function model.

In the oracle-function model, we are interested inthe minimum number of times the algorithm has to querythe players about their preferences. A query is defined asthe following operation: the algorithm presents a d-cut toa player and asks for the index (or indices) of the mostdesirable piece(s) in the cut.

If we assume that utility functions are arbitrary and con-tinuous, it may be impossible to present the envy-free solu-tion with a finite number of digits. Because of that, weintroduce the concept of approximate envy-free solutions:

Definition 6 (An �-Approximate Envy-Free Cut-ting). A partition is �-envy-free if different players�-approximately prefer different segments in S, i.e., thereis a permutation �2 I → S such that for all j 6= i,

ui4�4i55+ �¾ ui4�4j550

Therefore, the interesting question becomes how manyqueries or operations are needed for obtaining an�-approximate solution. The polynomial-time functionmodel captures the latter criterion. In this model, weassume that the utilities of the players are given explicitlyas functions or polynomial-time algorithms. Then we areinterested to see if there is a polynomial-time algorithm inthe number of players and the number of bits of the error(i.e., log41/�5) for finding an �-approximate envy-free cut.

Definition 7 (PTAS and FPTAS). Given a problem G,an algorithm A is called a polynomial-time approximatescheme (PTAS for short) if, for any � > 0, it finds an�-approximate solution in polynomial time in the originalinput size and 1/�. It is further called an FPTAS if it ispolynomial in the original input size and log 1/�.

2.4.1. Relevant Complexity Classes. Our results forthe oracle-function model are unconditional: we giveexplicit (and tight) bounds on the number of queries neededto compute an approximate envy-free allocation as a func-tion of number of players and the error. However, in thepolynomial-time function model, our results will build onthe extensive literature in computational complexity andconsists of proving inclusion or completion for certain com-plexity classes.

Problems such as finding envy-free cake cutting orapproximate envy-free cake cutting in our discussion arealways guaranteed to have a solution and they are definedto belong to the class TFNP, defined by Megiddo andPapadimitriou (1991). More formally, a binary relationF 4 · 1 · 5 is in the class TFNP if, for any x, there exists a y,�y� bounded by a polynomial in �x� such that P4x1 y5 holds,and P4x1 y5 can be verified to hold in polynomial time. Theproblem is to find out such a y in polynomial time.

An interesting subclass of TFNP, the class PPAD, isbased on an exponential size graph, consisting of directedpaths and cycles. For each instance of the problem, onestarting vertex is given for a directed path. We are requiredto find out the end vertex of any directed path or the start-ing vertex of another directed path different from the givenone. Here such a solution is guaranteed to exist becausethe graph consists of directed paths and cycles, with atleast a direct path (for the given starting vertex). Notethat as a further requirement to PPAD, given a vertex, theincoming edge and outgoing edge can be computed by agiven polynomial-time algorithm. Such a model is calledthe polynomial-time function model.


Figure 2. Barycentric triangulation on a two-dimensional triangle.

(a) 1st iteration (b) 2nd iteration (c) 3rd iteration

Alternatively, in the oracle model the function valuesare returned by an oracle function, which, given input x,outputs the function value f 4x5 in one oracle time. Notethat polynomial-time algorithms are a small faction of allthe algorithms. Therefore, the oracle function will covermore functions than the polynomial-time model. One unitof oracle time may in general translate into much longercomputer time if it is fully implemented. We also definethe query complexity of an algorithm to be the numberof times a function value is queried in the oracle model.Our discussion in the oracle model will focus on the querycomplexity.

3. Complexity Analysis for DiscreteCake Cutting

For the purpose of complexity analysis, let us introduce adefinition for a discrete envy-free cake-cut for d+1 playersusing d cuts.

For the discrete version, we will use Xi, i = 0111 0 0 0 1 dfor the ith barycentric coordinate of a cut in its integerrepresentation. That is, we partition the cake (as a line60117) into N equal segments. In the following parts, thecake is treated as 601N 7 and only integral cut X is con-sidered. Therefore, Xi ∈ 80111 0 0 0 1N 9, as opposed to 0 ¶�i ¶ 1 in the previous discussion.

We extend a utility function to the barycentric coordinateas follows:

Definition 8. Given a d-cut represented by the barycen-tric coordinate X, let cj4X5 = 41/N5

∑ji=0 Xi, j = 01

11 0 0 0 1 d with the convention that c−14X5= 0. Let the cor-responding intervals be Ij4X5 = 6cj−14X51 cj4X57, j = 0111 0 0 0 1 d. We extend a utility function u on the intervalsto the barycentric coordinate: u4X5= max8u4Ij4X551 j = 0111 0 0 0 1 d9.

Two d-cuts X and Y are adjacent to each other if ∀ i ∈

80111 0 0 0 1 d92 �Xi − Yi�¶ 1.

Definition 9 (Discrete Cut). A discrete cake cut isdefined to be a set 8X4051X4151X4251 0 0 0 1X4d59 of d + 1d-cuts such that for each pair of j and k, the two d-cutsX4j5 and X4k5 are adjacent.

When we choose finer and finer scale with larger andlarger N , the discrete cut structure becomes finer and finer(their nodes become closer and closer to each other). Thediscrete cut here is defined to be a set of cuts that areadjacent to each other, i.e., any two cuts are very closeto each other. This requirement is the essential point forthe complexity analysis of the approximate cake cutting,which overcomes the problem of large aspect-ratio in Su’sbarycentric triangulation.

Definition 10 (Discrete Envy-Free Cake Cut). A dis-crete cake cut is an envy-free solution if there is a permuta-tion � of 80111 0 0 0 1 d9 such that player i prefers the Ï4i5thsegment for some d-cut X4j5 in the set. We denote it byPi4X

4j55=Ï4i5.

Intuitively, in the limit point along a convergentsequence, the preference of players will be on different seg-ments of the limit point, representing a d-cut, and thereforeresults in an envy-free solution. The convergence will beguaranteed when the utility functions are Lipschitz.

Corollary 1. If utilities are Lipschitz with the Lipschitzconstant K, then any cut in the DISCRETE ENVY-FREECAKE CUT set is an �-approximate envy-free solution for� =O41/N5.

Proof. Assume X1Y ∈ DISCRETE ENVY-FREE CAKECUT set. Note that Ij4X5 = 6cj−14X51 cj4X57, we have forany j , �u4Ij4X55 − u4Ij4Y55� ¶ K × �4Ij4X5 � Ij4Y55 ¶K × 42/N5 by Definition 9. The statement follows.

The rest of this section is organized as follows. We startby introducing the Kuhn’s triangulation. We will show thatKuhn’s triangulation is balanced and admits Simmons-Sulabeling, i.e., a DISCRETE ENVY-FREE CAKE CUT cor-responds to an �-approximate envy-free solution for asmall �. Then we derive the complexity of finding a DIS-CRETE ENVY-FREE CAKE CUT for Kuhn’s triangula-tions in both oracle and polynomial-time function models.

3.1. Kuhn’s Triangulation

Kuhn’s triangulation was first defined on a unit hypercubein d dimensions. It partitions the d-cube into d! sim-plices. Specifically, let V0 = 40101 0 0 0 1051×d be one corner


Figure 3. An example of Kuhn’s triangulation on aunit cube in three dimension.

of the cube. The diagonal vertex to it would be Vd+1 =

41111 0 0 0 1151×d. Let � 2= 4�4151�4251 0 0 0 1�4d55 be anypermutations of the integers 1121 0 0 0 1 d. Each permutation� corresponds to one base simplex ãd

� whose vertices aregiven by vi� = vi−1

� + e�4i5, where v0� = V0 and ei is a unit

vector with ith component being 1. These simplices all havedisjoint interiors, and their union is the d-cube. Therefore,the unit hypercube is triangulated to d! base simplices.

For example, consider a unit 3-cube. Let 4010105 bethe base point. By Kuhn’s triangulation, the unit cube ispartitioned into six tetrahedrons according to the differentpermutations of 1, 2, 3. See Figure 3.

�1=41121352 ã�1 =8401010514110105141111051411111593

�2=41131252 ã�2 =8401010514110105141101151411111593

�3=42111352 ã�3 =8401010514011105141111051411111593

�4=42131152 ã�4 =8401010514011105140111151411111593

�5=43111252 ã�5 =8401010514010115141101151411111593

�6=43121152 ã�6 =8401010514010115140111151411111590

Next, let us extend Kuhn’s triangulation to a simplex.Let N be an integer bigger than 1. Take a unit d-cube anduse parallel cuts of equal distance to partition it into N d

smaller d-cubes of side length 1/N . Then, partition eachsmall cube into d! simplices by Kuhn’s triangulation. Now,observe that the unit cube can also be partitioned into d! bigsimplices first, and each big simplex contains N d smallersimplices or base cells. The proof of the consistency of thetwo processes is discussed as follows.

We scale the unit cube into one of side length N . Wecall it the big cube, and each of the N d subcubes the smallcubes.

It is enough to show that for a simplex in the refined gridstarting from any grid point x∗ = 4x∗

11 x∗21 0 0 0 1 x

∗d5 to y∗ =

4x∗1 +11 x∗

2 +11 0 0 0 1 x∗d +15, defined by a permutation �, all

of its d + 1 vertices are in the same big simplex derivedby some permutation � with base point 40101 0 0 0 105 to thepoint 4N 1N 1 0 0 0 1N 5.

Let x0 = x∗, and xi is the same as xi−1 except for its�4i5th coordinate xi

�4i5 which is xi−1�4i5 + 1. Then we have

1. If x∗i > x∗

j (which equivalent to x∗i ¾ x∗

j + 1), xki ¾ xk

j

for all k = 1121 0 0 0 1 d.

2. If x∗i = x∗

j and �−14i5 < �−14j5, xki ¾ xk

j for all k =

1121 0 0 0 1 d.Hence, there is a permutation � such that xk

�415 ¾ xk�425 ¾

· · · ¾ xk�4d5 for all k = 011121 0 0 0 1 d, which guarantees

the base simplex inside the large simplex derived bypermutation � with base point 40101 0 0 0 105 to the point4N 1N 1 0 0 0 1N 5.

Based on the equivalency of the two partitioning pro-cesses, we choose one of the big simplices, for examplethe one corresponding to � = 41121 0 0 0 1 d5, named ãd, andtriangulate it by Kuhn’s triangulation. Note that a vertexx ∈ ãd can also be represented by barycentric coordinates4�01�11 0 0 0 1�d5. For simplicity of presentation, we use theintegers X = 4X01X11 0 0 0 1Xd5 = 4N�01N�11 0 0 0 1N�d5 tobe a base point. Here we have

∑di=0 Xi = N . Set ai =

4ai01 ai11 0 0 0 1 aid5 where aij = 0 for all i 6= j−11 j and aii =

−1, ai1 i+1 = 1 for ∀ i ∈ 801 0 0 0 1 d − 19. Then the verticesof the base simplex according to permutation � based onX are given by vi� = vi−1

� + a�4i5−1 for ∀ i ∈ 811 0 0 0 1 d9 andv0� =X. For example, consider a simplex with N = 1 based

on vertex x = 40101 0 0 0 105d and permutation 41121 0 0 0 1 d5.We first set v0 = 41101 0 0 0 105d+1, then by the above trans-formation v1 = 40111 0 0 0 105d+1, v2 = 4010111 0 0 0 105d+1.Please refer to Scarf (1977, p. 42) for more details.

The main advantage of Kuhn’s triangulation is that,unlike the barycentric triangulation considered by Su, it isbalanced. In other words, the corner points of every basecell in this triangulation are close to each other.

Lemma 1. For given X = 4X01X11 0 0 0 1Xd5 and Y = 4Y01Y11 0 0 0 1 Yd5, define

�X−Y = max∀i∈80111 0001 d9

8�Xi − Yi�90

If X and Y are in the same base cell in Kuhn’s triangula-tion, then �X−Y = 1.

Proof. Without loss of generality, assume the base cellcorresponds to a permutation � and Y = X +

∑ki=l a�4i5−1

for some 0 ¶ l¶ k¶ d. Fix i, because aii = −1, ai1 i+1 = 1are the only nonzero coordinates in ai, and let b =

4b01 b11 0 0 0 1 bd5 =∑k

i=l a�4i5−1, then ∀ i, bi ∈ 8−110119.Therefore, �X−Y equals to either 0 or 1 and because X, Yare two different vertices, we must have �X−Y = 1.

Lemma 1 implies that the vertices of any base cell ofKuhn’s triangulation are adjacent to each other and thus:

Corollary 2. Any base cell of Kuhn’s triangulation formsa discrete cut.

3.2. Simmons-Su Simplex in Kuhn’s Triangulation

We can now present an algorithm for finding a DISCRETEENVY-FREE CAKE CUT in a Kuhn’s triangulation. Wewill crucially use the fact that Kuhn’s triangulation admitsa Simmons-Su labeling. The proof will be given later inthis section.


Figure 4. Kuhn’s triangulation for a simplex withN = 2.

For the simplicity of exposition, consider the problemfor four players: A1B1C1D. In this case, the closed set ofall possible cuts is a simplex of side length 1. To obtain anapproximate discrete envy-free cut, we partition the sim-plex by Kuhn’s triangulation with N = 2 and obtain N d = 8base simplices (see Figure 4).

Let V be the set of vertices of all base cells, i.e.,V = 84X01X11X21X352 X01X11X21X3 ∈�+1X0 +X1 +X2 +

X3 = 29, where �+ is the nonnegative integer set.The algorithm has two stages: labeling and coloring. We

partition V into four control subsets V01 V11 V21 V3, each cor-responding to one player. Starting by assigning 421010105to V0, 411110105 to V1, 411011105 to V2, and 411010115to V3. The rest of V is partitioned in such a way that thefour vertices of each base simplex belong to different sub-sets Vts.

Next, we should color each vertex in V . For a vertex Xlabeled by t, we let player t choose, among four segments,601X0/N 7, 6X0/N 1 4X0 +X15/N 7, 64X0 +X15/N 1 4X0 +

X1 + X25/N 7 and 64X0 +X1 +X25/N 117 of I , one thatmaximizes its utility. For simplicity of presentation, weshould assume a nondegenerate condition that the choiceis unique. The general case can be handled with a care-ful tie-breaking rule. If its optimal segment is the one oflength Xs/N , 0 ¶ s ¶ 3, we assign color s to the vertexX= 4X01X11X21X35. See Figure 5 for details of the exam-ple, where A2 0 means player A prefers the first piece atthat cut.

Because of the nonnegativity property of the utility func-tions, the above coloring is a valid Sperner coloring. Thefour vertices of each base simplex belong to four differentsubsets, or we say four different players. If we find a fullycolored base simplex, then on the four vertices of this sim-plex, different players prefer different pieces (see the shad-owed base simplex in Figure 5). It results in a Simmons-SuSimplex, which is a DISCRETE ENVY-FREE CAKE CUTunder our definition.

Figure 5. Four-players envy-free cake cutting withKuhn’s triangulation.

C:1

A:2

C:3

A:0

D:1B:2

A:1

B:1D:3

C:0

The case of d + 1 players for any d can be handled ina similar way if there exists a valid Simmons-Su labelingfor the triangulated d-simplex. We start by proving thatKuhn’s triangulation does admit a Simmons-Su labeling forall dimensions.

Claim 1. The Kuhn’s triangulation of a simplex admits aSimmons-Su labeling.

Proof. For a Kuhn’s triangulated simplex ãd with vertexset V , a weight function W is defined as:

W4X5=

d∑

i=0

iXi1

where X= 4X01X11 0 0 0 1Xd5 ∈ V 0Now, consider a labeling rule L:

L4X5= Vt1 if W4X5≡ t4modd+ 150

Consider a base simplex ã with base point X and permu-tation �.

Recall that the vertices of ã can be represented as vi� =

vi−1� + a�4i5−1 for ∀ i ∈ 811 0 0 0 1 d9, where aij = 0 for all i 6=j−11 j and aii = −1, ai1 i+1 = 1 for ∀ i ∈ 801 0 0 0 1 d−19 andv0� =X. Then,

W4vi�5=W4vi−1� 5+

(

44�4i5− 15+ 15− 4�4i5− 15)

=W4vi−1� 5+ 10

Therefore, all vertices of ã are labeled differently.

3.3. Complexity in the Polynomial Function Model

The problem of finding a Sperner Simplex (a fully col-ored base cell) in a Sperner coloring simplex is calledSPERNER.


Lemma 2 (Papadimitriou 1994). SPERNER is PPAD-complete.

Labeling of every vertex of the Kuhn’s triangulation canbe done in polynomial time, and so is the coloring, becausethe utility functions are given by a polynomial-time algo-rithm according to the polynomial-function model. Eachbase cell can be constructed and their colors can be ver-ified in polynomial time. In addition, all 4d + 15 verticesof every base cell in Kuhn’s triangulation have differentlabels. A fully colored base cell in this triangulation willbe a solution to the envy-free cake-cutting solution. There-fore, the problem reduces to SPERNER. Hence, we havethe following lemma:

Lemma 3. Finding a discrete d-cut set for envy-free cake-cutting problem for d+ 1 people is in PPAD.

In the rest of the section, we will prove that the envy-free cake-cutting problem is in fact PPAD-complete. Thisis done by applying a reduction from Brouwer’s fixed-pointtheorem (Daskalakis et al. 2006, Chen and Deng 2009a).Let us start by defining the problem.

Consider a d-dimensional hypergrid of side length N anddenote its node set by:

V dN =

{

p= 4p11 p21 0 0 0 1 pd5 ∈�d� ∀ i2 0 ¶ pi ¶N

}

0

Its boundary BdN consists of point p ∈ V d

N with pi ∈ 801N 9for some i:

BdN = 8p ∈ V d

N 2 pi ∈ 801N 9 for some i90

For each p ∈�d, let

Kp = 8q2 qi ∈ 8pi1 pi + 1990

Define m4p5 = 0 if ∀ i1 pi > 0; and m4p5 =

min8i2 pi = 09, otherwise. A coloring function g2 V dN ⇒

80111 0 0 0 1 d9 is valid if g4p5=m4p5, ∀p ∈ BdN .

Definition 11 (BROUWER). Let g be a valid coloringon V d

N . Given an algorithm that outputs g4p5 for every p ∈

V dN , find a point p ∈ V d

N such that Kp is fully colored, thatis, Kp has all d+ 1 colors.

Lemma 4 (Daskalakis et al. 2006, Chen and Deng2009a). BROUWER is PPAD-complete.

We are now ready to prove the theorem.

Theorem 4. Finding an approximate solution for envy-free cake-cutting for d + 1 people with d cuts is PPAD-complete.

Proof. Lemma 3 shows that the problem is in PPAD. Wewill prove the PPAD-hardness through a reduction fromBROUWER.

Let us start with the two-dimension case. Given aninput of 2D BROUWER, a grid 4G1V 2

N 5, and a color-ing function g2 N × N → 8011129, we embed it into a

Figure 6. 2D reduction.

Kuhn’s triangulated 2-simplex defined by three vertices:�401051 42N1051 42N12N5� by the mapping: T 4x1 y5 =

42N − x1N − y5.We define the preference functions for i = 01112 as

follows: Pi42N − x1N − y5 = g4x1 y5 for 0 ¶ x1 y ¶ N ;Pi4x1 y5 = 0 for 0 ¶ x < N ; Pi4x1 y5 = 2 for N < y ¶2N . Therefore, the preference functions form a DISCRETEENVY-FREE CAKE-CUT problem. From the origin, alongthe boundary in counterclockwise direction, we will walkthrough three pieces of consecutive vertices of all 0, 1,and 2 back to the origin. See Figure 6.

Therefore, there is a Sperner base cell in the triangulatedKuhn’s triangle �401053 42N1053 42N12N5�, which is at thesame time a ENVY-FREE CAKE CUT by our choices ofpreference functions. Therefore ENVY-FREE CAKE CUTdoes have a solution. Once we find one, it must be in theregion bounded by N ¶ x ¶ 2N , 0¶ y ¶N . The inversemapping of T will give us the required BROUWER’ssolution.

The above argument generalizes to higher dimensionsas follows. Let edi denote the i-coordinate unit vector ind-dimensions. Let 0d be a d-dimensional vector of all 0s.Define Xd

i = N4ed1 + ed2 + · · · + edi 5 for i = 1121 0 0 0 1 d, andXd

0 = 0d. Consider a Kuhn’s triangle Kd with vertices gen-erated by the identity permutation. Then Kd is the convexhull of Xd

0 1Xd1 1 0 0 0 1X

dd .

Now, we need to show that a d-dimensional BROUWERHd can be embedded into Kd, and the coloring of a validHd can be extended to a Sperner coloring of vertices in theKuhn’s triangulated simplex. This is done by the followingtwo claims, which are both proved by induction.

In the following discussion, we use the convention that,for a set S, tS = 8tx2 x ∈ S9 for any number t and S ⊆Rn.

Claim 2. A hypercube Hd of side length N/2d−1 canbe completely embedded into a Kuhn’s triangulated sim-plex Kd.

Proof of Claim 2. We prove it by induction. When d = 1,the line segment 601N 7 itself is a hypercube of side lengthN in dimension 1. Therefore, the claim holds.

Suppose the claim holds for d and consider Kd+1 =

8Xd+10 3Xd+1

1 3 0 0 0 3Xd+1d 3Xd+1

d+19 in 4d+ 15 dimensions. Notethat Kd+1 is the convex combination of the hyperplane4Kd105 and the point Xd+1

d+1 . By the induction hypothe-sis, there is a hypercube 4Hd105 of side length N/2d−1


Figure 7. Step I: Color reorienting.

in 4Kd105. Then the convex combination 41/25Xd+1d+1 +

1/24Hd105 denoted by H1 is a hypercube of side lengthN/42 ∗ 2d−15, which is also contained in the simplex Kd+1.We can rewrite H1 as

H1 =12 4X

dd1N 5+

12 4H

d105=12 4X

dd +Hd1N 5⊂Kd+10

On the other hand, because 4Hd105 ⊂ 4Kd105 and4Xd

d105 ∈ 4Kd105, the convex combination

H3 =12 4H

d105+12 4X

dd105=

12 4X

dd +Hd105

⊂ 4Kd105⊂Kd+10

Note that H3 is also a hypercube of side lengthN/42 ∗ 2d−15.

Let H2 = 41 − 1/2d−15H1 + 41/2d−15H3 = 41/254Xdd +

Hd1 41 − 1/2d−15N 5. Then H2 ∈Kd+1. Let

Hd+1= �H1 + 41 −�5H21 0 ¶ �¶ 10

Then Hd+1 ⊂ Kd+1, and it is a hypercube of side lengthN/42 ∗ 2d−15.

The claim follows.

To complete the reduction, we should extend the colorsof the hypergrid Hd to a Sperner coloring of vertices in theKuhn’s triangulated simplex Kd.

Claim 3. It is possible to extend the colors of an embeddedBROUWER hypergrid Hd to a Sperner coloring Kuhn’striangulated simplex Kd such that there is no fully coloredbase cell outside Hd.

Proof of Claim 3. We consider the problem for thed-dimension (d ¾ 2) case and apply induction to reduce itto the d− 1-dimensional problem.

Recall in the BROUWER coloring, a node on a bound-ary face at xi = 0 is colored i and on its opposite face iscolored 0. According to the above rule, the color of a nodeon the boundary of a boundary face would be chosen frommore than one color, but its final color is determined by theorder of colors: 1 ≺ 2 ≺ 3 ≺ · · · ≺ d ≺ 0. That is, if i ≺ jand they are colored on the same node, only i is the colorplaced at that node. All colors have priority over 0. Forsimplicity of the proof, we first reorient the hypercube andits colors: changing x to N − x (see Figure 7). Therefore,now a node on a boundary face at xi = 0 is colored 0.

Figure 8. Step II: Embedding and partition.

A boundary face is called Ni if its interior nodes arecolored i 6= 0. A boundary face is called Zi if its interiornodes are colored 0 while its opposite face is colored i.

Recall that a Kuhn’s simplex Kd is the convex hull ofXd

0 1Xd1 1 0 0 0 1X

dd . By the proof of Claim 2, we embed the

hypercube Hd with side length n into Kd by placing theface Nd on the hyperplane H2 xd = h and the opposite faceZd on the hyperplane H ′2 xd = h′ = h−n. Note that Nd, Zd

are parallel to Kd−1. We color vertices x ∈ Kd ∩ Hd withtheir original colors in Hd. (See Figure 8.)

We will expand the colors on the boundary of Hd to allthe remaining vertices in the Kd as follows. First we cutKd into three parts by two parallel hyperplanes H1H ′.

For a vertex x ∈Kd,1. x ∈ Part I, if xd >h.2. x ∈ Part II, if h¾ xd ¾ h′.3. x ∈ Part III, if xd ¶ h′.All vertices in Part I are colored d.For vertices in Part II, we color them inductively on

the hyperplane parallel to Kd−1 that contains them. On thehyperplane H ′, the colors of vertices will be determined bythe 4d − 15-dimensional solution obtained in recursion bythe embedding of Zd, which is a BROUWER hypercubein d− 1-dimension, into 4d− 15-dimensional simplex. Thevertices x with h> xd >h′ can be colored in a similar way.For the layer H , it corresponds to the embedding of Nd intothe d−1-dimension simplex. The color d on the boundaryof Nd plays a role similar to 0 in the lower-dimensionalcase; therefore, vertices on this layer can also be coloredrecursively.

For vertices in Part III, the colors on the boundary ofthe d−1-dimension simplex on the hyperplane 0 ¶ xd <h′

follow the colors on the boundary of the layer xd = h′.All the remaining vertices in Part III are colored 0. (SeeFigure 9.)

When d = 1, we simply extend the colors of the extremespoints of H 1 away from the BROUWER line segment toother vertices of K1 and obtain a Sperner coloring.

Next we prove that the above coloring rule is a properSperner coloring and does not add any new fully coloredbase cell outside Hd.

In general, at dimension d, the boundary face Zd of Hd isa 4d−15-dimensional BROUWER hypercube, and none of


Figure 9. Step III: Coloring.

its vertices are colored d. Similarly, the opposite boundaryface Nd has no color 0. By using hyperplanes parallel toZd between Zd and Nd to cut Hd, the boundaries of allthe obtained 4d−15-dimensional hypercube have the samecolors as the boundary of Zd.

By inductive hypothesis, the vertices of the d − 1-dimension simplex on any hyperplane between H and H ′

are SPERNER colored. The boundary vertices of xd = 0are colored the same as the boundary vertices on xd = h′,therefore, Xd

i is colored i for 0 ¶ i¶ d− 1. The vertex Xdd

is a vertex in Part I that is colored d.To prove that none of the base cubes outside Hd has all

the colors, we note that outside Hd vertices with color donly exist in Part I and on the hyperplane H . On the otherhand, all the vertices in Part I are colored d, and there is nocolor 0 on the hyperplane H . Therefore, the claim follows.

Thus, we reduce a given BROUWER instance 4G1V 4H55to one of finding Sperner’s simplex in Kuhn’s triangulatedsimplex. Because Kuhn’s triangulated simplex is 4d + 15-labeled, this gives us a Simmons-Su simplex.

By assigning each player the same preference function asthe given Sperner coloring for a Sperner input, we reduceBROUWER to ENVY-FREE CAKE CUTTING.

3.4. Complexity in the Oracle Function Model

Similarly, under the oracle model for utility functions, weshould show ENVY-FREE CAKE CUTTING has the samecomplexity as BROUWER.

Theorem 5. Solving the ENVY-FREE CAKE-CUT problemfor d + 1 players for the oracle functions requires timecomplexity �4N d−15.

Proof. In the proof of Lemma 3, the Kuhn’s triangu-lation in the last section reduces the problem of find-ing a Simmons-Su simplex to the problem of finding aSperner simplex. Because a Sperner simplex can be found

in O4N d−15 time (Deng et al. 2011), this reduction allowsus to find a Simmons-Su simplex in time O4N d−15. There-fore, the solution corresponds to a DISCRETE ENVY-FREE CAKE CUT.

We note here that the algorithm is based on a bin-ary search (on index) paradigm in fixed-point computa-tion (Chen and Deng 2008). This framework, with furtherinsight with respect to monotone utility function allows usto develop the FPTAS for three players discussed in the nextsection.

To prove the lower bound, we apply the same reduc-tion as the one in the last section. The lower boundì4N/2d5d−1 of ENVY-FREE CAKE CUT follows imme-diately from a lower bound ì4N d−15 of BROUWER ofsize length N (Deng et al. 2011). Because d is a constanthere, we obtain the lower bound ì4N d−15 for ENVY-FREECAKE CUT.

4. An FPTAS for Three Players withMonotone Utility Functions

For general oracle utility functions, the above results implya �41/�5 matching bound when the number of players isthree. Further improvement can only be possible when wehave further restrictions on the utility functions. In this sec-tion, we assume the utility functions in addition to satisfy-ing the nonnegativity condition and the Lipschitz conditionare also monotone.

Even in this case, the celebrated Stromquist’s movingknife (Stromquist 1980) for envy-free cake cutting of threeplayers can be shown to have an exponential lower bound.The main result here is to give an algorithm with a run-ning time polynomial in the number of bits of 1/�. Herea matching lower bound can be derived from very simpleutility functions.

4.1. Complexity of Finding Stromquist’s Solution

The celebrated Stromquist’s solution (Stromquist 1980)involves a referee who moves her sword from left to right.The three players each has a knife at the point that wouldcut the right piece to the sword in half, according to theirown valuation. Although the referee’s sword moves right,the three knives all move right in parallel but possibly atdifferent speeds. At all times, each player evaluates thepiece to the left of the sword, and the two pieces thatwould result if the middle knife cuts. If any of them seesthe left piece of the sword is the largest, he would shout“cut.” Then the sword cuts. The leftmost piece is assignedto the player who shouted. For the two players who did notshout, the one whose knife is to the left of the middle knifereceives the middle piece, and the one whose knife is onthe right of the middle knife receives the rightmost piece.

We will show that Stromquist’s moving knife procedurecannot be turned into a polynomial-time algorithm for find-ing an �-envy-free solution. We will do this by showing that


the particular fixed point found by the Stromquist’s proce-dure cannot be found with polynomial number of queries.We assume a query can ask about the valuation of a playerfor an interval 4x1 y5.

Suppose we have three players A, B, and C. The cakeis represented as an interval 60117. Players A and B havethe same utility function. Their utility functions can bedescribed as follows (we should normalize them for con-sistency but did not for simplicity of presentation).

uA4x1 y5= uB4x1 y5=

0 for x = 01 y = 1/10

2 for x = 1/101 y = 3/10

100 for x = 3/101 y = 4/10

2 for x = 4/101 y = 8/10

100 for x = 8/101 y = 9/10

0 for x = 9/101 y = 10

(1)

The value of both players for any other interval can becomputed, assuming that their valuation is uniform acrossall of the above intervals. For example uA41/2013/205 =

005. uC can be described in a similar way as follows:

uC4x1 y5=

100 − � for x = 01 y = 1/10

2 for x = 1/101 y = 3/10

98 for x = 3/101 y = 4/10

2 for x = 4/101 y = 5/10

0 for x = 5/101 y = 10

(2)

Assume that the value of � is very small. Now considerthe Stromquist’s moving knives. Suppose the referee startsby moving her knife from 0 towards 1. A and B have thesame utility function, so their knives are going to be at thesame place. Moreover, the “middle knife” will be alwaysA’s or B’s. Observe that when the referee’s knife reachespoint 1/10, the middle knife is at point 4/10. No one willshout cut yet. Now, observe that because according to Aand B the density of the interval 41/1013/105 is twice theinterval 44/1015/105, the speed of the knives of A and Bwill be exactly the same as the referee’s knife. For a similarreason, according to C the value of the middle piece (fromreferee’s knife to the knife of A or B) will remain just� above the value of the leftmost piece until the referee’sknife goes slightly beyond 3/10. At that time, C will shoutcut and receive the leftmost piece. Players A and B happilysplit the rest of the cake equally.

Now, we will slightly perturb the utility function of C.At a point 1/10 ¶ x ¶ 3/10 perturb the utility function ofC in the following way: increase the utility of C for theinterval 4x − �1x5 by �/2 and decrease his utility for theinterval 4x1 x+�5 by the same amount. Call the player withthis perturbed utility function Cx.

Now, it is easy to see that if we are running theStromquist’s method, player Cx will shout cut at the timethe referee’s knife reaches x. It is also not hard to see that

any value query asking for the value of an interval cannotdistinguish C from Cx unless one end point of the inter-val for which it is querying the value is in 4x − �1x + �5.Therefore, it is not possible to distinguish Cx from C withqueries of order polylogarithmic in 1/�.

4.2. An FPTAS for Three Players

Theorem 6. When the utility functions satisfy the abovethree conditions: nonnegativity condition, Lipschitz condi-tion, and monotonicity, an � envy-free solution can be foundin time O4log 1/�52 when the number of players is three.

For three players, the closed set of all possible cuts is atriangle. To obtain an approximate discrete envy-free solu-tion, the triangle is partitioned by Kuhn’s triangulation. LetV be the set of vertices of all base cells. Recall the label-ing and coloring rules defined in the previous sections. Wepartition V into three control subsets V01 V11 V2, each cor-responding to one player. According to the labeling rulegiven in the proof of Claim 1, a vertex X is assigned toVt if

∑di=0 iXi ≡ t4modd + 15. By this labeling rule, the

three vertices of each base simplex belong to different sub-sets Vt’s.

Next, we should color each vertex in V . For a vertex Xlabeled by t, we let player t choose. If its optimal segmentis the one of length Xs/N , 0 ¶ s ¶ 2, we assign color s tothe vertex X.

We proved a �41/�5 time solution for three players withgeneral utility functions in the previous section.

The improvement has been made possible by an effi-cient way to compute the index of a triangulated polygon,i.e., counting the sum of the signs of base intervals on theboundary. The main idea is an observation that at someappropriate cuts, the colors along the cut are monotone.Therefore, we can find the boundary of the three differentcolors on this cut to calculate the sum quickly.

More specifically, suppose the cake is cut into threepieces: a fixed interval 601 c/N 7, an interval varying in itsright end 6c/N 1 4c+ x5/N 7, and an interval varying in itsleft end 64c+ x5/N 117, where c is a fixed integer. For anyplayer, as x increases, the utility of the first piece willnot change, the second increases, and the third decreasesbecause of the monotonicity. Therefore, given the player, itspreferred pieces along the cuts in the form 4c1 x1N −x−c5with x varying will be monotone with only two possiblechanges in its colorings of those nodes. Note that such cutscorrespond to nodes in a line segment in the barycentriccoordinates, parallel to one of the boundaries of the trian-gle. If we can find the two possible changing points of thecolors, we know the colors of all vertices along the linesegments that belong to that player. We should do it oneby one for each of the three players. We should, in theend, know all the colors along this line segment. There-fore, we can calculate the number of base segments withcolors 0 and 1 on their different end vertices. That allowsus to calculate the number of such base segments along


the boundaries of the two polygons obtained by the linesegment crossing the previous polygon. This allows us todesign the following algorithm.

At any time in the algorithm, we maintain a subsetof V , with a nonzero index value, V 4i11 i21 k11 k25 =

84i1 j1 k52 i1 j1 k¾ 01 i+j+k =N , i1 ¶ i¶ i2, k1 ¶ k¶ k29,delimited by i = i1, i = i2 and k = k1, k = k2.

Algorithm 1. 1. If i2 − i1 = 1 and k2 −k1 = 1, find a fullycolored base triangle of V 4i11 i1 + 11 k11 k1 + 15, terminate.

2. Choose max8i2 − i11 k2 − k19 (and assume it is i2 − i1w.l.o.g.).

3. Let i3 = �4i1 + i25/2�.4. Calculate index4V 4i11 i31 k11 k255 and index4V 4i31 i21

k11 k255.5. Recurse on one of the two subpolygons with a

nonzero index.

Note that Step 4 is based on the above analysis and willbe explained in more detail later. By definition, index4V 4i11i21 k11 k255 = index4V 4i11 i31 k11 k255 + index4V 4i31 i21k11 k2550 At least one of the index4V 4i11 i31 k11 k255 andindex4V 4i31 i21 k11 k255 is nonzero because index4V 4i11 i21k11 k255 is nonzero. The algorithm always keep a poly-gon of nonzero index because the initial polygon has anonzero index. At the base case i2 = i1 +1 and k2 = k1 +1,V 4i11 i1 + 11 k11 k1 + 15 is either already a base triangle(with a nonzero index), or a diamond shape consisting oftwo base triangles, one of which must be of index nonzero.The correctness follows because the only possibility for itsindex being nonzero is when it is colored with all threecolors.

For complexity analysis, we first characterize the bound-ary conditions:

Property 1. There are up to three types of boundaries forV 4i11 i21 k11 k25

1. Bi=c = 84i1 j1 k5 ∈ V 2 i constant1 k1 ¶ k ¶ k21 i + j +

k =N1 i1 j1 k¾ 092. Bk=c = 84i1 j1 k5 ∈ V 2 k constant1 i1 ¶ i ¶ i21 i + j +

k =N1 i1 j1 k¾ 093. Bj=0 = 84i101 k5 ∈ V 2 i+ k =N1 i1 k¾ 09.

See Figure 10 as an example.Second, we establish the monotonicity property. Note

that for the third type of boundaries listed above, the mono-tonicity property does not hold for Bj=c in general, but onlyfor Bj=0.

Property 2. The colors of V0 ∩ Bi=c are monotone in k,and so are V1 ∩ Bi=c and V2 ∩ Bi=c. The same hold forVt ∩Bk=c, as well as Vt ∩Bj=0, t = 01112.

Proof. Observe that at V0 ∩ Bi=c, the color is determinedby Player 0 by finding the maximum of u04601 c/N 75,u046c/N 1 4c+ j5/N 75, and u0464c+ j5/N 1175, where 0 ¶j ¶ N − c. The first item is fixed. The seconditem is increasing in j because 6c/N 1 4c+ j5/N 7 ⊆

Figure 10. An example of three types of boundaries.

(5, 0, 0)

(0, 5, 0) (0, 0, 5)

(4, 1, 0)

(1, 4, 0)

(4, 0, 1)

(3, 0, 2)

(2, 0, 3)

(1, 0, 4)

(0, 4, 1) (0, 3, 2) (0, 2, 3) (0, 1, 4)

TYPE-2

TYPE-2

TYPE-3

TYPE-1

TYPE-1

6c/N 1 4c+ j ′5/N 7 for j ¶ j ′ and ui is assumed to be mono-tone. For the same reason, the last item is decreasing in j .The color will be 0 if u04601 c/N 75 is the maximum, 1 ifu046c/N 1 4c+ j5/N 75 is the maximum, or 2 otherwise. Ingeneral, as j increases, the color in V0 ∩Bi=c will start with2, then 0, and finally 1, assuming nondegeneracy. However,any of those colors may be missing.

The same analysis holds for other players and for thecase when k is fixed. However, it does not hold when j isfixed, except when j = 0.

Property 2 allows us to find the colors of all the verticeson the boundaries Vt ∩ Bi=i1

, Vt ∩ Bi=i2, Vt ∩ Bj=0, Vt ∩

Bk=k1and Vt ∩Bk=k2

, t = 01112, in time proportional to thelogarithm of the length of the sides, by finding all verticesat which the color changes along it. Once the changingpoints of the colors are found, using monotonicity, the totalnumber of positively and negatively signed base intervalsalong the boundary can be calculated in constant time.

The recursive algorithm reduces the size ofV 4i11 i21 k11 k25 geometrically. Let L = max8�i1 − i2�1 �k1 −

k2�5. L halves in two rounds of the algorithm. It takes2 logN steps to reduce L to 1.

At each round of the algorithm, we need to find out, foreach of three players, for each side of V 4i11 i21 k11 k25 (upto five in all), the boundary of the player’s color changes(two boundaries). That takes time logL, bounded by logN ,to derive a total of 3 × 5 × 2 logN .

Therefore, the time complexity is O4logN52.

5. Discussion and ConclusionIt remains open whether the approximate envy-free cake-cutting problem for four or more players would allow for afully polynomial-time approximation scheme as it does inthe three-player case, when we are dealing with monotonefunctions.


Utility functions may often not be monotone, and super-additive utility functions are abundant in real-life situations.However, the significant improvement in the three-playercase for monotone functions would by itself pose the ques-tion whether such progress can be carried over to higherdimensions.

Because the envy-free cake-cutting problems have exten-sive applications in political and social sciences, explicitprocedures solving such problems have been developed toimplement this fairness solution concept. Our current com-plexity study for cake cutting in the general setting couldbecome more practical in those applications. We leavethose as open problems for interested readers to furtherexplore the complexity of algorithms developed in thoseareas, such as consensus halving (Simmons and Su 2003)and cutting multiple cakes (Cloutier et al. 2010).

Acknowledgments

The authors thank the referees for their comments and suggestionsthat significantly improved the previous versions of this paper inthe presentation, organization, and possible extensions, as well asreferences. The third author thanks Arash Asadpour, as well as theorganizers and participants of Dagstuhl’s fair division workshopfor valuable discussions. The work of the first author is supportedby a grant from the University of Liverpool, by the ResearchGrants Council of Hong Kong [Project RGC CityU 112909], andby the National Science Foundation of China [Grant 61173011].

ReferencesAsadpour A, Saberi A (2007) An approximation algorithm for max-min

fair allocation of indivisible goods. Proc. 39th Annual ACM Sympos.Theory Comput. (ACM, New York), 114–121.

Bansal N, Sviridenko M (2006) The Santa Claus problem. Proc. 38thAnnual ACM Sympos. Theory Comput. (ACM, New York), 31–40.

Blum A, Jackson J, Sandholm T, Zinkevich M (2004) Preference elicita-tion and query learning. J. Machine Learn. Res. 5:649–667.

Blumrosen L, Nisan N (2007) Combinatorial auctions. Nisan N,Roughgarden T, Tardos E, Vazirani V, eds. Algorithmic Game Theory(Cambridge University Press, London), 267–298.

Brams SJ, Taylor AD (1995) An envy-free cake division protocol. Amer.Math. Monthly 102(1):9–18.

Brouwer LEJ (1910) Ueber eineindeutige, stetige Transformationen vonFlächen in sich. Math. Ann. 69:176–180.

Chen X, Deng X (2008) Matching algorithmic bounds for finding aBrouwer fixed point. J. ACM 55(3).

Chen X, Deng X (2009a) On the complexity of 2D discrete fixed pointproblem. Theoret. Comput. Sci. 410(44):4448–4456.

Chen X, Deng X (2009b) A simplicial approach for discrete fixed pointtheorems. Algorithmica 53(2)250–262.

Cloutier J, Nyman KL, Su FE (2010) Two-player envy-free multi-cakedivision. Math. Soc. Sci. 59(1):26–37

Crescenzi P, Silvestri R (1998) Sperner’s lemma and robust machines.Comput. Complexity 7(2):163–173.

Daskalakis C, Goldberg PW, Papadimitriou CH (2006) The complexityof computing a Nash equilibrium. Proc. 38th Annual ACM Sympos.Theory Comput. (ACM, New York), 71–78.

Deng X, Qi Q, Saberi A, Zhang J (2011) Discrete fixed points: Models,complexities and applications. Math. Oper. Res. 36(4):636–652.

Dobzinski S, Nisan N, Schapira M (2010) Approximation algorithms forcombinatorial auctions with complement-free bidders. Math. Oper.Res. 35(1):1–13.

Edmonds J, Pruhs K (2006a) Cake cutting really is not a piece of cake.Proc. 17th Annual ACM-SIAM Sympos. Discrete Algorithms (ACM,New York), 271–278.

Edmonds J, Pruhs K (2006b) Balanced allocations of cake. Proc. 47thAnnual IEEE Sympos. Foundations Comput. Sci. (IEEE, New York),623–634.

Elimam AA, Girgis M, Kotob S (1996) The use of linear programmingin disentangling the bankruptcies of Al-Manakh stock market crash.Oper. Res. 44:665–676.

Even S, Paz A (1984) A note on cake-cutting. Discrete Appl. Math.7:285–286.

Friedl K, Ivanyos G, Santha M, Verhoeven F (2005) On the black-boxcomplexity of Sperner’s lemma. Proc. 15th Internat. Sympos. Fun-damentals Comput. Theory (Springer, Lübeck, Germany), 245–257.

Halland A, Zeckhauser R (1979) The efficient allocation of individuals topositions. J. Political Econom. 87(2):293–314.

Hirsch PM (1986) From ambushes to golden parachutes: Corporatetakeover as an instance of cultural framing and institutional integra-tion. Amer. J. Sociol. 9(4):800–837.

Kelly FP, Maulloo AK, Tan DHK (1998) Rate control in communicationnetworks. Shadow prices, proportional fairness, and stability. J. Oper.Res. Soc. 49:237–252.

Kleinberg J, Rabani Y, Tardos E (1999) Fairness in routing and load bal-ancing. Proc. 40th IEEE Sympos. Foundations Comput. Sci. (IEEE,New York).

Kuhn HW (1960) Some combinatorial lemmas in topology. IBM J. Res.Development 4:518–524.

Kumar A, Kleinberg JM (2006) Fairness measures for resource allocation.SIAM J. Comput. 36(3):657–680.

Lipton RJ, Markakis E, Mossel E, Saberi A (2004) On approximately fairallocations of indivisible goods. ACM Conf. Electronic Commerce(ACM, New York), 125–131.

Megiddo N, Papadimitriou CH (1991) A note on total functions, exis-tence theorems and computational complexity. Theoret. Comput. Sci.81:317–324.

Meunier F (2008) Discrete splittings of the necklace. Math. Oper. Res.33(August):678–688.

Moulin H (2007) Minimizing the worst slowdown: Offline, online. Oper.Res. 55(5):876–889.

Papadimitriou CH (1994) On the complexity of the parity argumentand other inefficient proofs of existence. J. Comput. Systems Sci.48(3):498–532.

Pratt JW, Zeckhauser RJ (1990) The fair and efficient division of theWinsor family silver. Management Sci. 36(11):1293–1301.

Scarf H (1977) The computation of equilibrium prices: An exposition.H. Cowles Foundation for Research in Economics, Yale University,New Haven, CT.

Sherstyuk K (1998) How to gerrymander: A formal analysis. PublicChoice 95:27–49.

Simmons FW (1980) Private communication to Michael Starbird.Simmons FW, Su FE (2003) Consensus-halving via theorems of

Borsuk-Ulam and Tucker. Math. Soc. Sci. 45(1):15–25.Sperner E (1928) Neuer Beweis für die invarianz der dimensionszahl und

des gebietes. Abh. Math. Sem. Univ. Hamburg 6:265–272.Steinhaus H (1948) The problem of fair division. Econometrica

16:101–104.Steinhaus H (1949) Sur la division pragmatique. Econometrica (supple-

ment) 17:315–319.Stromquist W (1980) How to cut a cake fairly. Amer. Math. Monthly

87(8):640–644. Addendum 88(8, 1981):613–614.Su FE (1999) Rental harmony: Sperners lemma in fair division. Amer.

Math. Monthly 106:930–942.Thomson W (2007) Children crying at birthday parties. Why? Econom.

Theory 31:501–521.Todd MJ (1976) The Computation of Fixed Points and Applications.

Lecture Notes in Economics and Mathematical Systems (Springer-Verlag, New York).

Woeginger GJ, Sgall J (2007) On the complexity of cake cutting. DiscreteOptim. 4(2):213–220.


Xiaotie Deng is a chair professor in economics and computa-tion in the Department of Computer Science, University of Liv-erpool. His research focus is on algorithmic game theory, whichdeals with computational issues on fundamental economic prob-lems such as Nash equilibrium, resource pricing and allocationprotocols such as auction and market equilibrium, as well as the-ory and practice in Internet market design.

Qi Qi is an assistant professor in the Department of Indus-trial Engineering and Logistics Management at Hong Kong

University of Science and Technology. Her research interestsinclude game theory and mechanism design, Internet and com-putational economics, online auctions, resource allocation, andoptimization.

Amin Saberi is an associate professor and 3COM FacultyScholar in the Department of Management Science and Engi-neering at Stanford University. His research interests includealgorithms, approximation algorithms, and algorithmic aspects ofgames, markets, and networks.

algorithmic solutions for envy-free cake cutting · ﬁnding an envy-free cake cutting is the same...

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