Discrete Envy-free Division of Necklaces and Maps
Roberto Barrera, Kathryn Nyman, Amanda Ruiz, Francis Edward Su, Yan X. Zhang
aa r X i v : . [ m a t h . C O ] O c t Envy-free and Approximate Envy-free Divisionsof Necklaces and Grids of Beads
Roberto Barrera , Kathryn Nyman , Amanda Ruiz , FrancisEdward Su and Yan X Zhang Department of Mathematics, Texas State University Department of Mathematics, Willamette University Department of Mathematics, University of San Diego Department of Mathematics, Harvey Mudd College Department of Mathematics, San Jose State UniversityOctober 2, 2018
Abstract
We study discrete versions of the envy-free cake-cutting problem, in-volving one-dimensional necklaces of beads and two-dimensional grids ofbeads. In both cases beads are indivisible, in contrast to cakes which arecontinuously divisible, but a general theme of our methods is an appealto continuous cake-cutting ideas to achieve envy-free and approximateenvy-free divisions of discrete objects. (This problem is distinct from the“necklace-splitting problem” more commonly studied, which does not in-volve envy-freeness.) Our main result in two dimensions is an envy-freedivision of a grid of beads under certain conditions on the preferences,with near-vertical cuts.
Keywords : Discrete cake-cutting, Envy-free divisions, Necklace splitting
The archetypal fair-division problem considers a division of cake among severalplayers and seeks to find allocations that players consider “fair”. One suchnotion of fairness is that of envy-freeness . An allocation is called envy-free ifevery player is happy with the piece they are assigned and would not prefer totrade with another player. Under mild assumptions, the cake cutting problemhas a solution (see e.g., [11]):
Theorem 1 (Envy-Free Cake-Cutting) . For any set of n players who prefer caketo no cake and have closed preference sets, there exists an envy-free allocation f cake using ( n − cuts. Furthermore, there exists a finite ǫ -approximatealgorithm. The existence of envy-free divisions has been known since Neyman [8], withrecent attention paid to finding constructive proofs with potential for applica-tions. The first constructive n -player envy-free solution is due to Brams andTaylor [4]. It is a finite but unbounded procedure— meaning that it will termi-nate but, depending on the preferences, the number of steps may be arbitrarilylarge! Moreover, the cake could be divided into a huge number of pieces. Azizand Mackenzie [2] recently developed a bounded n -person procedure, thougheven for small n it can take an astronomical number of steps and cuts to re-solve.Other methods [9, 11] produce an approximate envy-free division, i.e., a di-vision in which each player feels their piece is within ǫ of being the best piece intheir estimation. An advantage to approximate procedures is that the number ofsteps and cuts is more manageable, and so has the possibility of being practical.One such method, due to Simmons and described in [11], uses Sperner’s lemmato accomplish a division by a minimal number of pieces. Hence it requires only( n −
1) cuts.In spirit of work such as [7], we consider a discrete analogue of the classicalcake-cutting problem in which a number of indivisible beads along a line isto be allocated in an envy-free way. We call this the problem of envy-freenecklace-cutting . Problems involving distributing indivisible goods has receivedconsiderable attention in the economics and fair division literature (see, e.g., [5]and [6]), but with less emphasis on geometric constraints. Barbanel’s work [3]studies and organizes fair division under a geometric framework, but assumesdivisible goods. Thus, we think our problem represents a potentially fruitfulavenue of research as it combines having indivisible goods and having geometricconstraints. Another representative of such problems is the necklace-splitting problem, in which a string of k types of beads is to be split among n thieves sothat each thief receives the same number of each type of bead. (See e.g., [1]).This problem is different from our problem as it focuses on balancing the piecesfor the k types simultaneously and does not involve the thieves having differentpreferences for the beads.The necklace cutting problem is like the traditional cake-cutting problem,but the discreteness of the problem means an envy-free division may not nec-essarily exist. The simplest example is when we have a single bead valued byevery player. If we give the bead to one player, all other players will be envious. Problem.
It is natural to consider the following as fundamental problems forstudying envy-free necklace-cutting: • What additional assumptions do we need to make to guarantee an envy-freedivision exists? Strictly speaking, [3] cares about non-atomic preferences, which is defined as when aplayer cannot put nontrivial value on a subset A of a geometric space (analogous to a pieceof cake) unless A has a subset B with strictly less but nonzero value to the player. However,this definition captures the intuition of divisible goods. What algorithm produces such divisions? • When envy-free is impossible, what is the best we can do? For example,what is the smallest ǫ for which we can guarantee an ǫ -envy-free division? Marenco and Tetzlaff [7] initiated the study of envy-free necklace-cutting byconsidering the case where each bead is valuable to exactly one player. (Theydon’t use the language of necklace-cutting, but speak of atoms arranged on aninterval.) Appealing to the techniques in [11] for traditional cake-cutting, theyuse a combinatorial result known as Sperner’s lemma to find “nearby” divisionsof the necklace, one for each player, in which each player prefers a different pieceof the necklace in their division. The location of cuts in these divisions differsby only one bead for each knife, and the idea is to use the location of the cutsto determine a final division that will be envy-free for all players. The positionof each knife in the final division is given by the position of that knife in thesplitting associated to the player who values the contested bead.We build on their work by examining the questions above for different classesof constraints, including more general preferences where several players mayvalue each bead (Section 3), and a two-dimensional arrangement of the beads(Section 4).
We wish to divide an open necklace of indivisible beads, but a recurring tech-nique of this paper is to cut the cake analogue of a necklace (so that beads arepotentially divided) and then slide the cuts so they do not divide beads.We model this by considering the necklace as an interval [0 , ℓ ], in whichthe beads are now intervals between consecutive integers. Viewed as a cake,cuts may be made at any point, but to be a necklace-cutting, cuts must lie atintegers. See Figure 1.Figure 1: Two equivalent ways of picturing a discrete necklace cut into threestrings of beads.We assume each player P has a nonnegative, additive, real-valued valuation function v P defined on any possible string of beads, but because we will becutting cake analogues of the necklace, we define v P on any fraction of a bead3o be that same fraction of the valuation of a full bead. In other words, weimagine the valuation of a bead to be uniformly distributed over the bead.This gives rise to a preference relation between each two possible strings ofbeads, where player P prefers string A to B if and only if v P ( A ) ≥ v P ( B ).We stress that a player may prefer several strings of beads if they value themequally. If A is preferred to B but not vice-versa, we say that the player strictlyprefers A to B . Marenco and Tetzlaff [7] showed that we can achieve an envy-free division ofthe necklace in the case where each bead is valued by only one player. We sayplayers have monolithic preferences in this case, and we can label beads by theidentities of the players who value them. See Figure 2 for an example.A B C C AA B CFigure 2: A discrete necklace where players have monolithic preferences, cut by2 cuts into 3 strings of beads. The labels in the necklace denote the players whovalue those beads; the labels above the necklace denotes the allocation. Thisparticular allocation is envy-free if A has the same valuation for both A -beads. Theorem 2. (Marenco–Tetzlaff ) If n players with monolithic preferences are todivide a necklace of beads, then there exists an envy-free division of the necklaceusing only ( n − cuts. We wish to study the case where each bead may be valued by more than oneplayer. After all, the problem of fair division is most interesting when playersbicker over items that are mutually desired. Thus, unless otherwise stated, weassume valuation functions v P ( b ) can be positive for any player P and for anybead b . We’ve already seen an example in Section 1 that we cannot guaranteean envy-free division in all cases. A less trivial example is given in Figure 3.However, just because we cannot guarantee an envy-free division does notmean we cannot come close. As in [11], call a division ǫ -envy-free if for eachplayer, the other strings are worth no more than ǫ plus the value of the stringassigned to them. Note that a division is 0-envy-free if and only if it is envy-free.We achieve the following result: Theorem 3.
For a necklace of beads, where the value of each bead is at most s to every player, there always exists an ǫ -envy-free division of the necklace among A and B ; they will fight over the middlebead. n players using only ( n − cuts with ǫ < s . In particular, there is always a (2 s ) -envy-free division.Proof. Represent our necklace of ℓ beads as the interval [0 , ℓ ] such that beadsare the intervals between consecutive integers and players’ valuations of eachbead are uniformly distributed over the bead’s interval. If we allow the cutsto be made at non-integer points, Theorem 1 guarantees that there exists anenvy-free division, D , of the interval. We begin with the division D and shiftany cuts that divide a bead by rounding each such cut to the nearest integralpoint. For example, if a cut is at x = 1 .
2, shift it so it is at x = 1 .
0. If a cut isat a half-integer, then we always round to the right.Suppose player P were assigned a string of beads with value v in D . Thenthis rounding process decreases P ’s valuation by at most s/ s/ P ’s valuation of her piece decreases bystrictly less than s . Furthermore, from the perspective of P , the value of anyother piece in D increases by at most s by the same logic. Thus, P ’s envy inthe final division, where all the cuts are now at integer points, is strictly lessthan 2 s .If we specialize to the situation where the valuations are integral, the strict-ness of the inequality above can be quite useful. One such situation is when theonly opinion a player has towards a bead is that they either desire the bead orare indifferent to the bead. We model this situation by assigning each bead’sworth to a player as either exactly 1 or 0 respectively. In this case, we obtain: Corollary 1.
For n players dividing a necklace of beads with valuations takingvalues either 0 or 1, there always exists a -envy-free division using only ( n − cuts.Proof. Theorem 3 shows that we can get the maximum envy to be strictly lessthan 2. However, since the valuations are sums of 1’s and 0’s, the envy must bean integer and so the envy is bounded above by 1.5
Two Dimensional Divisions: a Grid of Beads
In this section, we consider a 2-dimensional grid of indivisible beads that wewish to divide in an envy-free way. We can imagine that there are strings inthe 4 cardinal directions connecting the beads; we can also just imagine thatthe beads are on a valueless “quilt” which is to be cut. This is a generalizationof cake-cutting problem in two-dimensions, which has been studied (e.g., [10]),but for divisible goods. An important consideration in two-dimensional divisionis the geometry of the pieces. Our main theorem is Theorem 4, which uses a“sliding” method remniscent of network flow problems from computer science.Segal-Halevi (personal communication, Nov. 2015) has pointed out that anenvy-free division of a grid of beads can be accomplished by the Marenco-Tetzlaffresult by cutting the grid into a one-dimensional necklace along a snake-like paththat winds back and forth across the rows of the grid, from top to bottom. (Infact, any Hamiltonian path along the grid graph would do.) This would achievean an envy-free division with pieces that are connected in the original grid. Healso notes that in the case of preferences that are not necessarily monolithic,our Theorem 3 would yield an (2 s )-envy-free division where s is the maximumvaluation of any bead by any player.However, we desire our pieces to align with the geometry of the grid innear-vertical cuts, which could be useful if, for instance, the rectangular gridrepresented a grid of resources to be split up by players, and for some geometricreason, we need the pieces to “connect” vertically to resources at the top and thebottom of the grid. Perhaps the top of the grid is access to a waterway usefulfor trade, and the bottom of the grid is access to a canyon of raw materials.This situation may be contrived, but we believe the geometric constraint isinteresting and the methods generated by this paper useful for further analysis.If we demand actual vertical cuts, Theorem 3 yields an immediate corollary: Corollary 2.
Consider a -dimensional grid of beads. Suppose s is the highesttotal valuation of the beads in a column by any player. Then there exists an ǫ -envy-free division among n players using ( n − vertical cuts, where ǫ < s .Proof. We model our grid of beads as the region [0 , k ] × [0 , ℓ ] such that eachsquare is a bead and the players’ valuations are spread uniformly across eachbead. We can now treat our grid as a 1 × ℓ necklace by summing the valuationsof the i -th column to obtain the valuation of the i -th bead. Applying Theorem 3gives the result.However, it seems that we can do significantly better if we relax our require-ment so that cuts are allowed to be “near-vertical”. For what follows, define a near-vertical cut of a grid to be a path of edges, all lying within one verticalcolumn of the grid, that separate the squares into two sets (as a Jordan curve).We consider two cuts to be non-intersecting if the cuts do not intersect eachother transversely, that is, there do not exist two points of a cut such that theyare in the interiors of the two different sets created by the other cut.6 heorem 4. Suppose n players have monolithic preferences over a 2-dimensionalgrid of beads with valuations taking values either 0 or 1. Then there exists anenvy-free division of the grid using ( n − non-intersecting near-vertical cuts. Figure 4: The grid is cut via 2 cuts into 3 pieces, each (barely) path-connectedif we allow movement along the cuts themselves. The right is a visualization ofthis division interpreted as beads on a valueless “quilt.”
Proof.
Again, we model our grid of beads as the region [0 , k ] × [0 , ℓ ]. By Theo-rem 1, there exists an envy-free division of the grid using ( n −
1) parallel verticalcuts. We think of each of these vertical cuts as a union of k vertical edges. Wecall such a vertical edge integral if its x -coordinate is integral and non-integral otherwise. Our strategy is to slide non-integral vertical edges left or right so theybecome integral while keeping them connected via horizontal edges along theintegral y -coordinates. We do this while holding the relative left-to-right orderof the ( n −
1) vertical edges on each horizontal strip constant, thus generatingthe desired cuts for the discrete grid of squares.We say that a vertical edge borders players P and Q if the two adjacentpieces allocated by the original envy-free allocation belong to P and Q . First,consider any bead S desired by player P such that at least one vertical edgegoing through S borders P . In this case, we can allocate the bead completelyto player P (by moving all edges in S left of P ’s piece to S ’s left boundaryand all edges in S right of P ’s piece to S ’s right boundary). Due to monolithicpreferences, this gives player P a strictly more valuable piece and does notaffect the preferences and envy of other players, as only P cares about the bead.Doing this for all beads ensures two conditions now hold for our (still envy-free)division D : 7 for every player, P , D assigns an integral amount of beads that player P values to player P (if not, then there must be some non-integral verticaledge bordering P somewhere); • if a bead desired by P contains a non-integral vertical edge bordering P and P , then neither P nor P can be P .Our strategy to move the remaining non-integral edges is as follows. Sup-pose we have at least one non-integral vertical edge somewhere in our envy-freedivision. We describe a sliding process such that: • we stay envy-free at all times, • any time a non-integral edge slides into another, we consider the twoedges to have merged into a single edge (and sliding the resulting edgecorresponds to the underlying edges moving together as a group), and • any time a non-integral edge becomes integral, we no longer slide it.When a non-integral edge slides into another edge or becomes integral, thenumber of non-integral edges decreases by one; we can then repeat this processuntil all non-integral edges become integral. Thus, it suffices to show that wecan always perform this sliding process without losing the envy-free property.Consider any bead S with non-integral vertical edges going through it, de-sired by player P . Say that S is contested by Q if Q ’s piece contains part of S .Let the set of players contesting S contain P and P (recall that neither canbe P ). We say we donate from P to P through S if we slide the non-integralvertical edges through S in a (unique) way that the P piece in S decreases ata constant speed, the P piece in S increases at the same constant speed, andthe other pieces in S stay at constant size. Donating from P to P through S keeping the division envy-free results in one of the following: • the P part of S becomes empty, which causes the number of non-integraledges to decrease, or • P values P ’s piece exactly equal to her own, in which case any furtherdonation to P from P would make P envious of P .Suppose there exists at least one remaining bead desired by player P withat least one non-integral edge going through it. We define an auxiliary graph G with vertices indexed by players who are not P . For each bead S desiredby player P with at least one non-integral edge, draw an edge in G labeled by S with endpoints P and P for every pair such that a non-integral edge in S borders P and P (we allow multiple edges in G ). We have two cases: • If we have a cycle (with no repeated vertices) in G of the form P → P → · · · → P k → P , then we can simultaneously donate from P to P (through the bead cor-responding to the edge between P and P in G ), P to P , etc. through8 k to P . Since P ’s valuation of all the P i ’s pieces are constant (each pieceis donating and being donated to at the same rate), at some point at leastone of the parts belonging to some P i in one of these beads becomes 0,corresponding to a decrease in the number of non-integral vertical edges. • If we do not have a cycle, then some vertex must have degree 1. Thismeans there is some bead S desired by P and contested by Q = P where Q does not have a fractional piece of a bead desired by P anywhere else.This means P ’s valuation of Q ’s piece has fractional part exactly equal to Q ’s amount in S . Because P ’s evaluation of P ’s own piece is currentlyintegral, we can give the entire bead S to Q without fear that P willbecome envious of Q .For an example of this process, see Figure 5.CC AA BB ABC B A CC AA BB ABC B AFigure 5: Left: an envy-free cut where A values all 3 pieces equally. We wish tomove the non-integral vertical edges, but doing either in isolation would cause A envy. Thus, we slide the top edge left at the same rate that the bottom edgeslides right, keeping A envy-free, until they slide to integral points.In both cases, we strictly decrease the number of non-integral vertical edges.Thus, we are able to perturb the vertical edges until all edges are integral sothat the cut is near-vertical, in which case we have an envy-free allocation ofthe grid of beads.From the proof, we observe that a vertical edge through a bead S in theoriginal division can move to the left or right boundary of S . This means inthe worst case, we can get some disjoint-looking pieces, with each vertical cuthaving a possible horizontal deviation of distance at most 1; for an example, seeFigure 4. However, we maintained the vertical nature of the pieces, as desired.This proof only uses the 0-1 constraint in the second bullet point above whenresolving the degree 1 nodes. It would be interesting to see if this proof couldbe extended to more general preferences.9 Acknowledgments
This work began at the 2014 AMS Mathematics Research Community - Al-gebraic and Geometric Methods in Applied Discrete Mathematics, which wassupported by NSF DMS-1321794. It was finished while two of the authors werein residence at at the Mathematical Sciences Research Institute in Berkeley,California, during the Fall 2017 semester where they were supported by the Na-tional Science Foundation under Grant No. DMS-1440140. Su was supportedpartially by NSF Grant DMS-1002938. We thank Boris Alexeev for valuableconversation. We also thank Erel Segal-Halevi for his valuable comments inSection 4.
References [1] Noga Alon,
Splitting necklaces , Advances in Mathematics (1987), no. 3, 247–253.[2] Haris Aziz and Simon Mackenzie, A discrete and bounded envy-free cake cutting protocolfor any number of agents , CoRR abs/1604.03655 (2016).[3] J.B. Barbanel and A.D. Taylor,
The geometry of efficient fair division , Cambridge Uni-versity Press, 2005.[4] Steven J Brams and Alan D Taylor,
An envy-free cake division protocol , American Math-ematical Monthly (1995), 9–18.[5] ,
Fair division: From cake-cutting to dispute resolution , Cambridge UniversityPress, 1996.[6] R.J. Lipton, E. Markakis, E. Mossel, and A. Saberi,
On approximately fair allocationsof indivisible goods , Proceedings of the 5th ACM Conference of Electronic Commerce(2004), 125–131.[7] Javier Marenco and Tom´as Tetzlaff,
Envy-free division of discrete cakes , Electronic Notesin Discrete Mathematics (2011), 231–236.[8] J Neyman, Un theoreme d’existence , CR Acade. Sci. Paris (1946), 843–845.[9] J. Robertson and W. Webb,
Cake-cutting algorithms: Be fair if you can , Ak Peters Series,Taylor & Francis, 1998.[10] Erel Segal-Halevi, Shmuel Nitzan, Avinatan Hassidim, and Yonatan Aumann,
Fair andsquare: Cake-cutting in two dimensions , Journal of Mathematical Economics (2017),1–28.[11] Francis Edward Su, Rental harmony: Sperner’s lemma in fair division , American Math-ematical Monthly (1999), 930–942., American Math-ematical Monthly (1999), 930–942.