Achieving rental harmony with a secretive roommate
AACHIEVING RENTAL HARMONY WITH A SECRETIVE ROOMMATE
FLORIAN FRICK, KELSEY HOUSTON-EDWARDS, AND FRÉDÉRIC MEUNIER
Abstract.
Given the subjective preferences of n roommates in an n -bedroom apartment, onecan use Sperner’s lemma to find a division of the rent such that each roommate is content witha distinct room. At the given price distribution, no roommate has a strictly stronger preferencefor a different room. We give a new elementary proof that the subjective preferences of only n − of the roommates actually suffice to achieve this envy-free rent division. Our proof, inparticular, yields an algorithm to find such a fair division of rent. The techniques also givegeneralizations of Sperner’s lemma including a new proof of a conjecture of the third author. Introduction
As the rent in Larry’s new two-bedroom apartment is $1000 he will have to look for a roommate tosplit the cost. The two rooms are not the same size, and each has their own advantages, yet fairly dividingthe rent between the rooms is Larry’s primary concern. Moe is interested in moving in with Larry, andLarry feels that splitting the rent $600 – $400 between the two rooms is fair — for this division of therent the disadvantages of the second room are offset by its reduced cost. Now, when Larry offers the tworooms to Moe at these prices, it will not matter to him which room Moe chooses; Larry will be contentwith the other room. The two new roommates will not be envious of one another and live in a state of rental harmony . Larry accomplished this envy-free rent division without taking Moe’s preferences intoaccount.This is not a lucky accident of the two person – two bedroom situation: for a three-bedroom apartmentLarry and Moe can fairly divide the rent among the rooms without taking the preferences of a thirdroommate, Curly, into account. There is a division of the total rent such that Curly can decide on anarbitrary room, and this will leave Larry and Moe with sufficiently many options to accomplish rentalharmony among the three of them. In general, it suffices if n − roommates know each others preferencesto fairly divide the rent of an n -bedroom apartment. We give an algorithm for producing such a fairdivision of rent; see Asada et. al. [1] for the recent nonconstructive topological proof of this result.That rental harmony can always be achieved in an n -bedroom apartment (under mild conditions) ifthe subjective preferences of all n future roommates are known was shown by Su [7], partially reportingon work of Simmons. The proof uses a combinatorial-geometric lemma about labelings of simplices dueto Sperner [6]. This makes the proof, especially for low n , accessible to a nonexpert audience. Here ourgoal is to adapt Su’s arguments for n = 3 and then give a separate elementary proof of the existence ofa fair division of rent for n roommates, where the preferences of one roommate are unknown.We first recall the mild conditions stipulated by Su to guarantee the existence of an envy-free rentdivision:1. In any division of the rent, each tenant finds at least one room acceptable.2. Each person prefers a room which costs no rent (i.e., a free room) to a non-free room. Date : May 12, 2017. a r X i v : . [ m a t h . C O ] M a y FLORIAN FRICK, KELSEY HOUSTON-EDWARDS, AND FRÉDÉRIC MEUNIER
3. If a person prefers a room for a convergent sequence of prices, then that person also prefers the roomfor the limiting price.Here we will slightly change condition 3 to simplify the proof and eliminate the need to take limits: Theroommates do not care about one-cent error margins. We remark that Su’s second condition togetherwith the third condition imply that the roommates are indifferent among free rooms, that is, if multiplerooms are free then each roommate is content with any of them. This is because for any division of therent where multiple rooms are free, there is always a sequence of prices converging to this rent divisionwhere only one specific room is free. And thus, we will also assume that the roommates are indifferentamong free rooms. In summary, our conditions are:1. In any division of the rent, each tenant finds at least one room acceptable.2. Each person prefers a room which costs no rent (i.e., a free room) to a non-free room, and each personis indifferent among free rooms.3. The roommates do not care about one-cent error margins.Under these conditions, we can give a new, elementary (and now constructive) proof of our main theorem:
Theorem 1.
For an n -bedroom apartment it is sufficient to know the subjective preferences of n − roommates to find an envy-free division of rent. In Section 2 we recall Sperner’s lemma and provide two proofs. The first proof is based on a classical“trap-door” argument and the second proof introduces a piecewise linear map which is used in the proofof our main theorem. Section 3 gives two proofs of Theorem 1 in the case where n = 3 , the first ofwhich was originally presented by the second author in the PBS Infinite Series episode “Splitting Rentwith Triangles” [4]. Section 4 generalizes the second proof given in Section 3 to show the main theorem.Section 5 explains how this proof yields an algorithm to find the fair division of rent. Section 6 utilizesthe piecewise linear map introduced in previous sections to prove two generalizations of Sperner’s lemmain an elementary way — one of these generalizations had been conjectured by the third author and wasrecently proven by Babson [2] with different methods.2.
Sperner’s lemma
Begin with a triangle which is subdivided into several smaller triangles. Label the three vertices ofthe original triangle from the set { , , } so each vertex receives a distinct label. Then, label each vertexon the edges of the original triangle by either of the labels at the endpoints. Finally, label the interiorvertices , , or arbitrarily. This is a Sperner labeling of a triangle, as in Figure 1.The result known as
Sperner’s lemma for a triangle states that there exists some small triangle thatexhibits each of the three labels on its vertices. Even stronger, there will be an odd number of suchfully labeled triangles. Here we outline two proofs of Sperner’s lemma. The first, known as the trap-doorargument goes back to Cohen [3] and Kuhn [5]. The second is based on a piecewise linear map betweenthe vertex labels. We use this map in Section 3 to prove there exists a fair division of rent for threeroommates, one of whose preferences are secret.
Proof 1, Trap Door Argument:
Select two distinct vertices of the main triangle. To illustrate, we will usethose labeled and . Imagine each small triangle is a room and any edge with endpoints labeled and is a door. Notice that each room has zero, one or two doors; it is impossible to have three doors. Thecrucial observation is that a room has one door if and only if it is labeled with all three labels. To proveSperner’s lemma, we will find a room with exactly one door. CHIEVING RENTAL HARMONY WITH A SECRETIVE ROOMMATE 3
Figure 1.
Options for a Sperner labelingObserve that there must be an odd number of doors along the boundary of the original triangle. Allthe boundary doors will fall on one edge of the original triangle, between the main vertices labeled and . Starting from the vertex labeled and going down to the vertex labeled at the bottom, eachtime we encounter a smaller edge where the label switches from to or from to , it’s a door. It must“switch” an odd number of times, so there are an odd number of exterior doors.Now, pick one of these exterior doors and walk through it. Either that was the only door in the room,or there is exactly one other door. If there is another door, walk through that. Keep walking throughdoors – without backtracking – until you are stuck. This procedure leads to a room with only door, i.e.,a fully labeled room. See Figure 2.However, following this procedure, it is possible to wander back out through a door on the boundaryof the main triangle. This is why it is important that there are an odd number of doors on the boundary.Entering through one door and exiting through another still leaves an odd number of boundary doors –namely, at least one – to continue this procedure.Because there is an odd number of boundary doors, this procedure will result in an odd number offully labeled rooms. However, there may be other fully labeled rooms which are not accessible from theboundary. This will happen if and only if the single door of the fully labeled room leads to anotherinaccessible fully labeled room. Since these inaccessible fully labeled rooms always come in pairs, thetotal number of fully labeled rooms must be odd. (cid:3) The second proof will use the fact that a piecewise linear self map λ : ∆ −→ ∆ of the triangle ∆ thatmaps the boundary to itself and has degree one on the boundary of the triangle (i.e., going around theboundary of ∆ the image points wander once around the boundary as well) must be surjective. Whilethis is still intuitive, we will use a similar statement in higher dimensions when generalizing this proof: apiecewise linear self map λ of the n -simplex ∆ n that preserves faces setwise (i.e., λ ( σ ) ⊆ σ for any face σ of ∆ n ) has degree one on the boundary and thus is surjective. We will give a simple path-following proofof the fact that any such map is surjective in Section 5. In particular, any topological fact used in ourargument will be proved in an elementary fashion. Proof 2, Piecewise Linear Map:
A Sperner labeling of a subdivided triangle can naturally be thought ofas a piecewise linear map λ : ∆ −→ ∆ from the triangle to itself, defined as follows. For each small FLORIAN FRICK, KELSEY HOUSTON-EDWARDS, AND FRÉDÉRIC MEUNIER
Figure 2.
Procedure for finding fully labeled roomtriangle, λ maps each vertex to the vertex of the original triangle with the same label. Then extend thismap linearly within the small triangle. For example, the barycenter of a small triangle labeled , , maps to the barycenter of the original triangle, while the barycenter of a triangle labeled , , mapsto the point on the edge with endpoints labeled and that separates the edge in a two-to-one ratio.Note that, for any edge e of the original triangle λ ( e ) ⊆ e (thus λ also fixes the vertices), and that λ iscontinuous.Moving around the boundary of the original triangle and tracing the image points of λ , we can seethat λ has degree one on the boundary of the original triangle. This implies that it is surjective and,in particular, there exists a point x ∈ ∆ such that λ ( x ) is the barycenter of the original triangle, i.e., λ ( x ) = ( , , ) . Then τ , the smaller triangle containing x , must be fully labeled. If τ had only two ofthe labels, then λ ( τ ) would be contained in the edge of the original triangle spanned by the vertices withthe two labels of τ . This contradicts that λ ( x ) is the barycenter. (cid:3) Both proofs of Sperner’s lemma easily generalize to higher dimensions. Given an ( n − -dimensionalsimplex ∆ n − that is subdivided into smaller ( n − -dimensional simplices, a Sperner labeling is a labelingof the vertices with { , , ..., n } such that (1) the n vertices of the original ( n − -simplex receive distinctlabels, and (2) a vertex subdividing a k -face of ∆ n − is labeled by one of the k + 1 labels of the k -face. Sperner’s lemma for higher-dimensional simplices states that any subdivided ( n − -simplex witha Sperner labeling contains an odd number of smaller ( n − -simplices that exhibit all n labels on theirvertices. More precisely, by a subdivision of ∆ n − into smaller ( n − -simplices, we mean that ∆ n − iscovered by a collection of ( n − -simplices whose interiors are disjoint and such that any two of themintersect in a (possibly empty) face of both of them. We call such a subdivision a triangulation of ∆ n − .The first proof of Sperner’s lemma, using the trap-door argument, generalizes by induction. Assumingthe lemma holds up to dimensions n − , one can prove it holds for n -dimensional simplices by treatingthe ( n − -dimensional faces labeled (1 , , ..., n ) as doors. The proof of the existence of a fully labeled n -simplex then follows by direct analogy with the n = 2 case above. A fully labeled ( k − -dimensional faceis a boundary door for a k -dimensional face, so one can link together the paths created in the trap-doorargument to obtain the fully labeled room. CHIEVING RENTAL HARMONY WITH A SECRETIVE ROOMMATE 5
The second proof generalizes to higher-dimensional simplices more directly. The map λ : ∆ n −→ ∆ n from the n -simplex to itself is defined nearly identically: Map each vertex of a smaller n -simplex to thevertex of the original simplex with the same label and then extend the map linearly inside each smallersimplex. Again, λ has degree one on the boundary: the rules of a Sperner labeling impose that for anyface σ of the simplex ∆ n the image λ ( σ ) is contained in σ , since subdivision vertices contained in σ areonly labeled by labels found at the vertices of σ . Thus λ | ∂ ∆ n is homotopic to the identity by induction onskeleta, which implies it has degree one. One can build the homotopy dimension by dimension using thefact that homotopies extend from lower-dimensional skeleta of a simplicial complex. There must exist apoint x ∈ ∆ n such that λ ( x ) = ( n +1 , n +1 , ..., n +1 ) is the barycenter. (As mentioned earlier, we will givean elementary proof of this fact in Section 5.) The smaller n -simplex containing x must be fully labeled.3. Rental harmony in the absence of full information for three roommates
Here we give two combinatorial proofs of the n = 3 case of Theorem 1, which mirror the two proofsof Sperner’s lemma in Section 2. The first proof is constructive and yields an algorithm to find the fairdivision of rent, since it reduces Theorem 1 to Sperner’s lemma for n = 3 . As presented in this section,the second proof only gives the existence of a fair division with a secret preference, but Section 5 explainshow this method also yields an algorithm. Proof 1:
Form a triangle in R whose vertices are given by (1 , , , (0 , , , and (0 , , . Because this liesin the plane x + y + z = 1 , we can interpret each point in the triangle as a division of rent. For example, ( , , ) indicates that rooms 1 and 2 cost one-quarter of the total rent and room 3 costs one-half of thetotal rent.Subdivide this triangle into many smaller triangles, i.e., triangulate the original triangle. If thistriangulation is fine enough, then the vertices of a single small triangle represent divisions of the rentthat only differ from one another by a penny or so. At each vertex we survey both Larry and Moe, askingwhich room they would prefer if the rent were split in this specific way. We record their preferences as atuple of two integers ( a, b ) with ≤ a, b ≤ , where a is the number of Larry’s preferred room and b is thenumber of Moe’s preferred room. There are three vertices of the triangulation – the original vertices ofthe triangle – where two rooms are free. For those rent divisions we kindly ask Larry and Moe to makedistinct choices, that is, the original vertices of the triangle all have labels ( a, b ) with a (cid:54) = b .At every vertex of the triangulation transform the label ( a, b ) into a single-digit label according to thefollowing rules: label the vertex if the previous label was (1 , , (1 , , or (2 , , label for (2 , , (2 , , or (3 , , and for (3 , , (3 , , or (1 , . The three original vertices of the triangle receive pairwise distinctlabels since Larry and Moe decided for different rooms whenever two rooms were free simultaneously.On every edge of the triangle the vertices of the triangulation have the same label — only one room isfree. And this label is one of the two labels we find at the endpoints of that edge. Therefore we havedefined a Sperner labeling. Thus we can find a small triangle that has vertices with all three labels. Thecorresponding rent divisions are all within a small margin of error. It is simple to check that seeing allthree labels for this rent division (with small error margin) implies that Larry and Moe each prefer atleast two distinct rooms, and each room is preferred by at least one of them. For example, it is impossiblethat Larry only likes room since one vertex is labeled ; and it is impossible that both Larry and Moedislike room since one vertex is labeled . This means that regardless of which room Curly chooses,both Larry and Moe will be left with one of their favorite rooms. (cid:3) The reduction in the proof above actually proves a generalization of Sperner’s lemma: given twoSperner labelings of a subdivided triangle that match up on the original vertices of the triangle, use the
FLORIAN FRICK, KELSEY HOUSTON-EDWARDS, AND FRÉDÉRIC MEUNIER rules above to transform them into one Sperner labeling; a fully-labeled triangle, which exists by Sperner’slemma, now exhibits all three labels across the two Sperner labelings and both Sperner labelings exhibitat least two labels. Higher-dimensional generalizations of Sperner’s lemma to multiple labelings wereconjectured by the third author and proven by Babson [2]. We will treat these and other extensions withnew and simple proofs in Section 6.
Proof 2:
This second proof of Theorem 1 for n = 3 will easily generalize to the case of n roommates andstill yield an algorithm to find the fair division of rent. It is based on the same piecewise linear map usedin the second proof of Sperner’s lemma.As in the first proof, construct a triangulated standard simplex in R , and at each vertex, survey Larryand Moe about their room preferences. Instead of recording this as a tuple, construct two piecewise linearmaps λ , λ : ∆ −→ ∆ which reflect the preferences of Larry and Moe, respectively. That is, for eachvertex v of the triangulation, λ maps v to the vertex of the original triangle with the same label asLarry’s preferred room at the price division given at vertex v . The map λ is defined in the same wayusing Moe’s preferences. Then λ i is defined within each smaller triangle as the linear extension of thevalues at its vertices.Let λ = ( λ + λ ) denote their average, which again is a piecewise linear map λ : ∆ −→ ∆ . The map λ maps vertices of the subdivision of ∆ either to one of the three original vertices of ∆ (if the vertexreceives the same label by both Larry and Moe) or to one of the three midpoints of edges (if the labelingsdo not agree on the vertex).Observe that for every subdivision vertex v the vector λ ( v ) counts how often each label is exhibitedin v , that is, if λ ( v ) = (1 , , then λ ( v ) = e and λ ( v ) = e or vice versa. We can suppose thaton each of the original vertices of the triangle Larry and Moe decide for the same room, and that theydecide for each room precisely once on one of the three original vertices. Then as before we check that λ has degree one on the boundary of ∆ and thus there is a point x ∈ ∆ with λ ( x ) = ( , , ) . The point x lies in some small triangle τ .We claim that (1) τ exhibits all three labels, and (2) both λ and λ exhibit at least two labels. Tosee claim (1), note that if τ exhibited only two of the labels, then one of the coordinates of λ ( x ) wouldbe zero. To see claim (2) assume for contradiction that either λ or λ only exhibits one of the labels.Assume that label is 1. Then the first coordinate of λ ( v ) will be greater than or equal to for eachvertex v of τ . And so the first coordinate of λ ( y ) for any y ∈ τ must be at least , which contradicts λ ( x ) = ( , , ) . After the secret roommate selects their room, two remain. Since τ exhibits all threelabels, each of the remaining labels will be exhibited by λ or λ . Moreover, since λ and λ each exhibitat least two distinct labels, it is impossible that one person prefers neither room. Therefore, the remainingtwo rooms can be assigned in an envy-free way. (cid:3) The general case
Generalizing the last section, here we give a proof of the main result, Theorem 1: One can alwaysfind a fair division of the rent for an n -bedroom apartment given the subjective preferences of only n − roommates. The proof is a generalization of the second proof given in Section 3. Proof:
For n roommates, we consider the standard ( n − -simplex in R n . Its vertices lie on e , ..., e n ,the standard basis of R n , and x + x + · · · + x n = 1 for any points ( x , . . . , x n ) in the simplex. Similarto the n = 3 case, each point in the simplex is a distribution of the rent and the fraction of the rentcorresponding to the i th room is given by x i . Triangulate the simplex finely enough so that rent divisionin the same subdivision simplex are within a one-cent error-margin. CHIEVING RENTAL HARMONY WITH A SECRETIVE ROOMMATE 7
For each of the n − given subjective preferences, we define a map λ j : ∆ n − −→ ∆ n − from thetriangulated ( n − -simplex to itself, defined as follows. For each vertex v of the triangulated simplex, λ j ( v ) maps to a vertex of the original simplex, recording the j th roommate’s preference. For example, if λ j ( v ) = e then person j prefers room 3 at the price distribution given at v . After λ j is specified on eachvertex of the triangulation, define λ j within each smaller simplex as the linear extension of its values onthe vertices of the smaller simplex.We use that no roommate strictly prefers a free room over another free room to argue that we canimpose that the choices of each roommate define a Sperner labeling of the triangulation of ∆ n − . Theeffect of this is, that up to a permutation of the vertices, each map λ j satisfies λ j ( σ ) ⊆ σ for each face σ of ∆ n − .The vertices e , e , . . . , e n of ∆ n − correspond to the rooms , , . . . , n , where room i is the only non-free room at e i . We ask each roommate to decide for room π ( i ) := i + 1 at e i for i < n and room π ( n ) := 1 at e n . This labeling of the vertices extends to a Sperner labeling such that the label of any vertex onthe boundary of ∆ n − corresponds to a free room. To see this notice that the rooms that are free forrent divisions in some face σ of ∆ n − are precisely those rooms that correspond to vertices of ∆ n − notcontained in σ . Thus for any proper face σ a vertex of σ must be labeled by a free room, since π has onlyone orbit, and hence such a free room is a valid label for vertices inside σ .We have shown that we can assume that the choices of each roommate define a Sperner labeling.Thus, up to renaming vertex e i in the domain of λ j as e π ( i ) , each map λ j satisfies λ j ( σ ) ⊆ σ for eachface of ∆ n − . More precisely, let f : ∆ n − −→ ∆ n − be the affine map that is defined on vertices by f ( e i ) = e i +1 for i < n and f ( e n ) = e . Then f − ( λ j ( σ )) ⊆ σ for every face σ of ∆ n − . We will fromhere on tacitly assume this renaming of the vertices so that each λ j satisfies λ j ( σ ) ⊆ σ for every face σ of ∆ n − .Let λ : ∆ n − −→ ∆ n − denote the average, λ := n − ( λ + λ + · · · + λ n − ) . Since each λ j fixes thefaces of ∆ n − setwise, so does their average λ . Thus, as before, λ is surjective. Let x ∈ ∆ n − such that λ ( x ) = ( n , . . . , n ) . That is, x is mapped to the barycenter. Let τ be the simplex containing x .Fix k ∈ { , . . . , n − } . We claim that any k -subset of the labelings λ i will exhibit at least k + 1 labelswithin τ . Assume for contradiction that Λ is some k -subset of the λ i that only exhibits the k labels e j , . . . , e j k . Then λ ( τ ) will be shifted toward the vertices e j i and thus will not contain the barycenter.More precisely, for any vertex v of τ and any i ∈ Λ , (cid:104) λ i ( v ) , e j + · · · + e j k (cid:105) = 1 and hence (cid:104) λ ( v ) , e j + · · · + e j k (cid:105) ≥ kn − . Since this holds for all vertices v of τ , it also holds for any point inside τ . But, (cid:104) λ ( x ) , e j + · · · + e j k (cid:105) = (cid:104) ( 1 n , . . . , n ) , e j + · · · + e j k (cid:105) = kn , which is a contradiction.This means any subset of k roommates prefers at least k + 1 rooms. This implies there is a fair divisionindependently of which room is picked by the secretive roommate given by the labels of τ . This is becauseno matter which room the secretive roommate picks, any subset of k (nonsecretive) roommates has k rooms to pick from; construct a bipartite graph with vertices corresponding to the n − roommates onthe one hand and vertices corresponding to the n − untaken rooms on the other. We add an edge fora pair of roommate and room if the roommate prefers this particular room. A fair rent division nowcorresponds to a perfect matching, which exists by Hall’s marriage theorem: in any bipartite graph with FLORIAN FRICK, KELSEY HOUSTON-EDWARDS, AND FRÉDÉRIC MEUNIER bipartite sets A and B there exists a matching that entirely covers A if and only if for every subset W ⊆ A its neighborhood N ( W ) satisfies | N ( W ) | ≥ | W | . (cid:3) Algorithmic aspects
Our proof in the previous section described the construction of a piecewise linear map λ : ∆ n − −→ ∆ n − that fixes the faces of ∆ n − setwise: λ ( σ ) ⊆ σ . Such a map is necessarily surjective. A fair divisionof rent corresponds to a simplex τ of the triangulation of ∆ n − such that there is an x ∈ τ that λ mapsto the barycenter of ∆ n − . Here we describe a simple algorithm how to find the face τ . This algorithmdoes not use the surjectivity of λ , but only that λ ( σ ) ⊆ σ for each face σ of ∆ n − . Thus our algorithmalso gives an elementary proof of the surjectivity of λ .It is instructive to first consider low-dimensional cases. The algorithm for n = 2 roommates justtraverses the interval ∆ until for some edge e of (a triangulation of) ∆ the image λ ( e ) contains thebarycenter, which must exist by the Intermediate Value Theorem.In general, our algorithm is a trap-door argument, which we first describe for the case of a triangle ∆ = conv { e , e , e } . We are given a triangulation T of ∆ and a map λ : ∆ −→ ∆ that interpolateslinearly on every face of T . The map λ fixes the vertices and edges of ∆ setwise. We will constructa path that starts in the vertex e of ∆ and ends in a triangle σ of T such that λ ( σ ) contains thebarycenter ( e + e + e ) of ∆ . We will now describe the rooms and doors for our trap-door argument,or equivalently the vertices and edges of a graph G such that following paths in this graph will lead toa triangle mapped to the barycenter by λ . To build this graph we assume that λ is generic in the sensethat no vertex of T gets mapped to the segment connecting the barycenter of [ e , e ] to the barycenterof ∆ . The vertices of G , or the rooms, are: • the vertex e of ∆ , • any edge e of T that subdivides the edge [ e , e ] of ∆ and such that λ ( e ) ∩ [ e , ( e + e )] (cid:54) = ∅ ;that is, the image of e under λ intersects the segment from vertex e to the barycenter of [ e , e ] , • any triangle σ of T such that λ ( σ ) intersects the segment [ ( e + e ) , ( e + e + e )] connectingthe barycenter of the edge [ e , e ] to the barycenter of ∆ .The edges of G , or the doors, are: • between e and the vertex corresponding to the edge of T that contains e and subdivides [ e , e ] , • between any two vertices corresponding to boundary edges of T that share a common vertex v with λ ( v ) ∈ [ e , ( e + e )] , • between any two vertices corresponding to triangles of T that share a common edge e with λ ( e ) ∩ [ ( e + e ) , ( e + e + e )] (cid:54) = ∅ , • between a boundary edge e of T with ( e + e ) ∈ λ ( e ) and the incident triangle σ of T , i.e., theunique triangle σ that contains e as an edge.We claim that a connected component of the graph G is a path from e to some triangle σ with ( e + e + e ) ∈ λ ( σ ) . Notice that e is incident to one other vertex in G (corresponding to theunique edge of T that has e as a vertex and subdivides [ e , e ] ). If e is a boundary edge of T with λ ( e ) ⊆ [ e , ( e + e )] then in G it is connected to two other boundary faces of T (the ones that lie to theleft and right of e on the edge [ e , e ] ). If e is a boundary edge of T with ( e + e ) ∈ λ ( e ) then preciselyone of its vertices gets mapped to [ e , ( e + e )] , so e is connected to one other edge of T in G . The otherneighbor of e in G is the unique triangle σ of T that contains e as an edge. Now since generically linesegments intersect (boundaries of) triangles in either one or two edges, the vertices of G correspondingto triangles have precisely two neighbors unless the segment of points in the triangle σ that λ maps to [ ( e + e ) , ( e + e + e )] intersect σ in only one edge. In that case ( e + e + e ) ∈ λ ( σ ) . Thus G is CHIEVING RENTAL HARMONY WITH A SECRETIVE ROOMMATE 9 a graph where all vertices have degree one or two, and have degree one if and only if they correspond toa triangle that gets mapped to the barycenter of ∆ or correspond to our starting point e . Starting inthe vertex e and following the edges of G we must end up in such a triangle, as desired.To summarize the algorithm, we start walking in the vertex e and traverse along the edge [ e , e ] until we hit an edge of T that is mapped to the barycenter of [ e , e ] . From there we walk inwards intothe triangle ∆ following a path of triangles whose image under λ intersects [ ( e + e ) , ( e + e + e )] .This either ends in a triangle of T that gets mapped to the barycenter ( e + e + e ) , or we return toedges subdividing [ e , e ] . However, in the latter case we must leave the edge [ e , e ] and follow a path oftriangles again. After finitely many trips back to the edge [ e , e ] we must end up in a triangle mappedto the barycenter.This construction and algorithm easily generalize to higher dimensions. We work with the faces e , conv { e , e } , conv { e , e , e } , . . . , conv { e , . . . , e n } = ∆ n − and their barycenters b k = (cid:80) ki =1 1 k e i . All faces σ of the triangulation T of ∆ n − that subdivide oneof these faces, say conv { e , . . . , e k } , and such that λ ( σ ) intersects the segment [ b k − , b k ] that joins thebarycenter of conv { e , . . . , e k − } to that of conv { e , . . . , e k } make up the vertices of G . The vertex e of ∆ n is a vertex of G as well. Two such faces σ and σ of dimension k are connected by an edge in G if they share a common ( k − -face τ such that λ ( τ ) intersects [ b k , b k +1 ] . We assume that if λ ( τ ) intersects [ b k , b k +1 ] , then there is a point x in the relative interior of τ such that λ ( x ) ∈ [ b k , b k +1 ] . Thiscan be achieved by slightly perturbing the barycenters b k . Moreover, there is an edge between k -face σ and ( k − -face τ if τ is a face of σ in T and b k ∈ λ ( τ ) .A line segment generically cannot intersect a k -face in more than two of its ( k − -faces and it intersectsin precisely one ( k − -face if it ends inside the k -face. Thus our reasoning for ∆ also applies to thishigher-dimensional construction and starting in the vertex e of G we can follow edges of G to end up inan ( n − -face σ of T with n (cid:80) ni =1 e i ∈ λ ( σ ) .6. Generalizations of Sperner’s Lemma
The methods of Section 4 actually yield generalizations of Sperner’s lemma to multiple labelings. Fix atriangulation of the n -simplex and several Sperner labelings λ , . . . , λ m of it. We will always assume thatthese labelings match up on the original n + 1 vertices of ∆ n . By Sperner’s lemma each of these labelingshas a fully-labeled simplex. It is simple to come up with examples where no pair of these respectivefully-labeled simplices coincide. In attempting to understand how many labels a single simplex mustexhibit across the m Sperner labelings, there are two natural questions:1. How can we constrain ( n + 1) -tuples ( k , . . . , k n ) of nonnegative integers such that there is a simplex τ that exhibits the i th label k i times across the m Sperner labelings λ , . . . , λ m ?2. Dually, how can we constrain m -tuples ( k , . . . , k m ) of nonnegative integers such that there is a simplex τ on which λ i exhibits k i pairwise distinct labels?We will relate the first question to convex hulls of lattice points , i.e., points with integer coordinates,in m · ∆ n = { x ∈ R n +1 | (cid:80) x i = m, x i ≥ } , the n -simplex scaled by m . We will show that the labelmultiplicities ( k , . . . , k n ) that must occur are given by sets of n + 1 lattice points in m · ∆ n whose convexhulls all intersect in a common point y that are maximal with this property, that is, any other convexhull of n + 1 lattice points will not contain y .For example, given two Sperner labelings λ and λ of a triangulation of the triangle ∆ , we need tounderstand intersections of convex hulls of three lattice points in · ∆ . The relevant lattice points arethe vertices (2 , , , (0 , , , (0 , , and midpoints of edges (0 , , , (1 , , , (1 , , . Say we consider the convex hulls of lattice points that capture y = (2 − ε, ε, ε ) for some small ε > . The point y isclose to the vertex (2 , , and even closer to the edge between (2 , , and (0 , , without being on it.All possible choices of three lattice points whose convex hulls contain y have the lattice point (2 , , ,another lattice point on the edge between (2 , , and (0 , , , i.e., one of (0 , , or (1 , , , and onelattice point that is not on that edge, i.e., one of (1 , , , (0 , , , or (0 , , . So there are exactlysix choices of three lattice points whose convex hulls capture y . This geometric fact translates into thefollowing combinatorial fact about Sperner labelings λ and λ of a triangle: there always is a smallertriangle with vertices v , v , v such that • v exhibits the first label twice ( λ ( v ) = λ ( v ) = 1 ) corresponding to the lattice point (2 , , , • v either exhibits the second label twice or the first and second label once corresponding to (0 , , or (1 , , , • and v must exhibit the third label at least once corresponding to (1 , , , (0 , , , or (0 , , .A priori, the simplex that is labeled by λ , . . . , λ m and the scaled simplex that encodes all possible ( n + 1) -tuples ( k , . . . , k n ) of label multiplicities are entirely different objects. While we should expect nodirect relation between these two simplices, it turns out that we get constraints on the k i by thinking ofthe simplices as the same geometric object. Theorem 2.
Let λ , . . . , λ m be Sperner labelings of a triangulation T of ∆ n . Let y ∈ m · ∆ n be somepoint that is not in the convex hull of any n lattice points in m · ∆ n . Then there is a facet σ of T andan ordering of its vertices v , . . . , v n +1 such that the point y is contained in conv { y , . . . , y n +1 } , where y i ∈ m · ∆ n denotes the lattice point whose j th coordinate is the number of times j labels v i .Proof: Let λ = λ + · · · + λ m : ∆ n −→ m · ∆ n . As before since the λ i are Sperner labelings the map λ | ∂ ∆ n has degree one as a map to the boundary of m · ∆ n ; the map λ satisfies λ ( σ ) ⊆ m · σ for any face σ of ∆ n . Just as before the average m λ fixes faces setwise. Thus there is an x ∈ ∆ n with λ ( x ) = y . Let τ be a face of the triangulation of ∆ n that contains x . The map λ maps vertices of the triangulation of ∆ n to lattice points of m · ∆ n . Since y is not in the convex hull of fewer than n + 1 lattice points in m · ∆ n ,the vertices of τ must be mapped precisely to the elements of a set of n + 1 lattice points in m · ∆ n whoseconvex hull captures y . (cid:3) Question 2 can be approached in much the same way. Instead of defining the map λ as the sum oraverage of the piecewise linear extensions of the Sperner labelings as before, we now take a biased averagewith weights according to how many labels each Sperner labeling is supposed to exhibit. The third authorconjectured in his dissertation that (cid:80) k i = n + m is a valid constraint for question 2. This was recentlyproven by Babson [2]. We give a different proof below in the spirit of the other proofs of this manuscript. Theorem 3.
Let λ , . . . , λ m be m Sperner labelings of a triangulation of ∆ n and let k , . . . , k m be m positive integers summing up to n + m . Then there exists a simplex τ such that λ j exhibits at least k j pairwise distinct labels on τ for all j .Proof: Let α j = n +1 ( k j + m − for ≤ j ≤ m . Then since (cid:80) j k j = n + m we have that (cid:80) j α j = 1 .Thus λ = (cid:80) j α j λ j is a map ∆ n −→ ∆ n , and λ satisfies λ ( σ ) ⊆ σ for each face σ of ∆ n as usual.Let x ∈ ∆ n with λ ( x ) = ( n +1 , . . . , n +1 ) and let τ be a facet of the triangulation of ∆ n containing x .Denote the vertices of τ by v , . . . , v n and let x = (cid:80) i µ i v i for nonnegative µ i with (cid:80) i µ i = 1 . Define for i = 1 , . . . , n + 1 and j = 1 , . . . , m β ij = α j · (cid:88) { k | λ j ( v k )= e i } µ k . CHIEVING RENTAL HARMONY WITH A SECRETIVE ROOMMATE 11
Since (cid:80) i µ i = 1 we have that (cid:80) i β ij = α j for every j . The choice of x , definition of λ , and piecewiselinearity of the λ j imply that (cid:18) n + 1 , . . . , n + 1 (cid:19) = λ ( x ) = (cid:88) j α j λ j ( x ) = (cid:88) j α j n (cid:88) k =0 µ k λ j ( v k ) and thus (cid:80) j β ij = n +1 . Since in particular ≤ β ij ≤ n +1 and we already know that (cid:80) i β ij = α j , foreach j the number of indices i such that β ij > is at least α j ( n + 1) > k j − . Now β ij > implies thatthere is a vertex v of τ with λ j ( v ) = e i , and thus τ receives at least k j distinct labels by λ j . (cid:3) Acknowledgments
Figure 1 and Figure 2 are reproduced with permission from [4] and copyright 2017, PBS Infinite Seriesand Ray Lux.
References
1. Megumi Asada, Florian Frick, Vivek Pisharody, Maxwell Polevy, David Stoner, Ling Hei Tsang, and Zoe Wellner,
Fairdivision and generalizations of Sperner- and KKM-type results , arXiv preprint arXiv:1701.04955 (2017).2. Eric Babson,
Meunier Conjecture , arXiv preprint arXiv:1209.0102 (2012).3. Daniel I. A. Cohen,
On the Sperner lemma , J. Combin. Theory (1967), no. 4, 585–587.4. Kelsey Houston-Edwards, Splitting Rent with Triangles , 2017, Episode of
PBS Infinite Series . Video: .5. Harold W. Kuhn,
Simplicial approximation of fixed points , Proc. Natl. Acad. Sci. (1968), no. 4, 1238–1242.6. Emanuel Sperner, Neuer Beweis für die Invarianz der Dimensionszahl und des Gebietes , Abh. Math. Seminar Univ.Hamburg, vol. 6, Springer, 1928, pp. 265–272.7. Francis E. Su,
Rental harmony: Sperner’s lemma in fair division , Amer. Math. Monthly (1999), no. 10, 930–942.(FF, KHE)
Department of Mathematics, Cornell University, Ithaca, NY 14853, USA
E-mail address : {ff238,kah282}@cornell.edu (FM) Université Paris Est, CERMICS, 77455 Marne-la-Vallée CEDEX, France
E-mail address ::