Extending EFX Allocations to Further Multi-Graph Classes
Abstract
The existence of EFX allocations is one of the most significant open questions in fair division. Recent work by Christodoulou, Fiat, Koutsoupias, and Sgouritsa (“Fair allocation in graphs,” EC 2023) establishes the existence of EFX allocations for graphical valuations, when agents are vertices in a graph, items are edges, and each item has zero value for all agents other than those at its endpoints. Thus, in this setting, each good has non-zero value for at most two agents, and there is at most one good valued by any pair of agents. This marks one of the few cases when an exact and complete EFX allocation is known to exist for more than three agents.
In this work, we partially extend these results to multi-graphs, when each pair of vertices can have more than one edge between them. The existence of EFX allocations in multi-graphs is a natural open question given their existence in simple graphs. We show that EFX allocations exist, and can be computed in polynomial time, for agents with cancelable valuations in the following cases: (i) bipartite multi-graphs, (ii) multi-trees with monotone valuations, and (iii) multi-graphs with girth , where is the chromatic number of the multi-graph. The existence of EFX in cycle multi-graphs follows from (i), (iii), and the known existence of EFX for three agents.
Keywords and phrases:
Fair Division, EFX, Multi-graphsCopyright and License:
2012 ACM Subject Classification:
Theory of computation Algorithmic game theoryFunding:
Both authors acknowledge the support of the Department of Atomic Energy, Government of India, under project no. RTI4001.Editors:
C. Aiswarya, Ruta Mehta, and Subhajit RoySeries and Publisher:
Leibniz International Proceedings in Informatics, Schloss Dagstuhl – Leibniz-Zentrum für Informatik
1 Introduction
In discrete fair division, we are given a set of indivisible items and a set of agents, with the goal of allocating the items fairly among the agents. Various formalisations of fairness have been studied in the literature, of which the most popular is envy-freeness, which requires that no agent prefers another’s allocation to its own. However, envy-free allocations may not always exist, leading to extensive work on relaxations of envy-freeness. In this paper, we study the existence and computation of allocations that are envy-free up to any item, or EFX, in instances that can be represented as multi-graphs. The question of whether EFX allocations exist in general instances is possibly the most tantalizing open question in the fair allocation of resources, and many recent papers have focused on this question, leading to proofs of its existence in ever-larger classes of instances.
In our work, a fair division instance consists of a set of items that is to be allocated to a set of agents, with each item allocated to a single agent. Each agent has a monotone nondecreasing value function over the set of items. Given an allocation, agent envies agent if agent ’s value for its own allocation is less than its value for the set of items allocated to . An envy-free allocation is one where no agent envies another. Envy-free allocations do not exist in general, and the closest relaxation then studied is envy-freeness up to any good, or EFX. An EFX allocation allows an agent to envy another, but the removal of any item from the envied agent’s allocation should eliminate the envy as well.
EFX allocations were first studied by Caragiannis et al. [14], who established its relation to other notions of fairness, but left the question of existence open. If all agents have the same valuation, then an EFX allocation exists [25]. This can be extended to show the existence for two agents with distinct valuations. This was extended to multiple agents with two distinct additive valuations [24], then to three agents when at least one has an MMS-feasible valuation [3, 16], and recently to multiple agents with three distinct additive valuations [21]. A number of papers also study partial EFX allocations, when some items may remain unallocated, e.g., [17, 10], as well as approximately EFX allocations [5, 6, 15, 25].
Another widely studied relaxation of envy-freeness, weaker than EFX, is envy-freeness up to one item (EF1), which requires that any envy be resolved by removing some item from the envied agent’s bundle [13, 23]. Unlike EFX, EF1 allocations are known to exist for both monotone and for some non-monotone valuations [23, 7, 12, 9, 11]. Other fairness notions have also been extensively studied in the literature [4].
Given the difficulties in establishing the existence of EFX allocations, researchers study special cases of structured valuations. EFX allocations are known to exist if agents have lexicographic preferences [20]. Another such class of structured valuations is graphical valuations, when the instance can be represented as a graph [18]. Here the agents correspond to the vertices and each item is an edge. The restriction is that each item has nonzero marginal value for only the agents at its end-points, and has zero value for all other agents. Thus if , then for every other agent , and any subset of items, . Further, the graph is simple, and thus there is at most one good valued by every pair of agents. For this case, when agents have monotone valuations over incident edges, Christodoulou et al. [18] show that EFX allocations exist, marking one of the rare cases where EFX allocations are shown to exist for more than three agents.
As mentioned in [18], graphical valuations model real-world scenarios with geographical constraints, where an item is only valued by nearby agents. For example, fishing rights in waters that are adjacent to two nations are such an item, valuable to the two nations but useless to others. Afshinmehr et al. [1] give the example of football matches in various formats; each match must be played at the home of one of the two teams in the match, and hence can be thought of as an item to be allocated to one of the two teams playing.
A natural question with graphical valuations is if an EFX orientation exists – an EFX allocation where goods are only allocated to agents at their endpoints. Christodoulou et al. [18] show that an EFX orientation may not exist, and it is NP-hard to determine if a given instance has an EFX orientation. Zeng and Mehta [28] characterize graphs that permit EFX orientations. Zhou et al. [29] study variants of EFX in the graphical setting, where an edge may be valued positively or negatively by its endpoints.
Given the existence of EFX in simple graphs, a natural question is if these results extend to multi-graphs when there are multiple edges between vertices (thus, pairs of agents can value multiple goods). Here a -approximate EFX allocation exists [5], beating the earlier bound of for general additive valuations [6]. EFX allocations also exist in multi-graphs with symmetrical additive valuations, i.e., when agents have additive valuations, and both agents at the endpoints of a good have the same value for the good [22]. In independent and concurrent work, Afshinmehr et al. [1, 2] show that exact EFX allocations exist in bipartite multi-graphs and cycle multi-graphs. They also characterize when EFX orientations exist in bipartite graphs, and show it is NP-hard to decide if a given instance admits an EFX orientation. Sgouritsa and Sotiriou [26] also independently present a number of results for multi-graphs when agents have general monotone valuations over their incident goods. We discuss these papers in Section 1.2.
1.1 Our Contribution
We study EFX allocations in multi-graphs and show that EFX allocations exist in large and important subclasses of multi-graphs. Our results are algorithmic, and for agents with cancelable valuations (defined in the next section), we show that an EFX allocation in these subclasses can be obtained in polynomial time. A feature of our algorithms is that they are iterative and fairly simple to state, based on the well-known cut-and-choose paradigm. While the algorithms are natural, the analysis is complicated and somewhat subtle.
-
1.
Firstly, we show that for bipartite multi-graphs with cancelable valuations, an EFX allocation can be obtained in polynomial time.
-
2.
Secondly, we show that for multi-trees, EFX allocations exist for general monotone valuations. Since multi-trees are also bipartite multi-graphs, for multi-trees with cancelable valuations, a polynomial time algorithm exists from the previous result.
-
3.
Lastly, we generalize the result for bipartite multi-graphs, and show that for multi-graphs with girth at least , where is the chromatic number of the graph, EFX allocations can be obtained in polynomial time for agents with cancelable valuations, given the colour classes.111Note that approximating the chromatic number of a graph within for all is NP-hard [30]. Our algorithm here is a natural extension of the algorithm for bipartite multi-graphs, though the analysis is significantly more complicated.
Note that the chromatic number of a cycle multi-graph is for even cycles and for odd cycles. The existence of EFX allocations in even-length cycle multi-graphs follows from Result 1. The existence of EFX allocations in cycle multi-graphs of length 3 follows from prior work [16]. Result 1 implies the existence of EFX allocations in odd-length cycle multi-graphs of lengths at least 5. Hence, our work also shows the existence of EFX allocations for cycle multi-graphs of all lengths.
1.2 Comparison with Recent Work
Here, we compare our results with recent work on EFX existence in multi-graphs.
As noted earlier, Afshinmehr et al. [1, 2] independently and concurrently show the existence of EFX allocations in bipartite and cycle multi-graphs under additive valuations. Similar to earlier work [18], their algorithm for bipartite multi-graphs operates in two phases: the first constructs an EFX orientation with certain properties, and the second transforms it into a complete allocation using “safe” vertices. While their approach and analysis are somewhat involved, we present a simpler and efficient algorithm with a more straightforward analysis for cancelable valuations, which generalize additive valuations. Moreover, our algorithm for bipartite multi-graphs naturally extends to show the existence of EFX allocations in multi-graphs with chromatic number and girth at least . This extension avoids the challenging case – highlighted in Section 5 of [1, 2] – where there are two envied vertices with the multi-edge between them unallocated, which poses a key difficulty in extending results to general multi-graphs. In contrast, the algorithm, as stated in Afshinmehr et al., does not immediately lend itself to any such natural extension beyond bipartite multi-graphs. Afshinmehr et al. also establish the existence of EFX in cycle multigraphs, which follows as a special case of our result for -chromatic multigraphs with girth at least .
Another independent and concurrent work by Sgouritsa and Sotiriou [26] proves EFX existence for certain multi-graph classes under monotone valuations. They show that EFX allocations exist for: (i) multi-graphs where each vertex has at most neighbours (where is the total number of vertices), and (ii) multi-graphs with girth at least . In follow-up work on Arxiv, they extend these results to bipartite multi-graphs [27]. These results extend many of ours to monotone valuations, except for multi-graphs with chromatic number and girth (i.e., ), where we establish EFX existence. As in other papers [18, 1, 2], the algorithms by Sgouritsa and Sotiriou follow a multi-phase approach: the first phase computes an initial allocation (orientation) satisfying certain properties, the second aims to reduce the number of envied vertices, and the third assigns the remaining unallocated bundles to non-envied (safe) vertices who value them at zero. Our algorithms offer a different and simpler approach to those proposed in these other papers.
2 Notation and Preliminaries
Multi-graphs.
A multi-graph is a graph with possibly multiple edges between each pair of vertices. A function maps each edge to an unordered pair of vertices, which are its endpoints. We use for the set of parallel edges between the vertices , (hence ). We define and . For a vertex , is the set of neighbours
Given a multi-graph, we can define paths and cycles as in simple graphs without parallel edges. A path is a sequence of distinct vertices so that each pair of consecutive vertices are neighbours. The length of a path is one less than the number of vertices. A cycle is a sequence of distinct vertices so that each pair of consecutive vertices are neighbours, and , are neighbours.
Fair division.
In a fair division instance on a multi-graph , each vertex corresponds to an agent , and each edge corresponds to a good. Throughout this article, we will use the terms agents and vertices interchangeably, and the terms items and edges interchangeably. Each agent has a valuation function that maps subsets of goods to an integer value. A bundle is simply a subset of goods.
The multi-graph restricts the valuation functions in the following way: Each agent has zero marginal value for any good not adjacent to it. Thus for a good if , then for any subset of goods , . We say that agent values a good if , i.e., the edge is adjacent to , else it does not value the good . Extending this definition, we say that agent values a bundle if contains some good that values, else does not value the bundle .
An allocation is a partition of the set of items (or edges) among the agents. A partial allocation is a partition of a subset of items. An allocation is envy-free (EF) if for all agents , , . An allocation is envy-free up to any good (EFX) if for all agents , , either , or , for all . An orientation is an allocation where each item is allocated to an agent that values it (thus if , then ).
Given an allocation (or a partial allocation), we can define an envy graph among the agents: The vertices of are the agents, and there is a directed edge if agent envies agent , i.e., . Note that an envy graph is simple.
Given an allocation and the corresponding envy graph , let be a directed cycle in the envy graph. We use the term resolving the envy cycle to refer to the modification that exchanges bundles along the cycle: agent gets the bundle currently allocated to , and each agent gets the bundle currently allocated to , for . The modified allocation is called .
Valuation functions.
We will be interested in a few different classes of valuation functions. A valuation function is monotone nondecreasing (or just monotone) if for every and , .
A monotone valuation function is cancelable if for , and , if then . These functions were defined in prior work on EFX allocations for four agents [10]. We only consider monotone cancelable functions, and hence say cancelable to mean monotone cancelable functions. We in particular need the following two properties of cancelable functions.
Proposition 1.
Let be a cancelable function. Then given disjoint sets of goods , , , , and , , it follows that .
Proof.
By cancelability, and , and hence .
For two agents with monotone valuations, EFX allocations are known to exist [25]. However, even for two agents with submodular valuations, computing an EFX allocation is PLS-complete and requires an exponential number of queries [19]. For cancelable valuations, we can compute an EFX allocation in polynomial time [8, 19].222The actual result holds for the more general class of weakly well-layered valuations.
Proposition 2.
For two identical agents with cancelable valuations, an EFX allocation can be computed in polynomial time.
Cut and Choose.
Cut and choose is a standard procedure to obtain an EFX allocation for two agents. Given agents and and a set of items , one agent “cuts” and the other agent “chooses”. If agent “cuts”, we assume there are two identical agents with valuation function , and use the algorithm from Proposition 2 to obtain a partition of that is EFX for identical agents with valuation function . Then agent “chooses” the bundle with a higher value for it, while agent is assigned the other bundle. The allocation is clearly envy-free for agent , and is EFX for agent . Note that for two agents with cancelable valuations, this gives us a polynomial time algorithm to obtain an EFX allocation.
The cut-and-choose procedure is central to our algorithm. In Section 3, when we deal with bipartite multi-graphs , for allocating a set of parallel edges with , , agent will always cut (thus the agent in the right bipartition always cuts). In Section 5, when we deal with -chromatic multi-graphs, and partition the vertex set where vertices in are the same colour, we think of the vertices in as being to the left of . For a set of parallel edges , as in the bipartite case, the vertex to the right will always cut.
We use to refer to the use the algorithm from Proposition 2 to obtain a partition of that is EFX for identical agents with valuation function .
3 EFX Allocations in Bipartite Multi-graphs
We first show that for bipartite multi-graphs and agents with cancelable valuations, an EFX allocation always exists, and can be computed in polynomial time. Let be the bipartition of the vertex set. We first define some notation we will use for this case.
Structures
For each vertex , we define the structure as the subgraph induced by . Note that since , all edges in are between and vertices in . We say is the root of the structure. Each iteration in our algorithm will resolve the structure , for some , by which we mean that it will assign all the edges in . We also say that a vertex is resolved to mean that is resolved.
For allocating a set of parallel edges with , , agent will always cut (thus the agent in the right bipartition always cuts). For each , let be the partition of returned when agent cuts. Define as the bundle preferred by , and as the bundle preferred by . Ties are broken arbitrarily to ensure that and are single bundles. If either agent is indifferent between the two bundles, we break ties so that . Further, we define and .
For a structure , we further define a favourite neighbour as follows:
We break ties arbitrarily to ensure that is a single vertex. Then is a neighbour who offers the highest-value bundle after cutting the adjacent edges. The other neighbours are simply called ordinary neighbours.
Our algorithm is then simple to describe, and is formally given as Algorithm 1. We consider the vertices in in turn. For vertex , we consider the structure and resolve it, as follows. Each neighbour cuts the bundle . Each ordinary neighbour gets their higher-value bundle . Let be the union of the left-over goods from the ordinary neighbours of . For the favourite neighbour , if – both and prefer different bundles when cuts – then gets , and gets . In this case, everyone gets their largest value bundle, and there is no envy. On the other hand, if – both and prefer the same bundle when cuts – then is offered the choice of or . If prefers the former, then it gets , and gets . In this case, may envy , but this is EFX. If prefers , it gets this bundle, and gets . Again, in this case, there is no envy.
We note the following properties of the algorithm. In each iteration of the outer for loop, a vertex is chosen, and the goods in are assigned to the agents in . The allocation to all other agents remains unchanged. Since does not contain any agent from other than , an agent has no allocated goods until it is resolved, and the allocation to agent is not changed after it is resolved. Finally, no goods are removed from an agent once assigned.
Theorem 3.
Algorithm 1 runs in polynomial time, and returns an EFX allocation.
The polynomial running time is easily seen, and is because the procedure runs in polynomial time for cancelable valuations. The EFX property follows immediately from the next lemma.
Lemma 4.
Fix an agent . Let be the partial allocation after resolving . Then the only possible envy in allocation is that a resolved agent is envied by its favourite agent . Further, the partial allocation is EFX.
Proof.
The proof is by induction on the iterations of the outer for loop. The claim is clearly true for the initial empty allocation. Let be the allocation before is resolved. Then by the induction hypothesis, since is not yet resolved in , no agents in are envied.
In the current iteration, when is resolved, only agents in get goods, while agents not in retain their allocation. Hence, any new envy edges must be to agents in . Further since only agents in value the goods in , any new envy edges must also be from agents in .
Suppose agent envies agent in (as established, agents in are not envied in ). Thus . In the current iteration, the allocation to does not change, while does not lose any goods. Hence any such envy remains EFX. Thus to prove the lemma, we only need to show that any new envy between agents in is from to , and is EFX.
Consider first an ordinary neighbour . Of the goods allocated in this iteration, only values . Of this set, it gets its preferred partition . Then since did not envy anyone in earlier, for any agent , . Then
The first inequality is because valuations are cancelable, and prefers to , and to . The second inequality is because in this iteration, any agent other than either does not receive any good valued by , or receives the bundle .
Consider , agent ’s favourite neighbour. Again of the goods allocated in this iteration, is only interested in . Of this set, it either gets its preferred bundle (in either Line 14 or in Line 17), or it gets the bundle in Line 11. In the former case, since the valuations are cancelable and agent did not earlier envy any agent in , agent does not envy any agent in in allocation either. In the latter case, gets the bundle , and hence may envy since agent gets . But note that then , and this is an EFX partition since cuts the set of items . Thus, the only possible new envy edge from is to , and this is EFX.
Lastly, consider agent . We show that will not envy any agent in . If gets (Line 17), then by definition is preferred over all the other bundles allocated to the other agents in , and additionally gets all the left-over bundles. Otherwise, gets a choice between and ; the other bundle is given to . Then clearly does not envy . The bundle has a higher value for than for any neighbour . Since chooses the bundle with the highest value, does not envy any ordinary neighbour either.
4 EFX Allocations for Monotone Valuations in Tree Multi-graphs
We now show that for monotone valuations, an EFX allocation exists in tree multi-graphs. Note that even for two agents with monotone submodular valuations, computing an EFX allocation is PLS-complete [19], and hence a polynomial-time algorithm for tree multi-graphs may not exist. Since trees are also bipartite, a polynomial time algorithm for cancelable valuations follows from the previous section.
Our algorithm for trees is recursive and again utilizes the procedure , that obtains an EFX partition of for two identical agents with valuation . Let be the tree multi-graph, and be a leaf with parent . Then are the edges between and , and no other agent values these goods. Inductively, let be an EFX allocation of goods to agents . Our objective is then to extend to an EFX allocation of goods .
For this, let be the partition obtained when cuts . We define and . Note that ties are broken arbitrarily, ensuring that is a single bundle. Agent gets its preferred bundle . If is not envied in allocation , then it gets , and the algorithm terminates. If is already envied in allocation , it cannot get additional goods. Let be a source in the envy graph for with a path to . We add to agent ’s allocation. If does not envy now, the algorithm terminates. Else, there is an envy cycle where envies . Resolving the envy cycle gives us an EFX allocation.
Theorem 5.
Given a tree multi-graph, with monotone valuations, Algorithm Tree-EFX returns an EFX allocation .
Proof.
By induction, assume that is an EFX allocation on . Note that given an EFX allocation, resolving envy cycles does not destroy this property, since each agent’s value does not decrease, and bundles are merely reassigned. Hence, the partial allocation in Line 11 is EFX for agents , and .
On allocating to agent , clearly does not envy any agent. The only agent that could envy is agent , since only agents and value the bundle . Further, since gets to cut the bundle , as long as possesses a bundle with value at least , its envy towards will be EFX.
If is a source in the envy graph, then the condition in Line 14 is satisfied and gets the bundle . No agent will envy , since agents do not value the bundle , and already gets its preferred bundle. Thus the allocation is EFX.
If is not a source, then since the envy-graph is acyclic (and does not envy any agent), there is a source with a path to in the envy graph. Let be this path. We assign to agent . The only agent that could now envy is . If envies , there is an envy cycle, which we resolve. We claim that the resulting allocation is EFX. This is because agent continues to not envy any agent. Agent ’s value for its bundle is now at least , and hence any envy towards agent is EFX. Further agent ’s value for its bundle is also at least its earlier value; since resolving the cycle only shifted bundles around, any envy towards agents is also EFX by induction. Finally, for other agents, their value does not decrease, and any envy is unaffected by the shifting of bundles. Hence the allocation is EFX.
5 EFX Allocations in -Chromatic Multi-graphs with Girth at Least
We now extend our algorithm and analysis to show that EFX allocations exist for agents with cancelable valuations in multi-graphs with girth at least , where is the chromatic number of the graph. A multi-graph is -chromatic if is the minimum number of colours required to colour the vertices, so no edge has both end-points the same colour. The girth of a multi-graph is the length of the shortest simple cycle, without using parallel edges. For example, bipartite multi-graphs are 2-chromatic with girth at least 4, while the Petersen graph (with parallel edges) is 3-chromatic with girth 5. We will assume that we are given a -colouring of the multi-graph.
Our algorithm proceeds as follows. Given a multi-graph coloured with colours, let be the colour of each vertex . Let be the independent set consisting of all vertices coloured . We think of the vertices as ordered from left to right by their colour, and say a vertex is to the left of if . For , define , and .
We proceed in phases. In phase , we consider the bipartite multi-graph where an edge is in if , i.e., each edge is from a vertex in to a vertex in . Note that each vertex appears in a left bipartition in exactly one phase, when we consider all edges to vertices to the right of . Thus each edge appears in exactly one phase.
Now in phase , we run Algorithm 1 on the bipartite multi-graph . The only change we make is that in Algorithm 1, initially all vertices – in particular, vertices – had empty allocations. This may not be the case in the current algorithm in the second phase onwards. Hence, we remove Line 1 from Algorithm 1, which initializes the allocation, within the outer for loop. Further in Lines 16 and 19 in Algorithm 3, agent retains its earlier allocation. In Line 13, when is possibly envied by , its earlier allocation passes to . Algorithm 3 formally states the algorithm.
A phase is an iteration of the outer loop. In phase , the vertices in are picked sequentially in the inner for loop.
Structures
For our analysis, we slightly need to redefine structures. For a vertex , we now define the structure as the subgraph induced by . That is, a structure now consists of the subgraph induced by and all neighbours to its right. As before, is the root of the structure. Now each iteration in phase of our algorithm resolves the structure for some , and thus assigns all the edges in . A vertex is resolved if is resolved.
Fix agent . As before, for each , is the partition of returned when agent cuts. Define as the bundle preferred by , and as the bundle preferred by . As earlier, ties are broken arbitrarily to ensure that and are single bundles. If either agent is indifferent between the two bundles, we break ties so that . Further, for , we define . Then ’s favourite neighbour , with ties broken arbitrarily.
This is the right neighbour that offers the highest-value bundle after cutting the adjacent edges. The other right neighbours are simply called ordinary neighbours.
We note the following properties of our algorithm. In phase of the algorithm, the inner for loop picks an agent , and resolves . The goods in are assigned to the agents in . The allocation to other agents is unchanged. Hence in particular, the allocation to a resolved agent does not change subsequently. Furthermore, while is resolved, agents in do not lose any goods, except possibly itself. Agent may however give its entire set of goods to its favourite neighbour (Line 13). However, in this case, agent gets , which it prefers over its earlier allocation and all the left-over goods. Thus, the value of each agent for its bundle is nondecreasing.
Much of our effort is towards proving Lemma 11, which shows that any envy is within a structure, and is from a favourite neighbour to the root . In particular, we want to avoid the following. Say for some interim allocation , agent values goods in both and . Then if is transferred to in Line 13, agent should not envy .
We show this by contradiction: that if agent does indeed start envying , then it must have short paths to both and , and hence, taken with any edge between and , there is a cycle of length less than . We do this in two steps: Claim 7 shows that if envies the bundle of goods , then these must contain a good from . Claims 9 and 10 then establish bounds on the distance from for and .
Next, we state the claims required to prove Lemma 11. Our first claim is that an unresolved agent does not envy the entire set of goods held by all other unresolved vertices.
Claim 6.
Let and be the allocation and the set of resolved vertices at Line 5. Let be an unresolved agent. Then
Proof.
Fix an unresolved agent . We prove this by induction on the iterations of the inner loop. Initially, the allocation is empty, and the claim is trivially satisfied. Assume this is true prior to the loop when a vertex is resolved. Let , be the allocation and set of resolved vertices before is resolved, and , be the allocation and set of resolved vertices after. Note that the allocation to resolved agents does not change subsequently. Hence if now envies the set of goods held by unresolved agents, new goods that values must be allocated to unresolved agents.
In this loop, only the allocation to agents in is modified. Further, none of the agents in are resolved prior to the loop, i.e., . Hence by the induction hypothesis. Now if , then no additional goods it values are allocated in this loop. Hence the claim holds by induction. Thus, since (since is unresolved after the loop), it must be that is a neighbour to the right of . Of the goods allocated in this loop, only values .
If is an ordinary neighbour, then additionally gets its preferred bundle , while the bundle is allocated to other agents. Hence, since valuations are cancelable, the claim remains true. On the other hand, if , then could lose its preferred bundle to agent . But in this case agent is resolved, hence . Since additionally gets and does not lose any goods, the claim remains true after the loop also.
The next claim builds on the previous claim to show that if any vertex envies the union of goods held by some set of unresolved agents, then must be resolved, and some agent in must hold a good from .
Claim 7.
Let and be the allocation and the set of resolved vertices at Line 5. Let be any agent, and is a subset of unresolved vertices so that
Then is resolved, and for some agent , contains some good .
Proof.
By Claim 6, if is unresolved, then . Hence must in fact be resolved. We prove the contrapositive by induction, that if a subset of unresolved agents do not hold any good from , then . Clearly, the claim holds in any iteration before is resolved.
Consider the iteration where is resolved. Every agent in receives goods from , hence, we only consider subsets consisting of agents not in . But the allocation for these agents does not change, and ’s value for its bundle is nondecreasing, hence the claim holds in this iteration.
Consider any later iteration, where an agent is resolved. Then , since is resolved. Let be the allocation before resolving , and the allocation after. Again, only the allocation for agents in changes, hence by induction we only need to consider subsets that contain agents in . Note that since agent is already resolved, does not value any goods in . Hence the allocation of goods in does not affect the right hand side of the inequality in the claim.
Other than the allocation of goods in , the only change that happens in the algorithm is that the prior bundle may be transferred to . Thus if is not transferred to , or , then ’s value for the aggregate bundle is unchanged, and the claim holds.
Thus, , , and the only relevant change that occurs in the iteration is that the bundle is transferred to agent . Note that since , does not hold any goods from , and hence does not contain any goods from . Now in the allocation , consider the subset . Then
proving the claim.
Proposition 8.
For an agent , let and be the allocations before and after is resolved. If for some agent and good , but , then and . Thus in a phase, any good allocated before the phase began can only be transferred from the root of a structure to its favourite neighbour.
Proof.
As noted, when is resolved, the goods in the structure are allocated to the agents in the structure. If Line 13 is executed, then additionally the bundle is transferred to agent (and agent gets the bundle . Thus the only way that a good that is assigned to an agent before is resolved can be moved, is if it was assigned to and then is transferred to agent . The proposition follows.
Since a root only appears once in a phase, a good once transferred in a phase cannot be transferred again, and hence any good allocated prior to a phase can only be transferred once, from the root of a structure to its favourite neighbour.
The next few results show that a good valued by an agent cannot be very far from .
Claim 9.
Let be the allocation at Line 5. If agent values a good and , then along allocated edges.
Proof.
By Proposition 8, a good can only be transferred rightward, from an agent being resolved to its favourite neighbour. If , then once is resolved, ’s allocation does not change. Thus if and , received this good in phase 1 when was resolved. Since values , must be in , and there is a path of length 1 between and along the edge set satisfying the claim.
Now suppose . Consider the phase when is initially allocated, say to an agent when is resolved. Since values , . Hence and have a path of length 2 along the edges of . From Proposition 8, in each subsequent phase, is transferred from the root of a structure to its favourite neighbour. Suppose is allocated to in phase . Then must be a right neighbour for the agent being resolved, and hence the colour . There are at most phases after the phase is initially allocated. Thus there is a path of length along allocated edges from to .
If not only values , but and is held by agent , then we can get a tighter bound on .
Claim 10.
Let be the allocation at Line 5. Let there be agents , , and good . Then along allocated edges.
Proof.
Since , when is resolved, it is assigned either to or to a neighbour to the right. Further any reassignment of transfers it to a neighbour further to the right. Hence if , then must be to the right of , hence .
Now when is initially allocated in phase , it is allocated to an agent at a distance 1 from along the edges . Suppose is allocated to agent in phase . Then , and there are at most phases after the phase when is initially allocated. Thus there is a path of length along allocated edges from to .
We now prove our main lemma.
Lemma 11.
Let be the allocation at Line 5. Then if envies , then is a resolved vertex, and (i.e., is ’s favourite vertex in ).
Proof.
We will prove the lemma by induction. For the base case, the initial allocation is empty and trivially satisfies the two properties. Now fix any vertex . Let be the allocation before the structure rooted at is resolved, and be the allocation after resolving .
None of the vertices in are resolved in , and hence by the induction hypothesis, none of these are envied in . When is resolved, only the allocation to agents in is modified, while the allocation to other agents is unchanged. The value of agents in does not decrease. Hence, any new envy edge must be towards agents in .
Consider an ordinary neighbour in . Agent is not envied in . It receives its preferred bundle ; this is only valued by and , hence no agent other than will envy . To see that also will not envy , we first claim that , i.e., does not value ’s bundle in . To prove this, assume for a contradiction that values ’s bundle. Then by Claim 9, there exists a - path along allocated edges of length at most . Including the (unallocated) edges , this gives a cycle of length . If , this gives us a 3-cycle, which would mean that the graph is not 2-colourable. If , then , but the graph has girth at least . In either case we get a contradiction. Thus, .
Further, . This is because agent prefers over , and always has the choice of keeping . Hence, it ends up with a bundle of value at least that of . Then since and , since valuations are cancelable, . Thus, an ordinary neighbour is not envied.
Now consider agent . As shown, agent is not envied in . While resolving , the only goods agent can possibly receive are goods in , and hence no agent outside will envy . Within , all ordinary neighbours receive their preferred bundle, and hence will not envy . Thus the only agent that could possibly envy after resolving is .
Now if gets its preferred bundle (in Lines 16 and 19), it does not envy . If agent gets (in Line 13), then this is an EFX partition for (since cut s), and agent only gets . In this case, envies , but the envy is EFX.
Lastly, consider agent . This case is more complicated than the earlier cases, since the bundle previously held by could be transferred to in Line 13. We will show however that no agent envies .
First, let’s consider agents in . Again, the ordinary neighbours did not envy earlier, and from the goods they get their preferred bundle . Hence, since valuations are cancelable, they will not envy .
Agent , as argued previously, by Claim 9, does not value , else there is a cycle of length strictly smaller than . Further agent gets a bundle of value at least that of , and hence will not envy .
For the last case, consider any agent . In allocation , does not envy or . Since does not value goods in , if it envies , clearly (i) it must value both and , and (ii) it must envy .
Suppose . Then from Claim 7, either or holds a good from (say ). By Claim 10, is then at distance from along allocated edges. From (i) in the previous paragraph, and Claim 9, is at distance at most from along allocated edges. Thus, there is - path along allocated edges (via ) of length (since and ). But then including the (unallocated) edges , there is a cycle of length , which is a contradiction, since we assume the graph has girth at least .
Finally, suppose . Then note that all the goods that values are in . Since both and possess goods that values, by Claim 10, they are at distance and respectively from along allocated edges. Since , there is a path of length from to (via ) along allocated edges. But then including the (unallocated) edges , there is a cycle of length , which is again a contradiction. Thus, agent is not envied in the allocation . This completes the proof.
We now can easily prove that the allocation obtained is EFX.
Theorem 12.
Algorithm 3 obtains an EFX allocation for agents with cancelable valuations in polynomial time.
Proof.
The proof for the running time for Algorithm 3 is obtained easily, since for cancelable valuations runs in polynomial time, and the other steps are straightforward.
We now show that the algorithm obtains an EFX allocation. As before, the proof is by induction on the iterations of the inner for loop. For the base case, the statement clearly holds for the empty allocation. Then consider an agent , and let and be the allocations before and after is resolved.
Consider first an envy edge from agent to agent in allocation . We will show that this envy remains EFX in allocation as well. By Lemma 11, agent must be resolved. No agent in is resolved in , hence . Since only agents in have their allocations modified, agent ’s allocation is not modified. Agent is either also not in (in which case its allocation also does not change), or is in (in which case its value does not decrease). In either case, either does not envy in , or the envy remains EFX.
By Lemma 11, the only envy edge that may be in but not in is from to . This happens in Line 11, when gets . But then agent gets . Since agent cuts the bundle , and agent has no goods besides , the envy is EFX.
Finally, we show that EFX allocations exist in cycle multi-graphs, from the earlier results.
Theorem 13.
For agents with cancelable valuations in a cycle multi-graph, EFX allocations exist.
Proof.
We consider three separate cases, depending on the length of the cycle. For even-length cycles, the result follows from Theorem 3, since such a cycle is a bipartite multi-graph. If the length is 3, then the result follows from EFX existence for three agents [16]. Finally, if the cycle has odd length of at least 5, the chromatic number is 3, and the existence follows from Theorem 12, setting , since the girth is least .
6 Conclusion
Our paper extends the work on EFX existence to important classes of multi-graphs. We view our results as a significant extension over the work by Christodoulou et al. [18] since their results are for simple graphs. Our work also shows that the well-known cut-and-choose paradigm can be leveraged to obtain results in fairly general settings. Our algorithm is another tool in the toolbox for EFX allocations, different, for example, from the multiple-phase algorithms used in many other papers.
We note that if edges are chores instead of good, i.e., an edge is negatively-valued by both adjacent vertices, an envy-free allocation can be obtained by assigning each edge to a vertex other than its endpoints. The existence of EFX allocations for general multi-graphs with positively valued edges remains an open problem. We also find it interesting that the case of a cycle multi-graph with 3 agents does not yet have a simpler algorithm than for general 3 agents. Coming up with a simpler algorithm here may be crucial to extending our work to general multi-graphs.
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