The Convergence of Iterative Delegations in Liquid Democracy in a Social Network
Abstract
Liquid democracy is a collective decision making paradigm which lies between direct and representative democracy. One of its main features is that voters can delegate their votes in a transitive manner such that: A delegates to B and B delegates to C leads to A indirectly delegates to C. These delegations can be effectively empowered by implementing liquid democracy in a social network, so that voters can delegate their votes to any of their neighbors in the network. However, it is uncertain that such a delegation process will lead to a stable state where all voters are satisfied with the people representing them. We study the stability (w.r.t. voters preferences) of the delegation process in liquid democracy and model it as a game in which the players are the voters and the strategies are their possible delegations. We answer several questions on the equilibria of this process in any social network or in social networks that correspond to restricted types of graphs.
We show that a Nashequilibrium may not exist, and that it is even NPcomplete to decide whether one exists or not. This holds even if the social network is a complete graph or a bounded degree graph. We further show that this existence problem is W[1]hard w.r.t. the treewidth of the social network. Besides these hardness results, we demonstrate that an equilibrium always exists whatever the preferences of the voters iff the social network is a tree. We design a dynamic programming procedure to determine some desirable equilibria (e.g., minimizing the dissatisfaction of the voters) in polynomial time for tree social networks. Lastly, we study the convergence of delegation dynamics. Unfortunately, when an equilibrium exists, we show that a best response dynamics may not converge, even if the social network is a path or a complete graph.
Sorbonne Université, CNRS, LIP6 UMR 7606,
4 place Jussieu, 75005 Paris, France.
Institut Universitaire de France,
Gran Sasso Science Institute,
Viale Francesco Crispi 7, 67100, L’Aquila, Italy
EDF R&D, 7 boulevard Gaspard Monge, 91120 Palaiseau.
1 Introduction
Liquid Democracy (LD) is a voting paradigm which offers a middleground between direct and representative democracy. One of its main features is the concept of transitive delegations, i.e., each voter can delegate her vote to some other voter, called representative or proxy, which can in turn delegate her vote and the ones that have been delegated to her to another voter. Consequently, a voter who decides to vote has a voting weight corresponding to the number of people she represents, i.e., herself and the voters who directly or indirectly delegated to her. This voter is called the guru of the people she represents. This approach has been advocated recently by many political parties as the German’s Pirate party or the Sweden’s Demoex party and is implemented in several tools as the online platforms LiquidFeedback [BKN+14] and liquid.us or GoogleVotes [HL15]. One main advantage of this framework is its flexibility, as it enables voters to vote directly for issues on which they feel both concerned and expert and to delegate for others. In this way, LD provides a middleground between direct democracy, which is strongly democratic but which is likely to yield high abstention rates or uninformed votes, and representative democracy which is more practical but less democratic [GRE15, WIE13]. Importantly, LD can be conveniently used in a Social Network (SN), where natural delegates are connected individuals. These choices of delegates are also desirable as they ensure that delegations rely on a foundation of trust. For these reasons, several works studying LD in the context of an SN enforce the constraint that voters may only delegate directly to voters that are connected to them [BBC+09, KMP18, BGL19].
Aim of this paper.
In this work, we tackle the problem of the stability of the delegation process in the LD setting. Indeed, it is likely that the preferences of voters over possible gurus will be motivated by different criteria and possibly contrary opinions. Hence, the iterative process where each voter chooses her delegate may end up in an unstable situation, i.e., a situation in which some voters would change their delegations. A striking example to illustrate this point is to consider an election where the voters could be positioned on the real line in a way that represents their rightwing leftwing political identity. If voters are ideologically close enough, each voter, starting from the leftside, could agree to delegate to her closest neighbor on her right. By transitivity, this would lead to all voters, including the extremeleft voters, having an extremeright voter for guru. These unstable situations raise the questions: “Under what conditions do the iterative delegations of the voters always reach an equilibrium? Does such an equilibrium even exist? Can we determine equilibria that are more desirable than others?”.
We assume that voters are part of an SN represented by an undirected graph so that they can only delegate directly to one of their neighbors. The preferences of voters over possible gurus are given by unrestricted linear preference orders. In this setting, the delegation process yields a game where players are the voters involved in the election and each player seeks to minimize the rank of her guru in her preference order. We answer the questions raised above with a special emphasize on the SN.
Our results.
We first show that in a complete SN, the problem of determining if an equilibrium exists is equivalent to the NPcomplete problem of determining if a digraph admits a kernel. Then we strengthen this result by giving hardness results on two parameters of the SN, the maximum degree of a node and the treewidth. Secondly, we obtain positive results. Indeed, we show that an equilibrium is guaranteed to exist iff the SN is a tree. We study the case of trees for which we design efficient algorithms for the computation of several desirable equilibria. Lastly, we study the convergence of delegation dynamics. Unfortunately, when an equilibrium exists, we show that a best response dynamics may not converge, even if the SN is a path or is complete.
2 Related Work
The stability of the delegation process is one of the several algorithmic issues raised by LD. These issues have recently raised attention in the AI literature. We review some of these questions here.
Are votes in an LD setting “more correct”? The idea underlying LD is that its flexibility should allow each voter to make an informed vote either by voting directly, or by finding a suitable guru. Several works have investigated this claim as the ones of GreenArmytage and Kahng et al. [KMP18, GRE15].
In the former work [GRE15], the author proposes a setting of spacial voting in which voters’ opinions on different issues of a vote are given by their position on the real line (one position per issue). Moreover, the competence level of each voter is expressed as a random variable which adds noise to the estimates that she has about voters’ positions (hers included). Votes and delegations are then prescribed by these noisy estimates and by the competence levels of the voters. GreenArmytage defines the expressive loss of a voter as the squared distance between her vote and her position and proves that transitive delegations decrease on average this loss measure. In the latter work [KMP18], the authors study an election on a binary issue, for which there is one “correct” answer. Each voter has a competence level, i.e., a probability of being correct. They further assume that the voters belong to an SN and that each voter only accepts to delegate to her neighbors who are “sufficiently” more expert than her. The authors investigate delegation procedures which take as input the SN and the voters’ competence levels and output the probabilities with which each voter should vote or delegate to her approved neighbors. Their key result consists in showing that no “local” procedure (i.e., procedures s.t. the probability distribution for each voter only depends on her neighborhood) can guarantee that LD is, at the same time, never (in large enough graphs) less accurate and sometimes strictly more accurate than direct voting.
How much delegations should a guru get? This question is twofold. On the one hand, gurus should have incentives to obtain delegations, but on the other hand it would be undesirable if a guru was to become too powerful. This problem has led to two recent papers [GKM+18, KR18]. In the work of Kotsialou and Riley [KR18], voters have both preferences over candidates and over possible gurus. Given the preferences over gurus, a delegation rule function decides who should get the delegations, and then, a voting rule function decides who wins the election given the preferences and the voting power of each guru. Focusing on two delegation rules named depth first delegation rule and breadth first delegation rule, they show that the latter one guarantees that a guru is always better of w.r.t. the outcome of the election when receiving a delegation whereas it is not the case for the former one. This shows that incentivising participation (voting and delegating) can be a concern in LD. In the setting of Gölz et al. [GKM+18], each voter can decide to vote or to specify multiple delegation options. Then a centralized algorithm should select delegations to minimize the maximum voting power of a voter. The authors give a approximation algorithm ( being the number of voters) and show that approximating the problem within a factor is NPhard. Lastly, they gave evidence that allowing voters to specify multiple possible delegation options (instead of one) leads to a dramatic decrease of the maximum voting power of a voter.
Are votes in an LD setting rational?
In the work of Christoff and Grossi [CG17], the authors study the potential loss of a rationality constraint when voters should vote on different issues that are logically linked and for which they delegate to different proxies. In the continuation of this work, Brill and Talmon [BT18] considered an LD framework in which each voter should provide a linear order over possible candidates. To do so each voter may delegate different binary preference queries to different proxies. This setting allows more flexibility but at the price of obtaining incomplete (because of delegation cycles) and even intransitive preference orders (hence the violation of a rationality constraint as studied by [CG17]). Notably, the authors showed that it is NPhard to decide if an incomplete ballot obtained in this setting can be completed to obtain complete and transitive preferences while respecting the constraints induced by the delegations.
Are delegations in an LD setting rational? Lastly, the work of Bloembergen et al. [BGL19] gives a different perspective on the study of voters’ rationality in LD. The authors consider an LD setting where voters are connected in an SN and can only delegate to their neighbors in the network. The election is on a binary issue for which some voters should vote for the 0 answer and the others should vote for the 1 answer (voters of type or ). Each voter does not know exactly her type which is modeled by an accuracy representing the probability with which voter makes the correct choice. Similarly, each pair of voters do not know if they are of the same type which is also modeled by a probability . Hence, a voter which has as guru has a probability that makes the correct vote (according to ) which is a formula including the and values. The goal of each voter is to maximize the accuracy of her vote/delegation. Importantly, each voter that votes incurs a loss of satisfaction representing the work required to vote directly. This modeling leads to a class of games, called delegation games. The authors proved the existence of pure Nash equilibria in several types of delegation games and gave upper and lower bounds on the price of anarchy, and the gain (i.e., the difference between the accuracy of the group after the delegation process and the one induced by direct voting) of such games.
Our approach is closest to this last work as we consider the same type of delegation games. However, our model of preferences over guru is more general as we assume that each voter has a preference order over her possible gurus. These preference orders may be dictated by competence levels and types of voters as in [BGL19]. However, we do not make such hypothesis as the criteria to choose a delegate are numerous: geographic locality, cultural, political or religious identity, et caetera. Considering this more general framework strongly modifies the resulting delegation games.
3 Notations and Settings
3.1 Notations and Nashstable delegation functions
We denote by a set of voters that take part in a vote
Given a delegation function , let denote the set of gurus, i.e., . The guru of a voter , denoted by , can be found by following the successive delegations starting from . Formally, if there exists a sequence of voters such that for every , , and . However, it may happen that no such exists because the successive delegations starting from end up in a circuit, i.e., delegates to , who delegates to , and so on up to who delegates to with . In this case, we consider that the voters abstain, as none of them takes the responsibility to vote, i.e., we set for all . Such a definition of gurus allows to model the transitivity of delegations: if , , and , then the guru of will be , even if has not delegated directly to . Hence a voter can delegate directly to one of its neighbors in , but she can also delegate indirectly to another voter through a chain of delegations. Note that because voters can only delegate directly to their neighbors, such a chain of delegations coincides with a path in .
Given a voter and a delegation function , we now consider how voter may change her delegation to get a guru different from her current guru . She may decide to vote herself, to abstain, or to delegate to a neighbor with a different guru, and in the latter case she would get as a guru. We denote by the set of gurus of the neighbors of in . Then the gurus that can get by deviating from is exactly the set .
Example 1.
Consider the SN represented in Figure 1 with the delegation function defined by , , , , , , , and . For this example, , with , and . If she wants, voter 1 can change her delegation. If she delegates to voter 3, then this will not change her guru as voter 3 is delegating to guru 4. However, if she delegates to voter 2, then her new guru will be voter 5. In any case, she can also decide to modify her delegation to declare intention to vote or abstain. Note that in this example voter 1 cannot change unilaterally her delegation in order to have voter 6 as guru. Indeed, voter 6 does not belong to , as there is no delegation path from a neighbor of voter 1 to voter 6.
We assume that each voter has a preference order over who could be her guru in . For every voter , and for every , we have that if prefers to have as guru (or to vote herself if , or to abstain if ) rather than to have as guru (or to vote herself if , or to abstain if ). The collection of preference orders in turn defines a preference profile . To illustrate these notations, we consider in Example 2 a simple instance with 3 voters. Note that this instance will be reused as building block in several places in the paper under the name of 3cycle.
Example 2 (3cycle).
For illustration purposes, consider the following instance in which there are voters connected in a complete SN and with the following preference profile :
Put another way, each voter prefers to delegate to rather than to vote directly and each voter prefers to vote rather than to abstain.
As a consequence of successive delegations, a voter might end up in a situation in which she prefers to vote or to abstain or to delegate to another guru that she can reach than to maintain her current delegation. Such a situation is regarded as unstable as this voter would modify unilaterally her delegation. This is for instance the case in the previous example if , and : by successive delegations, the guru of is , but would prefer to vote instead. More formally, a delegation function is Nashstable for voter if
A delegation function is Nashstable if it is Nashstable for every voter in . A Nashstable delegation function is also called an equilibrium in the sequel.
It may seem difficult in practice that voters give a complete linear order over the set . We now highlight that the computation of equilibria does not require the whole preference profile. We say that voter is an abstainer in if she prefers to abstain rather than to vote, i.e., if ; she is a nonabstainer otherwise. We will denote by the set of abstainers. Note that an abstainer never votes directly in an equilibrium; similarly, a nonabstainer never abstains in an equilibrium. Given a preference profile , we define the set of acceptable gurus for . Note that this set is likely to be quite small compared to . A necessary condition for a delegation function to be Nashstable is that , or and , or and . Hence when looking for equilibria, the preferences of voter below (if ) or (if ) can be dropped. In the sequel, we may define a preference profile only by giving, for every voter , if she is an abstainer or not, and her preference profile on .
3.2 Existence and optimization problems investigated
Stable situations are obviously desirable. Sadly, there are instances for which there is no equilibrium.
Observation 1.
The instance described in Example 2 (3cycle) admits no equilibrium.
Proof.
Assume by contradiction that there exists a Nashstable delegation function . First note that as the SN is complete, any voter can delegate to any other voter. Second, note that for any pair of voters in , there is always one voter that approves the other as possible guru, i.e., she prefers to delegate to this voter rather than to vote or abstain. Hence, cannot be greater than 1, otherwise one of the guru would rather delegate to another guru than vote directly. On the other hand, note that there is no voter that is approved as possible guru by all other voters. Hence, cannot be less than 2, otherwise one of the voter in would rather vote than delegate to one of the gurus (as in this example ). We obtain the desired contradiction. ∎
Hence the first problem, called EXISTENCE, that we will investigate in this article is the one of the existence of an equilibrium.
EXISTENCE (abbreviated by EX)
INSTANCE: A preference profile and a social network .
QUESTION: Does there exist an equilibrium?
Note that problem EX is in NP, as given a delegation function, we can easily find the guru of each voter and then check the Nashstability condition in polynomial time.
For instances for which we know that some equilibrium exists, we will investigate if we can compute equilibria verifying particular desirable properties.
Firstly, given a voter , we will try to know if there exists a Nashstable delegation function for which is a guru, i.e., . We call this problem MEMBERSHIP.
MEMBERSHIP (abbreviated by MEMB)
INSTANCE: A preference profile , a social network and a voter .
QUESTION: Does there exist a Nashstable delegation function for which ?
Secondly, we will try to find a Nashstable delegation function that optimizes some objective function either to a) minimize the dissatisfaction of the voters, b) minimize the maximum voting power of a guru, or c) minimize the number of voters who abstain. More formally, we will study the three following optimization problems:

MIN DISSATISFACTION (MINDIS for short) minimizes the sum of dissatisfactions of the voters where the dissatisfaction of a voter is given by , being the rank of in the preference profile of ;

MIN MAX VOTING POWER (MINMAXVP for short) minimizes the maximal voting power of a guru, where the voting power of a guru is defined by ;

MIN ABSTENTION (MINABST for short) minimizes the number of voters that abstain in an equilibrium, i.e, it minimizes .
Problems MINDIS, MINMAXVP and MINABST
INSTANCE: A preference profile and a social network .
SOLUTION: A Nashstable delegation function .
MEASURE for MINDIS: (to minimize).
MEASURE for MINMAXVP: (to minimize).
MEASURE for MINABST: (to minimize).
The last questions that we investigate capture the dynamic nature of delegations.
3.3 Convergence problems investigated
In situations where an equilibrium exists, a natural question is whether a dynamic delegation process (necessarily) converges towards such an equilibrium. As classically done in game theory (see for instance [NSV+11]), we will consider dynamics where iteratively one voter has the possibility to change her delegation
Given a delegation function and a voter , let:

be the set of improved “moves” for : where is the same delegation as up to the fact that delegates to (or votes if , or abstains if ). Note that we do not consider moves where voter delegates to a voter that abstains or that creates a cycle. In this case, she would rather abstain herself, i.e., set .

be the set of maximizing ’s outcome. If there is no possible improvement () then ( will not change her delegation).
In a dynamics, we are given a starting delegation function and a token function . The most simple choice of starting delegation function is the one for which every voter declares intention to vote. Note however that the convergence of a dynamics depends on , hence we will specify in our results the starting delegation functions under consideration.
The token function specifies that voter has the token at step : she has the right to change her delegation. This gives a sequence of delegation functions where for any , if then . A dynamics is said to converge if there is a such that for all . We will assume, as usual, that each voter has the token an infinite number of times. A classical way of choosing such a function is to consider a permutation over the voters in , and to repeat this permutation over the time to give the token (if then ). We will call these dynamics permutation dynamics.
Given and , a dynamics is called:

A better response dynamics or Improved Response Dynamics (IRD) if for all , chooses an improved move if any, otherwise does not change her delegation: if then , otherwise .

A Best Response Dynamics (BRD) if for all , chooses a move in .
Note that a BRD is also an IRD.
The last problems that we investigate, denoted by IRCONVERGENCE (IRCONV for short) and BRCONVERGENCE (BRCONV for short), can be formalized as:
IRCONV (resp. BRCONV) INSTANCE: A preference profile and a social network . QUESTION: Does a dynamic delegation process under IRD (resp. BRD) necessarily converge whatever the token function ?
3.4 Summary of results and outline of the paper
Our results are presented in Table 1. In Section 4, we investigate the complexity of problem EX. We will show that when is complete, this problem is equivalent to the problem of determining if a digraph admits a kernel (i.e., an independent set of nodes such that for every other node not in , there exists an arc with ) which is NPcomplete [CHV73]. We then strengthen this result by showing that EX is also NPcomplete when the maximum degree of is bounded by 5 and is W[1]hard w.r.t. the treewidth of . These results are summarized in the left tabular of Table 1. Hence, deciding if an instance admits an equilibrium is an NPcomplete problem. However we will identify specific SNs that ensure that an equilibrium exists whatever the preference profile . More precisely, we will see that an equilibrium exists whatever the preferences of the voters iff the SN is a tree. Hence in Section 5, we investigate the class of tree SNs and we design a dynamic programming scheme which allows to solve problems MEMB, MINDIS, MINMAXVP and MINABST in polynomial time. The polynomial complexity results we obtain are given in the central tabular of Table 1. Lastly, in Section 6, we study the convergence of delegations dynamics in LD. Unfortunately, when an equilibrium exists, we show that a BRD may not converge even if is complete or is a path. For a star SN, we obtain that a BRD will always converge, whereas an IRD may not. These results are summarized in the right tabular of Table 1.
Type of or parameter  EX 

Complete  NPC 
Maximum degree  NPC 
Treewidth  W[1]hard 
Tree  AE 
Problem  Star  Tree 

MEMB  
MINDIS  
MINMAXVP  
ABST 
Problem  IR  BR  

with an equilibrium  CONV  CONV  
Star  NA  Always  
Path  NA  NA  
Complete  NA  NA 
4 Existence of equilibria: hardness results
4.1 Complete social networks
We focus in this subsection on the case where the SN is a complete graph. We mainly show that determining whether an equilibrium exists or not is an NPcomplete problem (Theorem 1), by showing an equivalence with the problem of finding a kernel in a graph. This equivalence is also helpful to find subcases where an equilibrium always exist.
We define the delegationacceptability digraph by its arcset . Stated differently, there is one vertex per nonabstainer and there exists an arc from to if accepts as a guru. For example, in Figure 2, we give a partial preference profile involving 5 voters and the corresponding delegationacceptability digraph .
The main result of this subsection, stated in Proposition 1, is a characterization of all sets of gurus of equilibria, as specific subsets of vertices of the delegationacceptability digraph. Let us introduce additional graphtheoretic definitions. Given a digraph , a subset of vertices is independent if there is no arc between two vertices of . It is absorbing if for every vertex , there exists s.t. (we say that absorbs ). A kernel of is an independent and absorbing subset of vertices.
Proposition 1.
Assume is complete, then given a preference profile and a subset of voters , the following propositions are equivalent:

there exists an equilibrium s.t. ;

contains no abstainer and is a kernel of .
Proof.
(i) (ii). Let be a Nashstable delegation function for . Let us prove that its set of gurus satisfy condition (ii). It was noted previously that Nashstability implies the absence of abstainer in . Assume that is not independent in . Then, there exists such that is an arc of , that is, . It implies that prefers to delegate to rather than remaining a guru. As is complete, and hence is not Nashstable for . Assume now that is not absorbing for all nonabstainers. Then there exists a nonabstainer such that for every guru in , is not an arc of , that is, . Such a voter prefers to vote herself rather than delegate to any guru in . Therefore is not Nashstable for . This proves that is a kernel of .
(ii) (i). Consider a subset of nonabstainers such that is a kernel of the delegationacceptability digraph . We define a delegation function by: if ; and if where is the voter that prefers in . Note that as is complete, each voter in can delegate directly to any voter in . It follows that for every , and the set of gurus is . Let us check that the delegation function is Nashstable. First, it holds that is Nashstable for every abstainer . Indeed, the guru of is , which is her preferred guru in . Also, we show that is Nashstable for any . We need to prove that does not prefer to delegate to any other guru, or to abstain. By assumption the set contains no abstainer. Since is independent, for every other guru , the arc does not exist in the delegation acceptability digraph, meaning that and . Finally, we prove that is Nashstable for any nonabstainer . Since has already chosen her preferred guru in , it is only necessary to check that does not prefer to vote herself. Because the set is absorbing, there exists such that is an arc of . Hence such is a guru in such that , hence , and . ∎
Note that any digraph is the delegationacceptability digraph of a preference profile . Indeed, given the digraph, it suffices to build a preference profile so that every voter prefers to delegate to its outneighbors, then to vote, then to delegate to other voters, then to abstain. Consequently, in a complete SN, determining if a preference profile admits an equilibrium is equivalent to the problem of determining if a digraph admits a kernel, which is an NPcomplete problem [CHV73].
Theorem 1.
EX is NPcomplete even when the social network is a complete graph.
Consequences. As a direct consequence of the results of this section, even when is complete, optimization problems MINDIS, MINMAXVP and MINABST are NPhard as it is NPhard to decide if their set of admissible solutions is empty or not. We also directly obtain that the decision problem MEMB is NPcomplete, by a direct reduction from EX. Indeed, solving problem MEMB for each voter in yields the answer to problem EX.
As mentioned above, we also point out that this equivalence is useful to find some interesting subcases. Let us consider for instance the case where there is a symmetry in the preferences in the sense that if and only if . In this case of symmetrical preference profiles, the delegationacceptability digraph has the arc iff it has the arc (it is symmetrical). Then, any inclusion maximal independent set is a kernel. Hence, for any nonabstainer there exists an equilibrium in which is a guru (take a maximal independent set containing ).
Proposition 2.
In a complete social network, there always exists an equilibrium when preferences are symmetrical. Moreover, the answer to MEMB is always yes.
More generally when preferences are symmetrical, given a set of nonabstainers it is easy to decide if there exists an equilibrium in which every voter in this set is a guru: we just have to check whether the set is independent or not in .
4.2 Sparse social networks
As the problem EX is NPcomplete when the social network is a complete graph, it remains NPcomplete in any class of graphs that contain cliques, such as interval graphs, split graphs, dense graphs,…In this section, we focus on classes of graphs that do not contain large cliques. We first deal with bounded degree graphs, and show that EX remains NPhard in social networks of degree bounded by 5 (Theorem 2). We then focus on graphs of bounded treewidth. Interestingly, while we will see in Section 5 that EX is polynomial if the social network is a tree (actually, an equilibrium always exists in trees), we prove here that EX is W[1]hard when parameterized by the treewidth of the social network (Theorem 3).
We refer the reader to [BK08] for the standard notion of treewidth of graphs. A problem is said to be fixed parameter tractable (FPT) with respect to some parameter (the treewidth of the graph for us) if it can be solved by an algorithm whose complexity is for some function and constant (where is the size of the instance). A W[1]hard problem is not FPT unless FPT=W[1]. We refer the reader to [DF12] for notions of parameterized complexity.
Theorem 2.
Problem EX is NPcomplete even if the maximum degree of the SN is at most 5.
Proof.
We make a reduction from the NPcomplete problem 3SAT4 [TOV84], which is the restriction of the 3SAT problem where each variable appears at most 4 times. Two voters and are created for each truth variable . These voters accept each other as possible gurus and are connected in the SN. Three voters are created for clause : . These voters are connected in the SN in a 3cycle such that accepts as a guru but rejects as a guru. Moreover, is connected to the three vertices or corresponding to the literals of the clause. Note that all these voters are nonabstainers. We illustrate the reduction by showing how a clause is handled (see also Fig.3):

and

and

and
The induced instance has maximum degree bounded by 5. We conclude the proof by proving that the 3SAT4 instance is satisfiable iff there exists an equilibrium.
Suppose first that there is a truth assignment. Then consider the delegation function where: 1) a voter (resp. ) votes if the corresponding literal is true, otherwise she delegates to (resp. ); 2) a voter delegates to a voter corresponding to a true literal of this clause (the best one if there are several of them); 3) both and delegate to .
Voters and have the same guru as . We can easily see that this is an equilibrium.
Conversely, suppose that we have an equilibrium. For each , exactly one voter among and votes. We define the truth assignment where is true (resp. false) if votes (resp. votes). In an equilibrium, the voter has to delegate to some voter or (otherwise the 3 voters of the clause cannot be in a stable state, as they would form a 3cycle). Then, the guru of must be a voter or that she accepts, i.e., this guru must correspond to a literal of . Hence, the truth assignment satisfies this clause. ∎
Theorem 3.
Problem EX is W[1]hard when parameterized by the treewidth of the social network.
Proof.
We make a reduction from the list coloring problem. This problem takes as input a graph , together with an assignment to each vertex of a set of colors . The problem is to determine whether it is possible to choose a color for vertex from the set of permitted colors , for each vertex, so that the obtained coloring is proper. This problem is W[1]hard, parameterized by the treewidth of [FFL+11].
To perform our reduction we show how to create the SN starting from graph . The SN contains all nodes of . Let be an edge of . For each color create three voters and (we omit the reference to to keep notation readable) such that is connected to , is connected to and and are connected in a triangle. For each color create a voter which is only connected to in the SN. This ends the construction of the SN. We illustrate this construction in Figure 4, representing the SN obtained for two adjacent vertices with and . Colors and are common to and , corresponding to the 2 triangles and .
We now argue that the treewidth of this SN is the maximal value between 4 and the treewidth of . Indeed, consider a tree decomposition of . For each color , create a node that contains the bag and connect it to a node of that contains . For each color create a node that contains the bag and connect it to a node of that contains and . The claim comes from the fact that we obtain in this way a tree decomposition of the SN.
We now detail the preferences of the voters. First note that there are no abstainers. Each vertex accepts the vertices as possible gurus. Each vertex accepts the vertices , for in , as possible gurus. Let us now consider vertices and resulting from a color for some edge . Voter accepts as possible guru, voter accepts as possible guru and voter accepts as possible guru. Voter accepts also the voters for all , and similarly voter accepts the voters for all . Voter prefers any () to .
We now show that the list coloring instance admits a proper coloring iff the created instance admits an equilibrium.
Given a proper coloring of we obtain an equilibrium as follows. First, for each vertex , if is colored in with color , then votes and delegates to . Then, each vertex with delegates to in order to have as guru. Let us now consider vertices and resulting from a color for some edge . If and are not assigned color ( and ), then delegates to , delegates to and votes. In this case, is the guru of (note that this is an acceptable guru since ) and is the guru of . Otherwise, either or is assigned color (but not both). Let us assume first that it is . In this case, votes, delegates to , and delegates to . Voter is then the (acceptable) guru of . Finally, let us consider the case where is assigned color . Then delegates to , votes and delegates to . The guru of is , the one of is . We have in all cases an equilibrium.
Given an equilibrium of the instance, we first notice that for each , there is exactly one vertex that votes. Indeed, there cannot be more than one, because in this case, 1) would delegate to one of them and then 2) the other gurus in would change to delegate to in order to have the same guru. There cannot be none: indeed voters are nonabstainers, hence if they do not vote they must delegate in order to be represented by an acceptable guru. And yet the only acceptable gurus for them are the other members of . Then each vertex in delegates to this particular voter . We show that coloring in each vertex with color is a proper list coloring. On the contrary, suppose that two adjacent vertices and receive the same color . We focus on the vertices (corresponding to this color for this edge ) in SN. cannot delegate to in the equilibrium since would be her guru, and she does not accept as a guru. Similarly, cannot delegate to in the equilibrium. Then, form a 3cycle which cannot reach a stable state, contradiction. ∎
Another parameter that is worth being considered is the maximal cardinal of a set of acceptable gurus, where the maximum is taken over all voters: . This number is likely to be small in practice. Would this assumption help for solving problem EX? Unfortunately, in the proof of Theorem 2, is bounded above by 4, so the problem remains NPhard when both and the maximum degree are bounded. An interesting question would be to determine whether the problem becomes FPT when parameterized by the treewidth and . We leave this as an open question.
5 Algorithms on Tree Social Networks
5.1 Equilibria and trees
In this section, we first answer the question of characterizing social networks in which an equilibrium always exists. It turns out that such social networks are exactly trees.
Theorem 4.
If is a tree, then for any preference profile there exists an equilibrium.
Proof.
We proceed by induction. The case of a tree of 1 voter is trivial. Consider the result true up to voters and consider a tree of voters. Root the tree at some arbitrary vertex. Consider a leaf and the parent of this leaf. First if is an abstainer we build an equilibrium in the following way. There exists an equilibrium in the tree without , add to this equilibrium by giving her her preferred option between abstaining and delegating to the guru of .
Now, we assume is not an abstainer. If it is easy to construct an equilibrium. Again, there exists an equilibrium in the tree without , add to this equilibrium by giving her her preferred option between voting and delegating to the guru of .
We now assume is not an abstainer and . In this case, we will assume that can always delegate to as last resort. Hence, we can consider that the acceptability set of can be restrained to the voters at least as preferred as . Secondly, this assumption means that other voters cannot hope to have as guru. More precisely, the only gurus that they can reach through are the voters at least as preferred as , and of course .
To materialize these constraints, we consider the tree without and where takes the place of in all voters preference list (including the one of ). There exists an equilibrium in this tree. If delegates in the equilibrium (then she prefers her guru to otherwise would vote), add to this equilibrium by giving her her preferred option between voting and delegating to the guru of . Otherwise, make delegate to and votes. ∎
The other direction is also true, as shown in the following theorem.
Theorem 5.
If is a social network such that for any preference profile there exists an equilibrium, then is a tree.
Proof.
Suppose that is not a tree, and let us consider a chordless cycle (with ). We consider the following preferences, with no abstainer:

A voter not in prefers to vote.

Voter has the following preference: consists of all the voters of up to (and ), and we have: .
Suppose that there is an equilibrium. Each voter not in votes, and no voter in delegates to a voter not in . We focus on voters in . Nobody abstains since there is no abstainer. If nobody votes, then there is a delegation cycle, so a subset of voters abstain, which is impossible in an equilibrium (they would rather vote). So let us consider one voter in , say , that declares intention to vote. Voter does not vote (otherwise would delegate to ), does not delegate to , so she delegates to . Similarly does not vote (otherwise would delegate to so that would be her guru), and does not delegate to , so she delegates to . By an easy recurrence, we get that delegates to for . Now since votes, delegates to her. But in this case would be the guru of everyone, including who does not approve her, so would vote, a contradiction. ∎
5.2 Solving optimization problems in trees
Let us address the complexity of the problems MEMB, MINDIS, MINMAXVP and MINABST in a tree. It will be shown that when the SN is a tree, a dynamic programming approach is successful in building and optimizing equilibria.
We introduce some additional tools. Let the social network be a tree . Assume that is rooted at some vertex , and let us define accordingly, for every vertex , the parent of and the set of its children. Let denote the subtree of rooted at .
Let and let be an equilibrium such that the guru of is some . Note that can be in : then delegates downwards, votes or abstains, i.e. ; or : then delegates to her parent, i.e., . Consider the chain of delegations starting from a voter : either it reaches , and then the guru of will be , or it does not reach , hence it is fully determined by the restriction of to the subtree . Using this remark we will show that an equilibrium can be built inductively by combining delegation functions in subtrees.
Let us now define local equilibria to formalize restrictions of equilibria in a subtree. Let be a delegation function over . We define gurus associated with in a similar way as we defined gurus for delegation function over all voters. Let the guru of be: the first voter such that reached by the chain of delegations starting from if delegates downwards; if ; or some if . Given a voter , let the guru of be: the first voter such that reached by the chain of delegations starting from if the chain does not reach ; if ; or otherwise.
Given and , we say that is a local equilibrium on with label if it satisfies :

either and , or there is a chain of delegations in going from to .

for every , voter does not want to change her delegation, given that the guru of is , i.e., with it holds that .

the root does not want to change her delegation to any of her children, or to vote or to abstain, given that her current guru is , i.e., .
Note that condition (i) means that the label is consistent with the delegations, i.e., it is indeed possible that is the guru of in an equilibrium that coincides with on . Condition (ii) corresponds to Nashstability for voters in , and condition (iii) is a relaxed Nashstability for . This definition slightly generalizes the definition of equilibrium: an equilibrium is exactly a local equilibrium on with label .
Proposition 3.
Let , and .
Then is a local equilibrium on with label if and only if the following assertions are satisfied:

(resp. , ) if (resp. , ) and if for some ;

for every ;

For every , (c1) or (c2) is satisfied:

and there exists such that , , and is a local equilibrium on with label ;

is a local equilibrium on with label .

Proof.
Let be a local equilibrium on with label . It is clear that (i) implies (a), and (iii) implies (b). Let us check (c) for a given . Assume first . Then is a local equilibrium on with label and (c2) is satisfied. Assume now that , hence is some with . It comes since otherwise would also have as guru because the only path from to goes through . Also is a local equilibrium on with label . Finally since satisfies (ii) and (iii), neither nor wants to change for the guru of each other, i.e., and . Hence (c1) is satisfied.
Conversely, let us assume that satisfies (a), (b), and (c). First condition (i) is satisfied. Indeed it is implied by (a) when or or . Otherwise for some , and since is a local equilibrium on with label , there exists a chain of delegations from to in . By (a) hence there exists a chain of delegations from to . Condition (ii) is satisfied for every because is a local equilibrium in every . Let us check (ii) for : it is sufficient to check that whenever . This holds since