Social Network Games with Obligatory Product Selection

Transcription

1 Social Netork Games ith Obligatory Product Selection Krzysztof R. Apt Centre for Mathematics and Computer Science (CWI), ILLC, University of Amsterdam, The Netherlands Sunil Simon Centre for Mathematics and Computer Science (CWI) Recently, e introduced in [1] a model for product adoption in social netorks ith multiple products, here the agents, influenced by their neighbours, can adopt one out of several alternatives (products). To analyze these netorks e introduce social netork games in hich product adoption is obligatory. We sho that hen the underlying graph is a simple cycle, there is a polynomial time algorithm alloing us to determine hether the game has a Nash equilibrium. In contrast, in the arbitrary case this problem is NP-complete. We also sho that the problem of determining hether the game is eakly acyclic is co-np hard. Using these games e analyze various types of paradoxes that can arise in the considered netorks. One of them corresponds to the ell-knon Braess paradox in congestion games. In particular, e sho that social netorks exist ith the property that by adding an additional product to a specific node, the choices of the nodes ill unavoidably evolve in such a ay that everybody is strictly orse off. 1 Introduction Social netorks became a huge interdisciplinary research area ith important links to sociology, economics, epidemiology, computer science, and mathematics. A flurry of numerous articles, notably the influential [11], and books, e.g., [7, 3], helped to delineate better this area. It deals ith many diverse topics such as epidemics, spread of certain patterns of social behaviour, effects of advertising, and emergence of bubbles in financial markets. Recently, e introduced in [1] social netorks ith multiple products, in hich the agents (players), influenced by their neighbours, can adopt one out of several alternatives (products). To study the situation hen the product adoption is obligatory e introduce here social netork games in hich product adoption is obligatory. An example of a studied situation is hen a group of people chooses an obligatory product, for instance, an operating system or a mobile phone provider, by taking into account the choice of their friends. The resulting games exhibit the folloing join the crod property: the payoff of each player eakly increases hen more players choose his strategy. that e define more precisely in Subsection 2.3. The considered games are a modification of the strategic games that e recently introduced in [14] and more fully in [15], in hich the product adoption as optional. The insistence on product selection leads to a different analysis and different results than the ones reported there. In particular, Nash equilibria need not exist already in the case hen the underlying graph is a simple cycle. We sho that one can determine in polynomial time hether for such social netorks a Nash equilibrium exists. We prove that for arbitrary netorks, determining hether a Nash equilibrium exists is NP-complete. We also sho that for arbitrary netorks and for netorks hose underlying graph has no source nodes, determining hether the game is eakly acyclic is co-np hard. Gabriele Puppis, Tiziano Villa (Eds.): Fourth International Symposium on Games, Automata, Logics and Formal Verification EPTCS 119, 2013, pp , doi: /eptcs c Krzysztof R. Apt & Sunil Simon This ork is licensed under the Creative Commons Attribution License.

2 Krzysztof R. Apt & Sunil Simon 181 The considered social netorks allo us to analyze various paradoxes that ere identified in the literature. One example is the paradox of choice first formulated in [13]. It has been summarised in [6, page 38] as follos: The more options one has, the more possibilities for experiencing conflict arise, and the more difficult it becomes to compare the options. There is a point here more options, products, and choices hurt both seller and consumer. The point is that consumers choices depend on their friends and acquaintances preferences. Another example is a bubble in a financial market, here a decision of a trader to sitch to some ne financial product triggers a sequence of transactions, as a result of hich all traders involved become orse off. Such paradoxes are similar to the renoned Braess paradox hich states that in some road netorks the travel time can actually increase hen ne roads are added, see, e.g., [12, pages ] and a dual version of Braess paradox that concerns the removal of road segments, studied in [4, 5]. Both paradoxes ere studied by means of congestion games. Hoever, in contrast to congestion games, Nash equilibria do not need to exist in the games e consider here. Consequently, one needs to rely on different arguments. Moreover, there are no to ne types of paradoxes that correspond to the situations hen an addition, respectively, removal, of a product can lead to a game ith no Nash equilibrium. For each of these four cases e present a social netork that exhibits the corresponding paradox. These paradoxes ere identified first in [2] in the case hen the adoption of a product as not obligatory. In contrast to the case here considered the existence of a strongest paradox ithin the frameork of [2] remains an open problem. 2 Preliminaries 2.1 Strategic games A strategic game for n>1 players, ritten as (S 1,...,S n, p 1,..., p n ), consists of a non-empty set S i of strategies and a payoff function p i : S 1 S n R, for each player i. Fix a strategic game G :=(S 1,...,S n, p 1,..., p n ). We denote S 1 S n by S, call each element s S a joint strategy, denote the ith element of s by s i, and abbreviate the sequence(s j ) j i to s i. Occasionally e rite (s i,s i ) instead of s. We call a strategy s i of player i a best response to a joint strategy s i of his opponents if s i S i p i (s i,s i ) p i (s i,s i). We call a joint strategy s a Nash equilibrium if each s i is a best response to s i. Further, e call a strategy s i of player i a better response given a joint strategy s if p i(s i,s i)> p i (s i,s i ). By a profitable deviation e mean a pair (s,s ) of joint strategies such that s =(s i,s i) for some s i and p i (s )> p i (s). Folloing [10], an improvement path is a maximal sequence of profitable deviations. Clearly, if an improvement path is finite, then its last element is a Nash equilibrium. A game is called eakly acyclic (see [16, 9]) if for every joint strategy there exists a finite improvement path that starts at it. In other ords, in eakly acyclic games a Nash equilibrium can be reached from every initial joint strategy by a sequence of unilateral deviations. Given to joint strategies s and s e rite s>s if for all i, p i (s)> p i (s ). When s>s holds e say that s is strictly orse than s.

3 182 Social Netork Games ith Obligatory Product Selection 2.2 Social netorks We are interested in strategic games defined over a specific type of social netorks introduced in [1] that e recall first. Let V ={1,...,n} be a finite set of agents and G=(V,E,) a eighted directed graph ith i j [0,1] being the eight of the edge (i, j). Given a node i of G, e denote by N(i) the set of nodes from hich there is an incoming edge to i. We call each j N(i) a neighbour of i in G. We assume that for each node i such that N(i) /0, j N(i) ji 1. An agent i V is said to be a source node in G if N(i) = /0. Given a (to be defined) netork S e denote by source(s) the set of source nodes in the underlying graph G. By a social netork (from no on, just netork) e mean a tuple S =(G,P,P,θ), here G is a eighted directed graph, P is a finite set of alternatives or products, P is function that assigns to each agent i a non-empty set of products P(i) from hich it can make a choice, θ is a threshold function that for each i V and t P(i) yields a value θ(i,t) (0,1]. {t 4 } {t 2 } 0.5 {t 1 1,t 2,t 4 } {t 2,t 3 } {t 1,t 3 } Figure 1: A social netork {t 3 } Example 1. Figure 1 shos an example of a netork. Let the threshold be 0.3 for all nodes. The set of products P is {t 1,t 2,t 3,t 4 }, the product set of each agent is marked next to the node denoting it and the eights are labels on the edges. Each source node is represented by the unique product in its product set. Given to social netorks S and S e say that S is an expansion of S if it results from adding a product to the product set of a node in S. We say then also that S is a contraction of S. 2.3 Social netork games Next, introduce the strategic games over the social netorks. They form a modification of the games studied in [14, 15] in that e do not admit a strategy representing the fact that a player abstains from choosing a product. Fix a netork S =(G,P,P,θ). With each netork S e associate a strategic game G(S). The idea is that the agents simultaneously choose a product. Subsequently each node assesses his choice by comparing it ith the choices made by his neighbours. Formally, e define the game as follos: the players are the agents (i.e., the nodes), the set of strategies for player i is S i := P(i),

4 Krzysztof R. Apt & Sunil Simon 183 For i V, t P(i) and a joint strategy s, let Ni t (s) :={ j N(i) s j = t}, i.e., Ni t (s) is the set of neighbours of i ho adopted in s the product t. The payoff function is defined as follos, here c 0 is some given in advance positive constant: for i source(s), p i (s) := c 0, for i source(s), p i (s) := ji θ(i,t), here s i = t and t P(i). j Ni t (s) In the first case e assume that the payoff function for the source nodes is constant only for simplicity. The second case of the payoff definition is motivated by the folloing considerations. When agent i is not a source node, his satisfaction from a joint strategy depends positively from the accumulated eight (read: influence ) of his neighbours ho made the same choice as him, and negatively from his threshold level (read: resistance ) to adopt this product. The assumption that θ(i, t) > 0 reflects the vie that there is alays some resistance to adopt a product. We call these games social netork games ith obligatory product selection, in short, social netork games. Example 2. Consider the netork given in Example 1 and the joint strategy s here each source node chooses the unique product in its product set and nodes 1, 2 and 3 choose t 2, t 3 and t 2 respectively. The payoffs are then given as follos: for the source nodes, the payoff is the fixed constant c 0, p 1 (s)= =0.2, p 2 (s)= =0.1, p 3 (s)= =0.1. Let s be the joint strategy in hich player 3 chooses t 3 and the remaining players make the same choice as given in s. Then (s,s ) is a profitable deviation since p 3 (s ) > p 3 (s). In hat follos, e represent each profitable deviation by a node and a strategy it sitches to, e.g., 3 : t 3. Starting at s, the sequence of profitable deviations 3 : t 3,1 : t 4 is an improvement path hich results in the joint strategy in hich nodes 1, 2 and 3 choose t 4, t 3 and t 3 respectively and, as before, each source node chooses the unique product in its product set. By definition, the payoff of each player depends only on the strategies chosen by his neighbours, so the social netork games are related to graphical games of [8]. Hoever, the underlying dependence structure of a social netork game is a directed graph. Further, note that these games satisfy the join the crod property that e define as follos: Each payoff function p i depends only on the strategy chosen by player i and the set of players ho also chose his strategy. Moreover, the dependence on this set is monotonic. The last qualification is exactly opposite to the definition of congestion games ith player-specific payoff functions of [9] in hich the dependence on the above set is antimonotonic. That is, hen more players choose the strategy of player i, then his payoff eakly decreases. 3 Nash equilibria The first natural question e address is hether the social netork games have a Nash equilibrium.

5 184 Social Netork Games ith Obligatory Product Selection 3.1 Simple cycles In contrast to the case of games studied in [14] the anser is negative already for the case hen the underlying graph is a simple cycle. Example 3. Consider the netork given in Figure 2, here the product set of each agent is marked next to the node denoting it and the eights are all equal and put as labels on the edges. 1 {t 1,t 2 } {t 3,t 1 } {t 2,t 3 } 3 2 Figure 2: A simple cycle no Nash equilibrium Let the thresholds be defined as follos: θ(1,t 1 )=θ(2,t 2 )=θ(3,t 3 )=r 1 and θ(1,t 2 )=θ(2,t 3 )= θ(3,t 1 )=r 2 here r 1 > r 2. We also assume that >r 1 r 2. Hence for all s 2 and s 3 p 1 (t 1,s 2,t 1 )> p 1 (t 2,s 2,s 3 )> p 1 (t 1,s 2,t 3 ) and similarly for the payoff functions p 2 and p 3. So it is more profitable for player i to adopt strategy t i provided its neighbour also adopts t i. It is easy to check that the game associated ith this netork has no Nash equilibrium. Indeed, here is the list of all the joint strategies, here e underline the strategy that is not a best response to the choice of other players: (t 1,t 2,t 1 ),(t 1,t 2,t 3 ),(t 1,t 3,t 1 ),(t 1,t 3,t 3 ),(t 2,t 2,t 1 ),(t 2,t 2,t 3 ),(t 2,t 3,t 1 ),(t 2,t 3,t 3 ). This example can be easily generalized to the case of an arbitrary simple cycle. Belo, i 1 and i 1 stand for addition and subtraction defined cyclically over the set {1,...,n}. So n 1=1 and 1 1=n. Indeed, consider a social netork ith n nodes that form a simple cycle and assume that each player i has strategies t i and t i 1. Choose for each player i the eights i 1 i and the threshold function θ(i,t) so that i 1 i θ(i,t i )> θ(i,t i 1 )> θ(i,t i ), so that (e put on first to positions, respectively, the strategies of players i 1 and i, hile the last argument is a joint strategy of the remaining n 2 players) p i (t i,t i,s)> p i (t,t i 1,s )> p i (t i 1,t i,s ), here t,s,s and s are arbitrary. It is easy to check then that the resulting social netork game has no Nash equilibrium. A natural question is hat is the complexity of determining hether a Nash equilibrium exists. First e consider this question for the special case hen the underlying graph is a simple cycle. Theorem 4. Consider a netork S hose underlying graph is a simple cycle. It takes O(n P 4 ) time to decide hether the game G(S) has a Nash equilibrium. Proof. Suppose S =(G,P,P,θ). When the underlying graph of S is a simple cycle, the concept of a best response of player i 1 to a strategy of player i is ell-defined. Let R i :={(t i,t i 1 ) t i P(i),t i 1 P(i 1),t i 1 is a best response to t i },

6 Krzysztof R. Apt & Sunil Simon 185 I :={(t,t) t P}, and let stand for the composition of binary relations. The question hether G(S) has a Nash equilibrium is then equivalent to the problem hether there exists a sequence a 1,...,a n such that (a 1,a 2 ) R 1,...,(a n 1,a n ) R n 1,(a n,a 1 ) R n. In other ords, is (R 1 R n ) I non-empty? To anser this question e first construct successively n 1 compositions R 1 R 2,(R 1 R 2 ) R 3,..., (...(R 1 R 2 ) R n 1 ) R n. Each composition construction can be carried out in P 4 steps. Indeed, given to relations A,B P P, to compute their composition A B requires for each pair (a,b) A to find all pairs (c,d) B such that b=c. Finally, to check hether the intersection of R 1 R n ith I is non-empty requires at most P steps. So to anser the original question takes O(n P 4 ) time. Note that this proof applies to any strategic game in hich there is a reordering of players π(1),...,π(n) such that the payoff of player π(i) depends only on his strategy and the strategy chosen by player π(i i). It is orthhile to note that for the case of simple cycles, the existence of Nash equilibrium in the associated social netork game does not imply that the game is eakly acyclic. 1 {t 1,t 2,t 4 } {t 3,t 1,t 4 } {t 2,t 3,t 4 } 3 2 (a) (t 1,t 3,t 1 ) (t 1,t 3,t 3 ) (t 2,t 3,t 3 ) (t 1,t 2,t 1 ) (t 2,t 2,t 1 ) (t 2,t 2,t 3 ) (b) Figure 3: A simple cycle and an infinite improvement path Example 5. Consider the netork in Figure 3(a) hich is a modification of the netork in Figure 2. We add a ne product t 4 to the product set of all the nodes i ith θ(i,t 4 ) > r 1. We also assume that θ(i,t 4 )> r 2. Then the joint strategy (t 4,t 4,t 4 ) is a Nash equilibrium. Hoever, Figure 3(b) shos the unique improvement path starting in (t 1,t 3,t 1 ) hich is infinite. For each joint strategy in the figure, e underline the strategy that is not a best response. This shos that the game is not eakly acyclic. In Section 4 e shall study the complexity of checking hether a social netork game is eakly acyclic. 3.2 Arbitrary social netorks In this section e establish to results hich sho that deciding hether a social netork has a Nash equilibrium is computationally hard. Theorem 6. Deciding hether for a social netork S the game G(S) has a Nash equilibrium is NPcomplete. To prove the result e first construct another example of a social netork game ith no Nash equilibrium and then use it to determine the complexity of the existence of Nash equilibria. Example 7. Consider the netork given in Figure 4, here the product set of each agent is marked next to the node denoting it and the eights are labels on the edges. Nodes ith a unique product in the product set is simply represented by the product.

7 186 Social Netork Games ith Obligatory Product Selection {t 1 } {t 2 } 1 {t 1 1,t 2 } 2 2 {t 2,t 3 } {t 1,t 3 } {t 3 } Figure 4: A netork ith no Nash equilibrium We assume that each threshold is a constant θ, here θ < 1 < 2. So it is more profitable to a player residing on a triangle to adopt the product adopted by his neighbour residing on a triangle than by the other neighbour. The game associated ith this netork has no Nash equilibrium. It suffices to analyze the joint strategies involving nodes 1, 2 and 3 since the other nodes have exactly one product in their product sets. Here e provide a listing of all such joint strategies, here e underline the strategy that is not a best response to the choice of other players: (t 1,t 1,t 2 ),(t 1,t 1,t 3 ),(t 1,t 3,t 2 ),(t 1,t 3,t 3 ),(t 2,t 1,t 2 ),(t 2,t 1,t 3 ), (t 2,t 3,t 2 ), (t 2,t 3,t 3 ). In contrast, hat ill be of relevance in a moment, if e replace {t 1 } by {t 1 }, then the corresponding game has a Nash equilibrium, namely the joint strategy corresponding to the triple (t 2,t 3,t 3 ). Proof of Theorem 6: As in [1], to sho NP-hardness, e use a reduction from the NP-complete PAR- TITION problem, hich is: given n positive rational numbers (a 1,...,a n ), is there a set S such that i S a i = i S a i? Consider an instance I of PARTITION. Without loss of generality, suppose e have normalised the numbers so that n i=1 a i = 1. Then the problem instance sounds: is there a set S such that i S a i = i S a i = 1 2? To construct the appropriate netork e employ the netorks given in Figure 4 and in Figure 5, here for each node i {1,...,n} e set ia = ib = a i, and assume that the thresholds of the nodes a and b are constant and equal 1 2. {t 1,t 1 } {t 1,t a 1b } {t 1,t 1 } n 2b nb 2a {t 1,t 2 1 } a na b {t 1,t 1 } Figure 5: A netork related to the PARTITION problem To finalize the construction e use to copies of the netork given in Figure 4, one unchanged and the other in hich the product t 1 is replaced everyhere by t 1, and construct the desired netork S by identifying ith the node marked by {t 1 } in the netork from Figure 4, the node a of the netork from Figure 5 and ith the node marked by{t 1 } in the modified version of the netork from Figure 4 the node b. Suppose that a solution to the considered instance of the PARTITION problem exists, i.e., for some set S {1,...,n} e have i S a i = i S a i = 1 2. Consider the game G(S) and the joint strategy s formed by the folloing strategies: t 1 assigned to each node i S in the netork from Figure 5, t 1 assigned to each node i {1,...,n}\S in the netork from Figure 5,

8 Krzysztof R. Apt & Sunil Simon 187 t 1 assigned to the nodes a and t 1 to the node b, t 2 assigned to node 1 and t 3 assigned to the nodes 2, 3 in both versions of the netorks from Figure 4, t 2 and t 3 assigned respectively to the nodes marked by {t 2 } and {t 3 }. We claim that s is a Nash equilibrium. Consider first the player (i.e., node) a. The accumulated eight of its neighbours ho chose strategy t 1 is 1 2. Therefore, the payoff for a in the joint strategy s is 0. The accumulated eight of its neighbours ho chose strategy t 1 is 1 2, as ell. Therefore t 1 is indeed a best response for player a as both strategies yield the same payoff. For the same reason, t 1 is a best response for player b. The analysis for the other nodes is straightforard. Conversely, suppose that a strategy profile s is a Nash equilibrium in G(S). From Example 7 it follos that s a = t 1 and s b = t 1. This implies that t 1 is a best response of node a to s a and therefore i {1,...,n} si =t 1 ia i {1,...,n} si =t 1 ia. By a similar reasoning, for node b e have i {1,...,n} si =t 1 ib i {1,...,n} si =t 1 ib. Since n i=1 a i = 1 and for i {1,...,n}, ia = ib = a i, and s i {t 1,t 1 } e have for S :={i {1,...,n} s i = t 1 }, i S a i = i S a i. In other ords, there exists a solution to the considered instance of the partition problem. Theorem 8. For a netork S hose underlying graph has no source nodes, deciding hether the game G(S) has a Nash equilibrium is NP-complete. Proof. The proof extends the proof of the above theorem. Given an instance of the PARTITION problem e use the folloing modification of the netork. We tin each node i {1,..., n} in Figure 5 ith a ne node i ith the product set{t 1,t 1 }, by adding the edges(i,i ) and(i,i). We also tin nodes marked {t 2 } and {t 3 } in Figure 4 ith ne nodes ith the product set {t 2 } and {t 3 } respectively. Additionally, e choose the eights on the ne edges ii, i i and the corresponding thresholds so that hen i and i adopt a common product, their payoff is positive. Then the underlying graph of the resulting netork does not have any source nodes and the above proof remains valid for this ne netork. 4 Weakly acyclic games In this section e study the complexity of checking hether a social netork game is eakly acyclic. We establish to results that are analogous to the ones established in [15] for the case of social netorks in hich the nodes may decide not to choose any product. The proofs are based on similar arguments though the details are different. Theorem 9. For an arbitrary netork S, deciding hether the game G(S) is eakly acyclic is co-np hard. Proof. We again use an instance of the PARTITION problem in the form of n positive rational numbers (a 1,...,a n ) such that n i=1 a i = 1. Consider the netork given in Figure 6. For each node i {1,...,n} e set P(i) ={t 1,t 2 }. The product set for the other nodes are marked in the figure. As before, e set ia = ib = a i. Since for all i {1,...,n}, a i is rational, it has the form a i = l i 1 r i. Let τ = 2 r 1... r n. The folloing property holds. Property 1. Given an instance (a 1,...,a n ) of the PARTITION problem and τ defined as above, for all S {1,...,n}

9 188 Social Netork Games ith Obligatory Product Selection 1 2 n {t 1,t 4 } a b {t 2,t 5 } 1 {t 1,t 2 } c 2 1 {t 2,t 3 } d 2 2 e{t 1,t 3 } 1 {t g 3 } Figure 6: A netork related to eakly acyclic games (i) if i S a i < 1 2, then i S a i 1 2 τ, (ii) if i S a i > 1 2, then i S a i τ. Proof. By definition, each a i and 1 2 is a multiple of τ. Thus i S a i = x τ and 1 2 = y τ here x and y are integers. (i) If x τ < y τ, then x τ (y 1) τ. Therefore i S a i 1 2 τ. The proof of (ii) is analogous. Note that given (a 1,...,a n ), τ can be defined in polynomial time. Let the thresholds be defined as follos: θ(a,t 1 ) = θ(b,t 2 )= 1 2 and 0<θ(a,t 4)=θ(b,t 5 ) < τ. The threshold for nodes c,d and e is a constant θ 1 such that θ 1 < 1 < 2. Thus, like in the netork in Figure 4, it is more profitable to a player residing on a triangle to adopt the product adopted by his neighbour residing on a triangle than by the other neighbour. Suppose that a solution to the considered instance of the PARTITION problem exists. That is, for some set S {1,...,n} e have i S a i = i S a i = 1 2. In the game G(S), take the joint strategy s formed by the folloing strategies: t 1 assigned to each node i S and the nodes a and c, t 2 assigned to each node i {1,...,n}\S and the nodes b and d, t 3 assigned to the nodes e and g. Any improvement path that starts in this joint strategy ill not change the strategies assigned to the nodes a, b and g. So if such an improvement path terminates, it produces a Nash equilibrium in the game associated ith the netork given in Figure 4 of Example 7. But e argued that this game does not have a Nash equilibrium. Consequently, there is no finite improvement path in the game G(S) that starts in the above joint strategy and therefore G(S) is not eakly acyclic. No suppose that the considered instance of the PARTITION problem does not have a solution. Then e sho that the game G(S) is eakly acyclic. To this end, e order the nodes of S as follos (note the positions of the nodes c,d and e): 1,2,...,n,g,a,b,c,e,d. Given a joint strategy, consider an improvement path in hich at each step the first node in the above list that did not select a best response

10 Krzysztof R. Apt & Sunil Simon 189 sitches to a best response. After at most n steps the nodes 1,2,...,n all selected a product t 1 or t 2. Let s be the resulting joint strategy. First suppose that i {1,...,n} s i =t 1 ia > 1 2. This implies that i {1,...,n} s i =t 2 ib < 1 2. By Property 1, i {1,...,n} s i =t 2 ib 1 2 τ. The payoff of the node b depends only on the choices made by the source nodes 1,2,...,n, so e have p b (t 2,s b ) τ. Since θ(b,t 5 ) < τ, e also have p b (t 5,s b ) > τ and therefore t 5 is a best response for node b. Let s b be the resulting strategy in hich node b selects t 5. Consider the prefix of ξ starting at s b (call it ξ b ). We argue that in ξ b, t 2 is never a better response for node d. Suppose that s b d = t 3. We have the folloing to cases: s b e = t 3: then p d (s b )= 2 θ 1 and so t 3 is the best response for node d. s b e = t 1: then p d (s b ) = θ 1 and if node d sitches to t 2 then p d (t 2,s b b ) = θ 1 (since s b b = t 5). Thus t 2 is not a better response. Using the above observation, e conclude that there exists a suffix of ξ b (call it ξ d ) such that node d never chooses t 2. This means that in ξ d the unique best response for node c is t 1 and for node e is t 1. This shos that ξ d is finite and hence ξ is finite, as ell. The case hen i {1,...,n} s i =t 2 ib > 1 2 is analogous ith all improvement paths terminating in a joint strategy here node a chooses t 4 and node c chooses t 2. Theorem 10. For a netork S hose underlying graph has no source nodes, deciding hether the game G(S) is eakly acyclic is co-np hard. Proof. The proof extends the proof of the above theorem. Given an instance of the PARTITION problem e use the folloing modification of the netork given in Figure 6. We tin each node i {1,...,n} ith a ne node i, also ith the product set {t 1,t 2 }, by adding the edges (i,i ) and (i,i). We also tin the node g ith a ne node g, also ith the product set {t 3 }, by adding the edges (g,g ) and (g,g). Additionally, e choose the eights ii and i i and the corresponding thresholds so that hen i and i adopt a common product, their payoff is positive. Suppose that a solution to the considered instance of the PARTITION problem exists. Then e extend the joint strategy considered in the proof of Theorem 9 by additionally assigning t 1 to each node i such that i S, t 2 to each node i such that i {1,...,n}\S and t 3 to the node g. Then, as before, there is no finite improvement path starting in this joint strategy, so G(S) is not eakly acyclic. Suppose no that no solution to the considered instance of the PARTITION problem exists. Take the folloing order of the nodes of S : 1,1,2,2,...,n,n,g,g,a,b,c,e,d, and as in the previous proof, given a joint strategy, e consider an improvement path ξ in hich at each step the first node in the above list that did not select a best response sitches to a best response. Note that each node from the list 1,1,2,2,...,n,n,g,g is scheduled at most once. So there exists a suffix of ξ in hich only the nodes a,b,c,e,d are scheduled. Using no the argument given in the proof of Theorem 9 e conclude that there exists a suffix of ξ that is finite. This proves that G(S) is eakly acyclic. 5 Paradoxes In [2] e identified various paradoxes in social netorks ith multiple products and studied them using the social netork games introduced in [14]. Here e carry out an analogous analysis for the case hen

11 190 Social Netork Games ith Obligatory Product Selection the product selection is obligatory. This qualification, just like in the case of social netork games, substantially changes the analysis. We focus on the main four paradoxes that e successively introduce and analyze. 5.1 Vulnerable netorks The first one is the folloing. We say that a social netork S is vulnerable if for some Nash equilibrium s in G(S), an expansion S of S exists such that each improvement path in G(S ) leads from s to a joint strategy s hich is a Nash equilibrium both in G(S ) and G(S) such that s > s. So the nely added product triggers a sequence of changes that unavoidably move the players from one Nash equilibrium to another one that is strictly orse for everybody. The folloing example shos that vulnerable netorks exist. Here and elsehere the relevant expansion is depicted by means of a product and the dotted arro pointing to the relevant node. Example 11. Consider the directed graph given in Figure 7, in hich the product set of each node is marked next to it. 1 {t 1,t 3,t 4 } 2 {t 1,t 4 } t 2 {t 2,t 3 } 3 {t 2,t 3 } 4 Figure 7: A directed graph We complete it to the desired social netork belo. Let _ stand for an arbitrary strategy of the relevant player. We stipulate that p 2 (_,t 2,_,t 2 )> p 2 (t 1,t 1,_,_), p 1 (t 3,t 2,_,_)> p 1 (t 1,t 2,_,_)> p 1 (t 4,t 2,_,_), p 3 (t 3,_,t 3,_)> p 3 (_,_,t 2,t 2 ), p 4 (_,_,t 3,t 3 )> p 4 (_,_,t 3,t 2 ), p 2 (_,t 4,_,_)> p 2 (_,t 2,_,t 3 ), p 1 (t 4,t 4,_,_)> p 1 (t 3,_,_,_)> p 1 (t 1,t 4,_,_), so that 2 : t 2,1 : t 3,3 : t 3,4 : t 3,2 : t 4,1 : t 4 is a unique improvement path that starts in(t 1,t 1,t 2,t 2 ) and ends in (t 4,t 4,t 3,t 3 ). Additionally e stipulate that p 1 (t 1,t 1,_,_)> p 1 (t 4,t 4,_,_), p 2 (t 1,t 1,_,_)> p 2 (t 4,t 4,_,_), p 3 (_,_,t 2,t 2 )> p 3 (_,_,t 3,t 3 ), p 4 (_,_,t 2,t 2 )> p 4 (_,_,t 3,t 3 ), so that (t 1,t 1,t 2,t 2 )> s (t 4,t 4,t 3,t 3 ). These requirements entail constraints on the eights and thresholds that are for instance realized by 12 = 0, 21 = 0.2, 42 = 0.3, 13 = 0.2, 34 = 0.2, 43 = 0, and θ(1,t 1 )=0.2, θ(1,t 3 )=0.1, θ(1,t 4 )=0.3, θ(2,t 1 )=0.1, θ(2,t 2 )=0.3, θ(2,t 4 )=0.2, θ(3,t 2 )=0.1, θ(3,t 3 )=0.2, θ(4,t 2 )=0.1, θ(4,t 3 )=0.2.

12 Krzysztof R. Apt & Sunil Simon 191 It is useful to note that in the setup of [2], in hich for each player the abstain strategy is alloed, it remains an open problem hether vulnerable netorks (called there because of various other alternatives s-vulnerable netorks) exist. 5.2 Fragile netorks Next, e consider the folloing notion. We say that a social netork S is fragile if G(S) has a Nash equilibrium hile for some expansion S of S, G(S ) does not. The folloing example shos that fragile netorks exist. Example 12. Consider the netork S given in Figure 8, here the product set of each node is marked next to it. {t 2 } 1 t 1 {t 3,t 1 } {t 2,t 3 } 3 2 Figure 8: A fragile netork Let the thresholds be defined as follos: θ(2,t 2 )=θ(3,t 3 )=r 1 and θ(1,t 2 )=θ(2,t 3 )=θ(3,t 1 )=r 2 here r 1 > r 2. We also assume that >r 1 r 2. Consider the joint strategy s, in hich nodes 1, 2 and 3 choose t 2, t 2 and t 1 respectively. It can be verified that s is a Nash equilibrium in G(S). No consider the expansion S of S in hich product t 1 is added to the product set of node 1 and let θ(1,t 1 )=r 1. Then S is the netork in Example 3 hich, as e sa, does not have a Nash equilibrium. 5.3 Inefficient netorks We say that a social netork S is inefficient if for some Nash equilibrium s in G(S), a contraction S of S exists such that each improvement path in G(S ) starting in s leads to a joint strategy s hich is a Nash equilibrium both in G(S ) and G(S) such that s > s. We note here that if the contraction as created by removing a product from the product set of node i, e impose that any improvement path in G(S ), given a starting joint strategy from G(S), begins by having node i making a choice (e allo any choice from his remaining set of products as an improvement move). Otherise the initial payoff of node i in G(S ) is not ell-defined. Example 13. We exhibit in Figure 9 an example of an inefficient netork. The eight of each edge is assumed to be, and e also have the same product-independent threshold, θ, for all nodes, ith >θ. Consider as the initial Nash equilibrium the joint strategy s=(t 2,t 2,t 1,t 1 ). It is easy to check that this is indeed a Nash equilibrium, ith the payoff equal to θ for all nodes. Suppose no that e remove product t 1 from the product set of node 3. We claim that the unique improvement path then leads to the Nash equilibrium in hich all nodes adopt t 2. To see this, note that node 3 moves first in any improvement path and it has a unique choice, t 2. Then node 4 moves and necessarily sitches to t 2. This yields a Nash equilibrium in hich each node selected t 2 ith the payoff of 2 θ, hich is strictly better than the payoff in s.

13 192 Social Netork Games ith Obligatory Product Selection {t 2 } {t 2 } {t 1,t 2 } {t 1,t 2 } 4 Figure 9: An example of an inefficient netork 5.4 Unsafe netorks Finally, e analyze the folloing notion. We call a social netork S unsafe if G(S) has a Nash equilibrium, hile for some contraction S of S, G(S ) does not. The folloing example shos that unsafe netorks exist. Example 14. Let S 1 be the modification of the netork S given in Figure 2 here node 1 has the product set {t 1,t 2,t 4 }, here θ(1,t 4 ) < r 2. Then the joint strategy (t 4,t 3,t 3 ) is a Nash equilibrium in G(S 1 ). No consider the contraction S 1 of S 1 here product t 4 is removed from node 1. Then S 1 is the netork S, hich as e sa in Example 3 has no Nash equilibrium. 6 Conclusions In this paper e studied dynamic aspects of social netorks ith multiple products using the basic concepts of game theory. We used the model of social netorks, originally introduced in [1] that e subsequently studied using game theory in [14], [15] and [2]. Hoever, in contrast to these three references the product adoption in this paper is obligatory. This led to some differences. For example, in contrast to the case of [14], a Nash equilibrium does not need to exist hen the underlying graph is a simple cycle. Further, in contrast to the setup of [2], e ere able to construct a social netork that exhibits the strongest form of the paradox of choice. On the other hand, some complexity results, namely the ones concerning eakly acyclic games, remain the same as in [14], though the proofs had to be appropriately modified. References [1] K. R. Apt & E. Markakis (2011): Diffusion in Social Netorks ith Competing Products. In: Proc. 4th International Symposium on Algorithmic Game Theory (SAGT11), Lecture Notes in Computer Science 6982, Springer, pp , doi: / _20. [2] K. R. Apt, E. Markakis & S. Simon (2013): Paradoxes in Social Netorks ith Multiple Products. Manuscript, CWI, Amsterdam, The Netherlands. Computing Research Repository (CoRR), org/abs/ [3] David Easley & Jon Kleinberg (2010): Netorks, Crods, and Markets. Cambridge University Press. [4] D. Fotakis, A. C. Kaporis, T. Lianeas & P. G. Spirakis (2012): On the Hardness of Netork Design for Bottleneck Routing Games. In: SAGT, pp , doi: / _14. [5] D. Fotakis, A. C. Kaporis & P. G. Spirakis (2012): Efficient methods for selfish netork design. Theor. Comput. Sci. 448, pp. 9 20, doi: /j.tcs [6] G. Gigerenzer (2008): Gut Feelings: The Intelligence of the Unconscious. Penguin. Reprint edition.

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