Parameterized Complexity of Immunization in the Threshold Model
PParameterized Complexity of Immunization in the Threshold Model
Gennaro CordascoDepartment of Psychology,University of Campania “L.Vanvitelli”, Italy Luisa GarganoDepartment of Computer Science,University of Salerno, ItalyAdele Anna RescignoDepartment of Computer Science,University of Salerno, Italy
Abstract
We consider the problem of controlling the spread of harmful items in networks, such as the conta-gion proliferation of diseases or the diffusion of fake news. We assume the linear threshold model ofdiffusion where each node has a threshold that measures the node resistance to the contagion. We studythe parameterized complexity of the problem: Given a network, a set of initially contaminated nodes,and two integers k and (cid:96) , is it possible to limit the diffusion to at most k other nodes of the networkby immunizing at most (cid:96) nodes? We consider several parameters associated to the input, including: thebounds k and (cid:96) , the maximum node degree ∆ , the treewidth, and the neighborhood diversity of the net-work. We first give W [1] or W [2] -hardness results for each of the considered parameters. Then we givefixed-parameter algorithms for some parameter combinations. Keywords:
Parameterized Complexity, Contamination minimization, Threshold model
The problem of controlling the spread of harmful items in networks, such as the contagion proliferation ofdiseases or the diffusion of fake news, has recently attracted much interest from the research community.The goal is to try to limit as much as possible the spreading process by adopting immunization measures.One such a measure consists in intervening on the network topology either blocking some links so that theycannot contribute to the diffusion process [28] or by immunizing some nodes [14]. In this paper we focus onthe second strategy: Limit the spread to a small region of the network by immunizing a bounded number ofnodes in the network. We study the problem in the linear threshold model where each node has a threshold,measuring the node resistance to the diffusion [27]. A node gets influenced/contaminated if it receives theitem from a number of neighbors at least equal to its threshold. The diffusion proceeds in rounds: Initiallyonly a subset of nodes has the item and is contaminated. At each round the set of contaminated nodesis augmented with each node that has a number of already contaminated neighbors at least equal to itsthreshold.In the presence of an immunization campaign, the immunization operation on a node inhibits the con-tamination of the node itself. Thus, given a network and a subset of its nodes, called spreader set, that has1 a r X i v : . [ c s . CC ] F e b he malicious item to be diffused to the other nodes in the network, at each round the set of contaminatednodes is augmented only with the nodes for which the number of already contaminated neighbors is at leastequal to the node threshold.Under this diffusion model, we perform a broad parameterized complexity study of the following prob-lem: Given a network, a spreader set, and two integers k and (cid:96) , is it possible to limit the diffusion to at most k other nodes of the network by immunizing at most (cid:96) nodes? During the past decade the study of spreading processes in complex networks have experienced a particularsurge of interest across many research areas from viral marketing, to social media, to population epidemics.Several studies have focused on the problem of finding a small set of individuals who, given the item tobe diffused, allow its diffusion to a vast portion of the network, by using the links among individuals inthe network to transmit the item itself to their contacts [32]. Threshold models are widely adopted bysociologists to describe collective behaviours [24] and their use to study of the propagation of innovationsthrough a network was first considered in [27]. The linear threshold model has then been widely used in theliterature to study the problem of influence maximization, which aims at identifying a small subset of nodesthat can maximize the influence diffusion [4, 6, 7, 9, 13, 27].Recently, some attention has been devoted to the important issue of developing strategies for reducingthe spread of negative things through a network. In particular several studies considered the problem ofwhat structural changes can be made to the network topology in order to block negative diffusion processes.Contamination minimization in linear threshold model by blocking some links has been studied in [16, 28].Strategies for reducing the spread size by immunizing/removing nodes has been considered in several paper.As an example [2, 33] consider a greedy heuristic that immunize nodes in decreasing order of out-degree.When all the node thresholds are 1, the immunization can be obtained by a (multi)cut of the network.Some papers dealing with this problem are [5, 25, 26] in case of edge cuts and [19] in case of node cuts.
Parameterized complexity is a refinement to classical complexity theory in which one takes into accountnot only the input size, but also other aspects of the problem given by a parameter p . We recall that aproblem with input size n and parameter p is called fixed parameter tractable (FPT) if it can be solved intime f ( p ) · n c , where f is a computable function only depending on p and c is a constant.We study the parameterized complexity of the studied problem, formally defined in Section 2. Weconsider several parameters associated to the input: the bounds k and (cid:96) , the number ζ related to ini-tially contaminated nodes, and some parameters of the underlying network: The maximum degree ∆ ,the treewidth tw [35], and the neighborhood diversity nd [31]. The two last parameters, formally de-fined in Sections 3.4 and 3.5 respectively, are two incomparable parameters of a graph that can be viewedas representing sparse and dense graphs respectively [31]; they received much attention in the literature[1, 3, 4, 7, 8, 10, 13, 18, 23, 20, 21, 30]. 2 v v v v v v
02 13 02 2 (a) D G [1] = { v , v } , D G [2] = { v , v , v } ,D G [3] = { v , v , v , v , v , v } , D G [4] = D G = V (b) X = { v , v , v } , D G [ X ] = { v , v } (cid:1) = XD G, { v } = { v , v } (c) X ′ = { v , v } , D G [ X ′ ] = X ′ = { v , v } , Y ( X ′ ) = { v } Figure 1:
A graph G (node thresholds appear in red). (a) The diffusion process in G . (b) An example of X whose G [ X ] includes nodes not influeced. (c) An example of immunizing set Y ( X (cid:48) ) = { v } , which enables to confine thediffusion to X (cid:48) = { v , v } . In Section 2, we formally define the studied immunization problem and summarize our findings. In Section3, we give hardness results for the considered parameters. In Section 4, we give fixed-parameter algorithmsfor some parameter combinations.
Denote by G = ( V, E, t ) a undirected graph where V is the nodes set, E is the set of edges, and t : V → N is a node threshold function. We use n and m to denote the number of nodes and edges in the graph,respectively. The degree of a node v is denoted by d G ( v ) . The neighborhood of v is denoted by Γ G ( v ) = { u ∈ V | ( u, v ) ∈ E } . In general, the neighborhood of a set V (cid:48) ⊆ V is denoted by Γ G ( V (cid:48) ) = { u ∈ V | ( u, v ) ∈ E, v ∈ V (cid:48) , u / ∈ V (cid:48) } . The graph induced by a node set V (cid:48) in G is denoted G [ V (cid:48) ] = ( V (cid:48) , E (cid:48) , t (cid:48) ) where E (cid:48) = { ( u, v ) : u, v ∈ V (cid:48) , ( u, v ) ∈ E } and t (cid:48) ( v ) = t ( v ) for each v ∈ V (cid:48) .Given the network and a spreader set S , after one diffusion round, the influenced nodes are all thosewhich are influenced by the nodes in S , that is, have a number of neighbors in S at least equal to theirthreshold. Noticing that nodes in S are already contaminated and cannot be immunized, we can then modelthe diffusion process as in a graph which represents the network except the spreader set. Namely, weconsider the graph G = ( V, E, t ) where: V is the set of nodes of the network excluding those in thespreader set, E ⊆ V × V is the edge set, and t is the threshold function t : V → N with t ( v ) equal to theoriginal threshold of the node v in the network decreased by the number of its neighbors in S . Definition 1.
The diffusion process in G = ( V, E, t ) in the presence of a set Y ⊆ V of immunized nodes isa sequence of node subsets D G,Y [1] ⊆ . . . ⊆ D G,Y [ τ ] ⊆ . . . ⊆ V with– D G,Y [1] = { u | u ∈ V − Y, t ( u ) = 0 } , and– D G,Y [ τ ] = D G,Y [ τ − ∪ (cid:110) u | u ∈ V − Y, (cid:12)(cid:12) Γ G ( u ) ∩ D G [ τ − (cid:12)(cid:12) ≥ t ( u ) (cid:111) .The process ends at τ ∗ such that D G,Y [ τ ∗ ] = D G [ τ ∗ + 1] . We set D G,Y = D G,Y [ τ ∗ ] . We omit the subscript Y when no node is immunized, that is, D G = D G, ∅ . Moreover, we assumethat for the input graph it holds D G = V ; indeed, we could otherwise remove all the nodes that cannot beinfluenced, since they are irrelevant to the immunization problem. In particular, each remaining node v ∈ V t ( v ) ≤ d G ( v ) , otherwise it could not be influenced. An example is given in Fig.1 (a). We are nowready to formally define our problem. I NFLUENCE -I MMUNIZATION B OUNDING ( IIB ) : Given a graph G = ( V, E, t ) and bounds k and (cid:96) , is there a set Y such that | Y | ≤ (cid:96) and | D G,Y | ≤ k ?For a given set Y we are partitioning the nodes into three subsets: The set D G,Y which contains the nodesthat get influenced, the immunizing set Y , which has the property that, if all its nodes are immunized thenthe diffusion process is circumscribed to D G,Y , and the set V − Y − D G,Y of the nodes that, by immunizing Y , are not influenced.We will refer to the nodes in the above subsets as influenced, immunized and safe , respectively.In some cases it will be easier to deal with a different formulation of IIB that starts from the set of nodesto which one wants to confine the diffusion. Given a set X ⊆ V , we define the immunizing set Y ( X ) of X as the set that contains all the nodes in V − X that can be influenced in one round by those in D G [ X ] , thatis, the nodes that get influenced in X when X is isolated from the rest of the graph, namely Y ( X ) = { u | u ∈ V − X, | Γ G ( u ) ∩ D G [ X ] | ≥ t ( u ) } . (1)By the above definitions, we have D G [ X ] = D G,Y ( X ) = D G [ V − Y ( X )] ⊆ X ; (2)hence, the influenced, immunized and safe node sets are D G [ X ] , Y ( X ) , V − Y ( X ) − D G [ X ] .For some X , some nodes in G [ X ] may be not influenced, even though they would in the whole graph G (seeFig.1 (b)). However, it is easy to see that for each X the set X (cid:48) = D G [ X ] ⊆ X is such that D G [ X (cid:48) ] = X (cid:48) and Y ( X (cid:48) ) = { u | u ∈ V − X (cid:48) , | Γ G ( u ) ∩ D G [ X (cid:48) ] | ≥ t ( u ) } = Y ( X ) . In the following, we will refer as minimal to a set X such that D G [ X ] = X (see Fig.1 (c)). Fact 1. (IIB equivalent) (cid:104)
G, k, (cid:96) (cid:105) is a
YES instance iff there is a minimal X ⊆ V s.t. | X | = | D G [ X ] | ≤ k and | Y ( X ) | ≤ (cid:96). (3) In this paper we prove that I
NFLUENCE -I MMUNIZATION B OUNDING is:i) W[1]-hard with respect to any of the parameters k , tw or nd ii) W[2]-hard with respect to the pairs ( (cid:96) , ∆ ), or ( (cid:96), ζ ) ;iii) FPT with respect to any of the pairs ( k, (cid:96) ) , ( k, ζ ) , ( k, tw ) , (∆ , tw ) , ( k, nd ) , ( (cid:96), nd ) ,where tw and nd denote the tree width and the neighborhood diversity of the input graph and ζ = |{ v | v ∈ V, t ( v ) = 0 }| is the number of nodes with threshold 0.4 Hardness
In this section we give W [1] or W [2] hardness results for the considered parameters. k Theorem 1.
IIB is W [1] -hard with respect to k .Proof. We give a reduction from the
CUTTING AT MOST k VERTICES WITH TERMINAL (CVT- k ) problemstudied in [19]: Given a graph H = ( V ( H ) , E ( H )) , s ∈ V ( H ) , and two integers k and (cid:96) , is there a set X H ⊆ V ( H ) such that s ∈ X H , | X H | ≤ k , and | Γ H ( X H ) | ≤ (cid:96) ? To this aim, construct the instance (cid:104)
G, k − , (cid:96) (cid:105) of IIB where G = H [ V ( H ) − { s } ] and t ( v ) = 0 foreach node v ∈ Γ H ( s ) and t ( v ) = 1 for each node v ∈ V ( H ) − { s } − Γ H ( s ) .Suppose (cid:104) G, k − , (cid:96) (cid:105) admits a solution. By (3), there exists a minimal set X such that | X | = | D G [ X ] | ≤ k − and | Y ( X ) | ≤ (cid:96) . Noticing that Γ H ( s ) ⊆ X ∪ Y ( X ) , one gets that for X H = X ∪ { s } it holds Γ H ( X ∪ { s } ) = Y ( X ) . Hence X H = X ∪ { s } satisfies the inequalities | X H | ≤ k and | Γ H ( X H ) | ≤ (cid:96) andis a solution to CVT- k .Suppose now X H = X ∪ { s } is a minimum size solution to CVT- k . Then H [ X H ] is connected,otherwise the connected component containing s would be a smaller solution. Recalling that in G all thresh-olds are at most 1, we have that all the nodes in the connected component of a node with threshold 0 getinfluenced. Hence, Y ( X ) = { u | u ∈ V − X, | Γ G ( u ) ∩ D G [ X ] | ≥ t ( u ) } = { u | u ∈ V − X, t ( u ) = 0 } ∪ { u | u ∈ V − X, | Γ G ( u ) ∩ X | ≥ } = Γ H ( { s } ∪ X ) . As a consequence, X is a solution to IIB. The theorem follows, since Theorem 3 in [19] proves that thelatter problem is W [1] -hard whit respect to k .The same reduction, recalling that Theorem 5 in [19] proves that CVT- k is W [1] -hard with respect to (cid:96) ,also gives that IIB is W [1] -hard with respect to (cid:96) ; however, a stronger result is given in the next section. ζ and (cid:96) Theorem 2.
IIB is W [2] -hard with respect to the pair of parameters ζ , the number of nodes with threshold0, and (cid:96) .Proof. We give a reduction from H
ITTING S ET (HS), which is W [2] -complete in the size of the hitting set: Given a collection { S , . . . , S m } of subsets of a set A = { a , . . . , a n } and an integer h > , is there a set H ⊆ A such that H ∩ S i (cid:54) = ∅ , for each i ∈ [ m ] and | H | ≤ h ? Given an instance (cid:104){ S , . . . , S m } , A = { a , . . . , a n } , h (cid:105) of HS, we construct an instance (cid:104) G, n + 1 , h (cid:105) of IIB. The graph G = ( V, E, t ) has node set V = I ∪ A ∪ S, For a positive integer a , we use [ a ] to denote the set of the first a integers, that is [ a ] = { , , . . . , a } . I = { v , . . . , v h } is a set of h + 1 independent nodes, A = { a , . . . , a n } is the ground set, and S = { s , . . . , s m } (each s j represents the set S j ), edge set E = { ( v i , a j ) | v i ∈ I, a j ∈ A } ∪ { ( a j , s t ) | a j ∈ A, s t ∈ S, a j ∈ S t } , and threshold function defined by t ( v ) = if v ∈ I if v ∈ A | S t | = d G ( s t ) if v = s t ∈ S .Trivially, D G [1] = I , D G [2] = I ∪ A , and D G [3] = I ∪ A ∪ S = V . We prove now that (cid:104){ S , . . . , S m } , A, h (cid:105) is a YES instance of HS iff (cid:104)
G, n + 1 , h (cid:105) is a
YES instance of IIB.Suppose first there exists H ⊆ A such that | H | ≤ h and H ∩ S t (cid:54) = ∅ , for each t ∈ [ m ] . If we considerin G the set of nodes ˜ Y ⊆ A corresponding to the elements of H then each node s t ∈ S is connected witha node in ˜ Y . Consequently, if all the nodes in ˜ Y are immunized, then the number of influenced neighborsof s t cannot reach its threshold t ( s t ) = d G ( s t ) . Hence, no node in S can get influenced. Let then Y be theset obtained by padding ˜ Y with nodes in A − ˜ Y , so to have | Y | = h . Clearly, D G,Y = I ∪ ( A − Y ) with | D G,Y | = n + 1 .Assume now there exists a solution Y of IIB. We notice that:a) I ⊆ D G,Y ∪ Y (having all the nodes in I threshold 0, they are immunized or influenced);b) If there exists v i ∈ I ∩ Y , we can update Y to Y (cid:48) = Y ∪ { a } − { v i } , for any a ∈ A − Y (this implies that D G,Y (cid:48) ⊆ D G,Y ∪ { v i } − { a } ).c) If there exists s t ∈ S ∩ Y we can update Y to Y (cid:48) = Y ∪ { a }−{ s t } , for any a ∈ A ∩ S t (this implies that D G,Y (cid:48) ⊆ D G,Y − { a } ).Using a) and iterating b) and c), we can assume that Y consists of at most h nodes in A . As a consequence I ∪ ( A − Y ) ⊆ D G,Y . If we assumed that S ∩ D G,Y (cid:54) = ∅ , then we would have | D G,Y | ≥ | I | + | A − Y | + | S ∩ D G,Y | > h + 1 + ( n − | Y | ) ≥ n + 1 . Being S ∩ D G,Y = ∅ implies each node in S has some neighbor in Y .Hence, the set H of elements corresponding to the h nodes in Y satisfies H ∩ S t (cid:54) = ∅ , for each t ∈ [ m ] . ∆ and (cid:96) Theorem 3.
IIB is W [2] -hard with respect to the pair of parameters ∆ , the maximum node degree, and (cid:96) . Given an instance (cid:104){ S , . . . , S m } , A = { a , . . . , a n } , h (cid:105) of HS, we construct an instance (cid:104) G, k, (cid:96) (cid:105) ofIIB, where the maximum node degree is . We start the construction of G by inserting the nodes in A ∪ W ∪ U ∪ S where A = { a , . . . , a n } is the ground set and S = { s , . . . , s m } (each s j representsthe set S j ), while W and U are two auxiliary sets, of at most nm nodes each, that will be used to keep thedegree bounded and, at the same time, simulating a complete bipartite connection between A and S . Wethen add the following expansion , reduction and path gadgets.6 i w i,i a i ∈ S i j ∀ j = 1 , , . . . , δ i w i,i w i,i δ S j = { a j , a j , . . . a j γ } j w j ,j w j ,j w j ,j w j γ ,j j u j ,j u j ,j u i γ − ,j j s j
222 1 i (a) (b) a a a n s s s m P j ,|P j | = n +2nm (c)11
11 111
A W U S
Figure 2: (a) The expansion gadget. (b) The reduction gadget. (c) The graph G . Expansion gadgets.
For each i ∈ [ n ] , if the sets containing a i are exactly S i , S i , . . . , S i δi then we encodethis relationships with a gadget, which includes four new nodes for each s i j , for j ∈ [ δ i ] . Namely, we add δ i nodes { w i,i , w i,i , . . . w i,i δi } and the edges ( a i , w i,i ) and ( w i,i j , w i,i j +1 ) for j ∈ [ δ i − . Reduction gadgets.
For each j ∈ [ m ] , if S j = { a j , a j , . . . , a j γj } then we encode this relationships with agadget. Namely, we add γ j − nodes { u j ,j , u j ,j , . . . , u j γj − ,j } and the edges ( w j r +1 ,j , u j r ,j ) , ( u j r ,j , u j r +1 ,j ) ,for r ∈ [ γ j − and ( w j ,j , u j ,j ) , ( w j γj ,j , u j γj − ,j ) and ( u j γj − ,j , s j ) . The reduction gadget is presented inFig.2 (b).
Path gadgets.
A path P j of p = n + 2 nm nodes departs from each s j ∈ S . See Fig.2 (c). Noticethat, by construction the degree of nodes is upper bounded by . We set now the thresholds of the nodes in G as: t ( v ) = 0 for each node v ∈ A , t ( v ) = 2 for each node v ∈ U and t ( v ) = 1 for all the remainingnodes. Lemma 1. (cid:104){ S , . . . , S m } , A, h (cid:105) is a YES instance of HS iff (cid:104)
G, p, h (cid:105) is a
YES instance of
IIB . Proof.
Suppose that there exists H ⊆ A such that | H | ≤ h and H ∩ S j (cid:54) = ∅ for each j ∈ [ m ] . Consider in G the set of nodes Y corresponding to the elements of H . Since H ∩ S j (cid:54) = ∅ , for each j ∈ [ m ] , we havethat each node s j ∈ S is connected, through a reduction gadget, with a node in w i,j such that a i ∈ S j ∩ Y .Consequently, if all the nodes in Y are immunized, then at least one node in the reduction gadget associatedto s j cannot reach the threshold and consequently s j will not be influenced. Hence, no node in S as wellas in the associated path gadgets can get influenced. We have | Y | ≤ h and | D G,Y | < p , where the lastinequality follows noticing that p = n + 2 nm is greater than the number of nodes that remain in G once weeliminate the nodes in S and in the path gadgets.Assume now there exists a solution Y to IIB such that | Y | ≤ h and | D G,Y | ≤ p . Without loss ofgenerality, we can assume that Y ⊆ A . Indeed, if Y contains either of the nodes w i,i j , u i,i j , s i j or a nodein the path P i j , for some i ∈ [ n ] , we could replace such a node by a i ∈ A without increasing neither thesize of Y nor D G,Y . Hence, we have that Y consists of at most h nodes in A . We argue that the set H ⊆ A of the elements corresponding to the nodes in Y satisfies H ∩ S j (cid:54) = ∅ , for each j ∈ [ m ] . Indeed, assumeby contradiction that there is a set S j such that H ∩ S j = ∅ . This implies that in G the node s j will beinfluenced. Indeed, s j is connected through gadgets, to all the nodes in S j . Moreover each node in S j belongs to A − Y and has threshold . It follows that s j and, as a consequence, all the p nodes on theassociated path get influenced and we obtain the desired contradiction because this violate the bound on thesize of D G,Y . 7 .4 Graphs of bounded treewidth
Definition 2.
A tree decomposition of a graph G = ( V, E ) is a pair ( T, { W u } u ∈ V ( T ) ) , where T is a tree inwhich each node u is assigned a node subset W u ⊆ V such that:1. (cid:83) u ∈ V ( T ) W u = V .2. For each edge e = ( v, w ) ∈ E, there exists u in T such that W u contains both v and w .3. For each v ∈ V , the set T v = { u ∈ V ( T ) : v ∈ W u } , induces a connected subtree of T . The width of a tree decomposition ( T, { W u } u ∈ V ( T ) ) of a graph G , is max u ∈ V ( T ) | W u | − . The treewidthof G , denoted by tw ( G ) , is the minimum width of a tree decomposition of G . Theorem 4.
IIB is W [1] -hard with respect to the treewidth of the input graph. In order to prove Theorem 4, we present a reduction from M
ULTI -C OLORED CLIQUE (MQ):
Given agraph G = ( V, E ) and a proper vertex-coloring c : V → [ q ] for G , does G contain a clique of size q ? Given an instance (cid:104)
G, q (cid:105) of MQ, we construct an instance (cid:104) G (cid:48) = ( V (cid:48) , E (cid:48) ) , k, (cid:96) (cid:105) of IIB. We denote by n (cid:48) = | V (cid:48) | the number of nodes in G (cid:48) . For a color c ∈ [ q ] , we denote by V c the class of nodes in G of color c and for a pair of distinct c, d ∈ [ q ] , we let E cd be the subset of edges in G between a node in V c and one in V d . Our goal is to guarantee that any solution of IIB in G (cid:48) encodes a clique in G and vice-versa. Followingsome ideas in [4], we construct G (cid:48) using the following gadgets: Parallel-paths gadget:
A parallel-paths gadget of size h , between nodes x and y , consists of h disjointpaths each made up by a connection node which is adjacent to both x and y . In order to avoid cluttering, wedraw such a gadget as an edge with label h (cf. Fig. 3 (a)). Selection gadgets:
The selection gadgets encode the selection of nodes (node-selection gadgets) and edges(edge-selection gadgets):
Node-selection gadget:
For each c ∈ [ q ] , we construct a c -node-selection gadget which consists of anode x v for each v ∈ V c ; these nodes are referred as node-selection nodes. We then add a guard node g c that is connected to all the other nodes in the gadget; thus the gadget is a star centered at g c . Edge-selection gadget:
For each c, d ∈ [ q ] with c (cid:54) = d , we construct a { c, d } -edge-selection gadgetwhich consists of a node x u,v for every edge ( u, v ) ∈ E cd ; these nodes are referred as edge-selectionnodes. We then add a guard node g cd that is connected to all the other nodes in the gadget; thus thegadget is a star centered at g cd .Overall there are n node-selection nodes with q guard nodes and m edge-selection nodes with (cid:0) q (cid:1) guardnodes (cf. Fig. 3 (b)). Validation gadgets:
We assign to every node v ∈ V ( G ) two unique identifier numbers, low ( v ) and high ( v ) , with low ( v ) ∈ [ n ] and high ( v ) = 2 n − low ( v ) . For every pair of distinct c, d ∈ [ q ] , we constructtwo validation gadgets. One between the c -node-selection gadget and the { c, d } -edge-selection gadget andone between the d -node-selection gadget and the { c, d } -edge-selection gadget. We describe the validationgadget between the c -node-selection and { c, d } -edge-selection gadgets. It consists of two nodes. The first8 x yhx y V = {u, v}V = {z}E = {(u, z), (v, z)} (a) (b)Black-hole nodesGuard nodes1111 1 1 1g g g Node-selection Edge-selection1- {1,2}-validation pair2- {1,2}-validation pairx u x v x z x u,z x v,z
000 00 h i gh ( u ) h i g h ( u ) l o w ( u ) l o w ( u ) h i g h ( z ) h i g h ( z ) l o w ( z ) l o w ( z ) Figure 3: (a) Parallel-paths gadget. (b) Representation of the graph G (cid:48) for a trivial instance of the MQproblem (cid:104) G = ( V ∪ V , E , ) , (cid:105) .one is connected to each node x v , for v ∈ V c , by parallel-paths gadgets of size high ( v ) , and to each edge-selection node x u,v , for ( u, v ) ∈ E cd and v ∈ V c , by parallel-paths gadgets of size low ( v ) . The other node isconnected to each node x v , for v ∈ V c , by parallel-paths gadgets of size low ( v ) , and to each edge-selectionnode x u,v , for ( u, v ) ∈ E cd and v ∈ V c , by parallel-paths gadgets of size high ( v ) . Overall, there are q ( q − validation gadgets, each composed by two nodes. Black-hole gadget:
We add a set B of | B | = ( n − q )(2 nq − n + 1) + (cid:0) m − (cid:0) q (cid:1)(cid:1) (4 n + 1) indepen-dent nodes and a complete bipartite graph between nodes in B and the guard nodes.To complete the construction, we specify the thresholds of the nodes in G (cid:48) t ( x ) = if x is a selection node if x is a connection node or x ∈ Bd G (cid:48) ( x ) − n + 1 if x is a validation node | V c | if x = g c is a guard node for some c ∈ [ q ] | E cd | if x = g cd is a guard node for some c, d ∈ [ q ] The complete construction of G (cid:48) for an instance of the MQ problem appears in Fig. 3 (b). Lemma 2. (cid:104)
G, q (cid:105) is a
YES instance of MQ if and only if (cid:104) G (cid:48) , k, (cid:96) (cid:105) , where k = ( n − q )(2 nq − n + 1) + (cid:0) m − (cid:0) q (cid:1)(cid:1) (4 n + 1) and (cid:96) = q + (cid:0) q (cid:1) is a YES instance of
IIB .Proof.
We first notice that a node v can belong to the desired clique only if { v } ∪ Γ G ( v ) contains at least onenode from each color class. Hence, we can remove from G all the nodes that do not satisfy such a property,since they are irrelevant to the problem. 9uppose that K = ( V ( K ) , E ( K )) is a multi-colored clique in G of size q . Let C denote the set ofconnection nodes and X K = { x v : v / ∈ V ( K ) } ∪ { x u,v : ( u, v ) / ∈ E ( K ) } . We set X = X K ∪ { c ∈ C : Γ G (cid:48) ( c ) ∩ X K (cid:54) = ∅} . We show that Y = { x v : v ∈ V ( K ) } ∪ { x u,v : ( u, v ) ∈ E ( K ) } is the immunizing set of X , i.e., Y = Y ( X ) . Notice that | Y | = q + (cid:0) q (cid:1) . We first observe that D G (cid:48) [ X ] = X. Indeed, nodes in { x v : v / ∈ V ( K ) } ∪ { x u,v : ( u, v ) / ∈ E ( K ) } havethreshold and their neighbors in C have threshold . Now we can easily evaluate the size of X . Indeed X is composed by:• n − q nodes in the set of node-selection nodes and their ( n − q )2 n ( q − neighbors in C . Indeed,each node-selection node is connected with q − validation pair and, for each node x u , we have low ( u ) + high ( u ) = 2 n .• m − (cid:0) q (cid:1) nodes in the set of edge-selection nodes and their ( m − (cid:0) q (cid:1) )4 n neighbors in C . Indeed,each edge-selection node is connected with two validation pair and for each node x u,v we have that low ( u ) + high ( u ) = low ( v ) + high ( v ) = 2 n. Overall the set X has size k = ( n − q )(2 nq − n + 1) + (cid:18) m − (cid:18) q (cid:19)(cid:19) (4 n + 1) . (4)It remains to show that Y = Y ( X ) . First of all, we observe that Y ⊆ Y ( X ) because all the nodes in Y belongs to V (cid:48) − X and have threshold , hence, by (1), each node in Y belongs to Y ( X ) . We show nowthat for any v ∈ V (cid:48) − X it holds | Γ G (cid:48) ( v ) ∩ X | < t ( v ) .• Each guard node g has a neighbor in Y and its threshold is equal to the number of its neighborsbelonging to its selection gadget. Hence, | Γ G (cid:48) ( g ) ∩ D G (cid:48) [ X ] | < t ( g ) .• For each b ∈ B , it holds | Γ G (cid:48) ( b ) ∩ X | = 0 < t ( b ) = 1 . • Consider now the validation nodes. Knowing that K is a multi-colored clique, we have that for eachvalidation pair there is exactly one node u and one edge ( u, v ) such that x u , x u,v ∈ Y . Hence, bothnodes have exactly low ( · ) + high ( · ) = 2 n neighbors which do not belong to X . Since the thresholdof each validation node x is t ( x ) = d G (cid:48) ( x ) − n + 1 , then | Γ G (cid:48) ( x ) ∩ X | = d G (cid:48) ( x ) − n < t ( x ) . • Finally, for each connection node c / ∈ X, we have | Γ G (cid:48) ( c ) ∩ X | = 0 < t ( c ) = 1 . Assume now there exists a solution Y to IIB such that | Y | ≤ (cid:96) = q + (cid:0) q (cid:1) and | D G (cid:48) ,Y | ≤ k = ( n − q )(2 nq − n + 1) + (cid:18) m − (cid:18) q (cid:19)(cid:19) (4 n + 1) . (5)Noticing that k < | B | + 1 and all the nodes in B get influences as soon as a guard node is, we havethat the immunization of Y saves all the guard nodes. Noticing that the number of guard nodes is exactly10 + (cid:0) q (cid:1) and each guard node is connected to a separate set of selection nodes, we have that | Y | = q + (cid:0) q (cid:1) and each node in Y can save one guard node. Recalling that the thresholds of guard nodes is equal to thenumber of neighbors belonging to the corresponding selection gadget, we have that in order to save a guardnode there are two options: Put the guard node in Y or put in Y one of its neighbors, belonging to thecorresponding selection gadget. Without loss of generality, we can assume that Y does not include anyguard node. Indeed, if Y contains a guard node we could replace such a node by one of its selection nodeneighbors without increasing neither the size of Y nor of D G (cid:48) ,Y .We can then assume that Y is composed by exactly q node-selection nodes and (cid:0) q (cid:1) edge-selection nodes.Let V Y ⊆ V be a set of q nodes in G , defined by V Y = { v ∈ V : x v ∈ Y } . We argue that G [ V Y ] is a clique.By contradiction suppose that G [ V Y ] is not a clique. There are two nodes u, v ∈ V Y such that ( u, v ) / ∈ E .Let c, d respectively the colors of v and u . Let x w,z the node in G (cid:48) which save the guard g cd associatedto the pair c, d . Since ( u, v ) / ∈ E we have that w (cid:54) = u or z (cid:54) = v or both. Without loss of generality, wecan assume that w (cid:54) = u. Consider now the validation pair between the c -node- and { c, d } -edge-selectiongadgets. Recalling that Y contains exactly one node for each selection gadget, we have that both the nodesin the validation pair have all the neighbors influenced, except for the connections of the nodes x u and x w,z .Since w (cid:54) = u , we have that one of the vertices in the validation pair will get influenced. This is because forany w (cid:54) = u either high ( w ) + low ( u ) < n or low ( w ) + high ( u ) < n . That is, there is a validation node x having less than n not influenced neighbors, while all the remaining neighbors get influenced. Recallingthat the threshold of x is d G (cid:48) ( x ) − n + 1 , we have that x get influenced.Hence, | D G (cid:48) ,Y | = k + 1 . Indeed k are due to non immunized selection nodes and their connectionneighbors (see (4)) plus at least one validation node. This contradicts (5). Lemma 3. G (cid:48) has treewidth O ( q ) .Proof. We show now that G (cid:48) admits a tree decomposition of width O ( q ) . The complete bipartite networkdefined by the guard nodes and the nodes in B has treewidth q + (cid:0) q (cid:1) . Let A be the set of the guard nodes ofsize q + (cid:0) q (cid:1) and b , b , . . . , b ˆ n the nodes in B . The decomposition tree has A as root and A ∪ b i as children.Then we can add to this network the q + (cid:0) q (cid:1) trees, rooted on the guard nodes and containing both selectionsand connection nodes, without increasing the treewidth. Finally we can add all O ( q ) validation nodes,getting a tree decomposition of width O ( q ) for G (cid:48) . Given a graph G = ( V, E ) , two nodes u, v ∈ V are said to have the same type if Γ G ( v ) \{ u } = Γ G ( u ) \{ v } .The neighborhood diversity of a graph G , introduced by Lampis in [31] and denoted by nd ( G ) , is theminimum number nd of sets in a partition V , V , . . . , V nd , of the node set V , such that all the nodes in V i have the same type, for i ∈ [ nd ] . The family { V , V , . . . , V nd } is called the type partition of G .Notice that each V i induces either a clique or an independent set in G . Moreover, for each V i , V j in the typepartition, we get that either each node in V i is a neighbor of each node in V j or no node in V i has a neighborin V j . Hence, between each pair V i , V j , there is either a complete bipartite graph or no edges at all. Theorem 5.
IIB is W[1]-hard with respect to the neighborhood diversity of the input graph.
In order to prove Theorem 5, we use a reduction from M
ULTI -C OLORED CLIQUE (MQ), defined inSection 3.4. As before, we refer to V c as a color class of G and to E cd as the set of edges between nodes11 s + 12rs s + 11,2r + 1, 4r + 1,· · · ,2rs + 1L c L c -pos M cd L cd -negL cd -pos M cd -posM cd -neg1,2r + 1, 4r + 1,· · · ,2rs + 12rs + 100 r + 1 s + 1 s + 12rs ℓ + 1L cd -guardrL c -neg0 ℓ + 1L c -guard rL d L d -pos0r + 1 rL d -neg0 ℓ + 1L d -guard s + 1I c:cd s + 1I c:cd -posI c:cd -neg I c:cd -guards + 1I d:cd s + 1I d:cd -posI d:cd -neg M cd -guard ℓ + 1 ℓ + 1s + 1I d:cd -guard ℓ + 1 Figure 4:
An overview of the reduction. Each circle represents a bag. The number inside a bag is the number ofnodes of the bag. The threshold of nodes in a bag is displayed in red. in the color classes V c and V d . Here we will use the fact that MQ remains W[1]-hard even if each colorclass has the same size and for each distinct colors c, d ∈ [ q ] , the set E cd has the same size [11]. We thendenote by r + 1 the size of each color class V c and by s + 1 the size of each set E cd , in particular we use thefollowing notation V c = { v c , v c , . . . , v cr } , E cd = { e cd , . . . , e cds } c, d ∈ [ q ] , c (cid:54) = d (6)and refer to v ci and e cdj as the i -th node in V c and the j -th edge in E cd , respectively.Let (cid:104) G, q (cid:105) be an instance of MQ. We describe a reduction from (cid:104)
G, q (cid:105) to an instance (cid:104) G (cid:48) , k, (cid:96) (cid:105) of IIBsuch that nd ( G (cid:48) ) is O ( q ) . The reduction runs in time poly ( | G | ) .In order to present the reduction we introduce some gadgets that are used in the construction of G (cid:48) . Theyare inspired by those used in [13]. The rationale behind the construction is the following. First, we createtwo sets of gadgets (Selection and Multiple gadgets), which encode in G (cid:48) the selection of nodes and edgesas part of a potential multicolored clique in G . Then we create another set of gadgets (Incidence gadgets)that is used to check whether the selected sets of nodes and edges actually represent a multicolored cliquein G . Our goal is to guarantee that any solution of IIB in G (cid:48) encodes a clique in G and vice-versa.In the following we call bag an independent set of nodes of a graph sharing all neighbors. So, a connec-tion between two bags points out a complete bipartite graph among the nodes in the bags. Fig. 4 shows thegadgets we are going to introduce and how they are connected. Selection Gadget.
For each c ∈ [ q ] , the selection gadget L c consists of three bags: L c -neg and L c -posof r nodes each, and L c -guard of (cid:96) + 1 nodes (the value (cid:96) , representing an upper bound on the numberof nodes to be immunized, will be determined later). The bag L c -guard is connected to both L c -neg and L c -pos. We set the threshold of each node g in L c -guard to t ( g ) = r + 1 and the threshold of each node v in L c -neg ∪ L c -pos to t ( v ) = 0 . The selection gadget L c is connected to the rest of the graph G (cid:48) using onlynodes from L c -neg ∪ L c -pos. 12 ultiple Gadget. For each c, d ∈ [ q ] with c (cid:54) = d , we create a multiple gadget M cd consisting of sixbags: L cd -pos and L cd -neg of rs nodes each, L cd -guard of (cid:96) + 1 nodes, M cd -pos and M cd -neg of s + 1 nodes each, and M cd -guard of (cid:96) + 1 nodes. M cd -guard is connected to the bags M cd -pos and M cd -neg. M cd -pos is connected to L cd -pos, and M cd -neg is connected to L cd -neg. Finally, the bag L cd -guard is con-nected to both L cd -pos and L cd -neg. The rest of graph G (cid:48) is connected only to the bags L cd -pos and L cd -neg.We set the threshold of each g ∈ M cd -guard to t ( g ) = s + 1 . For each node v ∈ L cd -pos ∪ L cd -neg, weset the threshold t ( v ) = 0 . Let M cd -pos = { x , . . . , x s } and M cd -neg = { y , . . . , y s } ; we set thresholds t ( x i ) = t ( y i ) = 2 ri + 1 . Finally, for each g ∈ L cd -guard, we set the threshold t ( g ) = 2 rs + 1 . Incidence Gadget.
For each pair of distinct c, d ∈ [ q ] , we construct two incidence gadgets: I c : cd (con-nected with the gadgets L c and M cd ) and I d : cd (connected with the gadgets L d and M cd ). In the followingwe present the gadget I c : cd which has the same structure of the gadget I d : cd . The incidence gadget I c : cd hasthree bags I c : cd -pos and I c : cd -neg of s +1 nodes each, and I c : cd -guard of (cid:96) +1 nodes. We connect I c : cd -guardto I c : cd -pos and I c : cd -neg. Furthermore, we connect I c : cd -pos to L c -pos and L cd -pos. Similarly, we connect I c : cd -neg to L c -neg and L cd -neg. We set the threshold of each g ∈ I c : cd -guard to t ( g ) = s + 1 . Recalling thatthere are s + 1 edges in the set E cd , and that there are s + 1 nodes in I c : cd -pos and I c : cd -neg, we create one-to-one correspondences between E cd and I c : cd -pos and between E cd and I c : cd -neg. Namely, for each j =0 , . . . s , we associate the j -th edge e cdj in E cd (cfr. (6)) to a node u j ∈ I c : cd -pos and to a node w j ∈ I c : cd -neg(with u j (cid:54) = u j (cid:48) and w j (cid:54) = w j (cid:48) , for j (cid:54) = j (cid:48) ). Moreover, if the endpoint of e cdj of color c is the i th node v ci of V c (cfr. (6)) then we set t ( u j ) = i + 1 + 2 rj, t ( w j ) = r − i + 1 + 2 r ( s − j ) . It is worth observing that the nodes in I c : cd -pos (respectively, I c : cd -neg) have different thresholds. Indeed,the numbers i + 1 + 2 rj (respectively, r − i + 1 + 2 r ( s − j ) ) are all different, for ≤ i ≤ r and ≤ j ≤ s . Black-hole Gadget.
Finally we add a gadget, which will force the immunizing set Y to contain a specificnumber of nodes for selection ( r nodes) and multiple gadgets ( rs nodes). We add a bag B of | B | = qr + (cid:0) q (cid:1) (2 r + 3) s nodes and connect it to the guard bags in all the selection, multiple and incidence gadgets.For each v ∈ B , we set t ( v ) = 1 . Lemma 4. (cid:104)
G, q (cid:105) is a
YES instance of MQ iff (cid:104) G (cid:48) , k, (cid:96) (cid:105) is a YES instance of
IIB , where k = qr + (cid:0) q (cid:1) (2 r +3) s and (cid:96) = qr + (cid:0) q (cid:1) rs. The proof of Lemma 4 will follow by Claims 1, 2 proved below.
Claim 1. If (cid:104) G, q (cid:105) is a
YES instance of MQ then (cid:104) G (cid:48) , k, (cid:96) (cid:105) is a YES instance of
IIB .Proof.
Let K = ( V ( K ) , E ( K )) be a multicolored clique of G . We will show how to select nodes to beadded to the immunizing set Y according to the nodes in K . First of all notice that, all the nodes in the bags L c -pos, L c -neg, L cd -pos, and L cd -neg belong to Y ∪ D G (cid:48) ,Y , as they all have threshold zero.For each c ∈ [ q ] , if the unique node of color c in K is v ci , the i -th node in V c , then we add i nodes of L c -neg and r − i nodes of L c -pos to Y . For each pair of distinct c, d ∈ [ q ] , if the unique edge with endpointsof colors c and d in K is e cdj , then we add rj nodes of L cd -neg and r ( s − j ) nodes of L cd -pos to Y .Overall, | Y | = (cid:96) = qr + (cid:0) q (cid:1) rs . We now prove that | D G (cid:48) ,Y | = k = qr + (cid:0) q (cid:1) (2 r + 3) s .Consider the diffusion process in V ( G (cid:48) ) − Y . At the first round, all non immunized nodes with thresholdzero are influenced; hence D G (cid:48) ,Y [1] contains: i nodes of L c -pos, for all c ∈ [ q ] and r − i nodes of L c -neg, rj nodes of L cd -pos, r ( s − j ) nodes of L cd -neg, for all c, d ∈ [ q ] with c (cid:54) = d .13e claim that, at the second round, the additional influenced nodes (in the neighborhood of D G (cid:48) ,Y [1] ) areexactly: s nodes in M cd -pos ∪ M cd -neg, s nodes in I c : cd -pos ∪ I c : cd -neg, and s nodes in I d : cd -pos ∪ I d : cd -neg,for each pair of distinct c, d ∈ [ q ] . Indeed, let M cd -pos = { x , . . . , x s } and M cd -neg = { y , . . . , y s } . Sinceat the end of the first round the nodes in M cd -pos have rj influenced neighbors in L cd -pos and the nodes in M cd -neg have r ( s − j ) influenced neighbors in L cd -neg, recalling that t ( x j ) = t ( y j ) = 2 rj + 1 , we havethat nodes x , . . . , x j − in M cd -pos and nodes y , . . . , y s − j − in M cd -neg get influenced. Overall s nodesin M cd -pos ∪ M cd -neg are influenced at the second round.Consider now the incidence gadgets. Since there are rj + i influenced nodes in L c -pos ∪ L cd -pos that arein neighborhood of the nodes in I c : cd -pos, recalling that the thresholds of nodes in I c : cd -pos are: t ( u j ) = 2 rj + i + 1 > rj + i and t ( u h ) = 2 rh + h (cid:48) + 1 for each ≤ h ≤ s, h (cid:54) = j , and ≤ h (cid:48) ≤ r, we have t ( u h ) ≤ rh + r + 1 ≤ r ( j −
1) + r + 1 = 2 rj − r + 1 ≤ rj + i if h < jt ( u h ) ≥ rh + 1 ≥ r ( j + 1) + 1 > rj + 2 r + 1 > rj + i if h > j .Hence, nodes u , . . . , u j − in I c : cd -pos are influenced at the second round.We now make a similar analysis for the nodes in I c : cd -neg. Since there are r − i + 2 r ( s − j ) influencednodes in L c -neg ∪ L cd -neg that are in neighborhood of the nodes in I c : cd -neg, recalling that the threshold ofnodes in I c : cd -pos are: t ( w j ) = 2 r ( s − j ) + r − i + 1 > r ( s − j ) + r − i and t ( w h ) = 2 r ( s − h ) + r − h (cid:48) + 1 for some ≤ h (cid:48) ≤ r, we have t ( w h ) ≥ r ( s − h ) + 1 ≥ r ( s − j ) + 2 r + 1 > r ( s − j ) + r − i for h < jt ( w h ) ≤ r ( s − h ) + n + 1 ≤ r ( s − j ) − r + 1 ≤ r ( s − j ) + r − i for h > j. Hence, nodes w j +1 , . . . , w s in I c : cd -neg are influenced at the second round. Overall, we have that s nodesin I c : cd -pos ∪ I c : cd -neg are influenced at the second round.Using exactly the same argument we can show that s nodes in I d : cd -pos ∪ I d : cd -neg are influenced at thesecond round.Finally, the nodes in L c -guard (resp. L cd -guard) have r (resp. rs ) influenced neighbors at the end ofthe first round and since all of them have threshold r + 1 (resp. rs + 1 ), we have that none of them getsinfluenced at the second round.We notice now that only the nodes in M cd -guard and I c : cd -guard have neighbors in D G (cid:48) ,Y [2] . However,they cannot be influenced (indeed, each of them has threshold s + 1 but it has only s influenced neighborsin D G (cid:48) ,Y [2] – in M cd -pos ∪ M cd -neg or in I c : cd -pos ∪ I c : cd -neg). We have that D G (cid:48) ,Y [3] = D G (cid:48) ,Y [2] and thediffusion process stops.Summarizing, D G (cid:48) ,Y contains: r influenced nodes for each of the q nodes in the clique K (those thatare influenced in the selection gadgets L c for c ∈ [ q ] ), rs + s influenced nodes for each of the (cid:0) q (cid:1) edges14n K (those in the multiple gadgets M cd , for c, d ∈ [ q ] ) and s influenced nodes, for each of the (cid:0) q (cid:1) edgesin K (those in the incidence gadgets I c : cd and I d : cd , for distinct c, d ∈ [ q ] ). Hence, the set D G (cid:48) ,Y contains k = qr + (cid:0) q (cid:1) (2 r + 3) s nodes.Let Y be an immunizing set such that | Y | ≤ (cid:96) = qr + (cid:0) q (cid:1) rs and | D G (cid:48) ,Y | ≤ k = qr + (cid:0) q (cid:1) (2 r + 3) s . Inthe following we derive some useful constraints on the nodes contained in Y and D G (cid:48) ,Y . Proposition 1.
For distinct c, d ∈ [ q ] , no node in L c -guard, L cd -guard, I c : cd -guard, I d : cd -guard, M cd -guardcan be in D G (cid:48) ,Y .Proof. Since the threshold of each v ∈ B is t ( v ) = 1 , it is sufficient that at least one guard node g ∈ L c -guard ∪ L cd -guard ∪ I c : cd -guard ∪ I d : cd -guard ∪ M cd -guard is influenced to influence the whole B .However this cannot be since | B | + 1 = k + 1 > | D G (cid:48) ,Y | . Proposition 2.
For distinct c, d ∈ [ q ] , both Y and D G (cid:48) ,Y contain(1) exactly r nodes of ( L c -pos ∪ L c -neg ) ,(2) exactly rs nodes of ( L cd -pos ∪ L cd -neg),(3) a multiple of r nodes of L cd -pos and L cd -neg.Proof. First of all consider that all the nodes in L c -pos, L c -neg, L cd -pos and L cd -neg have threshold zero,and so all of them are in Y ∪ D G (cid:48) ,Y . We claim that at most r of the nodes of ( L c -pos ∪ L c -neg ) can be in D G (cid:48) ,Y . Indeed, if D G (cid:48) ,Y contains at least r + 1 nodes in ( L c -pos ∪ L c -neg ) then each node g ∈ L c -guard(recall t ( g ) = r + 1 ) either is influenced (i.e., g ∈ D G (cid:48) ,Y ) or is immunized (i.e., g ∈ Y ). By Proposition 1,no node in L c -guard can be influenced. On the other hand, it cannot occur that all the nodes in L c -guard areimmunized, since | L c -guard | = (cid:96) + 1 > | Y | .Using the same argument we can prove that at most rs of the nodes of ( L cd -pos ∪ L cd -neg ) can be in D G (cid:48) ,Y . Assume on the contrary that | D G (cid:48) ,Y ∩ ( L cd -pos ∪ L cd -neg ) | ≥ rs + 1 . Having each node in L cd -guard threshold rs + 1 , we have that either the node is influenced or it must be immunized. However,by Proposition 1 we know that the nodes in L cd -guard are not influenced; moreover they cannot all beimmunized since | L cd -guard | = (cid:96) + 1 > | Y | .This allows to say that Y contains at least r nodes of ( L c -pos ∪ L c -neg ) and at least rs nodes of ( L cd -pos ∪ L cd -neg ) . However, if there exists a c ∈ [ q ] or a pair of distinct c, d ∈ [ q ] such that Y containsstrictly more than r nodes of ( L c -pos ∪ L c -neg ) or rs nodes of ( L cd -pos ∪ L cd -neg ) , then | Y | > qr + (cid:0) q (cid:1) rs and this is not possible. Hence, (1) and (2) follow.To prove (3) we proceed by contradiction. Suppose that D G (cid:48) ,Y contains ra + z nodes of L cd -pos,where a < s and < z < r . By (2) we have that D G (cid:48) ,Y contains r ( s − a ) − z nodes of L cd -neg. Write M cd -pos = { x , . . . , x s } and M cd -neg = { y , . . . , y s } . Recalling that the nodes in M cd -pos are neighborsof those in L cd -pos, the nodes in M cd -neg are neighbors of those in L cd -neg and t ( x i ) = t ( y i ) = 2 ri + 1 ,we have that nodes x , . . . , x a of M cd -pos and nodes y , . . . , y s − a − of M cd -neg get influenced. Since these s + 1 influenced nodes are neighbors of each node g ∈ M cd -guard, whose threshold is t ( g ) = s + 1 , we havethat either g is influenced or it is immunized. By Proposition 1, no node in M cd -guard can be influenced. Onthe other hand, it cannot occur that all the nodes in M cd -guard are immunized, since | M cd -guard | = (cid:96) + 1 > | Y | . Claim 2. If (cid:104) G (cid:48) , k, (cid:96) (cid:105) is a YES instance of
IIB then (cid:104)
G, q (cid:105) is a
YES instance of MQ. roof. Being (cid:104) G (cid:48) , k, (cid:96) (cid:105) a YES instance of IIB, there exists an immunizing set Y of size at most (cid:96) = qr + (cid:0) q (cid:1) rs such that | D G (cid:48) ,Y | ≤ k = qr + (cid:0) q (cid:1) (2 r + 3) s. We proceed by identifying the clique K of G according to the number of nodes that are in L c -neg ∩ Y for each c ∈ [ q ] and in L cd -neg ∩ Y , for each distinct c, d ∈ [ q ] . Namely, we select:– the node v ci ∈ V c , such that | L c -neg ∩ Y | = i , for some ≤ i ≤ r , and– the edge e cdj ∈ E cd such that | L cd -neg ∩ Y | = 2 rj , for some ≤ j ≤ s .The above selection is correct since, by Proposition 2, we know that | Y ∩ ( L c -pos ∪ L c -neg ) | = r and | Y ∩ ( L cd -pos ∪ L cd -neg ) | = 2 rs (in particular, Y contains a multiple of r nodes of both L cd -pos and L cd -neg).Let V ( K ) be the set of the q selected nodes and E ( K ) be the set of the (cid:0) q (cid:1) selected edges. We arguethat K = ( V ( K ) , E ( K )) is a clique. By contradiction assume there are two distinct colors c, d ∈ [ q ] suchthat v ci ∈ V ( K ) and e cdj ∈ E ( K ) but v ci is not an endpoint of e cdj . Consider the incidence gadget I c : cd .Let I c : cd -pos = { u , . . . , u s } and I c : cd -neg = { w , . . . , w s } . Assume that v ch is the endpoint of color c of e cdj . Recall that nodes u j and w j represent the edge e cdj and that, by the construction of G (cid:48) , it holds t ( u j ) = 2 rj + h + 1 and t ( w j ) = 2 r ( s − j ) + r − h + 1 . Since the nodes of I c : cd -pos have rj + i influencedneighbors (those in D G (cid:48) ,Y ∩ ( L c -pos ∪ L cd -pos ) ) and the nodes of I c : cd -neg have r ( s − j ) + r − i influencedneighbors, (those in D G (cid:48) ,Y ∩ ( L c -neg ∪ L cd -neg ) ) by an analysis similar to that in the proof of Lemma 1, wehave that nodes u , . . . , u j − in I c : cd -pos and nodes w j +1 , . . . , w s in I c : cd -neg all get influenced. It remainsto analyze the nodes u j and w j . We will prove that at least one of them gets influenced: If h < i then t ( u j ) = 2 rj + h + 1 ≤ rj + i and t ( w j ) = 2 r ( s − j ) + r − h + 1 > r ( s − j ) + r − i and u j is influenced;if h > i then t ( u j ) = 2 rj + h + 1 > rj + i and t ( w j ) = 2 n ( s − j ) + n − h + 1 ≤ r ( s − j ) + r − i and w j is influenced. This allows to say that if v ch ∈ e cdj then s + 1 nodes among those in I c : cd -pos and I c : cd -negare influenced. As a consequence, each node g ∈ I c : cd -guard, whose threshold is t ( g ) = s + 1 , must eitherbe influenced or immunized. By Proposition 1, no node in I c : cd -guard can be influenced. On the other hand,it cannot occur that all the nodes in I c : cd -guard are immunized, since | I c : cd -guard | = (cid:96) + 1 > | Y | . Lemma 5. G (cid:48) has neighborhood diversity O ( q ) .Proof. Since each bag in G (cid:48) is a type set in the type partition of G (cid:48) and, since for each c ∈ [ q ] , there arethree bags in L c and, for each c, d ∈ [ q ] with c (cid:54) = d there are six bags in M cd , and three bags in both I c : cd and I d : cd , we have that the neighborhood diversity of G (cid:48) is q + 12 (cid:0) q (cid:1) . In this section, we present FPT algorithm for several pairs of parameters. k and (cid:96) Theorem 6.
IIB can be solved in time k + (cid:96) ( k + (cid:96) ) O (log( k + (cid:96) )) · n O (1) .Proof. The fixed parameter tractability of IIB with respect to k + (cid:96) can be proved by the arguments usedin Theorem 1 in [19] for the problem CUTTING AT MOST k VERTICES WITH TERMINAL . For sake ofcompleteness, the complete proof is given in the following.16et (cid:104)
G, k, (cid:96) (cid:105) be the input instance of IIB. Consider a random labelling of the nodes of G , where eachnode is independently assigned either 0 or 1 with equal probability. Let now H = G [ V ] be the graphinduced by the set V of nodes having label 1. Consider the set D H of influenced nodes when we run thediffusion process on H . If | D H | ≤ k and | Y ( D H ) | ≤ (cid:96) then (3) holds for X = D H and we can answer YES .We estimate now the number of needed iterations of random labelling. Suppose G contains a set X satisfying (3). For such a set, it holds | X | = | D G [ X ] | ≤ k and | Y ( X ) | ≤ (cid:96) , then a random labellingidentifies a solution of IIB if and only if all the nodes in X are labelled 1 and all the nodes in Y ( X ) arelabelled 0, that is, X ⊆ V and Y ( X ) ∩ V = ∅ . Indeed, in such a case the above procedure identifies D H = X as a solution. This happens with probability − ( | D H | + | Y ( D H ) | ) ≥ − ( k + (cid:96) ) . Hence, the algorithm requires time k + (cid:96) n O (1) .A derandomization of the above process can be done using universal sets. A ( n, i ) -universal set is a col-lection of binary vectors of length n such that for each set of i indices, each of the i possible combinationsof values appears in some vector of the set. To run the algorithm, it suffices to try all labellings induced bya ( n, k + (cid:96) ) -universal set. Naor et al. [18] give a construction of ( n, i ) -universal sets of size i i O (log i ) log n that can be listed in linear time. k and ζ Theorem 7.
IIB can be solved in time O ( ζ k n ) , where ζ = |{ v ∈ V | t ( v ) = 0 }| .Proof. Let (cid:104)
G, k, (cid:96) (cid:105) be the input instance of IIB. Suppose v , . . . v ζ are the nodes in G having threshold 0and let ∆ denote the maximum degree of a node in G . Consider the graph G (cid:48) = ( V (cid:48) , E (cid:48) ) obtained from G by adding the internal nodes and the edges of a ∆ -ry tree whose leaves are v , . . . v ζ . Assume (cid:104) G, k, (cid:96), (cid:105) is a
YES instance of IIB. We notice that in G , the solution set X (cfr. (3)) can be disconnected but any ofits connected components must include at least one node of threshold 0. Hence, in G (cid:48) the nodes in X arenow connected through a path in the ∆ -ry tree. This implies that there exists X (cid:48) ⊆ V (cid:48) such that: X ⊆ X (cid:48) , ( X (cid:48) − X ) ⊆ V (cid:48) − V , and G (cid:48) [ X (cid:48) ] is connected. In particular, if s is the root of tree, we can assume that s ∈ X (cid:48) .In the worst case, all the paths within the ∆ -ry tree go through the root s , hence | X (cid:48) | ≤ | X | log ∆ ζ + 1 .Let k (cid:48) = k log ∆ ζ + 1 . We use the following result [29, Lemma 2]: There are at most k (cid:48) ∆ k (cid:48) connectedsubgraphs that contain s and have order at most k (cid:48) . Furthermore, these subgraphs can be enumerated in O (4 k (cid:48) ∆ k (cid:48) ( | V (cid:48) | + | E (cid:48) | )) time. We can then apply the result in [29] to enumerate all the connected subgraphsof G (cid:48) of size up to k (cid:48) . For each candidate set X (cid:48) (the node set of the current connected subgraph) one has todetermine whether X (cid:48) ∩ V is a solution according to (3), which can be done in O ( n ) time. k (or ∆ ) and Treewidth In this section we present a dynamic programming algorithm which exploiting the tree decomposition of agraph G enables to solve a minimization version of IIB, namely the I NFLUENCE D IFFUSION M INIMIZATION (IDM) : Given a graph G = ( V, E, t ) and a budget (cid:96) , find a set Y such that | Y | ≤ (cid:96) and | D G,Y | is minimized.We use the rooted tree decomposition named nice tree decomposition.17 efinition 3. A tree decomposition ( T, { W u } u ∈ V ( T ) ) is nice if conditions 1. and 2. hold:1. W r = ∅ for r the root of T and W v = ∅ for every leaf v of T .2. Every non-leaf node of T is of one of the following three types:Introduce: a node u with exactly one child u (cid:48) such that W u = W u (cid:48) ∪ { v } for a node v / ∈ W u (cid:48) .Forget: a node u with exactly one child u (cid:48) such that W u (cid:48) = W u ∪ { v } for a node v / ∈ W u .Join: a node u with two children u , u such that W u = W u = W u Lemma 6. [17] If a graph G admits a tree decomposition of width at most tw , then it admits a nice treedecomposition of width at most tw . Moreover, given a tree decomposition ( T, { W u } u ∈ V ( T ) ) of G of widthat most tw , one can compute in time O ( tw max {| V ( T ) | , | V ( G ) |} ) a nice tree decomposition of G of widthat most tw that has at most O ( tw | V ( G ) | ) nodes. Consider a graph G = ( V, E ) with treewidth tw and nice tree decomposition ( T, { W u } u ∈ V ( T ) ) . Let T be rooted at node r and denote by T ( u ) the subtree of T rooted at u , for any node u of T . Moreover, denoteby W ( u ) the union of all the bags in T ( u ) , i.e., W ( u ) = (cid:83) v ∈ T ( u ) W v . We will denote by s u = | W u | thesize of W u .We are going to recursively compute the solution of IDM. The algorithm exploits a dynamic program-ming strategy and traverses the input tree T in a breadth-first fashion. Moreover, in order to be able torecursively reconstruct the solution, we calculate optimal solutions under different hypothesis based on thefollowing considerations:– Fix a node u in T, for each node v ∈ W u we have three cases: v gets influenced, v is immunized, or v issafe. We are going to consider all the s u combinations of such states. We denote each combination with avector C of size s u indexed by the elements of W u , where the element indexed by v ∈ W u denotes the stateinfluenced ( ), immunized ( ), safe ( ) of node v . The configuration C = ∅ denotes the vector of length 0corresponding to an empty bag. We denote by C u the family of all the s u possible state vectors of the s u nodes in W u .– Let U be a subset of V ( G ) . Let us first notice that by 3) of Definition 2, all the edges between nodesin V − W ( u ) and W ( u ) connect a node in V − W ( u ) with a node in W u (the bag corresponding tothe root of T ( u ) ). We are going to consider all the possible contribution to the diffusion process, ofnodes in V − W ( u ) ; that is, for each v ∈ W u , we consider all the possible residual thresholds among t ( v ) , t ( v ) − , . . . , max { , t ( v ) − k } (recall that at most k nodes belong to X and can therefore reduce thethreshold of v ). We notice that, for each node v , it is possible to bound the number of residual thresholdsby the value min { t ( v ) , k } . Moreover, since no node with t ( v ) > d G ( v ) can be influenced and can be thenpurged from G in a preprocessing step, we can assume that in G it holds (max v ∈ V t ( v )) ≤ ∆ . Hence, wewill have up to µ s u threshold combinations, where µ = min { k, ∆ } . We will denote each possible thresholdcombination with a vector T , indexed by the s u elements in W u , where the element indexed by v belongsto { max { , t ( v ) − k } , . . . , t ( v ) } and denotes the residual threshold of v ∈ W u . The configuration T = ∅ denotes the vector of length 0 corresponding to an empty bag. We denote by T u the family of all the possiblethreshold combinations of nodes in W u .The following definition introduces the values that will be computed by the algorithm in order to keeptrack of all the above cases: 18 efinition 4. For each node u ∈ T, each j = 0 , . . . , (cid:96) , C ∈ C u and T ∈ T u we denote by X u ( j, C , T ) theminimum number of influenced nodes one can attain in G [ W ( u )] by immunizing at most j nodes in W ( u ) ,where the states and the thresholds of nodes in W u are given by C and T . Considering that the root r of a nice tree decomposition has W r = ∅ , we have that the solution of theIDM instance (cid:104) G, (cid:96) (cid:105) can be obtained by computing X r ( (cid:96), ∅ , ∅ ) . Claim 3.
For each u ∈ T , the computation of X u ( j, C , T ) , for each j ∈ { , . . . , (cid:96) } , state configuration C ∈ C u , and threshold configuration T ∈ T u comprises O ( (cid:96) tw µ tw ) values, where µ = min { k, ∆ } , eachof which can be computed recursively in time O (2 tw + (cid:96) ) .Proof. We show now how use a bottom–up strategy to compute all the values of X u ( j, C , T ) , for each u ∈ T , j = 0 , . . . , (cid:96) , state configuration C ∈ C u , and threshold configuration T ∈ T u . By Definition 4, weknow that such values are O ( (cid:96) tw µ tw ) , where µ = min { k, ∆ } .For each leaf u ∈ T and for each j = 0 , . . . , (cid:96) we have X u ( j, ∅ , ∅ ) = 0 . For any internal node u , we show how to compute each values X u ( j, C , T ) , for each j = 0 , . . . , (cid:96) , C ∈ C u ,and T ∈ T u in time O (2 tw + (cid:96) ) .We have three cases to consider according to the type of u (cf. Definition 3): u is an introduce node: In this case u has exactly one child u (cid:48) and we have that W u = W u (cid:48) ∪ { v } forsome node v / ∈ W u (cid:48) . For a given node u ∈ V ( T ) (introducing a node v ∈ V ) and state configuration C , we denote by S u ( C ) the set of influenced nodes (according to the configuration C ) that belongs to W u ∩ Γ G ( v ) . Given a threshold configuration T associated to a set of nodes W , and a set of nodes S ⊆ W we denote by T ( S ) the configuration obtained starting from T and decreasing by one thethreshold of each node in S. In the following we assume w.l.o.g. that the element indexed by v isthe last element of the vectors C and T . We have that for each j = 0 , . . . , (cid:96) , each C ∈ C u and each T ∈ T u .X u ( j, C =[ C (cid:48) , c ] , T =[ T (cid:48) , t ]) = min S ⊆ S u ( C ) , | S | = t (cid:0) X u (cid:48) ( j, C (cid:48) , T (cid:48) ( S u ( C ) − S )) (cid:1) +1 , if c = 0 AND t ≤ | S u ( C ) | X u (cid:48) ( j − , C (cid:48) , T (cid:48) ) , if c = 1 AND j > X u (cid:48) ( j, C (cid:48) , T (cid:48) ) , if c = 2 AND t > | S u ( C ) | + ∞ , otherwise. (7)It is worth to observe that the size of S u ( C ) is bounded by tw and for this reason the above value canbe computed in time O (2 tw ) u is a forget node: In this case u has exactly one child u (cid:48) and we have that W u (cid:48) = W u ∪ { v } for somenode v / ∈ W u . We have for each j = 0 , . . . , (cid:96) , each C ∈ C u , and each T ∈ T u X u ( j, C , T ) = min c ∈{ , , } { X u (cid:48) ( j, C (cid:48) = [ C , c ] , T (cid:48) = [ T , max { , t ( v ) − | S u ( C ) |} ]) } (8)19 lgorithm 1: IIB-k(
G, k, (cid:96) ) Input:
A graph G = ( V, E, t ) , integers k, (cid:96) and a type partition V , . . . , V nd of G . foreach f = 1 , . . . , k do foreach f = ( f , f , . . . , f nd ) such that (cid:80) nd i =1 f i = f do foreach i ∈ [ nd ] do let X i = { v i, , . . . , v i,f i } ⊆ V i Set X = (cid:83) nd i =1 X i if | Y ( X ) | ≤ (cid:96) then return YES return NO u is a join node: In this case u has exactly two child u , u such that W u = W u = W u . We have foreach j = 0 , . . . , (cid:96) , each C ∈ C u , and each T ∈ T u X u ( j, C , T ) = min ≤ a ≤ j − I ( C ) { X u ( a + I ( C ) , C , T ) + { X u ( j − a, C , T ) } , (9)where I ( C ) denotes the number of immunized nodes in the configuration state C . By induction on the tree, we can prove that the recursive formula presented in (7)-(9) coincides with thedefinition of X u ( · , · , · ) ; hence, the algorithm is correct.Hence, using [17, Lemma 18], we have that the desired value X r ( (cid:96), ∅ , ∅ )) can be computed in time O ( tw | V | (2 tw + (cid:96) ) (cid:96) tw µ tw ) . Standard backtracking techniques can be used to compute the optimal set X and Y ( X ) in the same time.As a consequence we have that IDM is FPT with respect to tw and ∆ or k . Theorem 8.
IDM is solvable in time O ( tw | V | (2 tw + (cid:96) ) (cid:96) tw µ tw ) , where µ = min { k, ∆ } . We present FPT algorithms for IIB with respect to both the pairs ( k, nd ) and ( (cid:96), nd ) .Let { V , V , . . . , V nd } be the type partition of G . Below, we assume that the nodes of each V i = { v i, , . . . , v i, | V i | } are sorted in non-decreasing order of thresholds, e.g. t ( v i,j ) ≤ t ( v i,j +1 ) . Parameters nd and k . We consider all the nd -ples ( f , . . . , f nd ) such that (cid:80) nd i =1 f i ≤ k . For each one,we construct a candidate set as detailed in Algorithm IIB-k below. Theorem 9.
Algorithm IIB-k solves
IIB in time O ( n k + nd − ) Proof.
We first show Algorithm IIB-k outputs
YES iff there exists X satisfying (3).If the output is YES then trivially the current set X has X ≤ k and | Y ( X ) | ≤ (cid:96) .Let now ˜ X be a minimal set satisfying (3), that is, ˜ X = D G [ ˜ X ] , | X | ≤ k , and | Y ( X ) | ≤ (cid:96) . Let ˜ X i = ˜ X ∩ V i for each i ∈ [ nd ] . Consider the iteration of the algorithm when f = ( f , f , . . . , f nd ) with f i = | ˜ X i | , for i ∈ [ nd ] . The algorithm selects a set X = (cid:83) nd i =1 X i such that | X i | = f i and t ( v ) ≤ t ( w ) foreach v ∈ X i and w ∈ V i − X i , for each i ∈ [ nd ] . We show that the algorithm outputs YES on X .20 lgorithm 2: IIB- (cid:96) ( G, k, (cid:96) ) Input:
A graph G = ( V, E, t ) , integers k, (cid:96) and a type partition V , . . . , V nd of G . foreach h = 1 , . . . , (cid:96) do foreach h = ( h , h , . . . , h nd ) such that (cid:80) nd i =1 h i = h do foreach i ∈ [ nd ] do let Y i = { v i, , . . . , v i,h i } ⊆ V i Set Y = (cid:83) nd i =1 Y i if | D G,Y | ≤ k then return YES return NO Fix any i ∈ [ nd ] . Knowing that | ˜ X i | = | X i | = f i , we have that if ˜ X i (cid:54) = X i , then there exists u ∈ ˜ X i − X i and v ∈ X i − ˜ X i such that t ( v ) ≤ t ( u ) . W.l.o.g assume that u is the node with maximum threshold in ˜ X i − X i . Since ˜ X = D G [ ˜ X ] , we have that u has at least t ( u ) neighbors in ˜ X . Furthermore, since v, u ∈ V i we have that u and v have the same neighbors. Hence, v has at least t ( u ) ≥ t ( v ) neighbors in ˜ X . As aconsequence, since v / ∈ ˜ X we have v ∈ Y ( ˜ X ) . Consider ˜ X (cid:48) = ˜ X − { u } ∪ { v } . By (i) in Proposition 3 (seeAppendix) we have that ˜ X (cid:48) = D G [ ˜ X (cid:48) ] with | ˜ X (cid:48) | = | ˜ X | and | Y ( ˜ X (cid:48) ) | = | Y ( ˜ X ) | .Hence, trading each node in ˜ X i − X i for one in X i − ˜ X i , for each i such that ˜ X i (cid:54) = X i , we can provethat | Y ( X ) | = | Y ( ˜ X ) | ≤ (cid:96) . Therefore, the algorithm returns YES .We now evaluate the running time. Fix f ∈ [ k ] , for each ( f , . . . , f nd ) with (cid:80) nd i =1 f i = f , one needstime O ( f ) to get X and O ( n ) to get Y ( X ) , moreover the number of all possible such nd -ple is (cid:0) f + nd − f (cid:1) .Summing on all f we get (cid:80) f ∈ [ k ] (cid:0) f + nd − f (cid:1) < k + nd − and the theorem holds. Parameters nd and (cid:96) . An idea similar to that in Algorithm 1 can be used to prove IIB is FPT with respectto nd and (cid:96) . Proposition 3.
Fix i ∈ [ nd ] .(i) Let X = D G [ X ] and Y = Y ( X ) be its immunizing set. Set u max = arg max u ∈ X ∩ V i t ( u ) . If thereexists v ∈ Y ∩ V i such that t ( v ) ≤ t ( u max ) then X (cid:48) = X − { u max } ∪ { v } satisfies X (cid:48) = D G [ X (cid:48) ] and | Y ( X (cid:48) ) | = | Y | .(ii) Let Y be an immunizing set. Set v max = arg max v ∈ Y ∩ V i t ( v ) . If there exists u ∈ D G,Y ∩ V i such that t ( u ) ≤ t ( v max ) then setting Y (cid:48) = Y − { v max } ∪ { u } it holds | D G,Y (cid:48) | ≤ | D G,Y | .Proof. Let us prove (i). Consider X (cid:48) = X − { u max } ∪ { v } and the diffusion process in G [ X (cid:48) ] . We havethat v is influenced at a round which is at most equal to that in which u max is influenced during the diffusionprocess in G [ X ] (recall t ( v ) ≤ t ( u max ) and that v and u max have the same neighbors). Furthermore, sinceall the neighbors of v and u max have the same number of neighbors in X (cid:48) as in X we have that all the nodesin X (cid:48) are influenced, that is X (cid:48) = D G [ X (cid:48) ] , and u max ∈ Y ( X (cid:48) ) . This allows to say that | Y ( X (cid:48) ) | = | Y | .Let us prove now (ii). If we consider the diffusion process in G [ V − Y (cid:48) ] we have that no node outside D G,Y −{ u } , except eventually for node v max , can be influenced. Hence, D G,Y (cid:48) ⊆ D G,Y −{ u }∪{ v max } . Theorem 10.
Algorithm IIB- (cid:96) solves
IIB in time O ( n (cid:96) + nd − ) roof. Given h ≤ (cid:96) , Algorithm IIB- (cid:96) ( G, k, (cid:96) ) considers all the possible nd -ples ( h , h , . . . , h nd ) with (cid:80) nd i =1 h i = h ; for each h = ( h , h , . . . , h nd ) we construct the set Y = (cid:83) nd i =1 Y i where Y i consists of thefirst (e.g. with the smallest thresholds) h i nodes in V i . We then consider the diffusion process in G and theset D G,Y of influenced nodes when the elements of Y are immunized. If | D G,Y | ≤ k then we answer YES .In case no h gives a set Y such that | D G,Y | ≤ k , we answer NO .If Algorithm IIB-nd- (cid:96) returns YES then the set Y constructed by algorithm IIB- (cid:96) has size at most (cid:96) andwe know that | D G,Y | ≤ k .Assume now that there exists ˜ Y such that | ˜ Y | = h ≤ (cid:96) and | D G, ˜ Y | ≤ k . Assume w.l.o.g. that no smallersolution exists, that is, for any Y such that | D G,Y | ≤ k it holds | Y | ≥ h .Define ˜ Y i = Y ( ˜ X ) ∩ V i and let | ˜ Y i | = h i , for i ∈ [ nd ] . Clearly, (cid:80) nd i =1 h i = h . Consider the nd -ple h = ( h , h , . . . , h nd ) and the set Y = (cid:83) nd i =1 Y i constructed at line 4 of algorithm IIB-nd- (cid:96) . Recall that | Y i | = h i and t ( v ) ≤ t ( w ) for each v ∈ Y i and w ∈ V i − Y i .Since | ˜ Y i | = | Y i | = h i , we have that if ˜ Y i (cid:54) = Y i , for some i , then there are v ∈ ˜ Y i − Y i and u ∈ Y i − ˜ Y i suchthat t ( u ) ≤ t ( v ) . W.l.o.g select u as the node with minimum threshold in Y i − ˜ Y i and v as the node withmaximum threshold in ˜ Y i − Y i . By the fact that v ∈ ˜ Y and ˜ Y is minimal, we know that v must have at least t ( v ) neighbors in D G, ˜ Y (otherwise, ˜ Y − { v } would be a smaller solution). Furthermore, since v, u ∈ V i we have that they have the same neighbors. As a consequence, also u has at least t ( v ) ≥ t ( u ) neighbors in D G, ˜ Y . Knowing that u (cid:54)∈ ˜ Y , we have that u ∈ D G, ˜ Y . Set Y (cid:48) = ˜ Y − { v } ∪ { u } . By (ii) in Proposition 3 wehave that D G,Y (cid:48) satisfies D G,Y (cid:48) ≤ D G, ˜ Y ≤ k . Hence, Y (cid:48) is also a solution.Starting from Y (cid:48) , we then can repeat the above reasoning until we get Y r = Y , the immunizing setconsidered in the algorithm for the tuple h . Hence, | D G,Y | ≤ k .Now we evaluate the running time of the algorithm. For each fixed h ∈ [ (cid:96) ] , the number of all the possible nd -ples ( h , h , . . . , h nd ) such that (cid:80) nd i =1 h i = h is (cid:0) h + nd − h (cid:1) ≤ (cid:0) (cid:96) + nd − h (cid:1) . Noticing that for each choice of ( h , . . . , h nd ) , one needs time O ( h ) to construct Y and O ( n ) to obtain D G,Y and that (cid:88) h ∈ [ (cid:96) ] (cid:18) (cid:96) + nd − h (cid:19) < (cid:96) + nd − , the desired result follows. We introduced the influence immunization problem on networks under the threshold model and analyzedits parameterized complexity. We considered several parameters and showed that the problem remains in-tractable with respect to each one. We have also shown that for some pairs (e.g., ( ζ , (cid:96) ) and ( ∆ , (cid:96) )) theproblem remains intractable.On the positive side, the problem was shown to be FPT for some other pairs: ( k, (cid:96) ) , ( k, ζ ) , ( k, tw ) , (∆ , tw ) , ( k, nd ) ,and ( (cid:96), nd ) .It would be interesting to asses the parameterized complexity of IIB with respect to the remaining pairs ofparameters; in particular with respect to k and ∆ . 22 eferences [1] F.N. Abu-Khzam, S. Li, C. Markarian, F. Meyer auf der Heide, P. Podlipyan. Modular-Width: AnAuxiliary Parameter for Parameterized Parallel Complexity. Proc. of Frontiers in Algorithmics. (FAW2017), LNCS, v. 10336. Springer, (2017).[2] R. Albert, H. Jeong, A.-L. Barab´asi. 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