Approximating the Sparsest k-subgraph in Chordal Graphs

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1 Approximating the Sparsest k-subgraph in Chordal Graphs R. Watrigant, M. Bougeret, and R. Giroudeau LIRMM, Université Montpellier, France Abstract. Given a simple undirected graph G = V, E and an integer k < V, the Sparsest k-subgraph problem asks for a set of k vertices which induces the minimum number of edges. As a generalization of the classical independent set problem, Sparsest k-subgraph is N P-hard and even not approximable unless P = N P in general graphs. hus, we investigate Sparsest k-subgraph in graph classes where independent set is polynomial-time solvable, such as subclasses of perfect graphs. Our two main results are the N P-hardness of Sparsest k-subgraph on chordal graphs, and a greedy tight -approximation algorithm. Finally, we also show how to derive a P AS for Sparsest k-subgraph on proper interval graphs. 1 Introduction 1.1 Related Problems Given a simple undirected graph G = V, E and an integer k < V, the Sparsest k-subgraph problem asks for a set of k vertices which induces 1 the minimum number of edges. It appears that this problem falls into the family of cardinality constrained optimization problems, introduced by [7], and is more precisely a generalization of the so-called independent set problem. his observation immediately implies that Sparsest k-subgraph is N P-hard and even not approximable in general graphs unless P = N P, as the optimal value is 0 whenever there is an independent set of size k. hus, we only consider Sparsest k-subgraph in graph classes where independent set is polynomial-time solvable. Let us first present some related problems, and then discuss their relation to Sparsest k-subgraph. Actually, the following three problems can all be considered as cardinality constrained versions of other well-known combinatorial optimization problems, namely vertex cover and max clique, both very close to independent set. In the maximum Quasi-Independent Set QIS problem [4] also called k-edge-in in [11], we are given a graph G and an integer C, and we ask for a set of vertices S of maximum size inducing at most C edges. In the minimum Partial Vertex Cover PVC problem [1], we are given a graph G and an integer C, and we ask for a set of vertices S of minimum size which covers 1 at least C edges. Finally, we can mention the corresponding maximization problem of Sparsest k-subgraph, namely Densest k-subgraph, which consists in finding a subset S of exactly k vertices inducing the maximum number of edges. he decision versions of QIS, P V C, and Sparsest k-subgraph are polynomially equivalent. Indeed, QIS could be considered as a dual version of Sparsest k-subgraph where the budget the number of edges in the solution of Sparsest k-subgraph is fixed. P V C and Sparsest his work has been funded by grant ANR 010 BLAN An edge {u, v} E is said to be induced resp. covered by a set S if u S and resp. or v S.

2 k-subgraph are also polynomially equivalent as for any S, the number of edges induced by S plus the number of edges covered by V \S equals E. hen, exact results for Densest k-subgraph on a graph class implies the same result for Sparsest k-subgraph on the corresponding complementary class, and conversely. Unlike exact results, approximation algorithms do not transfer directly between any of these problems. Considering these remarks and previous studies on these problems, Figure 1 presents known results and open problems about Sparsest k-subgraph SkS, Densest k-subgraph DkS and PVC in restricted graph classes. In each cell, the first line generally describes the general complexity N P-hard versus Polynomial, whereas other lines present some results concerning approximation or parameterized complexity. We recall that proper interval graphs interval graphs chordal graphs perfect graphs, as well as split graphs chordal and bipartite, cographs perfect graphs. Graphs classes DkS SkS P V C general N P-h N P-h, not approx. N P-h, W [1]-h [1] c.f. max clique c.f. indep. set -approx.[6] n 4 1 +ɛ -approx. [3] exact O 1, 4 C [14] chordal N P-h [9] N P-h [this paper] N P-h c.f. SkS 3-approx [15] -approx [this paper] interval OPEN OPEN, OPEN PAS [16] F P C [17] F P n-k c.f. SkS proper interval OPEN OPEN, OPEN PAS [16] P AS [this paper] bipartite N P-h [9] N P-h c.f. PVC N P-h [13] line OPEN P c.f. PVC P [1] planar OPEN N P-h c.f. indep. set N P-h c.f. SkS cographs, split, P [9] P [5] P c.f. SkS bounded treewidth max. degree P [9] P [5] P c.f. SkS max. degree 3 N P-h [9] OPEN OPEN Fig. 1: Main results for DkS, SkS and PVC in some restricted graph classes. 1. Contributions and Organization of the Paper According to Figure 1, Densest k-subgraph was already known to be N P-hard on chordal graphs. However, as the complement of a chordal graph and in particular the graph used in the reduction of [9] is a perfect graph and not necessarily a chordal graph, this result only provides the N P-hardness of Sparsest k-subgraph on perfect graphs. hus, our motivation is to study Sparsest k-subgraph on a classical subclass of perfect graphs. he main results of the paper are the N P-hardness of Sparsest k-subgraph in chordal graphs Section 3, and a tight -approximation greedy algorithm Section. Finally, we show in Section 4 how the arguments of [16] which provides a PAS for DkS in interval graphs can be adapted to SkS in proper interval graphs. Notice that our N P-hardness result implies the N P- hardness of PVC in chordal graphs, which supplements the recent N P-hardness of [, 13] for PVC in bipartite graphs. Due to space constraints, some details of the N P-hardness proof and the P AS in proper interval graphs were omitted and placed in the appendix.

3 3 1.3 Notations and Definitions All graphs studied in this paper are simple and without loop. For the remaining, G = V, E will denote the input graph of the problem, and we define as usually n = V, m = E. Chordal graphs are graphs with no induced cycle of length four or more. hey may also be defined equivalently in terms of simplicial elimination order [10]. A vertex v V is called simplicial if its neighbourhood Nv is a clique. A simplicial elimination order of G is an ordering v 1,..., v n of V such that for all i {1,..., n}, v i is simplicial in G[v i,..., v n ]. It is known that a graph G is chordal if and only if it admits a simplicial elimination order. In addition, such an ordering can be found in polynomial time for a chordal graph. Hence, we will suppose in the following that V = {v 1,..., v n } is sorted according to a simplicial elimination order of G. Similarly, for a subset of vertices S V, we will denote by mins resp maxs the first resp. last vertex of S in the simplicial elimination order chosen for the graph. Finally, since we have a total ordering on the vertices, we will use the notations x < y and x > y for two vertices x, y V. Given two sets S 1, S V, we denote by costs 1 the number of edges in the graph induced by vertices of S 1, and costs 1, S = {{v, v } E, v S 1, v S }. Given a set S V and x V, we denote by dx, S the degree of x in S. Finally, we refer the reader to the classical literature for definitions of approximation algorithms. -Approximation in Chordal Graphs.1 Idea of the Algorithm We now present a tight -approximation algorithm for chordal graphs. First, notice that any approximation algorithm for Sparsest k-subgraph must output a maximum independent set of size k if such a set exists, as in this case the optimal value is 0. Hence, a natural idea for computing a solution to Sparsest k-subgraph is to choose first a maximum independent set S this can be done in polynomial time in chordal graphs. If k vertices or more were picked, then the algorithm stops. Otherwise, several ideas may come up. A first idea would be to remove this independent set from the graph, and iteratively pick another one, until we get k vertices. his approach is the same as the 3-approximation of [15] for Densest k- Subgraph in chordal graphs computing maximum cliques instead of maximum independent sets. Unfortunately, as shown in Figure on the left, this algorithm has an unbounded approximation ratio for Sparsest k-subgraph even in interval graphs a subclass of chordal graphs. It still provides a -approximation in proper interval graphs [17]. hus, after picking the first maximum independent set, our idea is to assign weights on remaining vertices according to the size of their neighbourhood in the constructed solution. At each step, the algorithm just picks an independent set called a layer among the vertices of minimum weight, and then updates the weights of remaining vertices. he algorithm is more formally defined in the next subsection. In the next paragraph, we describe the key idea of the analysis. he idea of the proof of the -approximation ratio is to restructure an optimal solution S until we get S the output of the algorithm, while bounding the cost variation during the restructurations. Let us show what makes this restructuration work for the first layer. Let L be the independent set chosen by the algorithm at the first step. Roughly speaking, for each n j L which is not in S, we restructure S by removing y j, the first neighbour of n j which is after n j and in S, and adding n j instead. As depicted in Figure on the right, we see that the degree of a vertex

4 4 y 1 y z 1 x 1 x x t 1 x t x n 1 y 1 n y Fig. : On the left: example where picking successive independent sets gives an unbounded ratio. For k = t + the algorithm would take intervals {x 1,..., x t, y 1, y } of cost t whereas the solution {x 1,..., x t, y, z 1 } has cost 4. On the right: idea of the restructuration of a solution S. Circles denote vertices of S, and crosses denote vertex of L chosen by the algorithm. When replacing y 1 by n 1 and y by n, the degree of x can only increase by one. Indeed, x cannot be connected to n 1 and n, as L is an independent set. x S x / L will increase by at most 1. Concerning future layers, the analysis will become more complex, as we will have to take weights into account.. Algorithm and Analysis Presentation of the Algorithm. As described previously, Algorithm 1 picks successively an independent set among the vertices of lower weights. It also updates the weights according to the picked vertices. For technical reasons, the weights are not exactly equal to their degree in the constructed solution. Indeed, when restructuring an optimal solution to match L i we will see that the degree of almost all surviving vertices in the optimal solution increases by at most 1 this is why we add a bonus of 1 in the updated weight Line 13, and even sometimes cannot increases this is why there is no bonus Line 11. his modification will allow us to show that at the end of the algorithm, the value W returned by the algorithm is a lower bound of the optimal value Lemma 3. We will then show that the real value of the returned solution costs is less than two times W Lemma 4, and thus is a -approximation. Remark 1. he maximum independent set of line 3 is greedily constructed as follows: pick the first vertex of the simplicial elimination order in the independent set, delete its neighbourhood, and repeat the operation until the graph becomes empty. Even if we sometimes add 1 when updating the weights, we can observe that for a fixed x V, its successive weights are non decreasing: Lemma 1. For all i {0,..., t}, x V \ L 0 L i, w i+1 x i + 1. Proof. Let i and x be as in the statement. Suppose by induction that w i x i notice that w 0 x = 0. If w i x i + 1 then the results follows. Otherwise w i x = i, and by construction of the algorithm Line 11, if dx, L i 1, then w i+1 x i + 1. Finally, if dx, L i = 0, then x must belong to L i which contradicts the definition of x. Restructuration of Solutions. 1 Let S be an optimal solution for the Sparsest k-subgraph problem in chordal graphs. We will now show that we can modify this solution in order to obtain the output of the algorithm, while bounding the cost variation.

5 5 Algorithm 1 A -approximation for Sparsest k-subgraph in chordal graphs 1: S, W 0, i 0, w 0x = 0 x V : while S k do 3: L i a maximum independent set of the graph induced by {x V \L 0... L i 1 : w ix = i} 4: S S L i // or the k S L i leftmost vertices of L i if S L i > k 5: W W + i L i // we update the cost computed by the algorithm 6: for x V do 7: if x L 0... L i then 8: w i+1x = w ix 9: else 10: if dx, L i = 0 OR dx, L i = 1 AND w ix = i then 11: w i+1x = w ix + dx, L i 1: else 13: w i+1x = w ix + dx, L i 1 14: i i : t i 1 // L t is the last layer of the algorithm 16: return S, W Let us define by induction a sequence Si i= 1,0,...,t with S 1 = S and St = S the solution returned by the algorithm, such that Si V and S i = k for all i = 1...t. We also assure that L 0... L i Si for all i = 0...t. o that end, given i {0,..., t}, we show how to restructure the set Si 1 into a new set S i. Let us first introduce some notations. We partition the set L i defined in the algorithm into two sets of vertices, whether they belong to Si 1 or not: L i = M i N i, with M i = L i Si 1. he restructuration consists in adding all vertices of N i to Si 1, and removing a carefully chosen see Definition 1 subset D i Si 1 \L 0... L i with D i = N i. hen, we will define Si = Si 1 \D i N i, R i = Si 1 \D i L 0... L i and i = M i D i. Figure 3 summarizes the situation. o bound the cost variation, we show in Lemma that the degree of surviving vertices i.e. vertices in R i increases by at most one. he next definition shows how to choose properly the set D i. Definition 1. For i {0,..., t}, let D i be defined as follows: Let us suppose we are given a set Si 1 L 0... L i 1, and show how to choose properly the set D i Si 1 \ L 0... L i. Let N i = {n 1,..., n pi } and suppose that n 1 <... < n pi defines an ordering of N i according to the simplicial elimination order of the graph. For all j = 1...p i successively, we pick a vertex y j Si 1 \L 0... L i as follows: { minq y j = j if Q j maxsi 1 \L L i if Q j = with Q j = {y S i 1 \L 0... L i such that n j < y, and {n j, y} E} see Figure 3. Finally, we define D i = {y j : 1 j p i }. Now that D i is defined, recall that we have i = M i D i, and the surviving vertices R i = R i 1 \ i. Let us now upper bound the degree of vertices of R i. Lemma. Let R i = A i B i, with A i = {x R i : dx, L i = 0 or dx, L i = 1 and w i x = i} and B i = R i \A i. We have:

6 6 S i 1 D i i R i n j 1 y j 1 n j y j y j+1 n j+1 L i N i M i Li 1 Q j 1 Q j Q j+1 = L 0 Fig. 3: On the left: description of set S i 1. We obtain S i from S i 1 by removing D i and adding N i. Notice that R i 1 = R i i, and R i i =. On the right: example of sets Q j, together with y j. Circles represent vertices of the considered optimal solution, and crosses represent vertices chosen by the algorithm that are not in the optimal solution. Edges between vertices of the optimal solution have not been drawn for sake of simplicity. if x A i, dx, L i dx, i if x B i, dx, L i dx, i + 1 his immediatly implies that for all x R i we have w i+1 x dx, i + w i x. Proof. Let us show that if x A i, then dx, N i dx, D i, and if x B i, then dx, N i dx, D i + 1. Since L i = M i N i and i = M i D i these unions being disjoint, the desired inequalities follow immediatly. if x A i, then either dx, L i = 0, which obviously implies the result, or dx, L i = 1 and w i x = i. We thus only consider the second case. Here again if dx, N i = 0 then the result is straightforward, so let us suppose dx, N i = 1, i.e. suppose that there exists a vertex of N i, say n j0, such that x and n j0 are adjacent. wo cases are possible: First case: x < n j0. Recall that n j0 is the only neighbour of x in L i. Hence, x is not adjacent to all vertices of L i that are before n j0 in the simplicial elimination order. In addition, recall that w i x = i. hus, this case cannot happen since by definition of the algorithm, x would have been chosen in L i instead of n j0. Second case: n j0 < x. It is clear that in this case Q j0 as at least x Q j0. As by definition x / D i, we have y j0 < x. By definition of chordal graphs, since {n j0, y j0 } E and {n j0, x} E, we must have {x, y j0 } E. Hence dx, D i = 1 and the result follows. if x B i, then let N i = N i {y N i : y < x} and N + i = N i \ N i. For all n j N i such that {n j, x} E, then as previously Q j and by the definition of chordal graphs we have {y j, x} E with y j D i. hus, dx, N i dx, D i. Finally we claim that dx, N + i 1. Indeed, suppose that there exists 1 n j1, n j N + i such that {x, n j1 }, {x, n j } E. By definition of chordal graphs we must have {n j1, n j } E which contradicts the definition of L i which is an independent set. his proves that dx, N i dx, D i + 1 1

7 7 S i Let us now define the appropriate ζ function that computes the cost of an intermediate solution. For all i {0,..., t}, let ζs i = costr i + x R i w i+1 x + Notice that ζs 1 = costs and ζs t+1 = x S w tx = W. Lemma 3. For all i {0,..., t}, D i is such that ζs i ζs i 1. Proof. By definition, we have: x L 0 L i w i+1 x ζs i = costr i + x R i w i+1 x + x L 0 L i w i+1 x = costr i + x R i w i+1 x + i L i + x L 0 L i 1 w i+1 x costr i + x R i w i x + dx, i + i L i + x L 0 L i 1 w i x by Lemma In addition, since R i 1 = R i i and i = L i, we have: ζs i 1 = costr i 1 + x R i 1 w i x + x L 0 L i 1 w i x = costr i + costr i, i + x R i w i x + x i w i x + x L 0 L i 1 w i x costr i + costr i, i + x R i w i x + i L i + x L 0 L i 1 w i x which matches the upper bound for ζs i. he previous lemma implies that W = ζs t ζs 1 = costs. hus, to prove that Algorithm 1 is a -approximation we only need the following lemma. Lemma 4. costs W. Proof. Roughly speaking, when creating a layer L i and updating the cost W, the algorithm adds i L i, i.e. for all x L i the algorithm only adds i instead of dx, L 0 L i 1. hus, we will now prove for any x L i, dx, L 0 L i 1 i. Let x in L i. For any l, 0 l i, let x l = w l x be the weight of x before creating layer L l. hus, the successive weights of x is a sequence x 0,..., x i where x 0 = 0 and x i = i. Notice that after x is added in L i its weight will not be changed. Let us show by induction that for any l, dx, L 0 L l 1 x l + l. Let us suppose that the previous statement is true for l and prove it for l + 1. Let z = dx, L l. We have dx, L 0 L l = dx, L 0 L l 1 + z x l + l + z. As x l+1 x l + z 1, we get the desired inequality. hus, for any x L i we get dx, L 0 L i 1 x i +i = i, and thus costs = t i=1 x L i dx, L 0 L i 1 t i=1 i L i = W. heorem 1. here is a tight polynomial -approximation algorithm for SkS in chordal graphs. For the tightness result, consider the instance with n = 5, k = 4, and edges {x 1, x }, {x, x 3 } and {x 4, x 5 } notice that x 1, x, x 3, x 4, x 5 is a simplicial elimination order. he algorithm will first pick x 1, x 3 and x 4. hen, we have w 1 x = w 1 x 5 = 1 and the algorithm takes x instead of x 5.

8 8 3 N P-hardness in Chordal Graphs Main Arguments. he following N P-hardness proof is a reduction from the k-clique problem in general graphs. Roughly speaking, given an input instance G = V, E together with k N, we construct the split graph of adjacencies of G, i.e. we build a clique on a set A representing the vertices of G, and an independent set F representing the edges of G, connecting A and F with respect to the adjacencies of the graph. hen, we replace each vertex of the independent set corresponding to an edge e E by a gadget F e represented in Figure 4. Any solution will have to take the same number of vertices among each gadget. he key idea is that there is two ways to take these vertices in a gadget F e. he first way choosing X e and Z e encodes that the edge e belongs to the k-clique. It is cheaper than the second way, but is adjacent to the clique A. he second way choosing X e and Y e encodes that edge e does not belong to the k-clique. It induces more edges, but is not adjacent to the clique A. hus, as depicted in Figure 4, a k-clique is encoded by not picking the corresponding vertices in A, obtaining k gadgets of the first type, and m k of the second type. In this way, there is no edge in the solution between any gadget and the clique A. For technical reasons, each vertex of A is duplicated n times. Gadget. Let us define the gadget F mentioned above. F is composed of three sets X, Y and Z of vertices each we will set the value of later. We define X = {x 1,..., x }, Y = {y 1,..., y } and Z = {z 1,..., z }. the set X induces an independent set, while Z induces a clique, and there is a clique of size 1 on vertices {y,..., y }. For all i {1,..., }, x i is adjacent to y i, and y i is adjacent to all vertices of Z. Such a construction is depicted at the left of Figure 4. In the following we will force the solution to take vertices among each gadget. It is easy to see that the sparsest -subgraph of F is composed of the sets X and Z, which induces edges. In contrast, notice that choosing X and Y induces + 1 edges. heorem. Sparsest k-subgraph remains N P-hard in chordal graphs. Proof. We reduce from the classical k-clique problem in general graphs. Let G = V, E and k N. We note V = n, V = {v 1,..., v n }, E = m and = nn k. In the following we will define G = V, E together with k, C N such that G is a chordal graphs which can be constructed in polynomial time, and such that G contains a clique of size k if and only if one can find k vertices in G which induce C edges or less. he Construction. V is composed of two parts A and F : We first define a clique of size n over A = {a j i : i, j {1,..., n}}. For each u V, the column A u = {a j u : j {1,..., n}} represents the vertex u in G. For all e E, we construct a gadget F e composed of X e, Y e and Z e as defined previously. Let X e = {x e 1,..., x e }, Y e = {y1, e..., y e } and Z e = {z1, e..., z e }. Moreover, for all e = {v p, v q } E, all vertices of Z e are connected to A p and A q. We define k = m + and C = m + + m k. It is clear that the construction can be carried out in polynomial time. Let us briefly sketch that G is a chordal graph: for each gadget, X e, Y e, Z e is a simplicial elimination order. hen, the remaining vertices form a clique. See Appendix A for a more rigorous proof. Now we prove that G contains a clique of size k if and only if G contains k vertices inducing at most C edges.

9 9 gadget F e1 for e 1 = {u, v} E k gadgets m k gadgets X e1 F e1 F em Y e1 X e1 Y e1 Z e1 X em Y em Z em Z e1 n n A k n k n A u A v n Fig. 4: Schema of the reduction, with an example of a gadget F e1 on the left and its relations to A. Grey rectangles represent vertices of the solution. G contains a k-clique G contains k vertices inducing at most C edges. Let us suppose that K V is a clique of size k in G. W.l.o.g. we suppose K = {v 1,..., v k }. Moreover, we note E 0 = {{v p, v q } E such that v p, v q K} and E 1 = {{v p, v q } E such that v p / K or v q / K}. We construct K V as follows: For all i {k + 1,..., n} and all j = {1,..., n}, we add a j i to K. For all e E, we add all vertices of X e to K. For all e E 0, we add all vertices of Z e to K. For all e E 1, we add all vertices of Y e to K. One can verify that K is a set of k = m + vertices inducing exactly C = +m +m k edges. Indeed, we picked = nn k vertices from1 A which is a clique and thus induce edges. hen, for all e E, we picked vertices, which induce edges if e E0, and + 1 edges if e E 1. Since E 0 = k and thus E1 = m k, we have the desired number of edges. G contains a k-clique G contains k vertices inducing at most C edges. Suppose now that K is a set of k vertices of G which induces at most C edges. We re-define the sets E 0 and E 1 as follows: E 0 = {{v p, v q } E such that for all j {1,..., n} we have a j p / K and a j q / K }, and E 1 = E\E 0. Roughly speaking, E 0 denotes the set of gadgets not adjacent to vertices of the solution among A, and E 1 the set of gadgets adjacent to at least one vertex of the solution among A. For all R V, let trr = K R be the trace of K on R, and for all v V, let µv = trnv be the number of neighbours of v belonging to K.

10 10 Let u K and v V \K. We say that K \{u} {v} is a safe replacement if we have µv µu if {u, v} / E and µv 1 µu if {u, v} E. For sake of readability, we will keep and update the definitions of E 0 and E 1 when replacing vertices of A e.g. if we remove a vertex u A from K such that there exists e E 1 such that vertices of Z e were only adjacent to u among all vertices of tra, then e now belongs to E 0. he proof consists in replacing some vertices of K by other vertices not in K without increasing the number of induced edges, in order to obtain a solution that has the same structure as previously. We call such a replacement a safe modification or a safe replacement. he core of the proof is based on the three following lemmas. Lemma 5. Without loss of generality and optimality of K, we can suppose that for all e E we have X e K. Proof. Let S = e E X e. Since we have k > S, there always exists u K \S. Suppose that there exists e E and i {1,..., } such that x e i / K. If yi e / K, then we have µx e i = 0 and we can thus safely replace any other vertex of K \S by x e i. Now, if ye 1 K, then µx e i = 1. Since {x e i, ye i } E, K \{y1} e {x e i } is a safe replacement. Lemma 6. See App. A. K can be safely modified such that one of the two following holds: Case A1: for all e E 0 we have trz e = Z e. Case A: for all e E 0 we have try e =. Lemma 7. See App. A. K can be safely modified such that one of the two following holds: Case B1: for all e E 1 we have try e = Y e. Case B: for all e E 1 we have trz e =. Let us now define for each case and each e E the set of vertices D e Y e Z e that have to be replaced see Figure 5 in Appendix A.3: case A1: for all e E 0, D e = Y e K case A: for all e E 0, D e = Z e \ K case B1: for all e E 1, D e = Z e K case B: for all e E 1, D e = Y e \ K Notice that if D e = for all e E 0 resp. e E 1, then cases A1 and A resp. B1 and B collapse. If such a case happen for all e E, we can immediately conclude, as shown by the following lemma: Lemma 8. If D e = for all e E, then G contains a clique of size k. Proof. By construction, we have tra = and trf e = for all e E. hus, costtra = and costtrfe = + 1 if Ye K, and costtrf e = if Ze K. By construction, Y e K if and only if e E 1. hus, since costk + m + m k, we must have E 1 m k which is equivalent to E0 k. Hence, there exists at most A n = k vertices in G inducing at least k edges, i.e. G contains a clique of size k. We now have to analyse the four cases of Lemma 6 and 7. We only detail here the first one case A1 and B1, and refer the reader to Appendix A.3 for the other ones. Let us consider case A1 and B1. o summarize the situation, the solution K can be partitioned in K A = K A, and K F = K \ K A, the vertices selected in the gadgets. Let 0 = e E 0 D e be

11 11 the number of extra vertices allocated in all the gadgets F e, e E 0, and 1 = e E 1 D e be the number of extra vertices allocated in all the gadgets F e, e E 1. Let = Notice that we have K A =, as a regular solution that does not select any extra vertex in a gadget has to pick vertices in A. Moreover, vertices of K selected in gadgets of E 0 are not adjacent to K A by definition of E 0 each gadget of E 0 induces at least edges as we are in case A1 each gadget of E 1 induces at least + 1 edges as we are in case B1 each of the 0 vertices is adjacent to at least vertices in K such a vertex is in a set Y e, and thus is connected to the vertcies of Z e each of the 1 vertices is adjacent to at least + 1 vertices in K such a vertex is in a set Z e, and thus is connected to at least 1 vertex of K A and to the vertices of Y e Let us now lower bound the total cost of K. We have: costk E 0 + E = m + E m + m E Notice that in a bad structured solution, a large allows to select only a few vertices in A instead of, and thus to have many gadgets more than k in E0. Let us now consider the contrapositive, i.e. we consider that G does not contain a k-clique, and show that K induces more than C edges. Let q and r such that = qn + r, r < n. Let us upper bound E 0. As there are vertices in A, the number of empty columns column u is empty iff none of the a t u is selected is at most n n k + q. As G does not contain a k-clique, the k + q vertices corresponding to these k + q columns cannot induce a clique of size k + q, and thus E 0 < k+q. hus, we get: k + q + m costk > m q = C + kq hus, as > q + kq, we get the desired inequality. We refer now the reader to Appendix A.3 for the three other cases. 4 Approximation in Proper Interval Graphs Let us now discuss the status of Sparsest k-subgraph and Densest k-subgraph on interval graphs. First, notice that the complexity status N P-hardness versus P of Sparsest k-subgraph remains unknown in interval and proper interval graphs. We also recall that this question is a longstanding open problem for DkS, as well as its complexity in planar graphs. Indeed, the former paper [9] proves the N P-hardness of DkS in comparability, chordal graphs, and states the open

12 1 question of its complexity in planar and proper interval graphs. Since then, and despite a lot of effort, no major improvement has been done so far. As interval graphs are exactly the intersection of chordal graphs and co-comparability graphs, finding out the complexity status of Sparsest k-subgraph in interval graphs would determine the complexity of Densest k-subgraph in a subclass of comparability graphs, improving the results of [9]. Finally, as in [16] where the author design a P AS for Densest k-subgraph on interval graph despite the unknown complexity status, we show in Appendix B the following theorem. heorem 3. here is a P AS for SkS in proper intervals running in n O 1 ɛ. his result uses the same kind of arguments as in [16]: restructuring an optimal solution in each block of consecutive intervals, and using dynamic programing on these restructured blocks. References 1. N. Apollonio and A. Sebő. Minconvex factors of prescribed size in graphs. SIAM Journal of Discrete Mathematics, 33: , N. Apollonio and B. Simeone. he maximum vertex coverage problem on bipartite graphs. preprint, A. Bhaskara, M. Charikar, E. Chlamtac, U. Feige, and A. Vijayaraghavan. Detecting high log-densities: an On 1/4 approximation for densest k-subgraph. In Proceedings of the 4nd ACM symposium on heory of Computing, pages ACM, N. Bourgeois, A. Giannakos, G. Lucarelli, I. Milis, V. h. Paschos, and O. Pottié. he max quasiindependent set problem. Journal of Combinatorial Optimization, 31:94 117, H. Broersma, P. A. Golovach, and V. Patel. ight complexity bounds for FP subgraph problems parameterized by clique-width. In Proceedings of the 6th international conference on Parameterized and Exact Computation, IPEC 11, pages 07 18, Berlin, Heidelberg, 01. Springer-Verlag. 6. N. Bshouty and L. Burroughs. Massaging a linear programming solution to give a -approximation for a generalization of the vertex cover problem. In Proceedings of the 15th Annual Symposium on heoretical Aspects of Computer Science, pages Springer, Leizhen Cai. Parameterized complexity of cardinality constrained optimization problems. Computer Journal, 511:10 11, D. Chen, R. Fleischer, and J. Li. Densest k-subgraph approximation on intersection graphs. In Proceedings of the 8th international conference on Approximation and online algorithms, pages Springer, D.G. Corneil and Y. Perl. Clustering and domination in perfect graphs. Discrete Applied Mathematics, 91:7 39, D. Fulkerson and O. Gross. Incidence matrices and interval graphs. Pacific J. Math., 15: , O. Goldschmidt and D. S. Hochbaum. k-edge subgraph problems. Discrete Applied Mathematics, 74: , J. Guo, R. Niedermeier, and S. Wernicke. Parameterized complexity of vertex cover variants. heory of Computing Systems, 413:501 50, G. Joret and A. Vetta. Reducing the rank of a matroid. CoRR, abs/ , J. Kneis, A. Langer, and P. Rossmanith. In Proceedings of the 34th Workshop of Graph heoretic Concepts in Computer Science, pages Springer, M. Liazi, I. Milis, and V. Zissimopoulos. A constant approximation algorithm for the densest k-subgraph problem on chordal graphs. Information Processing Letters, 1081:9 3, Nonner. PAS for densest k-subgraph in interval graphs. In Proceedings of the 1th international conference on Algorithms and Data Structures, pages Springer, R. Watrigant, M. Bougeret, and R. Giroudeau. he k-sparsest subgraph problem. echnical Report RR-1019, LIRMM, 01.

13 13 A Missing Proofs of the N P-hardness A.1 Proof of claim G is a chordal graph We have the following simplicial elimination scheme: For all e E, we can remove X e since for all j {1,..., }, x e j is only connected to ye j. For all e E, we can remove Y e. Indeed the remaining neighbourhood of y1 e is Z e which is a clique. And the remaining neighbourhood of yj e with j is a subset Y e Z e \ {y1} e which induces a clique. For all e E, we can remove Z e since the remaining neighbourhood of zj e is a subset of Ze and vertices of A which induce a clique. he remaining vertices induce a clique on A and can thus be eliminated. A. Proof of Lemma 6 and 7 safe replacements Proof Proof of Lemma 6. Let us first restructure each gadget of E 0 separately. For all e E 0 such that try e and trz e Z e, let j 0 = max{j {1,..., } : yj e try e} and let j 1 be such that zj e 1 / trz e. Recall that Lemma 5 ensures that x e j 0 is in K. If j 0 1, then µyj e 0 = y + z + 1, where y = Nyj e 0 try e and z = Nyj e 0 trz e. On the other side, we have µzj e 1 y + z + 1 more precisely, µzj e 1 = y +z +1 if y1 e K, and µzj e 1 = y +z if y1 e / K. Roughly speaking, this switch ensures that we necessarily loose the edge due to the vertex of X e and we gain at most one edge due to y1. e Hence µzj e 1 µyj e 0 and K \{yj e 0 } {zj e 1 } is a safe replacement. If j 0 = 1, then it means that try e = {y1}. e Suppose that there exists j 1 such that zj e 1 / trz e. We have µy1 e = z + 1 where z = Ny1 e trz e, and µzj e 1 = z + 1. Here again K \{y1} e {zj e 1 } is a safe replacement. After all these replacements, given any e E 0, try e implies that trz e = Z e. hen, we proceed to replacements between gadgets F e, e E 0. If one can find a, b E 0 such that try a and trz b Z b, then let j 0 be such that yj a 0 try a and let j 1 be such that zj b 1 / trz b. We have µyj a and µzj b 1 1. hus, K \{yj a 0 } {zj b 1 } is a safe replacement. heses replacements end either when all the Y e are empty for all e E 0 or when all the Z e are full for all e E 0, which achieves the proof of Lemma 6. Proof Proof of Lemma 7. he proof is roughly based on the fact that replacing a vertex of Z e by a vertex of Y e permits to loose at least one edge with vertices A and gain one edge with a vertex of X e. Let us formally prove Lemma 7. Similarly to the proof of Lemma 6, we first restructure each gadget of E 1 separately: for all e E 1 such that trz e and try e Y e, let j 0 = max{j {1,..., } : yj e / K } and let j 1 be such that zj e 1 trz e. Recall that by definition of E 1, there exists i, j {1,..., n} such that zj e 1 is adjacent to a j i. We have µze j 1 y + z + 1, where y = Nzj e 1 Y e and z = Nzj e 1 Z e. On the other side, we have µyj e 0 z + y + indeed, Nyj e 0 Z e = z + 1, Nyj e 0 Y e y and Nyj e 0 X e = 1. Since {yj e 0, zj e 1 } E, it holds that K \{z j1 } {y j0 } is a safe replacement. After all these replacements, given any e E 1, trz e implies that try e = Y e. We now proceed to replacements between gadgets F e, e E 1. If one can find a, b E 1 such that trz a and try b Y b, then let j 0 be such that yj b 0 / try b and let j 1 be such that zj a 1 trz a. We have µzj a and µyj b 0 1. hus K \{z j1 } {y j1 } is a safe replacement.

14 14 A.3 End of N P-hardness proof: the three other cases e E 0 e E 0 e E 1 e E 1 X e Y e D e Z e D e D e case A1 case A case B1 D e case B Fig. 5: Schema of different cases. Shaded rectangles represent parts of K. Case A and B Let 0 = e E 0 D e, 1 = e E 1 D e and = recall that in this case, D e K for all e E. Here again we suppose > 0. Let us notice that for all u tra, µu. On the other hand, for all e E such that there exists v D e, we have µv remark that if e E 1, then D e Y e, and if e E 0, then v is not adjacent to tra by definition of E 0. hus K \{u} {v} is a safe replacement. Since before this replacement we had tra = +, it is clear that we can repeat this replacement i.e. K \{u} {v} where u tra and v D e for some e E times safely. At this point, the updated value of is 0, i.e. D e = for all e E. By Lemma 8, we must have a clique of size k in G. Case A and B1 If there exists e E 0 such that there exists u D e, then µu <. If such a vertex exists, then either tra > or there exists e E 1 such that there exists v D e. In the first case for all x tra we have µx, and K \ {x} {u} is a safe replacement. In the second case we have µv > and here again K \ {v} {u} is a safe replacement. After these replacements we must have D e = for all e E 0, and we can apply the same arguments as for case A1 and B1. Case A1 and B If there exists e E 1 such that there exists u D e, then µu <. If such a vertex exists, then either tra > or there exists e E 0 such that there exists v D e. In the first case for all x tra we have µx, and K \ {x} {u} is a safe replacement. In the second case we have µv > and here again K \ {v} {u} is a safe replacement. After these replacements we must have D e = for all e E 1, and we can apply the same arguments as for case A1 and B1. 1 B PAS for Proper Intervals Graphs In this section we design a PAS for Sparsest k-subgraph in proper interval graphs. We first assume that the instance has one connected component. We prove that we can re-structure an

15 15 optimal solution Opt into a near optimal solution Opt such that the pattern used in Opt in each block a block corresponds to a separator in the input graph is simple enough to be enumerated in polynomial time. hen, a dynamic programming algorithm will process the graph blocks by blocks from left to right and enumerate for each one all the possible patterns. Definitions Let us define some notations that will be used in the algorithm. Recall that we are given a set of proper intervals I = {I 1,..., I n } sorted by their right endpoints and by their left endpoints equivalently. First, we define by induction the following decomposition of the input graph see Figure 6. Let I m1 = I 1, L 1 = I m1, R 1 = {I j, j > m 1, I j overlaps I m1 }. hen, given any i 1 we define while there remains some intervals after R i : I mi+1 is the rightmost interval of the set X = {I / R i, I R i s.t. I overlaps I } X is well defined as the instance has a unique connected component L i+1 = {I j, j m i+1, I j overlaps I mi+1 and I j / R i } R i+1 = {I j, j > m i+1, I j overlaps I mi+1 }. I m1 B 1 R 1 L B I m R L 3 Fig. 6: Schema of the decomposition used in the P AS. Let a denote the maximum i such that I mi is defined. Notice that R a may be empty, and that I mi L i for all i {1,..., a}. For any i {1,..., a} we define the block i as B i = L i Ri. hus, the set of intervals is partitioned into blocs B i for 1 i a. For any 1 i a and any solution S a subset of k intervals, let L S i = L i S, R S i = R i S, and Bi S = B i S. Notice that for any S and i, intervals of Ri S do not intersect intervals of Ri 1 S, and intervals of L S i do not intersect I mi 1 nor intervals of L S i 1.

16 16 We can now write the cost of a solution S as the sum of the costs inside the blocks and the costs between the blocks. hus, we have costs = a i=1 costbs i + a 1 i=1 costrs i, LS i+1. Indeed, by definition, the only edges between blocks B i and B i+1 are edges between R i and L i+1. Compacting blocks Let Comp be an injective function from I to I. For any S I, we define CompS = I S CompI. he function Comp is called a compaction if for any S I and any 1 i a the following holds: for all I R S i we have CompI R i and rcompi ri. for all I L S i we have CompI L i and ri rcompi. Roughly speaking, a compaction pushes intervals of Bi S toward the center I mi. he idea is that a compaction may increase the cost of a solution inside the blocks, but cannot increase the costs between the blocks. hus, let us define a ρ-compaction as a compaction Comp such that for any S I and for all i {1,..., a} we have costcompbi S ρ.costbs i. Lemma 9. If Comp is a ρ-compaction, then for any solution S, costcomps ρ.costs. Proof. By definition of the decomposition, we have We now prove that costcomps = a i=1 a i=1 a 1 costcompbi S + costcompri S, CompL S i+1 i=1 a 1 ρ.costbi S + costcompri S, CompL S i+1 i=1 a 1 a 1 costcompri S, CompL S i+1 costri S, L S i+1 i=1 i=1 Indeed, let I R Ri S and I L L S i+1 such that I R and I L do not overlap. hen by definition of a compaction, we have rcompi R ri R and li L lcompi L. hus, intervals CompI R and CompI L do not overlap as well, which proves the result. According to the previous lemma, we only have now to find compactions that preserve costs inside the blocks. Given a fixed ɛ, the objective is now to define a 1 + ɛ-compaction that has a simple structure.

17 17 G L 1 I L 1 L i G L I L G L 3 I3 L I mi Fig. 7: Example of a compaction of a set X for a block L i, with P = 3, and x L = 7. Intervals marked with a cross represent X. Intervals marked with a circle represent CompX. Lemma 10. For any fixed P N, there is a P -compaction such that for any X, CompX can be described by P + 4 variables ranging in {0,..., n}. Proof. According to Lemma 9, we only describe CompX for X B i, given any 1 i a. Let X = X L X R with X L L i and X R R i. We define x L = X L, x R = X R. Moreover, we set x L = q L P + r L with r L < P and x R = q R P + r R with r R < P. Let us split X L into P subsets G L t 1 t P of consecutive intervals in the ordering of their right endpoints, with G L t = q L + 1 for t {1,..., r L } and G L t = q L for t {r L + 1,..., P } see Figure 7. Similarly, we split X R into P subsets G R t 1 t P of consecutive intervals, with G R t = q R + 1 for t {1,..., r R } and G R t = q R for t {r R + 1,..., P }. For all t {1,..., P }, let It L resp. It R be the rightmost resp. leftmost interval of G L t resp. G R t. he principle of the compaction is to flush every intervals of G L t resp. G R t to the right resp. left. hus, for t {1,..., r L }, CompG L t is defined as the q L + 1-rightmost intervals I such that ri rit L, and for t {r L + 1,..., P }, CompG L t is defined as the q L -rightmost intervals I such that ri rit L. Similarly, for t {1,..., r R }, CompG R t is defined as the q R + 1-leftmost intervals I such that rit R ri, and for t {r R + 1,..., P }, CompG R t is defined as the q R -leftmost intervals I such that rit R ri. he construction for a block L i is depicted in Figure 7. It is clear that the mapping Comp described above is a compaction. Moreover, given x L, r L, x R, r R and It L resp. It R for all 1 t P, we are clearly able to construct1compx in polynomial time. hus, it remains to prove that Comp is a P -compaction. One can easily show the two following key arguments: i all intervals of L i form a clique, as well as all intervals of R i. ii for any t 1, t {1,..., P } with t 1 P and t 1, if an interval of CompG L t 1 overlaps an interval of CompG R t, then for any s 1 {t 1 + 1,..., P } and any s {1,..., t 1}, all intervals of G L s 1 overlap all intervals of G R s. For all t {1,..., P }, we define x L t = G L t = CompG L t, x R t = G R t = CompG R t. By our construction and i, we have costcompx xl + xr + P costcompg L t, CompX R i t=1

18 18 costx xl + xr + P costg L t, X R i hen, for all t {1,..., P }, let λ t {0, 1,..., P } be the maximum s such that an interval of CompG L t overlaps an interval of CompG R s we set λ t = 0 if no interval of CompG L t overlaps an interval of CompG R 1. By ii, for all t {1,..., P }, we have costcompg L t, CompX R i x L λt t u=1 xr u and for all t {,..., P }, we have costg L t, X R i x L λt 1 1 t u=1 x R u since some intervals of G L t 1 overlap some intervals of G R λ t 1, it implies that all intervals of G L t overlap all intervals of G R λ. t 1 1 Combining the previous inequalities, we now have hus, we have costcompx costx xl xl + + xr t=1 xr + + P t= x L t P t=1 x L t λ t 1 1 u=1 λ t x R u u=1 x R u = costcompx costx x L 1 λ 1 x R u + P u=1 t= x L t λ t u=λ t 1 x R u As in our case we have x L t q L + 1, we get q L + 1 λ P u=1 xr u + P u=1 xr λ u q L + 1x R x L P + 1x R. It remains now to handle particular cases, according to the values of x L and x R. If x L P, then x L + 1x R 4 P x Lx R, and P costx 4 P x Lx R x L 1x L +x R 1x R 4 P x Lx R 1 4 x L +x R P we lower bounded x R 1 by x R as cases with x R 1 lead to even better ratio. If x L < P, then we set CompX L i = X L i.e. we keep the left part unchanged. If x R < P +1, then we set CompX = X and we get a 1-compaction. Notice that in these cases we are still able to construct CompX in polynomial time. Suppose now that x R P +1. One can improve the previous lower bound and write costx x L 1x L + x R 1x R + P t=1 xl t λ t 1 u=1 xr u. Indeed, for all t {1,..., x L } the set G L t is a singleton and G L t = for t {x L +1,..., P }, and thus the interval of G L t overlaps some intervals of G R λ t, which implies that it overlaps all intervals of G R λ. t 1 hus, we get P t=1 xl t x R λ t x L t=1 xr λ t x R, and costx x R x L 1x L +x R 1x R P, which terminates the proof of the lemma. Algorithm Let us now write a dynamic programming algorithm for the instances that have a unique connected component we will drop this hypothesis after. Let Opt be an optimal solution, P a fixed integer and Comp the previous P -compaction. he algorithm constructs a solution which is at least as good as CompOpt by enumerating for all blocks all the possible compacted patterns i.e. all the possible CompX.

19 19 Let us now define more formally the algorithm, starting with the parameters. he first parameter k k is the number of interval to choose. i is the starting block, meaning that the k interval must be chosen in a l=i B l. Finally, Bi 1 S represents the set of P + 4 variables that encode the set of intervals X i 1 chosen in block i 1. Since we can construct X i 1 from Bi 1 S in polynomial time, we will directly use Bi 1 S to denote X i 1, for the sake of readability. Algorithm DP k, i, B S i 1 // For the sake of clarity we drop the classical operations related to the marking // table that avoid multiple computations with same arguments // We also drop the base case i = a + 1 i.e. there are no more remaining intervals in the instance Ω all possible patterns for block i using less or equal than k intervals return arg min B Ω costb S i 1 B DP k B, i + 1, B Lemma 11. For any P, DP k, 1, outputs a 1+ 4 P -approximation for the Sparsest k-subgraph in n OP. Proof. he objective is to prove that costdp k, 1, costcompopt, where Comp is the previous 1+ 4 p -compaction. According to Lemma 10, it is sufficient to get a 1+ 4 p -approximation. For sake of readability, for all i {1,..., a}, we define B i = CompOpt B i and k i = a l=i B i. We prove by induction on i starting from i = a + 1 that costb i 1 DP k i, i, B i 1 costcompopt a l=i 1 B l. Let us suppose that the hypothesis is true for i + 1 and prove it for i. Considering the iteration where DP chooses B = B i. costb i 1 DP k i, i, B i 1 costb i 1 + costb i 1, B i + DP k i B i, i + 1, B i recall that costx 1, X = {I l, I l E, I l X 1, I l X }. Using the induction hypothesis we get the desired result. he dependency in P in the running time is due to the n P +O1 possible values for the set of parameters and the branching time in n P +O1 when enumerating sets Bi S. Finally, let us extend the previous result to instances having several connected component. We only sketch briefly the algorithm as it follows the same idea as [8] for the k-densest subgraph problem. Let us suppose that for any k k we have an algorithm Ak, X which is a ρ-approximation for k -sparsest subgraph on a instance X having one connected component. Let C i 1 i x denote the connected components of a general instance of Sparsest k-subgraph. It is sufficient to define a dynamic programming algorithm DP k, i that computes a ρ approximation of the k -sparsest subgraph on x t=i C t by keeping the best of all the Al, C i + DP k l, i + 1, for 1 l k. hus, we get the following result: heorem 4. here is a P AS for Sparsest k-subgraph on proper interval graphs running in n O 1 ɛ