A Parameterized Complexity View on Collapsing k-Cores

We study the NP-hard graph problem Collapsed k-Core where, given an undirected graph G and integers b, x, and k, we are asked to remove b vertices such that the k-core of remaining graph, that is, the (uniquely determined) largest induced subgraph with minimum degree k, has size at most x. Collapsed k-Core was introduced by Zhang et al. [AAAI 2017] and it is motivated by the study of engagement behavior of users in a social network and measuring the resilience of a network against user drop outs. Collapsed k-Core is a generalization of r-Degenerate Vertex Deletion (which is known to be NP-hard for all r>= 0) where, given an undirected graph G and integers b and r, we are asked to remove b vertices such that the remaining graph is r-degenerate, that is, every its subgraph has minimum degree at most r. We investigate the parameterized complexity of Collapsed k-Core with respect to the parameters b, x, and k, and several structural parameters of the input graph. We reveal a dichotomy in the computational complexity of Collapsed k-Core for k<= 2 and k>= 3. For the latter case it is known that for all x>= 0 Collapsed k-Core is W[P]-hard when parameterized by b. We show that Collapsed k-Core is W[1]-hard when parameterized by b and in FPT when parameterized by (b + x) if k<= 2. Furthermore, we show that Collapsed k-Core is in FPT when parameterized by the treewidth of the input graph and presumably does not admit a polynomial kernel when parameterized by the vertex cover number of the input graph.


Introduction
In recent years, modelling user engagement in social networks has received substantial interest [3,7,13,24,25,26]. A popular assumption is that a user engages in a social network platform if she has at least a certain number of contacts, say k, on the platform. Further, she is inclined to abandon the social network if she has less than k contacts [3,6,7,13,21,26]. In compliance with this assumption, a suitable graph-theoretic model for the "stable" part of a social network is the so-called k-core of the social network graph, that is, the largest induced subgraph with minimum degree k [23]. 1 Now, given a stable social network, that is, a graph with minimum degree k, the departure of a user decreases the degree of her neighbors in the graph by one which then might be smaller than k for some of them. Following our assumption these users now will abandon the network, too. This causes a cascading effect of users dropping out (collapse) of the network until a new stable state is reached.
From an adversarial perspective a natural question is how to maximally destabilize a competing social network platform by compelling b users to abandon the network. This problem was introduced as Collapsed k-Core by Zhang et al. [26] and the decision version is formally defined as follows.
Collapsed k-Core Input: An undirected graph G = (V, E), and integers b, x, and k. Question: Is there a set S ⊆ V with |S| ≤ b such that the k-core of G − S has size at most x?
In the mentioned motivation, one would aim to minimize x for a given b and k. Alternatively, we can also interpret this problem as a measure for resilience agains user drop outs of a social network by determining the smallest b for a given k and x.
Related Work. In 2017 Zhang et al. [26] showed that Collapsed k-Core is NP-hard for any k ≥ 1 and gave a greedy algorithm to compute suboptimal solutions for the problem. However, for x = 0 and any fixed k, solving Collapsed k-Core is equivalent to finding b vertices such that after removing said vertices, the remaining graph is (k − 1)-degenerate 2 . This problem is known as r-Degenerate Vertex Deletion and it is defined as follows.
r-Degenerate Vertex Deletion Input: An undirected graph G = (V, E), and integers b and r. Question: Is there a set S ⊆ V with |S| ≤ b such that G − S is r-degenerate?
It is easy to see that Collapsed k-Core is a generalization of r-Degenerate Vertex Deletion. In 2010 Mathieson [22] showed that r-Degenerate Vertex Deletion is NP-complete and W[P]-complete when parameterized by the budget b for all r ≥ 2 even if the input graph is already (r + 1)-degenerate and has maximum degree 2r + 1. In the mid-90s Abrahamson et al. [1] already claimed W[P]-completeness for r-Degenerate Vertex Deletion with r = 2 when parameterized by b under the name Degree 3 Subgraph Annihilator. For r = 1, the problem is equivalent to Feedback Vertex Set and for r = 0 it is equivalent to Vertex Cover, both of which are known to be NP-complete and fixed-parameter tractable when parameterized by the solution size [9,14]. The aforementioned results concerning r-Degenerate Vertex Deletion in fact imply the hardness results shown by Zhang et al. [26] for Collapsed k-Core.
Our Contribution. We complete the parameterized complexity landscape of Collapsed k-Core with respect to the parameters b, k, and x. Specifically, we correct errors in the literature [1,22] concerning the W[P]-completeness of r-Degenerate Vertex Deletion when parameterized by b for r = 2. We go on to clarify the parameterized complexity of Collapsed k-Core for k ≤ 2 by showing W[1]-hardness for parameter b and fixed-parameter tractability for the combination of b and x. Together with previously known results, this reveals a dichotomy in the computational complexity of Collapsed k-Core for k ≤ 2 and k ≥ 3.
We present two single exponential linear time FPT algorithms, one for Collapsed k-Core with k = 1 and one for k = 2. In both cases the parameter is (b+x). In particular, the algorithm for k = 2 runs in O(1.755 x+4b · n) time which means that it solves Feedback Vertex Set in O(9.487 b · n) time (here, b is the solution size of Feedback Vertex Set). To the best of our knowledge, despite of its simplicity our algorithm improves many previous linear time parameterized algorithm for Feedback Vertex Set [15,20]. However, recently a linear time computable polynomial kernel for Feedback Vertex Set was shown [16], which together with e.g. [18] yields an even faster linear time parameterized algorithm.
Furthermore, we conduct a thorough parameterized complexity analysis with respect to structural parameters of the input graph. On the positive side, we show that Collapsed k-Core is fixed-parameter tractable when parameterized by the treewidth of the input graph and show that it presumably does not admit a polynomial kernel when parameterized by either the vertex cover number or the bandwidth of the input graph. We also show that the problem is fixedparameter tractable when parameterized by the combination of the cliquewidth of the input graph and b or x. Further results include W[1]-hardness when parameterized by the clique cover number of the input graph and para-NP-hardness for the domination number of the input graph.

Hardness Results from the Literature
In this section, we gather and discuss known hardness results for Collapsed k-Core. Recall that that Collapsed k-Core with x = 0 is the same problem as r-Degenerate Vertex Deletion with r = k − 1. Hence, the hardness of Collapsed k-Core was first established by Mathieson [22] who showed that r-Degenerate Vertex Deletion is NP-complete and W[P]complete when parameterized by the budget b for all r ≥ 2 even if the input graph is already (r + 1)-degenerate and has maximum degree 2r + 1. However, in the proof of Mathieson [22] the reduction is incorrect for the case that r = 2. Abrahamson et al. [1] claim W[P]-completeness for r = 2 but their reduction is also flawed. In the following, we provide counterexamples for both cases. Then we show in the following how to adjust the reduction of Mathieson [22] for this case.
Counterexample for Mathieson's Reduction [22]. We refer to the original paper by Mathieson [22] for definitions, notation, and description of the gadgets and the reduction itself. Mathieson provides a reduction from Cyclic Monotone Circuit Activation to r-Degenerate Vertex Deletion [22,Theorem 4.4] showing that r-Degenerate Vertex Deletion is W[P]-complete when parameterized by the budget b for all r ≥ 2 even if the input graph is already (r + 1)-degenerate and has maximum degree 2r + 1. However, for the case of r = 2 it is easy to see that the OR gadget is already 2-degenerate. We illustrate the flawed OR gadget for r = 2 in Figure 1.
This means that whenever a Cyclic Monotone Circuit Activation instance is activated by the set of all its binary OR gates, the graph produced by the reduction (for r = 2) is already 2-degenerate, which clearly makes the reduction incorrect. An example of such an instance would be the one given by Mathieson [22]. We reproduce the example in Figure 2.
We describe how to repair the reduction (for r = 2) in the proof of Theorem 1.
Counterexample for Abrahamson et al.'s Reduction [1]. We refer to the original paper by Abrahamson et al. [1] for definitions, notation, and description of the gadgets and the reduction itself. The same reduction (using the same notation) can also be found in the book "Fundamentals of Parameterized Complexity" by Downey and Fellows [11]. Abrahamson et al. provide a reduction from Weighted Monotone Circuit Satisfiability to Degree 3 Subgraph Annihilator, which is equivalent to r-Degenerate Vertex Deletion with r = 2. With this reduction they claim to show that r-Degenerate Vertex Deletion with r = 2 is W[P]-complete when parameterized by the budget b [1, Theorem 3.7 (ii)].
The main idea of their reduction is that once a satisfying assignment is found, all variable gadgets corresponding to variables that are set to false are also removed. However, since in x [1] x [2] x [3] ∧ ∧ This follows from fan-out gadget not being removable from below. This allows us to create a counterexample with k = 1, which we illustrate in Figure 3. It is easy to check that = true is the only satisfying assignment that has at most one variable set to true. Furthermore, the AND gates in the red area have two outgoing connections each, hence the corresponding AND gadgets have fan-out gadgets attached to them. Initially, the output of these gates is false for the satisfying assignment so the fan-out gadget attached to them are only removed if the AND gadgets themselves are removed. An AND gadget is removed if both of its inputs are removed. However, note that the variable gadget for x [1] is not completely removed after v(3, 4) is removed from the graph. In particular, the vertex v(1, 4) has still degree m(k+1) > 3 after all vertices are removed since the AND gadgets it connects to are not removed. It follows that the graph is not 2-degenerate after removing v (3,4) and hence the reduction is not correct. We believe that the reduction can be corrected by replacing the high degree vertices v(i, 4) by fan-out gadgets that connect to the gate gadgets. However, we omit a proof of this claim.
Theorem 1 (Corrected from [22]). For any r ≥ 2 r-Degenerate Vertex Deletion is NPhard and W[P]-complete when parameterized by b, even if the degeneracy of the input graph is r + 1 and the maximum degree of the input graph is 2r + 1.
Proof. We refer to the original paper by Mathieson [22] for definitions, notation, and description of the gadgets and the reduction itself. Mathieson provides a reduction from Cyclic Monotone Circuit Activation to r-Degenerate Vertex Deletion [22,Theorem 4.4] showing that r-Degenerate Vertex Deletion is W[P]-complete when parameterized by the budget b for all r ≥ 2 even if the input graph is already (r + 1)-degenerate and has maximum degree 2r + 1. While Mathieson [22] claimed the result also for r = 2, the reduction is incorrect in this case. Similarly, while Abrahamson et al. [1] claimed the same result for r = 2 under the name Degree 3 Subgraph Annihilator, their reduction also seems to be flawed. However, there is a way to correct the proof of Mathieson [22]: The only problem is that his OR gadget is 2-degenerate, so the graph sometimes collapses without even deleting a single vertex. Before we introduce the correct gadget, note that the gadget is always used with exactly two inputs (predecessor gates). We can replace the OR gadget for case r = 2 with the graph illustrated in Figure 4. To collapse this gadget one can simply delete v o . The correctness now follows from an analogous argument as given by Mathieson [22].
The following observation shows that the hardness result by Mathieson [22] (Theorem 1) easily transfers to Collapsed k-Core (also in the cases where x = 0). Observation 1. Let x ′ > 0 be a positive integer. There is a reduction which transforms instances (G, b, x, k) of Collapsed k-Core with x = 0 into equivalent instances (G ′ , b, x ′ , k) of Collapsed k-Core.
Proof. We distinguish two cases, depending on the relation of x ′ to k.
If x ′ ≤ k, then we let G ′ = G. Obviously, if S is a solution for (G, b, x, k), then it is also a solution for (G ′ , b, x ′ , k). On the other hand, if S ′ is a solution for (G ′ , b, x ′ , k), then G ′ \ S ′ = G \ S ′ has a k-core with at most k vertices. However, any vertex of a graph with at most k has degree at most k − 1 and, thus, the k-core is empty. Therefore S ′ is a solution for (G, b, x, k).
If x ′ ≥ k + 1, then we obtain G ′ as a disjoint union of G and a clique C on x ′ vertices. Again obviously, if S is a solution for (G, b, x, k), then it is also a solution for (G ′ , b, x ′ , k), since the k-core of G ′ \ S is exactly C. Now let S ′ be a solution for (G ′ , b, x ′ , k). Note that if one vertex of C \ S ′ is part of the k-core of G \ S ′ , then all vertices of C \ S ′ are. If indeed C \ S ′ is a part of the k-core of G \ S ′ , then the k-core contains at most C ∩ S ′ other vertices. If Q is the set of vertices in the k-core of G \ (S ′ ∪ C) = G \ S ′ , then Q ∪ (S ′ \ C) is a solution for (G, b, x, k). Now if C \ S ′ is not a part of the k-core of G \ S ′ , then we know that |C ∩ S ′ | vertices are sufficient to collapse a clique of size x ′ . Since the k-core of G ′ \ S ′ , which is the same as the k-core of G \ S ′ is of size at most x ′ , there is a set Q of vertices at most |C ∩ S ′ | such that the k-core of G \ (S ′ ∪ Q) is empty. Hence (S ′ ∪ Q) is a solution for (G, b, x, k).
With that, we arrive at the following corollary. Corollary 1. Collapsed k-Core is NP-hard and W[P]-hard when parameterized by b for all x ≥ 0 and k ≥ 3, even if the degeneracy of the input graph is max{k, x − 1} and the maximum degree of the input graph is max{2k − 1, x − 1}.
Note that r-Degenerate Vertex Deletion is known to be NP-hard for all r ≥ 0. 3 Hence, we also know that Collapsed k-Core is NP-hard for k ≤ 2 and all x ≥ 0. However, the parameterized complexity with respect to b is open in this case. We settle this in the next section.
3 Algorithms for k = 1 and k = 2 In this section we investigate the parameterized complexity of Collapsed k-Core for the case that k ≤ 2. Since Corollary 1 only applies for k ≥ 3 we first show in the following that the problem is W[1]-hard with respect to the combination of b and (n−x) for all k ≥ 1. Furthermore, we present two algorithms; one that solves Collapsed k-Core with k = 1 and one for the k = 2 case. Both algorithm run in single exponential linear FPT-time with respect to the parameter combination (b + x).
We first give a parameterized reduction from Clique to Collapsed k-Core. Note that since this hardness result holds for the combination of b and the dual parameter of x, it is incomparable to Corollary 1 for k ≥ 3. Proof. We reduce from W[1]-hard problem Clique [9], where given a graph G = (V, E) and an integer p, the task is to decide whether G contains a clique of size at least p. Let (G, p) be an instance of Clique and k be a given constant. We build an instance (G ′ , b, x, k) of Collapsed k-Core as follows. We can assume that p ≤ |V (G)|, as otherwise we can output a trivial noinstance. We further assume that each vertex of G has degree at least p + 1. Vertex of degree less than p − 1 is not part of a clique of size at least p, while for all vertices of degree p − 1 or p we can check in O(n · p 3 ) time whether there is a clique of size at least p containing any of them. We We actually only introduce the sets W and E W if k ≥ 2.
Finally we set b = p and x = n − (p + (2k − 1) p 2 ), where n = |V (G ′ )| is the number of vertices of graph G ′ .
We claim that (G ′ , b, x, k) is a yes-instance of Collapsed k-Core if and only if (G, p) is a yes-instance of Clique.
⇒: If S is a clique of size at least p in G, then we claim that deleting S from G ′ results in a k-core of size at most x. Let e be an edge between two vertices of S in G. Then for any i ∈ {1, . . . , k} vertex u i e has only vertices w 1 e , . . . w k−1 e as neighbors in G ′ \ S. Hence, it has degree k − 1 in this graph and it is not part of the k-core of the graph. Since this holds for each i, vertices w 1 e , . . . w k−1 e do not have any neighbors in the k-core of G ′ \ S and, hence, they are also not in the k-core of the graph. This makes 2k − 1 vertices per each edge of the clique which are not deleted and not in the k-core, showing that the size of the k-core is at most x.
Indeed, for an arbitrary vertex v in V \ S the degree of v in G is at least p + 1 and, thus, there is at least one edge e incident to v which is not in S E . Hence v is in the k-core of G ′ \ S by the above argument.
For e ∈ S E possibly no vertex of U W e is in the k-core of G ′ \ S, effectively shrinking it by 2k − 1 vertices. However, this does not influence the other vertices in V , U , or W , since V \ S is in the k-core, as we already observed. If e = {x, y} and {x, y} ⊆ S, then also the whole U W e is not in the k-core as observed in the first implication. Thus, if |S ∩ V | = a and |S E | = c, then there are at most (2k − 1)( a 2 + c) vertices of G ′ which are neither in S nor in the k-core of G ′ \ S. As S is a solution, this number has to be at least (2k − 1) p 2 , while a + c ≤ b = p. It follows that a = p and S is a clique of size p in G.
Note that graph G ′ is bipartite and for k ≥ 2 it is also k-degenerate, as all vertices in W have degree k, after removing them the vertices of U have degree 2, and, finally, V forms an independent set in G ′ . Now we proceed with the algorithm for Collapsed k-Core with k = 1. While there is a simple algorithm with O(3 x+b (m + n)) running time 4 for this case, we prefer to present an algorithm with the slightly worse running time as stated, since we then generalize this algorithm to the case k = 2 with some modifications. Algorithm: We present a recursive algorithm (see Algorithm 1 for pseudocode) that maintains two sets S and Q. The recursive function is supposed to return a solution to the instance, whenever there is a solution B containing all of S and there is no solution containing S and anything of Q. If some of the conditions is not met, then the function should return "No solution". In other words, S is the set of deleted vertices and Q is the set of vertices the algorithm has decided not to delete in the previous steps but may be collapsed in the future. Hence, the solution to the instance, or the information that there is none, is obtained by calling the recursive function with both sets S and Q empty.
The algorithm first checks that S is of size at most b and Q is of size at most x + b. If any of these is not true, then it rejects the current branch. Then it computes the 1-core G ′ of the graph G \ S. If the 1-core is of size at most x, then it returns S as a solution. If G ′ is larger, but all its vertices are in Q, then we have no way to shrink the core and we again reject. Finally, the algorithm picks an arbitrary vertex v of largest degree in G ′ which is not in Q and recurses on both possibilities-either v is in the solution, modeled by adding it to S, or it is in no solution containing S, modeled by adding it to Q. We start by showing that Algorithm 1 has the claimed running time.
Proof. Assume the running time of Algorithm 1 is T (µ), where µ = |S| + |Q|. Let m and n be the number of edges and number of vertices in G respectively. Since when |S| > b or |Q| > b + x Algorithm 1 directly return "No solution" in line 2, we have T (µ) = O(1) for µ > 2b + x. Lines 2, 4, 5 and 6 can be done in O(n) time. In line 3 the 1-core G ′ can be found in O(n + m) time by first removing vertices in S and edges incident with them to get G \ S, and then removing isolated vertices in G \ S. Thus except for line 7 and 9, all steps can be done in O(m + n) time, and we have that Next we show the claimed conditional lower bound on the running time for any algorithm for Collapsed k-Core with k = 1.

Lemma 2. Assuming the Exponential Time Hypothesis, there is no
Proof. Since Collapsed k-Core with k = 1 and x = 0 is equivalent to Vertex Cover, and assuming the Exponential Time Hypothesis, there is no 2 o(k) n O(1) time algorithm for Vertex Cover [19], we have that assuming the Exponential Time Hypothesis, there is no 2 o(b)+f (x) n O(1) time algorithm for Collapsed k-Core with k = 1, where f can be an arbitrary function.
Before showing the correctness of Algorithm 1, we first show in the following lemma why in line 2 set Q should be bounded by x + b.
Lemma 3. If Q is of size more than b + x, then there is no solution containing whole S and no vertex from Q.
Proof. Suppose for contradiction that there is a set Q of size at least b + x + 1 and a solution B such that |B| ≤ b, S ⊆ B and B ∩ Q = ∅. Let v 1 , v 2 , . . . , v r ′ be the vertices of the set S ∪ Q in the order as they were added to the set by successive recursive calls. Moreover, if B \ S is empty, Our aim is to show that the number of edges lost by vertices of Q \ X is larger than the number of edges incident to the vertices of B, which would be a contradiction.
To this end, we construct an injective function f that maps the vertices in B to vertices of Since the set Q \ X contains at least b + 1 vertices, while B contains at most b, this way we find a mapping for every vertex in B. Moreover, there remains at least one vertex in Q \ X not being in the image of f , let us denote it v q 0 . For On the other hand, since deg which is a contradiction.
Otherwise, let i 0 be the largest i such that f (i) > i and j 0 = f (i 0 ). We consider the graph which is again a contradiction. Now we have all necessary pieces to prove Proposition 2.
Proof for Proposition 2. To show the correctness of the Algorithm 1, we first show that whenever the algorithm outputs a solution, then this solution is indeed correct. Then we show that whenever there exists a solution, the algorithm also finds a solution.
⇒: We show this part by induction on the recursion tree. If Algorithm 1 returns S as a solution in line 4, then it is of size at most b since line 2 does not apply and the 1-core of G \ S is of size at most x. This constitutes the base case of the induction.
If the solution is obtained from recursive calls on lines 7-9, then we know that it is correct by induction hypothesis.
⇐: Next, we show by induction on the recursion tree that if there is a solution then Algorithm 1 returns a solution. In particular, we show that if there is a recursive call of Algorithm 1 with sets S and Q such that there is a solution containing all of S and there is no solution containing the whole S and any vertex from Q, then the algorithm either directly outputs a solution or it invokes a recursive call with sets S ′ and Q ′ such that there is a solution containing all of S ′ and there is no solution containing the whole S ′ and any vertex from Q ′ . Let S and Q be two input sets of a recursive call of Algorithm 1 and let B be a solution such that S ⊆ B and B ∩ Q = ∅. First we show that none of lines 2 and 5 applies. Line 2 will not apply since we have |S| ≤ |B| ≤ b and by Lemma 3 we also know that we have |Q| ≤ b + x. If the 1-core of G[Q] is of size more than x, which is the case when line 5 applies and line 4 does not apply, then no set Since Algorithm 1 starts with S = Q = ∅ we have that for any solution B the conditions S ⊆ B and B ∩ Q = ∅ are initially fulfilled. Furthermore, it follows from the complexity analysis in Lemma 1 that the algorithm always terminates. Therefore the algorithm outputs a solution if one exists and hence is correct.
The running time bound for Algorithm 1 follows from Lemma 1 and the conditional running time lower bound for Collapsed k-Core with k = 1 follows from Lemma 2.
In the remainder of the section, we show how to adapt this algorithm for Collapsed k-Core with k = 2. The above theorem in particular yields an O(9.487 b · n) algorithm for Feedback Vertex Set. For the proof we need the following lemma, which shows that, except for some specific connected components, we can limit the solution to contain vertices of degree at least three.
Let C 1 , . . . , C r be the connected components of G ′ not containing vertices of Q, for i = 1, . . . , r ′ do Select an arbitrary vertex from C i and add it to S; Proof. Let v be any vertex with degree 2 in B. Let v ′ be a vertex of degree at least 3 in G such that there is a path P between v and v ′ with all internal vertices of degree exactly 2 in G. Since no component of G is a cycle, such a vertex must exist. Let We have the 2-core of G \ B 2 is the same as the 2-core of G \ B 1 and the 2-core of G \ B 1 is a subset of the 2-core of G \ B. So the 2-core of G \ B 2 is a subset of the 2-core of G \ B, and hence no larger than x. Therefore B 2 is also a solution for (G, b, x, 2). Following the same way, we can replace all degree 2 vertices in B with vertices which have degree larger than 2, and get a new solution B ′ for (G, b, x, 2).
Algorithm: Our algorithm for k = 2 is similar to Algorithm 1 with two main differences (see Algorithm 2 for pseudocode). First |Q| > b + x is replaced by |Q| > 3b + x. Second when selecting the maximum degree vertex from V (G ′ ) \ Q, we need to make sure that this vertex has degree greater than 2. Otherwise, either we can directly select vertices from V (G ′ ) \ Q to break cycles in G ′ and get a 2-core of size at most x, or the algorithm rejects this branch.
We start by showing that Algorithm 2 has the claimed running time. Proof. We first show by induction on the size of S ∪Q starting with the largest size achieved that a call with S and Q results in at most 2( 4b+x+2−|S|−|Q| b+1−|S| )−1 calls to the function in total. Indeed, the size of |Q| never exceeds 3b + x + 1 and the size of |S| never exceeds b + 1 since the sizes grow by one and if they achieve the bound, then line 2 applies. Hence, if any of the lines 2-12 applies, then we have only one call and |Q| ≤ 3b+x+1 and |S| ≤ b+1 implies 2( 4b+x+2−|S|−|Q| b+1−|S| )−1 ≥ 1, making the basic cases. If the call makes recursive calls, then in one of them S is one larger than in the current one, and if the second one is made, then Q is one larger in it. Hence the number of calls is at most Since we call the algorithm with sets S and Q empty, it follows that the total number of calls is at most 2( 4b+x+2 Next we show the claimed conditional lower bound on the running time for any algorithm for Collapsed k-Core with k = 2. Before showing the correctness of Algorithm 2, we prove the following lemmata which will be helpful in the correctness proof. The next lemma helps to show that line 12 is correct.

Lemma 7. If line 12 of Algorithm 2 applies, and there is no solution containing whole S and no vertex from Q, then there is no solution containing the whole S.
Proof. Suppose for contradiction that B is a solution containing S. Note that in this case B ′ = B \ S is a solution for (G ′ , b − |S|, x, 2). Let D be the graph formed by the union of connected components of G ′ containing vertices of degree at least 3 and C 0 be the graph formed by the union of connected components of G ′ which are cycles and contain vertices of Q, that is, If B C = B ∩ V (C 0 ) is nonempty, then let y be any vertex in B C and C y the connected component of C 0 containing y. By the definition of C 0 component C y contains a vertex of Q, let us denote it y ′ . Let B ′ = (B \ {y}) ∪ {y ′ }. The 2-cores of G \ B and G \ B are the same, since C y is a cycle. Thus B ′ is a solution to (G, b, x, 2) containing whole S and vertices of Q, contradicting our assumption. Hence also B C is empty. Figure 5: Illustration of function f . Vertices in Q are separated into two parts: gray vertices from X and blue vertices from Q ′ = Q \ X. Every vertex in S is mapped to a set of three consecutive blue vertices in Q ′ from the right to the left. Graph G j 0 with the property that i > max{j | j ∈ f (i)} for every i with v i ∈ B ∩ V (G j 0 ) is contained in the green box.
The components of G ′ neither in C 0 nor in D are cycles since G ′ is a 2-core, they contain no vertices of Q, and there are no other vertices of degree at least 3. Since B contains at most r ′ = min{r, b − |S|} vertices out of these components, it can destroy at most r ′ of these cycles. Since |V (C 1 )| ≥ |V (C 2 )| ≥ . . . ≥ |V (C r )|, this decreases the size of the 2-core by at most then there is no solution containing whole S.
Next, we show that the second part of line 2 of Algorithm 2 is correct.
Lemma 8. If Q is of size more than 3b + x, then there is no solution containing whole S and no vertex from Q.
Proof. Suppose for contradiction that there is a set Q of size at least 3b + x + 1 and a solution B such that |B| ≤ b, S ⊆ B and B ∩ Q = ∅. Let v 1 , v 2 , . . . , v r ′ be the vertices of the set S ∪ Q in the order as they were added to the set by successive recursive calls. Moreover, if B \ S is empty, then let r = r ′ . Otherwise, let B \ S = {v r ′ +1 , . . . , v r }, i.e., in both cases {v 1 , . . . , v r } = B ∪ Q. Let G ′ be the 2-core of G \ B and let X = V (G ′ ) ∩ Q. Since B is a solution, we know that |X| ≤ x.
We construct a function f that maps every vertex of B to a set of three consecutive vertices of Q \ X (see also Figure 5). First let v i be the vertex in B with the largest i. We let f (i) be the set {j 1 , j 2 , j 3 }, where j 1 = max{j | v j ∈ Q \ X}, j 2 = max{j < j 1 | v j ∈ Q \ X}, and j 3 = max{j < j 2 | v j ∈ Q \ X}. Now let v i be the vertex from B with the largest i such that f (i) was not set yet and v i ′ be the vertex from B with the least i ′ such that i ′ > i. We set f (i) = {j 1 , j 2 , j 3 }, where j 1 = max{j < min{k ∈ f (i ′ )} | v j ∈ Q \ X}, j 2 = max{j < j 1 | v j ∈ Q \ X}, and j 3 = max{j < j 2 | v j ∈ Q \ X}. Since the set Q \ X contains at least 3b + 1 vertices, while B contains at most b, this way we find a mapping for every vertex in B, keeping the images of different vertices disjoint. Moreover, denote p = |Q \ X| − 3|B| ≥ 1. There remains p vertices in Q \ X not being in the union of images of f , let us denote them v q 1 , . . . , v qp .
For t ∈ {1, . . . , r} let G t be the 2-core of the graph Figure 6: Illustration of the partial directed graph when considering the collapsing process. The set B contains the deleted vertices and X is the set of vertices remaining in the 2-core of G \ B. The set Q ′ contains vertices the algorithm has decided not to delete but eventually collapse and Y is the set of other collapsed vertices.
G into a partial directed graph by considering the collapsing process (see also Figure 6). More precisely, for every edge in G, except for edges which have two endpoints in X, we will assign it a direction. To this end, we just need to give an order of vertices in V (G)\X. We define this order based on the time vertices collapse. It may happens that several vertices in Q ′ or Y collapse at the same time. For this situation, we just order these vertices according to an arbitrary but fixed order. Since k = 2, every collapsed vertex in Q ′ ∪ Y has at most one outgoing edge. Let us consider the number, denoted by N , of edges of the form − − → v i v j such that the head v j is in Q ′ . Since every vertex in Q ′ has at most one outgoing edge, we have On the other hand, let • N−−→ BQ ′ be the number of edges going from B to Q ′ ; Denote the number of edges in G[Y ] by n Y . Since every vertex in Y has at least one incoming edge but at most one outgoing edge, we ) is the number of incoming (outgoing) edges of vertex v in G, respectively. This means finishing the proof of the claim.
Since the edges which have their heads in Q ′ have their tails from B ∪ Y ∪ Q ′ , Then we have that which is a contradiction. Otherwise, let i 0 be the largest i such that i < max{j | j ∈ f (i)} and j 0 = max{j | j ∈ f (i 0 )}. Now we consider the graph G j 0 and the function f restricted on B ∩ V (G j 0 ) (see also Figure  5). First note that in G j 0 vertex v j 0 is the only vertex which is not contained in any image Now similar to the first situation, we first transform graph G j 0 into a partial directed graph by considering the collapsing process.
and Y 0 is the set of collapsed vertices in V (G j 0 ) not contained in Q. Then consider the number, denoted by N 0 , of edges of the form − − → v i v j such that the head v j is in Q 0 ′ . Since every vertex in Q 0 ′ has at most one outgoing edge, we have On the other hand, let ′ be the number of edges going from B 0 to Q 0 ′ ; be the number of edges going from Q 0 ′ to Y 0 .
and get the upper bound for N 0 : Then we have that which is a contradiction.
Now we have all necessary pieces for the proof of Theorem 2.
Proof for Theorem 2. To show the correctness of the algorithm, it is again enough to show that for any pair of S and Q appearing in the recursive process, the set returned by the recursive function is a solution that contains whole S and no vertex from Q and if there is a solution containing whole S and there is no solution containing the whole S and any vertex from Q, then the function returns such a solution. For simplicity we assume without loss of generality that the input graph is a 2-core. ⇐: If the function returns S as a solution on line 4, then it is of size at most b since line 2 does not apply and the 2-core of G \ S is of size at most x. Hence it is obviously a solution and it is fulfilling the constraints. If the solution is obtained from recursive calls on lines 15-17, then it is returned without modification and S is subset of the solution, since S or its superset was passed to the recursive calls. Similarly, it contains no vertex from Q, since Q or its superset was passed to the recursive calls.
If the function returns set S ′ as a solution on line 10, then for each i ∈ {1, . . . , r ′ } set S ′ contains a vertex of the cycle C i . Hence the 2-core of G \ S ′ does not contain the cycle C i . Since the 2-core G ′′ of G \ S ′ differs from the 2-core G ′ of G \ S exactly in these missing cycle components, we have V ( Since this is at most x by line 8, line 2 does not apply, and Q ∩ r ′ i=1 V (C i ) = ∅, S is a solution fulfilling the constraints. Hence, if the function returns a solution, then the answer is correct.
⇒: Now we show by induction on |Q ∪ S| for every pair of S and Q appearing in the recursive process, starting from the largest |Q ∪ S| achieved, that if there is a solution containing whole S and there is no solution containing the whole S and any vertex from Q, then the algorithm returns such a solution.
If B is a solution such that S ⊆ B and B ∩ Q = ∅, then line 2 will not apply since |B| ≤ b and because of Lemma 8, line 12 does not apply according to Lemma 7. Let G ′ be the 2-core of G \ S. If B is a solution, then S ∪ ((B \ S) ∩ V (G ′ )) is also a solution, so we can assume that B \ (S ∪ V (G ′ )) = ∅. If B = S, then line 4 applies. If there are no vertices of degree at least 3 which are not in Q, then B contains no vertices of the components containing vertices of Q, as we have shown above. Hence B can decrease the size of the 2-core compared to G ′ by at most r ′ i=1 |V (C i )|, where r ′ is as in the algorithm. As B is a solution, x and a solution containing S is returned on line 10. This finishes the proof for the base cases of the induction. Now suppose that the claim already holds for all calls with larger |Q ∪ S| and v is the vertex selected by the algorithm on line 14. If there is a solution containing S ∪ {v} (in particular if v ∈ B), then the call SolveRec2(G, S ∪{v}, Q) must return a solution by induction hypothesis and otherwise there is no solution containing S and anything of Q ∪ {v} and the call SolveRec2(G, S, Q ∪ {v}) must return a solution by induction hypothesis. Thus the algorithm works correctly.
The running time bound for Algorithm 2 follows from Lemma 5 and the conditional running time lower bound for Collapsed k-Core with k = 1 follows from Lemma 6.

Structural Graph Parameters
In this section, we investigate the parameterized complexity of Collapsed k-Core with respect to several structural parameters of the input graph. Corollary 1 already implies hardness for constant values of several structural graph parameters. We expand this picture by observing that the problem remains NP-hard on graphs with a dominating set of size one and by showing that the problem is W[1]-hard when parameterized by the combination of b and the clique cover number of the input graph. On the positive side, we show that the problem is in FPT when parameterized by the treewidth of the input graph or the clique-width of the input graph and k combined with either b, x, n − x, or n − b. Lastly, we show that the problem presumably does not admit a polynomial kernel when parameterized by the combination of b and the vertex cover number of the input graph, or by the combination of b, k, and the bandwidth of the input graph.
We start with an easy observation that we will make use of in most of the hardness results in this section. (G, b, x, k) is an instance of Collapsed k-Core and vertex v is a part of the (k + b)-core of G, and S ⊆ V is of size at most b, then either v ∈ S or v is part of the k-core of G \ S.

Observation 2. If
Proof. Let C be the (k + b)-core of G. In C \ S the degree of each vertex is at least k + b − b, hence C \ S is a subgraph of the k-core of G \ S.
The following observation yields that we can reduce the size of a dominating set of any instance of Collapsed k-Core to one by introducing a universal vertex. Note that, for example, this only increases the degeneracy by one.
Observation 3. Let (G, b, x, k) be an instance of Collapsed k-Core and G ′ be the graph obtained from G by adding a universal vertex, then (G ′ , b + 1, x, k) is an equivalent instance of Collapsed k-Core.
Proof. Let (G, b, x, k) be an instance of Collapsed k-Core and let (G ′ , b ′ , x, k) be the instance formed by a graph G ′ which is obtained from G by adding an universal vertex u, b ′ = b + 1, and x and k from the original instance. We claim that the instances are equivalent.
First if S is a solution for (G, b, x, k), then S∪{u} is a solution for (G ′ , b ′ , x, k), as G\S = G ′ \S ′ . Second, let S ′ be a solution for (G ′ , b ′ , x, k). If S ′ contains u, then S ′ \ {u} is a solution for G. Now suppose that S ′ does not contain u and let v be an arbitrary vertex of S ′ . We claim that S ′′ = (S ′ \ {v}) ∪ {u} is also a solution to (G ′ , b ′ , x, k), since G ′ \ S ′′ is isomorphic to a subgraph C . . . of G ′ \ S ′ . Indeed, consider the bijection ϕ, which maps each vertex of V (G) \ S ′ to itself and v to u. To show that it is an isomorhism, it is enough to consider edges incident on v, however, as there is an edge between u and every vertex of V (G) \ S ′ , these definitely map to edges. Hence the k-core of G ′ \ S ′′ is at most as large as the k-core of G ′ \ S ′ and indeed S ′′ is a solution to (G ′ , b ′ , x, k). Now the equivalence of the instance follows from the case where u is in S ′ .
Considering a larger parameter than e.g. the size of the dominating set, namely the clique cover number 5 , we can show W[1]-hardness, even in combination with b. We do this by providing a parameterized reduction from Multicolored Clique parameterized by the solution size. Proof. We present a parameterized reduction from the W[1]-hard problem Multicolored Clique parameterized by the solution size [9]. In Multicolored Clique, we are given an integer s and a s-colorable graph with color classes V 1 , V 2 , . . . , V s , and the task is to find a clique of size s containing one vertex from each color. Let (G = (V, E), s) be an instance of Multicolored Clique. The edge set E can be partitioned into s 2 subsets: of Collapsed k-Core as follows.
• Denote k = max 1≤i<j≤s 2|E i,j |, n = |V | and set b = s, x = N − N ′ where N = 2n 4 + k + s + n + k s 2 is the number of vertices in G ′ we will construct and N ′ = s + k s 2 .
• For every V i , i = 1, 2, . . . , s, create a clique C i in G ′ , which contains all vertices in V i .
• For every E i,j , 1 ≤ i < j ≤ s, create a clique C i,j of size k in G ′ , which contains 2 copies of vertices v i,j , v i,j ′ for every edge v i v j in E i,j and k − 2|E i,j | more dummy vertices.
• Create a clique C of size 2n 4 + k + s. Add edges between vertices in C i , i = 1, 2, . . . , s and k + s vertices in C. For every dummy vertex in C i,j , 1 ≤ i < j ≤ s, we add edges between this vertex and two distinct vertices in C. The size of C is large enough such that no pair of edges between E i,j and C share the same end point in C. Notice that the clique cover number of G ′ is s+ s 2 +1. The construction is illustrated in Figure 7. We claim that there is a multicolored clique of size s in G if and only if (G ′ = (V ′ , E ′ ), b, x, k) is a yes-instance.
⇒: If there is a multicolored clique C ′ with vertex set S of size s in G, we show in the following that the k-core of G − S has size at most x. Since b = s and N ′ = s + k s 2 , it suffices to show that all edge cliques C i,j collapse. For any clique C i,j , every vertex in this clique has vertex degree k + 1, since it has k − 1 neighbors in C i,j and 2 neighbors in C i and C j (or C for dummy vertices). For any After deleting v i and v j , both v i,j and v i,j ′ will be collapsed, which then make all remaining vertices in C i,j collapse. Therefore all edge cliques C i,j will collapse after deleting S.
is a yes-instance, we need to show there is a multicolored clique of size s in G. Let S be the deleted vertex set of size b and let S ′ be the set of all collapsed vertices. Since N ′ = s + k s 2 , we have |S ′ | ≥ k s 2 . Notice that in the subgraph of G ′ induced by C and all C i 's all vertices in C i have vertex degree at least k + s and all vertices in C have degree at least 2n 4 + k + s − 1. Therefore, by Observation 2, these vertices will never collapse and only vertices in C i,j , 1 ≤ i < j ≤ s can collapse. Since the number of vertices in C i,j , 1 ≤ i < j ≤ s is exactly k s 2 , we have all vertices in C i,j , 1 ≤ i < j ≤ s will collapse and S only contains vertices from C and C i , 1 ≤ i ≤ s.
Suppose S contains t vertices from C and s − t vertices from C i , 1 ≤ i ≤ s. On one hand, for every clique C i,j , the first vertex to collapse must connect to two vertices from S, so overall there must be 2 s 2 edges between all such vertices and S. On the other hand, each vertex in S ∩ C can provide at most one such edge and each vertex in S from C i , 1 ≤ i ≤ s can provide at most s − 1 such edges, so overall the number is strictly less than 2 s 2 if t > 0. Since all cliques C i,j , 1 ≤ i < j ≤ s will collapse, we have t = 0 and S only contains vertices from C i .
So the firstly collapsed vertex v i,j in C i,j must connect to two vertices from S, one v i from C i and another v j from C j . This means S contains exactly one vertex v i from each C i and each pair of vertices v i and v j connect to at least one common vertex v i,j in C i,j , which means v i and v j are connected in G. Therefore, all vertices from S form a clique of size s in G.
On the positive side, we sketch a dynamic program on the tree decomposition of the input graph G which implies that Collapsed k-Core is in FPT when parameterized by the treewidth of the input graph. Proof Sketch. Observe that either k ≤ tw(G) or the k-core is (already) empty and we can answer Yes. Hence, for the rest of the proof we assume that k ≤ tw(G). We assume we are given a nice tree decomposition of G [5,17] and use dynamic programming on the nice tree decomposition of G. The indices of the table are formed for each bag of the decomposition by the number of vertices of the solution already forgotten, the number of vertices in the core already forgotten, a partition of the bag into three set B, X, and Q, an (elimination) order for the vertices in Q, and for each vertex in Q the number of its neighbors in X or higher in the order. This number is always in 0, . . . , k − 1, as otherwise it would not be possible to eliminate the vertex.
The set B represents the partial solution (or rather its intersection with the bag), i.e., the vertices to be deleted. The set X represents the vertices which (are free to) remain in the core. The vertices in Q should collapse after removing the vertices of the solution and the collapse of the vertices preceding them in the order.
Using monadic second order (MSO) logic formulas, we can show that for a smaller structural parameter, namely the cliquewidth of the input graph, we can also obtain positive results. Here however, we can only show fixed-parameter tractability for the combination of the cliquewidth of the input graph with k and either b, x, n − x, or n − b.
Proposition 5. Collapsed k-Core is in FPT when parameterized by the cliquewidth of the input graph combined with k and either b, x, n − x, or n − b.
Proof. We first develop, for a fixed k a formula core(G, B, X), which should express that the set X contains the whole k-core of the graph G − B. The formula thus says that no graph induced by a set larger than X, but not containing anything from B is a core. In other words, each such graph contains a vertex of degree at most k − 1, i.e., not having k distinct neighbors. For that purpose we use the following subformula: Now the sought formula is This formula is of length O(k). Combined with some of the following formulae it gives the result for all parameter combinations promised. The following formula bounds a set S passed to be of size at most s: The next formula bounds the size to at most n − s: Both these formulae have length O(s 2 ). Now the result follows from the theorem of Courcelle et al. [8].
selection gadget Figure 8: Illustration of the OR-cross composition from Cubic Vertex Cover to Collapsed k-Core with k = 5. The selection gadget is all the circles contained in the green box. Every gray edge in this figure means that all vertices in one endpoint of this edge are connected to all vertices in the other endpoint. Every vertex in V E i connects to two endpoints of its corresponding edge in G i . For example, the yellow vertex in V E 2 is connected to the blue and the red vertex in V , which represents the two endpoints of the corresponding edge in G 2 . The big clique C is separated into two parts. Every vertex in V E i is connected to k − 2 vertices in the upper part of C. Since k = 5, the yellow vertex in V E 2 is connected to three vertices in C. Vertices contained in thick outlined vertex sets form a vertex cover. To keep the picture simple, edges that contain vertices from V E i with i = 2 are not depicted.
In the remainder of this section, we show that Collapsed k-Core does not admit a polynomial kernel when parameterized by rather large parameter combinations. We first show an OR-cross composition [4,9] from Cubic Vertex Cover.
Theorem 3. For all k ≥ 2 Collapsed k-Core does not admit a polynomial kernel when parameterized by the combination of b and the vertex cover number of the input graph unless NP ⊆ coNP/poly.

Proof.
We apply an OR-cross composition [4,9] from the NP-hard problem Cubic Vertex Cover [14]. In Cubic Vertex Cover, we are given a 3-regular graph G an integer s, and the task is to find a vertex subset of size at most s which contains at least one endpoint of each edge of G.
We say an instance of Cubic Vertex Cover is malformed if the string does not represent a pair (G, s), where G is a 3-regular graph and s is a non-negative integer. It is trivial, if s ≥ |V (G)|. We define the equivalence relation R as follows: all malformed instances are equivalent, all trivial instances are equivalent and two well-formed non-trivial instances (G, s) and (G ′ , s ′ ) are R-equivalent if |V (G)| = |V (G ′ )| and s = s ′ . Observe that R is a polynomial equivalence relation.
Let the input consist of T R-equivalent instances of Cubic Vertex Cover. If the instances are malformed or trivial, we return a constant size no-or yes-instance of Collapsed k-Core, respectively. Let (G i , s) 0≤i≤T −1 be well-formed non-trivial R-equivalent instances of Cubic Vertex Cover. Since all instances have the same size of the vertex set, we can assume they share the same vertex set V = {v 1 , v 2 , . . . , v n } . We assume T to be a power of 2, as otherwise we can duplicate some instances. Now we create an instance (G, b, x, k) of Collapsed k-Core for some arbitrary but fixed k ≥ 2 as follows.
• Set b = s + 2ks log T and x = N − N ′ , where N = n + 3 2 nT + 4ks log T + k + s + 3 2 n(k − 2) is the number of all vertices in graph G we will construct and N ′ = s + 2ks log T + 3 2 n. • First for every vertex v i in V , create a vertex v i in G.
• For every edge set E(G i ), create a vertex set V E i in G, in which every vertex v p,q represents an edge v p v q in E(G i ). Then we have T of these vertex sets and each set has 3 2 n vertices. • For every edge v p v q in E(G i ), add 2 edges v p,q v p and v p,q v q in G.
• Now create the selection gadget in G. It contains log T pairs of cliques C d i (1 ≤ i ≤ log T, d ∈ {0, 1}), and all of them have the same size of 2ks. For every vertex set V E i , let i = (d log T −1 d log T −2 . . . d 0 ) 2 be the binary representation of the index i, where d j ∈ {0, 1} for 0 ≤ j ≤ log T − 1 and we add leading zeros so that the length of the representation is exactly log T , we add edges between all vertices in V E i and all vertices in C d j j (0 ≤ j ≤ log T − 1).
• Finally we create a clique C with |C| = k + b + 3 2 n(k − 2), which contains two parts of vertices. The first part contains k + b vertices and each of them connects to all vertices in V and all C d j with 0 ≤ j ≤ log T − 1 and d ∈ {0, 1}. In other words, all vertices in V and C d j connect to these k + b vertices in C. The second part of 3 2 n(k − 2) vertices are connected to vertices in V E i in the following way. For every vertex v p,q in V E i , add edges between v p,q and k − 2 vertices in C. We can make sure that all vertices in the same V E i connect to different vertices in C. In other words, every vertex in the second part of C connects to exactly one vertex in every V E i . Notice that the vertex cover number of G is n + 4ks log T + k + b + 3 2 n(k − 2). The construction is illustrated in Figure 8. We now show that at least one instance (G i , s) is a yes-instance if and only if the instance (G, b, x, k) of Collapsed k-Core constructed above is a yes-instance.
⇒: If instance (G i , s) is a yes-instance, which means there is a vertex subset V * of size s that covers all edges in G i , then we delete the corresponding s vertices in G and all vertices in C d j j (0 ≤ j ≤ log T − 1), where i = (d log T −1 . . . d 0 ) 2 is the binary representation of i. So far, we deleted s + 2ks log T vertices, and all vertices in V E i will collapse, since they just have at most k − 1 edges remaining, k − 2 of which connect to vertices in C and at most one to vertices in V . Therefore, the number of remaining vertices is x and instance (G, b, x, k) is a yes-instance.
⇐: If instance (G, b, x, k) is a yes-instance, we need to show there is at least one instance which has a vertex cover of size s. Let S be the deleted vertex subset of size b and let S ′ be the set of all collapsed vertices. Since N ′ = s + 2ks log T + 3 2 n, we have |S ′ | ≥ 3 2 n. In the subgraph of G induced by V , C and all C d i 's all vertices in V , C d i and C have degree larger than k + b. Hence, by Observation 2 they will not collapse, all collapsed vertices come from V E i (0 ≤ i ≤ T − 1). Then we show that all collapsed vertices can only come from one single V E i for some i. Suppose two vertices v and v ′ from different sets of V E i (0 ≤ i ≤ T − 1) collapse after deleting S, then there is at least one pair of cliques C 0 j 0 and C 1 j 0 such that v is connected to all vertices in C 0 j 0 and v ′ is connected to all vertices in C 1 j 0 . To make v 1 collapse, at least 2ks log T − (k − 1) vertices from the corresponding cliques in the selection gadget need to be deleted. Then to in G, then at most x vertices in G i remain. Therefore the k-core of G − S − V (C i ) has size at most x ′ .
⇐: If instance (G, b ′ , x ′ , k ′ ) is a yes-instance, then let S be the set of deleted vertices of size at most b ′ and let S ′ be the set of all collapsed vertices. Since the degree of vertices in C i and C ′ i is at least 2n 2 − 1 which is more than 2n ≥ b + k for n ≥ 3, these vertices will never collapse by Observation 2. So S ′ only contains vertices from T i=1 V (G i ). Furthermore, it is impossible for two vertices v i and v j from different sets V (G i ) and V (G j ) to collapse after deleting S. Indeed, suppose they do collapse, then |S ∩ C i | ≥ n 2 − k and |S ∩ C j | ≥ n 2 − k, which means |S| ≥ 2n 2 − 2k ≥ 2n 2 − 2n > n 2 + n ≥ n 2 + b = b ′ , where the middle inequality follows from n ≥ 3. Therefore S ′ only contains vertices from a single graph, say G i .
Since G contains T (n + 2n 2 ) vertices in total, x ′ = (n + 2n 2 )(T − 1) + n 2 + x, and b ′ = b + n 2 , we have |S ′ | ≥ n − b − x. To make vertices in G i collapse, it is always better to choose vertices from C i into S, as vertices from C i connect to all vertices in G i . Thus we can assume C i ⊆ S. Then S ∩ V (G i ) ≤ b. If |V (G i ) \ (S ∪ S ′ )| > x, then we can remove vertices from S \ (C i ∪ G i ) and add vertices from G i \ (S ∪ S ′ ) to S till S ∩ V (G i ) = b. This will not influence the collapsed vertices in S ′ . Then we get a vertex set S i = S ∩ V (G i ), and the k-core of G i − S i has size at most x.

Conclusion
Our results highlight a dichotomy in the computational complexity of Collapsed k-Core for k ≤ 2 and k ≥ 3. Along the way, we correct some inaccuracies in the literature concerning the parameterized complexity of Collapsed k-Core with k = 3 and x = 0 and give a simple single exponential linear time parameterized algorithm for Feedback Vertex Set. We further investigate the parameterized complexity with respect to several structural parameters of the input graph. As a highlight we show that Collapsed k-Core does not admit polynomial kernels for rather large parameter combinations. We leave the complexity of Collapsed k-Core when parameterized by the cliquewidth of the input graph open.