1 Introduction
For a family of graphs, the general Graph Modification problem asks whether we can modify a graph into a graph in by performing at most simple operations. Typical examples of simple operations wellstudied in the literature include vertex deletion, edge deletion, edge addition, or a combination of edge deletion and addition. We call these problems Vertex Deletion, Edge Deletion, Edge Addition, and Edge Editing, respectively. By a classical result by Lewis and Yannakakis [LewisY80], Vertex Deletion is NPcomplete for all nontrivial hereditary graph classes. The situation is quite different for the edge modification problems. Earlier efforts for edge deletion problems [ElmallahC88, Yannakakis81], though having produced fruitful concrete results, shed little light on a systematic answer, and it was noted that such a generalization is difficult to obtain.
Graph Modification problems have been extensively investigated for graph classes that can be characterized by a finite set of forbidden induced subgraphs. We say that a graph is free if it contains none of the graphs in as an induced subgraph. For this special case, the free Vertex Deletion is well understood. If contains a graph on at least two vertices, then all of these problems are NPcomplete, but admit a algorithm [Cai96], where is the size of the largest graph in (the algorithms with running time are called fixedparameter tractable (FPT) algorithms [CyganFKLMPPS15, DowneyFellows13]). On the other hand, the NPhardness proof of Lewis and Yannakakis [LewisY80] excludes algorithms with running time under the Exponential Time Hypothesis (ETH) [ImpagliazzoP01]. Finally, as observed by Flum and Grohe [FlumG06] a simple application of sunflower lemma [ErdosR60] gives a kernel with vertices, where is again the size of the largest graph in . A kernel is a polynomial time preprocessing algorithm which outputs an equivalent instance of the same problem such that the size of the reduced instance is bounded by some function that depends only on . We call the function the size of the kernel. It is wellknown that any problem that admits an FPT algorithm admits a kernel. Therefore, for problems with FPT algorithms one is interested in polynomial kernels, i.e., kernels whose size is a polynomial function.
For the edge modification problems, the situation is more complicated. While all of these problems also admit time algorithm, where is the maximum number of edges in a graph in [Cai96], the P vs NP dichotomy is still not known. Only recently Aravind et al. [AravindSS17b] gave the dichotomy for the special case when contains precisely one graph . From the kernelization point of view, the situation is also more difficult. The reason is that deleting or adding an edge to a graph can introduce a new copy of and this might further propagate. Hence, we cannot use the sunflower lemma to reduce the size of the instance. Cai asked the question whether free Edge Deletion admits a polynomial kernel for all graphs [bodlaender2006open]. Kratsch and Wahlström [KratschW13]
showed that this is probably not the case and gave a graph
on vertices such that free Edge Deletion and free Edge Editing does not admit a polynomial kernel unless . Consequently, it was shown that this is not an exception, but rather a rule [CaiCai15, GuillemotHPP13]. Indeed the result by Cai and Cai [CaiCai15] shows that free Edge Deletion, free Edge Addition, and freeEdge Editing do not admit a polynomial kernel whenever or its complement is a path or a cycle with at least edges or a connected graph with at least edges missing. Very recently, Marx and Sandeep [MarxSandeep20] gave a list of nine graphs, all on vertices such that if freeEdge Editing does not admit a kernel for any of these nine graphs under standard complexity assumptions, then freeEdge Editing admits a polynomial kernel for if and only if is either empty or complete graph. They also provided a similar characterization for free Edge Deletion and free Edge Editing. This suggests that actually the free edge modification problems with a polynomial kernels are rather rare and only for small graphs . Recently, Eiben, Lochet, and Saurabh [EibenLochetSaurabhArXiv] announced a polynomial kernel for the case when is a paw, which leaves only one last graph on vertices for which the kernelization of free edge modification problems remains open, namely known also as the claw.The class of clawfree graphs is a very well studied class of graphs with some interesting algorithmic properties. The most prominent example is probably the algorithm of Sbihi [SBIHI198053] for computing the maximum independent set in polynomial time. It also has been extensively studied from a structural point of view, and Chudnosky and Seymour proposed, after a series of papers, a complete characterization of clawfree graphs [CHUDNOVSKY2008839]. Because of such a characterization, it seems reasonable to believe that a polynomial kernel for Clawfree Edge Deletion exists. However, the characterization of Chudnosky and Seymour is quite complex, which makes it hard to use. For this reason, as noted by Cygan et al. [CyganPPLW17], trying to show the existence of a polynomial kernel in the cases of subclasses of clawfree graphs seems like a good first step to try to understand this problem. In this paper, we prove the result for the most famous such class, line graphs.
[] LineGraph Edge Deletion admits a kernel with vertices.
Overview of the Algorithm
As the first step of the kernelization algorithm, we use the characterization of line graphs by forbidden induced subgraphs to find a set of at most vertices such that for every vertex , is a line graph. This is simply done by a greedy edgedisjoint packing of forbidden induced subgraphs. Having the set , we use the algorithm by Degiorgi and Simon [DegiorgiSimon95] to find a partition of edges of into cliques such that each vertex is in precisely cliques. Let be the cliques in the partition. Since is also a line graph, it is a rather simple consequence of Whitney’s isomorphism theorem that the neighborhood of can be covered by constantly many cliques of . Furthermore, we will show that if a clique in has more than vertices then the optimal solution does not contain an edge in . Hence, we can partition the cliques in into two groups “large” and “small”. Note that if the optimal solution contains an edge in some small clique , then for this change to be necessary, it has to be propagated from by modifying small cliques on some cliquepath from to using only small cliques. We will therefore define the distance of a clique to , without going into too many details in here, to be basically the length of a shortest cliquepath from the clique to using only small cliques. Since there are only cliques in immediate neighborhood of and the number of cliques in the neighborhood of a small clique is bounded by its size, we obtain that there are at most cliques at distance at most . Our main contribution and most technical part of our proof is to show that we can remove the edges covered by cliques at distance at least from . This is covered in Section 4. Afterwards we end up with an instance with all cliques in at distance at least from being singletons. As discussed above there are only cliques at distance at most and because large cliques stay intact in any optimal solution, it suffices to keep vertices in each large clique, which leads to the desired kernel of size .
2 Preliminaries
We assume familiarity with the basic notations and terminologies in graph theory. We refer the reader to the standard book by Diestel [diestel] for more information. Given a graph and a set of edges , we denote by the graph whose set of vertices is and set of edges is the set . Given two vertices , we let the distance between and in , denoted , be the number of edges on a shortest path from to . Furthermore, for and we let . We omit the subscript , if the graph is clear from the context.
Parameterized Algorithms and Kernelization.
For a detailed illustration of the following facts the reader is referred to [CyganFKLMPPS15, DowneyFellows13]. A parameterized problem is a language , where is a finite alphabet; the second component of instances is called the parameter. A parameterized problem is fixedparameter tractable if it admits a fixedparameter algorithm, which decides instances of in time for some computable function .
A kernelization for a parameterized problem is a polynomialtime algorithm that given any instance returns an instance such that if and only if and such that for some computable function . The function is called the size of the kernelization, and we have a polynomial kernelization if is polynomially bounded in . It is known that a parameterized problem is fixedparameter tractable if and only if it is decidable and has a kernelization. However, the kernels implied by this fact are usually of superpolynomial size.
A reduction rule is an algorithm that takes as input an instance of a parameterized problem and outputs an instance of the same problem. We say that the reduction rule is safe if is a yesinstance if and only if is a yesinstance. In order to describe our kernelization algorithm, we present a series of reduction rules.
Line graphs.
Given a graph , its line graph is a graph such that each vertex of represents an edge of and two vertices of are adjacent if and only if their corresponding edges share a common endpoint (are incident) in . It is well known that if the line graphs of two connected graphs and are isomorphic then either and are and , respectively, or and are isomorphic as well (Whitney’s isomorphism theorem [Whitney32], see also Theorem 8.3 in [harary1969graph]). We say that a graph is a line graph, if there exists a graph such that . Note that in this paper we only consider simple graphs, i.e., the graphs without loops or multiple edges and in particular we also only consider line graphs of simple graphs. Formally, we then study the following parameterized problem:
0.98 LineGraphEdge Deletion
[5pt] Input: A graph and . Parameter: . Question: Is there a set of edges such that is a line graph and .
We call a set of edges such that is a line graph a solution for . A solution is optimal, if there does not exists a solution such that . To obtain our kernel, we will make use of several equivalent characterizations of line graphs.
[see, e.g., Theorem 8.4 in [harary1969graph]] The following statements are equivalent:

is a line graph.

The edges of can be partitioned into complete subgraphs in such a way that no vertex lies in more than two of the subgraphs.

does not have
as an induced subgraph, and if two odd triangles (triangles with the property that there exists another vertex adjacent to an odd number of triangle vertices) share a common edge, then the subgraph induced by their vertices is
. 
None of nine graphs of Figure 1 is an induced subgraph of .
3 Structure of Line Graphs
To obtain our kernel, we heavily rely on different characterizations of line graphs given by Theorem 2. The two main characterizations used throughout the paper are given in points (2) and (4) To ease the presentation of our techniques, we will define a notion of a clique partition witness for , whose existence is implied by the point (2) of Theorem 2. Let be a line graph, a clique partition witness for is a set be such that:

for all ,

is a complete graph for all , that is every is a clique in ,

for all ,

every is in exactly two sets in , and

for every edge there exists exactly one set such that .
Note that by Theorem 2, is a line graph if and only if there exists a clique partition witness for . The following three observations follow directly from the definition of clique partition witness and will be useful throughout the paper.
Observation .
If is clique partition witness for then every clique in is either a singleton, , or a maximal clique in .
Observation .
If is clique partition witness for , then every maximal clique in of size at least is in .
Observation .
If is clique partition witness for , then any clique of which is not a subclique of some element of is a triangle.
We would like to point out that given a line graph one can find a clique partition witness for for example by using an algorithm of Degiorgi and Simon [DegiorgiSimon95] for recognition of line graphs in polynomial time. In the following lemma, we sketch the main procedure of their algorithm together with necessary modifications to actually output a clique partition witness instead of the underlying graph such that , for completeness.
Given a graph , there is an algorithm that in time decides whether is a line graph and if so, constructs a clique partition witness for .
Proof.
The algorithm by Degiorgi and Simon construct the input graph by adding vertices one at a time, at each step it chooses a vertex to add that is already adjacent to at least one previouslyadded vertex. That is it construct graphs , , , such that is a connected subgraph of on vertices. At each step it maintains a graph such that is a line graph of . In here, we can actually keep a clique partition witness for such that there is a bijection between vertices of and clique in such that if and only if .
The algorithm heavily relies on the Whitney’s isomorphism theorem that implies that if the underlying graph of has at least vertices, then the underlying graph is unique up to isomorphism. When adding a vertex to a graph for , the algorithm simply bruteforces the possibilities for and .
When adding a vertex to when , let be the subgraph of formed by the edges that correspond to the neighbors of in . Check that has a vertex cover consisting of one vertex or two nonadjacent vertices, i.e., there are cliques and in with and . If there are two vertices in the cover, add an edge (corresponding to ) that connects these two vertices in and add to both and . If there is only one vertex in the cover, then add a new vertex to , adjacent to this vertex, add to the clique in and add a new clique to to create . ∎
3.1 Level Structure of Instances
For the rest of the paper, let be the input graph and let be a set of at most vertices such that for every the graph is a line graph. We let be a clique partition witness for . The goal of this subsection is to split the cliques in to levels such that 1) each level contains only bounded number of cliques (that are not singletons) and 2) if we do not remove any edge at level , then we do not need to remove any edge at level . We will later show that we do not need to remove any edges in cliques in level . The following lemma is useful to define/bound the number of cliques at the first level, i.e., cliques that interact with .
For every vertex there are at most two cliques such that is adjacent to all vertices in and to at most vertices in .
Proof.
By the choice of the set , it follows that is a line graph. Let be clique partition witness for . By definition, there are at most two cliques and in that contains and all its neighbors. If , for some , then by Observation 3, is a clique in and we can set to be . Else and contributes to at most neighbors of in . ∎
The following lemma shows that cliques of size at least can serve as kind of separators that will never be changed by a solution of size at most . Hence, we can remove all cliques separated from by large cliques. Moreover, it allows us to define the st level by only considering the cliques of size at most at level .
Let such that and let be an optimal solution for . Then . Moreover, the clique partition witness for contains a clique such that .
Proof.
Let such that . Clearly there are at most vertices in such that either or . Let be such that are edges in . Similarly, there are at most nonedges to in , so let be a vertex such that are edges in . Repeating the same argument once again, there is such that are edges in . However, the subgraph of induced on is minus an edge, which is one of the forbidden induced subgraphs in the characterization of line graphs.
The moreover part follows from the following argument. Since and, by Observation 3 it follows that the clique partition witness contains a maximal clique . It remains to show that no vertex in is in . Every vertex in is in two cliques , in that cover all its incident edges in . If none of these two cliques is , then intersect each of these two cliques in at most vertex. It follows that, because , there is no vertex in adjacent to all vertices of . ∎
Let us now partition the cliques in into two parts and such that contains precisely all the cliques in with less than vertices and contains the remaining cliques. We will refer to the cliques in as small cliques and the cliques in as large cliques. Intuitively, if we are forced to delete some edge in , then this change had to be propagated from only by changes in small cliques.
We are now ready to define the level structure on the cliques in . We divide the cliques in into levels , for some , that intuitively reflects on how far from the clique is if we consider a shortest path using only small cliques. We will define the levels recursively as follows. By Lemma 3.1 for every vertex there exists at most two cliques such that is adjacent to all vertices in and to at most vertices in . Now, for a vertex , let denote the set of cliques that contains and all the cliques in that contain at least one of the neighbors of in . We let be precisely the set . Note that vertices in can each appear in one other clique that is not in and in particular there are cliques that contain a vertex adjacent to a vertex in and are not in . For , we then let be the set of cliques in such that there is a small clique in the previous level (i.e., ) such that is not empty.
Observation .
Let and a vertex in . If has a neighbor in , then either or is in a large clique.
Proof.
Let be a neighbor of . Then contains a clique with . Clearly intersects in . Hence either is a large clique or by the definition of the clique is in . ∎
Let be such that and . While the following Reduction Rule is not completely necessary and would be subsumed by Reduction Rule 2, we include it to showcase some of the ideas needed for the proof in a simplified setting.
Reduction Rule 1.
Remove all vertices in that are not in a clique in .
Proof of safeness.
Let be the resulting graph and let be a set of cliques of obtained from , by taking all cliques in and for every clique in , contains , if it is nonempty. Since is an induced subgraph of and line graphs can be characterized by a set for forbidden induced subgraphs, it follows that for every , if is a line graph, then is a line graph. It remains to show that if there is a set of edges such that and is a line graph, then is also a line graph. Let be such a set of edges of minimum size and let be a clique partition witness for . It suffices to show that for every clique in , it holds that . If this is the case, we get a clique partition witness for by replacing the cliques of in by .
Now, means that all cliques intersecting are large. Moreover, because all vertices in are in some clique on some level, by Lemma 3.1, for each clique that intersect there is a clique in that is the union of and some vertices in . Hence, all vertices in are already in at least one clique in and all the edges incident to exactly one vertex in are already covered by these cliques. And hence every clique that contains a vertex in and intersects every other clique in in at most one vertex has to be a subset of . Moreover, the cliques in that are subsets of have to be vertex disjoint, since every vertex is in at most cliques in . Hence, if is not in , then some of the edges in have to be in , but replacing all the subsets of in by gives a clique partition witness for for some which contradicts the fact that is of minimum size. ∎
We will also say that is at distance from , denoted by , if is in . We note that still contains some cliques that are not in any of ’s. We will let for such a clique . We can now upper bound the number of cliques at distance from .
There are at most cliques in at level , i.e., in .
Proof.
By the definition of , where denote the set of cliques that contains and all the cliques in that contain at least one of the neighbors of in . By Lemma 3.1 for every vertex there exists at most two cliques such that is adjacent to all vertices in and to at most vertices in . Since every vertex appears in two cliques of , it follows that and consecutively contains at most cliques. Now by the definition of we know that for any a clique is at level if and only if it shares a vertex with a small clique at level . Since no three cliques in can share a vertex the number of cliques at level is at most the number of vertices in the small cliques at level and the lemma follows by a simple induction on . ∎
The remainder of the algorithm consists of two steps. First, in Section 4, we show that we can remove all edges from cliques that are at distance at least from . Afterwards, due to Lemma 3.1, we are left with only nonsingleton cliques in . To finish the algorithm in Section 5, for each clique that is not a singleton, we mark an arbitrary subset of vertices in and remove all unmarked vertices from . It is then rather straightforward consequence of Lemma 3.1 that this rule is safe and we get an equivalent instance with vertices.
4 Bounding the Distance from
The purpose of this section is to show that it is only necessary to keep the cliques in that are at distance at most from (and adding a singleton for vertices covered by exactly one clique at distance at most ). To do so, we need to show that there is always a solution that does not change the cliques at distance at all. For this purpose, we first need to understand the interaction of cliques at distance from with the solution. The first step will be to show that there is an optimal solution with clique partition witness such that all cliques in that share an edge with a clique in at distance at least from are actually subcliques of a clique in (when restricted to ). It is a simple consequence of Lemma 3.1 that this is true for any clique that intersect a large clique in an edge. Hence, we can only care about cliques in that intersect a small clique in an edge. By Observation 3.1, no vertex in has a neighbor in . It then follows by Observation 3 that any clique in that intersects in an edge and is not a subclique of a clique in is indeed a triangle. This leads us to the following definition.
[bad triangle] Let be such that is a line graph and let be a clique partition witness of . A triangle is said to be bad if it is not a subclique of a clique in , and one of the edges of the triangle, say , is an edge contained in a clique of distance at least from .
There exists an optimal solution without any bad triangle.
Proof.
Let be an optimal solution and the clique partition witness of . Suppose is a bad triangle and let and be the elements of containing the edges and respectively. See also Figure 2 for an illustration. Since is a bad triangle, no clique in is a superset of , and it is a simple consequence of Lemma 3.1 that is a small clique. By definition of bad triangle, at least one of , , and is at distance at least from and hence all of these cliques are at distance at least from . Let (resp. , ) denote the other clique of containing (reps. , ). Let us define and .
Let , and . Note that is a subclique of for . Now for every we will update as follows. As long as there exists an edge in such that belongs to , is a subclique of and , we set (see also Figure 1(b)). When this process stops, corresponds to the union of a set of elements of : which are subcliques of , and . Moreover, for any edge of which is strictly contained in another clique of (meaning this clique is not ), then this clique has to be a triangle by Observation 3, as the clique of containing is . Let denote the set of such edges and let be the triangles of containing these edges. Note that , as for any edge , either or has to be non adjacent to each extremity in or the edge would be in two cliques of (the same statement is also correct for and ). Let be the set obtained from by

Removing all the edges of , and .

Adding one of the two edges of different from for every and (see Figure 1(c) illustrating the replacement of in by its proper subclique in implied by this addition of an edge in .).
is a set of edges not larger than and such that is a line graph with fewer bad triangles than .
Proof.
The fact that follows from the fact that for all . To see that is a line graph, let us show that defined as follows is a clique partition witness for . Let be the set defined from by

Removing , , , , every for , , every for every and and every edge which are contained in one of the .

Adding for and for every and the edge of which has not been removed from , as well as singletons for vertices belonging to only one clique.
First it is clear that any set added to is a clique as does not contain any edge in , and and these sets are cliques of .
Now take and two cliques of . If and belong to , then clearly their intersection has size at most . If one belongs to and the other is the remaining edge of for and , then it is also clear as it is true for . For , and also intersect on one vertex, because and do and moreover, the cliques of intersecting on two vertices are exactly the , so if and , the intersection has also size at most , and we covered all the cases for .
Now for every vertex , if does not belong to and , then it belongs to the same cliques as in (where the have been reduced to an edge and a singleton). For the vertices of and different from , we replaced one subclique of by another. Finally belongs to and , to and and to and .
Suppose is an edge of . If belongs to one of the , then by definition of the and because we removed all these triangles, only belongs to one clique. For the other edges of , the fact that belongs to exactly one clique of follows from the fact that differs on those edges from only because we added some edges of the , and differs on these vertices only because we changed into the remaining edge outside .
Overall is indeed a clique partition for . Moreover, to obtain it, we removed at least one bad triangle from () without adding one. This ends the proof of the claim. ∎
Finally, we can repeat the process until is without any bad triangles, which ends the proof of the lemma. ∎
Before we show that indeed all cliques at distance at least from are intact in some optimal solution, we show another auxiliary lemma that is rather simple consequence of Lemma 4, namely that there is a clique partition witness for some optimal solution such that no two cliques that intersect the same clique at distance at least from in an edge can intersect. This is important later to show that indeed no vertex in a clique at distance from will be in two cliques in that are not subsets of .
There exists an optimal solution without any bad triangles and clique partition witness for such that for every of distance at least and every , if and are the two cliques in containing , then either or .
Proof.
Let be an optimal solution for without any bad triangles and clique partition witness for minimizing the number of pairs for which is at distance at least , and the two cliques, denoted and , in containing intersect in two vertices. Furthermore, it follows from Lemma 3.1 that is a small clique, as the clique containing as a subclique in would intersect in two vertices. Since there are no bad triangles and is at distance at least , it follows that and and in particular is a clique in . Indeed, our goal is to replace and by a clique such that . We start by setting . We will also keep a track of cliques we will remove from . This set will be and initialize it as .
As in the proof of Lemma 4, the only reason why we cannot replace and by and obtain a solution that removes a subset of edges of is because there exist two vertices and a clique with . Observe that by our assumption there is no bad triangle and . We let and and repeat until there is no such pair of vertices. Note that every vertex in is in at most two cliques of . Therefore, this process has to stop after at most steps.
When there are no two vertices in that appear together in a different clique, we remove from and replace it by and . For every vertex that appears in , we removed one clique that it appeared in. Hence, every vertex appears in at most cliques and we can always add a singleton to clique partition witness for vertices that are only in one clique. Moreover, no two cliques intersect in two vertices, since is the only clique we added, and we removed/changed all the cliques that intersected in at least two vertices. Finally, all edges in remain covered, we only potentially covered some additional edges in .
Note that this procedure does not introduce any bad triangles or new pair for which is at distance at least , and the two cliques in containing intersect in two vertices. As it also removes one such pair, we obtain a contradiction with the choice of . We can therefore deduce that does not contain such pair and the lemma follows. ∎
Finally, we can state the main lemma of this section.
There exists an optimal solution for and a clique partition witness for such that for every clique at distance at least it holds that .
Proof.
Let be an optimal solution without any bad triangles and clique partition witness for such that for every of distance at least and every , if and are the two cliques in containing , then either or . Note that existence of such a solution is guaranteed by Lemma 4. Moreover let be such an optimal solution satisfying properties in Lemma 4 that minimizes the number of cliques of distance at least such that . We claim that satisfies the properties of the lemma.
For a contradiction let be a clique at distance at least and let be the cliques in that intersects in at least vertices. Since there is no bad triangle, it follows that for all and by optimality of , (else is missing at least one edge). We claim that . Else let . Note that is a small clique and hence by Observation 3.1 does not have a neighbor in . In particular all neighbors of are covered by two cliques in , one of those cliques is and let the other clique be . Moreover, Let and be the two cliques in containing . Since both and are subsets of . However, is either a large clique and contains and the cliques and are and respectively, or is a small clique, in which case is at distance at least from , because it shares a vertex with the clique at distance at least from . It follows by the choice of that either or , but then again either or is the singleton . However then the clique partition witness defines a better solution. It follows that indeed for all cliques in at distance at least in . ∎
We are now ready to present our main reduction rule. Note that it would seem that we could remove just the vertices that do no appear in a clique at distance at most . However, because of the large cliques in at the first four levels, we would be potentially left with many cliques at distance infinity that we cannot remove because all of their vertices are in a large clique at distance at most from . While this case could have been dealt with separately, we can actually show a stronger claim, i.e., that we can remove all edges from that are covered by a clique at distance at least from . Note that in this case we cannot easily claim that if is YESinstance then so is the reduced instance and we crucially need the fact that cliques at distance at least are kept in clique partition witness of some optimal solution.
Reduction Rule 2.
Remove all edges such that for some clique with . Afterwards remove all isolated vertices from .
Let be the set of cliques at distance at least from , the set of vertices that appear in a clique in and in a clique in and be the graph obtained after applying the reduction rule and let . Note that is a clique partition witness for and that , for , is a clique at distance at least .
Proof of safeness.
Let , , , be as described above and let be an optimal solution for , that is is a line graph, and let be clique partition witness for . By Lemma 4, we can assume that . We will show that is a clique partition witness for . Clearly each edge in is either covered by or by . It is also easy to see that every vertex is in precisely two cliques. Moreover, two cliques in intersect in at most vertex, because and similarly two cliques in intersect in at most one vertex. Finally, let and . Clearly, . Moreover, for , the edge is not in and hence . Hence, .
On the other hand, let be an optimal solution for and a clique partition witness for such that for every clique at distance at least it holds that . Note that the existence of is guaranteed by Lemma 4. We claim that is a line graph. By the choice of , it follows that . Moreover, for every edge that is covered by a clique in it holds that . It follows rather straightforwardly that is indeed a clique partition witness for . ∎
5 Finishing the Proof
Suppose now that , , and correspond to the instance after applying Reduction Rules 1 and 2. Clearly all cliques in are either at distance at most from or there are singletons at distance or infinity, depending on whether the singleton intersects a small or a large clique, respectively. It follows from Lemma 3.1 that there are at most cliques at distance at most . We let be any minimal w.r.t. inclusion set of vertices such that for every clique in at distance at most it holds that . Such a set can be easily obtained by including arbitrary vertices from every clique at distance at most and then removing the vertices such that for all at distance at most . From this construction it is easy to see that .
Reduction Rule 3.
Remove all vertices in from .
Proof of safeness.
Let the clique partition witness for be . Since line graphs are characterized by a finite set of forbidden induced subgraphs, it is easy to see that if is a line graph, for some , then is also a line graph. For the other direction, let be such that is line graph. We will show that is a line graph. Let be a clique partition witness for