1 Introduction
The Independent Set problem, which asks for a maximum sized set of pairwise nonadjacent vertices in a graph, is one of the most wellstudied problems in algorithmic graph theory. It was among the first 21 problems that were proven to be NPhard by Karp [22], and is also known to be hopelessly difficult to approximate in polynomial time: Håstad [21] proved that under standard assumptions from classical complexity theory the problem admits no approximation, for any (by we always denote the number of vertices in the input graph). This was later strengthened by Khot and Ponnuswami [23], who were able to exclude any algorithm with approximation ratio , for any . Let us point out that the currently best polynomialtime approximation algorithm for Independent Set achieves the approximation ratio [17].
There are many possible ways of approaching such a difficult problem, in order to obtain some positive results. One could give up on generality, and ask for the complexity of the problem on restricted instances. For example, while the Independent Set problem remains NPhard in subcubic graphs [18], a straightforward greedy algorithm gives a 3approximation.
free graphs.
A large family of restricted instances, for which the Independent Set problem has been wellstudied, comes from forbidding certain induced subgraphs. For a (possibly infinite) family of graphs, a graph is free if it does not contain any graph of as an induced subgraph. If consists of just one graph, say , then we say that is free. The investigation of the complexity of Independent Set in free graphs dates back to Alekseev, who proved the following.
Theorem 1.1 (Alekseev [2]).
Let be a constant. The Independent Set problem is NPhard in graphs that do not not contain any of the following induced subgraphs:

a cycle on at most vertices,

the star , and

any tree with two vertices of degree at least 3 at distance at most .
We can restate Theorem 1.1 as follows: the Independent Set problem is NPhard in free graphs, unless is a subgraph of a subdivided claw (i.e., three paths which meet at one of their endpoints). The reduction also implies that for each such the problem is APXhard and cannot be solved in subexponential time, unless the Exponential Time Hypothesis (ETH) fails. On the other hand, polynomialtime algorithms are known only for very few cases. First let us consider the case when , i.e., we forbid a path on vertices. Note that the case of is trivial, as every free graph is a disjoint union of cliques. Already in 1981 Corneil, Lerchs, and Burlingham [11] showed that Independent Set is tractable for free graphs. For many years there was no improvement, until the breakthrough algorithm of Lokshtanov, Vatshelle, and Villanger [25] for free graphs. Their approach was recently extended to free graphs by Grzesik, Klimošova, Pilipczuk, and Pilipczuk [19]. We still do not know whether the problem is polynomialtime solvable in free graphs, and we do not know it to be NPhard in free graphs, for any constant .
Even less is known for the case if is a subdivided claw. The problem can be solved in polynomial time in clawfree (i.e., free) graphs, see Sbihi [32] and Minty [31]. This was later extended to free graphs, where is a claw with one edge once subdivided (see Alekseev [1] for the unweighted version and Lozin, Milanič [27] for the weighted one).
When it comes to approximations, Halldórsson [20] gave an elegant local search algorithm that finds a approximation of the maximum independent set in free graphs for any constant in polynomial time. Very recently, Chudnovsky, Thomassé, Pilipczuk, and Pilipczuk [10] designed a QPTAS (quasipolynomialtime approximation scheme) that works for every , which is a subgraph of a subdivided claw (in particular, a path). Recall that on all other graphs the problem is APXhard.
Parameterized complexity.
Another approach that one could take is to look at the problem from the parameterized perspective: we no longer insist on finding the maximum independent set, but want to verify whether some independent set of size at least exists. To be more precise, we are interested in knowing how the complexity of the problem depends on . The best type of behavior we are hoping for is fixedparameter tractability (FPT), i.e., the existence of an algorithm with running time , for some function (note that since the problem is NPhard, we expect to be superpolynomial).
It is known [12] that on general graphs the Independent Set problem is W[1]hard parameterized by , which is a strong indication that it does not admit an FPT algorithm. Furthermore, it is even unlikely to admit any nontrivial fixedparameter approximation (FPA): a FPA algorithm for the Independent Set problem is an algorithm that takes as input a graph and an integer , and in time either correctly concludes that has no independent set of size at least , or outputs an independent set of size at least (note that does not have to be a constant). It was shown in [6] that on general graphs no FPA exists for Independent Set, unless the randomized GapETH fails.
Parameterized complexity in free graphs.
As we pointed out, none of the discussed approaches, i.e., considering free graphs or considering parameterized algorithms, seems to make the Independent Set problem more tractable. However, some positive results can be obtained by combining these two settings, i.e., considering the parameterized complexity of Independent Set in free graphs. For example, the Ramsey theorem implies that any graph with vertices contains a clique or an independent set of size . Since the proof actually tells us how to construct a clique or an independent set in polynomial time [16], we immediately obtain a very simple FPT algorithm for free graphs. Dabrowski [13] provided some positive and negative results for the complexity of the Independent Set problem in free graphs, for various . The systematic study of the problem was initiated by Bonnet, Bousquet, Charbit, Thomassé, and Watrigant [4] and continued by Bonnet, Bousquet, Thomassé, and Watrigant [5]. Among other results, Bonnet et al. [4] obtained the following analog of Theorem 1.1.
Theorem 1.2 (Bonnet et al. [4]).
Let be a constant. The Independent Set problem is W[1]hard in graphs that do not contain any of the following induced subgraphs:

a cycle on at least 4 and at most vertices,

the star , and

any tree with two vertices of degree at least 3 at distance at most .
Note that, unlike in Theorem 1.1, we are not able to show hardness for free graphs: as already mentioned, the Ramsey theorem implies that Independent Set is FPT in free graphs. Thus, graphs for which there is hope for FPT algorithms in free graphs are essentially obtained from paths and subdivided claws (or their subgraphs) by replacing each vertex with a clique.
Let us point out that, even though it is not stated there explicitly, the reduction of Bonnet et al. [4] also excludes any algorithm solving the problem in time , unless the ETH fails.
Our results.
We study the approximation of the Independent Set problem in free graphs, mostly focusing on approximation hardness. Our first two results are related to Halldórsson’s [20] polynomialtime approximation algorithm for free graphs. First, in Section 3 we extend this result to free graphs, for any constants , showing the following theorem.
Theorem 1.3 ().
Given a free graph , an approximation can be computed in time.
Then, in Section 4 we show that the approximation ratio of the algorithm of Halldórsson [20] is optimal, up to logarithmic factors.
Theorem 1.4 ().
There is a function such that the Independent Set problem does not admit a polynomial time approximation algorithm in free graphs, unless ZPP = NP.
Note that the factor determining the approximation gap in Theorem 1.4 is expressed as an asymptotic function of , i.e., for growing . In our case however, it is an interesting question how small the degree can be so that we obtain an inapproximability result. We prove Theorem 1.4 by a reduction from the Label Cover problem, and a corresponding inapproximability result by Laekhanukit [24]. By calculating the bounds given in [24] (which heavily depend on the constant of Chernoff bounds) it can be shown that an inapproximability gap exists for in Theorem 1.4.
Then in Section 5 we study the existence of fixedparameter approximation algorithms for the Independent Set problem in free graphs. We show the following strengthening of Theorem 1.2, which also gives (almost) tight runtime lower bounds assuming the ETH or the randomized GapETH (for more information about complexity assumptions used in Theorem 1.5 see Section 2).
Theorem 1.5 ().
Let be a constant, and let be the class of graphs that do not contain any of the following induced subgraphs:

a cycle on at least 5 and at most vertices,

the star , and


the star , or

a cycle on 4 vertices and any tree with two vertices of degree at least 3 at distance at most .

The Independent Set problem on does not admit the following:

an exact algorithm with runtime , for any computable function , under the ETH,

a approximation algorithm with runtime for some constant and any computable function , under the deterministic GapETH,

a approximation algorithm with runtime for some constant and any computable function , under the randomized GapETH.
Finally, in Section 6 we study a slightly different setting, where the graph is not considered to be fixed. As mentioned before, Independent Set is known to be polynomialtime solvable in free graphs for . The algorithms for increasing values of get significantly more complicated and their complexity increases. Thus it is natural to ask whether this is an inherent property of the problem and can be formalized by a runtime lower bound when parameterized by .
We give an affirmative answer to this question, even if the forbidden family is not a family of paths: note that the independent set number of a path on vertices is .
Proposition 1 ().
Let be an integer and let be a family of graphs, such that for every . The Independent Set problem in free graphs is W[1]hard parameterized by and cannot be solved in time, unless the ETH fails.
We also study the special case when and consider the inapproximability of the problem parameterized by both and . Unfortunately, for the parameterized version we do not obtain a clearcut statement as in Theorem 1.4, since in the following theorem cannot be chosen independently of in order to obtain an inapproximability gap.
Proposition 2 ().
Let be any constant and . The Independent Set problem in free graphs has no approximation algorithm with runtime for any computable function , unless the deterministic GapETH fails.
Note that this in particular shows that if we allow to grow as a polynomial for any constant , then no approximation is possible for any (since ). This indicates that the approximation for free graphs [20] is likely to be best possible (up to subpolynomial factors), even when parameterizing by and . The proofs of Proposition 1 and Proposition 2 can be found in Section 6.
2 Preliminaries
All our hardness results for Independent Set are obtained by reductions from some variant of the Maximum Colored Subgraph Isomorphism (MCSI) problem. This optimization problem has been widely studied in the literature, both to obtain polynomialtime and parameterized inapproximability results, but also in its decision version to obtain parameterized runtime lower bounds. We note that by applying standard transformations, MCSI contains the wellknown problems Label Cover [24] and Binary CSP [26]: for Binary CSP the graph is a complete graph, while for Label Cover is usually bipartite.
Maximum Colored Subgraph Isomorphism (MCSI)  

Input:  A graph , whose vertex set is partitioned into subsets , and a graph on vertex set . 
Goal:  Find an assignment , where for every , that maximizes the number of satisfied edges, i.e.,

Given an instance of MCSI, we refer to the number of vertices of as the size of . Any assignment , such that for every it holds that , is called a solution of . The value of a solution is , i.e., the fraction of satisfied edges. The value of the instance , denoted by , is the maximum value of any solution of .
When considering the decision version of MCSI, i.e., determining whether or , a result by Marx [29] gives a runtime lower bound for parameter under the Exponential Time Hypothesis (ETH). That is, no time algorithm can solve MCSI for any computable function , assuming there is no deterministic time algorithm to solve the 3SAT problem. For the optimization version, an approximation is a solution with . When is a complete graph, a result by Dinur and Manurangsi [14, 15] states that there is no approximation algorithm (where for any constant ) with runtime for any computable function , unless the deterministic GapETH fails (see Theorem 6.3). This hypothesis assumes that there exists some constant such that no deterministic time algorithm for 3SAT can decide whether all or at most a fraction of the clauses can be satisfied. A recent result by Manurangsi [28] uses an even stronger assumption, which also rules out randomized algorithms, and in turn obtains a better runtime lower bound at the expense of a worse approximation lower bound: he shows that, when is a complete graph, there is no approximation algorithm for MCSI with runtime for any computable function and any constant , under the randomized GapETH. This assumes that there exists some constant such that no randomized time algorithm for 3SAT can decide whether all or at most a fraction of the clauses can be satisfied. (Note that the runtime lower bound under the stronger randomized GapETH does not have the factor in the polynomial degree as the runtime lower bound under ETH does.)
For our results we will often need the special case of MCSI when the graph has bounded degree. We define this problem in the following.
Degree Maximum Colored Subgraph Isomorphism (MCSI())  

Input:  A graph , whose vertex set is partitioned into subsets , and a graph on vertex set and maximum degree . 
Goal:  Find an assignment , where for every , that maximizes the number of satisfied edges, i.e.,

The bounded degree case has been considered before, and we harness some of the known hardness results for MCSI() in our proofs. First, let us point out that the lower bound for exact algorithms holds even for the case when , as shown by Marx and Pilipczuk [30]. We also use a polynomialtime approximation lower bound given by Laekhanukit [24], where can be set to any constant and the approximation gap depends on (see Theorem 4.1). The complexity assumption of this algorithm is that NPhard problems do not have polynomial time Las Vegas algorithms, i.e., NP ZPP. For parameterized approximations, we use a result by Lokshtanov et al. [26], who obtain a constant approximation gap for the case when (see Theorem 5.2). It seems that this result for parameterized algorithms is not easily generalizable to arbitrary constants so that the approximation gap would depend only on , as in the result for polynomialtime algorithms provided by Laekhanukit [24]: neither the techniques found in [24] nor those of [26] seem to be usable to obtain an approximation gap that depends only on but not the parameter . However, we develop a weaker parameterized inapproximability result for the case when (see Theorem 6.1 in Section 6), and use it to prove Proposition 2.
3 Approximation for free graphs
In this section we give a polynomialtime approximation algorithm for Independent Set on free graphs, where is the size of the maximum independent set in the input graph . The algorithm is a generalization of a known local search procedure. Note that it asymptotically matches the approximation factor of the approximation algorithm for free graphs of Halldórsson [20] by setting and . We note here that the following theorem was independently discovered by Bonnet, Thomassé, Tran, and Watrigant [3].
See 1.3
Proof.
The algorithm first computes a maximal independent set in the given graph , which can be done in linear time using a simple greedy approach. Since is maximal, every vertex in has at least one neighbor in . Now, we consider the vertices in that are neighbors to at most vertices of , and call this set . Let be a set of size , and let . If the graph induced by contains an independent set of size , then we can find it in time . Furthermore, is an independent set, since no vertex of is adjacent to any vertex of , and is larger by one than . Thus the algorithm replaces by in . The algorithm repeats this procedure until the largest independent set in each subgraph induced by a set (defined for the current ) is of size at most . At this point the algorithm outputs .
Let be the size of the output at the end of the algorithm. We claim that and this would prove the theorem, since then , which implies that is an approximation.
To show the claim, first note that the family is a partition into at most many sets. For each relevant , no subgraph induced by a set contains an independent set larger than , and thus if denotes the maximum independent set of , then . Thus,
Now consider the remaining set , and observe that every has at least neighbors in due to the definition of . For each with , we construct a set by fixing an arbitrary subset of size for every , and putting into if and only if . Observe that these sets form a partition of of size at most . We claim that each induces a subgraph of for which every independent set has size less than . Assume not, and let be an independent set in of size . But then induces a in , since every vertex of is adjacent to every vertex of . As this contradicts the fact that is free, we have , and consequently . Together with the above bound on the number of vertices of in we get
which concludes the proof. ∎
4 Polynomial time inapproximability in free graphs
In this section, we show polynomial time approximation lower bounds for Independent Set on free graphs.
See 1.4
For that, we reduce from the MCSI() problem, and leverage the lower bound by Laekhanukit [24, Theorem ]. Let us point out that the original statement of the lower bound by Laekhanukit [24] is in terms of the Label Cover problem, but, as we already mentioned, it is equivalent to MCSI.
Theorem 4.1 (Laekhanukit [24]).
Let be an instance of MCSI() where is a bipartite graph. Assuming ZPP NP, there exists a constant such that for any constant , there is no polynomial time algorithm that can distinguish between the two cases:

(YEScase) , and

(NOcase) .
We use a standard reduction from MCSI to Independent Set, which for instances of MCSI() of bounded degree gives the following lemma.
Lemma 1 ().
Let be an instance of MCSI(). Given , in polynomial time we can construct an instance of Independent Set such that

does not have as an induced subgraph for any ,

if then has an independent set of size at least , and

if then every independent set of has size at most .
Proof.
We first describe the construction of given , where we denote by the edge set between and for each edge . The graph has a vertex for each edge of , an edge between and if for some , and an edge between and if and and and do not share a vertex in for some three vertices of such that and . Note that the vertex set induces a clique in . This finishes the construction of . See Figure 1 for better understanding of the construction.
To see the first part of the lemma, for the sake of contradiction, let us suppose has a as an induced subgraph for . We know that for any the vertices in form a clique in , so the star can intersect with a fixed in at most two vertices of which one must be the center vertex of with degree . As has vertices, this means there are (at least) distinct vertex sets of that intersect the for some edges . Without loss of generality, let the center vertex of the come from . Note that the has an edge between a vertex from and a vertex from for each . Hence if , we have that either or for every by the construction of . This means that either or has at least neighbours in . That is, , which gives a contradiction to and .
Now, to see the second claim of the lemma, first we need to show that if , then has an independent set of size at least . To see that, let be a mapping that satisfies at least a fraction of the edges of . We claim that is an independent set of size at least in . Since satisfies at least fraction of edges, has size at least . So all we need to show is that is indeed an independent set. Suppose it was not the case, i.e., there exist that are adjacent in . By construction of there can be an edge between and only if and where possibly . Note that is a common endpoint of both and . If indeed , then is also a common endpoint of both and , so that , i.e., and are not distinct. Hence it must be that . But in this case, the construction of implies that and do not share a vertex, which contradicts the fact that they have as a common endpoint.
For the third part of the lemma, we prove the contrapositive: we claim that if has an independent set of size , then there exists an assignment satisfying at least edges in . To see that, first observe that the set can contain at most one vertex from as any two vertices in are adjacent. Let , for which we then have . We claim that all the edges in can be satisfied by an assignment defined as follows. For , let . Then we set and . We need to show that the function is welldefined. Suppose some vertex gets mapped to more than one vertex of by . This must mean that there exist two edges in that contain one endpoint in and are in . But this would mean that the two vertices in corresponding to these two edges in are adjacent due to the construction of . This is a contradiction to being an independent set. Also, for all , since for each we have , and we have set and . This concludes the proof. ∎
Now we are ready to prove Theorem 1.4.
Proof of Theorem 1.4.
Assume there was a polynomial time algorithm to approximate the Independent Set problem within a factor for some in free graphs, where , and is the constant given by Theorem 4.1. Given an instance of MCSI() and , we can reduce it to an instance of Independent Set in free graphs in polynomial time by using the reduction of Lemma 1. Now, setting and in the statement of Lemma 1, this gives that given an instance of MCSI() and , we can now use to differentiate between the YES and NOcases of Theorem 4.1 in polynomial time, which would mean that ZPP = NP. As , this implies Theorem 1.4. ∎
5 Parameterized approximation for fixed
In this section we prove Theorem 1.5. Let us define an auxiliary family of classes of graphs: for integers and , let denote the class of graphs that are free and free for any . Let be the class of trees with two vertices of degree at least 3 at distance at most . Let be the set of those , which are are also free. Actually, we will prove the following theorem, which implies Theorem 1.5.
Theorem 5.1 ().
Let be a constant. The following lower bounds hold for the Independent Set problem on graphs with vertices.

For any computable function , there is no time algorithm that determines if , unless the ETH fails.

There exists a constant , such that for any computable function , there is no time algorithm that can distinguish between the two cases: , or , unless the deterministic GapETH fails.

There exists a constant , such that for any computable function , there is no time algorithm that can distinguish between the two cases: , or , unless the randomized GapETH fails.
The proof of Theorem 5.1 consists of two steps: first we will prove it for graphs in , and then for graphs in . In both proofs we will reduce from the MCSI() problem. Let be an instance of MCSI(). For , by we denote the set of edges between and . Note that we may assume that has no isolated vertices, each is an independent set, and if and only if .
Lokshtanov et al. [26] gave the following hardness result (the first statement actually follows from Marx [29] and Marx, Pilipczuk [30]). We note that Lokshtanov et al. [26] conditioned their result on the Parameterized Inapproximability Hypothesis (PIH) and W[1] FPT. Here we use stronger assumptions, i.e., the deterministic and randomized GapETH, which are more standard in the area of parameterized approximation. The reduction in [26] yields the following theorem, when starting from [14, 15] and [28], respectively (see also [8, Corollary 7.9]).
Theorem 5.2 (Lokshtanov et al. [26]).
Consider an arbitrary instance of MCSI() with size .

Assuming the ETH, for any computable function , there is no time algorithm that solves .

Assuming the deterministic GapETH there exists a constant , such that for any computable function , there is no time algorithm that can distinguish between the two cases: (YEScase) , and (NOcase) .

Assuming the randomized GapETH there exists a constant , such that for any computable function , there is no time algorithm that can distinguish between the two cases: (YEScase) , and (NOcase) .
5.1 Hardness for free graphs
First, let us show Theorem 5.1 for , i.e., for free graphs for . Let be an instance of MCSI(3). We aim to build an instance of Independent Set, such that the graph .
For each , we introduce a clique of size , whose every vertex represents a different edge from . The cliques constructed at this step will be called primary cliques, note that their number is . Choosing a vertex from to an independent set of will correspond to mapping and to the appropriate endvertices of the edge from , corresponding to .
Now we need to ensure that the choices in primary cliques corresponding to edges of are consistent. Consider and suppose it has three neighbors (the cases if has fewer neighbors are dealt with analogously). We will connect the cliques using a gadget called a vertexcycle, whose construction we describe below. For each , we introduce copies of and denote them by , respectively. Let us call these copies secondary cliques. The vertices of secondary cliques represent the edges from analogously as the ones of . We call primary and secondary cliques as base cliques. We connect the base cliques corresponding to the vertex into vertexcycle . Imagine that secondary cliques, along with primary cliques , are arranged in a cyclelike fashion, as follows:
This cyclic ordering of cliques constitutes the vertexcycle, let us point out that we treat this cycle as a directed one. As we describe below we put some edges between two base cliques and only if they belong to some vertexcycle . See Figure 2 for an example of how we connect base cliques.
Now, we describe how we connect the consecutive cliques in . Recall that each vertex of each clique represents exactly one edge of , whose exactly one vertex, say , is in . We extend the notion of representing and say that represents , and denote it by .
Let us fix an arbitrary ordering on . Now, consider two consecutive cliques of the vertexcycle. Let be a vertex of the first clique and be a vertex from the second clique, and let and be the vertices of represented by and , respectively. The edge exists in if and only if . See Figure 3 how we connect two consecutive base cliques in a vertexcycle. This finishes the construction of .
We introduce a vertexcycle for every vertex of , note that each primary clique is in exactly two vertexcycles: and . The number of all base cliques is
This concludes the construction of . Since is partitioned into base cliques, is an upper bound on the size of any independent set in , and a solution of size contains exactly one vertex from each base clique.
We claim that the graph is in the class . Moreover, if , then the graph has an independent set of size and if the graph has an independent set of size at least for , then .
For two distinct base cliques , by we denote the set of edges with one endvertex in and another in . We say that are adjacent if .
Claim 5.1.
Let be two distinct base cliques in . Then the size of a maximum induced matching in the graph induced by is at most 1.
Proof.
If is empty, then the lemma holds trivially. Consider two disjoint edges and in , where and . We prove that there is an edge such that intersect both and .
By construction, and are consecutive cliques in a vertexcycle for some . Assume that is the successor of on this cycle. Recall that each represents some vertex . Since , we observe that and . Thus, either or , so one of the edges or exists in . ∎
Claim 5.2.
The graph is free.
Proof.
For contradiction, suppose that there exists an induced cycle in with consecutive vertices , where . Note that two consecutive vertices of might be in the same base clique, or two adjacent base cliques. Furthermore, no nonconsecutive vertices of may be in one base clique.
Note that each vertexcycle in has at least base cliques, so cannot intersect more than two base cliques. It cannot intersect one base clique, as , so suppose that intersects exactly two base cliques and . Observe that this means that and , while . However, by Claim 5.1, we observe that either and , or and , are adjacent in , so is not induced. ∎
Claim 5.3.
The graph is free.
Proof.
By contradiction suppose that the set induces a copy of in with being the central vertex. Let be the base clique containing . Since each of must be in a different base clique and is adjacent to at most four other base cliques, we conclude that one of ’s, say , belongs to . For , let be the base clique containing . Furthermore, note that must be a primary clique, since only those ones are adjacent to four base cliques. Therefore two of ’s, say and , must belong to the vertexcycle . Let precede , and succeed on this cycle. Consider the vertices and recall that since is adjacent to , we have . However, is nonadjacent to , which means that , which is a contradiction, since is transitive. ∎
Claim 5.4.
Let . Then, the graph is free.
Proof.
Suppose that contains as an induced subgraph. Let such that and . Note that primary cliques are at distance . Thus, and can not be both in primary cliques. Without loss of generality, let be in a secondary clique of a vertexcycle . There are only two base cliques and adjacent to the secondary clique . Let and be neighbors of in . Since and form an independent set in they have to be in different base cliques in . Thus, we can suppose and . However, by the same argument as in proof of Claim 5.3 these four vertices and can not exist. ∎
Claim 5.5.
If , then the graph has an independent set of size .
Proof.
Let be a solution of of value 1, i.e., for each holds that is an edge of . We will find an independent set in of size . For each we add to the set a vertex from the primary clique which represents the edge . Thus, we pick one vertex from each primary clique. Recall that each secondary clique is a copy of some primary clique . If we pick a vertex from then we add to also a copy of from . Thus, we add one vertex from each base clique to the set and therefore .
We claim that is independent. Suppose there exist such that . Let and for some base cliques and . First, suppose that and are copies of the same primary clique (or one of them is the primary clique itself and the second one is the copy)^{1}^{1}1The possibilities for are: or for .. Thus, the vertices and represent the same edge in and by construction, vertices in primary and secondary cliques representing the same edge in are not adjacent.
Therefore and (or vice versa) for some edges and in . Edges between and were added according to the ordering of vertices in . Note that the vertices and represent edges and . Thus, . Since and are adjacent in , it holds that
Comments
There are no comments yet.