Max Weight Independent Set in sparse graphs with no long claws111Parts of this work appeared in the proceedings of SODA 2021 [1] and STACS 2024 [3]
Abstract
For graphs and , we say that is -free if it does not contain as an induced subgraph. Already in the early 1980s Alekseev observed that the Max Weight Independent Set problem (MWIS) remains NP-hard in -free graphs, unless every component of is a path or a subdivided claw, i.e., a graph obtained from the three-leaf star by subdividing each edge some number of times (possibly zero). Since then determining the complexity of MWIS in these remaining cases is one of the most important problems in algorithmic graph theory.
In this paper we make an important step towards solving the problem by providing a polynomial-time algorithm for MWIS in graphs excluding a fixed graph forest of paths and subdivided claws as an induced subgraph, and a fixed biclique as a subgraph.
1 Introduction
A vertex-weighted graph is an undirected graph equipped with a weight function . For , we use as a shorthand for and for a subgraph of , is a shorthand for . By convention we use . Throughout the paper we assume that arithmetic operations on weights are performed in unit time.
For a graph , a set is independent or stable if there is no edge of with both endpoints in . By we denote the number of vertices in a largest independent set in . In the Max Independent Set (MIS) problem we are given an undirected graph , and ask for an independent set of size . In the Max Weight Independent Set (MWIS) problem we are given an undirected vertex-weighted graph , and ask for a maximum-weight independent set in . Note that MIS is a special case of MWIS where all weightes are equal. By we always denote the number of vertices of the instance graph.
MIS (and MWIS as its generalization) is a “canonical” hard problem: It was one of the first problems shown to be NP-hard [30], it is notoriously hard to approximate [29, 32], and it is W[1]-hard [20]. Many of these hardness results hold even if we restrict input instances to some natural graph classes [22, 9, 21]. Thus a very natural research direction is to consider restricted instances and try to capture the boundary between “easy” and “hard” cases.
State of the art.
The study of the complexity of MWIS in restricted graph classes is a central topic in algorithmic graph theory [2, 27, 53, 18, 5, 8]. Particular attention is given to classes that are hereditary, i.e., closed under vertex deletion. Among such classes a special role is played by ones defined by forbidding certain substructures. For graphs and , we say that is -free if it does not contain as an induced subgraph.
In what follows, for , by we denote the -vertex path. For integers , by we denote the graph obtained from the three-leaf star (i.e., the claw) by subdividing the three edges , , and times, respectively. Alternatively, we may think of as the graph obtained from paths , and by identifying one endvertex of each path. By we denote the graph with components, each of which is isomorphic to .
Let be the family of subcubic forests whose every component has at most one vertex of degree 3, i.e., whose every component is either a path or a subdivided claw.
The complexity study of MWIS in -free graphs dates back to the early 1980s and the work of Alekseev [4], who observed that for most graphs the problem remains NP-hard. Indeed, let us discuss the hard cases. First, MIS (and thus MWIS) is NP-hard in subcubic graphs [22], which are -free whenever has a vertex of degree at least 4. For the remaining cases we will use the so-called Poljak construction [50]: If is obtained from by subdividing one edge twice, then . Thus, if denotes the graph obtained from by subdividing each edge exactly times, then . Now observe that if has a cycle or two vertices of degree three in one component, then is -free. Consequently, for such graph , MIS is NP-hard in -free graphs. Let us point out that the above hardness reductions imply that MIS cannot even be solved in subexponential time unless the Exponential-Time Hypothesis (ETH) fails.
Summing up, the only graphs for which we may hope for a polynomial-time (or even subexponential-time) algorithm for MWIS in -free graphs are precisely the graphs in .
The complexity of MWIS in -free graphs when remains one of the most challenging and important problems in algorithmic graph theory. Despite significant attention received from the graph theory and the theoretical computer science communities, only partial results are known. Let us discuss them.
First, consider the case that for some . Since -free graphs, also known as cographs, have very rigid structure (in particular, they have clique-width at most 2), the polynomial-time algorithm for this class of graphs is rather simple [19]. However, already for -free graphs the situation is much more complicated. The existence of a polynomial-time algorithm for -free graphs was a long-standing open problem that was finally resolved in the affirmative in 2014 by Lokshtanov, Vatshelle, and Villanger [35] using the framework of potential maximal cliques. Later, using the same approach but with significantly more technical effort, Grzesik, Klimošová, Pilipczuk, and Pilipczuk [26] obtained a polynomial-time algorithm for -free graphs. The case of -free graphs remains open.
However, some interesting algorithmic results can be obtained if we relax our notion of an efficient algorithm. First, it was shown by Bacsó et al. [7] that for every fixed , MWIS can be solved in subexponential time for -free graphs (by we always denote the number of vertices of the instance graph). Another subexponential-time algorithm, with worse running time, was obtained independently by Brause [13]. While these results do not rule out the possibility that the problem is NP-hard, let us recall that, assuming the ETH, subexponential algorithms for MWIS in -free graphs cannot exist if . Later, Chudnovsky et al. [16, 15] showed that for every fixed , the problem admits a QPTAS in -free graphs. Finally, a very recent breakthrough result by Gartland and Lokshtanov [23] shows that for every fixed , the problem can be solved in quasipolynomial time ; see also a slightly simpler algorithm by Pilipczuk, Pilipczuk, and Rzążewski [49] with running time . Note that this means that if for some ,MWIS is NP-hard for -free graphs, then all problems in NP can be solved in quasipolynomial time. While this does yet not imply that P = NP, it still seems rather unlikely according to our current understanding of complexity theory.
Now let us turn to the case when is a subdivided claw. The simplest subdivided claw is the claw . Claw-free graphs appear to be closely related to line graphs [17] and thus a polynomial-time algorithm for MWIS in claw-free graphs can be obtained by a modification of the well-known augmenting path approach for finding a maximum-weight matching [51, 43] (i.e., a maximum-weight independent set in a line graph). Let us highlight the close relation of claw-free graphs and line graphs, as it will play an important role in our paper. The next smallest subdivided claw is the fork, i.e., . A polynomial-time algorithm for MIS in fork-free graphs was obtained by Alekseev [6]. Later it was extended to the MWIS problem by Lozin and Milanič [36]. For disconnected , it is known that MWIS is polynomial-time-solvable in -free graphs, for every fixed [10]. The existence of polynomial-time algorithms in the next simplest (connected) cases, i.e., and , is wide open.
Again, some interesting results can be obtained if we look beyond polynomial-time algorithms. Chudnovsky et al. [16, 15] proved that for every subdivided claw , the MWIS problem in -free graphs admits a QPTAS and a subexponential-time algorithm working in time . We point out that the arguments used for the case when is a subdivided claw are significantly more complicated and technically involved than their counterparts for -free graphs. These results were then simplified and improved by Majewski et al. [42]: They obtained another (faster) QPTAS and a subexponential-time algorithm with running time . Finally, very recently, Gartland et al. [24] announced a quasipolynomial-time algorithm for MWIS in -free graphs for every .
More tractability results can be obtained if we put some additional restrictions on the instance graph. In particular, there is a long line of research concerning graphs excluding a fixed (but still small) path or a subdivided claw and, simultaneously, some other small graphs, see e.g. [34, 11, 25, 39, 41, 45, 37, 38, 28, 44, 46, 47, 48, 12]. A slightly different direction was considered by Lozin, Milanič, and Purcell [37], who proved that for every fixed ,MWIS is polynomial-time solvable in subcubic -free graphs. Later, Lozin, Monnot, and Ries [38] showed a polynomial time algorithm for subcubic -free graphs. Finally, Harutyunya et al. [28] generalized both these results by providing a polynomial-time algorithm for subcubic -free graphs, for any fixed .
We remark that the case when is a subdivided claw (or, more precisely, is in and contains at least one component which is not a path) is the only case where the restriction to bounded degree graphs leads to an interesting problem. Indeed, the already mentioned hardness reduction of Alekseev [4] shows that if , then MIS is NP-hard even in subcubic -free graphs. On the other hand, if is a forest of paths, then -free graphs of bounded degree are of constant size and thus of little interest.
In this work, we continue and significantly extend the study of the complexity of MWIS in -free graphs with additional restrictions, where .
Our results.
As a warm-up, we present a polynomial-time algorithm for -free graphs of bounded degree, where .
Theorem 1.
There exists an algorithm that, given a vertex-weighted graph on vertices with maximum degree and integers in time either finds an induced or the maximum possible weight of an independent set in .
Note that by picking appropriate and , theorem 1 yields a polynomial-time algorithm for MWIS for bounded-degree graphs excluding a fixed graph from as an induced subgraph.
Then we proceed to the main result of the paper: we show that MWIS remains polynomial-time solvable in -free graphs, even if instead of bounding the maximum degree, we forbid a fixed biclique as a subgraph.
Theorem 2.
For every fixed integers , and there exists a polynomial-time algorithm that, given a vertex-weighted graph that does not contain as an induced subgraph nor as a subgraph, returns the maximum possible weight of an independent set in .
Let us remark that by the celebrated Kövári-Sós-Turán theorem [33], classes that exclude as a subgraph capture all hereditary classes of sparse graphs, where by “sparse” we mean that the graph has a subquadratic number of edges. Furthermore, by a simple Ramsey argument, for every positive integer there exists an integer such that if is -free and -free then does not contain as a subgraph. Hence, equivalently, theorem 2 yields a polynomial-time algorithm for MWIS for graphs that are simultaneously -free (for some ), -free, and -free.
Our techniques.
As in the previous works [16, 24], the crucial tool in handling -free graphs is an extended strip decomposition. Its technical definition can be found in preliminaries; for now, it suffices to say that it is a wide generalization of the preimage graph of a line graph (recall that line graphs are -free) that allows for recursion for the MWIS problem. An extended strip decomposition of a graph identifies some induced subgraphs of as particles and, knowing the maximum possible weight of an independent set in each particle, one can compute in polynomial time the maximum possible weight of an independent set in . (We remark that this computation involves advanced combinatorial techniques as it relies on a reduction to the maximum weight matching problem in an auxiliary graph.) In other words, finding an extended strip decomposition with small particles compared to is equally good for the MWIS problem as splitting the graph into small connected components.
The starting point is the following theorem of [42].
Theorem 3 ([42, Corollary 12] in a semi-weighted setting).
There exists an algorithm that, given an -vertex graph with a set and integers , in polynomial time outputs either:
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an induced copy of in , or
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a set of size at most and a rigid extended strip decomposition of with every particle containing at most vertices of .
(A rigid extended strip decomposition is an extended strip decomposition that does not have some unnecessary empty sets. By we denote the set consisting of and all vertices with a neighbor in .) Let us remark that the result stated in [42, Theorem 2] is for unweighted graphs (i.e., using the notation from theorem 3), but the statement of theorem 3 can be easily derived from the proof, see also [24].
Consider the setting of theorem 1, i.e., the graph has maximum degree . Apply theorem 3 to with . If we get the first outcome, i.e., an induced in , we return it and terminate. So assume that we get the second outcome, i.e., the set . Note that as , we have . It is now tempting to exhaustively branch on (i.e., guess the intersection of the sought independent set with ) and recurse on the particles of the extended strip decomposition of . However, implementing this strategy directly gives quasipolynomial (in ) running time bound of , as the branching step yields up to subcases and the depth of the recursion is .
Our main new idea now is to perform this branching lazily, by considering a more general border version of the problem, where the input graph is additionally equipped with a set of terminals and we ask for a maximum weight of an independent set for every possible behavior on the terminals.
A similar application of a border version of the problem to postpone branching in recursion appeared for example in the technique of recursive understanding [31, 14].
Let us return to our setting, where we have a set of size and an extended strip decomposition of with particles of size at most half of the size of . We would like to remove from the graph, indicate as terminals and solve Border MWIS in using the extended strip decomposition for recursion. Note that, thanks to the bounded degree assumption, the size of is bounded by .
This approach almost works: the only problem is that, as the recursion progresses, the set of terminals accummulates and its size can grow beyond the initial bound. Luckily, this can be remedied in a standard way: we alternate recursive steps where theorem 3 is invoked with with steps where theorem 3 is invoked with . In this manner, we can maintain a bound of on the number of terminals in every recursive call. Note that this bound also guarantees that the size of the domain of the requested function is of size , which is within the promised time bound.
Let us now move to the more general setting of theorem 2. Here, the starting points are the recent results of Weißauer [52] and Lozin and Razgon [40] that show that in the -free case, excluding a biclique as a subgraph is not that much different than bounding the maximum degree.
A -block in a graph is a set of vertices, no two of which can be separated by deleting fewer than vertices. The following result was shown by Weißauer (we refer to preliminaries for standard definitions of tree decompositions and torsos).
Theorem 4 (Weißauer [52]).
Let be a graph and be an integer. If has no -block, then admits a tree decomposition with adhesion less than , in which every torso has at most vertices of degree larger than .
Even though the statement of the result in [52] is just existential, the proof actually yields a polynomial-time algorithm to compute such a tree decomposition.
It turns out that -free graphs with no large bicliques have no large blocks.
Lemma 5.
For any , and there exists such that the following holds. Every -free graph with no subgraph isomorphic to has no -block.
Let us remark that Lozin and Razgon [40] showed lemma 5 for -free graphs. However, an extension of their argument applies to -free graphs; we include it in Appendix A.
Corollary 6.
For any , and there exists such that the following holds. Given a -free graph with no subgraph isomorphic to , in polynomial time one can compute a tree decomposition of with adhesion less than , in which every torso has at most vertices of degree larger than .
To prove theorem 2 using corollary 6 we need to carefully combine explicit branching on the (bounded number of) vertices of large degree in a single bag with — as in the bounded degree case — applying theorem 3 to the remainder of the graph and indicating as the terminal set of the border problem passed to the recursive calls. Finally, we combine these steps with the information passed over adhesions in the tree decomposition.
2 Preliminaries
Our algorithms take a vertex-weighted graph as an input. In the recursion, we will be working on various induced subgraphs of with vertex weight inherited from . Somewhat abusing notation, we will keep for the weight function in any induced subgraph of .
Tree decompositions.
Let be a graph. A tree decomposition of is a pair where is a tree and is a function satisfying the following: (i) for every there exists with , and (ii) for every the set induces a connected nonempty subtree of . For every and , the set is the bag at node and the set is the adhesion at edge . The critical property of a tree decomposition is that if and and are two connected components of that contain and , respectively, then separates from in .
The torso of a bag in a tree decomposition is a graph with and if or there exists a neighbor with . That is, the torso of is created from by turning the adhesion into a clique for every neighbor of in .
Extended strip decompositions.
We follow the notation of [42, 24]. For a graph , by we denote the set of triangles in . An extended strip decomposition of a graph is a pair that consists of:
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a simple graph ,
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a vertex set for every ,
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an edge set for every , and its subsets ,
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a triangle set for every ,
which satisfy the following properties:
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The family is a partition of .
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For every and every distinct , the set is complete to .
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Every is contained in one of the sets for , or is as follows:
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for some and , or
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for some , or
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and for some .
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An extended strip decomposition is rigid if for every , the sets , , and are nonempty, and for every isolated , the set is nonempty. Note that if is a rigid extended strip decomposition of , then is bounded by .
For an extended strip decomposition of a graph , we identify five types of particles.
vertex particle: | |||
edge interior particle: | |||
half-edge particle: | |||
full edge particle: | |||
triangle particle: |
As announced in the introduction, to solve MWIS in it suffices to know the solution to MWIS in particles. The proof of the following lemma follows closely the lines of proofs of analogous statement of [16] and is included for completeness in Appendix B.
Lemma 7.
Given a Border MWIS instance , an extended strip decomposition of , and a solution to the Border MWIS instance for every particle of , one can in time times a polynomial in compute the solution to the input .
We need the following simple observations.
Lemma 8.
Let be a -free graph and let be a rigid extended strip decomposition of . Then the maximum degree of is at most .
Proof..
Let . Observe that the sets are nonempty and complete to each other in . Hence, contains a clique of size equal to the degree of in .
Lemma 9.
Let be a graph and let be an extended strip decomposition of such that the maximum degree of is at most . Then, every vertex of is in at most particles.
Proof..
Pick and observe that:
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If for some , then is in the vertex particle of and in one half-edge and one full-edge particle for every edge of incident with . Since there are at most such edges, is in at most particles.
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If for some , then is in at most four particles for the edge .
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If for some , then is in the triangle particle for and in three full edge particles, for the three sides of the triangle .
3 Bounded-degree graphs: Proof of Theorem 1
This section is devoted to the proof of theorem 1.
Let be positive integers and let be the input vertex-weighted graph. We denote and to be the maximum degree of . Let
be an upper bound on the size of for any application of theorem 3 for any induced subgraph of .
We describe a recursive algorithm that takes as input an induced subgraph of with weights and a set of terminals of size at most and solves Border MWIS on . The root call is for and ; indeed, note that is the maximum possible weight of an independent set in .
Let be an input to a recursive call. First, the algorithm initializes for every .
If , the algorithm proceeds by brute-force: it enumerates independent sets and updates with whenever the previous value of that cell was smaller. As , this step takes time. This completes the description of the leaf step of the recursion.
If , the algorithm proceeds as follows. If , let , and otherwise, let . The algorithm invokes theorem 3 on and . If an induced is returned, then it can be returned by the main algorithm as it is in particular an induced subgraph of . Hence, we can assume that we obtain a set of size at most and an extended strip decomposition of whose every particle contains at most vertices of .
Observe that as and the maximum degree of is , we have . Let . Note that we have and . For every particle of , invoke a recursive call on , obtaining (or an induced , which can be directly returned). Use lemma 7 to obtain a solution to Border MWIS instance .
Finally, iterate over every (note that ) and, if is independent, update the cell with the value , if this value is larger than the previous value of this cell. This completes the description of the algorithm.
The correctness of the algorithm is immediate thanks to lemma 7 and the fact that is adjacent in only to which is a subset of .
For the complexity analysis, consider a recursive call to for a particle . If , then . Otherwise, and . As , we have . Hence, in the recursive call the invariant of at most terminals is maintained and, moreover:
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if , then and ;
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otherwise, and , hence the recursive call will fall under the first bullet.
We infer that the depth of the recursion is at most .
At every non-leaf recursive call, we spend time on invoking the algorithm from theorem 3, time to compute using lemma 7, and time for the final iteration over all subsets . Hence, the time spent at every recursive call is bounded by .
At every non-leaf recursive call, we make subcalls to for every particle of . Lemmas 8 and 9 ensure that the sum of over all particles is bounded by . Hence, the total size of all graphs in the -th level of the recursion is bounded by . Since the depth of the recursion is bounded by , the total size of all graphs in the recursion tree is bounded by . Since this also bounds the size of the recursion tree, we infer that the whole algorithm runs in time .
This completes the proof of theorem 1.
4 Graphs with no large bicliques: Proof of Theorem 2
This section is devoted to the proof of theorem 2.
Let be a positive integer and let be the constant depending on from corollary 6. Again, let be the input vertex-weighted graph, let , and let
be an upper bound on the size of for any application of theorem 3 for any induced subgraph of 222As the dependence of on is superpolynomial, for the sake of simplicity, we do not try to optimize the dependence of the complexity bound on ..
The general framework and the leaves of the recursion are almost exactly the same as in the previous section, but with different thresholds. That is, we describe a recursive algorithm that takes as input an induced subgraph of with weights and a set of terminals of size at most and solves Border MWIS on . The root call is for and and the algorithm returns as the final answer.
Let be an input to a recursive call. The algorithm initiates first for every .
If , the algorithm proceeds by brute-force: it enumerates independent sets and updates with whenever the previous value of that cell was smaller. As and is a constant depending on , and , this step takes polynomial time. This completes the description of the leaf step of the recursion.
Otherwise, if , we invoke corollary 6 on , obtaining a tree decomposition of . If , let , and otherwise, let .
For every , proceed as follows. For , let be the connected component of that contains and let . Clearly, separates from . Orient the edge towards with larger , breaking ties arbitrarily.
There exists of outdegree . Then, for every connected component of we have . Fix one such node and let and let be the set of connected components of . Let be a supergraph of obtained from by turning, for every , the neighborhood into a clique. Note that is a subgraph of the torso of . Hence, by the properties promised by corollary 6, for every we have (as this set is contained in a single adhesion of an edge incident with in ) and contains at most vertices of degree larger than . Let be the set of vertices of of degree larger than .
We perform exhaustive branching on . That is, we iterate over all independent sets and denote , , . For one , we proceed as follows.
We invoke theorem 3 to with set , obtaining a set of size at most and an extended strip decomposition of whose every particle has at most vertices of . Note that is an induced subgraph of , which is an induced subgraph of , so there is no induced in .
A component is dirty if and clean otherwise. Let
The following bounds will be important for further steps.
(1) |
To see (1) observe that a vertex has at most neighbors in (as every vertex of has degree at most in ) while every vertex has at most neighbors in , as every component of has at most neighbors in . This proves (1).
(2) |
To see (2), consider a dirty component . Observe that either contains a vertex of or contains a vertex of . There are at most dirty components of the first type, contributing in total at most vertices to . For the dirty components of the second type, although there can be many of them, we observe that if , then . Hence, for every dirty component of the second type, it holds that . Since the degree of each vertex of is at most , by (1) we have
The bound (2) follows.
A component is touched if it is dirty or contains a vertex of . Let
Using similar arguments as before, we can prove
(3) |
Indeed, if is touched, then contains a vertex (if is dirty, is contained in ), and then is contained in . Also, for we have . Hence, . Since the maximum degree of a vertex of is , this proves (3).
For every touched , denote and . Recurse on , obtaining .
Let
Note that, by the definition of dirty and touched, is an induced subgraph of . Hence, can be restricted to a (not necessarily rigid) extended strip decomposition of .
Let . For every particle of , recurse on , obtaining . Then, use these values with lemma 7 to solve a Border MWIS instance , obtaining .
Iterate over every independent set . Observe that admits an independent set with and weight:
Update the cell with this value if it is larger than the previous value of this cell. This finishes the description of the algorithm.
For correctness, it suffices to note that for every touched component , the whole is in the terminal set for the recursive call and the whole is in and thus in the terminal set for the Border MWIS instance .
For the sake of analysis, consider a recursive call on for a touched component . If and , then and . Otherwise, if and , then . Thus, the recursive call on will fall under the first case of at most terminals.
Analogously, consider a recursive call on for a particle of . If and , then due to (3). Furthermore, . Otherwise, if and , then again due to (3). Thus, the recursive call on will fall under the first case of at most terminals.
Finally, note that a recursive call without nonterminal vertices (i.e., with ) is a leaf call.
We infer that all recursive calls satisfy the invariant of at most terminals and the depth of the recursion tree is bounded by (as every second level the number of nonterminal vertices halves).
At each recursive call, we iterate over at most subsets . lemma 8 ensures that the maximum degree of is at most , while lemma 9 ensures that every vertex of is used in at most particles of . In a subcall for a touched component , vertices of are not used in any other call for the current choice of , while all vertices of are terminals. Consequently, every nonterminal vertex of is passed as a nonterminal vertex to a recursive subcall at most number of times (and a terminal is always passed to a subcall as a terminal). Furthermore, a recursive call without nonterminal vertices is a leaf call. As the depth of the recursion is , we infer that, summing over all recursive calls in the entire algorithm, the number of nonterminal vertices is bounded by and the total size of the recursion tree is .
At each recursive call, we iterate over all subsets and then we invoke theorem 3 and iterate over all independent sets in . Thanks to the invariant and bounds (2), and (3), this set is of size . Hence, every recursive call runs in time , where is a constant depending on and . As is a constant depending on , the final running time bound is polynomial.
This completes the proof of theorem 2.
5 Conclusion
While it is generally believed that MWIS is polynomial-time-solvable in -free (and even -free) graphs (with no further assumptions), such a result seems currently out of reach. Thus it is interesting to investigate how further can we relax the assumptions on instances, as we did when going from theorem 1 to theorem 2. In particular, we used the assumption of -freeness twice: once in lemma 5 and then to argue that (the pattern of an extended strip decomposition we obtain) is of bounded degree. On the other hand, the assumption of -freeness was used just once: in lemma 5. Thus it seems natural to try to prove the following conjecture.
Conjecture 10.
For every integers there exists a polynomial-time algorithm that, given an -free and -free vertex-weighted graph computes the maximum possible weight of an independent set in .
Acknowledgements
We acknowledge the welcoming and productive atmosphere at Dagstuhl Seminar 22481 “Vertex Partitioning in Graphs: From Structure to Algorithms,” where some part of the work leading to the results in this paper was done.
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Appendix A Appendix: Proof of Lemma 5
For positive integers , the Ramsey number of and , denoted by , is the smallest integer such that every graph on vertices contains either an independent set of size or a clique of size . It is well-known that .
For a graph , a subdivision of is any graph obtained from by subdividing each edge arbitrarly many times (possibly 0). By a -subdivision (resp., -subdivision) we mean a subdivision where each edge was subdivided exactly (resp., at most ) times. A proper subdivision is one where each edge was subdivided at least once. By we denote the -subdivision of the -leaf star. In particular, is exactly .
We will need the following two technical results shown by Lozin and Razgon [40].
Lemma 11 ([40, Lemma 2]).
There is a function with the following property. If a graph contains a collection of pairwise disjoint subsets of vertices, each of size at most and with at least one edge between any two of them, then contains as a subgraph.
Lemma 12 ([40, Theorem 3]).
There is a function with the following property. Every graph containing a -subdivision of as a subgraph contains either as a subgraph or a proper -subdivision of as an induced subgraph.
The following lemma is the main technical ingredient of the proof. It can be seen as a strengthening of Claim 2 in [40].
Lemma 13.
For all positive integers there exists such that the following holds. If contains a -block, then it contains:
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1.
a -subdivision of as a subgraph or
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2.
as a subgraph, or
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3.
as an induced subgraph.
Proof..
Let be the functions given by Lemma 11. Define the following constants
Suppose that has a -block but no subgraph isomorphic to nor induced subgraph isomorphic to . We aim to show that has a -subdivision of as a subgraph.
Consider a pair of distinct vertices from . Since and cannot be separated by deleting fewer than vertices, by Menger’s theorem there is a set of internally pairwise disjoint --paths in . Without loss of generality we can assume that each such path is induced.
Claim 1.
There is a subset of size with the following property. For each pair of distinct vertices there is a set of internally pairwise disjoint induced --paths in that do not contain any vertices from .
Proof of Claim..
Let be any set of vertices from . Consider any two distinct vertices . As the interiors of paths in are pairwise disjoint, we observe that at most paths might contain vertices from . Thus contains at least paths that do not intersect .
Fix a pair of distinct vertices from . A path in is long if it has at least internal vertices. The pair is distant if at least paths in are long.
Let be a minimum-size subset of that intersects all distant pairs. (It might be useful to think of as the minimum vertex cover in a graph with vertex set where the edges correspond to distant pairs.) We consider two cases, depending whether is “large” or “small”.
Case 1 (large ): .
In this case there is a set consisting of at least pairwise disjoint distant pairs (i.e., a matching in the mentioned graph). Let be such a set of size exactly .
Claim 2.
For each , there is a set contained in the union of paths in that induces in .
Proof of Claim..
Fix . For a long path , let be the set of first vertices of , starting from the side of , but excluding itself. So induces a -vertex path, and induces a -vertex path in .
Fix any set of long paths from , and define . Consider an auxiliary graph with vertex set , in which two elements are adjacent if and only if there is an edge between one set and the other (we emphasize here that the sets in are pairwise disjoint).
If contains a clique of size , then, by lemma 11, we obtain a subgraph isomorphic to in , a contradiction. Thus, since we obtain an independent set of size in . The paths forming this independent set, together with , induce the desired copy of .
Observe that even though the pairs in are pairwise disjoint, the sets might still intersect. In the next claim we extract from them induced copies of that are pairwise disjoint.
Claim 3.
For each there is an induced copy of contained in , such that for any distinct the corresponding copies of are disjoint.
Proof of Claim..
Fix an arbitrary order on pairs in . The proof is by induction.
The copy of for can be selected by picking any three out of paths in the copy of given by 2. So let and suppose that for each we have selected an induced copy of contained in . Note that in total we have selected vertices. Consider the pair and let be the -vertex paths obtained from by deleting . Note that at most of these paths might intersect . So, as , there is always a choice of three paths that are disjoint with . Recall that by 1 the vertex is not contained . Thus the vertices of the three selected paths, together with , form the set .
Now we proceed similarly as in the proof of 2. Let be the family consisting of induced copies of for all ; they are given by 3. Let be the graph with vertex set where two vertices are adjacent if and only if there is an edge from one set to another. Note that . An independent set in of size corresponds to an induced in . On the other hand, if has a clique of size at least , then by lemma 11 we obtain as a subgraph of . By the choice of one of these cases must happen, and both yield a contradiction. Thus we conclude that Case 1 cannot occur.
Case 2 (small ): .
Define ; note that . For any distinct , let be obtained from by removing all long paths. By the definition of we observe that for all we have .
Let be any -element subset of .
Claim 4.
For each pair of distinct vertices of there is a path such the paths selected for distinct pairs are internally disjoint.
Proof of Claim..
The proof is similar to the proof of 3. We use induction. Enumerate pairs of distinct vertices from as .
The path can be arbitrarily chosen from . Suppose that we have selected paths for some . Since each path is short, the selected paths have in total at most internal vertices.
Now consider the set . Recall that the paths in this set are pairwise disjoint. Since , we observe that there is a path in which is internally disjoint with all previously selected paths. We pick this path as .
Recall that for each distinct , the path does not contain vertices from . Thus the set with the paths given by 4 forms a -subdivision of which is a subgraph of . This concludes the claim.
Now we can proceed to the proof of lemma 5.
Proof. (of lemma 5.).
Define (i.e., the number of vertices of ). Define , where is as in lemma 12, and let be as in lemma 13. For contradiction, suppose that has a -block. By lemma 13 we conclude that contains a -subdivision of as a subgraph. By lemma 12 we observe that contains a proper -subdivision of as an induced subgraph.
It is straightforward to verify that this induced subgraph contains a subdivision of (in fact, of any graph with at most vertices and at most edges) as an induced subgraph. Therefore contains an induced subgraph isomorphic to , a contradiction.
Appendix B Appendix: Proof of Lemma 7
Iterate over every . For fixed , we aim at computing . If is not independent, we set . In the remainder of the proof, we show how to compute in polynomial time the value for fixed independent .
For a particle of , let and let be an independent set witnessing this value, that is, an independent set in of weight with . Note that as is independent, the value is not equal to and such an independent set exists.
We say that is forced if . Note that since is independent, if is forced, then for exactly one edge incident with . We call such an edge the enforcer of . Note that an edge may be the enforcer of both and .
The arguments now follow very closely the outline of Section 3.3 of [15].
We construct a set of particles and an edge-weighted graph as follows. We start with , , and .
For every that is not forced, add to . For every such that neither of the edges , , or is the enforcer of both its endpoints, add to . For every , proceed as follows.
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1.
If neither nor is forced, we add to a new vertex and edges , , , and set the edge weights as follows:
Furthermore, add to .
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2.
If exactly one of and is forced, say w.l.o.g. is forced and is not forced, proceed as follows.
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(a)
If is the enforcer of , then add to an edge with weight
Furthermore, add to .
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(b)
If is not the enforcer of , then add to a new vertex and an edge with weight
Furthermore, add to .
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(a)
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3.
If both and are forced, proceed as follows.
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(a)
If is neither the enforcer of nor of , add to .
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(b)
If is the enforcer of but not of add to .
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(c)
If is the enforcer of but not of add to .
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(d)
If is the enforcer of both and , add to .
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(a)
This finishes the description of the construction of and . In the next two paragraphs we make two observations that follow by a direct check from the definitions.
Observe that is independent in and has weight . Furthermore, for every , we have and . We think of as the “base” solution for .
Observe also that all weights of are nonnegative, as contains both and while contains , , , as well as all for all triangles containing the edge .
We will be asking for a maximum weight matching in . Intuitively, taking an edge to such a matching corresponds to replacing in the parts and with the part while taking an edge to such a matching corresponds to replacing in the parts , , and all parts for triangles containing the edge with part .
From another perspective, fix and recall that the sets for are complete to each other. Hence, any independent set in can contain vertices in at most one of such sets. For an edge , taking an edge or in a matching in corresponds to choosing that, among all neighbors of in , the neighbor is such that the set is allowed to contain vertices of the sought independent set. (Choosing to the matching corresponds to allowing both and to contain vertices of the sought independent set.)
However, there is a delicacy if contains a vertex of some interface . Then, in some sense already forces some choices in the corresponding matching in . This is modeled above by having alternate construction for vertices that are forced.
The following two claims prove that equals plus the maximum possible weight of a matching in and thus complete the proof of lemma 7. Their proofs follow exactly the lines of the proofs of Claims 3.7 and 3.8 of Section 3.3 of [15] and are thus omitted.
Claim 5.
Let be an independent set in with . Let be the set of edges of defined as follows: for every , if and , then , if and , then , and if and , then . Then, all the above edges indeed exist in and is a matching. Furthermore, the weight of is at most .
Claim 6.
Let be a matching in . Let be the set of particles of defined as follows. Start with and then for every edge ,
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•
if , insert into and remove from the following particles if present: , , , , , for any such that .
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•
if (resp. ), insert (resp. ) into , and remove from the following particles if present: and (resp. ).
Then is an independent set in with and of weight at least .