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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]

Tara Abrishami Department of Mathematics, University of Hamburg, Germany, [email protected]. Supported by the National Science Foundation Award Number DMS-2303251 and the Alexander von Humboldt Foundation. This work was performed in part while the author was at Princeton University and supported by NSF Grant DMS-1763817 and NSF-EPSRC Grant DMS-2120644.    Maria Chudnovsky Department of Mathematics, Princeton University, USA, [email protected]. Supported by NSF-EPSRC Grant DMS-2120644 and by AFOSR grant FA9550-22-1-008.    Cemil Dibek Department of Operations Research and Financial Engineering, Princeton University, USA, [email protected]. Supported by the National Science Foundation Award Number DMS-1763817.    Marcin Pilipczuk Institute of Informatics, University of Warsaw, Poland, [email protected]. Supported by Polish National Science Centre SONATA BIS-12 grant number 2022/46/E/ST6/00143.    Paweł Rzążewski Faculty of Mathematics and Information Science, Warsaw University of Technology and Institute of Informatics, University of Warsaw, Poland, [email protected]. This work is a part of the project BOBR that has received funding from the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme (grant agreement No. 948057)
Abstract

For graphs GG and HH, we say that GG is HH-free if it does not contain HH as an induced subgraph. Already in the early 1980s Alekseev observed that the Max Weight Independent Set problem (MWIS) remains NP-hard in HH-free graphs, unless every component of HH 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 GG equipped with a weight function 𝐰:V(G)\mathbf{w}:V(G)\to\mathbb{N}. For XV(G)X\subseteq V(G), we use 𝐰(X)\mathbf{w}(X) as a shorthand for xX𝐰(x)\sum_{x\in X}\mathbf{w}(x) and for a subgraph HH of GG, 𝐰(H)\mathbf{w}(H) is a shorthand for 𝐰(V(H))\mathbf{w}(V(H)). By convention we use 𝐰()=0\mathbf{w}(\emptyset)=0. Throughout the paper we assume that arithmetic operations on weights are performed in unit time.

For a graph GG, a set IV(G)I\subseteq V(G) is independent or stable if there is no edge of GG with both endpoints in II. By α(G)\alpha(G) we denote the number of vertices in a largest independent set in GG. In the Max Independent Set (MIS) problem we are given an undirected graph GG, and ask for an independent set of size α(G)\alpha(G). In the Max Weight Independent Set (MWIS) problem we are given an undirected vertex-weighted graph (G,𝐰)(G,\mathbf{w}), and ask for a maximum-weight independent set in (G,𝐰)(G,\mathbf{w}). Note that MIS is a special case of MWIS where all weightes are equal. By nn 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 GG and HH, we say that GG is HH-free if it does not contain HH as an induced subgraph.

In what follows, for t1t\geqslant 1, by PtP_{t} we denote the tt-vertex path. For integers a,b,c1a,b,c\geqslant 1, by Sa,b,cS_{a,b,c} we denote the graph obtained from the three-leaf star (i.e., the claw) by subdividing the three edges a1a-1, b1b-1, and c1c-1 times, respectively. Alternatively, we may think of Sa,b,cS_{a,b,c} as the graph obtained from paths Pa+1,Pb+1P_{a+1},P_{b+1}, and Pc+1P_{c+1} by identifying one endvertex of each path. By dSa,b,cdS_{a,b,c} we denote the graph with dd components, each of which is isomorphic to Sa,b,cS_{a,b,c}.

Let 𝒮\mathcal{S} be the family of subcubic forests HH 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 HH-free graphs dates back to the early 1980s and the work of Alekseev [4], who observed that for most graphs HH 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 HH-free whenever HH has a vertex of degree at least 4. For the remaining cases we will use the so-called Poljak construction [50]: If GG^{\prime} is obtained from GG by subdividing one edge twice, then α(G)=α(G)+1\alpha(G^{\prime})=\alpha(G)+1. Thus, if GpG^{p} denotes the graph obtained from GG by subdividing each edge exactly 2p2p times, then α(Gp)=α(G)+p|E(G)|\alpha(G^{p})=\alpha(G)+p\cdot|E(G)|. Now observe that if HH has a cycle or two vertices of degree three in one component, then G|V(H)|G^{|V(H)|} is HH-free. Consequently, for such graph HH, MIS is NP-hard in HH-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 HH for which we may hope for a polynomial-time (or even subexponential-time) algorithm for MWIS in HH-free graphs are precisely the graphs in 𝒮\mathcal{S}.

The complexity of MWIS in HH-free graphs when H𝒮H\in\mathcal{S} 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 H=PtH=P_{t} for some tt. Since P4P_{4}-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 P5P_{5}-free graphs the situation is much more complicated. The existence of a polynomial-time algorithm for P5P_{5}-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 P6P_{6}-free graphs. The case of P7P_{7}-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 tt, MWIS can be solved in subexponential time 2𝒪(nlogn)2^{\mathcal{O}(\sqrt{n\log n})} for PtP_{t}-free graphs (by nn 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 HH-free graphs cannot exist if H𝒮H\notin\mathcal{S}. Later, Chudnovsky et al. [16, 15] showed that for every fixed tt, the problem admits a QPTAS in PtP_{t}-free graphs. Finally, a very recent breakthrough result by Gartland and Lokshtanov [23] shows that for every fixed tt, the problem can be solved in quasipolynomial time n𝒪(log3n)n^{\mathcal{O}(\log^{3}n)}; see also a slightly simpler algorithm by Pilipczuk, Pilipczuk, and Rzążewski [49] with running time n𝒪(log2n)n^{\mathcal{O}(\log^{2}n)}. Note that this means that if for some tt,MWIS is NP-hard for PtP_{t}-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 HH is a subdivided claw. The simplest subdivided claw is the claw S1,1,1=K1,3S_{1,1,1}=K_{1,3}. 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., S2,1,1S_{2,1,1}. 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 HH, it is known that MWIS is polynomial-time-solvable in dS1,1,1dS_{1,1,1}-free graphs, for every fixed dd [10]. The existence of polynomial-time algorithms in the next simplest (connected) cases, i.e., H=S3,1,1H=S_{3,1,1} and H=S2,2,1H=S_{2,2,1}, 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 HH, the MWIS problem in HH-free graphs admits a QPTAS and a subexponential-time algorithm working in time n𝒪(n8/9)n^{\mathcal{O}(n^{8/9})}. We point out that the arguments used for the case when HH is a subdivided claw are significantly more complicated and technically involved than their counterparts for PtP_{t}-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 n𝒪(nlogn)n^{\mathcal{O}(\sqrt{n}\log n)}. Finally, very recently, Gartland et al.  [24] announced a quasipolynomial-time algorithm for MWIS in HH-free graphs for every H𝒮H\in\mathcal{S}.

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 tt,MWIS is polynomial-time solvable in subcubic St,t,1S_{t,t,1}-free graphs. Later, Lozin, Monnot, and Ries [38] showed a polynomial time algorithm for subcubic S2,2,2S_{2,2,2}-free graphs. Finally, Harutyunya et al. [28] generalized both these results by providing a polynomial-time algorithm for subcubic S2,t,tS_{2,t,t}-free graphs, for any fixed tt.

We remark that the case when HH is a subdivided claw (or, more precisely, is in 𝒮\mathcal{S} 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 H𝒮H\notin\mathcal{S}, then MIS is NP-hard even in subcubic HH-free graphs. On the other hand, if HH is a forest of paths, then HH-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 HH-free graphs with additional restrictions, where H𝒮H\in\mathcal{S}.

Our results.

As a warm-up, we present a polynomial-time algorithm for HH-free graphs of bounded degree, where H𝒮H\in\mathcal{S}.

Theorem 1.

There exists an algorithm that, given a vertex-weighted graph (G,𝐰)(G,\mathbf{w}) on nn vertices with maximum degree Δ\Delta and integers d,td,t in time 2𝒪(dtΔ2)n𝒪(tΔ2)2^{\mathcal{O}(dt\Delta^{2})}n^{\mathcal{O}(t\Delta^{2})} either finds an induced dSt,t,tdS_{t,t,t} or the maximum possible weight of an independent set in (G,𝐰)(G,\mathbf{w}).

Note that by picking appropriate dd and tt, theorem 1 yields a polynomial-time algorithm for MWIS for bounded-degree graphs excluding a fixed graph from 𝒮\mathcal{S} as an induced subgraph.

Then we proceed to the main result of the paper: we show that MWIS remains polynomial-time solvable in dSt,t,tdS_{t,t,t}-free graphs, even if instead of bounding the maximum degree, we forbid a fixed biclique as a subgraph.

Theorem 2.

For every fixed integers d,td,t, and ss there exists a polynomial-time algorithm that, given a vertex-weighted graph (G,𝐰)(G,\mathbf{w}) that does not contain dSt,t,tdS_{t,t,t} as an induced subgraph nor Ks,sK_{s,s} as a subgraph, returns the maximum possible weight of an independent set in (G,𝐰)(G,\mathbf{w}).

Let us remark that by the celebrated Kövári-Sós-Turán theorem [33], classes that exclude Ks,sK_{s,s} 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 rr there exists an integer ss such that if GG is KrK_{r}-free and Kr,rK_{r,r}-free then GG does not contain Ks,sK_{s,s} as a subgraph. Hence, equivalently, theorem 2 yields a polynomial-time algorithm for MWIS for graphs that are simultaneously HH-free (for some H𝒮H\in\mathcal{S}), KrK_{r}-free, and Kr,rK_{r,r}-free.

Our techniques.

As in the previous works [16, 24], the crucial tool in handling dSt,t,tdS_{t,t,t}-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 S1,1,1S_{1,1,1}-free) that allows for recursion for the MWIS problem. An extended strip decomposition of a graph GG identifies some induced subgraphs of GG 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 GG. (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 |V(G)||V(G)| 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 nn-vertex graph GG with a set UV(G)U\subseteq V(G) and integers d,td,t, in polynomial time outputs either:

  • an induced copy of dSt,t,tdS_{t,t,t} in GG, or

  • a set XX of size at most (d1)(3t+1)+(11logn+6)(t+1)(d-1)(3t+1)+(11\log n+6)(t+1) and a rigid extended strip decomposition of GN[X]G-N[X] with every particle containing at most |U|/2|U|/2 vertices of UU.

(A rigid extended strip decomposition is an extended strip decomposition that does not have some unnecessary empty sets. By N[X]N[X] we denote the set consisting of XX and all vertices with a neighbor in XX.) Let us remark that the result stated in [42, Theorem 2] is for unweighted graphs (i.e., U=V(G)U=V(G) 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 GG has maximum degree Δ\Delta. Apply theorem 3 to GG with U=V(G)U=V(G). If we get the first outcome, i.e., an induced dSt,t,tdS_{t,t,t} in GG, we return it and terminate. So assume that we get the second outcome, i.e., the set XX. Note that as |X|=𝒪(dt+tlogn)|X|=\mathcal{O}(dt+t\log n), we have |N[X]|=𝒪(dtΔ+tΔlogn)|N[X]|=\mathcal{O}(dt\Delta+t\Delta\log n). It is now tempting to exhaustively branch on N[X]N[X] (i.e., guess the intersection of the sought independent set with N[X]N[X]) and recurse on the particles of the extended strip decomposition of GN[X]G-N[X]. However, implementing this strategy directly gives quasipolynomial (in nn) running time bound of n𝒪(dtΔ+tΔlogn)n^{\mathcal{O}(dt\Delta+t\Delta\log n)}, as the branching step yields up to 2|N[X]|=2𝒪(dtΔ)n𝒪(tΔ)2^{|N[X]|}=2^{\mathcal{O}(dt\Delta)}\cdot n^{\mathcal{O}(t\Delta)} subcases and the depth of the recursion is 𝒪(logn)\mathcal{O}(\log n).

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.

Input: A vertex-weighted graph (G,𝐰)(G,\mathbf{w}) with a set TV(G)T\subseteq V(G) of terminals. Task: Compute fG,𝐰,T:2T{}f_{G,\mathbf{w},T}:2^{T}\to\mathbb{N}\cup\{-\infty\} defined for every ITTI_{T}\subseteq T as
fG,𝐰,T(IT)=max{𝐰(I)|IV(G)IisindependentIT=IT}.f_{G,\mathbf{w},T}(I_{T})=\max\{\mathbf{w}(I)~{}|~{}I\subseteq V(G)\wedge I\mathrm{\ is\ independent}\wedge I\cap T=I_{T}\}.
Border MWIS

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 XX of size 𝒪(dt+tlogn)\mathcal{O}(dt+t\log n) and an extended strip decomposition of GN[X]G-N[X] with particles of size at most half of the size of V(G)V(G). We would like to remove N[X]N[X] from the graph, indicate N(N[X])N(N[X]) as terminals and solve Border MWIS in (GN[X],𝐰,T:=N(N[X]))(G-N[X],\mathbf{w},T:=N(N[X])) using the extended strip decomposition for recursion. Note that, thanks to the bounded degree assumption, the size of T=N(N[X])T=N(N[X]) is bounded by 𝒪(dtΔ2+tΔ2logn)\mathcal{O}(dt\Delta^{2}+t\Delta^{2}\log n).

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 𝒪(dtΔ2+tΔ2logn)\mathcal{O}(dt\Delta^{2}+t\Delta^{2}\log n) bound. Luckily, this can be remedied in a standard way: we alternate recursive steps where theorem 3 is invoked with U=V(G)U=V(G) with steps where theorem 3 is invoked with U=TU=T. In this manner, we can maintain a bound of 𝒪(dtΔ2+tΔ2logn)\mathcal{O}(dt\Delta^{2}+t\Delta^{2}\log n) 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 fG,𝐰,Tf_{G,\mathbf{w},T} is of size 2𝒪(dtΔ2)n𝒪(tΔ2)2^{\mathcal{O}(dt\Delta^{2})}n^{\mathcal{O}(t\Delta^{2})}, 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 St,t,tS_{t,t,t}-free case, excluding a biclique as a subgraph is not that much different than bounding the maximum degree.

A kk-block in a graph is a set of kk vertices, no two of which can be separated by deleting fewer than kk 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 GG be a graph and k2k\geqslant 2 be an integer. If GG has no (k+1)(k+1)-block, then GG admits a tree decomposition with adhesion less than kk, in which every torso has at most kk vertices of degree larger than 2k(k1)2k(k-1).

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 dSt,t,tdS_{t,t,t}-free graphs with no large bicliques have no large blocks.

Lemma 5.

For any d,td,t, and ss there exists kk such that the following holds. Every dSt,t,tdS_{t,t,t}-free graph with no subgraph isomorphic to Ks,sK_{s,s} has no kk-block.

Let us remark that Lozin and Razgon [40] showed lemma 5 for St,t,tS_{t,t,t}-free graphs. However, an extension of their argument applies to dSt,t,tdS_{t,t,t}-free graphs; we include it in Appendix A.

Combining theorem 4 and lemma 5 we immediately obtain the following.

Corollary 6.

For any d,td,t, and ss there exists kk such that the following holds. Given a dSt,t,tdS_{t,t,t}-free graph GG with no subgraph isomorphic to Ks,sK_{s,s}, in polynomial time one can compute a tree decomposition of GG with adhesion less than kk, in which every torso has at most kk vertices of degree larger than 2k(k1)2k(k-1).

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 N(N[X])N(N[X]) 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 (G,𝐰)(G,\mathbf{w}) as an input. In the recursion, we will be working on various induced subgraphs of GG with vertex weight inherited from 𝐰\mathbf{w}. Somewhat abusing notation, we will keep 𝐰\mathbf{w} for the weight function in any induced subgraph of GG.

Tree decompositions.

Let GG be a graph. A tree decomposition of GG is a pair (𝒯,β)(\mathcal{T},\beta) where 𝒯\mathcal{T} is a tree and β:V(𝒯)2V(G)\beta:V(\mathcal{T})\to 2^{V(G)} is a function satisfying the following: (i) for every uvE(G)uv\in E(G) there exists tV(𝒯)t\in V(\mathcal{T}) with u,vβ(t)u,v\in\beta(t), and (ii) for every vV(G)v\in V(G) the set {tV(𝒯)|vβ(t)}\{t\in V(\mathcal{T})~{}|~{}v\in\beta(t)\} induces a connected nonempty subtree of 𝒯\mathcal{T}. For every tV(𝒯)t\in V(\mathcal{T}) and stE(𝒯)st\in E(\mathcal{T}), the set β(t)\beta(t) is the bag at node tt and the set σ(st):=β(s)β(t)\sigma(st):=\beta(s)\cap\beta(t) is the adhesion at edge stst. The critical property of a tree decomposition (𝒯,β)(\mathcal{T},\beta) is that if stE(𝒯)st\in E(\mathcal{T}) and VsV_{s} and VtV_{t} are two connected components of 𝒯{st}\mathcal{T}-\{st\} that contain ss and tt, respectively, then σ(st)\sigma(st) separates xVsβ(x)σ(st)\bigcup_{x\in V_{s}}\beta(x)\setminus\sigma(st) from xVtβ(x)σ(st)\bigcup_{x\in V_{t}}\beta(x)\setminus\sigma(st) in GG.

The torso of a bag β(t)\beta(t) in a tree decomposition (𝒯,β)(\mathcal{T},\beta) is a graph HH with V(H)=β(t)V(H)=\beta(t) and uvE(H)uv\in E(H) if uvE(G)uv\in E(G) or there exists a neighbor sN𝒯(t)s\in N_{\mathcal{T}}(t) with u,vσ(st)u,v\in\sigma(st). That is, the torso of β(t)\beta(t) is created from G[β(t)]G[\beta(t)] by turning the adhesion σ(st)\sigma(st) into a clique for every neighbor ss of tt in 𝒯\mathcal{T}.

Extended strip decompositions.

We follow the notation of [42, 24]. For a graph HH, by T(H)T(H) we denote the set of triangles in HH. An extended strip decomposition of a graph GG is a pair (H,η)(H,\eta) that consists of:

  • a simple graph HH,

  • a vertex set η(x)V(G)\eta(x)\subseteq V(G) for every xV(H)x\in V(H),

  • an edge set η(xy)V(G)\eta(xy)\subseteq V(G) for every xyE(H)xy\in E(H), and its subsets η(xy,x),η(xy,y)η(xy)\eta(xy,x),\eta(xy,y)\subseteq\eta(xy),

  • a triangle set η(xyz)V(G)\eta(xyz)\subseteq V(G) for every xyzT(H)xyz\in T(H),

which satisfy the following properties:

  1. 1.

    The family {η(o)|oV(H)E(H)T(H)}\{\eta(o)~{}|~{}o\in V(H)\cup E(H)\cup T(H)\} is a partition of V(G)V(G).

  2. 2.

    For every xV(H)x\in V(H) and every distinct y,zNH(x)y,z\in N_{H}(x), the set η(xy,x)\eta(xy,x) is complete to η(xz,x)\eta(xz,x).

  3. 3.

    Every uvE(G)uv\in E(G) is contained in one of the sets η(o)\eta(o) for oV(H)E(H)T(H)o\in V(H)\cup E(H)\cup T(H), or is as follows:

    • uη(xy,x),vη(xz,x)u\in\eta(xy,x),v\in\eta(xz,x) for some xV(H)x\in V(H) and y,zNH(x)y,z\in N_{H}(x), or

    • uη(xy,x),vη(x)u\in\eta(xy,x),v\in\eta(x) for some xyE(H)xy\in E(H), or

    • uη(xyz)u\in\eta(xyz) and vη(xy,x)η(xy,y)v\in\eta(xy,x)\cap\eta(xy,y) for some xyzT(H)xyz\in T(H).

An extended strip decomposition (H,η)(H,\eta) is rigid if for every xyE(H)xy\in E(H), the sets η(xy)\eta(xy), η(xy,x)\eta(xy,x), and η(xy,y)\eta(xy,y) are nonempty, and for every isolated xV(H)x\in V(H), the set η(x)\eta(x) is nonempty. Note that if (H,η)(H,\eta) is a rigid extended strip decomposition of GG, then |V(H)||V(H)| is bounded by |V(G)||V(G)|.

For an extended strip decomposition (H,η)(H,\eta) of a graph GG, we identify five types of particles.

vertex particle: Ax:=η(x) for each xV(H),\displaystyle\quad A_{x}:=\eta(x)\text{ for each }x\in V(H),
edge interior particle: Axy:=η(xy)(η(xy,x)η(xy,y)) for each xyE(H),\displaystyle\quad A_{xy}^{\perp}:=\eta(xy)\setminus(\eta(xy,x)\cup\eta(xy,y))\text{ for each }xy\in E(H),
half-edge particle: Axyx:=η(x)η(xy)η(xy,y) for each xyE(H),\displaystyle\quad A_{xy}^{x}:=\eta(x)\cup\eta(xy)\setminus\eta(xy,y)\text{ for each }xy\in E(H),
full edge particle: Axyxy:=η(x)η(y)η(xy)z:xyzT(H)η(xyz) for each xyE(H),\displaystyle\quad A_{xy}^{xy}:=\eta(x)\cup\eta(y)\cup\eta(xy)\cup\bigcup_{z~{}:~{}xyz\in T(H)}\eta(xyz)\text{ for each }xy\in E(H),
triangle particle: Axyz:=η(xyz) for each xyzT(H).\displaystyle\quad A_{xyz}:=\eta(xyz)\text{ for each }xyz\in T(H).

As announced in the introduction, to solve MWIS in GG 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 (G,𝐰,T)(G,\mathbf{w},T), an extended strip decomposition (H,η)(H,\eta) of GG, and a solution fG[A],𝐰,TAf_{G[A],\mathbf{w},T\cap A} to the Border MWIS instance (G[A],𝐰,TA)(G[A],\mathbf{w},T\cap A) for every particle AA of (H,η)(H,\eta), one can in time 2|T|2^{|T|} times a polynomial in |V(G)|+|V(H)||V(G)|+|V(H)| compute the solution fG,𝐰,Tf_{G,\mathbf{w},T} to the input (G,𝐰,T)(G,\mathbf{w},T).

We need the following simple observations.

Lemma 8.

Let GG be a KtK_{t}-free graph and let (H,η)(H,\eta) be a rigid extended strip decomposition of GG. Then the maximum degree of HH is at most t1t-1.

Proof..

Let xV(H)x\in V(H). Observe that the sets {η(xy,x)|yNH(x)}\{\eta(xy,x)~{}|~{}y\in N_{H}(x)\} are nonempty and complete to each other in GG. Hence, GG contains a clique of size equal to the degree of xx in HH. \square

Lemma 9.

Let GG be a graph and let (H,η)(H,\eta) be an extended strip decomposition of GG such that the maximum degree of HH is at most dd. Then, every vertex of GG is in at most max(4,2d+1)\max(4,2d+1) particles.

Proof..

Pick vV(G)v\in V(G) and observe that:

  • If vη(x)v\in\eta(x) for some xV(H)x\in V(H), then vv is in the vertex particle of xx and in one half-edge and one full-edge particle for every edge of HH incident with xx. Since there are at most dd such edges, vv is in at most 2d+12d+1 particles.

  • If vη(xy)v\in\eta(xy) for some xyE(H)xy\in E(H), then vv is in at most four particles for the edge xyxy.

  • If vη(xyz)v\in\eta(xyz) for some xyzT(H)xyz\in T(H), then vv is in the triangle particle for xyzxyz and in three full edge particles, for the three sides of the triangle xyzxyz.

\square

3 Bounded-degree graphs: Proof of Theorem 1

This section is devoted to the proof of theorem 1.

Let d,td,t be positive integers and let (G,𝐰)(G,\mathbf{w}) be the input vertex-weighted graph. We denote n:=|V(G)|n:=|V(G)| and Δ\Delta to be the maximum degree of GG. Let

:=(d1)(3t+1)+11logn+6(t+2)=𝒪(dt+tlogn)\ell:=(d-1)(3t+1)+\lceil 11\log n+6\rceil(t+2)=\mathcal{O}(dt+t\log n)

be an upper bound on the size of XX for any application of theorem 3 for any induced subgraph of GG.

We describe a recursive algorithm that takes as input an induced subgraph GG^{\prime} of GG with weights 𝐰\mathbf{w} and a set of terminals TV(G)T\subseteq V(G^{\prime}) of size at most 4Δ24\ell\Delta^{2} and solves Border MWIS on (G,𝐰,T)(G^{\prime},\mathbf{w},T). The root call is for G:=GG^{\prime}:=G and T:=T:=\emptyset; indeed, note that fG,𝐰,()f_{G,\mathbf{w},\emptyset}(\emptyset) is the maximum possible weight of an independent set in GG.

Let (G,𝐰,T)(G^{\prime},\mathbf{w},T) be an input to a recursive call. First, the algorithm initializes fG,𝐰,T(IT):=f_{G^{\prime},\mathbf{w},T}(I_{T}):=-\infty for every ITTI_{T}\subseteq T.

If |V(G)|4Δ2|V(G^{\prime})|\leqslant 4\Delta^{2}\ell, the algorithm proceeds by brute-force: it enumerates independent sets IV(G)I\subseteq V(G^{\prime}) and updates fG,𝐰,T(IT)f_{G^{\prime},\mathbf{w},T}(I\cap T) with 𝐰(I)\mathbf{w}(I) whenever the previous value of that cell was smaller. As =𝒪(dt+tlogn)\ell=\mathcal{O}(dt+t\log n), this step takes 2𝒪(dtΔ2)n𝒪(tΔ2)2^{\mathcal{O}(dt\Delta^{2})}n^{\mathcal{O}(t\Delta^{2})} time. This completes the description of the leaf step of the recursion.

If |V(G)|>4Δ2|V(G^{\prime})|>4\Delta^{2}\ell, the algorithm proceeds as follows. If |T|3Δ2|T|\leqslant 3\Delta^{2}\ell, let U:=V(G)U:=V(G^{\prime}), and otherwise, let U:=TU:=T. The algorithm invokes theorem 3 on GG^{\prime} and UU. If an induced dSt,t,tdS_{t,t,t} is returned, then it can be returned by the main algorithm as it is in particular an induced subgraph of GG. Hence, we can assume that we obtain a set XV(G)X\subseteq V(G) of size at most \ell and an extended strip decomposition (H,η)(H,\eta) of G:=GNG[X]G^{\ast}:=G^{\prime}-N_{G^{\prime}}[X] whose every particle contains at most |U|/2|U|/2 vertices of UU.

Observe that as |X||X|\leqslant\ell and the maximum degree of GG is Δ\Delta, we have |NG(NG[X])|Δ2|N_{G^{\prime}}(N_{G^{\prime}}[X])|\leqslant\Delta^{2}\ell. Let T:=(TV(G))NG(NG[X])T^{\ast}:=(T\cap V(G^{\ast}))\cup N_{G^{\prime}}(N_{G^{\prime}}[X]). Note that we have TV(G)T^{\ast}\subseteq V(G^{\ast}) and |T|5Δ2|T^{\ast}|\leqslant 5\Delta^{2}\ell. For every particle AA of (H,η)(H,\eta), invoke a recursive call on (GA:=G[A],𝐰,TA:=TA)(G^{\ast}_{A}:=G^{\ast}[A],\mathbf{w},T^{\ast}_{A}:=T^{\ast}\cap A), obtaining fGA,𝐰,TAf_{G^{\ast}_{A},\mathbf{w},T^{\ast}_{A}} (or an induced St,t,tS_{t,t,t}, which can be directly returned). Use lemma 7 to obtain a solution fG,𝐰,Tf_{G^{\ast},\mathbf{w},T^{\ast}} to Border MWIS instance (G,𝐰,T)(G^{\ast},\mathbf{w},T^{\ast}).

Finally, iterate over every ITTNG[X]I_{T}\subseteq T^{\ast}\cup N_{G^{\prime}}[X] (note that TTNG[X]T\subseteq T^{\ast}\cup N_{G^{\prime}}[X]) and, if ITI_{T} is independent, update the cell fG,𝐰,T(ITT)f_{G^{\prime},\mathbf{w},T}(I_{T}\cap T) with the value 𝐰(ITT)+fG,𝐰,T(ITT)\mathbf{w}(I_{T}\setminus T^{\ast})+f_{G^{\ast},\mathbf{w},T^{\ast}}(I_{T}\cap T^{\ast}), 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 NG[X]N_{G^{\prime}}[X] is adjacent in GG^{\prime} only to NG(NG[X])N_{G^{\prime}}(N_{G^{\prime}}[X]) which is a subset of TT^{\ast}.

For the complexity analysis, consider a recursive call to (GA,𝐰,TA)(G^{\ast}_{A},\mathbf{w},T^{\ast}_{A}) for a particle AA. If |T|3Δ2|T|\leqslant 3\Delta^{2}\ell, then |TA||T|4Δ2|T^{\ast}_{A}|\leqslant|T^{\ast}|\leqslant 4\Delta^{2}\ell. Otherwise, U=TU=T and |TA||T|/22Δ2|T\cap A|\leqslant|T|/2\leqslant 2\Delta^{2}\ell. As |NG(NG[X])|Δ2|N_{G^{\prime}}(N_{G^{\prime}}[X])|\leqslant\Delta^{2}\ell, we have |TA|3Δ2|T^{\ast}_{A}|\leqslant 3\Delta^{2}\ell. Hence, in the recursive call the invariant of at most 4Δ24\Delta^{2}\ell terminals is maintained and, moreover:

  • if |T|3Δ2|T|\leqslant 3\Delta^{2}\ell, then U=|V(G)|U=|V(G^{\prime})| and |V(GA)|=|A||V(G)|/2|V(G^{\ast}_{A})|=|A|\leqslant|V(G^{\prime})|/2;

  • otherwise, V(GA)V(G)V(G^{\ast}_{A})\subseteq V(G^{\prime}) and |TA|3Δ2|T^{\ast}_{A}|\leqslant 3\Delta^{2}\ell, hence the recursive call will fall under the first bullet.

We infer that the depth of the recursion is at most 2logn2\lceil\log n\rceil.

At every non-leaf recursive call, we spend n𝒪(1)n^{\mathcal{O}(1)} time on invoking the algorithm from theorem 3, 2𝒪(dtΔ2)n𝒪(tΔ2)2^{\mathcal{O}(dt\Delta^{2})}n^{\mathcal{O}(t\Delta^{2})} time to compute fG,𝐰,Tf_{G^{\ast},\mathbf{w},T^{\ast}} using lemma 7, and 2𝒪(dtΔ2)n𝒪(tΔ2)2^{\mathcal{O}(dt\Delta^{2})}n^{\mathcal{O}(t\Delta^{2})} time for the final iteration over all subsets ITTNG[X]I_{T}\subseteq T^{\ast}\cup N_{G^{\prime}}[X]. Hence, the time spent at every recursive call is bounded by 2𝒪(dtΔ2)n𝒪(tΔ2)2^{\mathcal{O}(dt\Delta^{2})}n^{\mathcal{O}(t\Delta^{2})}.

At every non-leaf recursive call, we make subcalls to (GA,𝐰,TA)(G^{\ast}_{A},\mathbf{w},T_{A}^{\ast}) for every particle AA of (H,η)(H,\eta). Lemmas 8 and 9 ensure that the sum of |V(GA)||V(G^{\ast}_{A})| over all particles AA is bounded by (2Δ+3)|V(G)|(2\Delta+3)|V(G^{\prime})|. Hence, the total size of all graphs in the ii-th level of the recursion is bounded by n(2Δ+3)in\cdot(2\Delta+3)^{i}. Since the depth of the recursion is bounded by 2logn2\lceil\log n\rceil, the total size of all graphs in the recursion tree is bounded by n𝒪(logΔ)n^{\mathcal{O}(\log\Delta)}. Since this also bounds the size of the recursion tree, we infer that the whole algorithm runs in time 2𝒪(dtΔ2)n𝒪(tΔ2)2^{\mathcal{O}(dt\Delta^{2})}n^{\mathcal{O}(t\Delta^{2})}.

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 tt be a positive integer and let kk be the constant depending on tt from corollary 6. Again, let (G,𝐰)(G,\mathbf{w}) be the input vertex-weighted graph, let n:=|V(G)|n:=|V(G)|, and let

:=(d1)(3t+1)+11logn+6(t+2)=𝒪(dtlogn)\ell:=(d-1)(3t+1)+\lceil 11\log n+6\rceil(t+2)=\mathcal{O}(dt\log n)

be an upper bound on the size of XX for any application of theorem 3 for any induced subgraph of GG222As the dependence of kk on d,t,sd,t,s is superpolynomial, for the sake of simplicity, we do not try to optimize the dependence of the complexity bound on d,t,sd,t,s..

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 GG^{\prime} of GG with weights 𝐰\mathbf{w} and a set of terminals TV(G)T\subseteq V(G^{\prime}) of size at most 32k532k^{5}\ell and solves Border MWIS on (G,𝐰,T)(G^{\prime},\mathbf{w},T). The root call is for G:=GG^{\prime}:=G and T:=T:=\emptyset and the algorithm returns fG,𝐰,()f_{G,\mathbf{w},\emptyset}(\emptyset) as the final answer.

Let (G,𝐰,T)(G^{\prime},\mathbf{w},T) be an input to a recursive call. The algorithm initiates first fG,𝐰,T(IT)=f_{G^{\prime},\mathbf{w},T}(I_{T})=-\infty for every ITTI_{T}\subseteq T.

If |V(G)|32k5|V(G^{\prime})|\leqslant 32k^{5}\ell, the algorithm proceeds by brute-force: it enumerates independent sets IV(G)I\subseteq V(G^{\prime}) and updates fG,𝐰,T(IT)f_{G^{\prime},\mathbf{w},T}(I\cap T) with 𝐰(I)\mathbf{w}(I) whenever the previous value of that cell was smaller. As =𝒪(dtlogn)\ell=\mathcal{O}(dt\log n) and kk is a constant depending on d,td,t, and ss, this step takes polynomial time. This completes the description of the leaf step of the recursion.

Otherwise, if |V(G)|>32k5|V(G^{\prime})|>32k^{5}\ell, we invoke corollary 6 on GG^{\prime}, obtaining a tree decomposition (𝒯,β)(\mathcal{T},\beta) of GG^{\prime}. If |T|24k5|T|\leqslant 24k^{5}\ell, let U:=V(G)TU:=V(G^{\prime})\setminus T, and otherwise, let U:=TU:=T.

For every t1t2E(𝒯)t_{1}t_{2}\in E(\mathcal{T}), proceed as follows. For i=1,2i=1,2, let 𝒯i\mathcal{T}_{i} be the connected component of 𝒯{t1t2}\mathcal{T}-\{t_{1}t_{2}\} that contains tit_{i} and let Vi=x𝒯iβ(x)σ(t1t2)V_{i}=\bigcup_{x\in\mathcal{T}_{i}}\beta(x)\setminus\sigma(t_{1}t_{2}). Clearly, σ(t1t2)\sigma(t_{1}t_{2}) separates V1V_{1} from V2V_{2}. Orient the edge t1t2t_{1}t_{2} towards tit_{i} with larger |UVi||U\cap V_{i}|, breaking ties arbitrarily.

There exists tV(𝒯)t\in V(\mathcal{T}) of outdegree 0. Then, for every connected component CC of Gβ(t)G^{\prime}-\beta(t) we have |CU||U|/2|C\cap U|\leqslant|U|/2. Fix one such node tt and let B:=β(t)B:=\beta(t) and let 𝒞\mathcal{C} be the set of connected components of GBG^{\prime}-B. Let GBG^{B} be a supergraph of G[B]G^{\prime}[B] obtained from G[B]G^{\prime}[B] by turning, for every C𝒞C\in\mathcal{C}, the neighborhood NG(C)N_{G^{\prime}}(C) into a clique. Note that GBG^{B} is a subgraph of the torso of β(t)\beta(t). Hence, by the properties promised by corollary 6, for every C𝒞C\in\mathcal{C} we have |NG(C)|<k|N_{G^{\prime}}(C)|<k (as this set is contained in a single adhesion of an edge incident with tt in 𝒯\mathcal{T}) and GBG^{B} contains at most kk vertices of degree larger than 2k(k1)2k(k-1). Let QQ be the set of vertices of GBG^{B} of degree larger than 2k(k1)2k(k-1).

We perform exhaustive branching on QQ. That is, we iterate over all independent sets JQJ\subseteq Q and denote GJ:=GQNG(J)G^{J}:=G^{\prime}-Q-N_{G^{\prime}}(J), TJ:=TV(GJ)T^{J}:=T\cap V(G^{J}), UJ:=UV(GJ)U^{J}:=U\cap V(G^{J}). For one JJ, we proceed as follows.

We invoke theorem 3 to GJG^{J} with set UJU^{J}, obtaining a set XJX^{J} of size at most \ell and an extended strip decomposition (HJ,ηJ)(H^{J},\eta^{J}) of GJNGJ[XJ]G^{J}-N_{G^{J}}[X^{J}] whose every particle has at most |UJ|/2|U|/2|U^{J}|/2\leqslant|U|/2 vertices of UU. Note that GJG^{J} is an induced subgraph of GG^{\prime}, which is an induced subgraph of GG, so there is no induced dSt,t,tdS_{t,t,t} in GJG^{J}.

A component C𝒞C\in\mathcal{C} is dirty if NGJ[XJ]NG[C]N_{G^{J}}[X^{J}]\cap N_{G^{\prime}}[C]\neq\emptyset and clean otherwise. Let

YJ:=(NGJ[XJ]B)C𝒞:Cisdirty(NG(C)V(GJ)).Y^{J}:=(N_{G^{J}}[X^{J}]\cap B)\cup\bigcup_{C\in\mathcal{C}:C\mathrm{\ is\ dirty}}(N_{G^{\prime}}(C)\cap V(G^{J})).

The following bounds will be important for further steps.

|NGJ[XJ]B|(2k(k1)+1)|XJ|.|N_{G^{J}}[X^{J}]\cap B|\leqslant(2k(k-1)+1)|X^{J}|. (1)

To see (1) observe that a vertex vXJBv\in X^{J}\cap B has at most 2k(k1)2k(k-1) neighbors in BB (as every vertex of BQB\setminus Q has degree at most 2k(k1)2k(k-1) in GBG_{B}) while every vertex vXJBv\in X^{J}\setminus B has at most kk neighbors in BB, as every component of GBG^{\prime}-B has at most kk neighbors in BB. This proves (1).

|YJ|(k+(2k(k1)+1)2)|XJ|4k4|XJ|4k4=𝒪(k4dtlogn).|Y^{J}|\leqslant(k+(2k(k-1)+1)^{2})|X^{J}|\leqslant 4k^{4}|X^{J}|\leqslant 4k^{4}\ell=\mathcal{O}(k^{4}dt\log n). (2)

To see (2), consider a dirty component C𝒞C\in\mathcal{C}. Observe that either CC contains a vertex of XJX^{J} or NG(C)V(GJ)N_{G^{\prime}}(C)\cap V(G^{J}) contains a vertex of NGJ[XJ]BN_{G^{J}}[X^{J}]\cap B. There are at most |XJ||X^{J}| dirty components of the first type, contributing in total at most k|XJ|k|X^{J}| vertices to YJY^{J}. For the dirty components of the second type, although there can be many of them, we observe that if vNG(C)NGJ[XJ]Bv\in N_{G^{\prime}}(C)\cap N_{G^{J}}[X^{J}]\cap B, then NG(C)V(GJ)NGB[v]N_{G^{\prime}}(C)\cap V(G^{J})\subseteq N_{G_{B}}[v]. Hence, for every dirty component of the second type, it holds that NG(C)V(GJ)NGB[NGJ[XJ]B]N_{G^{\prime}}(C)\cap V(G^{J})\subseteq N_{G_{B}}[N_{G^{J}}[X^{J}]\cap B]. Since the degree of each vertex of GBG_{B} is at most 2k(k1)2k(k-1), by (1) we have

|NGB[NGJ[XJ]B]|(2k(k1)+1)2|XJ|.\left|N_{G_{B}}\left[N_{G^{J}}[X^{J}]\cap B\right]\right|\leqslant(2k(k-1)+1)^{2}|X^{J}|.

The bound (2) follows.

A component C𝒞C\in\mathcal{C} is touched if it is dirty or NG(C)N_{G^{\prime}}(C) contains a vertex of YJY^{J}. Let

ZJ:=(NGJ[YJ]B)C𝒞:CistouchedNG(C)V(GJ).Z^{J}:=(N_{G^{J}}[Y^{J}]\cap B)\cup\bigcup_{C\in\mathcal{C}:C\mathrm{\ is\ touched}}N_{G^{\prime}}(C)\cap V(G^{J}).

Using similar arguments as before, we can prove

|ZJ|(2k(k1)+1)|YJ|8k5|XJ|8k5=𝒪(k5dtlogn).|Z^{J}|\leqslant(2k(k-1)+1)|Y^{J}|\leqslant 8k^{5}|X^{J}|\leqslant 8k^{5}\ell=\mathcal{O}(k^{5}dt\log n). (3)

Indeed, if CC is touched, then NG(C)N_{G^{\prime}}(C) contains a vertex vYJv\in Y^{J} (if CC is dirty, NG(C)V(GJ)N_{G^{\prime}}(C)\cap V(G^{J}) is contained in YJY^{J}), and then NG(C)N_{G^{\prime}}(C) is contained in NGB[v]N_{G_{B}}[v]. Also, for vYJv\in Y^{J} we have NGJ[v]BNGB[v]N_{G^{J}}[v]\cap B\subseteq N_{G_{B}}[v]. Hence, ZJNGB[YJ]Z^{J}\subseteq N_{G_{B}}[Y^{J}]. Since the maximum degree of a vertex of BQB\setminus Q is 2k(k1)2k(k-1), this proves (3).

For every touched C𝒞C\in\mathcal{C}, denote GC:=GJ[NG[C]V(GJ)]G_{C}:=G^{J}[N_{G^{\prime}}[C]\cap V(G^{J})] and TC:=((TC)NG(C))V(GJ)T_{C}:=((T\cap C)\cup N_{G^{\prime}}(C))\cap V(G^{J}). Recurse on (GC,𝐰,TC)(G_{C},\mathbf{w},T_{C}), obtaining fGC,𝐰,TCf_{G_{C},\mathbf{w},T_{C}}.

Let

GY:=GJYJC𝒞:CistouchedC.G^{Y}:=G^{J}-Y^{J}-\bigcup_{C\in\mathcal{C}:C\mathrm{\ is\ touched}}C.

Note that, by the definition of dirty and touched, GYG^{Y} is an induced subgraph of GJNGJ[XJ]G^{J}-N_{G^{J}}[X^{J}]. Hence, (HJ,ηJ)(H^{J},\eta^{J}) can be restricted to a (not necessarily rigid) extended strip decomposition (HJ,ηJ,Y)(H^{J},\eta^{J,Y}) of GYG^{Y}.

Let TY:=(TZJ)V(GY)T^{Y}:=(T\cup Z^{J})\cap V(G^{Y}). For every particle AA of (HJ,ηJ,Y)(H^{J},\eta^{J,Y}), recurse on (GY[A],𝐰,TYA)(G^{Y}[A],\mathbf{w},T^{Y}\cap A), obtaining fGY[A],𝐰,TYAf_{G^{Y}[A],\mathbf{w},T^{Y}\cap A}. Then, use these values with lemma 7 to solve a Border MWIS instance (GY,𝐰,TY)(G^{Y},\mathbf{w},T^{Y}), obtaining fGY,𝐰,TYf_{G^{Y},\mathbf{w},T^{Y}}.

Iterate over every independent set IT(TV(GJ))TYYJI_{T}\subseteq(T\cap V(G^{J}))\cup T^{Y}\cup Y^{J}. Observe that GG^{\prime} admits an independent set II with I(QTTYYJ)=JITI\cap(Q\cup T\cup T^{Y}\cup Y^{J})=J\cup I_{T} and weight:

𝐰(J)+𝐰(ITTY)+fGY,𝐰,TY(ITTY)+C𝒞:Cistouched(fGC,𝐰,TC(NG[C]IT)𝐰(ITNG(C))).\mathbf{w}(J)+\mathbf{w}(I_{T}\setminus T^{Y})+f_{G^{Y},\mathbf{w},T^{Y}}(I_{T}\cap T^{Y})+\sum_{C\in\mathcal{C}:C\mathrm{\ is\ touched}}\left(f_{G_{C},\mathbf{w},T_{C}}(N_{G^{\prime}}[C]\cap I_{T})-\mathbf{w}(I_{T}\cap N_{G^{\prime}}(C))\right).

Update the cell fG,𝐰,T((ITJ)T)f_{G^{\prime},\mathbf{w},T}((I_{T}\cup J)\cap T) 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 CC, the whole NG(C)V(GJ)N_{G^{\prime}}(C)\cap V(G^{J}) is in the terminal set for the recursive call (GC,𝐰,TC)(G_{C},\mathbf{w},T_{C}) and the whole NG(C)V(GY)N_{G^{\prime}}(C)\cap V(G^{Y}) is in ZJZ^{J} and thus in the terminal set for the Border MWIS instance (GY,𝐰,TY)(G^{Y},\mathbf{w},T^{Y}).

For the sake of analysis, consider a recursive call on (GC,𝐰,TC)(G_{C},\mathbf{w},T_{C}) for a touched component CC. If |T|24k5|T|\leqslant 24k^{5}\ell and U=V(G)TU=V(G^{\prime})\setminus T, then |TC||T|+k32k5|T_{C}|\leqslant|T|+k\leqslant 32k^{5}\ell and |V(GC)TC||CT||V(G)T|/2|V(G_{C})\setminus T_{C}|\leqslant|C\setminus T|\leqslant|V(G^{\prime})\setminus T|/2. Otherwise, if |T|>24k5|T|>24k^{5}\ell and U=TU=T, then |TC||T|/2+k16k5+k24k5|T_{C}|\leqslant|T|/2+k\leqslant 16k^{5}\ell+k\leqslant 24k^{5}\ell. Thus, the recursive call on (GC,𝐰,TC)(G_{C},\mathbf{w},T_{C}) will fall under the first case of at most 24k524k^{5}\ell terminals.

Analogously, consider a recursive call on (GY[A],𝐰,TYA)(G^{Y}[A],\mathbf{w},T^{Y}\cap A) for a particle AA of (HJ,ηJ,Y)(H^{J},\eta^{J,Y}). If |T|24k5|T|\leqslant 24k^{5}\ell and U=V(G)TU=V(G^{\prime})\setminus T, then |TYA||TY||T|+|ZJ|32k5|T^{Y}\cap A|\leqslant|T^{Y}|\leqslant|T|+|Z^{J}|\leqslant 32k^{5}\ell due to (3). Furthermore, |V(GY[A])TY||V(G)T|/2|V(G^{Y}[A])\setminus T^{Y}|\leqslant|V(G^{\prime})\setminus T|/2. Otherwise, if |T|>24k5|T|>24k^{5}\ell and U=TU=T, then |TYA||T|/2+|ZJ|16k5+8k524k5|T^{Y}\cap A|\leqslant|T|/2+|Z^{J}|\leqslant 16k^{5}\ell+8k^{5}\ell\leqslant 24k^{5}\ell again due to (3). Thus, the recursive call on (GY[A],𝐰,TYA)(G^{Y}[A],\mathbf{w},T^{Y}\cap A) will fall under the first case of at most 24k524k^{5}\ell terminals.

Finally, note that a recursive call (G,𝐰,T)(G^{\prime},\mathbf{w},T) without nonterminal vertices (i.e., with T=V(G)T=V(G^{\prime})) is a leaf call.

We infer that all recursive calls satisfy the invariant of at most 32k532k^{5}\ell terminals and the depth of the recursion tree is bounded by 2logn2\lceil\log n\rceil (as every second level the number of nonterminal vertices halves).

At each recursive call, we iterate over at most 2k2^{k} subsets JQJ\subseteq Q. lemma 8 ensures that the maximum degree of HJH^{J} is at most 2t12t-1, while lemma 9 ensures that every vertex of GYG^{Y} is used in at most 4t4t particles of (HJ,ηJ,Y)(H^{J},\eta^{J,Y}). In a subcall (GC,𝐰,TC)(G_{C},\mathbf{w},T_{C}) for a touched component CC, vertices of CC are not used in any other call for the current choice of JJ, while all vertices of V(GC)CV(G_{C})\setminus C are terminals. Consequently, every nonterminal vertex vv of GG^{\prime} is passed as a nonterminal vertex to a recursive subcall at most 2k4t2^{k}\cdot 4t 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 𝒪(logn)\mathcal{O}(\log n), we infer that, summing over all recursive calls in the entire algorithm, the number of nonterminal vertices is bounded by n𝒪(logt+k)n^{\mathcal{O}(\log t+k)} and the total size of the recursion tree is n𝒪(logt+k)n^{\mathcal{O}(\log t+k)}.

At each recursive call, we iterate over all 2k2^{k} subsets JQJ\subseteq Q and then we invoke theorem 3 and iterate over all independent sets ITI_{T} in (TV(GJ))TYYJ(T\cap V(G^{J}))\cup T^{Y}\cup Y^{J}. Thanks to the invariant |T|32k5|T|\leqslant 32k^{5}\ell and bounds (2), and (3), this set is of size 𝒪(k5)\mathcal{O}(k^{5}\ell). Hence, every recursive call runs in time n𝒪(k5t)+kcd,tn^{\mathcal{O}(k^{5}t)+kc_{d,t}}, where cd,tc_{d,t} is a constant depending on dd and tt. As kk is a constant depending on d,t,sd,t,s, 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 St,t,tS_{t,t,t}-free (and even dSt,t,tdS_{t,t,t}-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 KrK_{r}-freeness twice: once in lemma 5 and then to argue that HH (the pattern of an extended strip decomposition we obtain) is of bounded degree. On the other hand, the assumption of Kr,rK_{r,r}-freeness was used just once: in lemma 5. Thus it seems natural to try to prove the following conjecture.

Conjecture 10.

For every integers t,rt,r there exists a polynomial-time algorithm that, given an St,t,tS_{t,t,t}-free and KrK_{r}-free vertex-weighted graph (G,𝐰)(G,\mathbf{w}) computes the maximum possible weight of an independent set in (G,𝐰)(G,\mathbf{w}).

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 a,ba,b, the Ramsey number of aa and bb, denoted by 𝖱𝖺𝗆(a,b)\mathsf{Ram}(a,b), is the smallest integer rr such that every graph on rr vertices contains either an independent set of size aa or a clique of size bb. It is well-known that 𝖱𝖺𝗆(a,b)(a+b2a1)\mathsf{Ram}(a,b)\leqslant\binom{a+b-2}{a-1}.

For a graph HH, a subdivision of HH is any graph obtained from HH by subdividing each edge arbitrarly many times (possibly 0). By a tt-subdivision (resp., (t)(\leqslant t)-subdivision) we mean a subdivision where each edge was subdivided exactly (resp., at most tt) times. A proper subdivision is one where each edge was subdivided at least once. By S(p,t)S(p,t) we denote the tt-subdivision of the pp-leaf star. In particular, S(3,t)S(3,t) is exactly St,t,tS_{t,t,t}.

We will need the following two technical results shown by Lozin and Razgon [40].

Lemma 11 ([40, Lemma 2]).

There is a function c:×c:\mathbb{N}\times\mathbb{N}\to\mathbb{N} with the following property. If a graph GG contains a collection of c(a,s)c(a,s) pairwise disjoint subsets of vertices, each of size at most aa and with at least one edge between any two of them, then GG contains Ks,sK_{s,s} as a subgraph.

Lemma 12 ([40, Theorem 3]).

There is a function m:×m:\mathbb{N}\times\mathbb{N}\to\mathbb{N} with the following property. Every graph GG containing a (t)(\leqslant t)-subdivision of Km(h,t)K_{m(h,t)} as a subgraph contains either Ks,sK_{s,s} as a subgraph or a proper (t)(\leqslant t)-subdivision of Kh,hK_{h,h} 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 d,t,r,sd,t,r,s there exists k=k(d,t,r,s)k=k(d,t,r,s) such that the following holds. If GG contains a kk-block, then it contains:

  1. 1.

    a (t)(\leqslant t)-subdivision of KrK_{r} as a subgraph or

  2. 2.

    Ks,s,K_{s,s,} as a subgraph, or

  3. 3.

    dSt,t,tdS_{t,t,t} as an induced subgraph.

Proof..

Let cc be the functions given by Lemma 11. Define the following constants

q=\displaystyle q= 2𝖱𝖺𝗆(d,c(3t+1,s)),\displaystyle~{}2\;\mathsf{Ram}(d,c(3t+1,s)),
p=\displaystyle p= (q/21)(3t+1)+3,\displaystyle~{}(q/2-1)(3t+1)+3,
=\displaystyle\ell= 𝖱𝖺𝗆(p,c(t,s)),\displaystyle~{}\mathsf{Ram}(p,c(t,s)),
k=\displaystyle k^{\prime}= max(r+q,t(r2)+),\displaystyle~{}\max\left(r+q,t\binom{r}{2}+\ell\right),
k=\displaystyle k= 2k.\displaystyle~{}2k^{\prime}.

Suppose that GG has a kk-block BB but no subgraph isomorphic to Ks,sK_{s,s} nor induced subgraph isomorphic to dSt,t,tdS_{t,t,t}. We aim to show that GG has a (t)(\leqslant t)-subdivision of KrK_{r} as a subgraph.

Consider a pair {x,y}\{x,y\} of distinct vertices from BB. Since xx and yy cannot be separated by deleting fewer than kk vertices, by Menger’s theorem there is a set 𝒫(x,y)\mathcal{P}(x,y) of kk internally pairwise disjoint xx-yy-paths in GG. Without loss of generality we can assume that each such path is induced.

Claim 1.

There is a subset BBB^{\prime}\subseteq B of size k=k/2k^{\prime}=k/2 with the following property. For each pair of distinct vertices x,yBx,y\in B^{\prime} there is a set 𝒫(x,y)\mathcal{P^{\prime}}(x,y) of kk^{\prime} internally pairwise disjoint induced xx-yy-paths in GG that do not contain any vertices from B{x,y}B^{\prime}\setminus\{x,y\}.

Proof of Claim..

Let BB^{\prime} be any set of kk^{\prime} vertices from BB. Consider any two distinct vertices x,yBx,y\in B^{\prime}. As the interiors of paths in 𝒫(x,y)\mathcal{P}(x,y) are pairwise disjoint, we observe that at most kk^{\prime} paths might contain vertices from B{x,y}B^{\prime}\setminus\{x,y\}. Thus 𝒫(x,y)\mathcal{P}(x,y) contains at least kk=kk-k^{\prime}=k^{\prime} paths that do not intersect B{x,y}B^{\prime}\setminus\{x,y\}. \lrcorner

Fix a pair {x,y}\{x,y\} of distinct vertices from BB^{\prime}. A path in 𝒫(x,y)\mathcal{P^{\prime}}(x,y) is long if it has at least tt internal vertices. The pair {x,y}\{x,y\} is distant if at least \ell paths in 𝒫(x,y)\mathcal{P^{\prime}}(x,y) are long.

Let QBQ\subseteq B^{\prime} be a minimum-size subset of BB^{\prime} that intersects all distant pairs. (It might be useful to think of QQ as the minimum vertex cover in a graph with vertex set BB^{\prime} where the edges correspond to distant pairs.) We consider two cases, depending whether QQ is “large” or “small”.

Case 1 (large 𝑸Q): |𝑸|𝒒|Q|\geqslant q.

In this case there is a set consisting of at least |Q|/2𝖱𝖺𝗆(d,c(3t+1,s))|Q|/2\geqslant\mathsf{Ram}(d,c(3t+1,s)) pairwise disjoint distant pairs (i.e., a matching in the mentioned graph). Let MM be such a set of size exactly 𝖱𝖺𝗆(d,c(3t+1,s))\mathsf{Ram}(d,c(3t+1,s)).

Claim 2.

For each {x,y}M\{x,y\}\in M, there is a set S{x,y}S^{\{x,y\}} contained in the union of paths in 𝒫(x,y)\mathcal{P^{\prime}}(x,y) that induces S(p,t)S(p,t) in GG.

Proof of Claim..

Fix {x,y}M\{x,y\}\in M. For a long path P𝒫(x,y)P\in\mathcal{P^{\prime}}(x,y), let 𝗉𝗋𝖾𝖿𝗂𝗑(P)\mathsf{prefix}(P) be the set of tt first vertices of PP, starting from the side of xx, but excluding xx itself. So 𝗉𝗋𝖾𝖿𝗂𝗑(P)\mathsf{prefix}(P) induces a tt-vertex path, and 𝗉𝗋𝖾𝖿𝗂𝗑(P){x}\mathsf{prefix}(P)\cup\{x\} induces a (t+1)(t+1)-vertex path in GG.

Fix any set 𝒫long\mathcal{P^{\prime}}_{\text{long}} of \ell long paths from 𝒫(x,y)\mathcal{P^{\prime}}(x,y), and define Z={𝗉𝗋𝖾𝖿𝗂𝗑(P)|P𝒫long}Z=\{\mathsf{prefix}(P)~{}|~{}P\in\mathcal{P^{\prime}}_{\text{long}}\}. Consider an auxiliary graph 𝐙\mathbf{Z} with vertex set ZZ, 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 ZZ are pairwise disjoint).

If 𝐙\mathbf{Z} contains a clique of size c(t,s)c(t,s), then, by lemma 11, we obtain a subgraph isomorphic to Ks,sK_{s,s} in GG, a contradiction. Thus, since |Z|==𝖱𝖺𝗆(p,c(t,s))|Z|=\ell=\mathsf{Ram}(p,c(t,s)) we obtain an independent set of size pp in 𝐙\mathbf{Z}. The pp paths forming this independent set, together with xx, induce the desired copy of S(p,t)S(p,t). \lrcorner

Observe that even though the pairs in MM are pairwise disjoint, the sets S{x,y}S^{\{x,y\}} might still intersect. In the next claim we extract from them induced copies of St,t,tS_{t,t,t} that are pairwise disjoint.

Claim 3.

For each {x,y}M\{x,y\}\in M there is an induced copy of St,t,tS_{t,t,t} contained in S{x,y}S^{\{x,y\}}, such that for any distinct {x,y},{x,y}M\{x,y\},\{x^{\prime},y^{\prime}\}\in M^{\prime} the corresponding copies of St,t,tS_{t,t,t} are disjoint.

Proof of Claim..

Fix an arbitrary order {x1,y1},,{x|M|,y|M|}\{x_{1},y_{1}\},\ldots,\{x_{|M|},y_{|M|}\} on pairs in MM. The proof is by induction.

The copy of St,t,tS_{t,t,t} for {x1,y1}\{x_{1},y_{1}\} can be selected by picking any three out of pp paths in the copy of S(p,t)S(p,t) given by 2. So let i[1,|M|1]i\in[1,|M|-1] and suppose that for each 1ji1\leqslant j\leqslant i we have selected an induced copy SjS^{j} of St,t,tS_{t,t,t} contained in S{xj,yj}S^{\{x_{j},y_{j}\}}. Note that in total we have selected i(3t+1)i\cdot(3t+1) vertices. Consider the pair {xi+1,yi+1}\{x_{i+1},y_{i+1}\} and let P1,,PpP^{1},\ldots,P^{p} be the tt-vertex paths obtained from S{xi+1,yi+1}S^{\{x_{i+1},y_{i+1}\}} by deleting xi+1x_{i+1}. Note that at most i(3t+1)(|M|1)(3t+1)i(3t+1)\leqslant(|M|-1)(3t+1) of these paths might intersect S1SiS^{1}\cup\ldots\cup S^{i}. So, as p=(|M|1)(3t+1)+3p=(|M|-1)(3t+1)+3, there is always a choice of three paths that are disjoint with S1SiS^{1}\cup\ldots\cup S^{i}. Recall that by 1 the vertex xi+1x_{i+1} is not contained 1jiSj\bigcup_{1\leqslant j\leqslant i}S^{j}. Thus the vertices of the three selected paths, together with xi+1x_{i+1}, form the set Si+1S^{i+1}. \lrcorner

Now we proceed similarly as in the proof of 2. Let SS be the family consisting of induced copies of St,t,tS_{t,t,t} for all {x,y}M\{x,y\}\in M; they are given by 3. Let 𝐒\mathbf{S} be the graph with vertex set SS where two vertices are adjacent if and only if there is an edge from one set to another. Note that |S|=𝖱𝖺𝗆(d,c(3t+1,s))|S|=\mathsf{Ram}(d,c(3t+1,s)). An independent set in 𝐒\mathbf{S} of size dd corresponds to an induced dSt,t,tdS_{t,t,t} in GG. On the other hand, if 𝐒\mathbf{S} has a clique of size at least c(3t+1,s)c(3t+1,s), then by lemma 11 we obtain Ks,sK_{s,s} as a subgraph of GG. By the choice of |S||S| one of these cases must happen, and both yield a contradiction. Thus we conclude that Case 1 cannot occur.

Case 2 (small 𝑸Q): 𝑸<𝒒Q<q.

Define B′′=BQB^{\prime\prime}=B^{\prime}\setminus Q; note that |B′′|kqr|B^{\prime\prime}|\geqslant k^{\prime}-q\geqslant r. For any distinct x,yB′′x,y\in B^{\prime\prime}, let 𝒫′′(x,y)\mathcal{P}^{\prime\prime}(x,y) be obtained from 𝒫(x,y)\mathcal{P}^{\prime}(x,y) by removing all long paths. By the definition of QQ we observe that for all x,yx,y we have |𝒫′′(x,y)|kt(r2)|\mathcal{P}^{\prime\prime}(x,y)|\geqslant k^{\prime}-\ell\geqslant t\binom{r}{2}.

Let RR be any rr-element subset of B′′B^{\prime\prime}.

Claim 4.

For each pair {x,y}\{x,y\} of distinct vertices of RR there is a path P{x,y}𝒫′′(x,y)P^{\{x,y\}}\in\mathcal{P^{\prime\prime}}(x,y) 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 RR as {x1,y1},,{x(r2),y(r2)}\{x_{1},y_{1}\},\ldots,\{x_{\binom{r}{2}},y_{\binom{r}{2}}\}.

The path P{x1,y1}P^{\{x_{1},y_{1}\}} can be arbitrarily chosen from 𝒫′′(x1,y1)\mathcal{P^{\prime\prime}}(x_{1},y_{1}). Suppose that we have selected paths P{x1,y1},,P{xi,yi}P^{\{x_{1},y_{1}\}},\ldots,P^{\{x_{i},y_{i}\}} for some 1i<(r2)1\leqslant i<\binom{r}{2}. Since each path is short, the selected paths have in total at most it<t(r2)i\cdot t<t\binom{r}{2} internal vertices.

Now consider the set 𝒫′′(xi+1,yi+1)\mathcal{P^{\prime\prime}}(x_{i+1},y_{i+1}). Recall that the paths in this set are pairwise disjoint. Since |𝒫′′(xi+1,yi+1)|t(r2)|\mathcal{P^{\prime\prime}}(x_{i+1},y_{i+1})|\geqslant t\binom{r}{2}, we observe that there is a path in 𝒫′′(xi+1,yi+1)\mathcal{P^{\prime\prime}}(x_{i+1},y_{i+1}) which is internally disjoint with all previously selected paths. We pick this path as P{xi+1,yi+1}P^{\{x_{i+1},y_{i+1}\}}. \lrcorner

Recall that for each distinct x,yRx,y\in R, the path P{x,y}P^{\{x,y\}} does not contain vertices from R{x,y}R\setminus\{x,y\}. Thus the set RR with the paths given by 4 forms a (t)(\leqslant t)-subdivision of KrK_{r} which is a subgraph of GG. This concludes the claim. \square

Now we can proceed to the proof of lemma 5.

Proof. (of lemma 5.).

Define h=d(3t+1)h=d(3t+1) (i.e., the number of vertices of dSt,t,tdS_{t,t,t}). Define r=m(h,t)r=m(h,t), where mm is as in lemma 12, and let k=k(d,t,r,s)k=k(d,t,r,s) be as in lemma 13. For contradiction, suppose that GG has a kk-block. By lemma 13 we conclude that GG contains a (t)(\leqslant t)-subdivision of KrK_{r} as a subgraph. By lemma 12 we observe that GG contains a proper (t)(\leqslant t)-subdivision of Kh,hK_{h,h} as an induced subgraph.

It is straightforward to verify that this induced subgraph contains a subdivision of dSt,t,tdS_{t,t,t} (in fact, of any graph with at most hh vertices and at most hh edges) as an induced subgraph. Therefore GG contains an induced subgraph isomorphic to dSt,t,tdS_{t,t,t}, a contradiction. \square

Appendix B Appendix: Proof of Lemma 7

Iterate over every ITTI_{T}\subseteq T. For fixed ITI_{T}, we aim at computing fG,𝐰,T(IT)f_{G,\mathbf{w},T}(I_{T}). If ITI_{T} is not independent, we set fG,𝐰,T(IT)=f_{G,\mathbf{w},T}(I_{T})=-\infty. In the remainder of the proof, we show how to compute in polynomial time the value fG,𝐰,T(IT)f_{G,\mathbf{w},T}(I_{T}) for fixed independent ITTI_{T}\subseteq T.

For a particle AA of (H,η)(H,\eta), let a(A):=fG[A],𝐰,TA(ITA)a(A):=f_{G[A],\mathbf{w},T\cap A}(I_{T}\cap A) and let I(A)I(A) be an independent set witnessing this value, that is, an independent set in G[A]G[A] of weight 𝐰(a(A))\mathbf{w}(a(A)) with I(A)TA=ITAI(A)\cap T\cap A=I_{T}\cap A. Note that as ITI_{T} is independent, the value a(A)a(A) is not equal to -\infty and such an independent set exists.

We say that xV(H)x\in V(H) is forced if ITyNH(x)η(xy,x)I_{T}\cap\bigcup_{y\in N_{H}(x)}\eta(xy,x)\neq\emptyset. Note that since ITI_{T} is independent, if xx is forced, then η(xy,x)IT\eta(xy,x)\cap I_{T}\neq\emptyset for exactly one edge xyxy incident with xx. We call such an edge xyxy the enforcer of xx. Note that an edge xyxy may be the enforcer of both xx and yy.

The arguments now follow very closely the outline of Section 3.3 of [15].

We construct a set 𝒫\mathcal{P} of particles and an edge-weighted graph (H,𝐰)(H^{\prime},\mathbf{w}^{\prime}) as follows. We start with 𝒫=\mathcal{P}=\emptyset, V(H)=V(H)V(H^{\prime})=V(H), and E(H)=E(H^{\prime})=\emptyset.

For every xV(H)x\in V(H) that is not forced, add AxA_{x} to 𝒫\mathcal{P}. For every xyzT(H)xyz\in T(H) such that neither of the edges xyxy, yzyz, or xzxz is the enforcer of both its endpoints, add AxyzA_{xyz} to 𝒫\mathcal{P}. For every e=xyE(H)e=xy\in E(H), proceed as follows.

  1. 1.

    If neither xx nor yy is forced, we add to HH^{\prime} a new vertex tet_{e} and edges text_{e}x, teyt_{e}y, xyxy, and set the edge weights 𝐰\mathbf{w}^{\prime} as follows:

    𝐰(tex)\displaystyle\mathbf{w}^{\prime}(t_{e}x) :=a(Axyx)a(Axy)a(Ax),\displaystyle:=a(A_{xy}^{x})-a(A_{xy}^{\bot})-a(A_{x}),
    𝐰(tey)\displaystyle\mathbf{w}^{\prime}(t_{e}y) :=a(Axyy)a(Axy)a(Ay),\displaystyle:=a(A_{xy}^{y})-a(A_{xy}^{\bot})-a(A_{y}),
    𝐰(xy)\displaystyle\mathbf{w}^{\prime}(xy) :=a(Axyxy)a(Axy)a(Ax)a(Ay)z, s.t. xyzT(H)a(Axyz).\displaystyle:=a(A_{xy}^{xy})-a(A_{xy}^{\bot})-a(A_{x})-a(A_{y})-\sum_{z,\text{ s.t. }xyz\in T(H)}a(A_{xyz}).

    Furthermore, add AxyA_{xy}^{\bot} to 𝒫\mathcal{P}.

  2. 2.

    If exactly one of xx and yy is forced, say w.l.o.g. xx is forced and yy is not forced, proceed as follows.

    1. (a)

      If xyxy is the enforcer of xx, then add to HH^{\prime} an edge xyxy with weight

      𝐰(xy):=a(Axyxy)a(Axyx)a(Ay)z, s.t. xyzT(H)a(Axyz).\mathbf{w}^{\prime}(xy):=a(A_{xy}^{xy})-a(A_{xy}^{x})-a(A_{y})-\sum_{z,\text{ s.t. }xyz\in T(H)}a(A_{xyz}).

      Furthermore, add AxyxA_{xy}^{x} to 𝒫\mathcal{P}.

    2. (b)

      If xyxy is not the enforcer of xx, then add to HH^{\prime} a new vertex tet_{e} and an edge teyt_{e}y with weight

      𝐰(tey):=a(Axyy)a(Axy)a(Ay).\mathbf{w}^{\prime}(t_{e}y):=a(A_{xy}^{y})-a(A_{xy}^{\bot})-a(A_{y}).

      Furthermore, add AxyA_{xy}^{\bot} to 𝒫\mathcal{P}.

  3. 3.

    If both xx and yy are forced, proceed as follows.

    1. (a)

      If xyxy is neither the enforcer of xx nor of yy, add AxyA_{xy}^{\bot} to 𝒫\mathcal{P}.

    2. (b)

      If xyxy is the enforcer of xx but not of yy add AxyxA_{xy}^{x} to 𝒫\mathcal{P}.

    3. (c)

      If xyxy is the enforcer of yy but not of xx add AxyyA_{xy}^{y} to 𝒫\mathcal{P}.

    4. (d)

      If xyxy is the enforcer of both xx and yy, add AxyxyA_{xy}^{xy} to 𝒫\mathcal{P}.

This finishes the description of the construction of 𝒫\mathcal{P} and (H,𝐰)(H^{\prime},\mathbf{w}^{\prime}). In the next two paragraphs we make two observations that follow by a direct check from the definitions.

Observe that I0:=A𝒫I(A)I_{0}:=\bigcup_{A\in\mathcal{P}}I(A) is independent in GG and has weight a0:=A𝒫a(A)a_{0}:=\sum_{A\in\mathcal{P}}a(A). Furthermore, for every A𝒫A\in\mathcal{P}, we have I0AT=ITAI_{0}\cap A\cap T=I_{T}\cap A and I0T=ITI_{0}\cap T=I_{T}. We think of I0I_{0} as the “base” solution for fG,𝐰,T(IT)f_{G,\mathbf{w},T}(I_{T}).

Observe also that all weights 𝐰\mathbf{w}^{\prime} of HH^{\prime} are nonnegative, as AxyxA_{xy}^{x} contains both AxyA_{xy}^{\bot} and AxA_{x} while AxyxyA_{xy}^{xy} contains AxyA_{xy}^{\bot}, AxA_{x}, AyA_{y}, as well as all AxyzA_{xyz} for all triangles xyzxyz containing the edge xyxy.

We will be asking for a maximum weight matching in (H,𝐰)(H^{\prime},\mathbf{w}^{\prime}). Intuitively, taking an edge text_{e}x to such a matching corresponds to replacing in I0I_{0} the parts I(Axy)I(A_{xy}^{\bot}) and I(Ax)I(A_{x}) with the part I(Axyx)I(A_{xy}^{x}) while taking an edge xyxy to such a matching corresponds to replacing in I0I_{0} the parts I(Axy)I(A_{xy}^{\bot}), I(Ax)I(A_{x}), I(Ay)I(A_{y}) and all parts I(Axyz)I(A_{xyz}) for triangles xyzxyz containing the edge xyxy with part I(Axyxy)I(A_{xy}^{xy}).

From another perspective, fix xV(H)x\in V(H) and recall that the sets η(xy,x)\eta(xy,x) for yNH(x)y\in N_{H}(x) are complete to each other. Hence, any independent set in GG can contain vertices in at most one of such sets. For an edge e=xyE(H)e=xy\in E(H), taking an edge xyxy or text_{e}x in a matching in HH^{\prime} corresponds to choosing that, among all neighbors of xx in HH, the neighbor yy is such that the set η(xy,x)\eta(xy,x) is allowed to contain vertices of the sought independent set. (Choosing xyE(H)xy\in E(H^{\prime}) to the matching corresponds to allowing both η(xy,x)\eta(xy,x) and η(yx,y)\eta(yx,y) to contain vertices of the sought independent set.)

However, there is a delicacy if ITI_{T} contains a vertex of some interface η(xy,x)\eta(xy,x). Then, in some sense ITI_{T} already forces some choices in the corresponding matching in HH^{\prime}. This is modeled above by having alternate construction for vertices xV(H)x\in V(H) that are forced.

The following two claims prove that fG,𝐰,T(IT)f_{G,\mathbf{w},T}(I_{T}) equals a0a_{0} plus the maximum possible weight of a matching in (H,𝐰)(H^{\prime},\mathbf{w}^{\prime}) 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 II be an independent set in GG with IT=ITI\cap T=I_{T}. Let MM be the set of edges of HH^{\prime} defined as follows: for every e=xyE(H)e=xy\in E(H), if η(xy,x)I\eta(xy,x)\cap I\neq\emptyset and η(xy,y)I\eta(xy,y)\cap I\neq\emptyset, then xyMxy\in M, if η(xy,x)I\eta(xy,x)\cap I\neq\emptyset and η(xy,y)I=\eta(xy,y)\cap I=\emptyset, then texMt_{e}x\in M, and if η(xy,x)I=\eta(xy,x)\cap I=\emptyset and η(xy,y)I\eta(xy,y)\cap I\neq\emptyset, then teyMt_{e}y\in M. Then, all the above edges indeed exist in HH^{\prime} and MM is a matching. Furthermore, the weight of II is at most a0+eM𝐰(e)a_{0}+\sum_{e\in M}\mathbf{w}^{\prime}(e).

Claim 6.

Let MM be a matching in HH^{\prime}. Let 𝒫M\mathcal{P}_{M} be the set of particles of (H,η)(H,\eta) defined as follows. Start with 𝒫M:=𝒫\mathcal{P}_{M}:=\mathcal{P} and then for every edge e=xyE(H)e=xy\in E(H),

  • if xyMxy\in M, insert AxyxyA_{xy}^{xy} into 𝒫M\mathcal{P}_{M} and remove from 𝒫M\mathcal{P}_{M} the following particles if present: AxyxA_{xy}^{x}, AxyyA_{xy}^{y}, AxyA_{xy}^{\bot}, AxA_{x}, AyA_{y}, AxyzA_{xyz} for any zV(H)z\in V(H) such that xyzT(H)xyz\in T(H).

  • if texMt_{e}x\in M (resp. teyMt_{e}y\in M), insert AxyxA_{xy}^{x} (resp. AxyyA_{xy}^{y}) into 𝒫M\mathcal{P}_{M}, and remove from 𝒫M\mathcal{P}_{M} the following particles if present: AxyA_{xy}^{\bot} and AxA_{x} (resp. AyA_{y}).

Then IM:=A𝒫MI_{M}:=\bigcup_{A\in\mathcal{P}_{M}} is an independent set in GG with IMT=ITI_{M}\cap T=I_{T} and of weight at least a0+eM𝐰(e)a_{0}+\sum_{e\in M}\mathbf{w}^{\prime}(e).