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Fixed-Treewidth-Efficient Algorithms for Edge-Deletion to Interval Graph Classesthanks: This manuscript corrects some errors in [24].

Toshiki Saitoh Kyushu Institute of Technology, [email protected]    Ryo Yoshinaka Tohoku University, [email protected]    Hans L. Bodlaender Utrecht University, [email protected]
Abstract

For a graph class 𝒞\mathcal{C}, the 𝒞\mathcal{C}-Edge-Deletion problem asks for a given graph GG to delete the minimum number of edges from GG in order to obtain a graph in 𝒞\mathcal{C}. We study the 𝒞\mathcal{C}-Edge-Deletion problem for 𝒞\mathcal{C} the class of interval graphs and other related graph classes. It follows from Courcelle’s Theorem that these problems are fixed parameter tractable when parameterized by treewidth. In this paper, we present concrete FPT algorithms for these problems. By giving explicit algorithms and analyzing these in detail, we obtain algorithms that are significantly faster than the algorithms obtained by using Courcelle’s theorem.

1 Introduction

Intersection graphs are represented by geometric objects aligned in certain ways so that each object corresponds to a vertex and two objects intersect if and only if the corresponding vertices are adjacent. Intersection graphs are well-studied in the area of graph algorithms since there are many important applications and we can solve many NP-hard problems in general graphs in polynomial time on such graph classes. Interval graphs are intersection graphs which are represented by intervals on a line. Clique, Independent Set, and Coloring on interval graphs can be solved in linear time and interval graphs have many applications in bioinformatics, scheduling, and so on. See [14, 3, 26] for more details of interval graphs and other intersection graphs.

Graph modification problems on a graph class 𝒞\mathcal{C} are to find a graph in 𝒞\mathcal{C} by modifying a given graph in certain ways. 𝒞\mathcal{C}-Vertex-Deletion, 𝒞\mathcal{C}-Edge-Deletion, and 𝒞\mathcal{C}-Completion are to find a graph in 𝒞\mathcal{C} by deleting vertices, deleting edges, and adding edges, respectively, with the minimum cost. These problems can be seen as generalizations of many NP-hard problems. Clique is equivalent to Complete-Vertex-Deletion: we find a complete graph by deleting the smallest number of vertices. Modification problems on intersection graph classes also have many applications. For example, Interval-Vertex/Edge-Deletion problems have applications to DNA (physical) mapping [13, 12, 27]. Lewis and Yannakakis showed that 𝒞\mathcal{C}-Vertex-Deletion is NP-complete for any nontrivial hereditary graph class [17]. A graph class 𝒞\mathcal{C} is hereditary if for any graph in 𝒞\mathcal{C}, every induced subgraph of the graph is also in 𝒞\mathcal{C}. Since the class of intersection graphs are hereditary, 𝒞\mathcal{C}-Vertex Deletion is NP-complete for any nontrivial intersection graph class 𝒞\mathcal{C}. The problems 𝒞\mathcal{C}-Edge-Deletion are also NP-hard when 𝒞\mathcal{C} is the class of perfect, chordal, split, circular arc, chain [23], interval, proper interval [13], trivially perfect (a.k.a. nested interval) [25], threshold [20], permutation, weakly chordal, or circle graphs [4]. See the lists in [19, 4].

Parameterized complexity is well-studied in the area of computer science. A problem with a parameter kk is fixed parameter tractable, FPT for short, if there is an algorithm running in f(k)ncf(k)n^{c} time where nn is the size of input, ff is a computable function and cc is a constant. Such an algorithm is called an FPT algorithm. The treewidth 𝑡𝑤(G)\mathit{tw}(G) of a graph GG represents treelikeness and is one of the most important parameters in parameterized complexity concerning graph algorithms. For many NP-hard problems in general, there are tons of FPT algorithms with parameter 𝑡𝑤(G)\mathit{tw}(G) by dynamic programming on tree decompositions. Finding the treewidth of an input graph is NP-hard and it is known that Chordal-Completion with minimizing the size of the smallest maximum clique is equivalent to the problem. There is an FPT algorithm for computing the treewidth of a graph by Bodlaender [2] which runs in O(f(𝑡𝑤(G))(n+m))O(f(\mathit{tw}(G))(n+m)) time where nn and mm are the numbers of vertices and edges of a given graph: i.e., the running time is linear in the size of input. Courcelle showed that every problem that can be expressed in monadic second order logic (MSO2) has a linear time algorithm on graphs of bounded treewidth [9]. Some intersection graph classes, for example interval graphs, proper interval graphs, chordal graphs, and permutation graphs, can be represented by MSO2 [8] and thus there are FPT algorithms for Edge-Deletion problems on such graph classes. However, the algorithms obtained by Courcelle’s theorem have a very large hidden constant factor even when the treewidth is very small, since the running time is the exponential tower of the coding size of the MSO2 expression.

Our results: We propose concrete FPT algorithms for Edge-Deletion to interval graphs and other related graph classes, when parameterized by the treewidth of the input graph. Our algorithms virtually compute a set of edges SS with the minimum size such that GSG-S is in a graph class 𝒞\mathcal{C} by using dynamic programming on a tree-decomposition. We maintain possible alignments of geometric objects corresponding to vertices in the bag of each node of the tree-decomposition. Alignments of the objects of forgotten vertices are remembered only relatively to the objects of the current bag. If two forgotten objects have the same relative position to the objects of the current bag, we remember only the fact that there is at least one forgotten object at that position. In this way, we achieve the fixed-parameter-tractability, while guaranteeing that no object pairs of non-adjacent vertices of the input graph will intersect in our dynamic programming algorithm. Our algorithms run in O(f(𝑡𝑤(G))(n+m))O(f(\mathit{tw}(G))\cdot(n+m)) time where nn and mm are the numbers of vertices and edges of the input graph. Our explicit algorithms are significantly faster than those obtained by using Courcelle’s theorem. We also analyze the time complexity of our algorithms parameterized by pathwidth which is analogous to treewidth. The relation among the graph classes for which this paper provides 𝒞\mathcal{C}-Edge-Deletion algorithms is shown in Figure 1.

Circular ArcIntervalProper IntervalTrivially PerfectThresholdPermutation
Figure 1: The graph classes of which this paper presents algorithms for the edge-deletion problems.

Related works: Another kind of common parameters considered in parameterized complexity of graph modification problems is the number of vertices or edges to be removed or to be added. Here we review preceding studies on those problems for intersection graphs with those parameters.

Concerning parameterized complexity of 𝒞\mathcal{C}-Vertex-Deletion, Hof et al. proposed an FPT algorithm for Proper-Interval-Vertex-Deletion [28], and Marx proposed an FPT algorithm for Chordal-Vertex-Deletion [21]. Heggernes et al. showed Perfect-Vertex-Deletion and Weakly-Chordal-Vertex-Deletion are W[2]-hard [15]. Cai showed that CC-Vertex/Edge-Deletion are FPT when 𝒞\mathcal{C} is characterized by a finite set of forbidden induced subgraphs [5].

For modification problems on interval graphs, Villanger et al. presented an FPT algorithm for Interval-Completion [29], and Cao and Marx presented an FPT algorithm for Interval-Vertex-Deletion [7]. Cao improved these algorithms and developed an FPT algorithm for Edge-Deletion [6].

It is known that Threshold-Edge-Deletion, Chain-Edge-Deletion and Trivially-Perfect-Edge-Deletion are FPT, since threshold graphs, chain graphs and trivially perfect graphs are characterized by a finite set of forbidden induced subgraphs [5]. Nastos and Gao presented faster algorithms for the problems [22], and Liu et al. improved their algorithms to O(2.57k(n+m))O(2.57^{k}(n+m)) and O(2.42k(n+m))O(2.42^{k}(n+m)) using modular decomposition trees [18], where kk is the number of deleted edges. There are algorithms to find a polynomial kernel for Chain-Edge-Deletion and Trivially Perfect-Edge-Deletion [1, 11].

Organization of this article: Section 2 prepares the notation and definitions used in this paper. We propose an FPT algorithm for Interval-Edge-Deletion in Section 3. We then extend the algorithm related to the interval graphs in Section 4. We conclude this paper and provide some open questions in Section 6.

2 Preliminaries

For a set XX, its cardinality is denoted by |X||X|. A partition of XX is a tuple (X1,,Xk)(X_{1},\dots,X_{k}) of subsets of XX such that X=X1XkX=X_{1}\cup\dots\cup X_{k} and XiXj=X_{i}\cap X_{j}=\emptyset if 1i<jk1\leq i<j\leq k, where we allow some of the subsets to be empty. For entities x,y,zXx,y,z\in X, we let x[y/z]=yx[y/z]=y if x=zx=z, and x[y/z]=xx[y/z]=x otherwise. For a subset YXY\subseteq X, define Y[y/z]={x[y/z]xY}Y[y/z]=\{\,x[y/z]\mid x\in Y\,\}.

For a linear order π\pi over a finite set XX, the maximum and the minimum elements of XX w.r.t. π\pi are denoted by maxπX\max_{\pi}X and minπX\min_{\pi}X, respectively. We denote the successor of xXx\in X w.r.t. π\pi by 𝗌𝗎𝖼π(x)=minπ{yXx<πy}\mathsf{suc}_{\pi}(x)=\min_{\pi}\{\,y\in X\mid x<_{\pi}y\,\}. Note that 𝗌𝗎𝖼π(maxπX)\mathsf{suc}_{\pi}(\max_{\pi}X) is undefined. Similarly 𝗉𝗋𝖾𝖽π(x)=maxπ{yXy<πx}\mathsf{pred}_{\pi}(x)=\max_{\pi}\{\,y\in X\mid y<_{\pi}x\,\} denotes the predecessor of xx.

A simple graph G=(V,E)G=(V,E) is a pair of vertex and edge sets, where each element of EE is a subset of VV consisting of exactly two elements.

A tree-decomposition of G=(V,E)G=(V,E) is a tree TT such that111We use the terms “vertices” for an input graph and “nodes” for a tree-decomposition.

  • to each node of TT a subset of VV is assigned,

  • if the assigned sets of two nodes of TT contain a vertex uVu\in V, then so does every node on the path between the two nodes,

  • for each {u,v}E\{u,v\}\in E, there is a node of TT whose assigned set includes both uu and vv.

The width of a tree-decomposition is the maximum cardinality of the assigned sets minus one and the treewidth of a graph is the smallest width of its tree-decompositions. A tree-decomposition is said to be nice if it is rooted, the root is assigned the empty set, and its nodes are grouped into the following four:

  • leaf nodes, which have no children and are assigned the empty set,

  • introduce nodes, each of which has just one child, where the set assigned to the node adds one vertex to the child’s set,

  • forget nodes, each of which has just one child, where the set assigned to the node removes one vertex from the child’s set,

  • join nodes, each of which has just two children, where the same vertex set is assigned to the node and its children.

It is known that every tree-decomposition has a nice tree-decomposition of the same width whose size is O(k|V|)O(k|V|), where kk is the treewidth of the tree-decomposition [16, 10]. Hereafter, under a fixed graph GG and a fixed nice tree-decomposition TT, we let XsX_{s} denote the subset of VV assigned to a node ss of a tree-decomposition and XsX_{\leq s} denote the union of all the subsets assigned to the node ss and its descendant nodes. We call vertices in XsX_{s} and in XsXsX_{\leq s}-X_{s} active and forgotten, respectively. Moreover, we define Es={{u,v}Eu,vXs}E_{s}=\{\,\{u,v\}\in E\mid u,v\in X_{s}\,\} and Es={{u,v}Eu,vXs}E_{\leq s}=\{\,\{u,v\}\in E\mid u,v\in X_{\leq s}\,\}. Given a tree decomposition of treewidth kk, we can compute a nice tree-decomposition with treewidth kk and O(kn)O(kn) nodes in O(k2(n+m))O(k^{2}(n+m)) time [10].

A tree-decomposition is called a path-decomposition if the tree is a path. The pathwidth of a graph is the smallest width of its path-decompositions. Every path-decomposition has a nice path-decomposition of the same pathwidth, which consists of leaf, introduce, forget, but not join nodes.

The problem we tackle in this paper is given as follows.

Definition 1.

For a graph class 𝒞\mathcal{C}, the 𝒞\mathcal{C}-Edge-Deletion is a problem to find the minimum natural number cc such that there is a subgraph G=(V,E)G^{\prime}=(V,E^{\prime}) of GG with G𝒞G^{\prime}\in\mathcal{C} and |E||E|=c|E|-|E^{\prime}|=c for an input simple graph G=(V,E)G=(V,E).

In the succeeding sections, for different classes 𝒞\mathcal{C} of intersection graphs, we present algorithms for 𝒞\mathcal{C}-Edge-Deletion that run in linear time in the input graph size when the treewidth is bounded. We assume that the algorithm takes a nice tree-decomposition TT of GG in addition as input [2, 16]. Our algorithms are dynamic programming algorithms that recursively compute solutions (and some auxiliary information) for the subproblems on (Xs,Es)(X_{\leq s},E_{\leq s}) for each node ss in the given tree-decomposition from leaves to the root.

3 Finding a Largest Interval Subgraph

An interval representation π\pi over a set XX is a linear order over the set 𝐿𝑅X=LXRX{,}\mathit{LR}_{X}=\mathit{L}_{X}\cup\mathit{R}_{X}\cup\{\bot,\top\} with LX={lxxX}\mathit{L}_{X}=\{\,l_{x}\mid x\in X\,\} and RX={rxxX}\mathit{R}_{X}=\{\,r_{x}\mid x\in X\,\} such that <πlx<πrx<π\bot<_{\pi}l_{x}<_{\pi}r_{x}<_{\pi}\top for all xXx\in X. An interval is a pair (p,q)𝐿𝑅X×𝐿𝑅X(p,q)\in\mathit{LR}_{X}\times\mathit{LR}_{X}. We say that two intervals (p1,q1)(p_{1},q_{1}) and (p2,q2)(p_{2},q_{2}) intersect, denoted by (p1,q1)/\π(p2,q2)(p_{1},q_{1})\mathrel{\text{$/$\hbox to0.0pt{\hss$\backslash$}}}_{\pi}(p_{2},q_{2}) if p1<πq2p_{1}<_{\pi}q_{2} and p2<πq1p_{2}<_{\pi}q_{1}. Otherwise, we write (p1,q1)//π(p2,q2)(p_{1},q_{1})\mathrel{/\!/\!}_{\pi}(p_{2},q_{2}).

The interval graph GπG_{\pi} of an interval representation π\pi on VV is (V,π)(V,\mathcal{E}_{\pi}) where

π={{u,v}V(lu,ru)/\π(lv,rv) and uv}.\mathcal{E}_{\pi}=\{\,\{u,v\}\subseteq V\mid\text{$(l_{u},r_{u})\mathrel{\text{$/$\hbox to0.0pt{\hss$\backslash$}}}_{\pi}(l_{v},r_{v})$ and $u\neq v$}\,\}.

Figure 2 (a) and (b) show an example of an interval representation and the defined interval graph, respectively.

(a)ρ\rholxl_{x}rxr_{x}xxlyl_{y}ryr_{y}yylzl_{z}rzr_{z}zzlul_{u}rur_{u}uulvl_{v}rvr_{v}vvlwl_{w}rwr_{w}ww(c)(π,I)(\pi,I)lxl_{x}rxr_{x}xxlyl_{y}ryr_{y}yylzl_{z}rzr_{z}zz
(b)xxyyzzuuvvww
Figure 2: (a) Visualization of interval representation ρ\rho. (b) Interval graph GρG_{\rho}. (c) For Xs={x,y,z}X_{s}=\{x,y,z\} and XsXs={u,v,w}X_{\leq s}-X_{s}=\{u,v,w\}, we have [u]ρs={lu,ru}[u]_{\rho}^{s}=\{l_{u},r_{u}\}, [v]ρs=[w]ρs={lv,rv,lw,rw}[v]_{\rho}^{s}=[w]_{\rho}^{s}=\{l_{v},r_{v},l_{w},r_{w}\}, and 𝒜(ρ,s)=(π,I,0)\mathscr{A}(\rho,s)=(\pi,I,0) where I={(lx,lx),(rx,rz)}I=\{(l_{x},l_{x}),(r_{x},r_{z})\}.

This section presents an FPT algorithm for the interval edge deletion problem w.r.t. the treewidth. Let G=(V,E)G=(V,E) be an input graph and ss a node of a nice tree-decomposition TT of GG. On each node ss of TT, for each interval representation ρ\rho over XsX_{\leq s} that gives an interval subgraph of (Xs,Es)(X_{\leq s},E_{\leq s}), we would like to remember some pieces of information about ρ\rho, which we call the “abstraction” of ρ\rho. The abstraction is the triple (π,I,c)(\pi,I,c), where the linear order π\pi over 𝐿𝑅Xs\mathit{LR}_{X_{s}} is the restriction of ρ\rho to XsX_{s}, forgotten vertices are represented in II by anchoring the ends of intervals formed by forgotten vertices to active vertices, and cc counts the number of edges of EEρE-E_{\rho} such that at least one of their ends is forgotten. For each forgotten vertex wXsXsw\in X_{\leq s}-X_{s}, let the intersection closure of ww (w.r.t. ρ\rho and ss) be the smallest set [w]ρs𝐿𝑅XsXs[w]_{\rho}^{s}\subseteq\mathit{LR}_{X_{\leq s}-X_{s}} such that

  • lw,rw[w]ρsl_{w},r_{w}\in[w]_{\rho}^{s} and

  • if lu,ru[w]ρsl_{u},r_{u}\in[w]_{\rho}^{s}, vXsXsv\in X_{\leq s}-X_{s} and (lu,ru)/\ρ(lv,rv)(l_{u},r_{u})\mathrel{\text{$/$\hbox to0.0pt{\hss$\backslash$}}}_{\rho}(l_{v},r_{v}), then lv,rv[w]ρsl_{v},r_{v}\in[w]_{\rho}^{s}.

For an interval representation ρ\rho over XsX_{\leq s} such that GρG_{\rho} is a subgraph of (Xs,Es)(X_{\leq s},E_{\leq s}), we define the abstraction 𝒜(ρ,s)\mathscr{A}(\rho,s) of ρ\rho for ss to be the triple (π,I,c)(\pi,I,c) such that222Actually presence of intervals (,)(\bot,\bot) and (𝗉𝗋𝖾𝖽π(),𝗉𝗋𝖾𝖽π())(\mathsf{pred}_{\pi}(\top),\mathsf{pred}_{\pi}(\top)) in II does not matter.

  • π\pi is the restriction of ρ\rho to 𝐿𝑅Xs\mathit{LR}_{X_{s}},

  • I={(p,q)𝐿𝑅Xs×𝐿𝑅Xsthere is wXsXs such thatI=\{\,(p,q)\in\mathit{LR}_{X_{s}}\times\mathit{LR}_{X_{s}}\mid\text{there is $w\in X_{\leq s}-X_{s}$ such that}
                          p<ρminρ[w]ρs<ρ𝗌𝗎𝖼π(p)p<_{\rho}\min_{\rho}[w]_{\rho}^{s}<_{\rho}\mathsf{suc}_{\pi}(p) and q<ρmaxρ[w]ρs<ρ𝗌𝗎𝖼π(q)q<_{\rho}\max_{\rho}[w]_{\rho}^{s}<_{\rho}\mathsf{suc}_{\pi}(q) },

  • c=|EsρEs|c=|E_{\leq s}-\mathcal{E}_{\rho}-E_{s}|
    cc=|{{u,v}EsuXs{}=|\{\,\{u,v\}\in E_{\leq s}\mid u\notin{X_{s}} and (lu,ru)//ρ(lv,rv)}|(l_{u},r_{u})\mathrel{/\!/\!}_{\rho}(l_{v},r_{v})\,\}|.

We call elements of II forbidden intervals because introducing a new vertex interval intersecting a forbidden interval means making the new vertex and a forgotten vertex adjacent, which is forbidden. Figure 2 (c) shows an example of II. Since forbidden intervals are formed by intersection closures, no distinct forbidden intervals may intersect. If ss is the root of a nice tree-decomposition, the abstraction will be 𝒜(ρ,s)=(o,{(,)},cρ)\mathscr{A}(\rho,s)=(o,\{(\bot,\bot)\},c_{\rho}) where oo is the trivial order on {,}\{\bot,\top\} such that <o\bot<_{o}\top and cρ=|EEρ|c_{\rho}=|E-E_{\rho}|. That is, the smallest cρc_{\rho} for an interval subgraph GρG_{\rho} is the solution to our problem.

However, we do not have to compute 𝒜(ρ,s)\mathscr{A}(\rho,s) of all the possible interval representations ρ\rho over XsX_{\leq s} on each node ss. We say that (π,I,c)(\pi,I,c) dominates (π,I,c)(\pi^{\prime},I^{\prime},c^{\prime}) if and only if π=π\pi=\pi^{\prime}, IπII\sqsubseteq_{\pi}I^{\prime} and ccc\leq c^{\prime}, where we write IπII\sqsubseteq_{\pi}I^{\prime} if every forbidden interval of II is inside some forbidden interval of II^{\prime}: i.e., for every (p,q)I(p,q)\in I, there is (p,q)I(p^{\prime},q^{\prime})\in I^{\prime} such that pπpp^{\prime}\leq_{\pi}p and qπqq\leq_{\pi}q^{\prime}. In this case, every possible way of introducing new intervals to ρ\rho^{\prime} is also possible for ρ\rho by cheaper or equivalent cost. Therefore, it is enough to remember 𝒜(ρ,s)\mathscr{A}(\rho,s) discarding 𝒜(ρ,s)\mathscr{A}(\rho^{\prime},s). For a set of abstractions, the process of removing ones which are dominated by others is called reduction. We call a set of abstractions reduced if it has no pair of distinct elements such that one dominates the other.

Lemma 1.

Suppose 𝒜(ρ,s)=(π,I,c)\mathscr{A}(\rho,s)=(\pi,I,c) dominates 𝒜(ρ,s)=(π,I,c)\mathscr{A}(\rho^{\prime},s)=(\pi,I^{\prime},c^{\prime}). For any extension σ\sigma^{\prime} of ρ\rho^{\prime} such that σEs\mathcal{E}_{\sigma^{\prime}}\subseteq E_{\leq s}, there is an extension σ\sigma of ρ\rho such that σσEs\mathcal{E}_{\sigma^{\prime}}\subseteq\mathcal{E}_{\sigma}\subseteq E_{\leq s}.

Our algorithm calculates a reduced set s\mathscr{I}_{s} of abstractions of interval representations of interval subgraphs of (Xs,Es)({X}_{\leq s},{E}_{\leq s}) for each node ss of TT which satisfies the following invariant.

Condition 1.
  • Every element (π,I,c)s(\pi,I,c)\in\mathscr{I}_{s} is the abstraction of some interval representation of an interval subgraph of (Xs,Es)(X_{\leq s},E_{\leq s}) for XsX_{s},

  • Any interval representation ρ\rho of any interval subgraph of (Xs,Es)(X_{\leq s},E_{\leq s}) has an element of s\mathscr{I}_{s} that dominates its abstraction 𝒜(ρ,Xs)\mathscr{A}(\rho,X_{s}).

Since s\mathscr{I}_{s} is reduced, if Xs=X_{s}=\emptyset, we have s={(o,I,c)}\mathscr{I}_{s}=\{(o,I,c)\} for some I{(,)}I\subseteq\{(\bot,\bot)\} and cc\in\mathbb{N}, where oo is the trivial order such that <o\bot<_{o}\top. Particularly for the root node ss, the number cc is the least number such that one can obtain an interval subgraph by removing cc edges from GG. That is, cc is the solution to our problem. If ss is a leaf, s={(o,,0)}\mathscr{I}_{s}=\{(o,\emptyset,0)\} by definition. It remains to show how to calculate s\mathscr{I}_{s} from the child(ren) of ss, while preserving the invariant (Condition 1).

Introduce Node:

Suppose that ss is an introduce node. It has just one child tt such that Xs=Xt{x}X_{s}=X_{t}\cup\{x\}. For an extension π\pi^{\prime} of π\pi for (π,I,c)t(\pi,I,c)\in\mathscr{I}_{t}, let us anchor the new points lxl_{x} and rxr_{x} to points p0p_{0} and q0q_{0} in 𝐿𝑅Xt\mathit{LR}_{X_{t}}:

p0\displaystyle p_{0} =𝗉𝗋𝖾𝖽π(lx),\displaystyle=\mathsf{pred}_{\pi^{\prime}}(l_{x})\,,
q0\displaystyle q_{0} ={𝗉𝗋𝖾𝖽π(rx)if 𝗉𝗋𝖾𝖽π(rx)lx,𝗉𝗋𝖾𝖽π(lx)otherwise.\displaystyle=\begin{cases}\mathsf{pred}_{\pi^{\prime}}(r_{x})&\text{if $\mathsf{pred}_{\pi^{\prime}}(r_{x})\neq l_{x}$,}\\ \mathsf{pred}_{\pi^{\prime}}(l_{x})&\text{otherwise.}\end{cases}

We say π\pi^{\prime} respects EE and II if

  • {x,u}E\{x,u\}\notin E for uXtu\in X_{t} implies (lx,rx)//π(lu,ru)(l_{x},r_{x})\mathrel{/\!/\!}_{\pi^{\prime}}(l_{u},r_{u}),

  • (p,q)I(p,q)\in I implies (p0,q0)//π(p,q)(p_{0},q_{0})\mathrel{/\!/\!}_{\pi}(p,q),

respectively. If π\pi^{\prime} does not respect EE (resp. II), then it means that we are creating an edge between xx and a vertex uu in XsX_{s} (resp. in XsXsX_{\leq s}-X_{s}) which are not adjacent in the input graph GG. Figure 3 shows examples of extensions that do or do not respect II. For each interval representation π\pi^{\prime} extending π\pi to 𝐿𝑅Xs\mathit{LR}_{X_{s}} that respects EE and II, we put one or two elements into s\mathscr{I}_{s}^{\prime} by the following manner. Some forbidden intervals that had an anchor p0p_{0} or q0q_{0} in t\mathscr{I}_{t} may be re-anchored to lxl_{x} or rxr_{x}. Forbidden intervals are partitioned into three, where the ones in ILI_{\mathrm{L}} and IRI_{\mathrm{R}} are certainly left to and right to xx, respectively:

IL\displaystyle I_{\mathrm{L}} ={(p,q)Ip<πq0},\displaystyle=\{\,(p,q)\in I\mid p<_{\pi}q_{0}\,\}\,,
IR\displaystyle I_{\mathrm{R}} ={(p,q)Ip0<πq},\displaystyle=\{\,(p,q)\in I\mid p_{0}<_{\pi}q\,\}\,,
IM\displaystyle I_{\mathrm{M}} ={(p,q)Ip=q=p0=q0}.\displaystyle=\{\,(p,q)\in I\mid p=q=p_{0}=q_{0}\,\}\,.

Note that IMI_{\mathrm{M}} is either empty or a singleton and that ILI_{\mathrm{L}} and IRI_{\mathrm{R}} are mutually exclusive, since π\pi^{\prime} respects II. If a forbidden interval has a point anchored to q0q_{0} and is right to xx, then it will be re-anchored:

IR\displaystyle I_{\mathrm{R}}^{\prime} ={(p[rx/q0],q[rx/q0])(p,q)IR},\displaystyle=\{\,(p[r_{x}/q_{0}],q[r_{x}/q_{0}])\mid(p,q)\in I_{\mathrm{R}}\,\}\,,
IM\displaystyle I_{\mathrm{M}}^{\prime} ={(rx,rx)(q0,q0)IM}.\displaystyle=\{\,(r_{x},r_{x})\mid(q_{0},q_{0})\in I_{\mathrm{M}}\,\}\,.

Corresponding to the possible choices, we put both (π,ILIMIR,c)(\pi^{\prime},I_{\mathrm{L}}\cup I_{\mathrm{M}}\cup I_{\mathrm{R}}^{\prime},c) and (π,ILIMIR,c)(\pi^{\prime},I_{\mathrm{L}}\cup I_{\mathrm{M}}^{\prime}\cup I_{\mathrm{R}}^{\prime},c) into s\mathscr{I}^{\prime}_{s}. We will not add (π,ILIMIMIR,c)(\pi^{\prime},I_{\mathrm{L}}\cup I_{\mathrm{M}}\cup I_{\mathrm{M}}^{\prime}\cup I_{\mathrm{R}}^{\prime},c) since it is dominated by the other two.

(π,I)(\pi,I)mmnnppqqx1x_{1}x2x_{2}x3x_{3}x4x_{4}x5x_{5}
Figure 3: Introduce Node. When I={(m,n),(p,p),(p,q)}I=\{(m,n),(p,p),(p,q)\}, new intervals like x1,x2,x3x_{1},x_{2},x_{3} respect II, while x4,x5x_{4},x_{5} do not. If π\pi^{\prime} introduces a new interval (lx,rx)(l_{x},r_{x}) so that p<πlx<πrx<πqp<_{\pi^{\prime}}l_{x}<_{\pi^{\prime}}r_{x}<_{\pi^{\prime}}q, one can assume that (lx,rx)(l_{x},r_{x}) is either left to or right to the forbidden interval anchored to (p,p)(p,p). Those two possibilities are illustrated as x1x_{1} and x2x_{2}. The respective cases give {(m,n),(rx,rx),(rx,q)}\{(m,n),(r_{x},r_{x}),(r_{x},q)\} and {(m,n),(p,p),(rx,q)}\{(m,n),(p,p),(r_{x},q)\} as new forbidden interval sets.
Example 1.

Figure 3 shows several possibilities of extending π\pi (m<πn<πp<πqm<_{\pi}n<_{\pi}p<_{\pi}q) to π\pi^{\prime} by introducing lxl_{x} and rxr_{x}. Defining π\pi^{\prime} by letting m<πn<πlx<πp<πrx<πqm<_{\pi^{\prime}}n<_{\pi^{\prime}}l_{x}<_{\pi^{\prime}}p<_{\pi^{\prime}}r_{x}<_{\pi^{\prime}}q is illustrated by the interval of x3x_{3}, where IL={(m,n)}I_{\mathrm{L}}=\{(m,n)\}, IR={(p,p),(p,q)}I_{\mathrm{R}}=\{(p,p),(p,q)\}, and IM=I_{\mathrm{M}}=\varnothing, in which case the anchors (p,p),(p,q)IR(p,p),(p,q)\in I_{\mathrm{R}} of forbidden intervals will be updated to (rx,rx),(rx,q)(r_{x},r_{x}),(r_{x},q). Defining π\pi^{\prime} by letting m<πn<πp<πlx<πrx<πqm<_{\pi^{\prime}}n<_{\pi^{\prime}}p<_{\pi^{\prime}}l_{x}<_{\pi^{\prime}}r_{x}<_{\pi^{\prime}}q gives IL={(m,n)}I_{\mathrm{L}}=\{(m,n)\}, IR={(p,q)}I_{\mathrm{R}}=\{(p,q)\}, and IM={(p,p)}I_{\mathrm{M}}=\{(p,p)\}. We consider two ways to put the interval (lx,rx)(l_{x},r_{x}) relative to the forbidden interval anchored to (p,p)IM(p,p)\in I_{\mathrm{M}}. One is to put (lx,rx)(l_{x},r_{x}) left to those forbidden intervals, like the interval of x1x_{1} shows, in which case we update (p,p)IM(p,p)\in I_{\mathrm{M}} to (rx,rx)(r_{x},r_{x}), so we obtain I1=ILIMIR={(m,n),(rx,rx),(rx,q)}I_{1}=I_{\mathrm{L}}\cup I_{\mathrm{M}}^{\prime}\cup I_{\mathrm{R}}^{\prime}=\{(m,n),(r_{x},r_{x}),(r_{x},q)\}. The other is to put (lx,rx)(l_{x},r_{x}) right to them, as x2x_{2} shows, in which case we keep (p,p)IM(p,p)\in I_{\mathrm{M}}, so we obtain I2=ILIMIR={(m,n),(p,p),(rx,q)}I_{2}=I_{\mathrm{L}}\cup I_{\mathrm{M}}\cup I_{\mathrm{R}}^{\prime}=\{(m,n),(p,p),(r_{x},q)\}. We note that there can be several non-intersecting forbidden intervals between pp and 𝗌𝗎𝖼π(p)\mathsf{suc}_{\pi}(p), and one may locate (lx,rx)(l_{x},r_{x}) between them. This interpretation gives I={(m,n),(p,p),(rx,rx),(rx,q)}I^{\prime}=\{(m,n),(p,p),(r_{x},r_{x}),(r_{x},q)\} but we ignore this possibility, due to I1,I2II_{1},I_{2}\sqsubseteq I^{\prime}.

We then obtain s\mathscr{I}_{s} by reducing s\mathscr{I}^{\prime}_{s}.

Forget Node:
(π,I)(\pi,I)lxl_{x}rxr_{x}IXI_{\mathrm{X}}ppp0p_{0}q0q_{0}qqxx
Figure 4: Forget Node. I={(p,p0),(p0,p0),(p0,lx),(lx,lx),(lx,q0),(q0,rx),(rx,q)}I=\{(p,p_{0}),(p_{0},p_{0}),(p_{0},l_{x}),(l_{x},l_{x}),(l_{x},q_{0}),(q_{0},r_{x}),(r_{x},q)\} will become I={(p,p0),(p0,p0),(p0,q0),(q0,q)}I^{\prime}=\{(p,p_{0}),(p_{0},p_{0}),(p_{0},q_{0}),(q_{0},q)\} after xx has been forgotten.

Suppose that ss is a forget node. It has just one child tt such that Xt=Xs{x}X_{t}=X_{s}\cup\{x\}. For each (π,I,c)(\pi,I,c) in t\mathscr{I}_{t}, in accordance with the definition of abstractions, we add to s\mathscr{I}^{\prime}_{s} the triple (π,I,c)(\pi^{\prime},I^{\prime},c) where

  • π\pi^{\prime} is the restriction of π\pi,

  • c=c+|{{x,u}E(lu,ru)//π(lx,rx)c^{\prime}=c+|\{\,\{x,u\}\in E\mid(l_{u},r_{u})\mathrel{/\!/\!}_{\pi}(l_{x},r_{x}) and uXs}|u\in X_{s}\,\}|.

We make II^{\prime} from II by

  • making a new forbidden interval involving (lx,rx)(l_{x},r_{x}) and

  • re-ahcoring forbidden intervals if they have an anchor lxl_{x} or rxr_{x}.

Let us anchor the points lxl_{x} and rxr_{x} to points p0p_{0} and q0q_{0} in XsX_{s}:

p0\displaystyle p_{0} =maxπ{pXsp<πlx}=𝗉𝗋𝖾𝖽π(lx),\displaystyle=\max_{\pi^{\prime}}\{\,p\in X_{s}\mid p<_{\pi}l_{x}\,\}=\mathsf{pred}_{\pi}(l_{x})\,,
q0\displaystyle q_{0} =maxπ{pXsp<πrx}={𝗉𝗋𝖾𝖽π(rx)if 𝗉𝗋𝖾𝖽π(rx)lx,𝗉𝗋𝖾𝖽π(lx)otherwise.\displaystyle=\max_{\pi^{\prime}}\{\,p\in X_{s}\mid p<_{\pi}r_{x}\,\}=\begin{cases}\mathsf{pred}_{\pi}(r_{x})&\text{if $\mathsf{pred}_{\pi}(r_{x})\neq l_{x}$,}\\ \mathsf{pred}_{\pi}(l_{x})&\text{otherwise.}\end{cases}

The set IXII_{X}\subseteq I of forbidden intervals that intersect with (lx,rx)(l_{x},r_{x}) will be given below. They will be merged into one in II^{\prime}.

IX\displaystyle I_{\mathrm{X}} ={(p,q)I(p0,rx)/\π(p,q)}\displaystyle=\{\,(p,q)\in I\mid(p_{0},r_{x})\mathrel{\text{$/$\hbox to0.0pt{\hss$\backslash$}}}_{\pi}(p,q)\,\}
p\displaystyle p_{*} =minπ({p0}{p(p,q)IX})\displaystyle=\min_{\pi}(\{p_{0}\}\cup\{\,p\mid(p,q)\in I_{\mathrm{X}}\,\})
q\displaystyle q_{*} =maxπ({rx}{q(p,q)IX})\displaystyle=\max_{\pi}(\{r_{x}\}\cup\{\,q\mid(p,q)\in I_{\mathrm{X}}\,\})

Figure 4 may help understanding why we take (p,q)I(p,q)\in I “intersecting” with (p0,rx)(p_{0},r_{x}) rather than with (lx,rx)(l_{x},r_{x}). This comes from the gap of the meanings of position pairs (lx,rx)(l_{x},r_{x}) and (p,q)I(p,q)\in I. While the interval (lx,rx)(l_{x},r_{x}) of xx starts exactly at lxl_{x} and ends at rxr_{x}, the forbidden interval anchored to (p,q)I(p,q)\in I starts after pp and ends after qq. Although forbidden intervals anchored to (p,lx)(p,l_{x}) for some pπlxp\leq_{\pi}l_{x} must intersect with the interval (lx,rx)(l_{x},r_{x}) of xx, we have (p,lx)//π(lx,rx)(p,l_{x})\mathrel{/\!/\!}_{\pi}(l_{x},r_{x}) according to the definition. On the other hand, forbidden intervals anchored to (rx,q)(r_{x},q) for some qπrxq\geq_{\pi}r_{x} do not intersect with (lx,rx)(l_{x},r_{x}) and (rx,q)//π(lx,rx)(r_{x},q)\mathrel{/\!/\!}_{\pi}(l_{x},r_{x}) holds. The new forbidden interval will be

I\displaystyle I^{\prime} =(IIX{(p,q)})[q0/rx].\displaystyle=(I-I_{\mathrm{X}}\cup\{(p_{*},q_{*})\})[q_{0}/r_{x}]\,.

Then we obtain s\mathscr{I}_{s} by reducing s\mathscr{I}^{\prime}_{s}.

Join Node:
π\piI1I_{1}I2I_{2}mmnnppqq
Figure 5: Join Node. A1=(π,I1,c1)A_{1}=(\pi,I_{1},c_{1}) and A2=(π,I2,c2)A_{2}=(\pi,I_{2},c_{2}) with I1={(n,n),(p,p)}I_{1}=\{(n,n),(p,p)\} and I2={(m,p),(p,p),(p,q)}I_{2}=\{(m,p),(p,p),(p,q)\} are not compatible, which is witnessed by the pair of (n,n)I1(n,n)\in I_{1} and (m,p)I2(m,p)\in I_{2}. If one of them was absent, they were compatible.

Suppose that ss has two children t1t_{1} and t2t_{2}, where Xs=Xt1=Xt2X_{s}=X_{t_{1}}=X_{t_{2}}. We say that A1=(π1,I1,c1)t1A_{1}=(\pi_{1},I_{1},c_{1})\in\mathscr{I}_{t_{1}} and A2=(π2,I2,c2)t2A_{2}=(\pi_{2},I_{2},c_{2})\in\mathscr{I}_{t_{2}} are compatible if π1=π2\pi_{1}=\pi_{2} and there are no (p1,q1)I1(p_{1},q_{1})\in I_{1} and (p2,q2)I2(p_{2},q_{2})\in I_{2} such that (p1,q1)/\π1(p2,q2)(p_{1},q_{1})\mathrel{\text{$/$\hbox to0.0pt{\hss$\backslash$}}}_{\pi_{1}}(p_{2},q_{2}). Figure 5 illustrates an incompatible pair. If A1A_{1} and A2A_{2} are not compatible, any interval representation ρ\rho on XsX_{\leq s} which extends ρ1\rho_{1} and ρ2\rho_{2} will make two vertices adjacent which are not adjacent in the input graph GG for any interval representations ρi\rho_{i} on XtiX_{\leq t_{i}} of which AiA_{i} is the abstraction for i=1,2i=1,2. For each compatible pair (A1,A2)(A_{1},A_{2}), one can find an interval representation ρ\rho on XsX_{\leq s} that forms a subgraph of (Xs,Es)(X_{\leq s},E_{\leq s}) which extends some ρ1\rho_{1} and ρ2\rho_{2} whose abstractions are A1A_{1} and A2A_{2}, respectively. We add the triple (π1,I1I2,c1+c2)(\pi_{1},\,I_{1}\cup I_{2},\,c_{1}+c_{2}) to s\mathscr{I}^{\prime}_{s}. We obtain s\mathscr{I}_{s} by reducing s\mathscr{I}^{\prime}_{s}.

Theorem 1.

The edge deletion problem for interval graphs can be solved in O(|V|(2k)!27.76kpoly(k))O(|V|(2k)!\cdot 2^{7.76k}\mathrm{poly}(k)) time for the treewidth kk of GG and some polynomial function poly\mathrm{poly}. If kk is the pathwidth, it can be solved in O(|V|(2k)!23.38kpoly(k))O(|V|(2k)!\cdot 2^{3.38k}\mathrm{poly}(k)) time.

Proof.

Let kk be the maximum size of the assigned set XsX_{s} to a node of a nice tree-decomposition. We first estimate the number NN of possible forbidden interval sets II for a fixed π\pi in (π,I,c)s(\pi,I,c)\in\mathscr{I}_{s}. Recall that II must satisfy that

  • (p,q)I(p,q)\in I implies pπqp\leq_{\pi}q,

  • (p,q1),(p,q2)I(p,q_{1}),(p,q_{2})\in I and pq1,q2p\neq q_{1},q_{2} implies q1=q2q_{1}=q_{2},

  • (p1,q),(p2,q)I(p_{1},q),(p_{2},q)\in I and qp1,p2q\neq p_{1},p_{2} implies p1=p2p_{1}=p_{2}.

That is, there are at most three intervals in II that involves each p𝐿𝑅Xs{}p\in\mathit{LR}_{X_{s}}-\{\top\}:

  1. (1)

    ending a forbidden interval started earlier: (q,p)I(q,p)\in I for some q<πpq<_{\pi}p,

  2. (2)

    starting and ending a minimal forbidden interval: (p,p)I(p,p)\in I,

  3. (3)

    starting a new forbidden interval: (p,q)I(p,q)\in I for some q>πpq>_{\pi}p.

Those possibilities are neither mutually exclusive nor independent. To count the number of possible forbidden intervals inductively, hereafter we will also count interval sets that may contain a right-unbounded interval. Let Ab(n)A_{\mathrm{b}}(n) and Au(n)A_{\mathrm{u}}(n) be the numbers of possible ways of drawing forbidden intervals with nn points in addition to \bot where the last intervals are right-bounded and right-unbounded, respectively. Solving the recurrence equations

Ab(n+1)\displaystyle A_{\mathrm{b}}(n+1) =2Ab(n)+2Au(n),\displaystyle=2A_{\mathrm{b}}(n)+2A_{\mathrm{u}}(n)\,, Ab(0)=2,\displaystyle A_{\mathrm{b}}(0)=2\,,
Au(n+1)\displaystyle A_{\mathrm{u}}(n+1) =2Ab(n)+3Au(n),\displaystyle=2A_{\mathrm{b}}(n)+3A_{\mathrm{u}}(n)\,, Au(0)=1,\displaystyle A_{\mathrm{u}}(0)=1\,,

we obtain Ab(n)O(22.19n)A_{\mathrm{b}}(n)\in O(2^{2.19n}). Therefore, since we have 2k2k points, there are at most NO(24.38k)N\in O(2^{4.38k}) possible sets of forbidden intervals. There can be at most (2k)!/2k(2k)!/2^{k} varieties of π\pi. Recall that s\mathscr{I}_{s} is reduced, where if s\mathscr{I}_{s} has two elements of the form (π,I,c)(\pi,I,c) and (π,I,c)(\pi,I,c^{\prime}), then c=cc=c^{\prime}. We conclude that each s\mathscr{I}_{s} may contain at most (2k)!/2kNO(23.38k(2k)!)(2k)!/2^{k}N\in O(2^{3.38k}(2k)!) elements.

Suppose ss is a forget node, whose child is tt. Then, from each element (π,I,c)t(\pi,I,c)\in\mathscr{I}_{t}, we calculate just one element (π,I,c)s(\pi^{\prime},I^{\prime},c^{\prime})\in\mathscr{I}^{\prime}_{s}. Therefore, the calculation can be done in O(N(2k)!/2kPoly(k))O(N(2k)!/2^{k}\mathrm{Poly}(k)) time.

Suppose ss is an introduce node, whose child is tt. Then, each element (π,I,c)s(\pi^{\prime},I^{\prime},c^{\prime})\in\mathscr{I}_{s} is derived from a unique element (π,I,c)t(\pi,I,c)\in\mathscr{I}_{t}. Therefore, the calculation can be done in O(N(2k)!/2kPoly(k))O(N(2k)!/2^{k}\mathrm{Poly}(k)) time.

Suppose ss is a join node with children t1t_{1} and t2t_{2}. Recall that (π1,I1,c1)t1(\pi_{1},I_{1},c_{1})\in\mathscr{I}_{t_{1}} and (π2,I2,c2)t2(\pi_{2},I_{2},c_{2})\in\mathscr{I}_{t_{2}} are compatible only when π1=π2\pi_{1}=\pi_{2}. Checking the compatibility of (π,I1,c1)(\pi,I_{1},c_{1}) and (π,I2,c2)(\pi,I_{2},c_{2}) and computing their “join” (π,I1I2,c1+c2)(\pi,I_{1}\cup I_{2},c_{1}+c_{2}) takes Poly(k)\mathrm{Poly}(k) time. Since we have at most N2N^{2} pairs to examine for each π\pi, it takes O((2k)!/2kN2Poly(k))O((2k)!/2^{k}N^{2}\mathrm{Poly}(k)) time.

Since the nice tree-decomposition has O(|V|)O(|V|) nodes, we obtain the conclusion. ∎

(a)pp(p,p)(p,p)(p,)(p,-)(b)pp(q,p)(q,p)(p,p)(p,p)(p,)(p,-)
Figure 6: Possible ways to extend an interval set over nn points to one over n+1n+1 points. (a) If the last interval is right-bounded, there are four ways to add pp to the set. (b) If the last interval is right-unbounded, there are five ways to add pp to the set.

4 Finding a Largest Permutation Subgraph

A permutation representation over a set XX is a pair π=(π1,π2)\pi=(\pi_{1},\pi_{2}) of linear orders on X+=X{,}X^{+}=X\cup\{\top,\bot\} such that <πix<πi\bot<_{\pi_{i}}x<_{\pi_{i}}\top for all xXx\in X and i{1,2}i\in\{1,2\}. For two points uu and vv in X+X^{+}, we write u//πvu\mathrel{/\!/\!}_{\pi}v if π1\pi_{1} and π2\pi_{2} agree on the order of uu and vv: that is, either uπ1vu\leq_{\pi_{1}}v and uπ2vu\leq_{\pi_{2}}v or v<π1uv<_{\pi_{1}}u and v<π2uv<_{\pi_{2}}u. Otherwise, we write u/\πvu\mathrel{\text{$/$\hbox to0.0pt{\hss$\backslash$}}}_{\pi}v and say that uu and vv intersect in π\pi. The permutation graph GπG_{\pi} of a permutation representation π\pi on VV is (V,π)(V,\mathcal{E}_{\pi}) where

π={{u,v}Vu and v intersect in π}.\mathcal{E}_{\pi}=\{\,\{u,v\}\subseteq V\mid\text{$u$ and $v$ intersect in $\pi$}\,\}\,.

Figure 7 (a) shows an example of a permutation representation ρ\rho and (b) shows the induced permutation graph GρG_{\rho}. For each vertex uu of VV, we put a point on each of the two parallel lines respecting the linear orders π1\pi_{1} and π2\pi_{2} and draw a line, called the uu-line, between the points. Then we make an edge between uu and vv if and only if the uu-line and the vv-line intersect. This section gives an FPT algorithm for the edge deletion problem for permutation graphs.

(a)u2u_{2}w2w_{2}w1w_{1}u1u_{1}w3w_{3}w4w_{4}u3u_{3}u1u_{1}w1w_{1}w3w_{3}w2w_{2}w4w_{4}u3u_{3}u2u_{2}ρ1\rho_{1}ρ2\rho_{2}
(b)u1u_{1}u2u_{2}u3u_{3}w1w_{1}w2w_{2}w3w_{3}w4w_{4}
(c)u2u_{2}w2w_{2}w1w_{1}u1u_{1}w3w_{3}w4w_{4}u3u_{3}u1u_{1}w1w_{1}w3w_{3}w2w_{2}w4w_{4}u3u_{3}u2u_{2}ρ1\rho_{1}ρ2\rho_{2}
Figure 7: (a) Permutation representation ρ\rho. (b) Permutation graph GρG_{\rho}. (c) Illustration of the abstraction 𝒜(ρ,s)=(π,I,c)\mathscr{A}(\rho,s)=(\pi,I,c) where Xs={u1,u2,u3}X_{s}=\{u_{1},u_{2},u_{3}\} and XsXs={w1,w2,w3,w4}X_{\leq s}-X_{s}=\{w_{1},w_{2},w_{3},w_{4}\}. The forbidden areas are represented by I={(u2,u1,u1,u1),(u1,u1,u1,u1)}I=\{(u_{2},u_{1},u_{1},u_{1}),(u_{1},u_{1},u_{1},u_{1})\}, where the intersection closure [w1]ρs={w1,w2,w3}[w_{1}]_{\rho}^{s}=\{w_{1},w_{2},w_{3}\} is anchored to (u2,u1,u1,u1)(u_{2},u_{1},u_{1},u_{1}) and [w4]ρs={w4}[w_{4}]_{\rho}^{s}=\{w_{4}\} is anchored to (u1,u1,u1,u1)(u_{1},u_{1},u_{1},u_{1}) in 𝒜(ρ,s)\mathscr{A}(\rho,s).

4.1 Algorithm Invariant

We anchor the areas bordered with the lines of forgotten vertices XsXsX_{\leq s}-X_{s} to the points of the current bag XsX_{s}. For each wXsXsw\in X_{\leq s}-X_{s}, let the intersection closure of ww (w.r.t. ρ\rho and ss) be the smallest set [w]ρs[w]_{\rho}^{s} such that

  • w[w]ρsw\in[w]_{\rho}^{s} and

  • if w1[w]ρsw_{1}\in[w]_{\rho}^{s}, w2XsXsw_{2}\in X_{\leq s}-X_{s}, and w1/\ρw2w_{1}\mathrel{\text{$/$\hbox to0.0pt{\hss$\backslash$}}}_{\rho}w_{2}, then w2[w]ρsw_{2}\in[w]_{\rho}^{s}.

Each closure forms a forbidden area in the sense that we may introduce no new lines that intersect the area.

For each permutation representation ρ=(ρ1,ρ2)\rho=(\rho_{1},\rho_{2}) over XsX_{\leq s}, we define the abstraction 𝒜(ρ,s)=((π1,π2),I,c)\mathscr{A}(\rho,s)=((\pi_{1},\pi_{2}),I,c) of ρ\rho for ss as follows:

  • π1\pi_{1} and π2\pi_{2} are the restrictions of ρ1\rho_{1} and ρ2\rho_{2} to XsX_{s}, respectively,

  • I={(p1,q1,p2,q2)(Xs+)4I=\{\,(p_{1},q_{1},p_{2},q_{2})\in(X_{s}^{+})^{4}\mid there is wXsXsw\in X_{\leq s}-X_{s} s.t. for i{1,2}i\in\{1,2\},
    pi<ρiminρi[w]ρs<ρi𝗌𝗎𝖼πi(pi)p_{i}<_{\rho_{i}}\min_{\rho_{i}}[w]_{\rho}^{s}<_{\rho_{i}}\mathsf{suc}_{\pi_{i}}(p_{i}) and qi<ρimaxρi[w]ρs<ρi𝗌𝗎𝖼πi(qi)}q_{i}<_{\rho_{i}}\max_{\rho_{i}}[w]_{\rho}^{s}<_{\rho_{i}}\mathsf{suc}_{\pi_{i}}(q_{i})\,\},

  • c=|EsρEs|=|{{u,v}Es{u,v}Xs and u/\ρv}|c=|E_{\leq s}-\mathcal{E}_{\rho}-E_{s}|=|\{\,\{u,v\}\in E_{\leq s}\mid\text{$\{u,v\}\nsubseteq X_{s}$ and $u\mathrel{\text{$/$\hbox to0.0pt{\hss$\backslash$}}}_{\rho}v$}\,\}|.

Intuitively, II anchors occurrences of forgotten vertices in ρ\rho to occurrences of active vertices in π=(π1,π2)\pi=(\pi_{1},\pi_{2}). Figure 7 (c) illustrates an example of forbidden areas. We note that, two distinct elements (p1,q1,p2,q2)I(p_{1},q_{1},p_{2},q_{2})\in I and (p1,q1,p2,q2)I(p_{1}^{\prime},q_{1}^{\prime},p_{2}^{\prime},q_{2}^{\prime})\in I cannot “intersect”, due to the definition of intersection closures. In other words, there are no (p1,q1,p2,q2),(p1,q1,p2,q2)I(p_{1},q_{1},p_{2},q_{2}),(p_{1}^{\prime},q_{1}^{\prime},p_{2}^{\prime},q_{2}^{\prime})\in I such that pi<πiqip_{i}<_{\pi_{i}}q_{i}^{\prime} and pj<πjqjp_{j}^{\prime}<_{\pi_{j}}q_{j} for some i,j{1,2}i,j\in\{1,2\}. Elements of II will be linearly ordered by extending π\pi so that (p1,q1,p2,q2)<π(p1,q1,p2,q2)(p_{1},q_{1},p_{2},q_{2})<_{\pi}(p_{1}^{\prime},q_{1}^{\prime},p_{2}^{\prime},q_{2}^{\prime}) if and only if p1<q1p_{1}<q_{1}^{\prime} or p2<q2p_{2}<q_{2}^{\prime}. The integer cc is the number of the edges deleted from (Xs,Es)(X_{\leq s},E_{\leq s}) except those among active vertices.

We say that (π,I,c)(\pi,I,c) dominates (π,I,c)(\pi^{\prime},I^{\prime},c^{\prime}) if and only if π=π\pi=\pi^{\prime}, IπII\sqsubseteq_{\pi}I^{\prime} and ccc\leq c^{\prime}, where we write IπII\sqsubseteq_{\pi}I^{\prime} if every forbidden area of II is inside some forbidden area of II^{\prime}: i.e., for every (p1,q1,p2,q2)I(p_{1},q_{1},p_{2},q_{2})\in I, there is (p1,q1,p2,q2)I(p_{1}^{\prime},q_{1}^{\prime},p_{2}^{\prime},q_{2}^{\prime})\in I^{\prime} such that p1π1p1p_{1}^{\prime}\leq_{\pi_{1}}p_{1}, q1π1q1q_{1}\leq_{\pi_{1}}q_{1}^{\prime}, p2π2p2p_{2}^{\prime}\leq_{\pi_{2}}p_{2}, and q2π2q2q_{2}\leq_{\pi_{2}}q_{2}^{\prime}.

Lemma 2.

Suppose 𝒜(ρ,s)=(π,I,c)\mathscr{A}(\rho,s)=(\pi,I,c) dominates 𝒜(ρ,s)=(π,I,c)\mathscr{A}(\rho^{\prime},s)=(\pi^{\prime},I^{\prime},c^{\prime}). For any extension σ\sigma^{\prime} of ρ\rho^{\prime} such that σEs\mathcal{E}_{\sigma^{\prime}}\subseteq E_{\leq s}, there is an extension σ\sigma of ρ\rho such that σσEs\mathcal{E}_{\sigma^{\prime}}\subseteq\mathcal{E}_{\sigma}\subseteq E_{\leq s}.

We call a set of abstractions reduced if it has no pair of distinct elements of which one dominates the other. The reduced form of a set of abstractions is obtained by removing abstractions that are dominated by others. Our algorithm for Permutation-Edge-Deletion calculates, for each node ss of the tree-decomposition, a reduced set s\mathscr{I}_{s} of abstractions of permutation representations of permutation subgraphs of (Xs,Es)(X_{\leq s},E_{\leq s}) for XsX_{s} satisfying the following invariant properties.

Condition 2.
  • Every element of s\mathscr{I}_{s} is the abstraction of some permutation representation of a permutation subgraph of (Xs,Es)(X_{\leq s},E_{\leq s}) for XsX_{s},

  • Any permutation representation ρ\rho of any permutation subgraph of (Xs,Es)(X_{\leq s},E_{\leq s}) has an element of s\mathscr{I}_{s} that dominates its abstraction 𝒜(ρ,s)\mathscr{A}(\rho,s).

Clearly s\mathscr{I}_{s} satisfies the above condition if and only if so is its reduced form. We make each s\mathscr{I}_{s} reduced. Particularly if ss is the root node, we have s={((o,o),{(,,,)},c)}\mathscr{I}_{s}=\{((o,o),\{(\bot,\bot,\bot,\bot)\},c)\}, where oo is the trivial order over {,}\{\bot,\top\} such that <o\bot<_{o}\top and cc is the least number such that one can obtain a permutation subgraph by removing cc edges from GG. That is, the number cc will be the solution to our problem. If ss is a leaf, by definition s={((o,o),,0)}\mathscr{I}_{s}=\{((o,o),\emptyset,0)\}. Our algorithm computes those values s\mathscr{I}_{s} recursively from leaves to the root. In what follows, we show how to calculate s\mathscr{I}_{s} from t\mathscr{I}_{t} for child(ren) tt of ss, while preserving the invariant (Condition 2).

(a)p1p_{1}uup2p_{2}vvq2q_{2}x1x_{1}x1x_{1}x2x_{2}x2x_{2}x3x_{3}x3x_{3}x4x_{4}x4x_{4}
(b)ILI_{\mathrm{L}}IMI_{\mathrm{M}}IMI_{\mathrm{M}}IRI_{\mathrm{R}}y1y_{1}z1z_{1}y2y_{2}z2z_{2}xxxx
Figure 8: Introduce Node. (a) Extending π\pi by introducing x1x_{1} or x2x_{2} respects II and introducing x3x_{3} or x4x_{4} does not. We see 𝗌𝗎𝖼π1(p1)<π1x3<π2q2\mathsf{suc}_{\pi^{\prime}_{1}}(p_{1})<_{\pi^{\prime}_{1}}x_{3}<_{\pi^{\prime}_{2}}q_{2} and 𝗌𝗎𝖼π2(p2)<π2x4<π2q2\mathsf{suc}_{\pi^{\prime}_{2}}(p_{2})<_{\pi^{\prime}_{2}}x_{4}<_{\pi^{\prime}_{2}}q_{2}. (b) Relation between newly introduced vertex xx and forbidden areas. Since forbidden areas should not intersect with xx, they are grouped into three: ones (ILI_{\mathrm{L}}) that must be left to xx, ones (IRI_{\mathrm{R}}) that must be right to xx, and the other (IM{(y1,y1,y1,y1)}I_{\mathrm{M}}\subseteq\{(y_{1},y_{1},y_{1},y_{1})\}).

4.2 Algorithm

Leaf Node:

If ss is a leaf, we let s={((o,o),,0)}\mathscr{I}_{s}=\{((o,o),\emptyset,0)\}.

Introduce Node:

Suppose ss is an introduce node. It has just one child tt such that Xs=Xt{x}X_{s}=X_{t}\cup\{x\}. Figure 8 would help to understand the behavior of our algorithm in this case. For each (π,I,c)(\pi,I,c) in t\mathscr{I}_{t}, we say that an extension π\pi^{\prime} of π\pi to XsX_{s} respects EE and II precisely in the cases where

  • if {x,u}E\{x,u\}\notin E for uXtu\in X_{t}, then u//πxu\mathrel{/\!/\!}_{\pi^{\prime}}x,

  • there are no (p1,q1,p2,q2)I(p_{1},q_{1},p_{2},q_{2})\in I and i,j{1,2}i,j\in\{1,2\} such that 𝗌𝗎𝖼πi(pi)<πix\mathsf{suc}_{\pi^{\prime}_{i}}(p_{i})<_{\pi^{\prime}_{i}}x and x<πjqjx<_{\pi^{\prime}_{j}}q_{j},

respectively. If π\pi^{\prime} does not respect EE, we would wrongly make two non-adjacent vertices uXsu\in X_{s} and xx of GG intersect in π\pi^{\prime}, which gives a non-subgraph of the input graph. Otherwise, they will be non-adjacent. Since {x,w}E\{x,w\}\notin E for any forgotten vertex wXsXsw\in X_{\leq s}-X_{s}, we must avoid xx and ww intersect in an extension of ρ\rho. This requires π\pi^{\prime} to respect II. If π\pi^{\prime} respects II, there are ρ\rho such that 𝒜(ρ,t)=(I,π,c)\mathscr{A}(\rho,t)=(I,\pi,c) and an extension ρ\rho^{\prime} of both ρ\rho and π\pi^{\prime} such that ρEs\mathcal{E}_{\rho^{\prime}}\subseteq E_{\leq s}. We note that here we use 𝗌𝗎𝖼πi(pi)\mathsf{suc}_{\pi^{\prime}_{i}}(p_{i}) rather than pip_{i} to judge whether π\pi^{\prime} respects II. This is because the quadruple (p1,q1,p2,q2)I(p_{1},q_{1},p_{2},q_{2})\in I consists of anchors rather than the exact points of the forbidden area.

For each extension π\pi^{\prime} respecting EE and II, we add at most two triples to s\mathscr{I}^{\prime}_{s} as described below. Since XsXtX_{s}\supseteq X_{t}, cc need not be updated by definition.

Some forbidden areas that had an anchor yi=𝗉𝗋𝖾𝖽πi(x)y_{i}=\mathsf{pred}_{\pi_{i}^{\prime}}(x) may be re-anchored to xx. Forbidden areas are partitioned into three, where the ones in ILI_{\mathrm{L}} and IRI_{\mathrm{R}} are certainly left to and right to xx, respectively, while (y1,y1,y2,y2)(y_{1},y_{1},y_{2},y_{2}) can go to the left or right to xx:

IL\displaystyle I_{\mathrm{L}} ={(p1,q1,p2,q2)Ip1<π1y1 or p2<π2y2},\displaystyle=\{\,(p_{1},q_{1},p_{2},q_{2})\in I\mid p_{1}<_{\pi_{1}}y_{1}\text{ or }p_{2}<_{\pi_{2}}y_{2}\,\}\,,
IR\displaystyle I_{\mathrm{R}} ={(p1,q1,p2,q2)Iy1<π1q1 or y2<π2q2},\displaystyle=\{\,(p_{1},q_{1},p_{2},q_{2})\in I\mid y_{1}<_{\pi_{1}}q_{1}\text{ or }y_{2}<_{\pi_{2}}q_{2}\,\}\,,
IM\displaystyle I_{\mathrm{M}} ={(p1,q1,p2,q2)Ip1=y1=q1 and p2=y2=q2}.\displaystyle=\{\,(p_{1},q_{1},p_{2},q_{2})\in I\mid p_{1}=y_{1}=q_{1}\text{ and }p_{2}=y_{2}=q_{2}\,\}\,.

Note that IMI_{\mathrm{M}} is either empty or a singleton and that ILI_{\mathrm{L}} and IRI_{\mathrm{R}} are mutually exclusive, since π\pi^{\prime} respects II. If a forbidden area has a point anchored to yiy_{i} and is right to xx, it will be re-anchored:

IR\displaystyle I_{\mathrm{R}}^{\prime} ={(p1[x/y1],q1[x/y1],p2[x/y2],q2[x/y2])(p1,q1,p2,q2)IR},\displaystyle=\{\,(p_{1}[x/y_{1}],q_{1}[x/y_{1}],p_{2}[x/y_{2}],q_{2}[x/y_{2}])\mid(p_{1},q_{1},p_{2},q_{2})\in I_{\mathrm{R}}\,\}\,,
IM\displaystyle I_{\mathrm{M}}^{\prime} ={(p1[x/y1],q1[x/y1],p2[x/y2],q2[x/y2])(p1,q1,p2,q2)IM}.\displaystyle=\{\,(p_{1}[x/y_{1}],q_{1}[x/y_{1}],p_{2}[x/y_{2}],q_{2}[x/y_{2}])\mid(p_{1},q_{1},p_{2},q_{2})\in I_{\mathrm{M}}\,\}\,.

Corresponding to the possible choices, we put both (π,ILIMIR,c)(\pi^{\prime},I_{\mathrm{L}}\cup I_{\mathrm{M}}\cup I_{\mathrm{R}}^{\prime},c) and (π,ILIMIR,c)(\pi^{\prime},I_{\mathrm{L}}\cup I_{\mathrm{M}}^{\prime}\cup I_{\mathrm{R}}^{\prime},c) into s\mathscr{I}^{\prime}_{s}. We will not add (π,ILIMIMIR,c)(\pi^{\prime},I_{\mathrm{L}}\cup I_{\mathrm{M}}\cup I_{\mathrm{M}}^{\prime}\cup I_{\mathrm{R}}^{\prime},c) since it is dominated by the other two.

We then obtain s\mathscr{I}_{s} by reducing s\mathscr{I}^{\prime}_{s}.

Forget Node:

Suppose ss is a forget node. It has just one child tt such that Xt=Xs{x}X_{t}=X_{s}\cup\{x\}. For each (π,I,c)(\pi,I,c) in t\mathscr{I}_{t}, we add the following triple (π,I,c)(\pi^{\prime},I^{\prime},c^{\prime}) to s\mathscr{I}^{\prime}_{s} where

  • π\pi^{\prime} is the restriction of π\pi for XsX_{s},

  • c=c+|{uXs{x,u}Es and u//πx}|c^{\prime}=c+|\{\,u\in X_{s}\mid\text{$\{x,u\}\in E_{\leq s}$ and $u\mathrel{/\!/\!}_{\pi}x$}\,\}|.

Let

IX={(p1,q1,p2,q2)Ipi<πix and xπjqj with {i,j}={1,2}}I_{\mathrm{X}}=\{\,(p_{1},q_{1},p_{2},q_{2})\in I\mid p_{i}<_{\pi_{i}}x\text{ and }x\leq_{\pi_{j}}q_{j}\text{ with }\{i,j\}=\{1,2\}\,\}

be the set of forbidden areas of II intersecting with xx. We note that the asymmetry between pi<πixp_{i}<_{\pi_{i}}x and xπjqjx\leq_{\pi_{j}}q_{j} is due to the fact that the actual forbidden area starts after p1p_{1} and p2p_{2} and ends after q1q_{1} and q2q_{2}, while the points of xx are exact.

The new forbidden area formed by xx and IXI_{\mathrm{X}} will be α=(y1,z1,y2,z2)\alpha=(y_{1},z_{1},y_{2},z_{2}) where for x1=𝗉𝗋𝖾𝖽π1(x)x_{1}=\mathsf{pred}_{\pi_{1}}(x) and x2=𝗉𝗋𝖾𝖽π2(x)x_{2}=\mathsf{pred}_{\pi_{2}}(x),

y1\displaystyle y_{1} =minπ1({x1}{p1[x1/x](p1,q1,p2,q2)IX}),\displaystyle=\min_{\pi_{1}}(\{x_{1}\}\cup\{\,p_{1}[x_{1}/x]\mid(p_{1},q_{1},p_{2},q_{2})\in I_{\mathrm{X}}\,\})\,,
z1\displaystyle z_{1} =maxπ1({x1}{q1[x1/x](p1,q1,p2,q2)IX}),\displaystyle=\max_{\pi_{1}}(\{x_{1}\}\cup\{\,q_{1}[x_{1}/x]\mid(p_{1},q_{1},p_{2},q_{2})\in I_{\mathrm{X}}\,\})\,,
y2\displaystyle y_{2} =minπ2({x2}{p2[x2/x](p1,q1,p2,q2)IX}),\displaystyle=\min_{\pi_{2}}(\{x_{2}\}\cup\{\,p_{2}[x_{2}/x]\mid(p_{1},q_{1},p_{2},q_{2})\in I_{\mathrm{X}}\,\})\,,
z2\displaystyle z_{2} =maxπ2({x2}{q2[x2/x](p1,q1,p2,q2)IX}).\displaystyle=\max_{\pi_{2}}(\{x_{2}\}\cup\{\,q_{2}[x_{2}/x]\mid(p_{1},q_{1},p_{2},q_{2})\in I_{\mathrm{X}}\,\})\,.

We then define II^{\prime} by

I\displaystyle I^{\prime} ={(p1[x1/x],q1[x1/x],p2[x2/x],q2[x2/x])(p1,q1,p2,q3)IIX}{α}\displaystyle=\{\,(p_{1}[x_{1}/x],q_{1}[x_{1}/x],p_{2}[x_{2}/x],q_{2}[x_{2}/x])\mid(p_{1},q_{1},p_{2},q_{3})\in I-I_{\mathrm{X}}\,\}\cup\{\,\alpha\,\}

This is illustrated in Figure 9.

mmnnppqqx1x_{1}x2x_{2}xxxxIXI_{\mathrm{X}}IXI_{\mathrm{X}}mmnnppqqx1x_{1}x2x_{2}(x)(x)(x)(x)
Figure 9: Forget Node. The forbidden areas in IX={(m,m,p,p),(n,n,q,x)}I_{\mathrm{X}}=\{(m,m,p,p),(n,n,q,x)\} will be merged into the single area α=(x1,n,p,x2)\alpha=(x_{1},n,p,x_{2}) by forgetting xx.

We obtain s\mathscr{I}_{s} by reducing s\mathscr{I}^{\prime}_{s}.

Join Node:

Suppose ss has two children tt and tt^{\prime}, where Xs=Xt=XtX_{s}=X_{t}=X_{t^{\prime}}. We say that a pair of (π,I,c)t(\pi,I,c)\in\mathscr{I}_{t} and (π,I,c)t(\pi^{\prime},I^{\prime},c^{\prime})\in\mathscr{I}_{t^{\prime}} is compatible precisely in the case where

  • π=π\pi=\pi^{\prime}, say π=π=(π1,π2)\pi=\pi^{\prime}=(\pi_{1},\pi_{2}), and

  • no members of II and II^{\prime} intersect: that is, there are no pairs of (p1,q1,p2,q2)I(p_{1},q_{1},p_{2},q_{2})\in I and (p1,q1,p2,q2)I(p_{1}^{\prime},q_{1}^{\prime},p_{2}^{\prime},q_{2}^{\prime})\in I^{\prime} such that pi<πiqip_{i}<_{\pi_{i}}q_{i}^{\prime} and pj<πjqjp_{j}^{\prime}<_{\pi_{j}}q_{j} for some i,j{1,2}i,j\in\{1,2\}.

If they are compatible, we add the triple ((π1,π2),II,c+c)((\pi_{1},\pi_{2}),\,I\cup I^{\prime},\,c+c^{\prime}) to s\mathscr{I}^{\prime}_{s}. We then obtain s\mathscr{I}_{s} by reducing s\mathscr{I}^{\prime}_{s}.

Theorem 2.

The edge deletion problem for permutation graphs can be solved in O(|V|(k!)2N2poly(k))O(|V|(k!)^{2}N^{2}\mathrm{poly}(k)) time where N=27.34kN=2^{7.34k} for the treewidth kk of GG. If kk is the pathwidth, it can be solved in O(|V|(k!)2Npoly(k))O(|V|(k!)^{2}N\mathrm{poly}(k)) time.

Proof.

Let kk be the maximum size of the assigned set XsX_{s} to a node of the tree-decomposition. There are at most (k!)2(k!)^{2} variants of π\pi of abstractions (π,I,c)(\pi,I,c).

In order to count the number NN of possible forbidden interval sets II for a fixed π\pi in (π,I,c)s(\pi,I,c)\in\mathscr{I}_{s}, we give a transformation of II below. Let IiI_{i} for i{1,2}i\in\{1,2\} be the interval projections of II:

Ii={(pi,qi)(p1,q1,p2,q2)I}.I_{i}=\{\,(p_{i},q_{i})\mid(p_{1},q_{1},p_{2},q_{2})\in I\,\}\,.

The proof of Theorem 1 has shown that there can be at most 22.19k2^{2.19k} possibilities for each IiI_{i}. However, since there is no one-one corresponding between I1I_{1} and I2I_{2}, the pair (I1,I2)(I_{1},I_{2}) is not informative enough to recover II. For example, there can be distinct (p,q)(p,q) and (p,q)(p^{\prime},q^{\prime}) in I1I_{1} such that (p,q,n,n),(p,q,n,n)I(p,q,n,n),(p^{\prime},q^{\prime},n,n)\in I. We call an interval (p,p)I1(p,p)\in I_{1} a hub if there are two (or more) distinct intervals (p2,q2),(p2,q2)I2(p_{2},q_{2}),(p_{2}^{\prime},q_{2}^{\prime})\in I_{2} such that (p,p,p2,q2),(p,p,p2,q2)I(p,p,p_{2},q_{2}),(p,p,p_{2}^{\prime},q_{2}^{\prime})\in I. We call an interval (p,q)I1(p,q)\in I_{1} a terminal if it is not a hub and there is no nn such that (p,q,n,n),(𝗌𝗎𝖼π1(p,q),n,n)I(p,q,n,n),(\mathsf{suc}_{\pi_{1}}(p,q),n,n)\in I where 𝗌𝗎𝖼π1(p,q)I1\mathsf{suc}_{\pi_{1}}(p,q)\in I_{1} is the successor of (p,q)(p,q) in I1I_{1}. We use the same names for intervals in I2I_{2}. Let I1I_{1}^{\prime} enrich I1I_{1} so that each interval (p,q)I1(p,q)\in I_{1} has a 3-valued variable that indicate whether it is a hub, a terminal, or neither. Figure 10 illustrates how II will be transformed into I1I_{1}^{\prime} and I2I_{2}^{\prime}.

We will show that NN is at most the number of possible enriched sets I1I_{1}^{\prime} and I2I_{2}^{\prime}. Algorithm 1 recovers elements of II from left to right using I1I_{1}^{\prime} and I2I_{2}^{\prime}, where elements of I1I_{1}^{\prime} and I2I_{2}^{\prime} are sorted and stored in stacks. From the first intervals (pi,qi)=minπIi(p_{i},q_{i})=\min_{\pi}I_{i} of the stacks, we obtain the first forbidden area (p1,q1,p2,q2)I(p_{1},q_{1},p_{2},q_{2})\in I. If (pi,qi)(p_{i},q_{i}) is a hub, we keep it as the active hub. Note that it is impossible that both (p1,q1)(p_{1},q_{1}) and (p2,q2)(p_{2},q_{2}) are hubs simultaneously. Suppose we have reconstructed a forbidden area (p1,q1,p2,q2)I(p_{1},q_{1},p_{2},q_{2})\in I with no active hub. In this case, the next forbidden area will be (𝗌𝗎𝖼π1(p1,q1),𝗌𝗎𝖼π2(p2,q2))(\mathsf{suc}_{\pi_{1}}(p_{1},q_{1}),\mathsf{suc}_{\pi_{2}}(p_{2},q_{2})). Suppose (p1,q1)(p_{1},q_{1}) is the active hub. If (p2,q2)(p_{2},q_{2}) is neither a hub nor a terminal, then the next forbidden area will be (p1,q1,𝗌𝗎𝖼π2(p2,q2))(p_{1},q_{1},\mathsf{suc}_{\pi_{2}}(p_{2},q_{2})) and we keep the active hub (p1,q1)(p_{1},q_{1}). If (p2,q2)(p_{2},q_{2}) is a terminal, then the next forbidden area will be (𝗌𝗎𝖼π1(p1,q1),𝗌𝗎𝖼π2(p2,q2))(\mathsf{suc}_{\pi_{1}}(p_{1},q_{1}),\mathsf{suc}_{\pi_{2}}(p_{2},q_{2})) and the active hub will be null. If (p2,q2)(p_{2},q_{2}) is a hub, then the next forbidden area will be (𝗌𝗎𝖼π1(p1,q1),p2,q2)(\mathsf{suc}_{\pi_{1}}(p_{1},q_{1}),p_{2},q_{2}) and the new active hub is (p2,q2)I2(p_{2},q_{2})\in I_{2}.

p1p_{1}p2p_{2}p3p_{3}p4p_{4}p5p_{5}q1q_{1}q2q_{2}q3q_{3}q4q_{4}q5q_{5}q6q_{6}
Figure 10: The original forbidden areas of II can be recovered from I1={(p1,p1,𝚑),(p1,p2,),(p3,p3,𝚝),(p4,p4,𝚝)}I_{1}^{\prime}=\{(p_{1},p_{1},\mathtt{h}),(p_{1},p_{2},\mathtt{-}),(p_{3},p_{3},\mathtt{t}),(p_{4},p_{4},\mathtt{t})\} and I2={(q1,q1,),(q1,q2,),(q4,q4,𝚑),(q5,q5,𝚝)}I_{2}^{\prime}=\{(q_{1},q_{1},\mathtt{-}),(q_{1},q_{2},\mathtt{-}),(q_{4},q_{4},\mathtt{h}),(q_{5},q_{5},\mathtt{t})\}, where 𝚑\mathtt{h} and 𝚝\mathtt{t} represent a hub and a terminal, respectively.
1 II\leftarrow\emptyset and 𝖠𝖼𝗍𝗂𝗏𝖾𝖧𝗎𝖻𝙽𝚞𝚕𝚕\mathsf{ActiveHub}\leftarrow\mathtt{Null};
2 repeat
3      if 𝖠𝖼𝗍𝗂𝗏𝖾𝖧𝗎𝖻=𝙽𝚞𝚕𝚕\mathsf{ActiveHub}=\mathtt{Null} then
4            (p1,q1,r1)I1.𝚙𝚘𝚙(p_{1},q_{1},r_{1})\leftarrow I_{1}^{\prime}.\mathtt{pop} and (p2,q2,r2)I2.𝚙𝚘𝚙(p_{2},q_{2},r_{2})\leftarrow I_{2}^{\prime}.\mathtt{pop};
5             II{(p1,q1,p2,q2)}I\leftarrow I\cup\{(p_{1},q_{1},p_{2},q_{2})\};
6             if r1=𝚑r_{1}=\mathtt{h} then 𝖠𝖼𝗍𝗂𝗏𝖾𝖧𝗎𝖻(1,p1,q1)\mathsf{ActiveHub}\leftarrow(1,p_{1},q_{1});
7             else if r2=𝚑r_{2}=\mathtt{h} then 𝖠𝖼𝗍𝗂𝗏𝖾𝖧𝗎𝖻(2,p2,q2)\mathsf{ActiveHub}\leftarrow(2,p_{2},q_{2});
8            
9      else if 𝖠𝖼𝗍𝗂𝗏𝖾𝖧𝗎𝖻=(1,p1,q1)\mathsf{ActiveHub}=(1,p_{1},q_{1}) for some p1,q1p_{1},q_{1} then
10             (p2,q2,r2)I2.𝚙𝚘𝚙(p_{2},q_{2},r_{2})\leftarrow I_{2}^{\prime}.\mathtt{pop};
11             II{(p1,q1,p2,q2)}I\leftarrow I\cup\{(p_{1},q_{1},p_{2},q_{2})\};
12             if r2=𝚑r_{2}=\mathtt{h} then 𝖠𝖼𝗍𝗂𝗏𝖾𝖧𝗎𝖻(2,p2,q2)\mathsf{ActiveHub}\leftarrow(2,p_{2},q_{2});
13             else if r2=𝚝r_{2}=\mathtt{t} then 𝖠𝖼𝗍𝗂𝗏𝖾𝖧𝗎𝖻𝙽𝚞𝚕𝚕\mathsf{ActiveHub}\leftarrow\mathtt{Null};
14            
15      else if 𝖠𝖼𝗍𝗂𝗏𝖾𝖧𝗎𝖻=(2,p2,q2)\mathsf{ActiveHub}=(2,p_{2},q_{2}) for some p2,q2p_{2},q_{2} then
16             (p1,q1,r1)I1.𝚙𝚘𝚙(p_{1},q_{1},r_{1})\leftarrow I_{1}^{\prime}.\mathtt{pop};
17             II{(p1,q1,p2,q2)}I\leftarrow I\cup\{(p_{1},q_{1},p_{2},q_{2})\};
18             if r1=𝚑r_{1}=\mathtt{h} then 𝖠𝖼𝗍𝗂𝗏𝖾𝖧𝗎𝖻(1,p1,q1)\mathsf{ActiveHub}\leftarrow(1,p_{1},q_{1});
19             else if r1=𝚝r_{1}=\mathtt{t} then 𝖠𝖼𝗍𝗂𝗏𝖾𝖧𝗎𝖻𝙽𝚞𝚕𝚕\mathsf{ActiveHub}\leftarrow\mathtt{Null};
20            
21      
22until the stacks I1I_{1}^{\prime} and I2I_{2}^{\prime} are empty;
Algorithm 1 Reconstruction of II

A point pp may appear in IiI_{i} in the following ways:

  1. (1)

    ending a forbidden interval started earlier: (q,p)Ii(q,p)\in I_{i} for some q<πipq<_{\pi_{i}}p,

  2. (2)

    starting and ending a minimal forbidden interval: (p,p)Ii(p,p)\in I_{i}, which may be a hub or a terminal,

  3. (3)

    starting a new forbidden interval: (p,q)Ii(p,q)\in I_{i} for some q>πipq>_{\pi_{i}}p, which may be a terminal but not a hub.

We do not care whether or not it is a terminal in the first case, leaving counting different possibilities to the interval starting point qq. Solving the recurrence equations

Ab(n+1)\displaystyle A_{\mathrm{b}}(n+1) =4Ab(n)+4Au(n),\displaystyle=4A_{\mathrm{b}}(n)+4A_{\mathrm{u}}(n)\,, Ab(0)=4,\displaystyle A_{\mathrm{b}}(0)=4\,,
Au(n+1)\displaystyle A_{\mathrm{u}}(n+1) =8Ab(n)+9Au(n),\displaystyle=8A_{\mathrm{b}}(n)+9A_{\mathrm{u}}(n)\,, Au(0)=3,\displaystyle A_{\mathrm{u}}(0)=3\,,

we obtain Ab(n)O(23.67n)A_{\mathrm{b}}(n)\in O(2^{3.67n}). All in all, we have at most NO(27.34k)N\in O(2^{7.34k}) possibilities for II.

Arguments similar to the proof of Theorem 1 derive the theorem. ∎

5 Other Classes Related to Interval Graphs

The algorithm presented in the previous section can be applied to the 𝒞\mathcal{C}-Edge-Deletion for some subclasses and a superclass of interval graphs with a slight modifications.

5.1 Proper interval graphs

An interval representation π\pi is said to be proper if there are no u,vVu,v\in V such that lu<πlv<πrv<πrul_{u}<_{\pi}l_{v}<_{\pi}r_{v}<_{\pi}r_{u}. An interval graph is proper if it admits a proper interval representation. In accordance with the definition of the graph class, we simply require π\pi^{\prime} in Introduce Node of the algorithm in Section 3 to be a proper interval representation.

Corollary 1.

The edge deletion problem for proper interval graphs can be solved in O(|V|N2poly(k))O(|V|N^{2}\mathrm{poly}(k)) time where N=23.91k(2k)!(k+1)!N=2^{3.91k}\frac{(2k)!}{(k+1)!} for the treewidth kk of GG. If kk is the pathwidth, it can be solved in O(|V|Npoly(k))O(|V|N\mathrm{poly}(k)) time.

Proof.

Let kk be the maximum size of the assigned set XsX_{s} to a node of a nice tree-decomposition. The number of possible proper interval representations over kk vertices is k!Ckk!C_{k} for the kk-th Catalan number Ck=(2k)!(k+1)!k!C_{k}=\frac{(2k)!}{(k+1)!k!}. Let A(n)A(n) be the numbers of possible ways of drawing forbidden intervals with nn points in addition to \bot where the last intervals may be right-unbounded. We observed in the proof of Theorem 1 (Figure 6) that A(n+1)5A(n)A(n+1)\leq 5A(n). However, in the case of proper interval representations, we never have (lu,lu)I(l_{u},l_{u})\in I. If the last (n+1)(n+1)st point is lul_{u}, we have at most three times more possible forbidden interval sets than the number of possible forbidden interval sets over nn points. Therefore, the number of possible abstractions in each s\mathscr{I}_{s} is at most O((2k)!(k+1)!3k5k)O(23.91k(2k)!(k+1)!)O(\frac{(2k)!}{(k+1)!}\cdot 3^{k}\cdot 5^{k})\subseteq O(2^{3.91k}\frac{(2k)!}{(k+1)!}). ∎

5.2 Trivially perfect graphs

An interval representation π\pi is said to be nested if there are no u,vVu,v\in V, such that lu<πlv<πru<πrvl_{u}<_{\pi}l_{v}<_{\pi}r_{u}<_{\pi}r_{v}. A trivially perfect graph (a.k.a. nested interval graph) is an interval graph that admits a nested interval representation. The algorithm presented in Section 3 can easily be modified so that it solves the edge deletion problem for trivially perfect graphs. In accordance with the definition of the graph class, we simply require π\pi^{\prime} in Introduce Node of the algorithm in Section 3 to be a nested interval representation.

Corollary 2.

The edge deletion problem for trivially perfect graphs can be solved in O(|V|N2poly(k))O(|V|N^{2}\mathrm{poly}(k)) time where N=22.4k(2k)!k!N=2^{2.4k}\frac{(2k)!}{k!} for the treewidth kk of GG. If kk is the pathwidth, it can be solved in O(|V|Npoly(k))O(|V|N\mathrm{poly}(k)) time.

Proof.

In this case, forbidden intervals and active intervals are nested. Let DnD_{n} be the number of possible pairs (π,I)(\pi,I) in abstractions with nn active vertices modulo renaming of vertices. Then, the maximum cardinality of s\mathscr{I}_{s} is at most n!Dnn!\cdot D_{n}. Hereafter we assume that neither (,)(\bot,\bot) nor (𝗉𝗋𝖾𝖽π(),𝗉𝗋𝖾𝖽π())(\mathsf{pred}_{\pi}(\top),\mathsf{pred}_{\pi}(\top)) appears in II. Let us say that (π,I)(\pi,I) is splittable if 𝐿𝑅Xs{,}\mathit{LR}_{X_{s}}-\{\bot,\top\} can be partitioned into two non-empty subsets Y1Y_{1} and Y2Y_{2} so that

  • maxπY1<minπY2\max_{\pi}Y_{1}<\min_{\pi}Y_{2},

  • for any uXsu\in X_{s}, lul_{u} and rur_{u} belong to the same component Y1Y_{1} or Y2Y_{2},

  • if (p,q)I(p,q)\in I and pqp\neq q, 𝗌𝗎𝖼π(p)\mathsf{suc}_{\pi}(p) and qq belong to the same component Y1Y_{1} or Y2Y_{2}.

Otherwise, (π,I)(\pi,I) is non-splittable. Let DnD^{\prime}_{n} and Dn′′D^{\prime\prime}_{n} count the numbers of non-splittable and splittable pairs (π,I)(\pi,I), respectively, where n=|Xs|n=|X_{s}|. We have D0=D0=1D_{0}=D_{0}^{\prime}=1, D0′′=D1′′=0D_{0}^{\prime\prime}=D_{1}^{\prime\prime}=0, D1=D1=3D_{1}=D_{1}^{\prime}=3, and for n2n\geq 2,

Dn\displaystyle D_{n} =Dn+Dn′′,\displaystyle=D^{\prime}_{n}+D^{\prime\prime}_{n}\,, (1)
Dn\displaystyle D^{\prime}_{n} =Cn+4Dn1,\displaystyle=C_{n}+4D_{n-1}\,, (2)
Dn′′\displaystyle D^{\prime\prime}_{n} =i=1n12DiDni.\displaystyle=\sum_{i=1}^{n-1}2D^{\prime}_{i}D_{n-i}\,. (3)

The first term CnC_{n} of (2) is the nnth Catalan number, which counts the possible permutations π\pi when (,𝗉𝗋𝖾𝖽π())I(\bot,\mathsf{pred}_{\pi}(\top))\in I, i.e., when I={(,𝗉𝗋𝖾𝖽π())}I=\{(\bot,\mathsf{pred}_{\pi}(\top))\}. The second term 4Dn14D_{n-1} counts cases where (,𝗉𝗋𝖾𝖽π())I(\bot,\mathsf{pred}_{\pi}(\top))\notin I and (𝗌𝗎𝖼π(),𝗉𝗋𝖾𝖽π())=(lu,ru)(\mathsf{suc}_{\pi}(\bot),\mathsf{pred}_{\pi}(\top))=(l_{u},r_{u}) for some vertex uu. The coefficient 44 counts the number of subsets of {(lu,lu),(𝗉𝗋𝖾𝖽π(ru),𝗉𝗋𝖾𝖽π(ru))}\{(l_{u},l_{u}),(\mathsf{pred}_{\pi}(r_{u}),\mathsf{pred}_{\pi}(r_{u}))\}. Note that when n=1n=1, lu=𝗉𝗋𝖾𝖽π(ru)l_{u}=\mathsf{pred}_{\pi}(r_{u}) and thus we have D1=C1+2D0D_{1}^{\prime}=C_{1}+2D_{0}. Each term 2DiDni2D^{\prime}_{i}D_{n-i} of (3) counts the number of pairs where the first splitting point is at the iith active vertex interval. In other words, 2i2i is the minimum cardinality of the first splitting component Y1Y_{1}. The coefficient 22 counts the cases where (rv,rv)(r_{v},r_{v}) presents and absents in II where rv=maxπY1r_{v}=\max_{\pi}Y_{1}.

We show Dn2αnCnD_{n}\leq 2^{\alpha n}C_{n} for α=2.4\alpha=2.4 by induction on nn. One can confirm it is true for n22n\leq 22 by calculation. For n23n\geq 23,

Dn\displaystyle D_{n} =Cn+4Dn1+2i=1n1(Ci+4Di1)Dni4D0Dn1\displaystyle=C_{n}+4D_{n-1}+2\sum_{i=1}^{n-1}(C_{i}+4D_{i-1})D_{n-i}-4D_{0}D_{n-1}
Cn+2i=1n1(Ci+2α(i1)+2Ci1)2α(ni)Cni\displaystyle\leq C_{n}+2\sum_{i=1}^{n-1}(C_{i}+2^{\alpha(i-1)+2}C_{i-1})2^{\alpha(n-i)}C_{n-i}
=Cn+2i=1n12α(ni)CiCni+2α(n1)+3i=1n1Ci1Cni\displaystyle=C_{n}+2\sum_{i=1}^{n-1}2^{\alpha(n-i)}C_{i}C_{n-i}+2^{\alpha(n-1)+3}\sum_{i=1}^{n-1}C_{i-1}C_{n-i}
=Cn+i=1n1(2α(ni)+2αi)CiCni+2α(n1)+3(C0Cn1+Cn1)\displaystyle=C_{n}+\sum_{i=1}^{n-1}(2^{\alpha(n-i)}+2^{\alpha i})C_{i}C_{n-i}+2^{\alpha(n-1)+3}(C_{0}C_{n-1}+C_{n-1})
Cn+(2α(n1)+2α)Cn+2α(n1)+4Cn1\displaystyle\leq C_{n}+(2^{\alpha(n-1)}+2^{\alpha})C_{n}+2^{\alpha(n-1)+4}C_{n-1}
=(1+2α(n1)+2α+n+14n22α(n1)+4)Cn.\displaystyle=(1+2^{\alpha(n-1)}+2^{\alpha}+{\textstyle\frac{n+1}{4n-2}}2^{\alpha(n-1)+4})C_{n}\,.

Noting that n23n\geq 23 and α=2.4\alpha=2.4, we obtain

Dn\displaystyle D_{n} 22.4nCn.\displaystyle\leq 2^{2.4n}C_{n}\,.

Hence, the number of possible abstractions is at most (k+1)!Dk+1(k+1)!22.4(k+1)(2(k+1))!(k+2)!(k+1)!O(22.4k(2k)!k!)(k+1)!D_{k+1}\leq(k+1)!\cdot 2^{2.4(k+1)}\frac{(2(k+1))!}{(k+2)!(k+1)!}\in O(2^{2.4k}\frac{(2k)!}{k!}). ∎

5.3 Circular-arc graphs

Circular-arc graphs are a generalization of interval graphs which have an arc model, which can be seen as a “circular” interval representation. For a linear order π\pi over a set SS and four elements p1,q1,p2,q2Sp_{1},q_{1},p_{2},q_{2}\in S, we write (p1,q1)/\π(p2,q2)(p_{1},q_{1})\mathrel{\text{$/$\hbox to0.0pt{\hss$\backslash$}}}_{\pi}(p_{2},q_{2}) if either

  • p1<πm<πq1p_{1}<_{\pi}m<_{\pi}q_{1} for some m{p2,q2}m\in\{p_{2},q_{2}\}, or

  • q1<πp1<πmq_{1}<_{\pi}p_{1}<_{\pi}m or m<πq1<πp1m<_{\pi}q_{1}<_{\pi}p_{1} for some m{p2,q2}m\in\{p_{2},q_{2}\}.

Otherwise, we write (p1,q1)//π(p2,q2)(p_{1},q_{1})\mathrel{/\!/\!}_{\pi}(p_{2},q_{2}). A circular-arc graph is a graph Gπ=(V,E)G_{\pi}=(V,E) such that

E={{u,v}V(lu,ru)/\π(lv,rv)}E=\{\,\{u,v\}\subseteq V\mid(l_{u},r_{u})\mathrel{\text{$/$\hbox to0.0pt{\hss$\backslash$}}}_{\pi}(l_{v},r_{v})\,\}

for some linear order π\pi over LVRV\mathit{L}_{V}\cup\mathit{R}_{V}. Note that this set contains neither \top nor \bot. The algorithm presented in Section 3 can easily be modified so that it solves the edge deletion problem for circular-arc graphs by replacing the definitions of //π\mathrel{/\!/\!}_{\pi} and /\π\mathrel{\text{$/$\hbox to0.0pt{\hss$\backslash$}}}_{\pi} as above, and defining 𝗌𝗎𝖼π(maxπX)=minπX\mathsf{suc}_{\pi}(\max_{\pi}X)=\min_{\pi}X and 𝗉𝗋𝖾𝖽π(minπX)=maxπX\mathsf{pred}_{\pi}(\min_{\pi}X)=\max_{\pi}X. Since we allow ru<πlur_{u}<_{\pi}l_{u}, the number of admissible arc model is bigger than that of (ordinary) interval representations. This affects the computational complexity.

Corollary 3.

The edge deletion problem for circular-arc graphs can be solved in O(|V|N2poly(k))O(|V|N^{2}\mathrm{poly}(k)) time where N=(2k)!24.38kN=(2k)!\cdot 2^{4.38k} for the treewidth kk of GG. If kk is the pathwidth, it can be solved in O(|V|Npoly(k))O(|V|N\mathrm{poly}(k)) time.

Proof.

There can be (2k+2)!(2k+2)! varieties of π\pi. One can argue that each π\pi has at most O(24.38k)O(2^{4.38k}) sets of forbidden intervals similarly to the proof of Theorem 1. ∎

5.4 Threshold graphs

Threshold graphs are special cases of trivially perfect graphs, which can be defined in several different ways. Here we use a pair of a vertex subset WVW\subseteq V and a linear order π\pi over RV\mit{R}_{V} as a threshold interval representation. We say that vertices uu and vv intersect on (W,π)(W,\pi) if and only if uWu\in W and rv<πrur_{v}<_{\pi}r_{u} or the other way around. A threshold graph is a graph GW,π=(V,EW,π)G_{W,\pi}=(V,E_{W,\pi}) where (W,π)(W,\pi) is a threshold interval representation on VV and EW,π={{u,v}Vu and v intersect on (W,π)}E_{W,\pi}=\{\,\{u,v\}\subseteq V\mid\text{$u$ and $v$ intersect on $(W,\pi)$}\,\}. By extending π\pi to π\pi^{\prime} over 𝐿𝑅V\mathit{LR}_{V} so that lw<πrvl_{w}<_{\pi^{\prime}}r_{v} for all wWw\in W and vVv\in V and 𝗌𝗎𝖼π(lu)=ru\mathsf{suc}_{\pi^{\prime}}(l_{u})=r_{u} for all uVWu\in V-W, then the induced interval graph coincides with the threshold graph. To attain drastic improvement on the complexity, we design an algorithm for the edge deletion problem for threshold graphs from scratch, rather than modifying the one for interval graphs.

For a threshold representation (Y,ρ)(Y,\rho) of a subgraph GY,ρ=(Xs,EY,ρ)G_{Y,\rho}=(X_{\leq s},E_{Y,\rho}), we define its abstraction 𝒜((Y,ρ),s)=(Y,π,b,p,c)\mathscr{A}((Y,\rho),s)=(Y^{\prime},\pi,b,p,c) as follows: (1) π\pi is the restriction of ρ\rho to RXsR_{X_{s}}, (2) Y=YXsY^{\prime}=Y\cap X_{s}, (3) if Y=YY^{\prime}=Y, then b=0b=0 and p=maxπ{pRXsp<ρry for all yXsXs},p=\max_{\pi}\{\,p\in R_{X_{s}}\mid p<_{\rho}r_{y}\text{ for all }y\in X_{\leq s}-X_{s}\,\}, (4) if YYY^{\prime}\neq Y, then b=1b=1 and p=maxπ{pRXsp<ρry for some yYY}p=\max_{\pi}\{\,p\in R_{X_{s}}\mid p<_{\rho}r_{y}\text{ for some }y\in Y-Y^{\prime}\,\}\,, and (5) c=|EsEY,ρEs|=|{{u,v}E{u,v}Xs and u and v do not intersect on (Y,ρ)}|c=|E_{\leq s}-E_{Y,\rho}-E_{s}|=|\{\,\{u,v\}\in E\mid\{u,v\}\nsubseteq X_{s}\text{ and $u$ and $v$ do not intersect on $(Y,\rho)$}\,\}|.

We say that (Y,π,b,p,c)(Y^{\prime},\pi^{\prime},b^{\prime},p^{\prime},c^{\prime}) dominates (Y,π,b,p,c)(Y,\pi,b,p,c) if Y=YY^{\prime}=Y, π=π\pi^{\prime}=\pi, ccc^{\prime}\leq c, and either (a) b=b=0b^{\prime}=b=0 and pπpp^{\prime}\geq_{\pi}p, (b) b=b=1b^{\prime}=b=1 and pπpp^{\prime}\leq_{\pi}p, or (c) b=0b^{\prime}=0 and b=1b=1. Using the above invariant, we can provide an algorithm for Threshold-Edge-Deletion.

Our algorithm assigns a set s\mathscr{I}_{s} for each node ss of TT so that the threshold counterpart of Condition 1 holds. Accordingly s={(,o,0,,0)}\mathscr{I}_{s}=\{(\emptyset,o,0,\bot,0)\} for leaf nodes ss.

Introduce Node:

Suppose ss has just one child tt such that Xs=Xt{x}X_{s}=X_{t}\cup\{x\}. For each (Y,π,b,p,c)t(Y,\pi,b,p,c)\in\mathscr{I}_{t}, we add to s\mathscr{I}^{\prime}_{s} all tuples (Y,π,b,p,c)(Y^{\prime},\pi^{\prime},b,p^{\prime},c) fulfilling the following conditions:

  • π\pi^{\prime} is an extension of π\pi to RXsR_{X_{s}},

  • YYY{x}Y\subseteq Y^{\prime}\subseteq Y\cup\{x\}.

  • if {x,u}E\{x,u\}\notin E for uXtu\in X_{t}, then xx and uu do not intersect in (Y,π)(Y^{\prime},\pi^{\prime}).

  • if b=0b=0, then xYx\notin Y^{\prime} or rx<π𝗌𝗎𝖼π(p)r_{x}<_{\pi^{\prime}}\mathsf{suc}_{\pi}(p),
    and moreover p={rxif p<πrx<𝗌𝗎𝖼π(p),potherwise,p^{\prime}=\begin{cases}r_{x}&\text{if $p<_{\pi^{\prime}}r_{x}<\mathsf{suc}_{\pi}(p)$,}\\ p&\text{otherwise,}\end{cases}

  • if b=1b=1, then xYx\notin Y^{\prime} and p=p<πrxp^{\prime}=p<_{\pi^{\prime}}r_{x}.

We then obtain s\mathscr{I}_{s} by reducing s\mathscr{I}^{\prime}_{s}.

Forget Node:

Suppose ss has just one child tt such that Xt=Xs{x}X_{t}=X_{s}\cup\{x\}. For each (Y,π,b,p,c)t(Y,\pi,b,p,c)\in\mathscr{I}_{t}, we add the tuple (Y,π,b,p,c)(Y^{\prime},\pi^{\prime},b^{\prime},p^{\prime},c^{\prime}) to s\mathscr{I}^{\prime}_{s} where

  • π\pi^{\prime} is the restriction of π\pi for RXsR_{X_{s}},

  • Y=Y{x}Y^{\prime}=Y-\{x\},

  • b=1b^{\prime}=1 if xYx\in Y, and b=bb^{\prime}=b otherwise,

  • if b=0b^{\prime}=0, then p=min{p,𝗉𝗋𝖾𝖽π(rx)}p^{\prime}=\min\{p,\mathsf{pred}_{\pi}(r_{x})\},

  • if b=1b^{\prime}=1, then p={𝗉𝗋𝖾𝖽π(rx)if b=0 or p<πrxxY or rx=p,potherwise.p^{\prime}=\begin{cases}\mathsf{pred}_{\pi}(r_{x})&\text{if $b=0$ or $p<_{\pi}r_{x}\wedge x\in Y$ or $r_{x}=p$},\\ p&\text{otherwise.}\end{cases}

  • c=c+|{{u,x}EuXs and x do not intersect on (Y,π)}|c^{\prime}=c+|\{\,\{u,x\}\in E\mid\text{$u\in X_{s}$ and $x$ do not intersect on $(Y,\pi)$}\,\}|

We then obtain s\mathscr{I}_{s} by reducing s\mathscr{I}^{\prime}_{s}.

Join node:

Suppose ss has two children t1t_{1} and t2t_{2}, where Xs=Xt1=Xt2X_{s}=X_{t_{1}}=X_{t_{2}}. We add (Y,π,b,p,c)(Y,\pi,b,p,c) to s\mathscr{I}^{\prime}_{s} if there are (Y,π,b1,p1,c1)t1(Y,\pi,b_{1},p_{1},c_{1})\in\mathscr{I}_{t_{1}} and (Y,π,b2,p2,c2)t2(Y,\pi,b_{2},p_{2},c_{2})\in\mathscr{I}_{t_{2}} such that c=c1+c2c=c_{1}+c_{2}, and either

  • b=b1=b2b=b_{1}=b_{2} and p=minπ1{p1,p2}p=\min_{\pi_{1}}\{p_{1},p_{2}\},

  • b=b1=1b=b_{1}=1, b2=0b_{2}=0 and p=p1πp2p=p_{1}\leq_{\pi}p_{2}, or

  • b=b2=1b=b_{2}=1, b1=0b_{1}=0 and p=p2πp1p=p_{2}\leq_{\pi}p_{1}.

We then obtain s\mathscr{I}_{s} by reducing s\mathscr{I}^{\prime}_{s}.

Theorem 3.

The edge deletion problem for threshold graphs can be solved in O(|V|N2poly(k))O(|V|N^{2}\mathrm{poly}(k)) time where N=k!2kN=k!\cdot 2^{k} for the treewidth kk of GG. If kk is the pathwidth, it can be solved in O(|V|Npoly(k))O(|V|N\mathrm{poly}(k)) time.

6 Conclusion

We have proposed FPT algorithms for Edge-Deletion to some intersection graphs parameterized by treewidth in this paper. Our algorithms maintain partial intersection models on a node of a tree decomposition with some restrictions and extend the models consistently for the restrictions in the next step. We expect that the ideas in our algorithms can be applied to other intersection graphs whose intersection models can be represented as linear-orders, for example circle graphs, chain graphs and so on, and to Vertex-Deletion of intersection graphs.

We have the following questions as future work:

  • Do there exist single exponential time algorithms for the considered problems, that is, O(2𝑡𝑤(G))O^{*}(2^{\mathit{tw}(G)}) time, or can we show matching lower bounds assuming the Exponential Time Hypothesis?

  • Are there FPT algorithms parameterized by treewidth for 𝒞\mathcal{C}-Completion which is to find the minimum number of adding edges to obtain a graph in an intersection graph class 𝒞\mathcal{C}? We can naturally apply the idea of our algorithms to 𝒞\mathcal{C}-Completion problems. While 𝒞\mathcal{C}-Edge-Deletion algorithms do not allow introduced objects to intersect with forgotten objects, 𝒞\mathcal{C}-Completion algorithms do allow it with the cost of addition of new edges. Thus 𝒞\mathcal{C}-Completion algorithms based on this naive approach will be XP algorithms since we have to remember the number of forgotten objects in the representation to count the number of intersections between the introduced objects and forgotten objects.

  • Are there FPT algorithms for Edge-Deletion to intersection graphs defined using objects on a plane, like unit disk graphs? The intersection graph classes discussed in this paper are all defined using objects aligned on a line. Going up to a geometric space of higher dimension is a challenging topic.

Acknowledgement

We are very much grateful to Yasuaki Kobayashi for his valuable comment that has improved the complexity analyses. This work was supported in part by JSPS KAKENHI Grant Numbers JP18H04091 and JP19K12098.

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