Polarity and Monopolarity of 3 -colourable comparability graphs
aa r X i v : . [ c s . CC ] A p r Polarity and Monopolarity of 3-colourablecomparability graphs
Nikola Yolov ∗ [email protected] October 15, 2018
Abstract
We sharpen the result that polarity and monopolarity are NP-completeproblems by showing that they remain NP-complete if the input graph isrestricted to be a 3-colourable comparability graph.We start by presenting a construction reducing 1-3-SAT to monopo-larity of 3-colourable comparability graphs. Then we show that polarityis at least as hard as monopolarity for input graphs restricted to a fixeddisjoint-union-closed class. We conclude the paper by stating that bothpolarity and monopolarity of 3-colourable comparability graphs are NP-complete problems.
Keywords : Algorithms, Graph Theory, Complexity Theory, NP-completeness, Polar Graphs, Comparability Graphs
A partition (
A, B ) of the vertices of a graph G is called polar if G [ A ] and G [ B ]are unions of disjoint cliques. A polar partition ( A, B ) is called monopolar if A is an independent set, and unipolar if A is a clique. A graph G is called tobe polar , monopolar or unipolar if it admits a polar, a monopolar or a unipolarpartition respectively. If ( A, B ) is a polar partition of G , then ( B, A ) is polarpartition of G , hence a graph is polar iff its complement, G , is polar. Firststudied by Tyshkevich and Chernyak in [TC85b] and [TC85a], monopolar graphsare a natural generalisation of bipartite and split graphs, and polar graphsare a generalisation of monopolar and co-bipartite graphs. The problems ofdeciding whether a graph is polar, monopolar and unipolar are called polarity , monopolarity and unipolarity respectively. Polarity [CC86] and monopolarity[Far04] are NP-complete problems and enjoyed a lot of attention recently. In thiswork we sharpen the hardness result by showing that the problems remain NP-complete even if the input graph is restricted to be a 3-colourable comparabilitygraph. In contrast, unipolarity can be resolved in quadratic time [MY15]. ∗ The author is funded by the Engineering and Physical Sciences Research Council (EPSRC)Doctoral Training Grant and the Department of Computer Science, University of Oxford
1t is shown in [LN14] that polarity and monopolarity of 3-colourable graphsare NP-complete problems. On the other hand, there are polynomial time al-gorithms for monopolarity and polarity of permutation graphs [EHM09], and apolynomial time algorithm for monopolarity of co-comparability graphs [CH12].Note that a graph is a permutation graph if and only if it is both a comparabilityand a co-comparability graph, hence the latter algorithm is a generalisation ofthe prior, but less efficient. It is natural to ask whether there is a polynomialtime polarity or monopolarity algorithm for the other superclass of permutationgraphs – comparability graphs. In this paper we give a negative answer (pro-vided P = NP), in fact we show that even if the graph is restricted to a smallerclass – the class of 3-colourable comparability graphs – the problem remainsNP-hard. Polarity of comparability graphs and polarity of co-comparabilitygraphs are easily seen to be polynomially reducible to each other, since a graphis polar iff its complement is polar, hence we can deduce that polarity of co-comparability graphs is also NP-complete. The table below summarises ourdiscussion so far. (Here C and C c are used to denote the classes of comparabil-ity and co-comparability graphs respectively.)Input graph Monopolarity Polarity3 − colourable NP-c [LN14] NP-c [LN14] C ∩ C c P [EHM09] P [EHM09] C c P [CH12] NP-c, this paper C NP-c, this paper NP-c, this paper3 − colourable ∩ C NP-c, this paper NP-c, this paperWe note that Churchley announced an unpublished work proving that monopo-larity of comparability graphs is NP-complete. This paper is independent ofit. In comparison, we address both polarity and monopolarity, and consider asmaller class – the class of 3-colourable comparability graphs, in contrast to theclass of comparability graphs.In order to show the polarity result we present a brief lemma stating thatpolarity is not easier than monopolarity for graph classes closed under disjointunion. The question whether there exists a class for which polarity is easierthan monopolarity was asked in [LN14]. We contribute to the answer by statingthat such class is certainly not closed under disjoint union.
We use standard notation for graph theory and highlight that G [ S ] is usedto denote the induced subgraph of G by the vertices S ⊆ V ( G ). A union ofdisjoint cliques is a graph whose vertices can be partitioned into blocks, so thattwo vertices are joined by an edge if and only if they belong to the same block.A co-union of disjoint cliques is the complement of such graph.A graph G is k -colourable if its vertices can be covered by k independentsets. It is NP-hard to decide whether a graph is k -colourable for k ≥ G is a comparability graph if each edge can be oriented towards one ofits endpoints, so that the result orientation is transitive. There is an algorithmto test if a graph is a comparability graph in O (MM) time [Spi03], where MMis the time required for a matrix multiplication. A graph is perfect if χ ( H ) = ω ( H ) for every induced subgraph H . We note that there is an interestingconnection between unipolar and perfect graphs – almost all perfect graphs areeither unipolar or co-unipolar [PS92].Comparability graphs are easily seen to be perfect, and therefore a 3-colourablecomparability graph is simply a K -free comparability graph. The completegraph K is an example of a comparability graph which is not 3-colourable,and C is an example of a 3-colourable graph which is not comparability, hence3-colourable comparability graphs is a proper subset of the classes above. It this section we show that the problem of deciding whether a 3-colourablecomparability graph is monopolar is NP-complete. We call a CNF-formula φ = V i C i a positive C i contains exactly threenon-negated literals. We transform the following NP-complete problem (1-3-SAT) [GJ79] into the problem above: Instance:
A positive 3-CNF-formula φ with variables c . . . c n . Question:
Is there an assignment of the variables { c i } → { true, f alse } , suchthat each clause contains exactly one true literal.Recall that a partition ( A, B ) of G is monopolar if A is an independent setand G [ B ] is a union of disjoint cliques. For a fixed monopolar partition ( A, B ),we call a vertex v left if v ∈ A , and right otherwise. (Imagine that a partition( A, B ) is always drawn with A on the left-hand side and B on the right-handside). Observe that if v is a left vertex, the neighbours of v are right, hence theyinduce a union of disjoint cliques.Consider the graph Q in Figure 1. v v v v u Figure 1: Q Claim 3.1.
The only monopolar partition of Q is ( { v , v } , { v , v , u } ) .Proof. The neighbourhood of v and v is P (a path with three vertices), hencethey must be both right for all monopolar partitions of Q . As v and v arenot connected and they share a right neighbour, at least one of them is left.3herefore u has a left neighbour, so u must be right. However, if v , v and u are right, then v and v must be left.Consider the graph H in Figure 2. The orientation of the edges is transitive, v v v v v t t t v v v v v v v v v v Figure 2: H hence H is a comparability graph. Further, comparability graphs are perfect,hence χ ( H ) = ω ( H ) = 3, i.e. H is three-colourable. Lemma 3.2.
Exactly one of t , t and t is right in every monopolar partition of H . There are exactly three monopolar partitions of H and each one is uniquelydetermined by which vertex of { t , t , t } is right.Proof. Let V Q = { v , . . . v } and V Q = { v , v , v , v , v } . Observe that H [ V Q ] ∼ = H [ V Q ] ∼ = Q , and therefore in every monopolar partition v and v are right vertices and they cannot have other right neighbours from V Q and V Q respectively. The vertex v must be right, so at most one of { t , t , t } can be right. It is a routine to check that setting all t , t and t left cannotyield a monopolar partition. It is also a routine to check that setting each one of { t , t , t } right and the other two left defines a unique monopolar partition.We associate every clause of an arbitrary positive 3-CNF formula φ with anindependent copy of H . The selection of the right vertex among { t , t , t } ina monopolar partition will indicate which of the variables of the clause is true.Then we add extra synchronisation vertices, each uniquely associated with avariable of φ and joined by an edge to all vertices associated with the samevariable in the copies of H .More formally, let φ = V mi =1 C i be a positive 3-CNF formula with variables c . . . c n . Let G be the disjoint union of an independent set { x , . . . x n } and m disjoint copies of H , { H i } mi =1 . For each clause C i , say C i = { c k ∨ c l ∨ c p } ,connect x k with t i, , x l with t i, and x p with t i, , where t i, , t i, and t i, arerespectively t , t and t in H i . Definition 3.3.
For every positive 3-CNF formula φ define G φ to be the graphdescribed above. Lemma 3.4.
A positive 3-CNF formula φ is a “yes”-instance of 1-3-SAT iff G φ is monopolar. roof. ( ⇒ ) Assume that f : { c i } → { true, f alse } is an assignment of the vari-ables of φ such that every clause contains exactly one true literal. Create apartition ( A, B ) of V ( G φ ) as follows: x i ∈ A for each c i with f ( c i ) = true ,and x i ∈ B otherwise. The vertices { x . . . x n } have disjoint neighbourhoods,so we can extend the partition above with v ∈ A ⇔ x i ∈ B for each x i andneighbour v of x i . By Lemma 3.2 the partition above can be further extendeduniquely to each H i , and hence the entire graph G φ . We conclude that ( A, B )is a monopolar partition of G φ .( ⇐ ) Let ( A, B ) be a monopolar partition of G φ . Define f : { c i } → { true, f alse } as follows: f ( c i ) = true ↔ x i ∈ A . From Lemma 3.2 for each clause C i there is aunique t i,j ∈ B , which is adjacent to v i, ∈ B . Observe that v i, is non-adjacentto any x j , hence for each clause C there is a unique x j ∈ A with c j ∈ C . Lemma 3.5.
For every positive 3-CNF formula φ , G φ is a 3-colourable com-parability graph.Proof. To see that G φ is a comparability graph, orient each copy of H as ori-ented in Figure 2. Orient the remaining edges from x . . . x n towards t i,j . Thedescribed orientation of G φ is transitive. As G φ is perfect, we have χ ( G φ ) = ω ( G φ ) = 3. Corollary 3.6.
Monopolarity is NP-complete even if the input graph is re-stricted to be -colourable comparability graph.Proof. Monopolarity is in NP regardless of the restrictions on the input graph.Furthermore, NP-hardness follows from the reduction of 1-3-SAT in the state-ment of Lemma 3.4.It is worth noting that further restrictions can be imposed on problem. Wecan design G φ so that the different copies of H share the four triangles theycontain, and therefore build G φ with a constant number (four) of triangles. It this section we prove that it is NP-complete to decide whether a 3-colourablecomparability graph is polar or not. We do this by showing that polarity isat least as hard as monopolarity for classes of graphs which are closed underdisjoint union. Note that 3-colourable comparability graphs form such class.
Claim 4.1.
Suppose G is a complement of a union of disjoint cliques. Then G is either connected or empty.Proof. Since G is a union of disjoint cliques, G is either a clique of disconnected.The complement of a disconnected graph is connected, therefore G is eitherempty or connected.We use the notation G = 2 H to express that G is a union of two disjointcopies of H without edges inbetween. 5 emma 4.2. The following three statements are equivalent:1. H = 2 G is polar.2. G is monopolar3. H = 2 G is monopolar.Proof. (2 . ⇒ . ) Trivial.(3 . ⇒ . ) Trivial.(1 . ⇒ . ) Let ( A, B ) be a polar partition of V ( H ), and let ( V , V ) be apartition of V ( H ), such that H [ V i ] ∼ = G . Let A i = A ∪ V i . If A or A isempty, then G is a union of cliques and hence monopolar. Otherwise, H [ A ] isdisconnected because there are no edges between A and A . But H [ A ] is a co-union of cliques, hence H [ A ] is empty by Claim 4.1. We deduce that ( A i , V i \ A i )is a monopolar partition G [ V i ]. Lemma 4.3.
Let P be a class of graphs which is closed under disjoint union.Determining polarity of instances restricted to P is at least as hard as deter-mining monopolarity for the same class of instances.Proof. We reduce monopolarity for input graphs restricted to P to polarity forinput graphs restricted to P . To decide if G ∈ P is monopolar it is sufficient todecide whether 2 G ∈ P is polar by Lemma 4.2. Corollary 4.4.
The problem of deciding whether a -colourable comparabilitygraph is polar is an NP-complete problem.Proof. The problem is clearly in NP, and it is NP-complete to decide whethersuch graph is monopolar by Corollary 3.6, hence the statement follows fromLemma 4.3.
Corollary 4.5.
It is an NP-complete problem to decide whether a co-comparabilitygraph is polar.Proof.
A graph is polar if its complement is polar. In order to decide whether acomparability graph is polar, it is sufficient to decide whether its complement,a co-comparability graph, is polar. The prior decision problem is NP-completeby Corollary 4.4, hence the latter is also NP-complete.
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