Discrete Applied Mathematics ( ) –

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Discrete Applied Mathematics

journal homepage: www.elsevier.com/locate/dam

The complexity of forbidden subgraph sandwich problemsand the skew partition sandwich problem✩

Simone Dantas a, Celina M.H. de Figueiredo b,∗, Frédéric Maffray c,Rafael B. Teixeira d

a Instituto de Matemática e Estatística, Universidade Federal Fluminense, Brazilb COPPE, Universidade Federal do Rio de Janeiro, Brazilc CNRS, Laboratoire G-SCOP, INPG, Université Joseph Fourier, Grenoble, Franced Instituto de Ciências Exatas, Universidade Federal Rural do Rio de Janeiro, Brazil

a r t i c l e i n f o

Article history:Received 22 January 2013Received in revised form 14 August 2013Accepted 20 September 2013Available online xxxx

Keywords:Graph sandwich problemForbidden subgraphPerfect graphs

a b s t r a c t

The Π graph sandwich problem asks, for a pair of graphs G1 = (V , E1) and G2 = (V , E2)with E1 ⊆ E2, whether there exists a graph G = (V , E) that satisfies property Π andE1 ⊆ E ⊆ E2. We consider the property of being F-free, where F is a fixed graph. We showthat the claw-free graph sandwich and the bull-free graph sandwich problems are bothNP-complete, but the paw-free graph sandwich problem is polynomial. This completesthe study of all cases where F has at most four vertices. A skew partition of a graph Gis a partition of its vertex set into four nonempty parts A, B, C,D such that each vertexof A is adjacent to each vertex of B, and each vertex of C is nonadjacent to each vertexof D. We prove that the skew partition sandwich problem is NP-complete, establishing acomputational complexity non-monotonicity.

© 2013 Elsevier B.V. All rights reserved.

1. Introduction

All graphs considered here are finite and undirected. Given a graph property Π , the Π graph sandwich problem isdefined as follows:

Input: A pair (G1,G2) of graphs with G1 = (V , E1),G2 = (V , E2) and E1 ⊆ E2;Question: Is there a graph G = (V , E) that satisfies property Π and inclusion E1 ⊆ E ⊆ E2?

The graph sandwich problem was introduced by Golumbic and Shamir in [16] and further studied in [13,17]. Clearly,when G1 = G2 the problem is to decide whether G1 satisfies property Π . So the graph sandwich problem generalizesthe recognition problem of deciding whether a graph satisfies a given property. In particular, if the recognition problem isNP-complete, then the sandwich problem is also NP-complete. When the property Π is to belong to a class C of graphs,we may also speak of the C graph sandwich problem. Golumbic, Kaplan and Shamir [14,15] proved that the interval graph,unit interval graph, permutation graph and comparability graph sandwich problems are all NP-complete, while the splitgraph, threshold graph and cograph sandwich problems are in P. Graph sandwich problems have attracted much attention

✩ An extended abstract with some results of this paper was presented at LAGOS 2009, the Latin-American Algorithms, Graphs and OptimizationSymposium, and appeared in Electronic Notes in Discrete Mathematics 35 (2009) 9–14. Partially supported by CNPQ, FAPERJ and CAPES (Brazil) andCOFECUB (France) under joint project MA 622/08.∗ Corresponding author. Tel.: +55 21 25628678.

E-mail addresses: sdantas@im.uff.br (S. Dantas), celina@cos.ufrj.br, cmhfig@gmail.com (C.M.H. de Figueiredo), frederic.maffray@inpg.fr (F. Maffray),rafaelbt@ufrrj.br (R.B. Teixeira).

0166-218X/$ – see front matter© 2013 Elsevier B.V. All rights reserved.http://dx.doi.org/10.1016/j.dam.2013.09.004

2 S. Dantas et al. / Discrete Applied Mathematics ( ) –

Fig. 1. Claw, paw and bull.

because of many applications and as a natural generalization of recognition problems. See e.g. [1,7,6,8,10,18,19,25] wheremany results on sandwich problems for graph properties and classes were obtained.

We say that a graph G contains a graph F if some induced subgraph of G is isomorphic to F . A graph G is F-free if it doesnot contain F . Dantas, de Figueiredo, da Silva and Teixeira [6] initiated a study of the F-free graph sandwich problem, anddetermined the complexity status (in P or NP-complete) of the problem for several graphs F , including the cases when F isthe diamond (K4 \ e) and when F is the 4-hole C4. Here, we develop this study by determining the complexity status for thecases where F has four vertices that were not solved in [6]. We also consider some graphs on five vertices or more.

A long-standing open problem posed in the seminal paper by Golumbic, Kaplan and Shamir [15] is the complexity ofthe perfect graph sandwich problem. The celebrated proof of the Strong Perfect Graph Theorem by Chudnovsky et al. [3]established the characterization of perfect graphs by forbidden induced subgraphs; in that proof, the concept of skewpartition plays a key role. A skew partition of a graph G = (V , E) is a partition of its vertex set V into four nonempty partsA, B, C,D such that each vertex of part A is adjacent to each vertex of part B, and each vertex of part C is nonadjacent to eachvertex of partD. Note that if (A, B, C,D) is a skewpartition ofG then (C,D, A, B) is a skewpartition ofG and vice versa. It is thisself-complementarity which first suggested that these partitions might be important to an understanding of the structureof perfect graphs. Chvátal [4] introduced the concept of skew partition. He conjectured that no minimal imperfect graphadmits a skew partition and speculated that skew partitions might play a key role in a decomposition theoremwhich wouldimply the Strong Perfect Graph Conjecture. Chvátal [4] defined a star cutset as any cutset that contains a vertex adjacentto every other vertex of the cutset. If G contains a skew partition (A, B, C,D), then A ∪ B is a skew cutset of G separatingG[C] from G[D]. If A is a clique of G, then G has a star cutset. If both A and B are cliques of G, then G has a clique cutset.Chvátal [4] proved that no minimal imperfect graph has a star cutset and conjectured that no minimal imperfect graph hasa skew cutset. Polynomial-time algorithms for the recognition of clique cutset [28], star cutset [4], and skew cutset [9] wereestablished. Recent work has further studied the computational complexity of skew cutsets [20,22,27].

Regarding the corresponding sandwich problems clique cutset sandwich problem is NP-complete, but star cutsetsandwich problem is polynomial [26]. We prove that the further generalization skew partition sandwich problem is NP-complete establishing an interesting computational complexity non-monotonicity. Note that homogeneous set and cliquecutset are vertex partitions into three nonempty parts. The dichotomy polynomial time versus NP-completewas completelydetermined for the class of three nonempty part sandwich problems [24].

Let us recall some basic definitions and notation. For any integer k ≥ 4, a k-hole is a chordless cycle of length k and isdenoted by Ck. For any integer k ≥ 1, we let Kk and Pk respectively denote the complete graph and the path on k vertices.Let Kp \ e denote the complete graph on p vertices minus one edge. For any graph G and subset X of vertices of G, we let G[X]

denote the subgraph of G induced by X . The complementary graph of G is denoted by G. Given two vertex-disjoint graphsF1 and F2, their union is the graph F1 + F2 with vertex-set V (F1) ∪ V (F2) and edge-set E(F1) ∪ E(F2). The union of k copies ofone graph G is denoted by kG.

For an instance (G1,G2) of the sandwich problemwith G1 = (V , E1),G2 = (V , E2) and E1 ⊆ E2, we say that any memberof E1 is a forced edge, any member of Eopt = E2 \ E1 is an optional edge, and any pair in V × V \ E2 is a forbidden pair. Everygraph G = (V , E) with E1 ⊆ E ⊆ E2 is called a sandwich graph for the pair (G1,G2). In that case, E consists of all forcededges plus some (possibly zero) optional edges, and E contains no edge corresponding to a forbidden pair.

We recall some results from [6,15].

• Complementary graphs [15]. For any fixed graph F , if G is an F-free sandwich graph for the pair (G1,G2), then G is an F-freesandwich graph for the pair (G2,G1). It follows that the F-free graph sandwich problem and the F-free graph sandwichproblem have the same complexity.

• Complete graphs. If F is the complete graph Kp, then an instance (G1,G2) of the F-free graph sandwich problem has asolution if and only if G1 is Kp-free. This can be tested in polynomial time as p is a constant.

• Complete graph minus an edge [6]. For every fixed p, the (Kp \ e)-free sandwich problem is in P.• Holes [6]. For every fixed k ≥ 4, the k-hole-free sandwich problem is NP-complete.• P4-free graphs (cographs) [15]. The P4-free graph sandwich problem is in P.

Let F be any graph on four vertices. Up to isomorphism there are eleven such graphs. If F is either K4 (or its complement)or K4 \ e (or its complement) or P4 (which is self-complementary), then the F-free graph sandwich problem is in P by thepreceding results. If F is C4 (or its complement 2K2), then the problem is NP-complete as mentioned earlier. The remainingtwo cases are when F is the claw (or its complement) and the paw (or its complement), where the claw is the graphwith vertex-set {a, b, c, d} and edge-set {ab, ac, ad} and the paw is the graph with vertex-set {a, b, c, d} and edge-set{ab, ac, ad, bc}. See Fig. 1. These two cases are solved now as follows.

S. Dantas et al. / Discrete Applied Mathematics ( ) – 3

Fig. 2. Example of No instance (G1,G2) for the paw-free graph sandwich problem.

Fig. 3. Example of Yes instance (G1,G2) for the paw-free graph sandwich problemwith corresponding sandwich graph H at the middle.

Theorem 1. The paw-free graph sandwich problem is in P.

Theorem 2. The claw-free graph sandwich problem is NP-complete.

The bull is the graphwith vertex-set {a, b, c, d, e} and edge-set {ab, bc, cd, de, bd}. See Fig. 1.We establish the complexitystatus of two additional sandwich problems:

Theorem 3. The bull-free graph sandwich problem is NP-complete.

Theorem 4. The skew partition sandwich problem is NP-complete.

The proofs of these theorems are given in the next sections.

2. The paw-free graph sandwich problem

The structure of paw-free graphs is determined by the following theorem.

Theorem 5 ([21]). A graph G is paw-free if and only if every component of G is a K3-free graph or a complete multipartite graph.Moreover, G is complete multipartite if and only if it is P3-free.

Proof of Theorem 1. Let (G1,G2) be an instance of the paw-free graph sandwich problem. First suppose that G1 is notconnected. Let V1, . . . , Vk be the vertex-sets of the components of G1, with k ≥ 2. If (G1,G2) admits a paw-free sandwichgraph H , then, for each i in {1, . . . , k}, the graph H[Vi] is a paw-free sandwich graph for the pair (G1[Vi],G2[Vi]). Conversely,if, for each i in {1, . . . , k}, somegraphHi is a paw-free sandwich graph for the pair (G1[Vi],G2[Vi]), then the graphH1+· · ·+Hkis a paw-free sandwich graph for the pair (G1,G2). In other words, it suffices to solve the problem for each component of G1.

Wemay now assume that G1 is connected. By Theorem 5, the pair (G1,G2) admits a paw-free sandwich graph if and onlyif the pair (G1,G2) admits a sandwich graph H that is K3-free or P3-free, which in turn occurs if and only if the pair (G1,G2)admits a K3-free sandwich graph or the pair (G1,G2) admits a P3-free sandwich graph. In either case, the problem can besolved in polynomial time by the results given in Section 1. This completes the proof of the theorem. �

Please refer to Figs. 2 and 3 for examples of a No instance and a Yes instance for the paw-free graph sandwich problem.The solid edges are forced edges of E1, and the dotted edges are optional edges of Eopt.

3. The claw-free graph sandwich problem

Proof of Theorem 2. We give a reduction from the basic problem of 3-Satisfiability, proved NP-complete by Cook [5].Let the input of 3-Satisfiability be a Boolean formula ϕ on a set of n variables {x1, . . . , xn}, ϕ = C1

∧ · · · ∧ Cm, whereeach clause Cℓ is of the form zi ∨ zj ∨ zk for some distinct i, j, k in {1, . . . , n}, and each literal zh is either xh or xh. With ϕ weassociate an instance of the graph sandwich problem as follows.

For every variable xi, letmi be the number of clauses that contain xi (negated or not). Make a graph Xi with the followingsets of vertices:

Let Ai = {a1i , . . . , amii }, Bi = {b1i , . . . , b

mii },Di = {d1i , . . . , d

mii } and Qi = {q1i , . . . , q

mii } be four pairwise disjoint sets of

vertices, let ci, fi, gi be three more vertices, and let V (Xi) = Ai ∪ Bi ∪ Di ∪ Qi ∪ {ci, fi, gi}. Moreover, let Xi have the followingedges:

• Forced edges: every pair civ with v ∈ Ai ∪ Bi ∪ Di ∪ {fi}, every pair aℓi q

ℓi (ℓ = 1, . . . ,mi), the pair figi, and, for each R in

{Ai, Bi,Di}, every pair uv with u, v ∈ R (i.e., Ai, Bi,Di are cliques in the graph of forced edges);• Optional edges: every pair uv with u in Ai ∪ {fi} and v in Bi ∪ Di.

4 S. Dantas et al. / Discrete Applied Mathematics ( ) –

Fig. 4. Instance (V , E1, E2) obtained from the satisfiable instance of 3sat: I = (X, ϕ) = ({x1, x2, x3}, {(x1, x2, x3)}). The truth assignment x1 = x3 =

T , x2 = F satisfies (X, ϕ). The bold dashed edges are the optional edges added to G1 to obtain the claw-free sandwich graph H .

Now consider any clause Cℓ (ℓ ∈ {1, . . . ,m}) of ϕ, with Cℓ= zi ∨ zj ∨ zk, where each zh is either xh or xh. For each h in

{i, j, k}, we associate two vertices of Xh with Cℓ, as follows. If zh = xh, then we associate with Cℓ one pair {aℓh, b

ℓh}; if zh = xh,

then we associate with Cℓ one pair {aℓh, d

ℓh}. Each clause that contains xh or xh is associated with a pair that has a different

superscript ℓ in {1, . . . ,mh}. Any such pair is called a clause pair. Thus, for each h in {1, . . . , n} and each ℓ in {1, . . . ,mh},exactly one of {aℓ

h, bℓh} and {aℓ

h, dℓh} is a clause pair. To simplify the notation, the two vertices of Xh associated with Cℓ will be

called aℓh and βℓ

h (where the symbol β is either b or d); and we will call qℓh the vertex of Qh that is adjacent to aℓ

h. We add fourvertices wℓ, wℓ

i , wℓj , w

ℓk , and let V (Cℓ) = {wℓ, wℓ

i , wℓj , w

ℓk}. Moreover, we add the following edges:

• Forced edges: for each h in {i, j, k}, the pairs wℓhw

ℓ, wℓha

ℓh, w

ℓhβ

ℓh , w

ℓhq

ℓh;

• Optional edges: the three pairs in the set {wℓi , w

ℓj , w

ℓk}, and every pair uv with u in {wℓ, wℓ

i , wℓj , w

ℓk} and v in

h∈{i,j,k}{aℓh, β

ℓh , q

ℓh} such that uv is not forced. Let i < j < k, and add the pairs qℓ

i βℓj , q

ℓj β

ℓi , q

ℓkβ

ℓi .

Any pair of vertices that is not forced or optional as above is forbidden.Let V ϕ

=n

i=1 V (Xi) ∪m

ℓ=1 V (Cℓ). Let E1 be the set of forced edges, Eopt the set of optional edges, and E2 = E1 ∪ Eopt.Let G1 = (V ϕ, E1) and G2 = (V ϕ, E2). It is easy to see that |V ϕ

| = 3n + 16m, so the size of (G1,G2) is polynomial in the sizeof ϕ. See Fig. 4 for an example. The solid edges are forced edges of E1, and the dotted edges are optional edges of Eopt.

We claim that:

The formula ϕ is satisfiable if and only if the pair (G1,G2) admits a claw-free sandwich graph. (1)

Proof of (1). First suppose that H is a claw-free sandwich graph for (G1,G2). For each i, one of the two optional pairs fib1iand fid1i must be an edge of H , for otherwise {ci, fi, b1i , d

1i } induces a claw; and not both of them are edges, for otherwise

{fi, gi, b1i , d1i } induces a claw. This leads to the two cases (a) and (b) below, which are symmetric. (Here the word edge refers

to H .)

(a) fib1i is an edge and fid1i is not an edge. For each ℓ and ℓ′ in {1, . . . ,mi}, we observe that: fibℓi is an edge, for otherwise

{ci, fi, bℓi , d

1i } induces a claw; fidℓ

i is not an edge, for otherwise {fi, gi, b1i , dℓi } induces a claw; aℓ′

i dℓi is an edge, for otherwise

{ci, fi, aℓ′

i , dℓi } induces a claw; and aℓ′

i bℓi is not an edge, for otherwise {aℓ′

i , qℓ′

i , bℓi , d

ℓi } induces a claw. In summary, Ai ∪Di

and Bi ∪ {fi} are two cliques in H , and there is no edge between them.(b) fid1i is an edge and fib1i is not an edge. By the same arguments as in (a) and by symmetry, we obtain that Ai ∪Bi and Di ∪{fi}

are two cliques in H , and there is no edge between them.

In case (a), we set the variable xi to false; in case (b), we set it to true.Now consider any clause Cℓ (ℓ ∈ {1, . . . ,m}) of ϕ, say Cℓ

= zi ∨ zj ∨ zk, where zh is either xh or xh, for each h ∈ {i, j, k}.Suppose that all three literals zi, zj, zk are false. By (a) and (b), each of the three pairs {aℓ

h, βℓh} (h ∈ {i, j, k}) associatedwith Cℓ

is not an edge of H . Then wℓ must have a neighbor in {aℓh, β

ℓh}, for otherwise {wℓ

h, wℓ, aℓ

h, βℓh} induces a claw in H . This holds

for each of the three values of h ∈ {i, j, k}. But then wℓ and these three neighbors of wℓ induce a claw in H , a contradiction.So one of zi, zj, zk is true. Hence each clause is satisfied.

Conversely, suppose that ϕ admits a satisfying truth assignment (ξ1, . . . , ξn). Starting from G1, make a graphH by addingthe following optional edges for each i in {1, . . . , n} and each clause Cℓ:

– If ξi = true, add all edges between fi and Di and between Ai and Bi.

S. Dantas et al. / Discrete Applied Mathematics ( ) – 5

– If ξi = false, add all edges between fi and Bi and between Ai and Di.– For each clause Cℓ, say Cℓ

= zi ∨ zj ∨ zk, where zh is either xh or xh, choose exactly one literal that is true, say zi. Let j < k.Then add edges wℓqℓ

i , wℓβℓj , wℓβℓ

k , wℓi w

ℓj , wℓ

i βℓj , qℓ

i wℓj , qℓ

i βℓj .

– Do not add any other optional pair.

Let us prove that H is claw-free. For this purpose it suffices to examine every vertex u of H and to show that the closedneighborhood of u in H (i.e., the set NH(u) ∪ {u}) can be partitioned into two cliques R′

u and R′′u . In order to exhibit this

property concisely, we use the shorthand notation u → R′u ⊕ R′′

u (and if R′′u = ∅, then we write u → R′

u). Now we see that:

– If ξi = true, then ci → (Ai ∪ Bi) ⊕ ({ci, fi} ∪ Di), fi → ({ci} ∪ Di) ⊕ {fi, gi}, gi → {fi, gi}; if aℓi b

ℓi is not a clause pair, then

bℓi → Ai ∪ Bi ∪ {ci}; if aℓ

i dℓi is not a clause pair, then dℓ

i → Di ∪ {ci, fi}.– If ξi = false, similar properties hold by symmetry.– For each clause Cℓ

= zi ∨ zj ∨ zk, where zi is the chosen true literal and j < k, let K ℓ= {qℓ

i , wℓi , w

ℓ, wℓj , β

ℓj }. Note that K ℓ

is a clique. For each h in {i, j, k}, let B′

h be the set among Bh and Dh that contains βℓh and let B′′

h be the other set. Then wesee that:

aℓi → (Ai ∪ B′

i) ⊕ {qℓi , w

ℓi }, qℓ

i → {aℓi } ⊕ K ℓ, wℓ

i → {aℓi , β

ℓi } ⊕ K ℓ,

βℓi → (Ai ∪ B′

i) ⊕ {wℓi }, qℓ

j → {aℓj , w

ℓj }, wℓ

j → K ℓ⊕ {aℓ

j , qℓj },

wℓk → {wℓ, βℓ

k } ⊕ {aℓk, q

ℓk}, qℓ

k → {aℓk, w

ℓk}, wℓ

→ K ℓ⊕ {wℓ

k, βℓk };

moreover:If zj is true, then aℓ

j → (Aj ∪ B′

j) ⊕ {qℓj , w

ℓj } and βℓ

j → (Aj ∪ B′

j) ⊕ K ℓ;If zj is false, then aℓ

j → (Aj ∪ B′′

j ) ⊕ {qℓj , w

ℓj } and βℓ

j → (B′

j ∪ {fj}) ⊕ K ℓ;If zk is true, then aℓ

k → (Ak ∪ B′

k) ⊕ {qℓk, w

ℓk} and βℓ

k → (Ak ∪ B′

k) ⊕ {wℓ, wℓk};

If zk is false, then aℓk → (Ak ∪ B′′

k ) ⊕ {qℓk, w

ℓk} and βℓ

k → (B′

k ∪ {fk}) ⊕ {wℓ, wℓk}.

So the neighborhood of every vertex in H can be partitioned into at most two cliques, and consequently H is claw-free.Thus (1) holds. The validity of Claim (1) completes the proof of the theorem. �

As observed above, in the graphH built in the proof the neighborhood of every vertex can be partitioned into two cliques.Graphs with that property are called quasi line-graphs. (Not all claw-free graphs are quasi line-graphs: for example, addinga universal vertex to a C5 creates a claw-free graph that is not a quasi line-graph.) Thus the same proof as above shows thatthe quasi line-graph sandwich problem is NP-complete. It would be interesting to know the complexity of the line-graphsandwich problem.

4. The bull-free graph sandwich problem

Proof of Theorem 3. We give a reduction from the basic problem of 3-Satisfiability, proved NP-complete by Cook [5].Let the input of 3-Satisfiability be a Boolean formula ϕ on a set of n variables {x1, . . . , xn}, ϕ = C1

∧ · · · ∧ Cm, whereeach clause Cℓ is of the form zi ∨ zj ∨ zk for some distinct i, j, k ∈ {1, . . . , n}, and each zi is either xi or xi. With ϕ we associatean instance of the graph sandwich problem as follows.

For every variable xi, let Ci be the set of clauses that contain the literal xi, and let C i be the set of clauses that containthe literal xi. We may assume that Ci ∪ C i = ∅. Make a variable gadget Xi with vertices bi, ci, di, ei and vi, plus, for everyclause Cℓ in Ci, two vertices wℓ

i and f ℓi , and for every clause Cℓ in C i, two vertices wℓ

i and aℓi . LetWi = {wℓ

i | Cℓ∈ Ci ∪ C i}. If

Ci = ∅, let Fi = {f ℓi | Cℓ

∈ Ci}, else let Fi = {f ∗

i }, where f ∗

i is an additional vertex. Likewise, if C i = ∅ let Ai = {aℓi | Cℓ

∈ C i},else let Ai = {a∗

i }, where a∗

i is an additional vertex. Moreover, let Xi have the following edges, for every ai, fi, wi in Ai, Fi,Wirespectively:

• Forced edges: aibi, bici, cidi, diei, eifi, civi, divi, and viwi;• Optional edges: bivi, eivi, aiwi, biwi, eiwi, fiwi.

All other pairs between two vertices of Xi are forbidden.Consider any clause Cℓ of ϕ, with Cℓ

= zi ∨ zj ∨ zk (with i < j < k), where zh is either xh or xh. For each h in {i, j, k}, Xh hastwo vertices with superscript ℓ, which are wℓ

h and either f ℓh (if zh = xh) or aℓ

h (if zh = xh); call this second vertex uℓh, and call

{uℓh, w

ℓh} the h-pair of Cℓ. Then identify the six vertices uℓ

i , wℓi , u

ℓj , w

ℓj , u

ℓk, w

ℓk two by two as follows: let wℓ

i = uℓj , w

ℓj = uℓ

k ,wℓ

k = uℓi . Call these three vertices the triple of Cℓ.

If u, v are any two vertices that do not lie in a common variable gadget Xi, then the pair uv is forbidden.Let V =

ni=1 V (Xi). Let E1 be the set of forced edges, Eopt the set of optional edges, and E2 = E1 ∪ Eopt. Let G1 = (V , E1)

and G2 = (V , E2). It is easy to see that |V | ≤ 5n + 3m, so the size of (G1,G2) is polynomial in the size of ϕ. See Fig. 5 for anexample. The solid edges are forced edges of E1, and the dotted edges are optional edges of Eopt.

We claim that:

The formula ϕ is satisfiable if and only if the pair (G1,G2) admits a bull-free sandwich graph. (2)

6 S. Dantas et al. / Discrete Applied Mathematics ( ) –

Fig. 5. Instance (V , E1, E2) obtained from the satisfiable instance of 3sat: I = (X, ϕ) = ({x1, x2, x3}, {(x1, x2, x3)}). Note the instance contains just oneclause. The truth assignment x1 = x3 = T , x2 = F satisfies (X, ϕ). The bold dashed edges are the optional edges added to G1 to obtain the bull-freesandwich graph H .

Proof of (2). First suppose that H is a bull-free sandwich graph for (G1,G2). For each i, consider arbitrary vertices ai, fi, wi inAi, Fi,Wi respectively. One of the two optional pairs bivi and eivi must be an edge ofH , for otherwise {bi, ci, vi, di, ei} inducesa bull; and not both of them are edges, for otherwise {ai, bi, ci, vi, ei} induces a bull. This leads to the two cases below, whichare symmetric. (Here the word edge refers to H .)

(a) bivi is an edge and eivi is not an edge. Then, eiwi is an edge, for otherwise {ei, di, ci, vi, wi} induces a bull; biwi is not an edge,for otherwise either {ai, bi, vi, wi, ei} induces a bull (if aiwi ∈ E(H)) or {ci, bi, ai, wi, ei} induces a bull (if aiwi ∈ E(H));aiwi is an edge, for otherwise {ai, bi, ci, vi, wi} induces a bull; and fiwi is not an edge, for otherwise {di, ei, fi, wi, ai}induces a bull. Note that each vertex ofWi is adjacent to all of Ai and to no vertex of Fi.

(b) eivi is an edge and bivi is not an edge. By the same arguments as in (a), we obtain that biwi and fiwi are edges and eiwi andaiwi are not edges. Note that each vertex ofWi is adjacent to all of Fi and to no vertex of Ai.

In case (a), we set the variable xi to true; in case (b), we set it to false.Consider any clause Cℓ, say Cℓ

= zi∨zj∨zk, where zh is either xh or xh. If all three literals zi, zj, zk are false, then, by (b), thethree h-pairs of Cℓ (h = i, j, k) are edges of H , and the triple of Cℓ plus vertices vi and vj induce a bull in H , a contradiction.So each clause is satisfied.

Conversely, suppose that ϕ admits a satisfying truth assignment (ξ1, . . . , ξn). Starting from G1, make a graphH by addingthe following optional edges for each i in {1, . . . , n}:

– If ξi = true, add edges bivi, eiwi and aiwi, for all ai in Ai and wi in Wi;– If ξi = false, add edges eivi, biwi and fiwi for all fi in Fi and wi in Wi;– Do not add any other optional pair.

Suppose that H contains a bull. Let T be the K3 in this bull, and let t ′ and t ′′ be the other two vertices of the bull. SinceT is a K3 in H , it is also a K3 in G2. It is a routine matter to check that every K3 in G2 is of one of the following types: Type(i): {vi, bi, ci}, {vi, ci, di} or {vi, di, ei} for some i in {1, . . . , n}; Type (ii): {ai, bi, wi}, {bi, vi, wi}, or {fi, ei, wi}, {ei, vi, wi}, forsome i in {1, . . . , n} and ai ∈ Ai, fi ∈ Fi, wi ∈ Wi; Type (iii): the triple of a clause. Since every clause Cℓ is satisfied, someliteral zh that appears in Cℓ is true, and consequently the corresponding h-pair of Cℓ is not an edge. So H contains no K3 oftype (iii). Also, because of the way H is made, no K3 of type (ii) can be a K3 in H (whether ξi is true or false, in any such K3 onepair is not an edge of H). Therefore T is of type (i). Vertices t ′ and t ′′ have only one neighbor in T , and that is not the sameneighbor. But in each of the three cases, as detailed below, this condition implies that t ′t ′′ is an edge in H , a contradiction.The three cases are:

– T = {vi, bi, ci}, so (since bivi is an edge of H) ξi = true and {t ′, t ′′} = {ai, wi} for some ai in Ai and wi in Wi;– T = {vi, ci, di}, and either ξi = true and {t ′, t ′′} = {ei, wi} for some wi in Wi, or ξi = false and {t ′, t ′′} = {bi, wi} for

some wi in Wi;– T = {vi, di, ei}, so (since eivi is an edge of H) ξi = false and {t ′, t ′′} = {fi, wi} for some fi in Fi and wi in Wi.

Thus H is bull-free. Therefore (2) holds. The validity of Claim (2) completes the proof of the theorem. �

5. The skew partition sandwich problem

We prove that skew partition sandwich problem is NP-complete by reducing the NP-complete problem monotone1-in-3 3sat to skew partition sandwich problem. These two decision problems are defined as follows.

S. Dantas et al. / Discrete Applied Mathematics ( ) – 7

Fig. 6. Instance (V , E1, E2) obtained from the satisfiable instance of 1-in-3 3sat: I = (W ,Q ) = ({x1, x2, x3, x4, x5}, {(x1, x3, x4), (x2, x3, x4), (x1, x2, x5),(x2, x4, x5)}). The 1-in-3 truth assignment x3 = x5 = T satisfies (W ,Q ).

monotone 1-in-3 3satInput: A set W = {x1, . . . , xn} of variables, a collection Q = {q1, . . . , qm} of clauses over W such that each clause

q ∈ Q has three unnegated variables.Question: Is there a truth assignment forW such that each clause in Q has exactly one true variable?

Schaefer [23] proved that monotone 1-in-3 3sat is NP-complete.

skew partition sandwich problemInput: A pair (G1,G2) of graphs with G1 = (V , E1),G2 = (V , E2) and E1 ⊆ E2;Question: Is there a graph G = (V , E) such that E1 ⊆ E ⊆ E2 and G has a skew partition?

Proof of Theorem 4. We reducemonotone 1-in-3 3sat to skewpartition sandwich problem as follows. For every instance(W ,Q ) of monotone 1-in-3 3sat we construct an instance (V , E1, E2) of skew partition sandwich problem, with theproperty that Q is 1-in-3 satisfiable if, and only if, (V , E1, E2) admits a sandwich graph G = (V , E) that has a skew cutset,and the size of (V , E1, E2) is polynomial in the size of (W ,Q ).Construction of the instance

Given an instance I = (W ,Q ) ofmonotone 1-in-3 3sat, we construct an instance (V , E1, E2) of skewpartition sandwichproblem as follows. In order to exclude the trivial case of having just a single clause, we assume that m ≥ 2. See Fig. 6 foran example.

The vertex set V contains, for each clause qj = (xr , xs, xt) (1 ≤ j ≤ m), where r < s < t , a set of fourteen verticesQj = {kj, k′

j, xrj, x′

rj, yj, y′

j, xsj, x′

sj, zj, z′

j , xtj, x′

tj, lj, l′

j}. Thus V =m

j=1 Qj, and so |V | = 14m.Let Uj = {kj, xrj, yj, xsj, zj, xtj, lj} and U ′

j = {k′

j, x′

rj, y′

j, x′

sj, z′

j , x′

tj, l′

j}. Let U =m

j=1 Uj and U ′=

mj=1 U

′

j . Let K =mj=1{kj, k

′

j} and L =m

j=1{lj, l′

j}. For each i ∈ {1, . . . , n}, let

Xi = {xij, x′

ij | variable xi occurs in clause qj (j = 1, . . . ,m)}.

• The forced edge set E1 is defined as follows.– For each clause qj (j ∈ {1, . . . ,m}), let kjk′

j , xrjx′

rj, yjy′

j, xsjx′

sj, zjz′

j , xtjx′

tj, ljl′

j, kjx′

rj, k′

jxrj, xrjy′

j, x′

rjyj, yjx′

sj, y′

jxsj, xsjz′

j , x′

sjzj,zjx′

tj, z′

jxtj, xtjl′

j, x′

tjlj be forced edges (members of E1).– For all j, h ∈ {1, . . . ,m} with j = h, let kjk′

h, k′

jkh, ljl′

h, l′

jlh be forced edges.• The optional edge set E2 \ E1 is defined as follows. For any variable xp that occurs in two clauses qj and qh, let xpjx′

ph andx′

pjxph be optional edges. For any two variables xp and xo that do not occur together in the same clause, let xpjx′

oh and x′

pjxohbe optional edges.

• All remaining edges are forbidden. This completes the construction of the instance.

In Fig. 6 forced edges are represented as solid edges. Optional edges between vertices within the same set Xi (1 ≤ i ≤ 5)are not represented. We represent all other optional edges as dashed edges. Variables x3 and x5 do not occur in the sameclause.

8 S. Dantas et al. / Discrete Applied Mathematics ( ) –

Note that K and L induce complete bipartite subgraphs in both G1 and G2, and each Xi, 1 ≤ i ≤ n induces a matching inG1 and a complete bipartite subgraph in G2.

Suppose that there exists a 1-in-3 truth assignment that satisfies (W ,Q ). Let S =

{Xi | variable xi is true}. Let G be thesandwich graph obtained from G1 by adding all optional pairs whose two vertices are in S. We claim that S is a skew cutsetof G. In order to prove this claim, we make two observations.

(a) S induces a complete bipartite subgraph in G. Indeed, by the definition of a 1-in-3 truth assignment, the true variables donot occur together in the same clause. By the definition of the optional edges, S induces a complete bipartite subgraphin G2 (every vertex of S ∩ U is adjacent to every vertex of S ∩ U ′), and so in G too.

(b) K and L lie in distinct components of G \ S. Suppose that there exists a path P in G \ S from a vertex in K to a vertex in L.Note that, in G1, every edge is either between two vertices of K , or between two vertices of L, or between two verticesof Qj for some j. Moreover, it is easy to see that each set Xi is a cutset in G1[Qj], where variable xi occurs in clause qj, thatseparates K from L. So P must contain an optional edge uv. But then u and v must be in S, a contradiction. (For example,in Fig. 6, the 1-in-3 truth assignment x3 = x5 = T satisfies (W ,Q ). The set S = X3 ∪ X5 is indeed a cutset in the forcedgraph G1.)

Observations (a) and (b) imply that S is a skew cutset of G.Conversely, suppose that there exists a sandwich graphGwith a skewpartition (A, B, C,D). LetX =

ni=1 Xi and S = A∪B.

Note that, since G2 is bipartite, the skew cutset S induces a complete bipartite subgraph of G and A and B are stable sets.We call KL-link any path P such that one end of P is in K , the other is in L, and V (P) ⊆ Qh \ S for some h ∈ {1, . . . ,m}.

Given a vertex w of V \ L (w ∈ Qh, say), a K -link (for w) is any path Pw such that one end of Pw is w, the other is in K , andV (Pw) ⊆ Qh \ S. Given a vertex of V \ K , we define an L-link similarly.

We claim that for every clause qj (j ∈ {1, . . . ,m}), with qj = (xr , xs, xt), there exists p ∈ {r, s, t} such that Qj ∩ S =

{xpj, x′

pj}. For suppose the contrary. First suppose that Qj ∩ S contains two vertices u and v non-adjacent in G. Let u, v ∈ A.We distinguish three cases:

(i) B ∩ (Qj \ X) = ∅. Pick any b ∈ Qj \ X . If both u and v belong to X , then up to symmetry, b = yj and S ⊆ {yj, y′

j, x′

rj, x′

sj}

and it is easy to see that there exists a KL-link in G[Qh \ S] for any h = j and that every vertex w of V \ S (including ifw ∈ Qj) has a K -link or an L-link. This implies that G \ S is connected, a contradiction. If one of u, v, say v, is in Qj \ X ,then up to symmetry, {b, v} is equal to either {kj, k′

j} or {yj, y′

j}. It follows that S is a subset of either K , {kj, k′

j, xrj, x′

rj},{xrj, x′

rj, yj, y′

j} or {yj, y′

j, xsj, x′

sj}. In either case, V \ S contains L. If S ⊆ K , then it is easy to see that every vertex of V \ Shas an L-link, which implies that G \ S is connected, a contradiction. If S is a subset of the remaining three sets, then itis easy to see that there exists a KL-link in G[Qh \ S] for any h = j and that every vertex w of V \ S (including if w ∈ Qj)has a K -link or an L-link. This implies again that G \ S is connected, a contradiction.

(ii) B∩ (Qj \ X) = ∅ and B∩Qj ∩ X = ∅. Pick any b ∈ B∩Qj ∩ X , say b = x′

rj. One of u, v is not in X , and by the constructionof G2, we have B = {b}. It follows that A ⊆ NG(b) ⊆ NG2(b). In particular, for every h = j, we have A ∩ Qh ⊂ U . ThenG[Qh \S] contains a KL-link for every h = j, and every vertex of V \S has a K -link or an L-link. So again G\S is connected,a contradiction.

(iii) B ∩ Qj = ∅. Pick any b ∈ B. Then at least one of the edges ub and vb (ub, say) is optional, so u, b ∈ X , which impliesthat vb too is optional, so v ∈ X . Hence, up to symmetry, let u = xrj, v = xsj and b = x′

qh for some h = j. We haveQj ∩ A ⊆ {xrj, xsj, xtj} and B ⊆ X ∩ U ′. It follows that G[Qh \ S] contains a KL-link for every h, and every vertex of V \ Seither lies on a KL-link or has a neighbor on it. So again G \ S is connected, a contradiction.

Wemay now assume that Qj ∩S consists of vertices mutually adjacent in G. Therefore, for every j,Qj ∩S consists of eitherzero or one vertex or two adjacent vertices. In any case it follows that every vertex of Qj \ S has a K -link or an L-link.

Suppose that Qj ∩ S contains zero or one vertex. Then G[Qj \ S] contains a KL-link and every vertex of Qj \ S either lies onthe KL-link or has a neighbor on it. This implies again that G \ S is connected, a contradiction.

We may now assume that, for every j,Qj ∩ S consists of two adjacent vertices uj, vj. Since S induces a complete bipartitegraph, for j = h each of uj, vj must have one neighbor in {uh, vh}, which implies that either S = K , or S = L, or S ⊆ X .Since S is a cutset, we have that for each j, the adjacent pair uj, vj ∈ X . Thus we have established the claim that for everyclause qj, with qj = (xr , xs, xt), there exists p ∈ {r, s, t} such that Qj ∩ S = {xpj, x′

pj}. Then for every clause we set to true thecorresponding variable xp. The preceding claim implies that this is a 1-in-3 assignment that satisfies (W ,Q ). �

6. More NP-complete problems

Theorem 6. Let F and F∗ be two vertex-disjoint graphs. If the F-free graph sandwich problem is NP-complete, then the (F + F∗)-free graph sandwich problem is NP-complete.

Proof. For each instance (G1,G2) of the F-free graph sandwich problem we define an instance (G′

1,G′

2) of the (F + F∗)-freegraph sandwich problem as follows. Let (V ∗, E∗) be a copy of F∗, vertex-disjoint from G1 and G2. Let V (G′

1) = V (G′

2) =

S. Dantas et al. / Discrete Applied Mathematics ( ) – 9

Table 1NP-c: NP-complete, P: polynomial.

Cutset Recognition Sandwich

Clique P [28] NP-c [26]Star P [4] P [26]Skew P [9] NP-c [here]

V (G1) ∪ V ∗, E ′

1 = E1 ∪ E∗ and E ′

2 = E2 ∪ E∗ (thus in the instance (G′

1,G′

2) all edges of E∗ are forced, and every pair thatcontains a vertex of V ∗ and is not an edge of E∗ is forbidden). We claim that:

The pair (G1,G2) admits an F-free sandwich graph if and only if the pair (G′

1,G′

2)

admits an (F + F∗)-free sandwich graph. (3)

First suppose that G′ is an (F + F∗)-free sandwich graph for the pair (G′

1,G′

2). Let G be the subgraph of G′ induced by V . Theunion of those components of G′ that contain a vertex of V ∗ is equal to (V ∗, E∗), so it is a copy of F∗. It follows that G is F-free.Moreover, we have E1 ⊆ E(G) ⊆ E2. So G is an F-free sandwich graph for (G1,G2).

Conversely, suppose that G is an F-free sandwich graph for (G1,G2). Let G′= G + (V ∗, E∗). We have E ′

1 ⊆ E(G′) ⊆ E ′

2.Suppose that G′ contains a subgraph H0 isomorphic to F + F∗. We can write H0 = H +H∗ where H is a copy of F and H∗ is acopy of F∗. LetW andW ∗ respectively be the vertex-sets ofH andH∗. In G′, the four sets V ∩W , V ∩W ∗, V ∗

∩W and V ∗∩W ∗

are pairwise non-adjacent (because there is no edge between V and V ∗ by the definition of G′

2 and there is no edge betweenW andW ∗ by the definition of H0). Since G′

2[V∗] and G′

2[W∗] are both isomorphic to F∗ and they both contain G′

2[V∗∩W ∗

],there must be a subset X ofW ∗ such that G′

2[W ∩ V ∗] is isomorphic to G′

2[X]. It follows that G2[(V ∩ W ) ∪ X] is isomorphicto F ; but G2[(V ∩W ) ∪ X] is a subgraph of G, a contradiction. So G′ is an (F + F∗)-free sandwich graph for (G′

1,G′

2). Thus (3)holds.

By (3), every instance of the F-free graph sandwich problem can be reduced to an instance of the (F + F∗)-free graphsandwich problem in polynomial time so that the two instances either both have a solution or both have no solution. Thisproves the theorem. �

Using [6] and Theorems 2, 3 and 6,we can generate infinitelymany graphs F such that the F-free graph sandwich problemis NP-complete. This is the case in particular, when F has five vertices, if F is either C4 + K1, or 2K2 + K1, or claw + K1, orclaw + K1, or their complements.

7. Concluding remarks

We observe that in the proof of Theorem 2, the constructed graph H is a quasi line-graph, and so the same proof showsthat the quasi line-graph sandwich problem is NP-complete. We leave as an open problem the complexity status of theline-graph sandwich problem. Note that in the proof of Theorem 4, by contracting corresponding vertices u and u′, weobtain an alternative simpler proof that clique cutset sandwich problem is NP-complete [26].

The celebrated Strong Perfect Graph Theorem established the characterization of perfect graphs by forbidden odd holesand complements of odd holes. The properties Perfect Graph andOddHole give interesting related open sandwich problems.For every fixed k ≥ 4, the k-hole-free sandwich problem is NP-complete [6]. We remark that the perfect graph sandwichproblem is the only remaining open problem among those proposed by Golumbic et al. [15]. It is remarkable that therecognition of Berge trigraphs defined by Chudnovsky [2] corresponds to the graph sandwich problem for imperfectgraphs.

Polynomial sandwich problems are rare. Let F be any graph on five vertices. The F-free graph sandwich problem wasclassified if F is K2,3 (or its complement) as NP-complete, and if F is K5 \ e (or its complement) as polynomial [6], so far theonly known polynomial such problem, when considering forbidden graphs on five vertices. The present work classified asNP-complete the case F is a bull, a self-complementary graph, and by Theorem 6, several cases when F is not connected. Itis known from [12] that split graphs are precisely the {C4, C4, C5}-free graphs, and although the sandwich problem for splitgraphs is polynomial [15], for each of the three forbidden subgraphs C4, C4, and C5, the F-free graph sandwich problem isNP-complete [6].

It is interesting to observe the established complexity for the F-free graph sandwich problem: if F is the triangle, or ifF is the paw (the triangle plus one pendant vertex), then the problem is polynomial, but if F is the bull (the triangle plustwo non adjacent pendant vertices), or if F is Hajós graph (the complement of the triangle plus three nonadjacent pendantvertices), then the problem is NP-complete, the latter result was established in the context of clique graphs [11].

Table 1 summarizes the polynomial time versus NP-complete dichotomy obtained with graph sandwich problems forvertex cutsets. Remark the interesting established non-monotonicity.

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