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Complexity of the Exact Domatic Number Problem and of theExact Conveyor Flow Shop Problem

Tobias Riege∗ and Jorg Rothe†

Institut fur InformatikHeinrich-Heine-Universitat Dusseldorf

40225 Dusseldorf, Germany

December 2, 2003

Abstract

We prove that the exact versions of the domatic number problem are complete for the levelsof the boolean hierarchy over NP. The domatic number problem, which arises in the area ofcomputer networks, is the problem of partitioning a given graph into a maximum number ofdisjoint dominating sets. This number is called the domaticnumber of the graph. We prove thatthe problem of determining whether or not the domatic numberof a given graph isexactlyoneof k given values is complete for BH2k(NP), the2kth level of the boolean hierarchy over NP.In particular, fork = 1, it is DP-complete to determine whether or not the domatic number of agiven graph equals exactly a given integer. Note that DP= BH2(NP). We obtain similar resultsfor the exact versions of generalized dominating set problems and of the conveyor flow shopproblem, which arises in real-world applications in the wholesale business, where warehousesare supplied with goods from a central storehouse. Our reductions apply Wagner’s conditionssufficient to prove hardness for the levels of the boolean hierarchy over NP.

Key words: Computational complexity; completeness; domatic number problem;conveyor flow shop problem; boolean hierarchy

∗Email: riege@cs.uni-duesseldorf.de.†Email: rothe@cs.uni-duesseldorf.de.

1 Introduction

A dominating set in an undirected graphG is a subsetD of the vertex setV (G) such that everyvertex ofV (G) either belongs toD or is adjacent to some vertex inD. The domatic numberproblem is the problem of partitioning the vertex setV (G) into a maximum number of disjointdominating sets. This number, denoted byδ(G), is called the domatic number ofG. The domaticnumber problem arises in various areas and scenarios. In particular, this problem is related to thetask of distributing resources in a computer network, and also to the task of locating facilities in acommunication network.

Suppose, for example, that resources are to be allocated in acomputer network such thatexpensive services are quickly accessible in the immediateneighborhood of each vertex. If everyvertex has only a limited capacity, then there is a bound on the number of resources that canbe supported. In particular, if every vertex can serve a single resource only, then the maximumnumber of resources that can be supported equals the domaticnumber of the network graph. In thecommunication network scenario,n cities are linked via communication channels. A transmittinggroup is a subset of those cities that are able to transmit messages to every city in the network. Sucha transmitting group is nothing else than a dominating set inthe network graph, and the domaticnumber of this graph is the maximum number of disjoint transmitting groups in the network.

Motivated by these scenarios, the domatic number problem has been thoroughly investigated. Itsdecision version, denoted byDNP, asks whether or notδ(G) ≥ k for a given graphG and a positiveintegerk. This problem is known to be NP-complete (cf. [GJ79]), and itremains NP-completeeven if the given graph belongs to certain special classes ofperfect graphs including chordal andbipartite graphs; see the references in Section 2. Feige et al. [FHK00] established nearly optimalapproximation algorithms for the domatic number.

Expensive resources should not be wasted. Given a graphG and a positive integeri, howhard is it to determine whether or notδ(G) equalsi exactly? More generally, given a graphGand a listMk = i1, i2, . . . , ik of k positive integers, how hard is it to determine whether ornot δ(G) equals someij exactly? Motivated by suchexact versions ofNP-complete optimizationproblems, Papadimitriou and Yannakakis introduced in their seminalpaper [PY84] the class DP,which consists of the differences of any two NP sets. They also studied various other importantclasses of problems that belong to DP, includingfacet problems, unique solution problems, andcritical problems, and they proved many of them complete for DP. Cai and Meyer [CM87] showedthat Minimal-3-Uncolorability is DP-complete, a critical graph problem that asks whether agiven graph is not 3-colorable, but deleting any of its vertices makes it 3-colorable.

Generalizing DP, Cai et al. [CGH+88,CGH+89] introduced and studied BH(NP) =⋃

k≥1 BHk(NP), the boolean hierarchy over NP; see Section 2 for the definition. Note that DPis the second level of this hierarchy. Wagner [Wag87] identified a set of conditions sufficient toprove BHk(NP)-hardness for eachk, and he applied his sufficient conditions to prove a host of exactversions of NP-complete optimization problems complete for the levels of the boolean hierarchy. Tostate just one such result, Wagner [Wag87] proved that the problem of determining whether or notthe chromatic number of a given graph isexactlyone ofk given values is complete for BH2k(NP).The chromatic number of a graphG, denoted byχ(G), is the minimum number of colors needed

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to color the vertices ofG such that no two adjacent vertices receive the same color. Inparticular,for k = 1, Wagner showed that for any fixed integeri ≥ 7, it is DP-complete to determine whetheror not χ(G) = i for a given graphG. Recently, Rothe [Rot03] (see also [RSV02]) optimallystrengthened Wagner’s result by showing BH2k(NP)-completeness of the exact chromatic numberproblem using the smallest number of colors possible. In particular, it is DP-complete to determinewhether or notχ(G) = 4, yet the problem of determining whether or notχ(G) = 3 is in NP andthus cannot be DP-complete unless the boolean hierarchy over NP collapses to its first level.

Wagner’s technique was also useful in proving certain natural problems complete for PNP|| ,

the class of problems solvable in polynomial time via parallel (i.e., truth-table) access to NP.For example, the winner problem for Carroll elections [HHR97a,HHR97b] and for Youngelections [RSV02,RSV03] as well as the problem of determining when certain graph heuristicswork well [HR98,HRS02] each are complete for PNP

|| .In Section 2, we prove that determining whether or not the domatic number of a given graph

equals exactly one ofk given values is complete for BH2k(NP). In particular, fork = 1 and anyfixed integeri ≥ 5, it is DP-complete to determine whether or notδ(G) = i for a given graphG.Section 3 raises the results of Section 2 to prove various variants of the domatic number problemcomplete for PNP

|| . In Section 4, we study the exact versions of generalized dominating set problemsthat are parameterized by two sets of nonnegative integers,σ andρ, which restrict the number ofneighbors for each vertex in the partition. We obtain DP-completeness results for a number ofgraph problems based on this uniform approach proposed by Heggernes and Telle [HT98]. Finally,in Section 5, we prove similar results for the exact conveyorflow shop problem.

2 The Exact Domatic Number Problem

We start by introducing some graph-theoretical notation. For any graphG, V (G) denotes the vertexset ofG, andE(G) denotes the edge set ofG. All graphs in this paper are undirected, simple graphs.That is, edges are unordered pairs of vertices, and there areneither multiple nor reflexive edges (i.e.,for any two verticesu andv, there is at most one edge of the formu, v, and there is no edge of theform u, u). Also, all graphs considered do not have isolated vertices. For any vertexv ∈ V (G),thedegree ofv (denoted bydegG(v)) is the number of vertices adjacent tov in G; if G is clear fromthe context, we omit the subscript and simply writedeg(v). Let max-deg(G) = maxv∈V (G) deg(v)denote the maximum degree of the vertices of graphG, and letmin-deg(G) = minv∈V (G) deg(v)denote the minimum degree of the vertices of graphG.

A graphG is said to bek-colorableif its vertices can be colored with no more thank colors suchthat no two adjacent vertices receive the same color. Thechromatic number ofG, denoted byχ(G),is defined to be the smallestk such thatG is k-colorable. In particular, define the decision versionof the 3-colorability problem, which is one of the standard NP-complete problems (cf. [GJ79]), by:

3-Colorability = G | G is a graph withχ(G) ≤ 3.

For some of the reductions presented in this paper, we will need the following operations ongraphs.

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Definition 1 Thejoin operation on graphs, denoted by⊕, is defined as follows: Given two disjointgraphsA andB, their join A ⊕ B is the graph with vertex setV (A ⊕ B) = V (A) ∪ V (B) andedge setE(A⊕B) = E(A) ∪ E(B) ∪ a, b | a ∈ V (A) andb ∈ V (B).

Thedisjoint union of any two graphsA andB is defined as the graphA ∪ B with vertex setV (A) ∪ V (B) und edge setE(A) ∪ E(B).

Note that⊕ is an associative operation on graphs andχ(A ⊕ B) = χ(A) + χ(B). We nowdefine the domatic number problem.

Definition 2 For any graphG, a dominating set ofG is a subsetD ⊆ V (G) such that for eachvertexu ∈ V (G) − D, there exists a vertexv ∈ D with u, v ∈ E. Thedomatic number ofG,denoted byδ(G), is the maximum number of disjoint dominating sets. Define the decision versionof the domatic number problem by:

DNP = 〈G, k〉 |G is a graph andk is a positive integer such thatδ(G) ≥ k.

Note thatδ(G) ≤ min-deg(G) + 1. For general graphs and for each fixedk ≥ 3, DNP isknown to be NP-complete (cf. [GJ79]), and it remains NP-complete for circular-arc graphs [Bon85],for split graphs (thus, in particular, for chordal and co-chordal graphs) [KS94], and for bipartitegraphs (thus, in particular, for comparability graphs) [KS94]. In contrast,DNP is known to bepolynomial-time solvable for certain other graph classes,including strongly chordal graphs (thus,in particular, for interval graphs and path graphs) [Far84]and proper circular-arc graphs [Bon85].For graph-theoretical notions and special graph classes not defined in this extended abstract, werefer to the monograph by Brandstadt et al. [BLS99], which is a follow-up to the classic text byGolumbic [Gol80]. Feige et al. [FHK00] show that every graphG with n vertices has a domaticpartition with(1 − o(1))(min-deg(G) + 1)/ ln n sets that can be found in polynomial time, whichimplies a(1 − o(1)) ln n approximation algorithm for the domatic numberδ(G). This is a tightbound, since they also show that, for any fixed constantε > 0, the domatic number cannot beapproximated within a factor of(1 − ε) ln n, unless NP⊆ DTIME(nlog logn). Finally, Feige etal. [FHK00] give a refined algorithm that yields a domatic partition of Ω(δ(G)/ lnmax-deg(G)),which implies aO(lnmax-deg(G)) approximation algorithm for the domatic numberδ(G). Formore results on the domatic number problem, see [FHK00,KS94] and the references therein.

We assume that the reader is familiar with standard complexity-theoretic notions and notation.For more background, we refer to any standard textbook on computational complexity theory suchas Papadimitriou’s book [Pap94]. All completeness resultsin this paper are with respect to thepolynomial-time many-one reducibility, denoted by≤p

m . For setsA andB, defineA≤pmB if and

only if there is a polynomial-time computable functionf such that for eachx ∈ Σ∗, x ∈ A if andonly if f(x) ∈ B. A setB is C-hard for a complexity classC if and only ifA≤p

mB for eachA ∈ C.A setB is C-complete if and only ifB is C-hard andB ∈ C. To define the boolean hierarchy overNP, we use the symbols∧ and∨, respectively, to denote thecomplex intersectionand thecomplexunionof set classes. That is, for classesC andD of sets, define

C ∧ D = A ∩B |A ∈ C andB ∈ D;

C ∨ D = A ∪B |A ∈ C andB ∈ D.

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Definition 3 (Cai et al. [CGH+88]) Theboolean hierarchy over NPis inductively defined by:

BH1(NP) = NP,

BH2(NP) = NP∧ coNP,

BHk(NP) = BHk−2(NP) ∨ BH2(NP) for k ≥ 3, and

BH(NP) =⋃

k≥1

BHk(NP).

Note that DP= BH2(NP). In his seminal paper [Wag87], Wagner provided a set of conditionssufficient to prove hardness results for the levels of the boolean hierarchy over NP and for othercomplexity classes, respectively. His sufficient conditions were successfully applied to classify thecomplexity of a variety of natural, important problems, see, e.g., [Wag87,HHR97a,HHR97b,HR98,Rot03,HRS02,RSV02,RSV03]. Below, we state that one of Wagner’s sufficient conditions that isrelevant for this paper.

Lemma 4 (Wagner; see Thm. 5.1(3) of [Wag87])LetA be someNP-complete problem, letB bean arbitrary problem, and letk ≥ 1 be fixed. If there exists a polynomial-time computable functionf such that the equivalence

||i | xi ∈ A|| is odd ⇐⇒ f(x1, x2, . . . , x2k) ∈ B (2.1)

is true for all stringsx1, x2, . . . , x2k ∈ Σ∗ satisfying that for eachj with 1 ≤ j < 2k, xj+1 ∈ Aimpliesxj ∈ A, thenB is BH2k(NP)-hard.

Definition 5 LetMk ⊆ N be any set containingk noncontiguous integers. Define the exact versionof the domatic number problem by:

Exact-Mk-DNP = G | G is a graph andδ(G) ∈ Mk.

In particular, for each singletonM1 = t, we writeExact-t-DNP = G | δ(G) = t.

To apply Wagner’s sufficient condition from Lemma 4 in the proof of the main result of thissection, Theorem 7 below, we need the following lemma due to Kaplan and Shamir [KS94] thatgives a reduction from3-Colorability to DNP with useful properties. Since Kaplan and Shamir’sconstruction will be used explicitly in the proof of Theorem7, we present it below.

Lemma 6 (Kaplan and Shamir [KS94]) There exists a polynomial-time many-one reductiongfrom3-Colorability to DNP with the following properties:

G ∈ 3-Colorability =⇒ δ(g(G)) = 3; (2.2)

G 6∈ 3-Colorability =⇒ δ(g(G)) = 2. (2.3)

Proof. The reductiong maps any given graphG to a graphH such that the implications (2.2)and (2.3) are satisfied. Since it can be tested in polynomial time whether or not a given graph is2-colorable, we may assume, without loss of generality, thatG is not2-colorable. Recall that we

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also assume thatG has no isolated vertices; note that the domatic number of anygraph is always atleast2 if it has no isolated vertices (cf. [GJ79]). GraphH is constructed fromG by creating||E(G)||new vertices, one on each edge ofG, and by adding new edges such that the original vertices ofGform a clique. Thus, every edge ofG induces a triangle inH, and every pair of nonadjacent verticesin G is connected by an edge inH. Our construction in the proof of Theorem 7 below explicitlyuses this construction and, in particular, such triangles.

Let V (G) = v1, v2, . . . , vn. Formally, define the vertex set and the edge set ofH by:

V (H) = V (G) ∪ ui,j | vi, vj ∈ E(G);

E(H) = vi, ui,j | vi, vj ∈ E(G) ∪ vj , ui,j | vi, vj ∈ E(G)

∪ vi, vj | 1 ≤ i, j ≤ n andi 6= j.

Since, by construction,min-deg(H) = 2 andH has no isolated vertices, the inequalityδ(H) ≤min-deg(H) + 1 implies that2 ≤ δ(H) ≤ 3.

SupposeG ∈ 3-Colorability. Let C1, C2, andC3 be the three color classes ofG, i.e.,Ck = vi ∈ V (G) | vi is colored by colork, for k ∈ 1, 2, 3. Form a partition ofV (H) byCk = Ck ∪ ui,j | vi 6∈ Ck andvj 6∈ Ck, for k ∈ 1, 2, 3. Since for eachk, Ck ∩ V (G) 6= ∅andV (G) induces a clique inH, everyCk dominatesV (G) in H. Also, every trianglevi, ui,j , vjcontains one element from eachCk, so everyCk also dominatesui,j | vi, vj ∈ E(G) in H.Hence,δ(H) = 3, which proves the implication (2.2).

Conversely, supposeδ(H) = 3. Given a partition ofV (H) into three dominating sets,C1, C2,andC3, color the vertices inCk by colork. Every trianglevi, ui,j , vj is 3-colored, which impliesthat this coloring onV (G) induces a legal3-coloring ofG; soG ∈ 3-Colorability. Hence,χ(G) = 3 if and only if δ(H) = 3. Since2 ≤ δ(H) ≤ 3, the implication (2.3) follows.

Next, we state the main result of this section: For each fixed setMk containingk noncontiguousintegers not smaller than4k + 1, Exact-Mk-DNP is complete for BH2k(NP), the2kth level of theboolean hierarchy over NP.

Theorem 7 For fixedk ≥ 1, let Mk = 4k + 1, 4k + 3, . . . , 6k − 1. Then,Exact-Mk-DNPis BH2k(NP)-complete. In particular, fork = 1, Exact-5-DNP is DP-complete. In contrast,Exact-2-DNP is in coNP (and evencoNP-complete) and thus cannot beDP-complete unless theboolean hierarchy overNPcollapses.

Proof. To show thatExact-Mk-DNP is in BH2k(NP), partition the problem intok subproblems

Exact-Mk-DNP =⋃

i∈Mk

Exact-i-DNP.

Every setExact-i-DNP can be rewritten as

Exact-i-DNP = G | δ(G) ≥ i ∩ G | δ(G) < i+ 1.

Clearly, the setG | δ(G) ≥ i is in NP, and the setG | δ(G) < i+ 1 is in coNP. It follows thatExact-i-DNP is in DP, for eachi ∈ Mk. By definition,Exact-Mk-DNP is in BH2k(NP).

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In particular, supposek = 1 and consider the problem

Exact-2-DNP = G | δ(G) ≤ 2 ∩ G | δ(G) ≥ 2.

Since every graph without isolated vertices has a domatic number of at least2 (cf. [GJ79]), the setG | δ(G) ≥ 2 is in P. On the other hand, the setG | δ(G) ≤ 2 is in coNP, soExact-2-DNP isalso in coNP and, thus, cannot be DP-complete unless the boolean hierarchy over NP collapses toits first level. Note that the coNP-hardness ofExact-2-DNP follows immediately via the Lemma 6reductiong from 3-Colorability to DNP.

The proof thatExact-Mk-DNP is BH2k(NP)-hard draws on Lemma 4 with3-Colorabilitybeing the NP-complete setA and withExact-Mk-DNP being the setB from this lemma. Fix any2kgraphsG1, G2, . . . , G2k satisfying that for eachj with 1 ≤ j < 2k, if Gj+1 is in 3-Colorability,then so isGj . Without loss of generality, we assume that none of these graphsGj is 2-colorable, nordoes it contain isolated vertices, and we assume thatχ(Gj) ≤ 4 for eachj. Applying the Lemma 6reductiong from 3-Colorability to DNP, we obtain2k graphsHj = g(Gj), 1 ≤ j ≤ 2k, eachsatisfying the implications (2.2) and (2.3). Hence, for each j, δ(Hj) ∈ 2, 3, andδ(Hj+1) = 3impliesδ(Hj) = 3.

We now define a polynomial-time computable functionf that maps the graphsG1, G2, . . . , G2k

to a graphH such that Equation (2.1) from Lemma 4 is satisfied. The graphH is constructedfrom the graphsH1,H2, . . . ,H2k such thatδ(H) =

∑2kj=1 δ(Hj). Note that the analogous property

for the chromatic number (i.e.,χ(H) =∑2k

j=1 χ(Hj)) is easy to achieve by simply joining (seeDefinition 1) the graphsHj ([Wag87], see also [Rot03]). However, for the domatic number, theconstruction is more complicated. We first describe it for the special case thatk = 1, and thenexplain the general case. Fork = 1, we are given two graphs,H1 andH2, as above. Construct agadget connectingH1 andH2 as follows. Recalling the construction from Lemma 6, letT1 withV (T1) = vq, uq,r, vr be any fixed triangle inH1, and letT2 with V (T2) = vs, us,t, vt be anyfixed triangle inH2. ConnectT1 andT2 using the gadget that is shown in Figure 1. That is, add sixnew verticesa1, a2, . . . , a6, and add the following set of edges:

vq, a1, vq , a2, vq, a4, vq, a5, vq , a6,

uq,r, a1, uq,r, a3, uq,r, a4, uq,r, a5, uq,r, a6,

vr, a2, vr, a3, vr, a4, vr, a5, vr , a6,

vs, a1, vs, a2, vs, a3, vs, a4, vs, a5,

us,t, a1, us,t, a2, us,t, a3, us,t, a4, us,t, a6,

vt, a1, vt, a2, vt, a3, vt, a5, vt, a6.

Using pairwise disjoint copies of the gadget from Figure 1, connect each pair of triangles fromH1 andH2 and call the resulting graphH. Sincedeg(ai) = 5 for each gadget vertexai, wehaveδ(H) ≤ 6, regardless of the domatic numbers ofH1 andH2. We now show thatδ(H) =δ(H1) + δ(H2).

Let D1,D2, . . . ,Dδ(H1) be δ(H1) pairwise disjoint sets dominatingH1, and letDδ(H1)+1,Dδ(H1)+2, . . ., Dδ(H1)+δ(H2) be δ(H2) pairwise disjoint sets dominatingH2. Distinguish the

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vr

vq

vt

vs

us,t

T2T1

a4 a5 a6

a1 a2 a3

uq,r

Figure 1: Gadget connecting two trianglesT1 andT2.

following three cases.

Case 1: δ(H1) = δ(H2) = 3. Consider any fixedDj , where1 ≤ j ≤ 3. SinceDj

dominatesH1, every triangleT1 of H1 has exactly one vertex inDj . Fix T1, and supposeV (T1) = vq, uq,r, vr and, say,V (T1)∩Dj = vq; the other cases are analogous. For eachtriangleT2 of H2, sayT2 with V (T2) = vs, us,t, vt, let aT2

1 , aT2

2 , . . . , aT2

6 be the gadgetvertices connectingT1 andT2 as in Figure 1. Note that exactly one of these gadget vertices,aT2

3 , is not adjacent tovq. For each triangleT2, add the missing gadget vertex toDj , anddefineDj = Dj ∪ aT2

3 | T2 is a triangle ofH2. Since every vertex ofH2 is contained insome triangleT2 of H2 and sinceaT2

3 is adjacent to each vertex inT2, Dj dominatesH2.Also, Dj ⊇ Dj dominatesH1, and sincevq is adjacent to eachaT2

i exceptaT2

3 for eachtriangleT2 of H2, Dj dominates every gadget vertex ofH. Hence,Dj dominatesH. By asymmetric argument, every setDj, where4 ≤ j ≤ 6, dominatingH2 can be extended to a setDj dominating the entire graphH. By construction, the setsDj with 1 ≤ j ≤ 6 are pairwisedisjoint. Hence,δ(H) = 6 = δ(H1) + δ(H2).

Case 2: δ(H1) = 3 and δ(H2) = 2. As in Case 1, we can add appropriate gadget vertices tothe five given setsD1,D2, . . . ,D5 to obtain five pairwise disjoint setsD1, D2, . . . , D5 suchthat eachDi dominates the entire graphH. It follows that 5 ≤ δ(H) ≤ 6. It remains toshow thatδ(H) 6= 6. For a contradiction, suppose thatδ(H) = 6. Look at Figure 1 showingthe gadget between any two trianglesT1 andT2 belonging toH1 andH2, respectively. FixT1 with V (T1) = vq, uq,r, vr. The only way (except for renaming the dominating sets) to

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partition the graphH into six dominating sets, sayE1, E2, . . . , E6, is to assign to the setsEi

the vertices ofT1, of H2, and of the gadgets connected withT1 as follows:

• E1 containsvq andaT2

3 | T2 is a triangle inH2,

• E2 containsuq,r andaT2

2 | T2 is a triangle inH2,

• E3 containsvr andaT2

1 | T2 is a triangle inH2,

• E4 containsvs ∈ T2, for each triangleT2 of H2, andaT2

6 | T2 is a triangle inH2,

• E5 containsus,t ∈ T2, for each triangleT2 of H2, andaT2

5 | T2 is a triangle inH2,

• E6 containsvt ∈ T2, for each triangleT2 of H2, andaT2

4 | T2 is a triangle inH2.

Hence, all vertices fromH2 must be assigned to the three dominating setsE4, E5, andE6,which induces a partition ofH2 into three dominating sets. This contradicts the caseassumption thatδ(H2) = 2. Hence,δ(H) = 5 = δ(H1) + δ(H2).

Case 3: δ(H1) = δ(H2) = 2. As in the previous two cases, we can add appropriate gadgetvertices to the four given setsD1, D2, D3, andD4 to obtain a partition ofV (H) into foursetsD1, D2, D3, andD4 such that eachDi dominates the entire graphH. It follows that4 ≤ δ(H) ≤ 6. By the same arguments as in Case 2,δ(H) 6= 6. It remains to show thatδ(H) 6= 5. For a contradiction, suppose thatδ(H) = 5. Look at Figure 1 showing the gadgetbetween any two trianglesT1 andT2 belonging toH1 andH2, respectively. SupposeH ispartitioned into five dominant setsE1, E2, . . . , E5.

First, we show that neitherT1 norT2 can have two vertices belonging to the same dominatingset. Suppose otherwise, and let, for example,vq anduq,r be both inE1, and letvr be inE2;all other cases are treated analogously. This implies that the verticesvs, us,t, andvt in T2

must be assigned to the other three dominating sets,E3, E4, andE5, since otherwise one ofthe setsEi would not dominate all gadget verticesaj, 1 ≤ j ≤ 6. SinceT1 is connected witheach triangle ofH2 via some gadget, the same argument shows thatV (H2) can be partitionedinto three dominating sets, which contradicts the assumption thatδ(H2) = 2.

Hence, the vertices ofT1 are assigned to three different dominating sets, sayE1, E2, andE3.Then, every triangleT2 of H2 must have one of its vertices inE4, one inE5, and one ineither one ofE1, E2, andE3. Again, this induces a partition ofH2 into three dominating sets,which contradicts the assumption thatδ(H2) = 2. It follows thatδ(H) 6= 5, soδ(H) = 4 =δ(H1) + δ(H2).

By construction,δ(H2) = 3 impliesδ(H1) = 3, and thus the case “δ(H1) = 2 andδ(H2) = 3”cannot occur. The case distinction is complete.

Definef(G1, G2) = H. Note thatf is polynomial-time computable and, by the case distinction

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above,f satisfies Equation (2.1):

G1 ∈ 3-Colorability andG2 6∈ 3-Colorability

⇐⇒ δ(H1) = 3 andδ(H2) = 2

⇐⇒ δ(H) = δ(H1) + δ(H2) = 5

⇐⇒ f(G1, G2) = H ∈ Exact-5-DNP.

Applying Lemma 4 withk = 1, it follows thatExact-5-DNP is DP-complete.To prove the general case, fix anyk ≥ 1. Recall that we are given the graphsH1,H2, . . . ,H2k

that are constructed fromG1, G2, . . . , G2k. Generalize the above construction of graphH asfollows. For any fixed sequenceT1, T2, . . . , T2k of triangles, whereTi belongs toHi, add6k newgadget verticesa1, a2, . . . , a6k and, for eachi with 1 ≤ i ≤ 2k, associate the three gadget verticesa1+3(i−1), a2+3(i−1), anda3i with the triangleTi. For eachi with 1 ≤ i ≤ 2k, connectTi witheveryTj , where1 ≤ j ≤ 2k andi 6= j, via the same three gadget verticesa1+3(i−1), a2+3(i−1), anda3i associated withTi the same wayT1 andT2 are connected in Figure 1 via the verticesa1, a2,anda3. It follows thatdeg(ai) = 6k − 1 for eachi, soδ(H) ≤ 6k. An argument analogous to theabove case distinction shows thatδ(H) =

∑2kj=1 δ(Hj), and it follows that:

||i | Gi ∈ 3-Colorability|| is odd

⇐⇒ (∃i : 1 ≤ i ≤ k) [χ(G1) = · · · = χ(G2i−1) = 3 andχ(G2i) = · · · = χ(G2k) = 4]

⇐⇒ (∃i : 1 ≤ i ≤ k) [δ(H1) = · · · = δ(H2i−1) = 3 andδ(H2i) = · · · = δ(H2k) = 2]

⇐⇒ (∃i : 1 ≤ i ≤ k)

δ(H) =

2k∑

j=1

δ(Hj) = 3(2i − 1) + 2(2k − 2i+ 1)

⇐⇒ (∃i : 1 ≤ i ≤ k) [δ(H) = 4k + 2i− 1]

⇐⇒ δ(H) ∈ 4k + 1, 4k + 3, . . . , 6k − 1

⇐⇒ f(G1, G2, . . . , G2k) = H ∈ Exact-Mk-DNP.

Thus,f satisfies Equation (2.1). By Lemma 4,Exact-Mk-DNP is BH2k(NP)-complete.

3 Domatic Number Problems Complete for Parallel Access to NP

In this section, we consider the problem of deciding whetheror not the domatic number of a givengraph is an odd integer, and the problem of comparing the domatic numbers of two given graphs.Applying the techniques of the previous section, we prove inTheorem 10 below that these variantsof the domatic number problem are complete for PNP

|| , the class of problems that can be solved bya deterministic polynomial-time Turing machine making parallel (i.e., truth-table) queries to someNP oracle set. Among many other characterizations, PNP

|| is known to be equal to PNP[O(log)], the

9

class of problems solvable by logarithmically many Turing queries to an NP oracle; see [Hem89,Wag90,BH91,KSW87].

Definition 8 Define the following variants of the domatic number problem:

DNP-Odd = G | G is a graph such thatδ(G) is odd;

DNP-Equ = 〈G,H〉 |G andH are graphs such thatδ(G) = δ(H);

DNP-Geq = 〈G,H〉 |G andH are graphs such thatδ(G) ≥ δ(H).

Wagner [Wag87] provided a sufficient condition for proving PNP|| -hardness that is analogous to

Lemma 4 except that in Lemma 9 the value ofk is not fixed.1 The introduction gives a list of relatedPNP|| -completeness results for which Wagner’s technique [Wag87] was applied.

Lemma 9 (Wagner; see Thm. 5.2 of [Wag87])LetA be someNP-complete problem andB be anarbitrary problem. If there exists a polynomial-time computable functionf such that the equivalence

||i | xi ∈ A|| is odd ⇐⇒ f(x1, x2, . . . , x2k) ∈ B (3.4)

is true for eachk ≥ 1 and for all stringsx1, x2, . . . , x2k ∈ Σ∗ satisfying that for eachj with1 ≤ j < 2k, xj+1 ∈ A impliesxj ∈ A, thenB is PNP

|| -hard.

Theorem 10 DNP-Odd, DNP-Equ, andDNP-Geq each arePNP|| -complete.

Proof. It is easy to see that each of the problemsDNP-Odd, DNP-Equ, andDNP-Geq belongs to PNP|| ,

since the domatic number of a given graph can be determined exactly by parallel queries to the NPoracleDNP. It remains to prove that each of these problems is PNP

|| -hard. ForDNP-Odd, this followsimmediately from the proof of Theorem 7 using Lemma 9.

We now show thatDNP-Equ is PNP|| -hard by applying Lemma 9 withA being the NP-complete

problem3-Colorability andB beingDNP-Equ. Fix anyk ≥ 1, and letG1, G2, . . . , G2k be anygiven sequence of graphs satisfying that for eachj with 1 ≤ j < 2k, if Gj+1 is 3-colorable, then soisGj . Since PNP

|| is closed under complement, Equation (3.4) from Lemma 9 can be replaced by:

||i | Gi ∈ 3-Colorability|| is even ⇐⇒ f(G1, G2, . . . , G2k) ∈ DNP-Equ. (3.5)

As in the proof of Theorem 7, construct the graphsH1,H2, . . . ,H2k from the given graphsG1, G2, . . . , G2k according to Lemma 6, where eachHj = g(Gj) satisfies the implications (2.2)and (2.3). Let× denote the associative operation on graphs constructed in the proof of Theorem 7to sum up the domatic numbers of the given graphs, and define the graphs:

Godd = H1 ×H3 × · · · ×H2k−1,

Geven = H2 ×H4 × · · · ×H2k.

1Wagner actually states a criterion for PNPbf -hardness in [Wag87], but he later proved in [Wag90] that PNP

bf = PNP|| .

10

We now prove Equation (3.5). From left to right we have:

||i |Gi ∈ 3-Colorability|| is even

⇒ (∀i : 1 ≤ i ≤ k) [δ(H2i−1) = δ(H2i)]

⇒∑

1≤i≤k

δ(H2i−1) =∑

1≤i≤k

δ(H2i)

⇒ δ(Godd) = δ(Geven)

⇒ 〈Godd, Geven〉 = f(G1, G2, . . . , G2k) ∈ DNP-Equ.

From right to left we have:

||i |Gi ∈ 3-Colorability|| is odd

⇒ (∃i : 1 ≤ i ≤ k) [δ(H2i−1) = 3 ∧ δ(H2i) = 2 andδ(H2j−1) = δ(H2j) for j 6= i]

⇒ −1 +∑

1≤i≤k

δ(H2i−1) =∑

1≤i≤k

δ(H2i)

⇒ δ(Godd)− 1 = δ(Geven)

⇒ 〈Godd, Geven〉 = f(G1, G2, . . . , G2k) /∈ DNP-Equ.

Lemma 9 implies thatDNP-Equ is PNP|| -complete.

The above proof forDNP-Equ also gives PNP|| -completeness forDNP-Geq.

4 Exact Versions of Generalized Dominating Set Problems

Heggernes and Telle [HT98] proposed a general, uniform approach to define graph problems bypartitioning the vertex set of a graph into generalized dominating sets. Generalized dominatingsets are parameterized by two sets of nonnegative integers,σ andρ, which restrict the number ofneighbors for each vertex in the partition. We use this approach to define the exact versions of suchgeneralized dominating set problems and to study their computational complexity. In particular, weprove DP-completeness for a number of these problems. Usingthe techniques of Section 2, ourresults in the present section can easily be extended to moregeneral results involving slightly moregeneral problems and showing completeness in the higher levels of the boolean hierarchy.

4.1 Preliminaries

Let G be a graph. Theneighborhood of a vertexv in G is the set of all vertices adjacent tov, i.e.,N(v) = w ∈ V (G) | v,w ∈ E(G). A partition of V (G) into k pairwise disjoint subsetsV1, V2, . . . , Vk satisfiesV (G) =

⋃ki=1 Vi andVi ∩Vj = ∅ for 1 ≤ i < j ≤ k. LetN = 0, 1, 2, . . .

denote the set of nonnegative integers, and letN+ = 1, 2, 3, . . . denote the set of positive integers.

11

We now define the notions of(σ, ρ)-sets and(k, σ, ρ)-partitions introduced by Heggernes andTelle [HT98].

Definition 11 (Heggernes and Telle [HT98]) LetG be a given graph, letσ ⊆ N andρ ⊆ N begiven sets, and letk ∈ N

+.

1. A subsetU ⊆ V (G) of the vertices ofG is said to be a(σ, ρ)-setif and only if for eachu ∈ U ,||N(u) ∩ U || ∈ σ, and for eachu 6∈ U , ||N(u) ∩ U || ∈ ρ.

2. A(k, σ, ρ)-partition ofG is a partition ofV (G) into k pairwise disjoint subsetsV1, V2, . . . , Vk

such thatVi is a (σ, ρ)-set for eachi, 1 ≤ i ≤ k.

3. Define the problem

(k, σ, ρ)-Partition = G |G is a graph that has a(k, σ, ρ)-partition.

Heggernes and Telle [HT98] examined the(k, σ, ρ)-partitions of graphs for the parametersσandρ chosen among0, 1, 0, 1, N, andN+. In particular, they determined the precise cut-off points between tractability and intractability for these problems, that is, they determined theprecise value ofk for which the resulting(k, σ, ρ)-Partition problem is NP-complete, yet it can bedecided in polynomial time whether or not a given graph has a(k − 1, σ, ρ)-partition. An overviewof their (and previously known) results is given in Table 1.

ρ N N+ 1 0, 1

σ

N ∞− 3+ 2 ∞−

N+ ∞− 2+ 2 ∞−

1 2− 2 3 3−

0, 1 2− 2 3 3−

0 3− 3 4 4−

Table 1: NP-completeness cut-off points for(k, σ, ρ)-Partition

For example,(3,N,N+)-Partition is nothing else than the NP-complete domatic numberproblem: Given a graphG, decide whether or notG can be partitioned into three dominating sets.In contrast,(2,N,N+)-Partition is in P, and therefore the corresponding entry in Table 1 is3 forσ = N andρ = N

+. A value of∞ in Table 1 means that this problem is efficiently solvable forallvalues ofk. The value ofρ = 0 is not considered, since all graphs have a(k, σ, 0)-partition ifand only if they have the trivial partition intok disjoint (σ, 0)-setsV1 = V (G) andVi = ∅, foreachi ∈ 2, . . . , k.

The problems in Table 1 that are marked by a “+” are maximum problems, and the problemsthat are marked by a “−” are minimum problems. That is, for allk ≥ 1, we have that(k + 1, σ, ρ)-Partition ⊆ (k, σ, ρ)-Partition for the maximum problems, and we have that(k, σ, ρ)-Partition ⊆ (k + 1, σ, ρ)-Partition for the minimum problems. These properties arestated in the following fact.

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Fact 12 1. For each k ≥ 1 and for eachσ ∈ N,N+, 0, 0, 1, 1, it holds that(k, σ,N)-Partition ⊆ (k + 1, σ,N)-Partition.

2. For each k ≥ 1 and for each σ ∈ N,N+, 0, 0, 1, 1, it holds that(k, σ, 0, 1)-Partition ⊆ (k + 1, σ, 0, 1)-Partition.

3. For eachk ≥ 1 and for eachσ ∈ N,N+, it holds that(k + 1, σ,N+)-Partition ⊆(k, σ,N+)-Partition.

Proof. To see that all(k, σ, ρ)-Partition problems withρ = N are minimum problems, notethat we obtain a(k + 1, σ,N)-partition from a(k, σ,N)-partition by simply adding the empty setVk+1 = ∅. The proof for the caseρ = 0, 1 is similar.

To prove that the(k, σ, ρ)-Partition problems withρ = N+ are maximum problems, note that

once we have found a(k + 1, σ,N+)-partition intok + 1 pairwise disjoint setsV1, V2, . . . , Vk+1,the setsV1, V2, . . . , Vk−1, Vk with Vk = Vk ∪ Vk+1 are a(k, σ,N+)-partition as well.

Observe that those problems in Table 1 that are marked neither by a “+” nor by a“−” are neither maximum nor minimum problems in the sense defined above. That is,we have neither(k + 1, σ, ρ)-Partition ⊆ (k, σ, ρ)-Partition nor (k, σ, ρ)-Partition ⊆(k + 1, σ, ρ)-Partition, since for eachk ≥ 1, there exist graphsG with G ∈ (k, σ, ρ)-PartitionandG 6∈ (ℓ, σ, ρ)-Partition for any ℓ ≥ 1 with ℓ 6= k. For example, consider the problem(k, 1, 1)-Partition. By definition, (k, 1, 1)-Partition contains all those graphsGthat can be partitioned intok subsetsV1, V2, . . . , Vk such that, for eachi, if v ∈ Vi then||N(v) ∩ Vi|| = 1, and if v 6∈ Vi then ||N(v) ∩ Vi|| = 1. It follows that every graphin (k, 1, 1)-Partition must bek-regular; that is, every vertex has degreek. Hence, forall k ≥ 1, (k, 1, 1)-Partition and (k + 1, 1, 1)-Partition are disjoint, so neither(k, 1, 1)-Partition ⊆ (k + 1, 1, 1)-Partition nor (k + 1, 1, 1)-Partition ⊆(k, 1, 1)-Partition. In the case of(k, 0,N+)-Partition, the complete graphKn withn vertices is in(n, 0,N+)-Partition but not in(k, 0,N+)-Partition for anyk ≥ 1 withk 6= n. Almost the same argument applies to the caseσ = N andρ = 1, except that nowKn

is in (k,N, 1)-Partition for k ∈ 1, n but not in(ℓ,N, 1)-Partition for any ℓ ≥ 1 withℓ 6∈ 1, n. Similar arguments work in the other cases.

Therefore, when defining the exact versions of generalized dominating set problems, we confineourselves to those(k, σ, ρ)-Partition problems that are minimum or maximum problems in theabove sense. For a maximum problem, its exact version asks whetherG ∈ (k, σ, ρ)-Partitionbut G 6∈ (k + 1, σ, ρ)-Partition, and for a minimum problem, its exact version asks whetherG ∈ (k, σ, ρ)-Partition butG 6∈ (k − 1, σ, ρ)-Partition.

Definition 13 Let σ and ρ be chosen amongN,N+, 0, 0, 1, 1, let k ∈ N+, and let

(k, σ, ρ)-Partition be a minimum or a maximum problem. Define its exact version as follows:

• Exact-(k, σ, ρ)-Partition = (k, σ, ρ)-Partition∩ (k − 1, σ, ρ)-Partition if k ≥ 2 and(k, σ, ρ)-Partition is a minimum problem.

• Exact-(k, σ, ρ)-Partition = (k, σ, ρ)-Partition∩ (k + 1, σ, ρ)-Partition if k ≥ 1 and(k, σ, ρ)-Partition is a maximum problem.

13

For example, the problem(k, 0,N)-Partition is equal to thek-colorability problem, whichis a minimization problem: Given a graphG, find a partition into at mostk color classes such thatany two adjacent vertices belong to distinct color classes.In contrast,(k,N,N+)-Partition isequal toDNP, the domatic number problem, which is a maximization problem.

Clearly, since(k, σ, ρ)-Partition is in NP,Exact-(k, σ, ρ)-Partition is in DP.We now define two well-known problems that will be used later in our reductions. Both

problems were shown to be NP-complete by Schaefer [Sch78]. Note that1-3-SAT remains NP-complete even if all literals are positive.

Definition 14 LetX = x1, x2, . . . , xn be a finite set of variables.

• 1-3-SAT (“one-in-three satisfiability”): Let H be a boolean formula consisting of acollectionS = S1, S2, . . . , Sm of m sets of literals overX such that eachSi has exactlythree members.H is in 1-3-SAT if and only if there exists a subsetT of the literals overXwith ||T ∩ Si|| = 1 for eachi, 1 ≤ i ≤ m.

• NAE-3-SAT (“not-all-equal satisfiability”): Let H be a boolean formula consisting of acollection C = c1, c2, . . . , cm of m clauses overX such that eachci contains exactlythree literals. H is in NAE-3-SAT if and only if there exists a truth assignment forX thatsatisfies all clauses inC and such that in none of the clauses, all literals are true.

4.2 The caseρ = N

In this section, we are concerned with the minimum problemsExact-(k, σ,N)-Partition, whereσ is chosen fromN,N+, 0, 0, 1, 1. Depending on the valuek ≥ 2, we ask how hard it isto decide whether a given graphG has a(k, σ,N)-partition but not a(k − 1, σ,N)-partition.

The casesσ = N andσ = N+ are trivial, since for allk ≥ 1, (k,N,N)-Partition and(k,N+,N)-Partition are in P, which implies that the problemsExact-(k,N,N)-Partition andExact-(k,N+,N)-Partition are in P as well.

Now, suppose thatσ = 0. Recall that the problem(k, 0,N)-Partition is equal tothe k-colorability problem. The question about the complexity of the exact versions of thisproblem was optimally solved by Rothe [Rot03]:Exact-(4, 0,N)-Partition is DP-complete,yet Exact-(3, 0,N)-Partition is NP-complete and thus cannot be DP-complete unless theboolean hierarchy collapses to the first level.

Now, suppose thatσ = 0, 1.

Definition 15 For every graphG, define the minimum value ofk for whichG has a(k, 0, 1,N)-partition as follows:

α(G) = mink ∈ N+ |G ∈ (k, 0, 1,N)-Partition.

Theorem 16 1. Exact-(5, 0, 1,N)-Partition is DP-complete.

2. Exact-(2, 0, 1,N)-Partition is NP-complete.

14

Proof. 1. Since we have already noted thatExact-(5, 0, 1,N)-Partition is contained in DP,it remains to prove DP-hardness. Again, we apply Wagner’s Lemma 4 with1-3-SAT being the NP-complete problemA and withExact-(5, 0, 1,N)-Partition being the setB from this lemma.

In their paper [HT98], Heggernes and Telle presented a≤pm -reductionf from 1-3-SAT to

(2, 0, 1,N)-Partition with the following properties:

H ∈ 1-3-SAT =⇒ α(f(H)) = 2

H 6∈ 1-3-SAT =⇒ α(f(H)) = 3.

In short, reductionf works as follows. LetH be any given boolean formula that consists of acollectionS = S1, S2, . . . , Sm of m sets of literals overX = x1, x2, . . . , xn. Without loss ofgenerality, we may assume that all literals inH are positive. Reductionf mapsH to a graphG asfollows. For each setSi = x, y, z, there is a4-cliqueCi in G induced by the verticesxi, yi, zi,andai. For each literalx, there is an edgeex in G. For eachSi in whichx occurs, both endpointsof ex are connected to the vertexxi in Ci corresponding tox ∈ Si. Finally, there is yet another4-clique induced by the verticess, t1, t2, andt3. For eachi with 1 ≤ i ≤ m, vertexs is connectedto ai. This completes the reductionf . Figure 2 shows the graphG resulting from the reductionfapplied to the formulaH = (x ∨ y ∨ z) ∧ (v ∨ w ∨ x) ∧ (u ∨ w ∨ z).

t3

t1

t2

s

a1 a2

a3

x1

z1 y1 v2

x2

w2

w3z3

u3

ex

ewez

eu

evey

Figure 2: Heggernes and Telle’s reductionf from 1-3-SAT to (2, 0, 1,N)-Partition.

In order to apply Lemma 4, we need to find a reductiong satisfying

(H1 ∈ 1-3-SAT ∧H2 6∈ 1-3-SAT) ⇐⇒ α(g(H1,H2)) = 5 (4.6)

15

for any two given instancesH1 andH2 such thatH2 ∈ 1-3-SAT impliesH1 ∈ 1-3-SAT.Reductiong is constructed fromf as follows. LetG1,1 andG1,2 be two disjoint copies of the

graphf(H1), and letG2,1 andG2,2 be two disjoint copies of the graphf(H2). For i ∈ 1, 2,defineGi to be the disjoint union ofGi,1 andGi,2. Define the graphG = g(H1,H2) to be the joinof G1 andG2. That is,

g(H1,H2) = G = G1 ⊕G2 = (G1,1 ∪G1,2)⊕ (G2,1 ∪G2,2).

Figure 3 shows the graphG resulting from the reductiong applied to the formulas

H1 = (x ∨ y ∨ z) ∧ (v ∨ w ∨ x) ∧ (u ∨ w ∨ z) and

H2 = (c ∨ d ∨ e) ∧ (e ∨ f ∨ g) ∧ (g ∨ h ∨ i) ∧ (i ∨ j ∨ c).

G1,1 G1 G1,2

⊕

G2,1 G2 G2,2

Figure 3: GraphG = g(H1,H2) showing thatExact-(5, 0, 1,N)-Partition is DP-complete.

Let a = α(G1,1) = α(G1,2) andb = α(G2,1) = α(G2,2). Clearly,α(G1) = a, α(G2) = b,andα(G) ≤ a+ b. Simply partitionG the same way as graphsG1 andG2 were partitioned before.Note that we obtain8-cliques inG as a result of joining pairs of4-cliques fromG1 andG2. Thus,α(G) ≥ 4, since an8-clique has to be partitioned into at least four disjoint(0, 1,N)-sets.

To prove thatα(G) = α(G1) + α(G2) = a+ b, let k = α(G). Thus, we know4 ≤ k ≤ a+ b.For a contradiction, suppose thatk < a+ b. Distinguish the following cases.

Case 1: a = b = 2. Thenk < 4 is a contradiction tok ≥ 4.

Case 2: a = 2 and b = 3. Thenk = 4 < 5 = a + b. One of the four disjoint(0, 1,N)-setsconsists of at least one vertexu in G1 and one vertexv in G2. (Otherwise, it would inducea partition of less than two(0, 1,N)-sets inG1 or of less than three(0, 1,N)-sets inG2,which contradicts our assumptiona = 2 and b = 3.) Suppose that this set isV1. Then,

16

sinceσ = 0, 1 and sinceu is adjacent to every vertex inG2 andv is adjacent to everyvertex inG1, we haveV1 = u, v. But there is no way to assign the8-cliques, which donot containu or v, to the remaining three(0, 1,N)-sets in order to obtain a(4, 0, 1,N)-partition forG. This is a contradiction, and our assumptionk < a + b = 5 does not hold.Thus,k = 5.

Case 3: a = 3 and b = 2. This case cannot occur, since we have to prove Equation (4.6) onlyfor instancesH1 andH2 such thatH2 ∈ 1-3-SAT impliesH1 ∈ 1-3-SAT.

Case 4: a = b = 3. By the same argument used in Case 2,k = 4 does not hold. Supposek = 5.As seen before, one of the sets in the partition must contain exactly one vertexu fromG1 andexactly one vertexv from G2. Let V1 = u, v be this set. There are four sets left for thepartition, sayV2, V3, V4, andV5. Every setVi can have only vertices from eitherG1 or G2.This means that two of these sets cover all vertices inG1 except foru. Vertexu is eitherin G1,1 or in G1,2, which implies that one of these induced subgraphs (G1,1 or G1,2) has a(2, 0, 1,N)-partition. This is a contradiction toa = 3. Thus,k = 6.

Thus,α(G) = α(G1)+α(G2), which implies Equation (4.6) and thus fulfills Equation (2.1) ofLemma 4:

||i |Hi ∈ 1-3-SAT|| is odd ⇐⇒ H1 ∈ 1-3-SAT ∧H2 /∈ 1-3-SAT⇐⇒ α(G1) = 2 ∧ α(G2) = 3⇐⇒ α(G) = 5.

By Lemma 4,Exact-(5, 0, 1,N)-Partition is DP-complete.2. Exact-(2, 0, 1,N)-Partition is in NP, sinceExact-(2, 0, 1,N)-Partition = A ∩B

with A = (2, 0, 1,N)-Partition being in NP andB = (1, 0, 1,N)-Partition being in P.NP-hardness follows immediately via the reductionf :

H ∈ 1-3-SAT ⇐⇒ f(H) ∈ Exact-(2, 0, 1,N)-Partition.

Thus,Exact-(2, 0, 1,N)-Partition is NP-complete.

Finally, suppose thatσ = 1.

Definition 17 For every graphG, define the minimum valuek for which G has a (k, 1,N)-partition as follows:

β(G) = mink ∈ N+ |G ∈ (k, 1,N)-Partition.

Theorem 18 1. Exact-(5, 1,N)-Partition is DP-complete.

2. Exact-(2, 1,N)-Partition is NP-complete.

Proof. Clearly,α(G) ≤ β(G) for all graphsG. Conversely, we show thatα(G) ≥ β(G). It isenough to do so for all graphsG = f(H) resulting from any given instanceH of 1-3-SAT via thereductionf in Theorem 16. IfH ∈ 1-3-SAT, we haveα(G) = 2. Using the same partition, we even

17

get two(1,N)-sets forG. Every vertex ofG has exactly one neighbor, which is in the same set ofthe partition as the vertex itself. IfS 6∈ 1-3-SAT, thenα(G) = 3. We can then partitionG into three(1,N)-sets:V1 consists of the verticess andt1 plus the endpoints of each edgeex. V2 consists oft2 andt3, every vertexai, and one more vertex in the4-cliqueCi, for eachi with 1 ≤ i ≤ 2m. Thetwo remaining vertices in eachCi are then put into the setV3.

Hence,α(G) = β(G). The rest of the proof is analogous to the proof of Theorem 16.

4.3 The caseρ = N+

We consider the casesσ = N andσ = N+ only. These two are the only maximum problems

in Table 1. Recall that since(k,N,N+)-Partition and (k,N+,N+)-Partition are maximumproblems, their exact versions are defined as follows:

Exact-(k, σ,N+)-Partition =

GG ∈ (k, σ,N+)-Partition andG 6∈ (k + 1, σ,N+)-Partition

,

whereσ ∈ N,N+.Again, note thatExact-(k,N,N+)-Partition andExact-(k,N+,N+)-Partition are in DP.Suppose thatσ = N. Recall that the problem(k,N,N+)-Partition is equal toDNP, the

domatic number problem. The exact version of this problem was already analyzed in Section 2.In the notation used in the present section, Theorem 7 (for the special case ofk = 1) gives thefollowing.

Corollary 19 1. Exact-(5,N,N+)-Partition is DP-complete.

2. Exact-(2,N,N+)-Partition is coNP-complete.

Now, suppose thatσ = N+.

Definition 20 For every graphG, define the maximum valuek for whichG has a(k,N+,N+)-partition as follows:

γ(G) = maxk ∈ N+ |G ∈ (k,N+,N+)-Partition.

Theorem 21 1. Exact-(3,N+,N+)-Partition is DP-complete.

2. Exact-(1,N+,N+)-Partition is coNP-complete.

Proof. 1. Heggernes and Telle [HT98] presented a reduction from theproblemNAE-3-SAT tothe problem(2,N+,N+)-Partition to prove the latter problem NP-complete. We modify theirreduction as follows. Let two boolean formulasH1 = (X, C) andH2 = (Y, D) be given, withdisjoint variable sets,X = x1, x2, . . . , xn andY = y1, y2, . . . , yr, and with disjoint clausesets,C = c1, c2, . . . , cm and D = d1, d2, . . . , ds. If the variable sets consist of less thantwo variables, we put additional variables into the sets. Moreover, we may assume without loss ofgenerality that every literal appears in at least one clause, since otherwise we can easily alter the

18

given formulasH1 andH2, without changing membership inNAE-3-SAT, so that they are of thisform.

For any clausec = (x ∨ y ∨ z), definec = (x ∨ y ∨ z), wherex, y, andz, respectively, denotesthe negation of the literalx, y, andz. DefineC = c1, c2, . . . , cm andD = d1, d2, . . . , ds, anddefineC = C ∪ C andD = D ∪ D. Note that due to the not-all-equal property, we have:

(X,C) ∈ NAE-3-SAT ⇐⇒ (X, C) ∈ NAE-3-SAT

⇐⇒ (X, C) ∈ NAE-3-SAT

and

(Y,D) ∈ NAE-3-SAT ⇐⇒ (Y, D) ∈ NAE-3-SAT

⇐⇒ (Y, D) ∈ NAE-3-SAT.

We apply Lemma 4 withk = 1 and withNAE-3-SAT being the NP-complete problemA andwith Exact-(3,N+,N+)-Partition being the setB from this lemma. LetH1 andH2 be such thatH2 ∈ NAE-3-SAT impliesH1 ∈ NAE-3-SAT. Our polynomial-time reductionf transformsH1 andH2 into a graphG = f(H1,H2) with the property:

(H1 ∈ NAE-3-SAT ∧H2 6∈ NAE-3-SAT) ⇐⇒ γ(G) = 3. (4.7)

The reductionf is defined as follows. ForH1, we create an8-cliqueA8 with verticesa1, a2,. . ., a8. We do the same forH2, creating an8-cliqueB8 with verticesb1, b2, . . ., b8. For eachiwith 1 ≤ i ≤ n, we create two vertices,xi andxi, for the variablexi. For eachj with 1 ≤ j ≤ r,we create two vertices,yj andyj , for the variableyj. Every vertexxi andxi is connected to botha1 anda2, and every vertexyj andyj is connected to bothb1 andb2. For each pair of variables(xi, yj), we create one vertexui,j that is connected to the four verticesxi, xi, yj, andyj . Finally,for each clauseci ∈ C anddj ∈ D with 1 ≤ i ≤ m and1 ≤ j ≤ s, we create the two verticescianddj . Each such clause vertex is connected to the vertices representing the literals in that clause.Additionally, every vertexci is connected to botha1 anda2, and every vertexdj is connected tobothb1 andb2. This completes the construction of the graphG = f(H1,H2).

Figure 4 shows the graphG resulting from the reductionf applied to the two formulas

H1 = (x1 ∨ x2 ∨ x3) ∧ (x1 ∨ x2 ∨ x3) and

H2 = (y1 ∨ y2 ∨ y3) ∧ (y1 ∨ y2 ∨ y3).

Note thatγ(G) ≤ 4, since the degree of eachui,j is four. We have four cases to distinguish.

Case 1: H1 ∈ NAE-3-SAT and H2 ∈ NAE-3-SAT. Let t be a truth assignment satisfyingH1, andlet t be a truth assignment satisfyingH2. We can partitionG into four (N+,N+)-setsV1, V2,

19

a8 b7

x1

x1

x2

y1

y2

y3

y1

u1,2

u2,2

c2 d2 d1c1c2c1 d2 d1

u3,2x3

b8b6a6

a7

u1,1

u3,3

y2y3

u2,3

u3,1

u2,1

u1,3

b1

b2b3 b4

b5a3a4

a5 a2

a1

x3

x2

Figure 4: GraphG = f(H1,H2) showing thatExact-(3,N+,N+)-Partition is DP-complete.

V3, andV4 as follows:

V1 = C ∪ C ∪ a5, a6 ∪ b1, b3 ∪ x | x is a literal overX andt(x) = true,

V2 = ui,j | (1 ≤ i ≤ n− 1 ∧ j = 1) ∨ (i = n ∧ 2 ≤ j ≤ r) ∪ a7, a8 ∪ b2, b4

∪ x | x is a literal overX andt(x) = false,

V3 = D ∪ D ∪ a1, a3 ∪ b5, b6 ∪ y | y is a literal overY andt(y) = true,

V4 = ui,j | (i = n ∧ j = r) ∨ (1 ≤ i ≤ n− 1 ∧ 2 ≤ j ≤ r) ∪ a2, a4 ∪ b7, b8

∪ y | y is a literal overY andt(y) = false.

Thus,γ(G) ≥ 4. Sinceγ(G) ≤ 4, it follows thatγ(G) = 4 in this case.

Case 2: H1 ∈ NAE-3-SAT and H2 /∈ NAE-3-SAT. Let t be a truth assignment satisfyingH1. Wecan partitionG into three(N+,N+)-setsV1, V2, andV3 as follows:

V1 = C ∪ C ∪ a5, a6 ∪ b1, b3 ∪ x | x is a literal overX andt(x) = true,

V2 = ui,j | 1 ≤ i ≤ n ∧ 1 ≤ j ≤ r ∪ a7, a8 ∪ b2, b4

∪ x | x is a literal overX andt(x) = false,

V3 = D ∪ D ∪ a1, a2, a3, a4 ∪ b5, b6, b7, b8 ∪ y | y is a literal overY .

Thus,3 ≤ γ(G) ≤ 4. For a contradiction, suppose thatγ(G) = 4, with a partition ofG intofour (N+,N+)-sets, sayU1, U2, U3, andU4. Vertexu1,1 is adjacent to exactly four vertices,namely tox1, x1, y1 andy1. These four vertices must then be in four distinct sets of the

20

partition. Without loss of generality, suppose thatx1 ∈ U1, x1 ∈ U2, y1 ∈ U3, andy1 ∈ U4.For eachj with 2 ≤ j ≤ r, the verticesyj andyj are connected tox1 andx1 via vertexu1,j,so it follows that eitheryj ∈ U3 andyj ∈ U4, or yj ∈ U4 andyj ∈ U3.

Every clause vertexdj , 1 ≤ j ≤ r, is connected only to the vertices representing its literalsand to the verticesb1 andb2, which therefore must be in the setsU1 andU2, respectively.Thus, every clause vertexdj is connected to at least one literal vertex inU3 and to at least oneliteral vertex inU4. This describes a valid truth assignment forH2 in the not-all-equal sense.This is a contradiction to the case assumptionH2 /∈ NAE-3-SAT.

Case 3: H1 6∈ NAE-3-SAT and H2 ∈ NAE-3-SAT. This case is a contradiction to our assumptionthatH2 ∈ NAE-3-SAT impliesH1 ∈ NAE-3-SAT, and therefore it cannot occur.

Case 4: H1 6∈ NAE-3-SAT and H2 6∈ NAE-3-SAT. A valid partition ofG into two (N+,N+)-setsis:

V1 = ui,j | 1 ≤ i ≤ n ∧ 1 ≤ j ≤ r ∪ xi | 1 ≤ i ≤ n ∪ yj | 1 ≤ j ≤ r

∪ a1, a3, a5, a7 ∪ b1, b3, b5, b7,

V2 = C ∪ C ∪ D ∪ D ∪ xi | 1 ≤ i ≤ n ∪ yj | 1 ≤ j ≤ r

∪ a2, a4, a6, a7 ∪ b2, b4, b6, b8.

Thus,2 ≤ γ(G) ≤ 4. By the same argument as in Case 2,γ(G) 6= 4. For a contradiction,suppose thatγ(G) = 3, with a partition ofG into three(N+,N+)-sets, sayU1, U2, andU3.Without loss of generality, assume thatx1 andx1 belong to distinctUi sets,2 sayx1 ∈ U1 andx1 ∈ U2.

It follows that for eachj with 1 ≤ j ≤ r, at least one ofyj or yj has to be inU3. If bothvertices are inU3, then we have:

(∀i : 1 ≤ i ≤ n) [eitherxi ∈ U1 andxi ∈ U2, or xi ∈ U2 andxi ∈ U1]. (4.8)

SinceH1 6∈ NAE-3-SAT, for each truth assignmentt for H1, there exists a clauseci ∈ C suchthatci = (x∨y∨z) and the literalsx, y, andz are either simultaneously true or simultaneouslyfalse undert. Note that for the corresponding clauseci ∈ C, which contains the negationsof x, y, andz, the truth value of its literals is flipped undert. That is, t(x) = 1 − t(x),t(y) = 1− t(y), andt(z) = 1− t(z). Since the corresponding clause vertexci is adjacent tox, y, z, a1, anda2, it follows thatx, y, andz are in the same set of the partition, say inU1.Hence, eithera1 ∈ U2 anda2 ∈ U3, or a1 ∈ U3 anda2 ∈ U2. Similarly, since the clausevertex ci is adjacent tox, y, z, a1, anda2, the verticesx, y, z are in the same set of thepartition that must be distinct fromU1. LetU2, say, be that set. It follows that eithera1 ∈ U1

anda2 ∈ U3, or a1 ∈ U3 anda2 ∈ U1, which is a contradiction.

Each of the remaining subcases can be reduced to (4.8), and the above contradiction follows.Hence,γ(G) = 2.

2If x1 andx1 both belong to the same setUi, then eachyj andyj must belong to distinct setsUk andUℓ, k 6= ℓ, sinceu1,j is connected withx1, x1, yj , andyj . Thus, a symmetric argument works foryj andyj in this case.

21

Thus, we obtain:

||i |Hi ∈ NAE-3-SAT|| is odd ⇐⇒ H1 ∈ NAE-3-SAT ∧H2 6∈ NAE-3-SAT⇐⇒ γ(G) = 3,

which proves Equation (4.7). Thus, Equation (2.1) of Lemma 4is fulfilled, and it follows thatExact-(3,N+,N+)-Partition is DP-complete.

2. Exact-(1,N+,N+)-Partition is in coNP, sinceExact-(1,N+,N+)-Partition = A ∩Bwith A = (1,N+,N+)-Partition being in P andB = (2,N+,N+)-Partition being in NP. Notethat the coNP-hardness ofExact-(1,N+,N+)-Partition follows immediately via the originalreduction fromNAE-3-SAT to (2,N+,N+)-Partition presented in [HT98].

ρ N N+

σ

N ∞ 5∗

N+ ∞ 3∗

1 5∗ −0, 1 5∗ −0 4 −

Table 2: DP-completeness values forExact-(k, σ, ρ)-Partition

The numbers in Table 2 indicate the best DP-completeness results currently knownfor the exact versions of generalized dominatic set problems, where the results from thispaper are marked with an asterisk.3 That is, they give the best value ofk for whichthe problemExact-(k, σ, ρ)-Partition is known to be DP-complete. In many cases thisvalue is not yet optimal. For example,Exact-(5,N,N+)-Partition is known to be DP-complete andExact-(2,N,N+)-Partition is known to be coNP-complete. What aboutExact-(3,N,N+)-Partition andExact-(4,N,N+)-Partition? Only the DP-completeness ofExact-(4, 0,N)-Partition is known to be optimal [Rot03].

5 The Exact Conveyor Flow Shop Problem

The conveyor flow shop problem is a minimization problem arising in real-world applications in thewholesale business, where warehouses are supplied with goods from a central storehouse. Supposeyou are givenm machines,P1, P2, . . . , Pm, andn jobs,J1, J2, . . . , Jn. Conveyor belt systems areused to convey jobs from machine to machine at which they are to be processed in a “permutationflow shop” manner. That is, the jobs visit the machines in the fixed orderP1, P2, . . . , Pm, and themachines process the jobs in the fixed orderJ1, J2, . . . , Jn. An (n×m) task matrixM = (µj,p)j,pwith µj,p ∈ 0, 1 provides the information which job has to be processed at which machine:

3Again, a value of∞ in Table 2 means that this problem is efficiently solvable forall values ofk.

22

µj,p = 1 if job Jj is to be processed at machinePp, andµj,p = 0 otherwise. Every machine canprocess at most one job at a time. There is one worker supervising the system. Every machine canprocess a job only if the worker is present, which means that the worker occasionally has to movefrom one machine to another. If the worker is currently not present at some machine, jobs can bequeued in a buffer at this machine. The objective is to minimize the movement of the worker, wherewe assume the “unit distance” between any two machines, i.e., to measure the worker’s movement,we simply count how many times he has switched machines untilthe complete task matrix has beenprocessed.4 Let∆min(M) denote the minimum number of machine switches needed for theworkerto completely process a given task matrixM, where the minimum is taken over all possible ordersin which the tasks inM can be processed. Define the decision version of the conveyorflow shopproblem by:

CFSP = 〈M, k〉 |M is a task matrix andk is a positive integer such that∆min(M) ≤ k.

Espelage and Wanke [EW00,Esp01,EW01,EW03] introduced andstudied the problemCFSP,and variations thereof, extensively. We are interested in the complexity of the exact version ofCFSP.

Definition 22 Define the exact version of the conveyor flow shop problem by:

Exact-k-CFSP =

〈M, Sk〉M is a task matrix andSk ⊆ N is a set ofknoncontiguous integers with∆min(M) ∈ Sk

.

To show thatCFSP is NP-complete, Espelage [Esp01, pp. 27–44] provided, in a rather involved17 pages proof, a reductiong from the3-SAT problem toCFSP, via the intermediate problem offinding a “minimum valid block cover” of a given task matrixM. In particular, finding a minimumblock cover ofM directly yields a minimum number of machine switches. Espelage’s reductioncan easily be modified so as to have certain useful properties, which we state in the followinglemma. The details of this modification can be found in [Rie02]; in particular, prior to the Espelagereduction, a reduction from the (unrestricted) satisfiability problem to3-SAT is used that has theproperties stated as Equations (5.9) and (5.10) below.

Lemma 23 (Espelage and Riege; see pp. 27–44 of [Esp01] and pp.37–42 of [Rie02])There exists a polynomial-time many-one reductiong that witnesses3-SAT≤p

m CFSP and satisfies,for each given boolean formulaϕ, the following properties:

1. g(ϕ) = 〈Mϕ, zϕ〉, whereMϕ is a task matrix andzϕ ∈ N is an odd number.

2. ∆min(Mϕ) = zϕ + uϕ, whereuϕ denotes the minimum number of clauses ofϕ not satisfiedunder assignmentt, where the minimum is taken over all assignmentst of ϕ. Moreover,uϕ = 0 if ϕ ∈ 3-SAT, anduϕ = 1 if ϕ 6∈ 3-SAT.

In particular, ϕ ∈ 3-SAT if and only if∆min(Mϕ) is odd.

Theorem 24 For eachk ≥ 1, Exact-k-CFSP is BH2k(NP)-complete. In particular, fork = 1,Exact-1-CFSP is DP-complete.

4In this paper, we do not consider possible generalizations of the problemCFSP such as other distance functions,variable job sequences, more than one worker, etc. We refer to [Esp01] for results on such more general problems.

23

Proof. Analogously to the proof of Theorem 7, we can show thatExact-k-CFSP is in BH2k(NP).To prove BH2k(NP)-hardness ofExact-k-CFSP, we again apply Lemma 4, with some fixed NP-complete problemA and withExact-k-CFSP being the problemB from this lemma. The reductionf satisfying Equation (2.1) from Lemma 4 is defined by using twopolynomial-time many-onereductions,g andh.

We now define the reductionsg andh. Fix the NP-complete problemA. Letx1, x2, . . . , x2k bestrings inΣ∗ satisfying thatcA(x1) ≥ cA(x2) ≥ · · · ≥ cA(x2k), wherecA denotes the characteristicfunction ofA, i.e.,cA(x) = 1 if x ∈ A, andcA(x) = 0 if x 6∈ A. Wagner [Wag87] observed that thestandard reduction (cf. [GJ79]) from the (unrestricted) satisfiability problem to3-SAT can be easilymodified so as to yield a reductionh from A to 3-SAT (via the intermediate satisfiability problem)such that, for eachx ∈ Σ∗, the boolean formulaϕ = h(x) satisfies the following properties:

x ∈ A =⇒ sϕ = mϕ; (5.9)

x 6∈ A =⇒ sϕ = mϕ − 1, (5.10)

wheresϕ = maxtℓ | ℓ clauses ofϕ are satisfied under assignmentt, andmϕ denotes the numberof clauses ofϕ. Moreover,mϕ is always odd.

Let ϕ1, ϕ2, . . . , ϕ2k be the boolean formulas after applying reductionh to each givenxi ∈ Σ∗,i.e.,ϕi = h(xi) for eachi. For i ∈ 1, 2, . . . , 2k, let mi = mϕi

be the number of clauses inϕi,and letsi = sϕi

denote the maximum number of satisfiable clauses ofϕi, where the maximum istaken over all assignments ofϕi. For eachi, apply the Lemma 23 reductiong from 3-SAT to CFSP

to obtain2k pairs〈Mi, zi〉 = g(ϕi), where eachMi = Mϕiis a task matrix and eachzi = zϕi

isthe odd number corresponding toϕi according to Lemma 23. Use these2k task matrices to form anew task matrix:

M =

M1 0 · · · 0

0 M2. . .

......

. . . . . . 00 · · · 0 M2k

.

Every task of some matrixMi, where1 ≤ i ≤ 2k, can be processed only if all tasks of thematricesMj with j < i have already been processed; see [Esp01,Rie02] for arguments as to whythis is true. This implies that:

∆min(M) =

2k∑

i=1

∆min(Mi).

Let z =∑2k

i=1 zi; note thatz is even. Define the setSk = z+1, z+3, . . . , z+2k−1, and definethe reductionf by f(x1, x2, . . . , x2k) = 〈M, Sk〉. Clearly,f is polynomial-time computable.

Let ui = uϕi= mintℓ | ℓ clauses ofϕi are not satisfied under assignmentt. Equations (5.9)

and (5.10) then imply that for eachi:

ui = mi − si =

0 if xi ∈ A1 if xi 6∈ A.

Recall that, by Lemma 23, we have∆min(Mi) = zi + ui. Hence:

24

||i | xi ∈ A|| is odd

⇐⇒ (∃i : 1 ≤ i ≤ k) [x1, . . . , x2i−1 ∈ A andx2i, . . . , x2k 6∈ A]

⇐⇒ (∃i : 1 ≤ i ≤ k) [s1 = m1, . . . , s2i−1 = m2i−1 ands2i = m2i − 1, . . . , s2k = m2k − 1]

⇐⇒ (∃i : 1 ≤ i ≤ k) [∆min(M1) = z1, . . . ,∆min(M2i−1) = z2i−1 and

∆min(M2i) = z2i + 1, . . . ,∆min(M2k) = z2k + 1]

⇐⇒ (∃i : 1 ≤ i ≤ k)

∆min(M) =2k∑

j=1

∆min(Mj) =

2k∑

j=1

zj

+ 2k − 2i+ 1

⇐⇒ ∆min(M) ∈ Sk = z + 1, z + 3, . . . , z + 2k − 1

⇐⇒ f(x1, x2, . . . , x2k) = 〈M, Sk〉 ∈ Exact-k-CFSP.

Thus,f satisfies Equation (2.1). By Lemma 4,Exact-k-CFSP is BH2k(NP)-complete.

6 Conclusions and Open Questions

In this paper, we have shown that the exact versions of the domatic number problem and of theconveyor flow shop problem are complete for the levels of the boolean hierarchy over NP. Inparticular, fork = 1 and for each given integeri ≥ 5, it is DP-complete to determine whetheror not δ(G) = i for a given graphG. In contrast,Exact-2-DNP is coNP-complete, and thusthis problem cannot be DP-complete unless the boolean hierarchy collapses. Fori ∈ 3, 4, thequestion of whether or not the problemsExact-i-DNP are DP-complete remains an interesting openproblem. As mentioned in the introduction, the corresponding gap for the exact chromatic numberproblem was recently closed by Rothe [Rot03]; see also [RSV02]. His reduction uses both thestandard reduction from3-SAT to 3-Colorability (cf. [GJ79]) and a very clever reduction foundby Guruswami and Khanna [GK00]. The decisive property of theGuruswami–Khanna reduction isthat it maps each satisfiable formulaϕ to a graphG with χ(G) = 3, and it maps each unsatisfiableformulaϕ to a graphG with χ(G) = 5. That is, the graphs they construct are never4-colorable.To close the above-mentioned gap for the exact domatic number problem, one would have to find areduction from some NP-complete problem toDNP with a similarly strong property: the reductionwould have to yield graphs that never have a domatic number of3.

In Section 4, we have studied the exact versions of generalized dominating set problems. Basedon the uniform approach of Heggernes and Telle [HT98], we have considered problems of the formExact-(k, σ, ρ)-Partition, where the parametersσ andρ give the number of neighbors allowedfor each vertex in the partition. We have obtained DP-completeness results for the minimizationproblems Exact-(5, 0, 1,N)-Partition and Exact-(5, 1,N)-Partition, and for themaximization problemsExact-(5,N,N+)-Partition andExact-(3,N+,N+)-Partition. Theseresults are summarized in Table 2. Note thatExact-(k,N,N+)-Partition = Exact-k-DNP.Again, there arises the question of whether the value ofk = 3 for σ = ρ = N

+ and the value

25

of k = 5 in the other cases is optimal in all these results. We were only able to show these problemsNP-complete or coNP-complete for the value ofk = 1 if σ = ρ = N

+ and for the value ofk = 2 inthe other cases, thus leaving a gap between DP-completenessand membership in NP or coNP.

Another interesting open question is whether one can obtainsimilar results for the minimizationproblems Exact-(k, σ, 0, 1)-Partition for σ ∈ 0, 0, 1, 1. It appears that theconstructions that we used in Theorems 7, 16, 18, and 21 do notwork here.

In Section 5, we studied the exact conveyor flow shop problem using similar techniques. Weproved thatExact-1-CFSP is DP-complete andExact-k-CFSP is BH2k(NP)-complete. Note that indefining these problems, we do not specify a fixed setSk with k fixed values as problem parameters;see Definition 22. Rather, only the cardinalityk of such sets is given as a parameter, andSk is partof the problem instance ofExact-k-CFSP. The reason is that the actual values ofSk depend onthe input of the reductionf defined in the proof of Theorem 24. In particular, the numberzϕ fromLemma 23, which is used to define the numberz =

∑2ki=1 zi in the proof of Theorem 24, has the

following form (see [Esp01,Rie02]):

zϕ = 28nK + 27nK + 8nU + 90mt+ 99m,

where t is the number of variables andm is the number of clauses of the given booleanformulaϕ, andnK , nK , andnU denote respectively the number of “coupling, inverting coupling,and interrupting elements” of the “minimum valid block cover” constructed in the Espelagereduction [Esp01] from3-SAT to CFSP. It would be interesting to know whether one can obtainBH2k(NP)-completeness ofExact-k-CFSP even if a setSk of k fixed values is specified a priori.

Acknowledgments. We are grateful to Gerd Wechsung for his interest in this paper, for manyhelpful conversations and, in particular, for pointing outthe coNP-completeness ofExact-2-DNP.

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