Dominating Set - Tree decompositions of graphs and their eﬀect on algorithmic complexity

Adominating set of a graphGis a set S⊆V(G)such that V(G) =N[S]. In theDominating Set problem we are given a graph Gand the objective is to find a dominating set of minimum size.

The basic idea for this reduction is similar to the one forIndependent Set. However, we need one more new idea here, which will also be used in other reductions. We group variables into an appropriate number of groups of size at most β =blog 3^pc, wherepis a constant depending only on . Then, for every group we make a gadget such that an assignment on the group should correspond to a selection on the gadget. These group gadgets are then connected to clause gadgets so that every assignment on the group that satisfies the clause results in some desired outcome.

Theorem 2.7. If Dominating Set can be solved in (3−)^pw(G)·n^O(1) time for some >0 then SAT can be solved in (2−δ)ⁿ·n^O(1) time for someδ >0.

Construction. Given <1 and an instanceφ toSAT we construct a graphG as follows. We first choose an integer p depending only on . Exactly how p is chosen will be discussed in the

2.3. DOMINATING SET 25

p¹_p p³₁ p³_p gp

g₁⁰ g₁

p³₁ p³₁ p¹₁

x x⁰_S x_S

g_p⁰

Figure 2.2: Reduction to Dominating Set: group gadget B. The setb S is shown by the circled vertices.

proof of Theorem 2.7. We group the variables ofφinto groups F₁, F₂, . . . , F_t, each of size at most β=blog 3^pc. Hencet=dn/βe. We now proceed to describe a “group gadget” Bb, which is central in our construction.

To build the group gadget B, we introduceb p paths P1, . . . , Pp, where the pathPi contains the vertices p¹_i, p²_i and p³_i (see Figure 2.2). To each path Pi we attach two guards gi and g_i⁰, both of which are neighbours to p¹_i, p²_i andp³_i. When the gadgets are attached to each other, the guards will not have any neighbours outside of their own gadgetBb, and will ensure that at least one vertex out ofp¹_i,p²_i and p³_i are chosen in any minimum size dominating set ofG. LetP be the vertex set containing all the vertices on the pathsP1, . . . , Pp. For every subset S ofP that picksexactly one vertex from each pathP_i, we introduce two verticesx_S andx⁰_S, wherex_S is adjacent to all vertices of P\S (all those vertices that are on paths and not inS) andx⁰_S is only adjacent toxS. We conclude the construction ofBb by making all the verticesx⁰_S (for every set S) adjacent to each other, that is making them into a clique, and adding a guardx adjacent tox⁰_S for every setS. In other words, the x⁰_S’s together with x form a clique and all the neighbors ofx reside in this clique.

We construct the graph G as follows (see Figure 2.3). For every group F_i of variables, we introducem(2pt+ 1)copies of the gadgetB, call themb Bb_i^j for1≤j≤m(2pt+ 1). We can imagine these t·m(2pt+ 1)gadgets arranged in t rows and m(2pt+ 1) columns, with the columns being divided into 2pt+ 1 regions of m columns each. For every fixed i ≤ t, we connect the gadgets Bb_i¹,Bb_i². . . ,Bb_i^m(2pt+1) in a path-like manner. In particular, for everyj < m(2pt+ 1) and every`≤p we make an edge between p³_` in the gadget Bb_i^j withp¹_` in the gadgetBb_i^j+1. Now we introduce two

h⁰ h

Bb_t^x Bb₁^x bc^`_j

Figure 2.3: Reduction toDominating Set: arranging the group gadgets. Note thatx=m`+j, thus bc^`_j is attached to vertices in Bb₁^x,. . .,Bb_t^x.

new verticesh and h⁰, withh adjacent to h⁰,p¹_j inBb_i¹ for every i≤t,j≤pand to p³_j inBb_i^m(2pt+1) for every i≤t,j≤p. That is, for all1≤i≤t,h is adjacent to the first and last vertices of “long paths” obtained after connecting the gadgets Bb_i¹,Bb_i². . . ,Bb_i^m(2pt+1) in a path-like manner.

For every 1 ≤i≤t and to every assignment of the variables in the group F_i, we designate a subsetS ofP in the gadgetBb that picks exactly one vertex from each path P_j. Since there are at most2^β different assignments to the variables inFi, and there are3^p ≥2^β such setsS, we can assign a uniqueset to each assignment. Of course, the same setS can correspond to one assignment of the group F₁ and some other assignment of the groupF₂. Recall that the clauses ofφ areC₁, . . . , C_m. For every clause Cj we introduce2pt+ 1 verticesbc^`_j, one for each 0≤` < 2pt+ 1, corresponding to the2pt+ 1 regions. The vertexbc^`_j will be connected to the gadgetsBb_i^m`+j for every 1≤i≤t (which appear in the `-th region). In particular, for every assignment of the variables in the group F_i that satisfy the clause C_j, we consider the subset S ofP that corresponds to the assignment. For every0≤` <2pt+ 1, we makex⁰_S in Bb^m`+j_i adjacent tobc^`_j. The best way to view this is that every

2.3. DOMINATING SET 27 clauseC_j has 2pt+ 1 private gadgets in the i-row, Bb_i^j,Bb_i^m+j, . . . ,Bb^m2pt+j_i , one in each region. Now we have 2pt+ 1 vertices corresponding to the clauseC_j, each connected to one of these gadgets.

This concludes the construction ofG.

Lemma 2.8. Ifφ has a satisfying assignment, then G has a dominating set of size (p+ 1)tm(2pt+ 1) + 1.

Proof. Given a satisfying assignment toφ, we construct a dominating set Dof Gthat contains the vertex h andexactlyp+ 1 vertices in each gadgetBb_i^j. For each groupFi of variables we consider the set S that corresponds to the restriction of the assignment to the variables in F_i. From each gadgetBb^j_i we add the setS toD and also the vertexx⁰_S toD. It remains to argue thatD is indeed a dominating set. Clearly the size is bounded by(p+ 1)tm(2pt+ 1) + 1, as the number of gadgets is tm(2pt+ 1).

For a fixed i ≤ t and j consider the vertices on the path P_j in the gadgets Bb_i^` for every

`≤m(2pt+ 1). Together these vertices form a path of length 3m(2pt+ 1)and every third vertex of this path is in S. Thus, all vertices on this path are dominated by other vertices on the path, except perhaps for the first and last one. Both these vertices, however, are dominated by h.

Now, fix some i≤t and `≤m(2pt+ 1)and consider the gadget Bb_i^`. Since D contains some vertex on the path P_j, we have that for every j both g_j and g_j⁰ are dominated. Furthermore, for every set S^∗ not equal toS that picks exactly one vertex from eachP_j, vertex x_S^∗ is dominated by some vertex on somePj—namely by all vertices inS\S^∗ 6=∅. The last assertion follows since xS^∗

is connected to all the vertices on the paths exceptS^∗. On the other hand,x_S is dominated byx⁰_S, andx⁰_S also dominates all the other vertices x⁰_S∗ for S^∗ 6=S, as well as the guardx.

The only vertices not yet accounted for are the vertices bc^`_j for everyj ≤m and` <2pt+ 1. Fix a j and a `and consider the clause C_j. This clause contains a literal set to true, and this literal corresponds to a variable in the groupF_i for somei≤t. Of course, the assignment toF_i satisfiesC_j. LetS be the set corresponding to this assignment of Fi. By the construction ofD, the dominating set containsx⁰_S inBb_i^m`+j and x⁰_S is adjacent tobc^`_j. This concludes the proof.

Lemma 2.9. If G has a dominating set of size (p+ 1)tm(2pt+ 1) + 1, then φ has a satisfying assignment.

Proof. LetDbe a dominating set ofGof size at most(p+ 1)tm(2pt+ 1) + 1. SinceDmust dominate h⁰, without loss of generality we can assume thatDcontainsh. Furthermore, inside every gadgetBb_i^`, Dmust dominate all the guards, namelyg_j andg_j⁰ for every j≤p, and also x. Thus D contains at leastp+ 1vertices from each gadgetBb_i^` which in turn implies thatD contains exactlyp+ 1vertices from each gadgetBb_i^`. The only way Dcan dominategj andg_j⁰ for everyj and in addition dominate x with onlyp+ 1vertices if Dhas one vertex from each P_j,j≤p and in addition contains some vertex in N[x]. LetS be D∩P inBb_i^`. Observe thatx_S is not dominated byD∩S. The only vertex inN[x]that dominatesxS isx⁰_S and hence Dcontains x⁰_S.

Now we want to show that for every 1≤i≤tthere exists one0≤`≤2tp such that for fixedi, D∩P is same in all the gadgets Bb_i^m`+r for every1≤r≤m, i.e., it is the same in every gadget of the i-th row in the `-th region. Consider a gadgetBb_i^` and its follower, Bb^`+1_i . Let S beD∩P in Bb_i^` andS⁰ beD∩P inBb_i^`+1. Observe that if S containsp^a_j in Bb_i^` and p^b_j inBb_i^`+1 then we must have b≤a. We call a consecutive pair badif for some j ≤p, D containsp^a_j in Bb_i^` and p^b_j in Bb_i^`+1 and b < a. Hence for a fixed i, we can at most have 2pconsecutive bad pairs, spoiling at most2p regions.

Now we mark all the bad pairs that occur among the gadgets corresponding to someF_i. This way we can mark only2tp bad pairs. Thus, by the pigeon hole principle, there exists an`∈ {0, . . . ,2tp}

such that there are no bad pairs in Bb_i^m`+r for all 1≤i≤t and1≤r ≤m.

We make an assignmentφby reading offD∩P in each gadgetBb_i^m`+1. In particular, for every group F_i, we consider S=D∩P in the gadgetBb_i^m`+1. This set S corresponds to an assignment of Fi, and this is the assignment ofFi that we use. It remains to argue that every clauseCr is satisfied by this assignment.

Consider the vertexbc^r_`. We know that it is dominated by somex⁰_S in a gadgetBb_i^m`+r. The setS corresponds to an assignment of F_i that satisfies the clauseC_r. BecauseD∩P remains unchanged in all gadgets from Bb_i^m`+1 toBb_i^m`+r, this is exactly the assignment φrestricted to the group F_i. This concludes the proof.

Lemma 2.10. pw(G)≤tp+O(3^p)

Proof. We give a mixed search strategy to clean the graph withtp+O(3^p) searchers. For a gadget Bb we call the vertices p¹_j and p³_j, 1 ≤j ≤p, as entry vertices and exit vertices respectively. We search the graph in m(2tp+ 1) rounds. In the beginning of round `there are searchers on the entry vertices of the gadgets Bb_i^` for every i ≤t. Let 1 ≤a ≤m and0 ≤b < 2tp+ 1 be integers such that `=a+mb. We place a searcher on bc^b_a. Then, for each ibetween 1and p in turn we first put searchers on all vertices ofBb_i^` and then remove all the searchers fromBb_i^`except for the ones standing on the exit vertices. After all gadgetsBb^`₁. . .Bb_t^` have been cleaned in this manner, we can remove the searcher frombc^b_a. To commence the next round, the searchers slide from the exit positions ofBb_i^` to the entry positions of Bb_i^`+1 for everyi. In total, at most tp+|V(B)|b + 1≤tp+O(3^p) searchers are used simultaneously. This together with Proposition 2.2 give the desired upperbound on the pathwidth.

Proof (of Theorem 2.7). Suppose Dominating Set can be solved in(3−)^pw(G)·n^O(1)= 3^λpw(G)· n^O(1) time, whereλ= log₃(3−)<1. We chooseplarge enough such thatλ·_bp_{log 3c}^p = _{log 3}^δ⁰ for some δ⁰ <1. Given an instance of SAT, we construct an instance of Dominating Setusing the above construction and the chosen value of p. Then we solve theDominating Set instance using the 3^λpw(G)·n^O(1) time algorithm. Correctness is ensured by Lemmata 2.8 and 2.9. Lemma 2.10 yields that the total time taken is upper bounded by3^λpw(G)·n^O(1)≤3^λ(tp+f(λ))·n^O(1) ≤3^λ

bplog 3c)·n^O(1)≤ 3^δ⁰^{log 3}ⁿ ·n^O(1)≤2^δ⁰⁰ⁿ·n^O(1) =(2−δ)ⁿ·n^O(1), for some δ⁰⁰, δ <1. This concludes the proof.

In document Tree decompositions of graphs and their eﬀect on algorithmic complexity (Pldal 24-28)