Tropical Dominating Sets in Vertex-Coloured Graphs

# Tropical Dominating Sets in Vertex-Coloured Graphs

J.-A. Anglès d’Auriac, Cs. Bujtás, A. El Maftouhi, M. Karpinski,
Y. Manoussakis, L. Montero, N. Narayanan, L. Rosaz,
J. Thapper, Zs. Tuza
Université Paris-Sud, L.R.I., Bât. 650, 91405 Orsay Cedex, France.
Department of Computer Science and Systems Technology,
University of Pannonia, 8200 Veszprém, Egyetem u. 10, Hungary.
University of Bonn, Department of Computer Science,
Friedrich-Ebert-Allee 144, 53113 Bonn, Germany.
Université Paris-Est, Marne-la-Vallée, LIGM, Bât. Copernic,
5 Bd Descartes, 77454 Marne-la-Vallée Cedex 2, France.
Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences,
1053 Budapest, Reáltanoda u. 13–15, Hungary.
angles@lri.fr, bujtas@dcs.uni-pannon.hu, hakim.maftouhi@orange.fr,
marek@cs.uni-bonn.de, yannis@lri.fr, lmontero@lri.fr,
narayana@gmail.com, rosaz@lri.fr, thapper@u-pem.fr, tuza@dcs.uni-pannon.hu
###### Abstract

Given a vertex-coloured graph, a dominating set is said to be tropical if every colour of the graph appears at least once in the set. Here, we study minimum tropical dominating sets from structural and algorithmic points of view. First, we prove that the tropical dominating set problem is NP-complete even when restricted to a simple path. Then, we establish upper bounds related to various parameters of the graph such as minimum degree and number of edges. We also give an optimal upper bound for random graphs. Last, we give approximability and inapproximability results for general and restricted classes of graphs, and establish a FPT algorithm for interval graphs.

Keywords: Dominating set, Vertex-coloured graph, Approximation, Random graphs

## 1 Introduction

Vertex-coloured graphs are useful in various situations. For instance, the Web graph may be considered as a vertex-coloured graph where the colour of a vertex represents the content of the corresponding page (red for mathematics, yellow for physics, etc). Given a vertex-coloured graph , a subgraph (not necessarily induced) of is said to be tropical if and only if each colour of appears at least once in . Potentially, any kind of usual structural problems (paths, cycles, independent and dominating sets, vertex covers, connected components, etc.) could be studied in their tropical version. This new tropical concept is close to, but quite different from, the colourful concept used for paths in vertex-coloured graphs [1, 26, 27]. It is also related to (but again different from) the concept of colour patterns used in bio-informatics [18]. Here, we study minimum tropical dominating sets in vertex-coloured graphs. Some ongoing work on tropical connected components, tropical paths and tropical homomorphisms can be found in [13, 14, 19]. A general overview on the classical dominating set problem can be found in [23].

Throughout the paper let denote a simple undirected non-coloured graph. Let and . Given a set of colours , denotes a vertex-coloured graph where each vertex has precisely one colour from and each colour of appears on at least one vertex. The colour of a vertex is denoted by . A subset is a dominating set of (or of ), if every vertex either belongs to or has a neighbour in . The domination number () is the size of a smallest dominating set of (). A dominating set of is said to be tropical if each of the colours appears at least once among the vertices of . The tropical domination number is the size of a smallest tropical dominating set of . A rainbow dominating set of is a tropical dominating set with exactly vertices. More generally, a -element set with precisely one vertex from each colour is said to be a rainbow set. We let (respectively ) denote the minimum (maximum) degree of . When no confusion arises, we write , , and instead of , , and , respectively. We use the standard notation for the (open) neighbourhood of vertex , that is the set of vertices adjacent to , and write for its closed neighbourhood. The set and the number of neighbours of inside a subgraph is denoted by and by , independently of whether is in or in . Although less standard, we shall also write sometimes to abbreviate .

Note that tropical domination in a vertex-coloured graph can also be interpreted as “simultaneous domination” in two graphs which have a common vertex set. One of the two graphs is the non-coloured itself, the other one is the union of vertex-disjoint cliques each of which corresponds to a colour class in . The notion of simultaneous dominating set111Also known under the names ‘factor dominating set’ and ‘global dominating set’ in the literature. was introduced by Sampathkumar [32] and independently by Brigham and Dutton [9]. It was investigated recently by Caro and Henning [10] and also by further authors. Remark that is regularly assumed for each factor graph in the results of these papers that is not the case in the present manuscript, as we do not forbid the presence of one-element colour classes.

The Tropical Dominating Set problem (TDS) is defined as follows.

###### Problem 1.

TDS
Input: A vertex-coloured graph and an integer .
Question: Is there a tropical dominating set of size at most ?

The Rainbow Dominating Set problem (RDS) is defined as follows.

###### Problem 2.

RDS
Input: A vertex-coloured graph .
Question: Is there a rainbow dominating set?

The paper is organized as follows. In Section 2 we prove that RDS is NP-complete even when graphs are restricted to simple paths. In Section 3 we give upper bounds for related to the minimum degree and the number of edges. We give upper bounds for random graphs in Section 4. In Section 5 we give approximability and inapproximability results for TDS. We also show that the problem is FPT (fixed-parameter tractable) on interval graphs when parametrized by the number of colours.

## 2 NP-completeness

In this section we show that the RDS problem is NP-complete. This implies that the TDS problem is NP-complete too.

###### Theorem 2.1.

The RDS problem is NP-complete, even when the input is restricted to vertex-coloured paths.

###### Proof.

Clearly the RDS problem is in NP. The reduction is obtained from the 3-SAT problem. Let be an instance of 3-SAT where is a collection of clauses on a finite set of boolean variables. From this instance, we will define a vertex-coloured path such that contains a rainbow dominating set if and only if is satisfiable.

In order to define , we first construct a segment , and we colour its vertices as follows. Vertices and are coloured black. Vertices are each coloured with a unique colour. The remaining vertices, that will be henceforth called clausal, are coloured from to with colours . Figure 1 shows if .

Next, we define a number of gadgets as follows. If a pair of literals and satisfies that , we say that is antithetic to . For each literal , , we consider the list of all literals that are antithetic to . Now, to each literal , , we associate a constraint gadget on five vertices defined as follows. Vertex is an artificial one and is coloured with a unique colour. Vertex is the positive one of the gadget, it corresponds to literal being true and has colour . The middle vertex is coloured with black. Vertex is the negative one, it corresponds to being false and has colour . Vertex is the link vertex and represents the relation between and . If , then is coloured with colour , otherwise, with colour . See Figure 2. Finally, in order to obtain path , we concatenate all these gadgets to in a serial manner and we close the path with a final vertex of a unique colour. Clearly, this construction is polynomial as we have gadgets.

We first prove the "if case". Consider an assignment to the variables that satisfies the 3-SAT instance. From this assignment we obtain a rainbow dominating set for as follows:

1. We add and every vertex of a unique colour to .

2. For each true literal , we add the clausal vertex of colour to and for all (if any), we add the positive vertex and the link vertex of to .

3. For each false literal and all (if any), we add the negative vertex of to .

4. If there are vertices with some colour still not present in , we add them to the set.

We can check that each colour is present exactly once in . In fact, this conclusion is straightforward for black colour, for every unique colour and for each colour . For the colour on the link vertex of a gadget , as literals and are antithetic, then exactly one of them is true, so it stands that either is present once in if , or otherwise.

We show now that is a dominating set. Observe first that is dominated by (and also by ). Then, as every clause contains at least one true literal, every clausal vertex is either in or has some neighbour in . Last, every constraint gadget has either its negative vertex or both its positive and link vertex in . In both cases, all vertices of the gadget are covered by . Therefore is a rainbow dominating set as claimed. An example of the construction of and its corresponding rainbow dominating set for an arbitrary assignment to the variables is shown in Figure 3.

We now prove the "only if" case. Given the path constructed as before, let be a rainbow dominating set for . We consider first a partial assignment where for every clausal vertex of colour that is in , we assign the value to the corresponding variable such that is true. Suppose by contradiction that this assignment method leads to some incoherences, that is, some variable ends up being assigned both true and false. It implies that contains two clausal vertices of colour and , respectively, such that and are antithetic. Suppose without loss of generality that . As and are antithetic, there exist two gadgets , , and , , where the link vertices are both coloured (as ). We will show that both of those vertices are in lead to a contradiction. Consider (a similar proof works for ).

Suppose first that . If does not belong to , then must be in as being black, cannot belong to since must already contain either or that are also coloured black. This is a contradiction since the colour of is and already contains the clausal vertex on that colour. Therefore belongs to .

Suppose next that . By the same argument as before, if does not belong to then must be in . As the colour of is , then the vertex of cannot belong to since it has the same colour. Therefore, as must dominate the vertex of , we have that the vertex on colour of belongs to . Following the same argument, we have that the negative vertex belongs to for each , . This is a contradiction since the vertex of has colour and already contains the clausal vertex on that colour. Therefore belongs to .

As the same reasoning holds for as well, contains two vertices of colour . This is a contradiction to the hypothesis that is rainbow. Hence our partial assignment is coherent. We complete the assignment by setting to true every unassigned variable.

Now, it remains to see that the obtained assignment satisfies the 3-SAT instance. Indeed, as every clausal vertex is covered by , then for every clause we have one literal who was assigned to true, that is, the assignment is a solution to the 3-SAT instance. This completes the argument and the proof of the theorem. ∎

## 3 Upper Bounds

We begin with an easy observation which intends to avoid some trivial technical distinctions later on.

###### Proposition 3.1.

If is a vertex-coloured graph with colours on vertices, and , then every rainbow set is a rainbow dominating set of . As a consequence, holds.

###### Proof.

Let be any rainbow set, and any vertex. Then , thus dominates . ∎

Without further reference to this proposition, throughout the text below, we shall disregard whether or not any proof works for graphs of minimum degree at least , because we know the tropical domination number of those graphs exactly.

###### Proposition 3.2.

For any graph , . Furthermore, there are extremal graphs that attain this bound.

###### Proof.

The upper bound follows from the fact that taking a minimal dominating set of and then adding vertices from every colour not present in the dominating set gives a tropical dominating set. To construct extremal graphs, for , consider a cycle of length and add leaves to one vertex of the cycle. Colour the new leaves with unique colours and the rest of the graph with a single colour. Taking and every vertex which its distance to is a multiple of three, is a minimum tropical dominating set that attains the bound. The reader can easily check that . ∎

###### Proposition 3.3.

Given an integer , if and , then . Furthermore, there are graphs with that number of edges and .

###### Proof.

We can check that for , therefore, by the contrapositive of Vizing’s theorem stated in [34], we obtain a minimum dominating set on at most vertices. Then we may add at most other vertices to this dominating set to represent the colours that are absent. Thus obtain a tropical one.

To construct a graph that attains this bound, do the following. Consider a clique on vertices. Pick some vertices and colour them with unique colours. Every other vertex of the graph is coloured with the remaining colour. Let be the set of the remaining vertices. Add a new set of vertices. Now make a bipartite graph between these two sets such that each vertex in the set is adjacent to exactly one vertex of . It is easy to check that this graph has the necessary number of edges. Also, at least vertices are needed to dominate the vertices in . This set can represent only one colour. So we need to add the uniquely coloured vertices to the set to get the required tropical dominating set. ∎

###### Conjecture 3.4.

Let be a connected graph with minimum degree and . Then, .

We are motivated to raise this conjecture by the fact that its particular case for (i.e., simple graphs without colours) holds true. Its proof is a long story, however, taking nearly a half century, along the works by Ore [30] (1962, ), Blank [7] (1973), and independently McCuaig and Shepherd [29] (1989) (), Arnautov [4] (1974, with a stronger upper bound in general), Reed [31] (1996, ), Xing, Sun and Chen [22] (2006, ) and Sohn and Xudong [33] (2009, ).

###### Lemma 3.5.

If is a connected graph with vertices and minimum degree , then , with precisely seven exceptions if .

We first prove Conjecture 3.4 for .

###### Theorem 3.6.

If is a connected graph with , then .

###### Proof.

Let be a connected graph with minimum degree . Let be a subset of vertices of with each of the colours once. Let : has a neighbour in . Clearly as is connected and has . Consider the graph . If this graph is empty, then is a tropical dominating set for of size and we are done.

Now, suppose first that has no isolated vertices. Then, since the domination number of a graph without isolated vercites is at most half the order (cf. [23]), we obtain a dominating set for of size at most . Now, as this is less than or equal to , adding the vertices of to this dominating set we obtain a tropical one for of size at most as desired.

Suppose next that has isolated vertices. Let be the set of isolated vertices of . Let and . As above, we obtain a dominating set for of size . Now, if , then is a tropical dominating set of size . Otherwise, if , let be a vertex such that its colour appears on some vertex in . Then, is a tropical dominating set for as is dominated by a vertex either in or in . The size of this tropical dominating set is and this completes the proof. ∎

The validity of Conjecture 3.4, for , will be a consequence of the following theorem.

###### Theorem 3.7.

For any connected graph with minimum degree , either or holds.

###### Proof.

Given a graph satisfying the conditions, first we pick one vertex from each colour class. If this set dominates all vertices, then . Otherwise, there is a vertex which is undominated, that is ; and further, there is a vertex with . Then, the set is rainbow and dominates at least vertices from . Observe that in this case also and hold.

From now on, we apply the probabilistic method in a similar way as it is done on pages 4–5 of [3] concerning . Let be a subset of chosen at random, such that independently for each ,

 P(u∈D0)=ln(δ+1)δ+1.

The cardinality of is a random variable, which is the sum of the random variables defined as if and if . Therefore, the expected number of selected vertices is

 E(|D0|)=ln(δ+1)δ+1(n−c).

The set dominates all the at least vertices in . Consider the set consisting of those vertices which are not dominated by . For each vertex , we have

 P(u∈D1)≤(1−ln(δ+1)δ+1)δ+1

that implies

 E(|D1|)<1δ+1(n−c−δ).

Clearly, is a tropical dominating set of and is not greater than the expected value of its size. Therefore, we have

 γt(G) < c+ln(δ+1)δ+1(n−c)+1δ+1(n−c−δ) = 1+ln(δ+1)δ+1(n−c+1)+c−1+1δ+1−1+ln(δ+1)δ+1 < 1+ln(δ+1)δ+1(n−c+1)+c−1

which completes the proof. ∎

###### Corollary 3.8.

For any connected graph with minimum degree , either or holds.

###### Proof.

For every , holds. Then, Theorem 3.7 implies

 γt(Gc)<1+ln(δ+1)δ+1(n−c+1)+c−1<δ3δ−1(n−c+1)+c−1,

if . ∎

For we present the following result that is slightly weaker than Conjecture 3.4.

###### Theorem 3.9.

Let be a connected graph with minimum degree such that and . Then, .

###### Proof.

Let be a subset of vertices of with each of the colours once. Let  : . Clearly, if , then by Lemma 3.5 we obtain a dominating set for of size . Thus, adding to this set we obtain a tropical dominating set for of size . Assume therefore that and let . We intend to keep in the tropical dominating set to be constructed, by extending with a subset of that dominates .

First we can assume that for every vertex , . Otherwise, if there is a vertex with a unique neighbour , we can set to make the degree of at least and therefore any dominating set extending that contains is equivalent to one containing instead.

Second, if , then making a complete graph we obtain that for every vertex , . We can therefore apply Lemma 3.5 to obtain a dominating set for of size except for the case when and some components of are one of the seven exceptions (which are also listed in [23]). In this special case, if at least one of the vertices of a component is dominated from outside (as is the case in our problem since every vertex in is dominated by some vertex in ), then each of them satisfies the bound. Therefore we have that is a tropical dominating set for of size .

We can conclude that , and consequently holds. Consider now the following graph obtained from plus a complete graph on new vertices. First make a complete graph. Join every vertex in to every vertex in . Finally, for every vertex set for some vertex . Clearly this new graph has minimum degree therefore by [23] we obtain a dominating set of size at most . Now we obtain a tropical dominating set for the original graph as follows. If any of the vertices of belongs to we just delete them from and add instead the chosen vertex . Then add to the dominating set. This new set is dominating as every vertex in is dominated by some vertex in in and clearly it is tropical. Finally, its size is not greater than and this number is maximum when is as small as possible, that is, . We obtain then that the size of the tropical dominating set is which completes the proof. ∎

###### Proposition 3.10.

Let be super-dense, i.e., . Then .

###### Proof.

Since , then the complement of satisfies . Therefore, the diameter of is at least . Thus according to a theorem from [23], the domination number of is at most . Now, the result for follows from Proposition 3.2. ∎

## 4 Tropical dominating sets in random graphs

In this section we study the tropical domination parameter of a randomly vertex-colored random graph. Recall that the random graph is the graph on vertices where each of the possible edges appears with probability , independently. For more details on random graph theory, we refer the reader to [8] and [24]. Given a positive integer , let be the vertex-colored graph obtained from by coloring each vertex with one of the colors uniformly and independently at random. The choice of colors is independent of the existence of edges. In what follows, we will say that has a property asymptotically almost surely (abbreviated a.a.s.) if the probability it satisfies tends to as tends to infinity. For convenience, we will use the notation .

The domination number of the random graph has been well studied, see for example [16], [28] and [35]. In Particular, Wieland and Godbole [35] proved the following two-point concentration result.

###### Theorem 4.1.

([35])   Let be such that . Set . Then a.a.s. the domination number of the random graph is equal to

 ⌊logbn−logb[(logbn)(logn)]⌋+1   or   ⌊logbn−logb[(logbn)(logn)]⌋+2.

Recently Glebov, Liebenau and Szabó [21] strengthened this two-point concentration result by extending the range of down to .

We are interested here in the maximum number of colors that can be used so that a.a.s. has a tropical minimal dominating set. We only deal with the case when is fixed. It follows from Theorem 4.1 that the number of colors should not exceed . We show in Theorem 4.3 that this upper bound is achieved. This result can be extended to hold when tends to sufficiently slowly. This could be the subject of another study since, in this case, the proof is very technical.

For , let be the random variable counting the number of tropical dominating sets of size in .

 Xc=(nc)∑j=1Ij,

where is the indicator random variable indicating if the j-th -set is both tropical and dominating in .

In order to prove Theorem 4.3, we need the following lemma.

###### Lemma 4.2.

Let be fixed. Set . Denote by the number of tropical dominating sets of size in . Then

 E(Xc)→∞   as   n→∞,

for

###### Proof.

Clearly, for , we have

 E(Ij)=P[Ij=1]=(1−(1−p)c)n−cc!cc,

where is the probability that a given -set is dominating, and is the probability that is tropical. By the linearity of expectation, we have

 E(Xc) = (nc)(1−(1−p)c)n−cc!cc. = (n)ccc(1−(1−p)c)n−c. ≥ (n)ccc(1−(1−p)c)n.

Using the inequality and the estimate , we have

 E(Xc) ≥ (1−o(1))ncccexp[−n(1−p)c1−(1−p)c] = (1−o(1))exp[Ψ(c)],

where

 Ψ(c)=clogn−clogc−n(1−p)c1−(1−p)c.

Recall that . Thus,

 (1−p)c≤(1−p)logbnlogn

and

 n(1−p)c1−(1−p)c≤(1−p)logbnlogn+o(1).

It follows that

 Ψ(c)≥clogn−clogc−(1−p)logbnlogn+o(1).

Straightforward calculations show that

 Ψ(c)≥(p−o(1))logbnlogn.

Since is fixed, we conclude that , and thus also , tends to infinity as . ∎

###### Theorem 4.3.

Let be fixed and set . Let be the function defined by

Then a.a.s. contains a tropical dominating set of size .

###### Proof.

To prove the Theorem, we use the second moment method. For this, we need to estimate the variance of the number of tropical dominating sets of size .

 Var(Xc) = (nc)∑i=1(nc)∑j=1E(IiIj)−E2(Xc) (1) = c∑k=0(nc)(ck)(n−cc−k)E(I1Ik)−E2(Xc), = E(Xc)+c−1∑k=0(nc)(ck)(n−cc−k)E(I1Ik)−E2(Xc),

where is the indicator random variable of any generic -set that intersect the first -set in elements. We have

 E(I1Ik) = P[ A1~{}dominates~{}and~{}Ak~{}dominates~{}]×P[ A1~{}and~{% }Ak~{}are tropical~{}] (2) ≤ P[ A1~{}dominates~{}¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(A1∪Ak) ∧ Ak~{}dominates~{}¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(A1∪Ak) ]×P[ A1~{}and~{}Ak~{}% are tropical~{}] = (1−2(1−p)k+(1−p)2c−k)n−2c+k×c!(c−k)!c2c−k

Relations (1) and (2) imply

 Var(Xc)≤E(Xc)+E2(Xc)A+B, (3)

where

 A=(nc)−1(n−cc)(1−(1−p)c)−2c−1

and

 B=(nc)c−1∑k=1(ck)(n−cc−k)c!(c−k)!c2c−k(1−2(1−p)k+(1−p)2c−k)n−2c+k.

Using once again the inequality , we can bound as follows

 A ≤ e−c2nexp[2c(1−p)c1−(1−p)c]−1 ≤ exp[−c2n+2c(1−p)c(1+o(1))]−1.

Thus, for

 A ≤ (2c(1−p)c−c2n)(1+o(1)) (4) = O((logn)3n).

Now, we need to estimate . We have

 B≤(nc)c−1∑k=1g(k),

where

 g(k)=(c!)2k!(c−k)!nc−kc2c−kexp[(n−2c)((1−p)2c−k−2(1−p)c)].

We shall show that

• there exists such that is decreasing if and increasing if ,

• ,

which will imply that

 c−1∑k=1g(k)≤cg(1). (5)

Clearly,

 g(1)≥g(c−1)

iff

 nc−2cc−2exp[(n−2c)((1−p)2c−1−(1−p)c+1)]≥1

iff

 exp[(c−2)logn−(c−2)logc+(n−2c)n2b3(logbnlogn)2−(n−2c)nb3logbnlogn]≥1

iff

 exp[(1−1b3+o(1))logbnlogn]≥1.

Since the left-hand side of the last inequality tends to infinity as , the above condition is thus satisfied and (ii) is proved.

For , let

 h(k)=g(k+1)g(k)=c(c−k)n(k+1)exp[(n−c)p(1−p)2c−k−1].

It is straightforward to see that, for ,

 h(k)≥1

iff

iff

 (1−p)k−3logn(1+δ(k))≤(n−c)n2p(logbnlogn)2,

where . Therefore if and only if

 k≥logbn−2logb(1−cn)−2logblogbn−logblogn−logbp+logb(1+δ(k))+3.

The right-hand side of the above inequality is of the form , . Thus, there exists such that if and only if . We have thus shown (i). Combining (3), (4) and (5), it follows that

 Var(Xc)E2(Xc)≤1E(Xc)+O((logn)3n)+(nc)cg(1)E2(Xc).

Since, by Lemma 4, as , we will have if the last term in the right-hand side of the above inequality tends to zero as . We have

 (nc)cg(1)E2(Xc) = 1E2(Xc)×(n)cc!nc−1(c−1)!c2c−2exp[(n−2c)((1−p)2c−1−2(1−p)c)] = c3nc−1(n)c(1−(1−p)c)2n−2cexp[(n−2c)((1−p)2c−1−2(1−p)c)] = c