bounds related to domination in graphs with minimum degree two

Bounds Related toDomination in Graphswith Minimum Degree Two

Laura A. Sanchis*DEPARTMENT OF COMPUTER SCIENCE

COLGATE UNIVERSITYHAMILTON, NY 13346

E-mail: [email protected]

Received June 6, 1996

Abstract: A dominating set for a graph G = (V,E) is a subset of vertices V ′ ⊆ V suchthat for all v ∈ V − V ′ there exists some u ∈ V ′ for which {v, u} ∈ E. The dominationnumber of G is the size of its smallest dominating set(s). We show that for almostall connected graphs with minimum degree at least 2 and q edges, the dominationnumber is bounded by (q + 1)/3. From this we derive exact lower bounds for thenumber of edges of a connected graph with minimum degree at least 2 and a givendomination number. We also generalize the bound to k-restricted domination numbers;these measure how many vertices are necessary to dominate a graph if an arbitrary setof k vertices must be included in the dominating set. c© 1997 John Wiley & Sons, Inc. J Graph

Theory 25: 139–152, 1997

1. INTRODUCTION

A dominating set for a graph G = (V,E) is a subset of vertices V ′ ⊆ V such that for allv ∈ V − V ′ there exists some u ∈ V ′ for which {v, u} ∈ E. The domination number of G,denoted by γ(G), is the size of its smallest dominating set(s).

Let n be the number of vertices, q the number of edges, and d the domination number of agraph G. McCuaig and Shepherd show in [1] that, for connected graphs with minimum degreeat least 2, the bound d ≤ 2n/5 holds, except for a finite set of graphs. We show in this paper that

*Supported in part by NSF Grant CCR-9402334.

c© 1997 John Wiley & Sons, Inc. CCC 0364-9024/97/020139-14

140 JOURNAL OF GRAPH THEORY

for almost all connected graphs with minimum degree at least 2, d ≤ (q + 1)/3. This is a betterbound than 2n/5 for very sparse graphs, namely those for which q ≤ 6n/5 − 1.

From this result we then derive tight lower bounds for the number of edges of a connectedgraph with minimum degree at least 2 and a given domination number. These bounds complementprevious results regarding how many edges a graph with a given number of vertices and a givendomination number can have (see [2, 3]).

We also generalize the bound on the domination number to a variant concept of domination,which can be motivated by the following application of the dominating set problem. A networkof distributed computers is modelled as a graph where the vertices represent the processors and acommunication link between two processors is represented by an edge. A set of resources is to beplaced on the network, so that each processor should have to traverse at most one link to obtainone of the resources. Thus the processors to which the resources are assigned must constitute adominating set for the underlying graph, and the minimum number of resources to be assignedcorresponds to the domination number of the graph.

Now suppose that k of the processors in the network already own the specified resources. Theproblem then becomes, how to best choose the additional resource locations so that every site canbe efficiently serviced. In terms of dominating sets, the problem can be stated as follows: givena graph G = (V,E) and a subset U ⊆ V of size k, find a smallest possible dominating set forG which includes all vertices in U . This leads us to define the k-restricted domination numberof a graph G = (V,E) as the smallest number dk such that, given any subset U ⊆ V of size k,there exists a dominating set for G including U and having at most dk vertices. We show thatdk ≤ (q + 2k + 1)/3 for most connected graphs with minimum degree at least 2.

2. DEFINITIONS

If G = (V,E) and U ⊆ V , we let G − U denote the graph obtained by removing from G allvertices inU together with their incident edges. If e ∈ E,G−e is the graph obtained by removingthe edge e from G.

The notation x ≡3 y means x ≡ y (mod 3).

Definition 2.1. Let l, k be nonnegative integers. Def ine F (l, k) = b(l + 2k)/3c.

Observation 2.1. Let l, k be nonnegative integers.(1) F (l, k + 1) = F (l + 2, k).(2) F (l, k) + 1 = F (l + 3, k).(3) F (l1, k1) + F (l2, k2) ≤ F (l1 + l2, k1 + k2).

Definition 2.2. Let G = (V,E) be a graph, and U ⊆ V .

(1) Let γ(G,U) denote the cardinality of a smallest possible dominating set for G whichincludes all vertices in U .

(2) Def ine γk(G) = maxU⊆V,|U |=k γ(G,U).

We call γk(G) the k-restricted domination number of G.

Definition 2.3. Let G = (V,E) be a graph, and U ⊆ V, v ∈ V .

(1) Let γv(G,U) denote the cardinality of a smallest possible set of vertices which dominatesall vertices in G except possibly v, and includes all vertices in U .

DOMINATION IN GRAPHS 141

(2) Def ine γvk(G) = maxU⊆V,|U |=k γ

v(G,U).

We let Cn denote the cycle on n vertices, and Pn denote the path on n vertices. For m ≥3, n ≥ 1, define Lm,n to be the graph formed by joining with an edge any vertex in Cm to anendvertex of Pn. We call such a graph a key on n,m vertices.

Definition 2.4. Let w1 and w2 be the endvertices of Pn. Let 1 ≤ k ≤ n. Def ine γ̃k(Pn) to bethe maximum value of γ(Pn, U) over all subsets U of size k containing w1. For k ≥ 2, def ineγ̂k(Pn) to be the maximum value of γ(Pn, U) over all subsets U of size k containing both w1

and w2.

3. UNRESTRICTED DOMINATION NUMBERS

Theorem 3.1. Let q ≥ n ≥ 3. Let G = (V,E) be a connected graph with minimum degree atleast 2, n vertices, and q edges. If q > n or if q = n and n 6≡3 1, then γ(G) ≤ (q + 1)/3. Ifq = n and n ≡3 1, then γ(G) = (q + 2)/3.

Proof. If q = n and n ≡3 1, G is a cycle on n vertices and the result is clear. The proof forthe other cases is by induction on q, and the base case q = 3 is clear. Assume that q > 3 and Gis not a cycle of n vertices where n ≡3 1. Let W1 be the set of vertices of G whose degree is atleast 3, and let W2 = V −W1.

(1) Suppose that there exists an edge e in G joining two vertices in W1, and such that G− eis connected. If G − e is a cycle and n ≡3 1, then γ(G) ≤ γ(G − e) = (q + 1)/3. Otherwise,using the induction hypothesis, γ(G) ≤ γ(G− e) ≤ (q − 1 + 1)/3.

(2) Suppose that there exists an edge e inG joining two vertices inW1, such thatG−e has twocomponents G1, G2, and neither G1 nor G2 is a cycle of n′ vertices with n′ ≡3 1. Let q1 be thenumber of edges of G1, and q2 the number of edges of G2. Then using the induction hypothesis,

γ(G) ≤ γ(G1) + γ(G2) ≤ (q1 + 1)/3 + (q2 + 1)/3 = (q + 1)/3.

(3) Suppose that there exists an edge e in G joining two vertices x, y ∈ W1, such that G− ehas two components G1, G2, where G1 is a cycle of n1 vertices and G2 is a cycle of n2 vertices,with n1 ≡3 n2 ≡3 1. Then it is possible to find a dominating set D1 for G1 including x, of size(n1 + 2)/3, and a set D2 which dominates all vertices in G2 except for y, of size (n2 − 1)/3.Then D = D1 ∪D2 has size q/3 and dominates G.

(4) If G is a cycle of n vertices where n ≡3 0 or n ≡3 2, then G has a dominating set of sizeat most (q + 1)/3.

(5) Suppose that none of the previous conditions hold, and there exists a path v1, v2, v3, v4, v5

inG such that all five vertices are distinct, v2, v3, v4 ∈ W2, and if v1 and v5 are adjacent, then theyare both in W1. Form the graph G′ by adding (if necessary) the edge {v1, v5} to G−{v2, v3, v4}.Since conditions (1) and (4) do not hold, G′ is not a cycle. So by the induction hypothesis,γ(G′) ≤ (q − 3 + 1)/3. Let D′ be a minimum size dominating set for G′. If both v1, v5 are inD′, or if neither is in D′, let D = D′ ∪ {v3}. If v1 is in D′ but v5 is not, let D = D′ ∪ {v4}; andif v5 is in D′ but v1 is not, let D = D ∪ {v2}. Then D is a dominating set for G satisfying thegiven bound.

Note that if none of conditions (1)–(5) hold, then

(a) Every vertex in W2 is adjacent to a vertex in W1, unless it belongs to a cycle of size 4 or 5for which exactly one vertex is in W1.


FIGURE 1.

(b) No two vertices inW1 are adjacent, unless the edge joining them connects two components,exactly one of which is a cycle of size 4.

(6) If G is one of the graphs shown in Figure 1, then the result follows by observation.(7) Suppose that none of conditions (1)–(6) hold, and that there is a vertex v1 ∈ W1 attached to

one of the configurations formed by vertices v2, v3, · · · as shown in Figure 2. Let U consistof all labelled vertices, except for v1, in any of these configurations. Let G′ = G − U .Since condition (6) does not hold, G′ satisfies the induction hypothesis. The result followsby applying the induction hypothesis to G′ and noting that the subgraph spanned by U canbe dominated by bqU/3c vertices, where qU is the number of edges incident with verticesin U .

If none of conditions (1)–(7) hold, then if a W2 vertex is not adjacent to at least one W1 vertex,or if there are two adjacentW1 vertices, then these must appear in one of the configurations shownin Figure 3. In Figure 3 (a), (b), or (c), let H be the subgraph spanned by all vertices closer to xthan to y, including x itself (i.e., that part of the graph depicted ‘‘below’’x in the picture). Notethat H is not a cycle of size 4 since this case was dealt with in (3) for (a), and in (7) for (b) and (c)


(using the configurations in Figure 2 (a) and (c), respectively). It is therefore possible to definethe set D to consist of the following vertices:

(1) All vertices in W1 except for the vertices labelled y occurring in configurations shown inFigure 3(a).

(2) The vertices labelled z occurring in all configurations of the type shown in Figure 3.

It is not difficult to see that D is a dominating set for G. Let D1, D2 denote the vertices fromD given by each of (1), (2) above. Each vertex in D1 is incident with at least three edges notincident with any other vertex in D1, and there are at least three edges not counted in the previouscase, for each vertex in D2. Thus the size of D is bounded above by q/3.

4. LOWER BOUND FOR NUMBER OF EDGES

We can now specify the lower bounds on the number of edges for connected graphs with minimumdegree at least 2 and having a given domination number.

Theorem 4.1. Let G = (V,E) be a graph with n vertices, q edges, and γ(G) = d. Suppose Gis connected and has minimum degree at least 2.

FIGURE 2.


FIGURE 3.

(1) If d < n/3, then q ≥ d3(n− d)/2e.(2) If n/3 ≤ d ≤ (n + 2)/3, then q ≥ n.(3) If d > (n + 2)/3, then q ≥ 3d− 1.

Each of these bounds is tight for values of n and d for which there exist graphs with the givenproperties.

Proof. Let D be a minimum dominating set for G. Every vertex in V −D is adjacent to atleast one vertex in D, accounting for n− d edges. In addition, since every vertex in V −D hasdegree at least 2, there must be at least d(n− d)/2e additional edges. Thus, since clearly q ≥ n,all graphs with the given properties obey the first two bounds. By Theorem 3.1, the third boundholds for all the specified graphs. To complete the proof, it suffices to show that all bounds aretight for values of d within the specified ranges, by exhibiting graphs attaining the bounds.

For d < n/3, let H be a cycle having 3d vertices, let v be one of the vertices on the cycle, andlet w1, . . . , wn−3d be isolated vertices. If d = (n − 1)/3 (i.e., n − 3d = 1), form the graph Gby adding two edges connecting w1 to two adjacent vertices of H . If d < (n− 1)/3, form G byjoining with an edge each vertex wi to v, for 1 ≤ i ≤ n− 3d, and also joining w2i to w2i−1 for1 ≤ i ≤ b(n− 3d)/2c, and joining wn−3d to w1 if n− 3d is odd. It can be checked that G hasthe required properties.

If n/3 ≤ d ≤ (n + 2)/3, let G be a cycle.Assume that d > (n+ 2)/3. From reference [1] we know that if a graph is connected and has

minimum degree at least 2, then d ≤ 2n/5, unless the graph belongs to the class B defined in[1]. However, none of the graphs in B obeys the bound d > (n + 2)/3. So we may assume thatd ≤ 2n/5. In this case, let k1 = 3d− n and k2 = n− 2d. Note n = 2k1 + 3k2, k2 ≥ k1 ≥ 1,and d = k1 + k2. Let r1, . . . , rk1 be a sequence of integers such that ri ≥ 1 for each i and∑k1

i=1 ri = k2. For 1 ≤ i ≤ k1 let qi = 3ri + 2. For each i, let Hi be a cycle of qi vertices, andlet vi, wi be two vertices of Hi which are at distance 2 from each other. Form G by joining wi

to vi+1 for 1 ≤ i ≤ k1 − 1. Then the number of vertices of G is∑k1

i=1 qi = 2k1 + 3k2 = n,the domination number of G is

∑k1

i=1(ri + 1) = k1 + k2 = d, and the number of edges of G isn + k1 − 1 = 3d− 1, as required.


5. RESTRICTED DOMINATION NUMBERS

In this section we generalize the bound for domination numbers derived in Section 3 to restricteddomination numbers. Note that given the result in Section 3, we can bound the k-restricteddomination number by (q + 3k + 1)/3 (for most cases). We show that this can be improved to(q + 2k + 1)/3.

The idea is to first determine bounds for the restricted domination numbers of simple structuressuch as cycles, paths, and keys, and then analyze other graphs in terms of these components. Alsonote that the results in this section provide a different, albeit longer, proof for Theorem 3.1.

5.1. Cycles, Paths, and Keys

Lemma 5.1. Let n ≥ 3, 0 ≤ k ≤ n.(1) If n ≡3 0, then γk(Cn) = F (n, k).(2) If n 6≡3 0, then γ0(Cn) = F (n, 0) + 1, and γk(Cn) = F (n, k) for k > 0.

Proof. The cases where k = 0 can be checked directly. Assume that k > 0. Let v1, v2, . . . , vkbe distinct vertices on Cn, and assume that they are arranged consecutively around the cycle (butnot necessarily adjacent to each other). Let U = {v1, . . . , vk}. For i = 1, . . . , k − 1, let li bethe number of vertices between vi and vi+1 (not including vi and vi+1). Let lk be the number ofvertices between vk and v1. So we have

∑ki=1 li = n− k.

Any minimum size dominating set for Cn including U must contain exactly d(li − 2)/3e ofthe vertices between vi and vi+1. Hence

γ(Cn, U) = k +k∑

i=1

d(li − 2)/3e = k +k∑

i=1

bli/3c

≤ k + b(n− k)/3c = F (n, k).

So γk(Cn) ≤ F (n, k). For the equality, note that when v1, . . . , vk form a path, thenγ(Cn, U) = F (n, k).

Lemma 5.2. Let n ≥ 1, 0 ≤ k ≤ n. Then γk(Pn) ≤ F (n+2, k). Equality holds for k 6= n−1.

Proof. The case where k = 0 can be checked directly. As in the previous proof, let v1, . . . , vkbe distinct vertices ofPn, arranged consecutively (but not necessarily adjacent to each other) alongthe path. For i = 1, . . . , k − 1, let li be the number of vertices between vi and vi+1. Let l0 bethe number of vertices before v1, and lk the number of vertices after vk. Let U = {v1, . . . , vk}.Then

γ(Pn, U) = k + d(l0 − 1)/3e + d(lk − 1)/3e +k−1∑

i=1

d(li − 2)/3e

= k + b(l0 + 1)/3c + b(lk + 1)/3c +

k−1∑

i=1

bli/3c ≤ k + b(n− k + 2/3)c

= F (n + 2, k).

So γk(Pn) ≤ F (n + 2, k).


For the equality, when k = n, obviously γn(Pn) = n = F (n + 2, n). If k = n − 2, let Uconsist of all vertices in Pn except for the first 2. Then γ(Pn, U) = n − 1 = F (n + 2, n − 2),so γn−2(Pn) = F (n + 2, n− 2).

Assume k ≤ n − 3. Let U consist of all vertices in Pn except for the first 2 and the lastn− k − 2. Then

γ(Pn, U) = k + 1 + d(n− k − 3)/3e = k + 1 + b(n− k − 1)/3c= b(3k + 3 + n− k − 1)/3c = F (n + 2, k).

Lemma 5.3. Let w be one of the endvertices of Pn, 0 ≤ k ≤ n. Then γwk (Pn) = F (n + 1, k).

Proof. The case k = 0 for all n follows from the previous lemma. The case n = k = 1 isclear. The proof that γw

k (Pn) ≤ F (n+1, k) is by induction on n. Assume n > 1 and k > 0. LetU be a subset of vertices ofPn of size k, and letD be a smallest possible set of vertices dominatingall vertices in Pn except possibly w, and including U . If w 6∈ U , then we may without loss ofgenerality assume that w 6∈ D. Hence

|D| ≤ γk(Pn−1) ≤ F (n− 1 + 2, k) = F (n + 1, k).

If w ∈ U , let D1 = D − {w}. D1 dominates all vertices in Pn except possibly for w and itsneighbor, which is the endvertex of a Pn−1 graph. Hence by the induction hypothesis,

|D| = 1 + |D1| ≤ 1 + F (n− 1 + 1, k − 1) = F (n + 3, k − 1) = F (n + 1, k).

So γwk (Pn) ≤ F (n+1, k). To see the equality, let U consist of the first k consecutive vertices

of Pn, starting with and including w.

Corollary 5.4. Let 1 ≤ k ≤ n. Then γ̃k(Pn) = F (n + 1, k).

Lemma 5.5. Let 2 ≤ k ≤ n. Then γ̂k(Pn) = F (n, k).

Proof. The proof is by induction on n. The cases n = 2, 3, and 4 follow by observation.Assume that n > 4. Let w1, w2 be the endvertices of Pn, and v1, v2 be their neighbors in Pn,respectively. Let U be a set of k vertices including w1 and w2. Let U ′ = U − {w1, w2}. Ifneither v1 nor v2 is in U , then by Lemma 5.2

γ(Pn, U) ≤ 2 + γ(Pn−4, U′) ≤ 2 + F (n− 4 + 2, k − 2) = F (n + 4, k − 2) = F (n, k).

If exactly one of v1, v2 is in U , then by Corollary 5.4

γ(Pn, U) ≤ 2 + γ(Pn−3, U′) ≤ 2 + F (n− 3 + 1, k − 2) = F (n, k).

If both v1 and v2 are in U , then by induction,

γ(Pn, U) ≤ 2 + γ(Pn−2, U′) ≤ 2 + F (n− 2, k − 2) = F (n, k).

For the equality, let U consist of the first k − 1 consecutive vertices of Pn, starting with andincluding w1, and also w2.

Lemma 5.6. γk(Lm,n) ≤ F (m + n + 2, k).

Proof. It suffices to prove that for any subsetU consisting of k vertices ofLm,n, there exists adominating set D for Lm,n containing all vertices in U , and having size at most F (m+n+2, k).


Let U1 contain the vertices from U in Cm, and let U2 = U − U1. Let k1 = |U1|, k2 = |U2|. Letw1, w2 be the vertices in Cm and Pn, respectively, which are adjacent to each other.

Case m ≡3 0. Let D1 be a smallest possible dominating set for Cm containing all verticesin U1. Let D2 be a smallest possible dominating set for Pn containing all vertices in U2. So|D1| ≤ γk1

(Cm) ≤ F (m, k1) and |D2| ≤ γk2(Pn) ≤ F (n + 2, k2).

Let D = D1 ∪D2. D is a dominating set for Lm,n containing all vertices in U , and

|D| ≤ F (m, k1) + F (n + 2, k2) ≤ F (m + n + 2, k).

Case m ≡3 1. If k1 > 0, the argument is the same as above. Suppose that k1 = 0, sok2 = k. Let U ′ = U ∪{w2}. It is not hard to see that there exists a set D1 of size F (m, 0) whichdominates all vertices in Cm except for w1. Let D2 be a smallest possible dominating set for Pn

which includes all vertices in U ′. So |D2| ≤ γ̃k+1(Pn) ≤ F (n+ 1, k + 1). Then D = D1 ∪D2

dominates Lm,n and

|D| ≤ F (m, 0) + F (n + 1, k + 1) = bm/3c + b(n + 1 + 2k + 2)/3c= (m− 1)/3 + b(n + 3 + 2k)/3c= b(m + n + 2 + 2k)/3c= F (m + n + 2, k).

Case m ≡3 2. If k1 > 0, the argument is the same as above. Suppose that k1 = 0, so k2 = k.Let D1 be a smallest possible dominating set for Cm which includes w1. So |D1| ≤ F (m, 1). LetD2 be a smallest possible set which dominates all vertices in Pn except possibly w2 and whichincludes all vertices in U . So |D2| ≤ γw2

k (Pn) ≤ F (n + 1, k). Then D = D1 ∪D2 dominatesLm,n and

|D| ≤ F (m, 1) + F (n + 1, k) = b(m + 2)/3c + b(n + 1 + 2k)/3c= (m + 1)/3 + b(n + 1 + 2k)/3c = F (m + n + 2, k).

Lemma 5.7. Let m ≥ 3, n ≥ 1, and 0 ≤ k ≤ m + n. Let w be the endvertex of Lm,n. Thenγwk (Lm,n) ≤ F (m + n + 1, k).

Proof. The proof is by induction on n. Let n = 1. If k = 0, then clearly γwk (Lm,n) =

γk(Cm). If m ≡3 0, then this quantity equals F (m, 0) ≤ F (m + n + 1, 0). If m 6≡3 0, then

γk(Cm) = F (m, 0) + 1 = F (m− 1, 0) + 1 = F (m + 2, 0) = F (m + n + 1, 0).

For k > 0, let U be a subset of vertices of Lm,n of size k. If w 6∈ U , then as before

γw(Lm,n, U) = γ(Cm, U) ≤ γk(Cm) = F (m, k) ≤ F (m + n + 1, k).

If w ∈ U and k > 1, then

γw(Lm,n, U) ≤ 1 + F (m, k − 1) = F (m + 3, k − 1) = F (m + 1, k) ≤ F (m + n + 1, k).

The same argument works if w ∈ U, k = 1, and m ≡3 0. Suppose w ∈ U, k = 1, and m ≡3 1.Then there exists a set of vertices of Cm which dominates all the vertices of Cm except for thevertex adjacent to w, and which has size F (m, 0). Hence

γw(Lm,n, U) ≤ 1 + F (m, 0) ≤ F (m + n + 1, 1).


If w ∈ U, k = 1, and m ≡3 2, then

γw(Lm,n, U) ≤ F (m, 0) + 1 + 1 = F (m− 2, 0) + 2

= F (m + 4, 0) = F (m + 2, 1) = F (m + n + 1, 1).

So we have γwk (Lm,n) ≤ F (m + n + 1, k) when n = 1.

For the inductive step, assume that n > 1, and again let U be a subset of vertices of Lm,n ofsize k. If w 6∈ U , then

γwk (Lm,n, U) ≤ γk(Lm,n−1) ≤ F (m + n− 1 + 2, k) = F (m + n + 1, k).

If w ∈ U , let D be a smallest possible dominating set for Lm,n which includes all vertices in U ,and let D′ = D − {w}. So by the induction hypothesis,

|D| = 1 + |D′| ≤ 1 + γwk−1(Lm,n−1) ≤ 1 + F (m + n, k − 1) = F (m + n + 1, k).

5.2. Putting them Together

We now consider graphs constructed by putting together copies of the simple graphs consideredin the previous section. In order to do this we investigate the effect of adjoining a cycle, path, orkey, to another arbitrary graph.

Lemma 5.8. Let m ≥ 3, n ≥ 1, p ≥ 1, i ≥ 0. Let H be a graph of p vertices, for whichγk(H) ≤ F (p + i, k) for 0 ≤ k ≤ p. Let G be a graph formed by joining with an edge theendvertex ofLm,n to any vertex ofH . Thenγk(G) ≤ F (m+n+p+i+1, k) for0 ≤ k ≤ m+n+p.

Proof. LetU be a set of k vertices ofG. LetU1 be the subset ofU found in theLm,n subgraph,and U2 = U − U1. Let k1 = |U1|. Let w1 be the endvertex of Lm,n and w2 its neighbor in H .

A dominating set for G including all vertices in U can be formed in (at least) two ways. Oneway consists of taking the union of a dominating set for Lm,n containing all vertices in U1, anda dominating set for H containing all vertices in U2. Alternatively, one may take the union of aset of vertices dominating all vertices in Lm,n except possibly for w1, and containing all verticesin U1, together with a dominating set for H which includes all vertices in U2 as well as w2. Wethus have

γk(G) ≤ max0≤k1≤k

(min(γk1(Lm,n) + γk−k1

(H), γw1

k1(Lm,n) + γk−k1+1(H))

≤ max0≤k1≤k

(min(F (m + n + 2, k1) + F (p + i, k − k1),

F (m + n + 1, k1) + F (p + i, k − k1 + 1))).

We have

F (m + n + 2, k1) + F (p + i, k − k1) = b(m + n + 2 + 2k1)/3c + b(p + i + 2k − 2k1)/3c.If m + n + 2 + 2k1 6≡3 0, then this quantity is

b(m + n + 1 + 2k1)/3c + b(p + i + 2k − 2k1)/3c≤ b(m + n + p + i + 1 + 2k)/3c = F (m + n + p + i + 1, k).

If p + i + 2k − 2k1 6≡3 0, a similar argument can be made.


So suppose that m + n + 2 + 2k1 ≡3 p + i + 2k − 2k1 ≡3 0. We have

F (m + n + 1, k1) + F (p + i, k − k1 + 1)

= b(m + n + 1 + 2k1)/3c + b(p + i + 2k − 2k1 + 2)/3c.By the assumption, m + n + 1 + 2k1 ≡3 2, and p + i + 2k − 2k1 + 2 ≡3 2. Hence

F (m + n + 1, k1) + F (p + i, k − k1 + 1)

= b(m + n + 2k1 − 1)/3c + b(p + i + 2k − 2k1)/3c≤ b(m + n + p + i− 1 + 2k)/3c ≤ F (m + n + p + i + 1, k).

This completes the proof.

Lemma 5.9. Let m ≥ 3, n ≥ 1, i ≥ 0.Let H be a graph of n vertices, for which γk(H) ≤F (n+ i, k) for 0 ≤ k ≤ n. Let G be a graph formed by joining with an edge an endvertex of Pm

to any vertex of H . Then γk(G) ≤ F (m + n + i + 1, k) for 0 ≤ k ≤ m + n.

Proof. The proof is similar to that of the previous lemma.

Corollary 5.10. Let m ≥ 3, n ≥ 1, i ≥ 0. Let H be a graph of n vertices, for which γk(H) ≤F (n+ i, k) for 0 ≤ k ≤ n. Let G be a graph formed by joining the two endvertices of Pm to anytwo vertices of H (the two vertices need not be distinct). Then γk(G) ≤ F (m + n + i + 1, k)0 ≤ k ≤ m + n.

Lemma 5.11. Let m ≥ 3, n ≥ 1, i ≥ 1. Let H be a graph of n vertices, for which γk(H) ≤F (n + i, k) for 0 ≤ k ≤ n. Let G be a graph formed by joining with an edge any vertex of Cm

to any vertex of H . Then γk(G) ≤ F (m + n + i + 1, k) for 0 ≤ k ≤ m + n.

Proof. Let U be a set of k vertices of G. Let U1 be the subset of U found in the Cm subgraph,and U2 = U −U1. Let k1 = |U1|. Let w1, w2 be the adjacent vertices in Cm and H , respectively.

Case m ≡3 0. We have

γ(G,U) ≤ F (m, k1) + F (n + i, k − k1) ≤ F (m + n + i, k).

Case m ≡3 1. If k1 > 0, the argument is as above. If k1 = 0, a dominating set for Gcontaining all vertices in U can be formed by taking the union of a set of vertices dominating allvertices in Cm except for w1, and a dominating set for H containing U as well as w2. So

γ(G,U) ≤ F (m, 0) + F (n + i, k + 1)

= F (m− 1, 0) + F (n + i, k + 1)

≤ F (m + n + i + 1, k).

Case m ≡3 2. If k1 > 0, the argument is as above. If k1 = 0,

γ(G,U) ≤ F (m, 0) + 1 + F (n + i, k)

= F (m− 2, 0) + 1 + F (n + i, k)

≤ F (m + n + i + 1, k).

We need additional results for graphs formed by joining together just two of these simplegraphs.


Lemma 5.12. Let m,n ≥ 3. Let G be a graph formed by joining a vertex of Cm to a vertex ofCn. Then γk(G) ≤ F (m + n + 2, k) for 0 ≤ k ≤ m + n.

Proof. Let U be a set of k vertices of G. Let U1 be the subset of U found in Cm, andU2 = U − U1. Let k1 = |U1|, k2 = |U2|. Let w1, w2 be the adjacent vertices in Cm and Cn,respectively. We must show that γ(G,U) ≤ F (m + n + 2, k). If k1 > 0 or k2 > 0, then theresult follows by arguments similar to those used in the proof of the previous lemma. Likewiseif m ≡3 0 or n ≡3 0. So assume k = 0,m 6≡3 0, and n 6≡3 0.

Suppose that m ≡3 1. A dominating set for G can be formed from a set D1 which dominatesall vertices in Cm except possibly for w1, and a set D2 which dominates Cn and includes w2.Thus

γ(G,U) ≤ F (m, 0) + F (n, 1) ≤ F (m + n + 2, 0).

Suppose that m ≡3 2 and n ≡3 2. Then

γ(G,U) ≤ F (m, 0) + 1 + F (n, 0) + 1

= F (m− 2, 0) + F (n− 2, 0) + 2

= F (m + n + 2, 0).

By symmetry, this takes care of all cases.

Lemma 5.13. Let m,n ≥ 3, p ≥ 1. Let G be a graph formed by joining a vertexof Cm to an endvertex of Pp, and a vertex of Cn to the other endvertex of Pp. Then γk(G) ≤F (m + n + p + 2, k) for 0 ≤ k ≤ m + n + p.

Proof. Let U be a set of k vertices of G. Let U1 be the subset of U found in Cm, U2 thesubset of U found in Cn, and U3 = U − U1 − U2. Let k1 = |U1|, k2 = |U2|, k3 = |U3|. Letw1, w2 be the endvertices of Pp, v1 the neighbor of w1 in Cm, and v2 the neighbor of w2 in Cn.We must show that γ(G,U) ≤ F (m+n+ p+2, k). If k1 > 0 or k2 > 0, then the result followsby arguments similar to those used in the proof of Lemma 5.8. Likewise if m ≡3 0 or n ≡3 0.So assume k = k3,m 6≡3 0, and n 6≡3 0.

Suppose that m ≡3 n ≡3 1. Construct a dominating set for G by taking the union of a setwhich dominates all vertices in Cm except for v1, a set that dominates all vertices in Cn exceptfor v2, and a dominating set for Pp which includes U,w1, and w2. Then we have by Lemma 5.5

γ(G,U) ≤ F (m, 0) + F (n, 0) + F (p, k + 2)

= F (m− 1, 0) + F (n− 1, 0) + F (p + 4, k)

≤ F (m + n + p + 2, k).

Suppose that m ≡3 1 and n ≡3 n ≡3 2. Let D1 be a minimum size set of vertices dominatingall vertices in Cm except for v1. Let D2 be a minimum size set dominating all vertices in Cn andincluding v2. If w2 6∈ U , construct a dominating set for G by taking the union of D1, D2, and aset that dominates all vertices in Pp except possibly for w2, and which includes U and w1. Thenwe have using Corollary 5.4

γ(G,U) ≤ F (m, 0) + F (n, 0) + 1 + F (p− 1 + 1, k + 1)

= F (m− 1, 0) + F (n− 2, 0) + 1 + F (p + 2, k)

≤ F (m + n + p + 2, k).


If w2 ∈ U , construct a dominating set for G by taking the union of D1, D2, and a set thatdominates all vertices in Pp and includes w1 and w2. Then we have by Lemma 5.5

γ(G,U) ≤ F (m, 0) + F (n, 0) + 1 + F (p, k + 1)

= F (m− 1, 0) + F (n− 2, 0) + 1 + F (p + 2, k)

≤ F (m + n + p + 2, k).

Suppose that m ≡3 n ≡3 2. Let D1 be a minimum dominating set for Cm which includes v1,and let D2 be a minimum dominating set for Cn which includes v2. If both w1 and w2 belongto U , let D3 be a minimum dominating set for Pp including U . So |D3| ≤ F (p, k) by Lemma5.5. If w1 ∈ U , but w2 6∈ U , let D3 be a minimum size set dominating all vertices in Pp exceptpossibly w2. So |D3| ≤ F (p − 1 + 1, k) = F (p, k) by Corollary 5.4. If neither w1 nor w2 isin U , let D3 be a minimum set dominating all vertices in Pp except possibly for w1 and w2. So|D3| ≤ F (p− 2 + 2, k) = F (p, k) by Lemma 5.2. So in every case we have

γ(G,U) ≤ F (m− 2, 0) + 1 + F (n− 2, 0) + 1 + F (p, k) ≤ F (m + n + p + 2, k).

Lemma 5.14. Let m ≥ 3, n ≥ 1. Let G be a graph formed by joining each endvertex of Pn toa vertex of Cm. The vertices in Cm adjacent to the endvertices of Pn need not be distinct. Thenγk(G) ≤ F (m + n + 2, k) for 0 ≤ k ≤ m + n.

Proof. Let H be the graph formed by removing from G one of the edges joining an endvertexof Pn to a vertex of Cm. Note that H is a Lm,n graph. Since γk(G) ≤ γk(H), the result followsfrom Lemma 5.6.

5.3. Main Result

We now derive an upper bound for the k-restricted domination number of any graph which isconnected and has minimum degree at least 2.

Theorem 5.15. Let n ≥ 3, q ≥ n, 0 ≤ k ≤ n. Let G = (V,E) be a connected graph withminimum degree at least 2, n vertices, and q edges. Then if k = 0, n ≡3 1, and q = n, thenγk(G) ≤ F (q, k) + 1. In all other cases, γk(G) ≤ F (q + 1, k).

Proof. The proof is by induction on q−n. If q = n, then G is a cycle, and the result followsfrom Lemma 5.1. If q = n+1, then there must be exactly one vertex of degree 4, or two verticesof degree 3, with all other vertices having degree 2. So the graph is of the form considered in oneof the Lemmas 5.12, 5.13, or 5.14, and the result follows.

For the inductive step, suppose q ≥ n+2. Let V ′ ⊆ V consist of all vertices of degree greaterthan 2. We will consider the consequences of the following two conditions:

(1) There exists a vertex v ∈ V ′ for which there is a cycle vw1 · · ·wpv where p ≥ 2 andw1, . . . , wp are distinct vertices of degree 2.

(2) There exist two vertices v1, v2 ∈ V ′ and two paths v1x1 · · ·xrv2, v1y1 · · · ysv2 wherer, s ≥ 0, at least one of r, s is positive, and x1, · · ·xr, y1, · · · ys are all distinct vertices ofdegree 2.

Suppose that condition (1) holds. If deg(v) ≥ 4, let G1 = G − {w1, . . . , wp}. Then G1

satisfies the conditions of the theorem, and by the induction hypothesis, γk(G1) ≤ F (q − p −


1 + 1, k) = F (q − p, k) for 0 ≤ k ≤ n − p. So by Corollary 5.10, γk(G) ≤ F (q − p + p +1, k) = F (q + 1, k). On the other hand, if deg(v) = 3, then let z be the vertex in V ′ whichis closest to v. If v and z are adjacent, the result follows in a similar way by Lemma 5.11,using the induction hypothesis on G − {v, w1, . . . , wp}. Otherwise there is a path vz1 · · · ztz,where z1, · · · zt all have degree 2, and again the result follows by the induction hypothesis onG− {v, w1, . . . , vp, z1, . . . , zt}, using Lemma 5.8.

Suppose that condition (2) holds, and suppose without loss of generality that r > 0. Then theresult follows by the induction hypothesis applied to G− {x1, . . . , xr}, using Corollary 5.10.

So we may assume that neither condition (1) nor condition (2) holds. Form the graph G′ =(V ′, E′) as follows. For each pair v, w ∈ V ′, if there is a path in G from v to w all of whoseinternal vertices have degree 2, then v, w are joined by an edge in G′. By our assumption, G′

has no loops and each vertex in G′ has degree at least 3. Also, since G is connected, G′ isconnected. Suppose that removing any edge from G′ disconnects G′. Then G′ must be a tree,which is impossible since the minimum degree of G′ is at least 3. It follows that there existdistinct vertices v1, v2 ∈ V ′ such that removing the edge e = {v1, v2} does not disconnectG′. It follows that either v1, v2 are adjacent in G, or there is a path v1w1 · · ·wpv2 in G, wherew1, . . . , wp have degree 2. In the first case, let G1 = G − e, and let p = 0. In the second case,let G1 = G − {w1, . . . , wp}. In either case, G1 is connected and has minimum degree at least2. So we may again apply the induction hypothesis to G1: γk(G1) ≤ F (q − p − 1 + 1, k) =F (q− p, k). If p = 0, γk(G) ≤ γk(G1) ≤ F (q, k) ≤ F (q + 1, k). In the second case, the resultfollows from Lemma 5.10.

ACKNOWLEDGMENTS

The author is grateful to an anonymous reviewer for suggesting the existence of the shorter proofin Section 3, and other helpful comments.

References

[1] W. McCuaig and B. Shepherd, Domination in graphs with minimum degree two, J. Graph Theory13(6) (1989), 749–762.

[2] L. A. Sanchis, Maximum number of edges in connected graphs with a given domination number,Discrete Math. 87 (1991), 65–72.

[3] V. G. Vizing, A bound on the external stability number of a graph, Docl. akad. Nauk. 164 (1965),729–731.

bounds related to domination in graphs with minimum degree two

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