vertices contained in every minimum dominating set of a tree

Vertices Contained inEvery MinimumDominating Set of a Tree

C. M. MynhardtDEPARTMENT OF MATHEMATICS

UNIVERSITY OF SOUTH AFRICAP.O. BOX 392

PRETORIA0003 SOUTH AFRICA

[email protected]

Received November 26, 1997; revised November 28, 1998

Abstract: In this article we begin the study of the vertex subsets of a graphG which consist of the vertices contained in all, or in no, respectively, minimumdominating sets ofG. We characterize these sets for trees, and also obtain resultson the vertices contained in all minimum independent dominating sets of trees.c© 1999 John Wiley & Sons, Inc. J Graph Theory 31: 163–177, 1999

Keywords: domination, dominating set, tree

1. INTRODUCTION

We generally follow the definitions and notation of [5]. It is well known that thedomination number γ(G) is bounded above by the independent domination numberi(G) for all graphs G. If X is a minimum (minimum independent, respectively)dominating set of a graph G, we also refer to X as a γ-set (i-set, respectively) ofG. Define the vertex subsets A(G),Ai(G),N (G), and Ni(G) of the graph G by

A(G) = {v ∈ V (G): v is in every γ-set of G}

c© 1999 John Wiley & Sons, Inc. CCC 0364-9024/99/030163-15

164 JOURNAL OF GRAPH THEORY

Ai(G) = {v ∈ V (G): v is in every i-set of G}

N (G) = {v ∈ V (G): v is in no γ-set of G}

Ni(G) = {v ∈ V (G): v is in no i-set of G}.For an arbitrary graph G, neither A(G) and Ai(G), nor N (G) and Ni(G), are

necessarily related by inclusion. For example, if u and v are central vertices ofK1,mand K1,n, respectively, where 1 < m < n, and T is the tree obtained by joiningu and v, then A(T ) = {u, v} while Ai(T ) consists of v together with the leavesadjacent to u. N (T ) consists of the leaves of T , and Ni(T ) consists of u togetherwith the leaves adjacent to v. However, if γ(G) = i(G), then clearly every i-set ofG is a γ-set and, therefore,A(G) ⊆ Ai(G) andN (G) ⊆ Ni(G). For some graphs,for example all cycles, all four of these sets are empty. In this article we begin thestudy of these sets. We characterize A(T ) and N (T ), if T is a tree, and obtain acharacterization of Ai(T ) for trees T with γ(T ) = i(T ). Of course, if a graph Ghas a unique γ-setD, thenA(G) = D andN (G) = V (G)−D. Such graphs werestudied in [4], for example. The corresponding problem for independent sets wasconsidered in [7].

The results obtained here are used in [2] to characterize trees with γ = i — theso-called (γ, i)-trees. Other classes of graphs for which γ = i have been found in,for example, [1, 3, 6].

For any graph G = (V,E) with v ∈ V and X ⊆ V , we define the followingsets:

N(v) = {u ∈ V : uv ∈ E}

N [v] = N(v) ∪ {v}

N(X) =⋃v∈X

N(v)

N [X] =⋃v∈X

N [v]

and, if v ∈ X ,

pn(v,X) = N [v]−N [X − {v}].Note that if u ∈ pn(v,X), then either u = v and u is not adjacent to any vertex inX , or u 6∈ X and u is adjacent to v but not to any other vertex in X . We also callu an X-private neighbor, abbreviated X-pn, of v. Note that a dominating set X isminimal if and only if pn(v,X) 6= ∅ for each v ∈ X . If X dominates Y ⊆ V , wealso write X � Y , or X � G if Y = V , or X � y if Y = {y}.

We now turn our attention to trees. For ease of presentation, we mostly considerrooted trees. Strictly speaking, a rooted tree T is a directed tree in which there existsa vertex r with the property that there is a directed path in T from r to every othervertex of T . The vertex r is unique with respect to the above-mentioned property

VERTICES IN MINIMUM DOMINATING SETS 165

and is called the root of T . (Thus, if T is rooted at r, then all edges of T are directedaway from r.) For a vertex v of a rooted tree T , the parent p(v) of v is the uniquevertex such that there is a directed edge from p(v) to v, a child of v is a vertex usuch that p(u) = v, and a descendant of v is a vertex u such that there is a directedv − u path in T .

When we consider a rooted tree, we will assume its edges to be directed asexplained above, but will not mention this explicitly; we will also refer to the‘‘edges’’ of the tree, not ‘‘arcs’’ or ‘‘directed edges’’. We define the followingnotation:

C(v) = {u ∈ V : u is a child of v}c(v) = u if C(v) = {u}

D(v) = {u ∈ V : u is a descendant of v}D[v] = D(v) ∪ {v}.

If confusion is possible (for example, when two or more subtrees of a tree areconsidered), we also write CT (v), DT (v) and DT [v]. The subtree of T induced byD[v] is denoted by Tv; note that if T is rooted at v then T = Tv.

A vertex of T is said to be remote if it is adjacent to a leaf (i.e., an end-vertex),and to be a branch vertex if it has degree at least 3. The set of branch vertices ofT is denoted by B(T ) and the set of leaves by L(T ). A path P in T is said to bea v − L path, if P joins v to a leaf of T . Denote the length of P by l(P ), and forj = 0, 1, 2, define

Cj(v) = {u ∈ C(v): Tu contains a u− L path P with l(P ) ≡ j (mod 3)}.We sometimes write CjT (v) to emphasise the tree (or subtree) concerned.

2. PRELIMINARY RESULTS

We begin with some elementary observations about dominating sets of the path Pn.

Observation 1.

(a) γ(Pn) = i(Pn) = dn/3e.(b) Pn has a unique γ-set X if and only if n ≡ 0 (mod 3). In this case, neither

leaf of Pn is in X, and X is independent.

(c) If n ≥ 4 and n ≡ 1 (mod 3), then Pn has an independent γ-set that containsboth its leaves, and an independent γ-set that contains exactly one leaf. Eachγ-set that contains a leaf is independent.

(d) If n ≥ 5 and n ≡ 2 (mod 3), then Pn has an independent γ-set that containsneither leaf ofPn, an independent γ-set that contains either leaf of Pn, but noγ-set that contains both its leaves. Moreover, each γ-set ofPn is independent.


Our next purpose is to characterize A(T ),Ai(T ),N (T ), andNi(T ) in the casewhere |B(T )| ≤ 1. We first determine γ(T ) and i(T ) for such a tree T .

Proposition 1. Let T be a tree rooted at v such that deg u ≤ 2 for all u ∈V (T )− {v} 6= ∅. Then

γ(T ) = i(T ) =∑

w∈C(v)

γ(Tw)− α(v),

where

α(v) =

|C0(v)| − 1 if C0(v) 6= ∅0 if C0(v) = ∅, C1(v) 6= ∅−1 if C0(v) = C1(v) = ∅.

Proof. Since Tw is a path for each w ∈ C(v), γ(Tw) is easy to determine, andCj(v) ∩ Ck(v) = ∅ for j 6= k.

For eachw ∈ C(v) we denote c(w) byw′ and define an independent dominatingset Yw of Tw or Tw′ as described later, and then consider an independent set Y ′defined by

Y ′ =⋃

w∈C(v)

Yw.

We consider three cases.

Case 1. C0(v) 6= ∅.If w is a leaf, then define Yw = ∅. If the nonleaf w ∈ C0(v), note that Tw′ ∼= Pn,where n ≡ 0 (mod 3) and γ(Tw) = γ(Tw′) + 1. Let Yw be the unique γ-set of Tw′and note that Yw is independent and w′ 6∈ Yw, i.e., Yw 6� w. If w ∈ C1(v), thenTw ∼= Pn, where n ≡ 2 (mod 3). Let Yw be an independent γ-set of Tw such thatw 6∈ Yw — note that Yw exists by Observation 1(d). If w ∈ C2(v), then Tw ∼= Pnwith n ≡ 0 (mod 3). Let Yw be the unique (independent) γ-set of Tw, and note thatw 6∈ Yw. It is obvious that N [v] ∩ Y ′ = ∅, Y ′ dominates T − (C0(v) ∪ {v}), isindependent, and

|Y ′| =∑

w∈C(v)

γ(Tw)− |C0(v)|.

Therefore Y = Y ′ ∪ {v} is an independent dominating set of T with Σγ(Tw)− α(v) vertices, as required.

Suppose that γ(T ) < |Y | and let X be a γ-set of T . If v 6∈ X , then Xw =X ∩ V (Tw) � Tw for each w ∈ C(v). Hence |Xw| ≥ γ(Tw) and

|X| ≥∑

w∈C(v)

γ(Tw) ≥ |Y |,

a contradiction. If v ∈ X , then Xw � Tw′ for each w ∈ C(v), and so |Xw| ≥γ(Tw′). If w ∈ C1(v) ∪ C2(v), then Tw′ ' Pn with n ≡ 1 or 2 (mod 3) and thus


γ(Tw′) = γ(Tw). But then

|X| ≥ 1 +∑

w∈C0(v)

γ(Tw′) +∑

w∈C1(v)∪C2(v)

γ(Tw)

= 1 +∑

w∈C0(v)

[γ(Tw)− 1] +∑

w∈C1(v)∪C2(v)

γ(Tw)

= |Y |,a contradiction. Hence Y is an independent γ-set of T .

Case 2. C0(v) = ∅ and C1(v) 6= ∅.If w ∈ C1(v), let Yw be an independent γ-set of Tw such that w ∈ Yw — notethat Yw exists by Observation 1(d). If w ∈ C2(v), define Yw as in Case 1. SinceC0(v) = ∅, Y ′ � N(v), and since C1(v) 6= ∅, Y ′ � v. Thus Y = Y ′ is anindependent dominating set ofT with the required cardinality, and it follows similarto Case 1 that Y is a γ-set of T .

Case 3. C0(v) = C1(v) = ∅.For each w ∈ C2(v), define Yw as in Case 1 and note that Y ′ � T − v. ThusY = Y ′ ∪ {v} is an independent dominating set of T with the required cardinality.As in Case 1, Y is a γ-set of T .

Proposition 1 shows that if T is a tree with at most one branch vertex, thenγ(T ) = i(T ). We now show that A(T ) = Ai(T ) and N (T ) = Ni(T ) in thiscase also.

Theorem 1. Let T be a tree rooted at v such that deg u ≤ 2 for all u ∈V (T )− {v}. Then

(a) v ∈ A(T ) if and only if v ∈ Ai(T ) if and only if |C0(v)| ≥ 2;(b) v ∈ N (T ) if and only if v ∈ Ni(T ) if and only if C0(v) = ∅ and C1(v) 6= ∅;(c) each γ-set is an i-set if and only if C0(v) 6= ∅ and C1(v) = ∅, or C0(v) = ∅

and C1(v) 6= ∅.

Proof. (a) Suppose that |C0(v)| ≤ 1 and let Y be the independent γ-set of Tconstructed in the proof of Proposition 1. Recall that if C0(v) = ∅ and C1(v) 6= ∅,then v 6∈ Y . If C0(v) = ∅ = C1(v), then we have v ∈ Y and pn(v, Y ) = {v}, andhence for any fixed w ∈ C(v), if Y ′w = {c(u) : u ∈ Yw}, then Y ′ = (Y − (Yw ∪{v})) ∪ (Y ′w ∪ {w}) is an independent γ-set of T with v 6∈ Y ′. If C0(v) = {w},then pn(v, Y ) = {v, w} and hence Y ′ = (Y −{v})∪{w} is an independent γ-setof T with v 6∈ Y ′.

Conversely, suppose that |C0(v)| ≥ 2 and let X be any γ-set of T with v 6∈ X .Then, for eachw ∈ C(v), X∩D[w] � Tw and thus |X| ≥∑w∈C(v) γ(Tw) > γ(T )by Proposition 1.(b) If C0(v) 6= ∅ or C1(v) = ∅, then v ∈ Y , where Y is the independent γ-setconstructed in the proof of Proposition 1. Thus v 6∈ N (T ) ∪ Ni(T ). Conversely,


suppose that C0(v) = ∅ and C1(v) 6= ∅ and let X be any γ-set of T with v ∈ X .For each w ∈ C(v), X − {v} dominates D(w) and hence |X ∩D[w]| ≥ γ(Tw′),where w′ = c(w). But if w ∈ C1(v) ∪ C2(v), then |D(w)| ≡ 1 or 2 (mod 3) andhence γ(Tw′) = γ(Tw). Thus by Proposition 1,

|X| ≥ 1 +∑

w∈C(v)

γ(Tw) > γ(T ).

(c) If C0(v) = ∅ and C1(v) 6= ∅, then by (b), v ∈ N (T ) = Ni(t). Thus for anyγ-set X of T,X ∩ D[w] dominates Tw for each w ∈ C(v) and has cardinalityγ(Tw) (by Proposition 1). It follows immediately from Observation 1(b) and (d)that X is independent. Suppose that C0(v) 6= ∅ and C1(v) = ∅. Let X be a γ-setof T .

If v ∈ X , then for each w ∈ C(v) and w′ = c(w), X ∩ D[w] � Tw′ . But if(firstly) w ∈ C2(v), then |D[w]| ≡ 0 (mod 3) and γ(Tw) = γ(Tw′). Moreover, byObservation 1(b), X ∩D[w] is independent and does not contain w. If (secondly)w ∈ C0(v), then |D(w)| ≡ 0 (mod 3) and, by Proposition 1, |X ∩ D[w]| =γ(Tw′) = γ(Tw)− 1. By Observation 1(b), X ∩D[w] is independent and does notcontain w (or w′). Thus X is an i-set of T .

If v 6∈ X , then by (a), C0(v) = {w} for some w ∈ C(v) and X ∩D[u] � Tufor each u ∈ C(v). By Proposition 1 and the fact that no vertex in Tu dominatesany vertex of Tu′ , u 6= u′, it follows that |X ∩D[u]| = γ(Tu) for each u ∈ C(v).For u 6= w, Tu ∼= Pn with n ≡ 0 (mod 3) and, by Observation 1(b), X ∩ D[u]is independent and does not contain u. Since v is dominated, w ∈ X . Further,Tw ∼= Pn with n ≡ 1 (mod 3) and by Observation 1(c), X ∩D[w] is independent.This proves that X is an i-set of T .

For the converse, suppose firstly that C0(v) = C1(v) = ∅. Let Y ′ be the setconstructed in the proof of Proposition 1, and letY = Y ′∪{w} for somew ∈ C(v).Since c(w) ∈ Y ′, Y is not independent. Now suppose that C0(v) 6= ∅ 6= C1(v).Letw ∈ C1(v) and let Y be the γ-set of T constructed in the proof of Proposition 1;note that v ∈ Y andw 6∈ Y . Since Yw = Y ∩D[w] � Tw, it follows that c(w) ∈ Y .By Observation 1(d), we may assume without loss of generality that Yw containsthe leaf u 6= w of Tw. Let Y ∗w = {p(x) : x ∈ Yw}, and define Y ∗ = (Y − Yw) ∪Y ∗w . Then Y ∗ is a γ-set of T that contains both v and w ∈ C(v).

3. TREE PRUNING

We now describe a new technique called tree pruning, which will allow us to useTheorem 1 to characterize A(T ) and N (T ) for an arbitrary tree T . For any vertexu of a rooted tree T , denote the set of all u− L paths in Tu by Π(u) (or ΠTu(u) ifnecessary). For j = 0, 1, 2, define

Πj(u) = {P ∈ Π(u): l(P ) ≡ j (mod 3)}.


The pruning of T is performed with respect to the root. Hence suppose that T isrooted at v, i.e., T = Tv. Let u be a branch vertex at maximum distance from v; notethat |C(u)| ≥ 2 and deg x ≤ 2 for each x ∈ D(u). For each w ∈ C(u), allocatea priority to w or, equivalently, to the unique path P ∈ Π(w), where w0 ∈ C0(u)and P 0 ∈ Π0(u) have higher priority than w1 ∈ C1(u) and P 1 ∈ Π1(u), whichagain have higher priority than w2 ∈ C2(u) and P 2 ∈ Π2(u). Let z be a child of uof highest priority. For eachw ∈ C(u)−{z}, deleteD[w]. This step of the pruningprocess, where all but one child of u together with their descendants are deleted togive a tree in which u has degree 2, is called a pruning of Tv at u. Repeat the aboveprocess until a tree Tv is obtained with deg u ≤ 2 for each u ∈ V (Tv)−{v}. ThenTv is called a pruning (here used as a noun) of Tv.

The pruning process can be shortened by noting that if r is a remote vertex ofT adjacent to a leaf l, then l has highest priority in the pruning of Tv at r. Hencedefine

Λ(v) = {P ∈ Π(v): P contains exactly one remote vertex of T distinct from v}.For each P ∈ Λ(v), let r be the remote vertex (r 6= v) on P and let l be anyleaf adjacent to r. For each w ∈ C(r)− {l}, delete D[w]. Now continue with thepruning process as described above. Note that Tv need not be unique; however,if Tv and T ′v are two prunings of Tv, then Cj

Tv(v) = Cj

T ′v(v) for each j = 0,1, 2. Thus, to simplify notation, we write Cj(v) instead of Cj

Tv(v).

To illustrate the pruning process, consider the treeT in Fig. 1. The remote verticesu, x, n and z on paths in Λ(v) are indicated by solid black dots. Hence we deleteD[a], D[b] andD[c]. We still have to prune Tv atw. Now, with respect to the currenttree, x, z ∈ C1(w) and y ∈ C2(w); hence we delete D[y] and (say) D[z]. HenceTv consists of the path luvwxk, together with a leaf joined to v.

FIGURE 1.


4. CHARACTERIZATIONS OF A(T ) ANDN (T )

We now show that a vertex v of a tree T is in all γ-sets (respectively, no γ-set) ofT if and only if v is in all γ-sets (no γ-set) of the pruning Tv of T . Moreover, if Thas no vertices that appear in all its γ-sets, then T also does not have any verticesthat appear in all its i-sets, and, in this case, v is in no i-set of T if and only if v isin no i-set of Tv. To be more precise, we show that the membership of v ∈ V (T ) toA(T ) orN (T ) is determined exactly by the membership of v to A(Tv) orN (Tv).We show further that if A(T ) = ∅, then the membership of v to Ai(T ) or Ni(T )is determined exactly by the membership of v to Ai(Tv) or Ni(Tv). In particular,it follows that Ai(T ) = ∅. Define

A∗(T ) = {v ∈ V (T ): |C0(v)| ≥ 2}.Theorem 2. Let T be a tree rooted at v and let Tv be a pruning of T. For everyγ-set X of Tv there exists a γ-set X of T such that v ∈ X if and only if v ∈ X.Conversely, for every γ-set X of T there exists a γ-set X of Tv such that v ∈ X ifand only if v ∈ X. Further, if A∗(T ) = ∅, the same statements hold with respectto i-sets X and X.

Proof. The proof is by induction on λT =∑u∈B(T )−{v}(deg u−2). If λT = 0,

then Tv = T and the result follows trivially. Suppose that the theorem holds for alltrees with λ < k, and let T be a tree with λT = k. Let u be a vertex of B(T ) atmaximum distance from v and note that v 6= u. Let y be a child of u with lowestpriority, and let T ′ = T − D[y]. Then λT ′ < k and Tv is a pruning of T ′v. Notethat if A∗(T ) = ∅, then |C0

T (u)| ≤ 1. Since u is a branch vertex, there are at leasttwo children. Since y is a child of lowest priority, it follows that if y ∈ C0

T (u), then|C0T (u)| ≥ 2. Hence y 6∈ C0

T (u).Consider any γ-set (i-set, respectively) X of Tv. By the induction hypothesis

there exists a γ-set (an i-set) X ′ of T ′ such v ∈ X ′ if and only if v ∈ X .

Case 1. y ∈ C2T (u).

Then Ty ∼= Pn with n ≡ 0 (mod 3) and, by Observation 1(b), y is not contained inany γ-set or i-set S of Ty. Further, γ(Tc(y)) = i(Tc(y)) = γ(Ty). Thus, regardlessof whether u ∈ X ′ or not, X = X ′ ∪ S is a γ-set (an i-set) of T with v ∈ X if andonly if v ∈ X .


By the choice of y, C2(u) = ∅. By Observation 1(d), there exists an independentγ-set Sy of Ty with y 6∈ Sy and therefore X = X ′ ∪ Sy is a dominating set (anindependent dominating set, respectively) of T . If X is not a γ-set (an i-set) of T ,let Y be any dominating (independent dominating) set of T with |Y | < |X|. If Y ′ =Y −D[y] � T ′, then |Y ′| ≥ |X ′| and hence |Yy| < |Sy|, where Yy = Y ∩D[y].But Yy � Tc(y), hence |Yy| ≥ γ(Tc(y)) = i(Tc(y)) = γ(Ty), since Ty ∼= Pn withn ≡ 2 (mod 3). However, γ(Ty) = |Sy|, contradicting |Yy| < |Sy|.

Thus we may assume that Y ′ 6� T ′, i.e., y ∈ Y and u ∈ pn(y, Y ) (so u 6∈ Y ).Since u ∈ B(T ), there exists y′ ∈ (C0(u) ∪ C1(u)) − {y}. Since u 6∈ Y, Yy′ =


Y ∩D[y′] � Ty′ and therefore |Yy′ | ≥ γ(Ty′). But Ty′ ∼= Pn with n ≡ 1 or 2 (mod3) and so, by Observation 1(c) and (d), there exists a γ-setQy′ of Ty′ with y′ ∈ Qy′ .Now Y ′′ = (Y ′ − Yy′) ∪Qy′ � T ′ (since y′ � u) and |Y ′′| ≤ |Y |. Consequently,the case Y ′ 6� T ′ is the same as the case Y ′ � T ′ and X is a γ-set (an i-set).


By the choice of y, C(u) = C0(u) and u ∈ A∗(T ). Let y′ ∈ C(u)−{y}. If u 6∈ X ′,then since X ′ � y′ and no γ-set of Tc(y′) dominates y′, we may assume withoutloss of generality that y′ ∈ X ′ and that X ′ ∩ D(y′) � Tc(y′). Then y′ � u and{y′} ⊆ pn(y′, X ′) ⊆ {u, y′}. But this implies that X ′′ = (X ′ − {y′})∪ {u} � T ′and hence is a γ-set of T ′. Clearly also v ∈ X ′′ if and only if v ∈ X ′. Hence wemay assume without loss of generality that u ∈ X ′.

Therefore, if S is any γ-set of Tc(y), then X = X ′ ∪ S is a dominating setof T satisfying the requirements. Moreover, if Y is any γ-set of T , then u ∈ Ysince u ∈ A∗(T ). Therefore Y − D[y] � T ′ and Y ∩ D[y] � Tc(y), so that|Y | ≥ |X ′|+ |S| = |X|, and it follows that X is a γ-set of T .

Conversely, consider any γ-set (i-set, respectively) X of T and suppose firstlythatX ′ = X−D[y] 6� T ′. Then u 6∈ X, y ∈ X and u ∈ pn(y,X). Let y′ 6= y be achild of uwith highest priority. Since u 6∈ X , it follows thatXy′ = X∩D[y′] � Ty′and therefore |Xy′ | ≥ γ(Ty′). If y′ ∈ C0(u)∪C1(u), thenTy′ ∼= Pn withn ≡ 1 or 2(mod 3), and there exists an independent γ-set Yy′ of Ty′ with y′ ∈ Y . ConsequentlyY = (X − Xy′) ∪ Yy′ � T, |Y | ≤ |X| and Y ′ = Y − D[y] � T ′. Moreover, ifX is independent, then so is Y . In addition, v ∈ X if and only if v ∈ Y . Ify′ ∈ C2(u), then, by the choice of y′, C(u) = C2(u) and so Ty ∼= Pn, wheren ≡ 0 (mod 3). Therefore no γ-set of Ty contains y, from which it follows (sincey ∈ X) that |Xy| ≥ γ(Ty) + 1. Let Sy be any independent γ-set of Ty. ThenY = (X − Xy) ∪ Sy ∪ {u} � T, |Y | ≤ |X| and Y ′ = Y − D[y] � T ′. If Xis independent then Y is independent since u ∈ pn(y,X). Furthermore, v ∈ Y ifand only if v ∈ X . We may therefore assume that X ′ = X − D[y] � T ′ and soXy = X ∩D[y] � Tc(y).

We prove that X ′ is a γ-set (an i-set) of T ′. Suppose to the contrary that S′is a dominating (an independent dominating) set of T ′ with |S′| < |X ′|. If y ∈C1(u) ∪ C2(u), then Tc(y)

∼= Pn with n ≡ 1 or 2 (mod 3) and there exists anindependent γ-set Sy of Tc(y) that dominates y. Then S = S′ ∪ Sy � T and|S| = |S′| + |Sy| < |X ′| + γ(Tc(y)) ≤ |X| (since Xy � Tc(y)), a contradiction.If y ∈ C0(u), then C(u) = C0(u), and there exists y′ ∈ C0(u) − {y}, i.e.,u ∈ A∗(T ) and soA∗(T ) 6= ∅. Suppose that u 6∈ X . Then Xy � Ty. But Ty ∼= Pnwith n ≡ 1 (mod 3) and therefore |Xy| ≥ γ(Ty) = γ(Tc(y)) + 1. Similarly|Xy′ | ≥ γ(Ty′) = γ(Tc(y′)) + 1. Let Yy, Yy′ be γ-sets of Tc(y), Tc(y′) and defineX∗ = (X−(Xy∪Xy′))∪Yy∪Yy′ . ThenX∗ � T and |X∗| < |X|, a contradiction.Therefore u ∈ X, |Xy| = γ(Tc(y)) and |Xy′ | = γ(Tc(y′)).

Now, if u ∈ S′, then S = S′ ∪ Xy is a dominating set of T with |S| < |X|,a contradiction. Therefore u 6∈ S′ and it follows that Sy′ = S′ ∩ D[y′] � Ty′ ,where |Sy′ | ≥ γ(Ty′) = γ(Tc(y′)) + 1. Let S′′ = S′ − Sy′ and X ′′ = X ′ −Xy′ .


Then

|S′′| = |S′| − |Sy′ | ≤ |X ′| − 1− γ(Tc(y′))− 1

= |X ′| − |Xy′ | − 2

= |X ′′| − 2.

If we define S∗ = S′′ ∪ Yy ∪ Yy′ ∪ {u}, then S∗ � T and

|S∗| ≤ |X ′′|+ γ(Tc(y)) + γ(Tc(y′)) + 1− 2

< |X|,a contradiction.

The characterizations of A(T ) and N (T ) now follow immediately from Theo-rems 1 and 2.

Corollary 1. For any tree T and any vertex v of T, v ∈ A(T ) if and only if|C0(v)| ≥ 2, and v ∈ N (T ) if and only if C0(v) = ∅ and C1(v) 6= ∅.

Hence we see that A(T ) = A∗(T ) and we obtain the following corollary.

Corollary 2. Let A(T ) = ∅. Then Ai(T ) = ∅, and v ∈ Ni(T ) if and only ifC0(v) = ∅, C1(v) 6= ∅, i.e., Ni(T ) = N (T ).

Proof. If A(T ) = ∅, then by Theorem 2, v ∈ Ai(T ) if and only if v ∈ Ai(Tv)if and only if v ∈ A(Tv) (Theorem 1). But A(Tv) = ∅ by Theorem 2. The rest ofthe statement follows directly from Theorems 1 and 2.

Consider the tree T in Fig. 1. Since |C0(v)| = 1, v 6∈ A(T ) ∪ N (T ). Clearly,|C0(c)| = 2, hence c ∈ A(T ). Similarly, C0(u) = {a, l}, henceu ∈ A(T ). Finally,C0(w) = ∅ and x ∈ C1(w), hence w ∈ N (T ).

5. CHARACTERIZATION OF Ai(T ) IF γ(T ) = i(T )

It is obvious that if T is a tree such that v1 and v2 are adjacent vertices in A(T ),then each γ-set of T contains v1 and v2 and hence is not independent, so thati(T ) > γ(T ). We thus henceforth restrict our attention to trees T such thatA(T ) isindependent. We begin by showing that if u ∈ A(T ), where A(T ) is independent,and T ′ is the graph obtained by removingN [u] from T , then the remaining verticesof A(T ) are also in all minimum dominating sets of T ′.

Theorem 3. If {u, v} ⊆ A(T ),whereA(T ) is independent, and T ′ = T−N [u],then v ∈ A(T ′).

Proof. Suppose to the contrary that there exists a γ-setX ′ ofT ′ such that v 6∈ X ′,and letX = X ′∪{u}. ThenX dominatesT and v 6∈ X , henceX is not a γ-set ofT .Let Y be a γ-set of T and note that u ∈ Y and |Y | < |X|. If Y ∩N [u] = {u}, then,since u does not dominate any vertex of T ′, it follows that Y ′ = Y ∩ V (T ′) � T ′.But Y ′ = Y − {u} and so |Y ′| = |Y | − 1 < |X| − 1 = |X ′|, a contradiction.Hence Y ∩N(u) 6= ∅.


For each w ∈ Y ∩ N(u), let Tw be the subtree of T − u containing w andconsider Tw to be rooted at w. Note that C(w) = CTw(w) ⊆ V (T ′) and hence foreach x ∈ C(w), Tx is a subtree of T ′. For x ∈ C(w), define Xx (Yx, respectively)by Xx = X ∩ V (Tx) (Yx = Y ∩ V (Tx), respectively). Since X ′ is a γ-set of T ′,it follows that |Xx| = γ(Tx) for each x ∈ C(w). Also, if |Yx| < γ(Tx) for somex ∈ C(w), then |Yx| = γ(Tx)− 1 and x is the only vertex of Tx not dominated byYx, for otherwise Y does not dominate Tx and thus also not T . Note that

X = {u} ∪ ⋃w∈Y ∩N(u)

⋃x∈C(w)

Xx

and

Y = (Y ∩N [u]) ∪ ⋃w∈Y ∩N(u)

⋃x∈C(w)

Yx

.If each w ∈ Y ∩N(u) has at most one child x such that |Yx| = γ(Tx)− 1, then

|Y | = 1 + |Y ∩N(u)|+ ∑w∈Y ∩N(u)

∑x∈C(w)

|Yx|

≥ 1 + |Y ∩N(u)|+ ∑w∈Y ∩N(u)

∑x∈C(w)

γ(Tx)

− |Y ∩N(u)|

= 1 +

∑w∈Y ∩N(u)

∑x∈C(w)

|Xx|

= |X|,a contradiction. Hence there exists a vertex w ∈ Y ∩ N(u) with x1, x2 ∈ C(w)

such that |Yxj | < γ(Txj ), j = 1, 2. Abbreviate Txj , Xxj and Yxj to Tj , Xj and Yj ,respectively.

Now consider T to be rooted at w. (That is, we consider the whole tree T andnot just Tw, so that CT (w) = CTw(w) ∪ {u}.) We use the notation of the aboveparagraph and prove that {x1, x2} ⊆ C0(w), thus contradicting the hypothesis thatA(T ) is independent.

Firstly, since |Y1| < γ(T1), it follows that Y1 does not dominate x1 and thusx1 ∈ pn(w, Y ), while Y1 � V (T1) − {x1}, i.e., |Y1| ≥ γ(V (T1) − {x1}). Itfollows that |Y1| = γ(T1) − 1 and thus Z = Y1 ∪ {x1} is a γ-set of T1. Considerany y ∈ C(x1), the subtree Ty of T1 and Sy = Y1 ∩ V (Ty). Then Sy � Ty and so|Sy| ≥ γ(Ty). Now, if y ∈ S for some γ-set S of Ty, then Z ′ = (Z − (Sy ∪{x1}))∪ S dominates T1 and |Z ′| < |Z|, a contradiction. It follows that y ∈ N (Ty).Thus, by Corollary 1 applied to Ty, C0(y) = ∅ and C1(y) 6= ∅. Hence, in thepruning of T = Tw at y, C0(y) = ∅ and C1(y) 6= ∅. Since this holds for each


y ∈ C(x1), it follows that, in the pruning of Tw at x1, C(x1) = C2(x1) andtherefore x1 ∈ C0(w). Similarly, x2 ∈ C0(w) and the result follows.

The next result shows that if a vertex of A(T ) and its neighborhood is removedfrom a (γ, i)-tree T , then the resulting forest is also a (γ, i)-graph.

Proposition 2. If γ(T ) = i(T ), S ⊆ A(T ) and F = T − N [S], then γ(F ) =i(F ) = γ(T )− |S|.

Proof. If V (T ) ⊆ N [S], the result holds vacuously, so suppose V (T )−N [S] 6=∅. Since γ(T ) = i(T ),A(T ) is independent. Further, if X is any independentdominating set of T , thenX−S is an independent dominating set ofF . Conversely,if Y is a minimal (an independent, respectively) dominating set of F , then Y ∪ Sis a minimal (an independent) dominating set of T . Hence

i(F ) ≤ i(T )− |S|and

i(T ) ≤ i(F ) + |S|,from which it follows that

i(F ) = i(T )− |S|.Similarly,

γ(F ) ≤ i(F ) ≤ i(T )− |S| = γ(T )− |S|and

γ(T ) ≤ γ(F ) + |S|,hence

γ(F ) = i(T )− |S| = i(F ).

We used the pruning procedure described in Section 3 to determineA(T ). Nowwe useA(T ) to describe a different procedure, which can also be seen as a type ofpruning procedure, to determine Ai(T )−A(T ) for (γ, i)-trees.

For any tree T and v ∈ V (T )−A(T ), letAv(T ) = A(T )−N(v). We define thesubforest Tv of T recursively by means of a sequence of subgraphs T 0

v , T1v , . . . , T

kv

of T , where T 0v = T and, informally, if v 6∈ A(T jv ), then T j+1

v is obtained byremoving from T jv the closed neighborhood of each vertex in A(T jv ) which is notadjacent to v. That is, the sequence ends with T kv when either v ∈ A(T kv ) or wheneach vertex in A(T kv ) is adjacent to v, and we define Tv = T kv .

More formally, for any T jv , let T j+1v = T jv −N [Av(T jv )]. Since T is finite, there

exists an integer k such that Av(T kv ) = ∅ (i.e., A(T kv ) ⊆ N(v)) or v ∈ A(T kv ).Then Tv = T kv . Define the subset I(T ) of V (T ) by

I(T ) = {v ∈ V (T ): v ∈ A(Tv)}.The characterization of Ai(T )−A(T ) for (γ, i)-trees follows.


Theorem 4. If γ(T ) = i(T ), then I(T ) = Ai(T )−A(T ).

Proof. If v ∈ I(T ), then by definition of Tv, v 6∈ A(T ). Consider any in-dependent γ-set X0 of T . Then A(T ) ⊆ X0. Let R0 = A(T ) ∩ N(v), i.e.,R0 = A(T )−Av(T ). By Proposition 2, T 1

v = T −N [Av(T )] satisfies

γ(T 1v ) = i(T 1

v ) = γ(T )− |Av(T )|.By Theorem 3,

R0 ⊆ R1 = A(T 1v ) ∩N(v) = A(T 1

v )−Av(T 1v ).

Moreover, X1 = X0 − Av(T ) is an independent γ-set of T 1v and A(T 1

v ) ⊆ X1.Repeat the process until Tv = T kv is obtained. Note that

R0 ⊆ R1 ⊆ · · · ⊆ Rk = A(Tv) ∩N(v),

that

Xk = X0 −Av(T )−Av(T 1v )− · · · − Av(T k−1

v ) (1)

(where the deletion takes place in the given order) is an independent γ-set of Tvand that

A(Tv) ⊆ Xk. (2)

But v ∈ I(T ); hence by definition v ∈ A(Tv) and it follows from (1) and (2)that v ∈ X0. Since X0 was chosen arbitrarily and v 6∈ A(T ), it follows thatv ∈ Ai(T )−A(T ).

Conversely, consider v ∈ Ai(T ) − A(T ). We first prove that v ∈ Ai(Tv).Suppose to the contrary that Y k is an i-set of Tv not containing v. Recall that, bydefinition,

Tv = T k−1v −N [Av(T k−1

v )].

Further,A(T ) is independent since γ(T ) = i(T ) and henceAv(T k−1v ) is indepen-

dent since

Av(T k−1v ) ⊆ A(T k−1

v ) ⊆ A(T )

by repeated application of Theorem 3. It follows that

Y k−1 = Y k ∪ Av(T k−1v )

is independent, dominates T k−1v (since Av(T k−1

v ) dominates N [Av(T k−1v )]) and

|Y k−1| = |Y k|+ |Av(T k−1v )|

by Proposition 2. Hence, Y k−1 is an i-set of T k−1v . Note that v 6∈ Y k−1 since

v 6∈ Y k by choice and v 6∈ Av(T k−1v ) by definition of Av(T k−1

v ). Repeating theprocess we obtain an i-set Y 0 of T with v 6∈ Y 0, contradicting the assumption thatv ∈ Ai(T ). This proves that v ∈ Ai(Tv).


We now prove that either v ∈ A(Tv), in which case v ∈ I(T ), or A(Tv) = ∅,which would imply (by Corollary 2) that Ai(Tv) = ∅, a contradiction. Supposethat v 6∈ A(Tv). By the construction of Tv, each vertex in A(Tv) is adjacent to v.We prove that A(Tv) = ∅.

Note that by Proposition 2, γ(Tv) = i(Tv). Let S be any independent γ-set ofTv. Since v ∈ Ai(Tv), v ∈ S. Further, A(Tv) ⊆ S. But S is independent, thusno vertex of A(Tv) is adjacent to v, and it follows that A(Tv) = ∅. But then, byCorollary 2, Ai(Tv) = ∅, a contradiction. Thus v ∈ A(Tv) and so by definitionv ∈ I(T ), as required.

To illustrate the procedure described above to determine Ai(T ) for (γ, i)-trees,consider the tree T in Fig. 2. It is easy to see that A(T ) = {g, h}, that X ={a, b, c, d, g, h} is a γ-set ofT and thatY = {a, b, e, f, g, h} is an i-set ofT . Further,T ′ = T −N [{g, h}] consists of T 〈N [a, b]〉 together with the isolated vertices e andf . It is clear that A(T ′) = {a, b, e, f}. Therefore T ′ = Tv for v ∈ {a, b, e, f}, and{a, b, e, f} ⊆ I(T ). Moreover, by Theorem 4, I(T ) ⊆ Ai(T ), which implies thatI(T ) is independent. But V (T ′) = N [{a, b, e, f}] and thus each vertex of T ′ isadjacent to one of these vertices, hence I(T ) = A(T ′) = {a, b, e, f}. By Theorem4, Ai(T )−A(T ) = {a, b, e, f}, i.e., Ai(T ) = Y . Thus T has a unique i-set.

The problem of characterizingAi(T ) andNi(T ) for a tree T with i(T ) > γ(T )seems to be considerably more difficult and remains an open problem. Also, apartfrom some trivial cases such as cycles, complete graphs and complete bipartitegraphs, we have not investigated the sets A,N ,Ai, and Ni for cyclic graphs. Itmay be interesting to find other classes of graphs for which (some or all of) thesesets can be characterized.

The pruning procedure, or an adaptation thereof, may also be useful in otherstudies of domination in trees.

FIGURE 2.


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vertices contained in every minimum dominating set of a tree

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