domination in graphs with minimum degree two

Domination in Graphs with Minimum Degree Two -

William McCuaig Bruce Shepherd

DEPARTMENT OF COMBINATORICS AND OPTIMIZATION UNlVERSlN OF WATERLOO

WATERLOO, ONTARIO, CANADA

ABSTRACT

The domination number y ( G ) of a graph G = (V E ) is the minimum cardinality of a subset of Vsuch that every vertex is either in the set or is adjacent to some vertex in the set. We show that if a connected graph G has minimum2degree two and is not one of seven exceptional graphs, then y ( g ) I ~ l V l . We also characterize those connected graphs with y(G) = FJVI.

1. INTRODUCTION

In this paper we assume G = (V, E) is an undirected simple graph with finite vertex set V and edge set E. The edge joining distinct elements u and u of V is denoted by uu or uu. For a vertex u, its neighborhoodN(u) is {x E V : ux E E}, its closed neighborhood N [ u ] is N(u) U {u}, and its degree, d(u), is (N(u)l. The minimum (maximum) degree of G, denoted 6(G), (A(G)) is the minimum (maximum) degree of a vertex in V. If ~ w o graphs G and H are the same (up to labeling of the vertices), then we say that they are isomorphic, written G = H. For a subset S of V we let G(S) be the subgraph of G induced by S , and E ( S ) be the edge set of G(S). For J C {xy: x , y E V } , G - J(G U J ) is the graph with vertex set V and edge set E - J(E U J ) , and G - S - J is the graph G(V - S)-J. If J consists of just one element M U , then we often write G - uu(G + uu) in place of G - J(G U J ) . We call an edge uu a bridge when G - M U has more connected components than G. For two graphs G , = ( V , , E , ) and G, = ( V , , E , ) we define their union G I U G, to be (Vl U V,, E l U E,) . For n positive, we denote the cycle on n vertices by C,.

Journal of Graph Theory, Vol. 13, No. 6, 749-762 (1989) 0 1989 by John Wiley & Sons, Inc. CCC 0364-9024/89/060749-14$04.00

750 JOURNAL OF GRAPH THEORY

A subset D of V is a dominating set of G if each u in V - D has a neighbor in D. The minimum cardinality of a dominating set of G is the domination number y ( G ) of G, and for 0 < c 5 1, we say G is c-dominated if y(G) I clVI. Similarly, a set D is called c-dominating if it is a dominating set and ID1 5 clVI. For subsets S and T of V, we say that S dominates T if S is a dominating set of G(S U T).

We let B(G) = {u E V: d(u) # 2) and for u in B(G) we call the connected component of G - (B(G) - u) that contains u the 2-graph of u. Note that if S(G) 2 2, then each vertex of the 2-graph has degree 2 in G , except for u itself. The 2-graph will consist of edge disjoint cycles through u (2-graph cycles) and paths starting at u (Zgraph paths).

Our main result is that except for the collection PA of graphs in Figure 1, all connected graphs with minimum degree two are $dominated. We also characterize those connected graphs G with minimum degree two, satisfying y(G) = ilVl, which are edge-minimal or edge-maximal.

Our theorem is an upper bound for the domination number of a graph G in terms of S(G) and IVI. Payan has obtained the best such bound for arbitrary values of 6 ( G ) using the greedy algorithm:

Theorem 1. (Payan [4,5]). If G is connected, then

The upper bound of Theorem 1 is asymptotically equivalent (as S(G) + 00) to (log S(G)/S(G)) (VI. An early theorem of Ore gives a better bound of this form when S(G) 5 2.

Theorem 2. (Ore [3]). If G has minimum degree 1, then y ( G ) 5 IVl/2.

Proof. Let D be a minimal dominating set of G. Each vertex in D must be adjacent to some vertex of V - D, or D would not be minimal. Thus V - D is also a dominating set, and the theorem follows. I

H5 H6 H7

FIGURE 1. The collection 3 of "bad" graphs.

DOMINATION IN GRAPHS 751

If a graph G has a vertex u of degree 1, then there is a minimal dominating set of G that does not contain u. In this sense, these pendant leaves are unim- portant from a computational point of view. Under more restrictive conditions, our result improves the constant in the bound from to 3.

Finally, Gamble [2] has constructed an infinite class of graphs, with 6(G) 2 3 and y(G) = SIVl. He has conjectured the following:

Conjecture 1. (Gamble).* If G is c o ~ e ~ t e d and 6(G) 2 3, then y(G) I ilVl.

2. THE UPPER BOUND

We define a unit to be a graph that is isomorphic to either of the graphs of Figure 2. There are two types of units and we will call a unit type 1 or type 2 according to whether it is isomorphic to the graph of Figure 2(a) or Figure 2(b) respectively. Let 9 denote the collection of all graphs G of the form G = (UiEr Gi) U J where the Gi’s are pairwise disjoint units (called the units of C ) and J is a subset of edges uu satisfying the following:

(1) uu is a bridge of G ; and (2) each of u and u is either

(a) a pendant vertex in a type 1 unit, or (b) a vertex in a type 2 unit such that neither neighbor in that unit is

Since each unit is $dominated, so too is G. An attachment vertex of G is any vertex incident to an edge of J. An outer vertex of G is any vertex that satisfies one of 2(a) or (b). (See Figure 3.) Otherwise x is an inner vertex. The degree of the unit Gi is the number of edges of J that are incident with Gi.

incident with any edge of J.

(a I ( b )

FIGURE 2. (a) type 1 unit and (b) type 2 unit.

*Bruce Reed has recently given a proof of this conjecture.


FIGURE 3. A graph in the collection 9* with the outer vertices circled.

In the remainder of this paper we refer to a graph G = (V, E) as $minimal (maximal) if G is edge-minimal (maximal) with respect to satisfying S(G) 2 2, y(G) = fiVl, and connectedness.

Let 8. = {G E 9: S(G) 2 2 and G is connected} and F,, . . . , F6 be the graphs defined in Figure 4. The main theorem of this paper (Theorem 3) states that 9, U {F,}P=, is the collection of all $minimal graphs. We start by proving three lemmas.

Lemma 1. Let G be from 9. U {Fi}r=,. We have

(9 y ( G ) = $q. (ii) For any u in V, G has a minimum dominating set containing u. If

G E 9, and S is a set of vertices from distinct units of G, then G has a minimum dominating set containing S (in fact, this is true for any G in 9) *

(iii) F, and C, are the only $minimal graphs on 5 vertices. (iv) If IVI > 10, u is an inner vertex and u is a vertex from a unit not con-

taining u, then y ( ~ + uu) < 3 1 ~ 1 . (v) If G ’ = ( V ’ , E ’) is in 9, U {Fi}:=, and disjoint from G, u E V ’ , u E V

and Gt = (G‘ U G) + uu is $-minimal, then G , G ’ E 9, and u and u are outer vertices in G ‘ and G, respectively.


A m FIGURE 4. Some small graphs F, = (& ,€ i ) with y ( 6 ) = $l&(.

Proof.

It is straightforward to check this for F i , i = 1, . . . , 6. If G E $,, then y(G) I flVl and any dominating set must contain at least two vertices for each unit of G. Hence y (G) = $lV(. It is straightforward to check each F, . If G E $,, then y ( G ) = f IVI and for u and S we can clearly find another vertex that together with u will dominate the unit containing u; (ii) follows immediately. This statement is easily verified. Let U , U" be the units containing u and u, respectively. Let u' be an attachment vertex in U . If U is of type 2, we choose u' adjacent to u. If u ' is (a) adjacent to a vertex u' in a unit other than U or U", or (b) adjacent to u, or (c) adjacent to a vertex u' of U" that together with u dominates U", then we can construct a $dominating set of G - U that also dominates u and u ' . Hence we need only add a single vertex to this set in order to dominate all of G (and hence G + uu). So suppose none of (a), (b), or (c) occurs. Since (a) does not occur, u' has exactly one neighbor not in U , u' say. In fact u' must be in U". Since (b) and (c) do not occur, U" - {u,u'} can be dominated by a single vertex. Hence u is also an inner vertex and so by symmetry, (a), (b), and (c) do not occur for u and u' either. Since IVI > 10, one of U or U" has degree at least 2. By symmetry we may assume it is U . Thus U has some attachment vertex other then u ' (note that this implies U is of type 2). Now take any $dominating set of G - (U U U") which also dominates the second attachment vertex of U . In order to dominate G + uu we need only add to this set the vertices u, u' and an appropriate vertex of U". Thus G + uu has a dominating set containing $ 1 ~ 1 - 1 elements.


(v) First suppose G ' and G are in 9. and that u , say, is an inner vertex of G'. By definition, there is some attachment vertex u' in the same unit, U say, as u. If U is of type 2, then choose u' so that it is adjacent to u. Let u'' be a neighbor of u' that is not in U . We know that G has a :-dominating set containing u and G ' - U has a $dominating set containing u". We then need only add one more vertex in order to dominate U and so Gt could not be $minimal. By symmetry, u and u are both outer vertices. So suppose G ' E { F 2 , . . . , F6}; then it is straightforward to check that we need only 3 vertices to dominate V' - u in G', and since G has a $dominating set containing u, G' has a dominating set with fewer than t(1Vl + IV'() elements. A similar argument holds if G ' = F, and u is of degree 4. Finally if G ' = F, and d(u) = 3 (see Figure 3, then G' is not edge-minimal.

2

I

We now examine the graphs G, on a small number of vertices, with y(G) > $vl. Lemma 2. If G is connected, 6 ( G ) r 2, (VI I 7, and y(G) > t l V ( , then G E 99.

Proof. Let G = ( V , E ) satisfy the hypothesis of the lemma. It is straightforward to check that H I is the only graph on fewer than 7 vertices that satisfies the hypothesis. So suppose IVI = 7, then clearly A(G) s 4.

Case 1. A(G) = 2.

H, = C7 is the only connected graph G on 7 vertices with 6(G) = 2 = A(G).

Case2. A(G) = 3.

Let u, be a vertex of maximum degree for which (E(N(u,))I is minimized. Let N(uJ = {u,,ug,u4} and T = { u , , v ~ , u , } = V - N[u,] . Since y(G) = 3, no vertex can dominate T, in particular

FIGURE 5. G' - w' is connected and has minimum degree 2.


Thus since 6(G) 2 2, [{uu E E : u E N(u,), u E T}I 2 4. So we may assume u2 is adjacent to exactly 2 vertices of T, us and vg, say. (See Figure 6.) If {u3, u4} C N(u7), then {uz, u7} dominates G, and so 1N(u7) r ) N(u, ) [ = 1. Thus we may assume that N(u7) = {u6, u4} and so by (1) and our choice of uI we have

E(N(V,) ) = @ - (2)

From (2) we have that u3 is adjacent to one of u5 or V6. If u3u5 $ E , then UJV6 E E and u4u5 E E but then {u4,v6} is a dominating set, so u3us E E, and so G contains H4 as a subgraph. (See Figure 7.)

By (1) and (2), the only other possible edges of G are u4uS, u4V6, and V3u6.

If V4V6 E E, then {u4, u5} is a dominating set. As we have seen, at most one of u4us and u3u6 can be an edge, and if exactly one of them is an edge, then G = H,.

Case 3. A(G) = 4.

Let u , be a vertex of degree 4 and set N(u,) = {uz, u3, u4, u5}, T = V - N[uJ = {u6,u7}. T cannot be dominated by a single vertex and so q 9 7 E. Since 6(G) 1 2 we may assume N(u6) = {uz, u3} and N(u7) = {u4, u5}. (See Fig- ure 8.) Thus the only other edges of E are in E(N(u,)) and if either u p , or u4u5 is an edge, or uz and u3 are adjacent to u4 and u5, respectively, then y(G) 5 2. It follows that G(N(u,)) does not contain two independent edges. Thus if G is not isomorphic to H 3 , then for IE(N(u,))I = 1 we must have G 2 H6, and for [E(N(u,))I > 1, G(N(u,)) must have exactly one degree 2 vertex, u2 say, such that u2 is adjacent to u4 and us. Hence G = H,. I

FIGURE 6.

"5 "6

FIGURE 7.


"6 v7

FIGURE 8.

We will refer to a graph in the collection 93, as a bad graph. We make four observations about bad graphs before proceeding with the main lemma.

Observations.

(1) For any G in 93 and u in V, there is a minimum dominating set of G that

(2) For any G in 93 and u in V, only y ( G ) - 1 vertices are required to domi-

(3) For any G I = ( V l , E , ) , G, = (V,,E,) in 93

contains u.

nate G - v .

(4) If G E 93, then

and if G # H,, then

Lemma 3. If G = (V,E) satisfies

(i) 6(G) L 2, (ii) G is connected, (iii) y(G) 2 $ l V ) , and (iv) G is edge-minimal with respect to (i)-(iii),

then G E 9, U { H I , H,, H , , F I , F,, . . . ,F6} .

Proof. The proof is by induction on IVl. We have shown (Lemmas 1 and 2) the statement to be true for ( V J 5 7, so let G = ( V , E ) satisfy (i)-(iv) and 1Vl 2 8, and assume the induction hypothesis (1H)-that is the lemma is true for graphs on fewer than IVl vertices. We assume G # F, and so B ( G ) # 8.


Suppose u E B ( G ) and uu is a bridge whose removal yields components GI = (Vl ,El ) and G, = (V,,EJ, where G, contains u. Note that 6(Gl) 2 2. We show that we may assume G, is not one of the graphs in Figure 9. We use this later to be able to assume that B(G) is independent.

First suppose GI is bad. If G, is one of (b) or (e) in Figure 9, then we can find a $dominating set of G, containing u. By observation (2) it follows that we need only y(G,) - 1 more vertices to dominate G and so by observation (3a) we conclude y ( G ) < 5iVl. Suppose G2 is one of (a), (c), (d), or (f). If G # H I , then by observation (1) we can take a minimum dominating set of GI that contains u . By observation (4), y(GI + u u ) < $(lVll + 1) and since G, - u is $dominated, G violates (iii). If GI = H I , and G2 is either (a) or (d), then it is easy to check that y ( G ) < $lVl, and if G, is either (c) or (f), then G E 9,. From these arguments we assume that GI is not bad.

Now suppose y ( G , ) = $ l V l ( . By (IH), G I E 9, U {F] , . . . , F 6 } . By Lemma l(ii) it follows that G, cannot be one of the graphs (a)-(d) or (f) in Figure 9. Hence G, is the pendant 4-cycle (e) and by Lemma l(i) and (v), u and u must be outer vertices and so G E 9,. So suppose

?(GI) < w f (*)

Each of (a)--(c), (e), and (f) is $dominated, so G, must be (d). If GI has a i- dominating set containing u, then clearly G violates (iii). Thus

No $dominating set of GI contains u and so y(Gl - u) 5 y(Gl) . (**>

We will prove the following claim.

Claim. If GI = (Vl,El) and u satisfy (*), (**), IVll < IVI, and GI is edge- minimal with respect to satisfying (i) and (ii), then either (a) ?(GI - u ) < i(lVll - 1) or (b) G , - u is a collection of type 1 units and graphs in 9,.

Note that by the claim, GI - u can be $dominated and evidently G, + uu is 3 i-dominated and so G violates (iii). Thus we will be able to assume:

None of the configurations of Figure 9 occur as the 2-graph of a vertex in B ( G ) . Dl


(Proof of Claim). Suppose (a) is not true, and let w be a neighbor of u and J = {wx:x E N(u) - w}. Let G* = (GI - u) U J . Clearly G* satisfies (i) and (ii). Since each edge added is in u's neighborhood, we have

If D is a dominating set of G*, then D U {u} dominates GI . (***)

Now suppose D is a minimum dominating set of G*. Then by (***), D U {u} is a dominating set of G I but by (**) it is not $dominating. Thus by (*) ID( 2 y(G,) and so by (**). y(G*) 2 y(G1 - u). Hence y(G*) = ?(GI - u).

If y(G*) > i(lVll - l), then by (IH) G* contains one of H,, H,, or H 3 as a spanning subgraph. By observation (2) there is a subset D of size y(G*) - 1 that dominates G* - w. By (***) and observation (4), D U {u} $dominates G I , contradicting (**).

Thus y(G*) = ?J(lV,l - 1). We show that GI - u satisfies (b). By (IH), G* contains a spanning subgraph H that is in 9, U {2PJ}:=,. Suppose H E {9,}>p=,. Then it is straightforward to check that for any vertex x, in V(H), H - x can be dominated by three vertices. If H is F , , then since dG,(u) 2 2, u dominates at least two vertices of H. It is then easy to see that we can dominate H using u and at most one other vertex of H. In either case we have that if H E {?JJ}:=,, GI U J has a $dominating set containing u. Thus by (***), so does GI. This contradicts (**). Thus H E 9,. Now suppose some unit U of H contains an edge of J. If U is of type 1, then choose a set D that $dominates H - U and contains a neighbor of U's attachment vertex (we use Lemma l(ii) here). If U is of type 2, then take any $dominating set D of H - U. We now add to D, the vertex u, and an appropriate vertex from U to obtain a $dominating set of GI that contradicts (**). Thus no unit of H contains an edge of J and so H - J, which is a subgraph of GI - u, is a collection of type 1 units and graphs in 9,. It is then enough to show GI - u H - J. Suppose G I - u contains some edge xy that is not in H - J. This edge cannot be a cut edge of GI since H is connected, and so by minimality, one of x or y, say x, is of degree 2 in GI. Hence x has degree at most 1 in H - J . But S ( H ) 2 2, so x is incident to an edge of J and so x E N(u) . But since S(H - J) 5 1, this implies x has degree at least 3 in GI, a contradiction. We conclude that GI - u = H - J and hence satisfies (b). This completes the proof of the claim.

Now, if uu is an edge between 2 vertices in B(G), then by minimality of G , uu must be a bridge. Let GI = (VI,El), G, = (V.,E,) be the components of G - uu and assume u E V,. Each G, satisfies (i) and (ii). Thus if G I , say, satisfies (*), then G, must be bad. It follows that GI also satisfies (**). It is then easy to deduce from the claim that G violates (iii). Thus y(G,) 2 $lVJl for i = 1,2. Suppose G I is bad and hence G , is either bad or in 9. U {F,}:=,. Ei- ther Lemma l(ii) or observation (1) gives that G, has a minimum dominating set containing u. Now by observation (2), it follows that we need only add y(Gl) - 1 vertices to form a dominating set of G. Hence by observation (3) ((a) or (b) depending on whether G2 is bad), G violates (iii). By symmetry we assume y ( G , ) = $IV,\, i = 1 ,2 . By the induction hypothesis G, E 9, U {Fl,. . . ,F6} . Since G is $minimal it follows from Lemma l(v) that G E 9..


Thus we may assume

B(G) is independent. P I

Suppose u in B(G) starts a 2-graph path (or cycle) containing the path uu,u2u3. Let u' be the neighbor of u3 , other than u2. Set G ' = G(V - { u i ) ~ , , ) + w'. Suppose that G' is also simple. Then

since we need only add a single vertex to dominate G . Thus by (IH), G' E 93. Since B(G) is independent, G'(B(G'))can have at most one edge, uu'. Hence G ' H I , H,, H 3 , or H4, and so G = H,, F,, F,, or F,. Therefore we assume

For each u in B(G), the' 2-graph cycles of u have lengths 3 , 4, or 5 , and the 2-graph paths have lengths 1 or 2. 131

Suppose for k = 3 or 4, there is some u E B(G) that has a pendant cycle, C = w , u , . . . u p , in its 2-graph. If G 2 = G - (C - u) does not have minimum degree two, then u must be adjacent to uI, uk and exactly one other vertex, u say. By [2], u E B(G) and by [ 3 ] u is at most distance 2 from B ( G ) - u. Hence [ l ] is violated. Thus 6(G2) 2 2 and we claim:

If k = 3 , then y(G2) 5 $(lVl - k) implies G violates (iii). So suppose k = 4 and let G' = (G + u,u4) - w , (i.e., G 2 with a pendant 4-cycle). Clearly G' satisfies the lemma hypotheses if G does. Since G satisfies (iii), no $- dominating set of G 2 contians u. Thus if y(G2) C $(lV) - k), it is not hard to show using the claim, that G' E 9,. Hence G will be too. And by (IH) and Lemma l(ii), if y(G2) = $(IVl - k ) , then G 2 has a $dominating set containing u and this implies that G violates (iii). Thus (****) holds, and so by minimality G 2 is one of H I , H 2 , or H,. It is then easy to check that G must be one of H 3 , F,, F4, or F,. Thus we will assume that only triangles can exist as cycles in a 2-graph of G. Thus by [3], B(G) is a dominating set and so by (iii)

3b 2 2d,

where b = IB(G)I, d = Iq - b. But by simple counting arguments and [2],

Thus 2d = 3b and b = $Vl. Further, each x in B(G) has degree exactly 3 and each edge is incident with exactly one vertex of degree 2. Thus there are no 2-


graph cycles and each 2-graph path is of length one. But since 1V( 1 8, i.e., G # F,, there are vertices u and u in B ( G ) with exactly one neighbor w in common. Hence ( B ( G ) - {u ,u} ) U {w} is a dominating set with fewer than -j(V( elements. This completes the proof of the lemma. I 2

Theorem 3. If G = ( V , E ) is connected, 6(G) z 2 and G €E 93, then y ( G ) 5 i lVl. Furthermore, 9* U {Fi}f=) is the collection of all f-minimal graphs.

Proof. This follows immediately from Lemmas 1 through 3. I

3. EDGE-MAXIMAL GRAPHS

In this section we determine all $maximal graphs.

where G is of the form We now define a class 8,, consisting of all graphs G = ( V , E ) with IVI > 10

where the G:’s are mutually disjoint units of type 1 and 2 for j = 1 and 2, respectively, E, is a pair of edges joining two degree 2 vertices of Gf to the pendant vertex of that unit, and J is a set of edges joining all pairs of vertices from some subset of V that consists of exactly two nonadjacent vertices from each G j and the pendant vertex from each G; (see Figure 11).

We present our final theorem.

mmG!!!!!B Fgc F; F;

FIGURE 10. The $-maximal graphs on 5 and 10 vertices.


FIGURE 11. A graph G in 3”

Theorem 4. 9* U {F*}:=, is the collection of all $maximal graphs.

Proof. If G’ E 8* U {F*}:=,, then it is straightforward (with the help of Lemma 1) to check that G’ is $-maximal. So let G be any $-maximal graph. All such graphs on 5 and 10 vertices have been generated by computer (D. Apple- gate [ l ] (See Figure lo.)), so we assume G has at least 15 vertices. By Theorem 3, G has a spanning subgraph H in 9,. Let 0 be the set of all vertices that have neighbors in G, outside of the unit containing them in H.

By Lemma l(iv), each unit of H contains at most two elements of 0. If U = (V,,, E,,) is a unit of H containing u, , u2 in 0, then by Lemma l(iv) U is a type 2 unit and u1 and u2 are nonadjacent. Also by Lemma l(ii) we could dominate G in less than SlVl vertices if for i = 1 or 2 we can dominate V, - ui with a single vertex.

Thus G(V,,) c U + uIu2. Suppose U fl 0 = {u}. Again S = V,, - u cannot be dominated by a single vertex and so since G(S) has a path of length 3 as a subgraph, we must have G(S) C C,. Furthermore, u can have degree at most 3 in U, thus U is isomorphic to a subgraph of the graph in Figure 12.

The statements above imply the existence of some G’ in 8* such that G c G’. By maximality G = G’. I

ACKNOWLEDGMENTS

The authors would like to thank Professor W. R. Pulleyblank for his help with improving this paper and David Applegate for his corrections to our list of edge-maximal graphs on 10 vertices. The authors would also like to thank the referees for their careful reading of the paper and for excellent suggestions. The second author would like to thank Professor E. J. Cockayne for encouragement and an introduction to domination.

FIGURE 12.


REFERENCES

111 D. Applegate, private communication. 121 B. Gamble, private communication. [3] 0. Ore, Theory of Graphs. American Mathematics Society, Colloquium

Publications Vol. 38, American Mathematics Society, Providence, RI (1962).

[4] C. Payan, Sur le nombre d’absorption d’un graphe simple. Cahier du C.E.R.O. 17(2,3,4) (1975).

[5 ] C . Payan, Sur quelques probltmes de couverture et de couplage en cornbi- natoire. These Grenoble (1977).

[6] E. J. Cockayne, B. Gamble, and B. Shepherd, An upper bound for the k- domination Number of a Graph. J . Graph Theory 9 (1985) 533-534.

domination in graphs with minimum degree two

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