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An Induced Subgraph Characterization of Domination Perfect Graphs lgor E. Zverovich Vadim E. Zverovich DEPARTMENT OF MECHANICS AND MATHEMATICS BELARUS STATE UNIVERSITY REPUBLIC OF BELARUS E-mail: vadimQmmf.bsu.minsk. by
ABSTRACT Let y(G) and L(G)be the domination number and independent domination number of a graph G, respectively. A graph G is called domination perfect if y ( H ) = L ( H ) , for every induced subgraph H of G. There are many results giving a partial characterization of domination perfect graphs. In this paper, w e present a finite induced subgraph characterization of the entire class of domination perfect graphs. The list of forbidden subgraphs in the characterization consists of 17 minimal domination imperfect graphs. Moreover, the dominating set and independent dominating set problems are shown to be both NP-complete on some classes of graphs. 0 1995 John Wiley & Sons, Inc.
1. BASIC TERMINOLOGY All graphs will be finite and undirected, without loops or multiple edges. If G is a graph, V ( G ) denotes the set, and IGl the number, of vertices in G. We will denote the neighborhood of a vertex x by N ( x ) . More generally, N(X)= N ( x ) for X V ( G ) .We will write x l X (x '4 X) to indicate that a vertex x is adjacent to all vertices (no vertex) of X C V ( G ) . For a set of vertices X, (X) denotes the subgraph of G induced by X . N ( X ) U X. In If X, Y are subsets of V ( G ) ,then X dominates Y if Y particular, if X dominates V ( G ) ,then X is called a dominating set of G . An independent dominating set is a vertex subset that is both independent and dominating, or equivalently, is maximal independent. The domination
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Journal of Graph Theory, Vol. 20, No. 3, 375-395 (1995) CCC 0364-9024/95/030375-21 0 1995 John Wiley & Sons, Inc.
376 JOURNAL OF GRAPH THEORY
number y ( G ) is the minimum cardinality taken over all dominating sets of G, and the independent domination number L ( G )is the minimum cardinality taken over all maximal independent sets of vertices of G. Sumner and Moore [9] define a graph G to be domination perfect if y ( H ) = L ( H ) , for every induced subgraph H of G. A graph G is called minimal domination imperfect if G is not domination perfect and y ( H ) = L ( H ) ,for every proper induced subgraph H of G. In all our figures a dotted line will always mean that the corresponding edge may or may not be a part of the depicted graph. 2. SUMMARY OF RESULTS ON DOMINATION PERFECT GRAPHS
In this section we give a short summary of the known results concerning domination perfect graphs. Allan and Laskar [ l ] established that K,,3-free graphs are domination perfect:
Theorem 1 (Allan and Laskar). If G has no induced subgraph isomorphic to K I , l , then y ( G ) = L ( G ) . This theorem implies that line graphs are domination perfect (proved independently by Gupta, see Theorem 10.5 in [4]) and that middle graphs are domination perfect. Theorem 1 improves also the result of Mitchell and Hedetniemi [8] that the line graph L ( T ) of a tree T satisfies y ( L ( T ) )= L ( L ( T ) ) .A characterization of trees with y ( T ) = L ( T ) by Harary and Livingston [5] can be mentioned here. The following results are due to Sumner and Moore [9,10].
Theorem 2 (Sumner and Moore). A graph G is domination perfect iff y ( H ) = L ( H )for every induced subgraph H of G with y ( H ) = 2. Theorem 3 (Sumner and Moore). If G is chordal, then G is domination perfect iff G does not contain an induced subgraph isomorphic to the graph GI in Figure 3 . Let A
=
{ H : 1HI
5
8, y ( H ) = 2, L ( H ) > 2).
Theorem 4 (Sumner and Moore). If G does not contain any member of A as an induced subgraph, and also does not contain an induced copy of the graph S in Figure 1, then G is domination perfect. In connection with Theorem 4 we make the following remark. The list of forbidden subgraphs in the original version of this theorem in [ 101 consists of A and the graphs S and S\uu. However, the last graph is superfluous, since (S\uu)\{x,y} (see Fig. 1) is isomorphic to G3 in Figure 3 and G3 E A.
DOMINATION PERFECT GRAPHS 377
Y FIGURE 1.
Graph S.
Besides chordal graphs, Sumner and Moore [9,10] characterized planar domination perfect graphs.
Theorem 5 (Sumner and Moore). A planar graph is domination perfect iff it does not contain any graph from A as an induced subgraph. Bollobiis and Cockayne [2] generalized the result of Allan and Laskar (Theorem 1) as follows:
Theorem 6 (Bollobiis and Cockayne). If G has no induced subgraph isomorphic to K l , k ( k 2 3), then L ( G )5 y ( G ) ( k - 2) - ( k - 3 ) .
(1)
Zverovich and Zverovich proved in [12] that inequality (1) is actually true for a wider class of graphs, and found the first 4 minimal domination imperfect graphs of order 6.
Theorem 7 (Zverovich and Zverovich). Suppose G does not contain two induced subgraphs K l , k ( k 2 3 ) having different centers and exactly one edge in common. Then inequality (1) holds. For the case k
=
3 , we have the following result.
Corollary 1. If a graph G does not contain the graphs G I - G4 in Figure 3 and T I ,T2 in Figure 2 as induced subgraphs, then G is domination perfect.
*I P T 2 FIGURE 2.
ux J2 Graphs T I , T 2and U 1 , U z .
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It may be pointed out that Theorem 1 and Theorem 3 follow directly from Corollary 1. The following theorem corrects the formulation of Theorem 3 in [ 121 and gives a characterization of triangle-free domination perfect graphs.
Theorem 8 (Zverovich and Zverovich). A triangle-free graph G is domination perfect iff G does not contain any of the graphs G I - G4 in Figure 3 as an induced subgraph. The fifth and sixth minimal domination imperfect graphs Gs and G6 of order 7 were found by Fulman in [3].
Theorem 9 (Fulman). If a graph G does not contain the graphs G1 - G6 in Figure 3 and U1,U2 in Figure 2 as induced subgraphs, then G is domination perfect. Finally Topp and Volkmann [ll] distinguished the 13 minimal domination imperfect graphs from among the graphs in the family A (Theorems 4, 5).
Theorem 10 (Topp and Volkmann). If a graph G does not contain any of the graphs G I - G13in Figure 3 and S in Figure 1 as an induced subgraph, then G is domination perfect. Note that the original version of this theorem in [ 111 was stated with two superfluous graphs. In terms of [I I], the graph HS contains the induced H2 and the graph H6 contains the induced H3, i.e., the graphs H5 and H6 are not minimal domination imperfect graphs.
3. CHARACTERIZATION OF DOMINATION PERFECT GRAPHS Sumner [lo] supposed that it is impossible to provide a finite forbidden characterization of the entire class of domination perfect graphs. The following theorem gives a characterization of domination perfect graphs and shows that there are 17 minimal domination imperfect graphs only.
Theorem 11. A graph G is domination perfect if and only if G does not contain any of the graphs G I - GI7in Figure 3 as an induced subgraph. Proof. The necessity follows from the fact that y ( G i ) = 2 and L ( G ~=) 3 for 1 Ii 5 17. To prove the sufficiency, let F be a minimum counterexample, i.e., y ( F ) < L ( F ) ,the graph F does not contain any of the induced subgraphs G I - G17 in Figure 3, and F has a minimum order. Let D be a minimum dominating set of F such that the number of edges in ( D ) is minimum. Since ID1 = y ( F ) < L ( F )and D dominates F , it follows that
DOMINATION PERFECT GRAPHS 379
7
. J . I'
'3
'6
'4
mmm
'5
G7
'8
'10
'9
G12
'I1
I' 3
'I4
'I6
'I5
'I7 FIGURE 3. Minimal domination imperfect graphs G, - G17 the set D is dependent and hence contains some edge u u . Let
A
=
{a E V(F)D : N ( a ) fl D
= {u}},
B
=
{b E V ( F ) V ) : N ( b ) f l D
=
C
=
{c E V ( F ) V ): N ( c ) n D
= { u , v}}.
We need the following lemmas.
{v}},
380 JOURNAL OF GRAPH THEORY
Lemma 1. L ( F )> 2 and V ( F ) = A U B U C U { u , ~ } . Proof. We first prove that L ( H ) > 2, where H = (A U B U C U { u , ~ } ) Indeed, . if L ( H )= 1 and x is a dominating vertex of H , then D is not a minimum dominating set for F , since the set D\{u,u} U { x } dominates F . If L ( H ) = 2 and {x,y } is an independent dominating set of H , then D\{u, v} U { x , y } is a minimum dominating set of F whose induced subgraph contains fewer edges than ( D ) , contrary to hypothesis. Thus y ( H ) = 2, L ( H ) > 2 and H does not contain any of the induced subgraphs GI - GI7. Since F is a minimum counterexample, we obtain F=H. I Any of the graphs GI - G17 in Figure 3 can be represented in the form of Figure 4, since y(Gi) = 2 for all 1 5 i 5 17. For uniformity, the graph Gi in such a representation will be cited as follows: ( u , vertices of A; u , vertices of B ; vertices of C).
Lemma 2. Let a , b E A and c , d E B be arbitrary vertices. (i) If a Y c , then { a , c} dominates the graph (A U B ) ; (ii) If a Y b and c 4' d , then ( a , b, c, d ) = 2K2; (iii) The graph ( N ( a ) f l B ) ( ( N ( c ) r l A ) ) is complete.
Proof. (i) Assume to the contrary that { a , c } does not dominate ( A U B ) . Then there exists a vertex s E A U B such that s Y { a , c}. Without loss of generality, let s E A. By Lemma 1, the set { u , c } does not dominate F , hence there is a vertex t E B such that t Y c. Now consider the graph ( u , a , s; u , c , t ) shown in Figure 5(a). It is easy to see that, depending on the existence of dotted edges, this graph is isomorphic to G1,G2,or G3, a contradiction. (ii) Assume to the contrary that ( a ,b , c , d ) 2K2. Then, if ( a ,b , c , d ) contains 0,1, or 2 edges, we have a contradiction to (i). Therefore ( u , b , c , d ) = P4 or C4. If ( a , b , c , d ) z Cq then ( u , a , b ; v , c , d ) = G4, a contradiction. Thus ( a , b , c , d ) = P4, and let b Y c. By Lemma 1 , the
+
U
V
FIGURE 4. Graph F.
DOMINATION PERFECT GRAPHS
381
set { b , c } does not dominate F , i.e., there is a vertex f E V ( F ) such that f 4' { b ,c}. We further know that { b ,c } dominates ( A U B), so f E C. The resulting graph L is shown in Figure 5 (b). I f f Y { a , d } then L\u = G2. If f Y a and f l d ( f l u and f '4 d ) , then L\u = G3 (L\u G3). At last, if f l { a , d } then L = G6. In all cases a contradiction is obtained. (iii) Suppose that a l { c , d } and c '4 d. By Lemma 1, since { a , u } does not dominate F , there is a vertex b E A such that b Y a . Then ( a , b , c , d ) S 2K2, contrary to Lemma 2 (ii). Consequently, c l d . I
Lemma 3. Let a , b E A, c , d E B and f E C. If a Y b and c Y d , then f is adjacent to at least two vertices of the set { a , b , c , d } . Proof. Suppose to the contrary that f is adjacent to at most one vertex from the set { a ,b , c , d}. Without loss of generality, by Lemma 2 (ii), alc,bldandaYd,b Yc.IffY{a,b,c,d},then(u,a,b;u,c,d;~)~ G 5 , a contradiction. Hence f is adjacent to exactly one vertex from { a , b , c , d } . Without loss of generality, let f l b . By Lemma 1, the set { b , c } does not dominate F , i.e., there is a vertex g E V ( F ) such that g Y {b,c}. By Lemma 2 (i), g E C . If g Y { a , d } , then ( u , a , b ; u , c , d ; g ) = G=,, a contradiction. The resulting graph M is shown in Figure 6. We have
All cases yield a contradiction.
I
Lemma4. L e t a , b E A , c , d E B a n d f , g E C . I f ( a , b , c , d , f ) i s P 5 with the path ( a - c - f - b - d ) and g Y { b , c } , then g l { a , d } and g Y f (see Fig. 8).
f
U
V
U
FIGURE 5
V
382 JOURNAL OF GRAPH THEORY
€5
f b
d
a
C
V
U
FIGURE 6. Graph M
Proof. Apply Lemma 3 to the set { a , b , c , d } and the vertex g. Since g Y {b,c}, we have gl_{a,d}. If now g l f , then ( u , a , b ; u , c , d ; f , g ) is isomorphic to G13, a contradiction. Consequently, g Y f . I Lemma 5. Let a , b E A, c , d E B, f , g E C and h E V(F). If ( a , b , c , d , f , g ) is the graph of Figure 8 and h Y c f , g } , then only four cases are possible: (1) h E (2) h E (3) h E (4) h E
A , h l { a , b , c } ,h A, h l { a , b , d } , h B, h l { c , d , a } ,h B , h l { c , d , b}, h
Y Y Y Y
d;
c; b; a.
Proof. Let us assume that h E C . By Lemma 3 , h is adjacent to at least two vertices of { a ,b , c , d}. Taking symmetry into account, we have the 5 possibilities shown in Figure 7. In what follows, for simplicity, we will not draw the vertices u and u . It can be checked directly that (V(N,)\{c,g} U { u } ) G3, ( V ( N 2 )U {U, GI43 (V(N3) U {u,.I> GI52 ( V W 4 ) \ U 9 d } U { u } ) G3, f
h
f
g
h
f
g
b
d
b
d b
a
c
a
c a
NI
N2
g
N3
h
g
f
h
g
a b
d b
d
c a
c a
C
N4
FIGURE 7 . Graphs N , - N s .
B5
DOMINATION PERFECT GRAPHS 383
( V ( N 5 ) U { u , u } ) = GIG.Thus h E A U B. Now consider the case h E A . By Lemma 2 (i), the vertex h must be adjacent to at least one vertex of { c , d } .
Case 1. Suppose h l c . By Lemma 2 (iii), we have h Y d and h l a . Suppose that h Y b. Then by Lemma 3, g must be adjacent to at least two vertices of { h , b , c , d } . This is a contradiction, since g Y {h,b,c}. Consequently, h lb. Case 2. Suppose h l d . By Lemma 2 (iii), we have h Y c and h l b . Suppose that h Y a. Then by Lemma 3, f must be adjacent to at least two vertices of { a ,h , c , d}. This is a contradiction, since f 4' { a ,h, d}. Consequently, h I a . The case h E B is similar. I
Now we go on with the proof of Theorem 11. By Lemma 1, there exist vertices a , b E A and c , d E B such that a Y b and c Y d . By Lemma 2 (ii), ( a , b , c , d ) 2K2. Without loss of generality, let a l c and b l d . In accordance with Lemma 1, the set { a , d } does not dominate F . Therefore, there is a vertex f E V ( F ) such that f Y { a , d } . By Lemma 2 (i), f E C . Then f l { b , c } by Lemma 3 . Now consider the vertices b and c. Again by Lemma 1, there is a vertex g E V ( F ) such that g Y { b , c } . By Lemma 2 (i), we have g E C and by Lemma 4, g l { a , d } and g Y f . The resulting graph is shown in Figure 8. Consider the set { f , g } . By Lemma I , there exists a vertex h E V ( F )such that h Y { f , g}. In accordance with Lemma 5 , there are four cases. These cases are symmetrical, so we can assume that h E A and h I { a , 6 , c},h '4 d. By Lemma 1, since {h,u } does not dominate F , there exists a vertex k E A such that k Y h. By Lemma 2 (ii), ( h ,k , c , d ) 2K2, hence k l d and k Y c. Then k l b by Lemma 2 (iii). For the set { h , k , c , d } and f , g , we can apply Lemma 3, which produces k l { f , g}. Thus, we obtain the graph of Figure 9 (a). In accordance with Lemma 1, the set { k , c } does not dominate F . Therefore, there is a vertex Z € V ( F ) such that I Y { k , c}. By Lemma 2 (i), 1 E C. Applying Lemma 4 to the set { h ,k , c , d , f } and the vertex 1, we obtain Z l { h , d } and 1 Y f.Now apply Lemma 5. Assume that 1 Y g . Then
FIGURE 8
384 JOURNAL OF GRAPH THEORY f
f
{ h , k , c , d , f,1 } and the vertex g satisfy the hypotheses of Lemma 5. By the C , a contradiction. Thus, 1 lg. conclusion of Lemma 5 , we have g Suppose that 1 Y a . By applying Lemma 3 to the set { a ,b , c , d } and 1, we have l l b . Since 1 is adjacent to only one vertex of { a , k , c , d } , it follows that k l a , for otherwise we have a contradiction to Lemma 3. Thus, 1 Y a , l l b and k l a , hence ( u , a , b , h , k ; u , c , d ; f,g,Z) = G17, a contradiction. Therefore 1l a . Thus, we have a collection of graphs indicated in Figure 9 (b). By Lemma 1, the set { l , f } does not dominate F , so there exists a vertex m E V ( F ) such that m Y (1, f}. Now we can state the last lemma.
Lemma 6. Let a , b , h , k E A , c , d E B , f,g, 1 E C and m E V ( F ) .If ( a , h , 6 , k , c , d , f,g,I) is one of the graphs of Figure 9 (b) and m Y { f,Z}, then
1) m E A , m l { a ,b , h , k , d , g} and m Y c ; 2 ) b l l and a Y k . Proof. The set {h,k , c , d , f , I } and the vertex m satisfy the hypotheses of Lemma 5 , hence there are four cases to consider. Case I. m E A and m l { h , k , c } , m Y d . By Lemma 2 (iii), m l a . Assume that m Y 6 . Applying Lemma 3 to the set {m,b , c , d } and I, g,we have b l l and g l m . Then ( u , m , h , b , k ; u , c , d ; f,g,l) G17, a contradiction. Consequently, m l b . Further b l l , for otherwise ( h , I , c ; 6 , d , f ;m , u ) = G I ~a, contradiction. The resulting graph is shown in Figure 10 (a). By Lemma 1, the set {m, u } does not dominate F . Hence there is a vertex n E A such that n Y m. Then by Lemma 2 (ii), n l d and by Lemma 2 (iii), n l b . Now we can apply Lemma 3 to {m,n , c , d } and f,1, which produces n l { f,1}. We have (1, a , u ; b , m, f ; d , n ) Glo or G I depending ~ on the existence of an, a contradiction.
DOMINATION PERFECT GRAPHS 385
Case 2. m E A and m l { h , k , d}, m Y c. By Lemma 2 (iii), m l b . Assume that m 4' a. Then { a ,m, c , d } and f satisfy the hypotheses of Lemma 3 and f '4 { a , m, d}, contrary to the conclusion of Lemma 3. Hence m l a . The resulting graph is shown in Figure 10 (b). Furthermore alk b Y 1
-
( k , f , d ; a , c , l ; m , u ) G GI2
=j
3
a Yk,
G12 * b l l .
( b ,f , d ; h , c , l ; m , u )
By Lemma 1, there exists a vertex n E A such that n Y m. By Lemma 2 (ii), n l c and by Lemma 2 (iii), n l a . Applying Lemma 3 to { n , m , c , d } and 1, we have l l n . At last mY g
=j
(I, b , v ; a , m , c ; g , n )
G7,Gg,Glo or GI2
*mlg.
Thus, the conclusions of Lemma 6 are satisfied. Case 3. m E B and m l { c , d , h } , m Y k . Suppose that m Y a. By Lemma 2 (i), the set { m , a } dominates (A U B), hence a l k . We have ( a , k , 1; c , f , m ) G2, a contradiction. Therefore m l a . Then Lemma 2 (iii) implies m Y b , since a Y b. Further
a Y k, b Y 1 =j ( b , h , f ; d , m , l ) z G3 * b l l , alk
3
m Y g
( a , l , m ; k , d ,f )
G
* ( b , h ,f ; d , m , g )
G3
G2
=j
mlg
By Lemma 1, there is a vertex n E B such that n Y m. By Lemma 2 (ii), ( h ,k , m , n ) = 2K2, so n l k and n Y h. By Lemma 2 (iii), n l d . If n l a , then we have a contradiction to Lemma 2 (iii), since n l k and a Y k . f
f
d
d
C
1
C
1 (a)
g (b)
FIGURE 10. Illustration of the cases 1 and 2.
386 JOURNAL OF GRAPH THEORY Hence n 4' a . The set { b ,m } dominates (A U B ) by Lemma 2 (i), and so n l b. For the set {h,k , m, n } and 1, we can apply Lemma 3, which produces n l l . Assume that n Y g. Then { k ,h, m, n , I } and g satisfy the hypotheses of Lemma 4. Therefore g Y I, a contradiction. Consequently, n l g . The resulting graph is shown in Figure 11. By Lemma 1, the set { b , m } does not dominate F . Therefore, there is a vertex p Y { b , m } and p E C by Lemma 2 (i). Applying Lemma 3 to { a ,b , m, n } and p , we obtain p l { a , n}. Furthermore P Y g p Y 1
-
* (u,b,p;g,n,m)
G G3
( v , m , p ; l , a , b ) = G3
*pig,
*p l l ,
p+'f *(1,b,a;u,f,m;p,d)~G~gorG12*plf, p l d * ( l , b , a ; v , f , m ; p , d ) = G13 * p Y d . Now apply Lemma 3 to { a ,b , c , d } and p . We have p l c . Assume that p '4 h. The set { h ,k , c , d , I} and p satisfy the hypotheses of Lemma 4,and so p Y I , a contradiction. Thus p l h . Finally
f
1
43
FIGURE 11. Illustration of the case 3.
DOMINATION PERFECT GRAPHS 387
FIGURE 12. Illustration of the case 4
Case 4. m E B and m l { c , d , k } , m Y h. If m l { a , b}, contrary to Lemma 2 (iii). If m Y { a ,b}, contrary to Lemma 2 (i). If m Y a and m l b , then l l b , for otherwise ( b ,f , m ; h, c, 1 ) = G?. We have (v,f , m ; I, b , a ) G3, a contradiction. Thus m l a and m Y b. Since m l { a , k } , by Lemma 2 (iii), k l a . Now
* ( b ,f , h ; d , g , m )
GI * m l g , bYl*(c,f,m; h,b,l)=G2-bll.
rn Y g
The resulting graph is a subgraph of Figure 12. By Lemma 1, there is a vertex n E B such that n Y m. For the sets { a , b , m, n } and { h ,k , m, n} we can apply Lemma 2 (ii), which produces n l { b ,h } and n 4' { a ,k}. Then by Lemma 2 (iii), n l { c , d}. The application of Lemma 3 to the set { h , k , n , m } and the vertices 1 and f gives n l { l ,f}. Further
The resulting graph is a subgraph of Figure 12. In accordance with Lemma 1, the set { k , n } does not dominate F . Therefore, there is a vertex p E V ( F )such that p Y { k , n}. By Lemma 2 (i),
388 JOURNAL OF GRAPH THEORY
p E C . The application of Lemma 4 to { h ,k , n, m , f } and p gives p l { h , m } and p Y f . Suppose that Z Y p . Then { h ,k , n , m , f , p } and Z satisfy the hypotheses of Lemma 5 , and hence 1 @ C, a contradiction. Thus Z l p . Analogously, g l p . Further, p Y a implies ( h , p , a ; n , d , f ) G I or G I , hence p l a . Assume that p Y d. The application of Lemma 3 to {h,k , c , d } and p gives p l c , and we have ( v , d , m , c , n ; u , k , h ; f , Z , p ) G u , a contradiction. Hence p l d . The resulting graph is shown in Figure 12. By Lemma 1, the set { f , p } does not dominate F , and hence there is a vertex r E V ( F )such that r Y { f , p } . The set { h ,k , n , m,f, p } and r satisfy the hypotheses of Lemma 5. Therefore, there are four cases to consider. Subcase 4.1. r E A and r l { h , k , n}, r Y m. By Lemma 2 (iii), r l b , since n l { b , r } . If r Y a , contrary to Lemma 3 , since f is adjacent to one vertex of { a , r , m , n } . Hence r l a . If r Y d , then ( d , n , p ; k , f , r ) G3, a contradiction. Therefore r l d . Now r Y c, for otherwise r l { c , d } and c Y d , contrary to Lemma 2 (iii). We have p Y c a ( c , m , f;h , p , r ) r Y g ( p , c , d ; u,f , r ; g , a )
G2 Gg
*plc, rlg
By Lemma I , the vertex r does not dominate (A), and so there is a vertex s E A such that s Y r . By Lemma 2 (ii), s l { c , m } and s Y n. Then s l { h , k } by Lemma 2 (iii), since c l { h , s } and m l { k , s } . Consider the induced subgraph shown in Figure 13. Applying Lemma 3 to { r ,s, n , m } and f,p ,
g
P
FIGURE 13. Illustration of the subcase 4.1
DOMINATION PERFECT GRAPHS 389
we get sl{ f , p } . At last s l g =+ ( g , r , v ; s , h , f ; m , k ) z G12
*s Y g
and
(u,h,r,k,s; v,n,m; f , p , g )
=GI~,
a contradiction. Subcase 4.2. r E A and r l { h , k , m } , r Y n. We have ( h , p , r ; n , d , f ) = G2 or G3 depending on the existence of the edge rd, a contradiction. Subcase 4.3. r E B and r l { n ,m, h}, r Y k. Since f is adjacent to only one vertex of { h ,k , r , d } , it follows by Lemma 3 that r l d . We obtain ( d , p , r ; k , a , f ) G2 or G3 depending on the existence of the edge ra, a contradiction. Subcase 4.4. r E B and r l { n , m , k}, r Y h . Since k l { d , r } , it follows by Lemma 2 (iii) that r l d . We have ( d , p , k ; n , h , f ; r , v ) G12. This contradiction completes the consideration of the case 4, and with it the proof of Lemma 6. I Thus, applying Lemma 6 to the graph of Figure 9 (b) and m , we obtain the graph of Figure 14 without the vertex n. By Lemma 1, since m does not dominate ( A ) , there is a vertex n E A such that n Y m. By Lemma 2 (ii), n l c and n '4 d . Since c l { n , a , h}, it follows by Lemma 2 (iii) that n l { a ,h } . Apply Lemma 3 to {n,m, c, d } and f , l . We have n l u , l } . Further n Y b
nlg
3
* (b,l,m;f,u,n)
(l,b,v;a,m,c; g,n)
FIGURE 14
G3 G13
nlb, 3
n Y g.
390 JOURNAL OF GRAPH THEORY The resulting graph is shown in Figure 14. By Lemma 1, the set { n , d } does not dominate F . Hence there is a vertex p E V ( F )such that p Y { n , d } . By Lemma 2 (i), p E C . Applying Lemma 4 to {n,m, c , d , l } and p , we obtain p I { m , c } and p Y 1. Suppose that p Y f and apply Lemma 5 to {n,rn, c , d , p , I } and f .We have f @ C , a contradiction. Therefore p If.Assume that p Y k and consider the induced subgraph shown in Figure 15 (a). If n Y k , contrary to Lemma 3 , since p is adjacent to one vertex of { n ,k , c, d } . Applying Lemma 3 to {h,k , c , d } and p , we get p l h . Now ( u , k , m , h , n ; v , d , c ; I, f , p ) = G17, a contradiction. Thus p l k . The set { 1 , p } dominates the resulting graph. By Lemma 1, there is a vertex r E V ( F ) such that r Y { I , p } . Consider the subgraph shown in Figure 15 (b). Clearly that this graph and the vertex r satisfy the hypotheses of Lemma 6. By the conclusions of Lemma 6, r E A, r l { k , r n , h , n , c , f}, r '4 d and h l p , k Y n. Replace the vertex h in Figure 15 (b) by the vertex a and apply Lemma 6 again. In addition we obtain a l { p , r}. Now
If r Y b , then we have a contradiction to Lemma 3 , since p is adjacent to one vertex of { r , b , c , d}. The resulting graph is the graph W\s in Figure 16. By Lemma 1, the vertex r does not dominate (A), and hence there is a vertex s E A such that s Y r . By Lemma 2 (ii), s l d and s Y c . By Lemma 2 (iii), s l { b , k , r n } . Applying Lemma 3 to { r , s , c , d } and p , g , I ,
P
f
f
P
1
1
(a)
(b)
FIGURE 15
DOMINATION PERFECT GRAPHS 391
we get s l { p ,g , l}. We have
slf
( a , p , r ;I , u , s )
s+ a
3
sY h
* ( h ,p , r ; I, u , s)
* ( k , d , r ;p , u , a ; f , s )
= G3 =+
sla,
*slh, = G13 * s Y f . G3
The only edge undetermined is sn. It is straightforward to see that the graph ( n , r , m, s, c , d , p , 1 ) is isomorphic to the graph of Figure 9(a) with the correspondence of partition into the parts A , B and C . We have already proved that the case a l k for the graph of Figure 9(a) is impossible. Therefore we can assume that n Y s in our case. Thus, the graph F contains the induced subgraph W shown in Figure 16. For convenience, a general structure of the graph W is shown in Figure 16(a), and edges between the sets A and C are indicated in Figure 16(b). We remember that ( a ,h , n, r ) G K4 and ( b ,k , m , s) = K 4 . By Lemma 1, the set { b , p } does not dominate F , hence there is a vertex x such that x Y { b , p } .
Claim 1. x I { u , u , a , c , d , g , h, I, m, n , r } and x Y { b ,f , k , p , s}. If x E A , then we have a contradiction to Lemma 3, since p is adjacent to one vertex of { x , b , c , d } . Hence x G A and x l u . Suppose that x Y 1. Apply Lemma 6 to ( k , m, h , n , d , c , I, f , p ) and x . We have x E A , a contradiction. Therefore x l l . Now x '4 a
* ( a ,r , p ; 1, b , x ) = G2 or G3
xY n
3
( a , x , n ;p , k , u ) = GZ or G3
xlk
3
( k , x , b ; p , a , u ) = G3
x Y r orxls
3
( 1 , b , v ; a , r , p ; x , s)
u
-
xla, xln,
x Y k,
= G9, G L 2or G I 3 * x l r and x '4
v
(a)
FIGURE 16. Graph W.
s.
392 JOURNAL OF GRAPH THEORY
Let x E B, i.e., x Y u . If x '4 h , then { x ,h } does not dominate the vertex k , contrary to Lemma 2 (i). Hence x l h . Since x l n , we obtain x Y m by Lemma 2 (iii). We have xYd*(u,r,p; I,x,d)gG2*xxld,
* ( b , d , f ;h , x , p ; r , m ) = G
x Yf
( b ,d , h ; f , x , p ; n ) g
Ij ~ x l f , G6, a contradiction.
Let x E C, i.e., x l u . We get x Y c
* ( I , b, u ; a , m , c ; x , n )
G12
or GI3
*x l c .
Assume that x Y h . Applying Lemma 3 to { h ,k , c , d } and x , we have x l d . Now
( x , c , ~U;, f , h ) G3 * x l f , GL3,a contradiction. ( b ,d , h ; f , x , p ; r , k )
x Yf
j
Consequently, x lh. Furthermore
Thus, Claim 1 is satisfied. Now the set { g , r } dominates the resulting graph. By Lemma 1, there is a vertex y E V ( F ) such that y Y { g , r } .
Claim 2. y l { u , u , b , c , d ,f , h , k , I , n , p , s } and y Y { a , g , m , r , x } . If y E A, contrary to Lemma 3, since g is adjacent to one vertex of { r , y , c , d } . Hence y f$ A and y l u . Let y E B , i.e., y Y u. Since y Y r , it follows by Lemma 2 (i) that y l s . Now y Y n by Lemma 2 (iii). Then y l k by Lemma 2 (i) and y Y h by Lemma 2 (iii). Further
a contradiction.
DOMINATION PERFECT GRAPHS 393
Let y E C , i.e., y l u . Suppose that y Y f and apply Lemma 5 to { a , b , c , d . f , g l and y . We obtain y 4 C , a contradiction. Therefore y l f . Furthermore
Now assume that y 4' d and apply Lemma 3 to
{Y, s, c , d
} and y . We get
y I s . Further
Therefore y Id . We obtain
Thus, Claim 2 is satisfied. Now the set { g , n } dominates the resulting graph, By Lemma 1 , there is a vertex z E V ( F ) such that z 4' {g,n}. If z E A, contrary to Lemma 3 , since g is adjacent to one vertex of {n,z , c , d } . Hence y 4 A, i.e., z l u . We have zla
(a,z,n ; g, u , k )
z Y y =+ ( u , y , z ; g , k , a ) z '4 k * ( k , r , g ; y , n , z )
= G2 or G3 =+z G2 or G3
Y
a,
zly,
G2 or G3 =j z l k , ( k , r , g ; y , n, u ; s , z > = G7, G9, Glo, or GI2, a contradiction.
This contradiction completes the proof of Theorem 1 1 .
I
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4. COROLLARIES
We first note that Theorems 1, 2, 3, 4, 5 , 8, 9, and 10 and Corollary 1 follow directly from Theorem 1I.
Corollary 2. Suppose G does not contain the complements of the graphs G, - G17in Figure 3 as induced subgraphs and diam(G) > 2. Then G has an edge that does not belong to any triangle.
c
Proof. Let G be the complement of G. Clearly that does not contain the induced subgraphs G I - GI7 and y(G) = 2. By Theorem 11, we have L( G) = 2, i.e., the graph G has an edge that does not belong to any triangle.
I Now we prove that the dominating set (DS) and independent dominating set (IDS) problems are NP-complete on some classes of graphs. The DS problem is known (see [6]) to be NP-complete for 3-regualr planar graphs, for bipartite graphs and for split graphs. In contrast, the IDS problem is known to be NP-complete for general graphs only and, if arbitrary weights are allowed, for chordal graphs. We say that a graph G belongs to the class L if the following conditions hold: (i) (ii) (iii) (iv)
G G G G
is planar; is bipartite; has the maximum degree 3; has the girth g(G) 1 k , where k is fixed.
Corollary 3. The DS and IDS problems are both NP-complete on the class L .
Proof. We describe a polynomial time reduction from the DS problem for 3-regular planar graphs to the DS and IDS problems for the class L , which implies the required result. Define the operation of 3-partition of an edge as follows: replace an edge uu by the path P5 = ( u - x - y - z - u ) . Suppose H is obtained from G by 3-partition of an edge uu. Prove that y ( H ) = y(G) + 1 . Let D be a minimum dominating set of G. If { u , u } E D or { u , u } E D, then D U { y } dominates H . If u E D and u @ D , then D U { z } dominates H . The case u 4 D and u E D is similar. Therefore y ( H ) 5 y(G) + 1. Let E be a minimum dominating set of H . Clearly that S = E r l { u , x , y , z } # 0. If IS1 = 1 then S = { x } or S = (v}, and in the case S = {x} we have u E E. It is easily to see that E\S dominates G. If IS1 I2, then (E\S) U { u } dominates G. Thus y(G) Iy ( H ) - 1. Applying the operation of 3-partition to edges of a 3-regular planar graph G, we can obtain a graph with the girth at least k . The single 3-partition of each edge results in a bipartite graph. The resulting graph H is certainly
DOMINATION PERFECT GRAPHS 395
constructible in polynomial time, since the number k is fixed, and H belongs to L . Moreover, H does not contain the graphs G I - G17 in Figure 3 as induced subgraphs. By Theorem 11, L ( H ) = y ( H ) = y ( G ) + m, where m is the number of the 3-partitions of G. I
ACKNOWLEDGMENT We thank the referees for helpful suggestions.
References [ 11 R. B. Allan and R. Laskar, On domination and independent domination
numbers of a graph. Discrete Math. 23 (1978) 73-76. [2] B. Bollobis and E. J. Cockayne, Graph-theoretic parameters concerning domination, independence, and irredundance. J. Graph Theory 3 (1979) 241 -249. [3] J. Fulman, A note on the characterization of domination perfect graphs. J. Graph Theory 17 (1993) 47-51. [4] F. Harary, Graph Theory. Addison-Wesley, Reading, MA (1969). [5] F. Harary and M. Livingston, Characterization of trees with equal domination and independent domination numbers. Congr. Numer. 55 (1986) 121-150. [6] D. S. Johnson, The NP-completeness column: an ongoing guide. J. Algorithms 5 (1984) 147-160. [7] R. Laskar and H. B. Walikar, On domination related concepts in graph theory. Lecture Notes in Math. 885 (1981) 308-320. [8] S. Mitchell and S. Hedetniemi, Edge domination in trees. Congr. Numer. 19 (1977) 489-509. [9] D. Sumner and J.I. Moore, Domination perfect graphs. Notices Am. Math. Soc. 26 (1979) A-569. [lo] D. Sumner, Critical concepts in domination. Discrete Math. 86 (1990) 33-46. [ll] J. Topp and L. Volkmann, On graphs with equal domination and independent domination numbers. Discrete Math., 96 (1991) 75-80. [ 121 I. E. Zverovich and V. E. Zverovich, A characterization of domination perfect graphs. J. Graph Theory 15 (1991) 109-114.
Received August 27, 1993
ABSTRACT Let y(G) and L(G)be the domination number and independent domination number of a graph G, respectively. A graph G is called domination perfect if y ( H ) = L ( H ) , for every induced subgraph H of G. There are many results giving a partial characterization of domination perfect graphs. In this paper, w e present a finite induced subgraph characterization of the entire class of domination perfect graphs. The list of forbidden subgraphs in the characterization consists of 17 minimal domination imperfect graphs. Moreover, the dominating set and independent dominating set problems are shown to be both NP-complete on some classes of graphs. 0 1995 John Wiley & Sons, Inc.
1. BASIC TERMINOLOGY All graphs will be finite and undirected, without loops or multiple edges. If G is a graph, V ( G ) denotes the set, and IGl the number, of vertices in G. We will denote the neighborhood of a vertex x by N ( x ) . More generally, N(X)= N ( x ) for X V ( G ) .We will write x l X (x '4 X) to indicate that a vertex x is adjacent to all vertices (no vertex) of X C V ( G ) . For a set of vertices X, (X) denotes the subgraph of G induced by X . N ( X ) U X. In If X, Y are subsets of V ( G ) ,then X dominates Y if Y particular, if X dominates V ( G ) ,then X is called a dominating set of G . An independent dominating set is a vertex subset that is both independent and dominating, or equivalently, is maximal independent. The domination
UxtX
Journal of Graph Theory, Vol. 20, No. 3, 375-395 (1995) CCC 0364-9024/95/030375-21 0 1995 John Wiley & Sons, Inc.
376 JOURNAL OF GRAPH THEORY
number y ( G ) is the minimum cardinality taken over all dominating sets of G, and the independent domination number L ( G )is the minimum cardinality taken over all maximal independent sets of vertices of G. Sumner and Moore [9] define a graph G to be domination perfect if y ( H ) = L ( H ) , for every induced subgraph H of G. A graph G is called minimal domination imperfect if G is not domination perfect and y ( H ) = L ( H ) ,for every proper induced subgraph H of G. In all our figures a dotted line will always mean that the corresponding edge may or may not be a part of the depicted graph. 2. SUMMARY OF RESULTS ON DOMINATION PERFECT GRAPHS
In this section we give a short summary of the known results concerning domination perfect graphs. Allan and Laskar [ l ] established that K,,3-free graphs are domination perfect:
Theorem 1 (Allan and Laskar). If G has no induced subgraph isomorphic to K I , l , then y ( G ) = L ( G ) . This theorem implies that line graphs are domination perfect (proved independently by Gupta, see Theorem 10.5 in [4]) and that middle graphs are domination perfect. Theorem 1 improves also the result of Mitchell and Hedetniemi [8] that the line graph L ( T ) of a tree T satisfies y ( L ( T ) )= L ( L ( T ) ) .A characterization of trees with y ( T ) = L ( T ) by Harary and Livingston [5] can be mentioned here. The following results are due to Sumner and Moore [9,10].
Theorem 2 (Sumner and Moore). A graph G is domination perfect iff y ( H ) = L ( H )for every induced subgraph H of G with y ( H ) = 2. Theorem 3 (Sumner and Moore). If G is chordal, then G is domination perfect iff G does not contain an induced subgraph isomorphic to the graph GI in Figure 3 . Let A
=
{ H : 1HI
5
8, y ( H ) = 2, L ( H ) > 2).
Theorem 4 (Sumner and Moore). If G does not contain any member of A as an induced subgraph, and also does not contain an induced copy of the graph S in Figure 1, then G is domination perfect. In connection with Theorem 4 we make the following remark. The list of forbidden subgraphs in the original version of this theorem in [ 101 consists of A and the graphs S and S\uu. However, the last graph is superfluous, since (S\uu)\{x,y} (see Fig. 1) is isomorphic to G3 in Figure 3 and G3 E A.
DOMINATION PERFECT GRAPHS 377
Y FIGURE 1.
Graph S.
Besides chordal graphs, Sumner and Moore [9,10] characterized planar domination perfect graphs.
Theorem 5 (Sumner and Moore). A planar graph is domination perfect iff it does not contain any graph from A as an induced subgraph. Bollobiis and Cockayne [2] generalized the result of Allan and Laskar (Theorem 1) as follows:
Theorem 6 (Bollobiis and Cockayne). If G has no induced subgraph isomorphic to K l , k ( k 2 3), then L ( G )5 y ( G ) ( k - 2) - ( k - 3 ) .
(1)
Zverovich and Zverovich proved in [12] that inequality (1) is actually true for a wider class of graphs, and found the first 4 minimal domination imperfect graphs of order 6.
Theorem 7 (Zverovich and Zverovich). Suppose G does not contain two induced subgraphs K l , k ( k 2 3 ) having different centers and exactly one edge in common. Then inequality (1) holds. For the case k
=
3 , we have the following result.
Corollary 1. If a graph G does not contain the graphs G I - G4 in Figure 3 and T I ,T2 in Figure 2 as induced subgraphs, then G is domination perfect.
*I P T 2 FIGURE 2.
ux J2 Graphs T I , T 2and U 1 , U z .
378 JOURNAL OF GRAPH THEORY
It may be pointed out that Theorem 1 and Theorem 3 follow directly from Corollary 1. The following theorem corrects the formulation of Theorem 3 in [ 121 and gives a characterization of triangle-free domination perfect graphs.
Theorem 8 (Zverovich and Zverovich). A triangle-free graph G is domination perfect iff G does not contain any of the graphs G I - G4 in Figure 3 as an induced subgraph. The fifth and sixth minimal domination imperfect graphs Gs and G6 of order 7 were found by Fulman in [3].
Theorem 9 (Fulman). If a graph G does not contain the graphs G1 - G6 in Figure 3 and U1,U2 in Figure 2 as induced subgraphs, then G is domination perfect. Finally Topp and Volkmann [ll] distinguished the 13 minimal domination imperfect graphs from among the graphs in the family A (Theorems 4, 5).
Theorem 10 (Topp and Volkmann). If a graph G does not contain any of the graphs G I - G13in Figure 3 and S in Figure 1 as an induced subgraph, then G is domination perfect. Note that the original version of this theorem in [ 111 was stated with two superfluous graphs. In terms of [I I], the graph HS contains the induced H2 and the graph H6 contains the induced H3, i.e., the graphs H5 and H6 are not minimal domination imperfect graphs.
3. CHARACTERIZATION OF DOMINATION PERFECT GRAPHS Sumner [lo] supposed that it is impossible to provide a finite forbidden characterization of the entire class of domination perfect graphs. The following theorem gives a characterization of domination perfect graphs and shows that there are 17 minimal domination imperfect graphs only.
Theorem 11. A graph G is domination perfect if and only if G does not contain any of the graphs G I - GI7in Figure 3 as an induced subgraph. Proof. The necessity follows from the fact that y ( G i ) = 2 and L ( G ~=) 3 for 1 Ii 5 17. To prove the sufficiency, let F be a minimum counterexample, i.e., y ( F ) < L ( F ) ,the graph F does not contain any of the induced subgraphs G I - G17 in Figure 3, and F has a minimum order. Let D be a minimum dominating set of F such that the number of edges in ( D ) is minimum. Since ID1 = y ( F ) < L ( F )and D dominates F , it follows that
DOMINATION PERFECT GRAPHS 379
7
. J . I'
'3
'6
'4
mmm
'5
G7
'8
'10
'9
G12
'I1
I' 3
'I4
'I6
'I5
'I7 FIGURE 3. Minimal domination imperfect graphs G, - G17 the set D is dependent and hence contains some edge u u . Let
A
=
{a E V(F)D : N ( a ) fl D
= {u}},
B
=
{b E V ( F ) V ) : N ( b ) f l D
=
C
=
{c E V ( F ) V ): N ( c ) n D
= { u , v}}.
We need the following lemmas.
{v}},
380 JOURNAL OF GRAPH THEORY
Lemma 1. L ( F )> 2 and V ( F ) = A U B U C U { u , ~ } . Proof. We first prove that L ( H ) > 2, where H = (A U B U C U { u , ~ } ) Indeed, . if L ( H )= 1 and x is a dominating vertex of H , then D is not a minimum dominating set for F , since the set D\{u,u} U { x } dominates F . If L ( H ) = 2 and {x,y } is an independent dominating set of H , then D\{u, v} U { x , y } is a minimum dominating set of F whose induced subgraph contains fewer edges than ( D ) , contrary to hypothesis. Thus y ( H ) = 2, L ( H ) > 2 and H does not contain any of the induced subgraphs GI - GI7. Since F is a minimum counterexample, we obtain F=H. I Any of the graphs GI - G17 in Figure 3 can be represented in the form of Figure 4, since y(Gi) = 2 for all 1 5 i 5 17. For uniformity, the graph Gi in such a representation will be cited as follows: ( u , vertices of A; u , vertices of B ; vertices of C).
Lemma 2. Let a , b E A and c , d E B be arbitrary vertices. (i) If a Y c , then { a , c} dominates the graph (A U B ) ; (ii) If a Y b and c 4' d , then ( a , b, c, d ) = 2K2; (iii) The graph ( N ( a ) f l B ) ( ( N ( c ) r l A ) ) is complete.
Proof. (i) Assume to the contrary that { a , c } does not dominate ( A U B ) . Then there exists a vertex s E A U B such that s Y { a , c}. Without loss of generality, let s E A. By Lemma 1, the set { u , c } does not dominate F , hence there is a vertex t E B such that t Y c. Now consider the graph ( u , a , s; u , c , t ) shown in Figure 5(a). It is easy to see that, depending on the existence of dotted edges, this graph is isomorphic to G1,G2,or G3, a contradiction. (ii) Assume to the contrary that ( a ,b , c , d ) 2K2. Then, if ( a ,b , c , d ) contains 0,1, or 2 edges, we have a contradiction to (i). Therefore ( u , b , c , d ) = P4 or C4. If ( a , b , c , d ) z Cq then ( u , a , b ; v , c , d ) = G4, a contradiction. Thus ( a , b , c , d ) = P4, and let b Y c. By Lemma 1 , the
+
U
V
FIGURE 4. Graph F.
DOMINATION PERFECT GRAPHS
381
set { b , c } does not dominate F , i.e., there is a vertex f E V ( F ) such that f 4' { b ,c}. We further know that { b ,c } dominates ( A U B), so f E C. The resulting graph L is shown in Figure 5 (b). I f f Y { a , d } then L\u = G2. If f Y a and f l d ( f l u and f '4 d ) , then L\u = G3 (L\u G3). At last, if f l { a , d } then L = G6. In all cases a contradiction is obtained. (iii) Suppose that a l { c , d } and c '4 d. By Lemma 1, since { a , u } does not dominate F , there is a vertex b E A such that b Y a . Then ( a , b , c , d ) S 2K2, contrary to Lemma 2 (ii). Consequently, c l d . I
Lemma 3. Let a , b E A, c , d E B and f E C. If a Y b and c Y d , then f is adjacent to at least two vertices of the set { a , b , c , d } . Proof. Suppose to the contrary that f is adjacent to at most one vertex from the set { a ,b , c , d}. Without loss of generality, by Lemma 2 (ii), alc,bldandaYd,b Yc.IffY{a,b,c,d},then(u,a,b;u,c,d;~)~ G 5 , a contradiction. Hence f is adjacent to exactly one vertex from { a , b , c , d } . Without loss of generality, let f l b . By Lemma 1, the set { b , c } does not dominate F , i.e., there is a vertex g E V ( F ) such that g Y {b,c}. By Lemma 2 (i), g E C . If g Y { a , d } , then ( u , a , b ; u , c , d ; g ) = G=,, a contradiction. The resulting graph M is shown in Figure 6. We have
All cases yield a contradiction.
I
Lemma4. L e t a , b E A , c , d E B a n d f , g E C . I f ( a , b , c , d , f ) i s P 5 with the path ( a - c - f - b - d ) and g Y { b , c } , then g l { a , d } and g Y f (see Fig. 8).
f
U
V
U
FIGURE 5
V
382 JOURNAL OF GRAPH THEORY
€5
f b
d
a
C
V
U
FIGURE 6. Graph M
Proof. Apply Lemma 3 to the set { a , b , c , d } and the vertex g. Since g Y {b,c}, we have gl_{a,d}. If now g l f , then ( u , a , b ; u , c , d ; f , g ) is isomorphic to G13, a contradiction. Consequently, g Y f . I Lemma 5. Let a , b E A, c , d E B, f , g E C and h E V(F). If ( a , b , c , d , f , g ) is the graph of Figure 8 and h Y c f , g } , then only four cases are possible: (1) h E (2) h E (3) h E (4) h E
A , h l { a , b , c } ,h A, h l { a , b , d } , h B, h l { c , d , a } ,h B , h l { c , d , b}, h
Y Y Y Y
d;
c; b; a.
Proof. Let us assume that h E C . By Lemma 3 , h is adjacent to at least two vertices of { a ,b , c , d}. Taking symmetry into account, we have the 5 possibilities shown in Figure 7. In what follows, for simplicity, we will not draw the vertices u and u . It can be checked directly that (V(N,)\{c,g} U { u } ) G3, ( V ( N 2 )U {U, GI43 (V(N3) U {u,.I> GI52 ( V W 4 ) \ U 9 d } U { u } ) G3, f
h
f
g
h
f
g
b
d
b
d b
a
c
a
c a
NI
N2
g
N3
h
g
f
h
g
a b
d b
d
c a
c a
C
N4
FIGURE 7 . Graphs N , - N s .
B5
DOMINATION PERFECT GRAPHS 383
( V ( N 5 ) U { u , u } ) = GIG.Thus h E A U B. Now consider the case h E A . By Lemma 2 (i), the vertex h must be adjacent to at least one vertex of { c , d } .
Case 1. Suppose h l c . By Lemma 2 (iii), we have h Y d and h l a . Suppose that h Y b. Then by Lemma 3, g must be adjacent to at least two vertices of { h , b , c , d } . This is a contradiction, since g Y {h,b,c}. Consequently, h lb. Case 2. Suppose h l d . By Lemma 2 (iii), we have h Y c and h l b . Suppose that h Y a. Then by Lemma 3, f must be adjacent to at least two vertices of { a ,h , c , d}. This is a contradiction, since f 4' { a ,h, d}. Consequently, h I a . The case h E B is similar. I
Now we go on with the proof of Theorem 11. By Lemma 1, there exist vertices a , b E A and c , d E B such that a Y b and c Y d . By Lemma 2 (ii), ( a , b , c , d ) 2K2. Without loss of generality, let a l c and b l d . In accordance with Lemma 1, the set { a , d } does not dominate F . Therefore, there is a vertex f E V ( F ) such that f Y { a , d } . By Lemma 2 (i), f E C . Then f l { b , c } by Lemma 3 . Now consider the vertices b and c. Again by Lemma 1, there is a vertex g E V ( F ) such that g Y { b , c } . By Lemma 2 (i), we have g E C and by Lemma 4, g l { a , d } and g Y f . The resulting graph is shown in Figure 8. Consider the set { f , g } . By Lemma I , there exists a vertex h E V ( F )such that h Y { f , g}. In accordance with Lemma 5 , there are four cases. These cases are symmetrical, so we can assume that h E A and h I { a , 6 , c},h '4 d. By Lemma 1, since {h,u } does not dominate F , there exists a vertex k E A such that k Y h. By Lemma 2 (ii), ( h ,k , c , d ) 2K2, hence k l d and k Y c. Then k l b by Lemma 2 (iii). For the set { h , k , c , d } and f , g , we can apply Lemma 3, which produces k l { f , g}. Thus, we obtain the graph of Figure 9 (a). In accordance with Lemma 1, the set { k , c } does not dominate F . Therefore, there is a vertex Z € V ( F ) such that I Y { k , c}. By Lemma 2 (i), 1 E C. Applying Lemma 4 to the set { h ,k , c , d , f } and the vertex 1, we obtain Z l { h , d } and 1 Y f.Now apply Lemma 5. Assume that 1 Y g . Then
FIGURE 8
384 JOURNAL OF GRAPH THEORY f
f
{ h , k , c , d , f,1 } and the vertex g satisfy the hypotheses of Lemma 5. By the C , a contradiction. Thus, 1 lg. conclusion of Lemma 5 , we have g Suppose that 1 Y a . By applying Lemma 3 to the set { a ,b , c , d } and 1, we have l l b . Since 1 is adjacent to only one vertex of { a , k , c , d } , it follows that k l a , for otherwise we have a contradiction to Lemma 3. Thus, 1 Y a , l l b and k l a , hence ( u , a , b , h , k ; u , c , d ; f,g,Z) = G17, a contradiction. Therefore 1l a . Thus, we have a collection of graphs indicated in Figure 9 (b). By Lemma 1, the set { l , f } does not dominate F , so there exists a vertex m E V ( F ) such that m Y (1, f}. Now we can state the last lemma.
Lemma 6. Let a , b , h , k E A , c , d E B , f,g, 1 E C and m E V ( F ) .If ( a , h , 6 , k , c , d , f,g,I) is one of the graphs of Figure 9 (b) and m Y { f,Z}, then
1) m E A , m l { a ,b , h , k , d , g} and m Y c ; 2 ) b l l and a Y k . Proof. The set {h,k , c , d , f , I } and the vertex m satisfy the hypotheses of Lemma 5 , hence there are four cases to consider. Case I. m E A and m l { h , k , c } , m Y d . By Lemma 2 (iii), m l a . Assume that m Y 6 . Applying Lemma 3 to the set {m,b , c , d } and I, g,we have b l l and g l m . Then ( u , m , h , b , k ; u , c , d ; f,g,l) G17, a contradiction. Consequently, m l b . Further b l l , for otherwise ( h , I , c ; 6 , d , f ;m , u ) = G I ~a, contradiction. The resulting graph is shown in Figure 10 (a). By Lemma 1, the set {m, u } does not dominate F . Hence there is a vertex n E A such that n Y m. Then by Lemma 2 (ii), n l d and by Lemma 2 (iii), n l b . Now we can apply Lemma 3 to {m,n , c , d } and f,1, which produces n l { f,1}. We have (1, a , u ; b , m, f ; d , n ) Glo or G I depending ~ on the existence of an, a contradiction.
DOMINATION PERFECT GRAPHS 385
Case 2. m E A and m l { h , k , d}, m Y c. By Lemma 2 (iii), m l b . Assume that m 4' a. Then { a ,m, c , d } and f satisfy the hypotheses of Lemma 3 and f '4 { a , m, d}, contrary to the conclusion of Lemma 3. Hence m l a . The resulting graph is shown in Figure 10 (b). Furthermore alk b Y 1
-
( k , f , d ; a , c , l ; m , u ) G GI2
=j
3
a Yk,
G12 * b l l .
( b ,f , d ; h , c , l ; m , u )
By Lemma 1, there exists a vertex n E A such that n Y m. By Lemma 2 (ii), n l c and by Lemma 2 (iii), n l a . Applying Lemma 3 to { n , m , c , d } and 1, we have l l n . At last mY g
=j
(I, b , v ; a , m , c ; g , n )
G7,Gg,Glo or GI2
*mlg.
Thus, the conclusions of Lemma 6 are satisfied. Case 3. m E B and m l { c , d , h } , m Y k . Suppose that m Y a. By Lemma 2 (i), the set { m , a } dominates (A U B), hence a l k . We have ( a , k , 1; c , f , m ) G2, a contradiction. Therefore m l a . Then Lemma 2 (iii) implies m Y b , since a Y b. Further
a Y k, b Y 1 =j ( b , h , f ; d , m , l ) z G3 * b l l , alk
3
m Y g
( a , l , m ; k , d ,f )
G
* ( b , h ,f ; d , m , g )
G3
G2
=j
mlg
By Lemma 1, there is a vertex n E B such that n Y m. By Lemma 2 (ii), ( h ,k , m , n ) = 2K2, so n l k and n Y h. By Lemma 2 (iii), n l d . If n l a , then we have a contradiction to Lemma 2 (iii), since n l k and a Y k . f
f
d
d
C
1
C
1 (a)
g (b)
FIGURE 10. Illustration of the cases 1 and 2.
386 JOURNAL OF GRAPH THEORY Hence n 4' a . The set { b ,m } dominates (A U B ) by Lemma 2 (i), and so n l b. For the set {h,k , m, n } and 1, we can apply Lemma 3, which produces n l l . Assume that n Y g. Then { k ,h, m, n , I } and g satisfy the hypotheses of Lemma 4. Therefore g Y I, a contradiction. Consequently, n l g . The resulting graph is shown in Figure 11. By Lemma 1, the set { b , m } does not dominate F . Therefore, there is a vertex p Y { b , m } and p E C by Lemma 2 (i). Applying Lemma 3 to { a ,b , m, n } and p , we obtain p l { a , n}. Furthermore P Y g p Y 1
-
* (u,b,p;g,n,m)
G G3
( v , m , p ; l , a , b ) = G3
*pig,
*p l l ,
p+'f *(1,b,a;u,f,m;p,d)~G~gorG12*plf, p l d * ( l , b , a ; v , f , m ; p , d ) = G13 * p Y d . Now apply Lemma 3 to { a ,b , c , d } and p . We have p l c . Assume that p '4 h. The set { h ,k , c , d , I} and p satisfy the hypotheses of Lemma 4,and so p Y I , a contradiction. Thus p l h . Finally
f
1
43
FIGURE 11. Illustration of the case 3.
DOMINATION PERFECT GRAPHS 387
FIGURE 12. Illustration of the case 4
Case 4. m E B and m l { c , d , k } , m Y h. If m l { a , b}, contrary to Lemma 2 (iii). If m Y { a ,b}, contrary to Lemma 2 (i). If m Y a and m l b , then l l b , for otherwise ( b ,f , m ; h, c, 1 ) = G?. We have (v,f , m ; I, b , a ) G3, a contradiction. Thus m l a and m Y b. Since m l { a , k } , by Lemma 2 (iii), k l a . Now
* ( b ,f , h ; d , g , m )
GI * m l g , bYl*(c,f,m; h,b,l)=G2-bll.
rn Y g
The resulting graph is a subgraph of Figure 12. By Lemma 1, there is a vertex n E B such that n Y m. For the sets { a , b , m, n } and { h ,k , m, n} we can apply Lemma 2 (ii), which produces n l { b ,h } and n 4' { a ,k}. Then by Lemma 2 (iii), n l { c , d}. The application of Lemma 3 to the set { h , k , n , m } and the vertices 1 and f gives n l { l ,f}. Further
The resulting graph is a subgraph of Figure 12. In accordance with Lemma 1, the set { k , n } does not dominate F . Therefore, there is a vertex p E V ( F )such that p Y { k , n}. By Lemma 2 (i),
388 JOURNAL OF GRAPH THEORY
p E C . The application of Lemma 4 to { h ,k , n, m , f } and p gives p l { h , m } and p Y f . Suppose that Z Y p . Then { h ,k , n , m , f , p } and Z satisfy the hypotheses of Lemma 5 , and hence 1 @ C, a contradiction. Thus Z l p . Analogously, g l p . Further, p Y a implies ( h , p , a ; n , d , f ) G I or G I , hence p l a . Assume that p Y d. The application of Lemma 3 to {h,k , c , d } and p gives p l c , and we have ( v , d , m , c , n ; u , k , h ; f , Z , p ) G u , a contradiction. Hence p l d . The resulting graph is shown in Figure 12. By Lemma 1, the set { f , p } does not dominate F , and hence there is a vertex r E V ( F )such that r Y { f , p } . The set { h ,k , n , m,f, p } and r satisfy the hypotheses of Lemma 5. Therefore, there are four cases to consider. Subcase 4.1. r E A and r l { h , k , n}, r Y m. By Lemma 2 (iii), r l b , since n l { b , r } . If r Y a , contrary to Lemma 3 , since f is adjacent to one vertex of { a , r , m , n } . Hence r l a . If r Y d , then ( d , n , p ; k , f , r ) G3, a contradiction. Therefore r l d . Now r Y c, for otherwise r l { c , d } and c Y d , contrary to Lemma 2 (iii). We have p Y c a ( c , m , f;h , p , r ) r Y g ( p , c , d ; u,f , r ; g , a )
G2 Gg
*plc, rlg
By Lemma I , the vertex r does not dominate (A), and so there is a vertex s E A such that s Y r . By Lemma 2 (ii), s l { c , m } and s Y n. Then s l { h , k } by Lemma 2 (iii), since c l { h , s } and m l { k , s } . Consider the induced subgraph shown in Figure 13. Applying Lemma 3 to { r ,s, n , m } and f,p ,
g
P
FIGURE 13. Illustration of the subcase 4.1
DOMINATION PERFECT GRAPHS 389
we get sl{ f , p } . At last s l g =+ ( g , r , v ; s , h , f ; m , k ) z G12
*s Y g
and
(u,h,r,k,s; v,n,m; f , p , g )
=GI~,
a contradiction. Subcase 4.2. r E A and r l { h , k , m } , r Y n. We have ( h , p , r ; n , d , f ) = G2 or G3 depending on the existence of the edge rd, a contradiction. Subcase 4.3. r E B and r l { n ,m, h}, r Y k. Since f is adjacent to only one vertex of { h ,k , r , d } , it follows by Lemma 3 that r l d . We obtain ( d , p , r ; k , a , f ) G2 or G3 depending on the existence of the edge ra, a contradiction. Subcase 4.4. r E B and r l { n , m , k}, r Y h . Since k l { d , r } , it follows by Lemma 2 (iii) that r l d . We have ( d , p , k ; n , h , f ; r , v ) G12. This contradiction completes the consideration of the case 4, and with it the proof of Lemma 6. I Thus, applying Lemma 6 to the graph of Figure 9 (b) and m , we obtain the graph of Figure 14 without the vertex n. By Lemma 1, since m does not dominate ( A ) , there is a vertex n E A such that n Y m. By Lemma 2 (ii), n l c and n '4 d . Since c l { n , a , h}, it follows by Lemma 2 (iii) that n l { a ,h } . Apply Lemma 3 to {n,m, c, d } and f , l . We have n l u , l } . Further n Y b
nlg
3
* (b,l,m;f,u,n)
(l,b,v;a,m,c; g,n)
FIGURE 14
G3 G13
nlb, 3
n Y g.
390 JOURNAL OF GRAPH THEORY The resulting graph is shown in Figure 14. By Lemma 1, the set { n , d } does not dominate F . Hence there is a vertex p E V ( F )such that p Y { n , d } . By Lemma 2 (i), p E C . Applying Lemma 4 to {n,m, c , d , l } and p , we obtain p I { m , c } and p Y 1. Suppose that p Y f and apply Lemma 5 to {n,rn, c , d , p , I } and f .We have f @ C , a contradiction. Therefore p If.Assume that p Y k and consider the induced subgraph shown in Figure 15 (a). If n Y k , contrary to Lemma 3 , since p is adjacent to one vertex of { n ,k , c, d } . Applying Lemma 3 to {h,k , c , d } and p , we get p l h . Now ( u , k , m , h , n ; v , d , c ; I, f , p ) = G17, a contradiction. Thus p l k . The set { 1 , p } dominates the resulting graph. By Lemma 1, there is a vertex r E V ( F ) such that r Y { I , p } . Consider the subgraph shown in Figure 15 (b). Clearly that this graph and the vertex r satisfy the hypotheses of Lemma 6. By the conclusions of Lemma 6, r E A, r l { k , r n , h , n , c , f}, r '4 d and h l p , k Y n. Replace the vertex h in Figure 15 (b) by the vertex a and apply Lemma 6 again. In addition we obtain a l { p , r}. Now
If r Y b , then we have a contradiction to Lemma 3 , since p is adjacent to one vertex of { r , b , c , d}. The resulting graph is the graph W\s in Figure 16. By Lemma 1, the vertex r does not dominate (A), and hence there is a vertex s E A such that s Y r . By Lemma 2 (ii), s l d and s Y c . By Lemma 2 (iii), s l { b , k , r n } . Applying Lemma 3 to { r , s , c , d } and p , g , I ,
P
f
f
P
1
1
(a)
(b)
FIGURE 15
DOMINATION PERFECT GRAPHS 391
we get s l { p ,g , l}. We have
slf
( a , p , r ;I , u , s )
s+ a
3
sY h
* ( h ,p , r ; I, u , s)
* ( k , d , r ;p , u , a ; f , s )
= G3 =+
sla,
*slh, = G13 * s Y f . G3
The only edge undetermined is sn. It is straightforward to see that the graph ( n , r , m, s, c , d , p , 1 ) is isomorphic to the graph of Figure 9(a) with the correspondence of partition into the parts A , B and C . We have already proved that the case a l k for the graph of Figure 9(a) is impossible. Therefore we can assume that n Y s in our case. Thus, the graph F contains the induced subgraph W shown in Figure 16. For convenience, a general structure of the graph W is shown in Figure 16(a), and edges between the sets A and C are indicated in Figure 16(b). We remember that ( a ,h , n, r ) G K4 and ( b ,k , m , s) = K 4 . By Lemma 1, the set { b , p } does not dominate F , hence there is a vertex x such that x Y { b , p } .
Claim 1. x I { u , u , a , c , d , g , h, I, m, n , r } and x Y { b ,f , k , p , s}. If x E A , then we have a contradiction to Lemma 3, since p is adjacent to one vertex of { x , b , c , d } . Hence x G A and x l u . Suppose that x Y 1. Apply Lemma 6 to ( k , m, h , n , d , c , I, f , p ) and x . We have x E A , a contradiction. Therefore x l l . Now x '4 a
* ( a ,r , p ; 1, b , x ) = G2 or G3
xY n
3
( a , x , n ;p , k , u ) = GZ or G3
xlk
3
( k , x , b ; p , a , u ) = G3
x Y r orxls
3
( 1 , b , v ; a , r , p ; x , s)
u
-
xla, xln,
x Y k,
= G9, G L 2or G I 3 * x l r and x '4
v
(a)
FIGURE 16. Graph W.
s.
392 JOURNAL OF GRAPH THEORY
Let x E B, i.e., x Y u . If x '4 h , then { x ,h } does not dominate the vertex k , contrary to Lemma 2 (i). Hence x l h . Since x l n , we obtain x Y m by Lemma 2 (iii). We have xYd*(u,r,p; I,x,d)gG2*xxld,
* ( b , d , f ;h , x , p ; r , m ) = G
x Yf
( b ,d , h ; f , x , p ; n ) g
Ij ~ x l f , G6, a contradiction.
Let x E C, i.e., x l u . We get x Y c
* ( I , b, u ; a , m , c ; x , n )
G12
or GI3
*x l c .
Assume that x Y h . Applying Lemma 3 to { h ,k , c , d } and x , we have x l d . Now
( x , c , ~U;, f , h ) G3 * x l f , GL3,a contradiction. ( b ,d , h ; f , x , p ; r , k )
x Yf
j
Consequently, x lh. Furthermore
Thus, Claim 1 is satisfied. Now the set { g , r } dominates the resulting graph. By Lemma 1, there is a vertex y E V ( F ) such that y Y { g , r } .
Claim 2. y l { u , u , b , c , d ,f , h , k , I , n , p , s } and y Y { a , g , m , r , x } . If y E A, contrary to Lemma 3, since g is adjacent to one vertex of { r , y , c , d } . Hence y f$ A and y l u . Let y E B , i.e., y Y u. Since y Y r , it follows by Lemma 2 (i) that y l s . Now y Y n by Lemma 2 (iii). Then y l k by Lemma 2 (i) and y Y h by Lemma 2 (iii). Further
a contradiction.
DOMINATION PERFECT GRAPHS 393
Let y E C , i.e., y l u . Suppose that y Y f and apply Lemma 5 to { a , b , c , d . f , g l and y . We obtain y 4 C , a contradiction. Therefore y l f . Furthermore
Now assume that y 4' d and apply Lemma 3 to
{Y, s, c , d
} and y . We get
y I s . Further
Therefore y Id . We obtain
Thus, Claim 2 is satisfied. Now the set { g , n } dominates the resulting graph, By Lemma 1 , there is a vertex z E V ( F ) such that z 4' {g,n}. If z E A, contrary to Lemma 3 , since g is adjacent to one vertex of {n,z , c , d } . Hence y 4 A, i.e., z l u . We have zla
(a,z,n ; g, u , k )
z Y y =+ ( u , y , z ; g , k , a ) z '4 k * ( k , r , g ; y , n , z )
= G2 or G3 =+z G2 or G3
Y
a,
zly,
G2 or G3 =j z l k , ( k , r , g ; y , n, u ; s , z > = G7, G9, Glo, or GI2, a contradiction.
This contradiction completes the proof of Theorem 1 1 .
I
394 JOURNAL OF GRAPH THEORY
4. COROLLARIES
We first note that Theorems 1, 2, 3, 4, 5 , 8, 9, and 10 and Corollary 1 follow directly from Theorem 1I.
Corollary 2. Suppose G does not contain the complements of the graphs G, - G17in Figure 3 as induced subgraphs and diam(G) > 2. Then G has an edge that does not belong to any triangle.
c
Proof. Let G be the complement of G. Clearly that does not contain the induced subgraphs G I - GI7 and y(G) = 2. By Theorem 11, we have L( G) = 2, i.e., the graph G has an edge that does not belong to any triangle.
I Now we prove that the dominating set (DS) and independent dominating set (IDS) problems are NP-complete on some classes of graphs. The DS problem is known (see [6]) to be NP-complete for 3-regualr planar graphs, for bipartite graphs and for split graphs. In contrast, the IDS problem is known to be NP-complete for general graphs only and, if arbitrary weights are allowed, for chordal graphs. We say that a graph G belongs to the class L if the following conditions hold: (i) (ii) (iii) (iv)
G G G G
is planar; is bipartite; has the maximum degree 3; has the girth g(G) 1 k , where k is fixed.
Corollary 3. The DS and IDS problems are both NP-complete on the class L .
Proof. We describe a polynomial time reduction from the DS problem for 3-regular planar graphs to the DS and IDS problems for the class L , which implies the required result. Define the operation of 3-partition of an edge as follows: replace an edge uu by the path P5 = ( u - x - y - z - u ) . Suppose H is obtained from G by 3-partition of an edge uu. Prove that y ( H ) = y(G) + 1 . Let D be a minimum dominating set of G. If { u , u } E D or { u , u } E D, then D U { y } dominates H . If u E D and u @ D , then D U { z } dominates H . The case u 4 D and u E D is similar. Therefore y ( H ) 5 y(G) + 1. Let E be a minimum dominating set of H . Clearly that S = E r l { u , x , y , z } # 0. If IS1 = 1 then S = { x } or S = (v}, and in the case S = {x} we have u E E. It is easily to see that E\S dominates G. If IS1 I2, then (E\S) U { u } dominates G. Thus y(G) Iy ( H ) - 1. Applying the operation of 3-partition to edges of a 3-regular planar graph G, we can obtain a graph with the girth at least k . The single 3-partition of each edge results in a bipartite graph. The resulting graph H is certainly
DOMINATION PERFECT GRAPHS 395
constructible in polynomial time, since the number k is fixed, and H belongs to L . Moreover, H does not contain the graphs G I - G17 in Figure 3 as induced subgraphs. By Theorem 11, L ( H ) = y ( H ) = y ( G ) + m, where m is the number of the 3-partitions of G. I
ACKNOWLEDGMENT We thank the referees for helpful suggestions.
References [ 11 R. B. Allan and R. Laskar, On domination and independent domination
numbers of a graph. Discrete Math. 23 (1978) 73-76. [2] B. Bollobis and E. J. Cockayne, Graph-theoretic parameters concerning domination, independence, and irredundance. J. Graph Theory 3 (1979) 241 -249. [3] J. Fulman, A note on the characterization of domination perfect graphs. J. Graph Theory 17 (1993) 47-51. [4] F. Harary, Graph Theory. Addison-Wesley, Reading, MA (1969). [5] F. Harary and M. Livingston, Characterization of trees with equal domination and independent domination numbers. Congr. Numer. 55 (1986) 121-150. [6] D. S. Johnson, The NP-completeness column: an ongoing guide. J. Algorithms 5 (1984) 147-160. [7] R. Laskar and H. B. Walikar, On domination related concepts in graph theory. Lecture Notes in Math. 885 (1981) 308-320. [8] S. Mitchell and S. Hedetniemi, Edge domination in trees. Congr. Numer. 19 (1977) 489-509. [9] D. Sumner and J.I. Moore, Domination perfect graphs. Notices Am. Math. Soc. 26 (1979) A-569. [lo] D. Sumner, Critical concepts in domination. Discrete Math. 86 (1990) 33-46. [ll] J. Topp and L. Volkmann, On graphs with equal domination and independent domination numbers. Discrete Math., 96 (1991) 75-80. [ 121 I. E. Zverovich and V. E. Zverovich, A characterization of domination perfect graphs. J. Graph Theory 15 (1991) 109-114.
Received August 27, 1993
