- Email: [email protected]
On strongly asymmetric graphs
DISCRETE MATHEMATICS ELSEVIER
Discrete Mathematics 145 (1995) 315- 320
Note On strongly asymmetric graphs Mirko Lepovid Ul. Tihomira Vuksanovii'a 32, 34000 Kragujevac, Yugoslavia
Received 19 January 1993; revised 9 November 1993
Abstract Let G be an arbitrary simple graph of order n. G is called strongly asymmetric if all induced overgraphs of G of order (n + 1) are nonisomorphic. In this paper we give some properties of such graphs and prove that the class 6ec of all connected strongly asymmetric graphs is infinite.
In this paper we consider only finite graphs having no loops or multiple edges. The vertex set of a graph G is denoted by V(G), and its order (number of vertices) by IGI. The edge set of G is denoted by E(G), and its number of edges by IE(G)b. Let S be any (possibly empty) subset of the vertex set V(G). Denote by Gs the graph obtained from the graph G by adding a new vertex x(x q~ V(G)), which is adjacent exactly to the vertices from S. Graph G is obviously an induced subgraph of the graph Gs, and Gs is an induced overgraph of G. Varying the set S ~ V(G) we get 2161 such graphs Gs. The family of all such graphs is denoted by Cg(G).We shall briefly call (~(G) the overset of G. We say that fg(G) is connected if every graph Hef#(G) except Go so is. It is easily seen that this happens if and only if the graph G is connected. We now introduce the following definition. Definition 1. Let G be an arbitrary graph of order n. Graph G is called strongly asymmetric if for any two sets $1,$2 c_ V(G) (St ~ $2), the graphs Gsl,Gs2 are nonisomorphic. Let F(G) be the corresponding group of automorphisms of a graph G. As is well known, if F(G) is trivial, then the graph G is called asymmetric. Next we prove that every strongly asymmetric graph is asymmetric. However, we will later see that there exist asymmetric graphs which are not strongly asymmetric. * This research was supported by Science Fund of Serbia, through the Mathematical Faculty of Belgrade. Elsevier Science B.V. SSDI 0 0 1 2 - 3 6 5 X ( 9 4 ) 0 0 0 4 5 - K
316
M. Lepovib / Discrete Mathematics 145 (1995) 315-320
Proposition 1. Every strongly asymmetric graph is also asymmetric. Proof. Let G be an arbitrary strongly asymmetric graph, and assume F(G) is the corresponding group of automorphisms. If we suppose that G is not an asymmetric graph, then there exists an automorphism feF(G) and at least two distinct vertices xl,x2~V(G) such that f ( x l ) = x2. Let Gi (i = 1,2) be the overgraph of G whose additional vertex ~ is adjacent just to the vertex x~. Since G is a strongly asymmetric graph, G1 is not isomorphic to G2. Define a mapping f * : V(G1)~ V(G2) as
f*(x) =f(x),
xeV(G); f*(.~) = 2.
It is clear that f * is an isomorphism of graph G1 to G2, wherefrom we get the statement. [] Let 6e be the class of all strongly asymmetric graphs. Besides, in this paper we consider the class 5ec ___5a of all connected strongly asymmetric graphs. The class 5a~ # 0. That statement is contained in the next proposition, as follows.
Proposition 2. There exists at least one connected strongly asymmetric graph. A simple example with the least number of vertices is displayed in Fig. 1, and it has 6 vertices. By a straightforward calculation, we can check that there is no connected strongly asymmetric graph with 2, 3, 4 or 5 vertices. Now, we prove that class Sac is infinite. First, we prove some auxiliary propositions. If G is an arbitrary graph, let (~ be the complementary graph of G, whose vertex set is also V(G) and two distinct vertices x, y ~ V(G) are adjacent in G if and only if x and y are not adjacent in G.
Proposition 3. If G is strongly asymmetric, then so is G. Proof. Let fC(G) = {GslS --- V(G) = V(G)} be the corresponding overset of the graph (7. It is sufficient to prove that for any sets $1, $2 ~- V(G) (St # $2), the graphs dsl, tTs~ are nonisomorphic. Let Ti = V(G)\S~ (i = 1, 2). Then obviously Tx # T2 and t~s, = Gr, (i = 1, 2). Since the graphs Grl, Gr2 are nonisomorphic, we immediately get that t~s~,ds2 are also nonisomorphic. []
3
A 0
£
~0
1
2
4
Fig. 1
0
0
5
6
M. Lepovi~/ Discrete Mathematics 145 (1995) 315 320
317
Let G and H be two connected graphs without common vertices. The union G w H is the graph whose vertex set is V(G)w V(H) and whose edge set is E(G)wE(H).
Proposition 4. Let G and H be two connected strongly asymmetric graphs without common vertices. Then G u H is a strongly asymmetric graph if and only if G and H are nonisomorphic. Proof. If G w H is a strongly asymmetric graph, then G and H are obviously nonisomorphic. Conversely, assume that G w H is not strongly asymmetric and G , H are nonisomorphic graphs. Then there exist at least two isomorphic graphs (GwH)R~Sl, (G w H ) R 2 ~ s ~ ( G w H), where Ri ~- V(G), Si ~_ V(H)(i = 1, 2) and R1 w $1 ~ R2 L)S2. If we suppose that R1 = 0, then it immediately follows that (GuH)RtuS~ = (G w H)sl = G u H s~ is disconnected, and consequently (G w H)R~s~ is disconnected, too. F r o m this we get that at least one of the sets R2, S 2 is empty one. If we assume that R2 = 0 we get G u Hs~ ~- G u Hs~, and therefore Hsl is isomorphic with Hs~. Since Hs~, Hs~e C~(H) (St v~ $2) and H is a strongly isomorphic graph, we get a contradiction. If R~ = 0 and $2 = 0, we get that
GwH& ~- GR2UH.
(1)
By assumption G and H are nonisomorphic, and by relation ( 1 ) w e get that G --- G R2, what is contradiction since IGR2[ = [G[ + 1. Consequently, (G u H)gl~sl and (G u H)R2~s: are two connected graphs. Put Xi = (GuH)R,~S,, and V(Xi)= V ( G u H ) w { x } (i = 1,2), where x is the additional vertex of the graph Xi(i = 1,2). L e t f b e an isomorphism from X 1 to X2. Since xe V(Xi) (i = 1, 2) is the only vertex adjacent to some vertices of G and some vertices of H, it follows that f(x) = x. Next we have R1 ¢ R2 or S1 :/: $2, and without loss of generality we will assume that R1 ~ R2. Let Yo be a fixed vertex from V(H). We firstly assume that f(yo)~ V(H). We shall prove that then f(y)~ V(H) for any vertex ye V(H). Indeed, assume that there exists a y~ V(H) such t h a t f ( y ) ~ V(G). Consider the case when y is adjacent to Yo. Then, we get that (f(Yo),f(Y))~ E(G w H), wherefrom we have that G u H is a connected graph, what is a contradiction. If y~ V(H) is not adjacent to Yo, then since H is a connected graph, there exist Yl,Y2 ..... yk~V(H) such that (Yo,Yl), (Y~,Y2), ...,(yk,y)eE(H). Having in mind the previous case, we get f(y~)E(H). Similarly we find that f(y2) ..... f(Yk),f(Y)~ V(H). Consequently, we get that GR1 = X I \ H is isomorphic to GR2 = X2\H. Since GR~, GR2~f~(G) and G is a strongly asymmetric graph, we get a contradiction. Finally, assume thatf(yo)~ V(G). Then for any vertex ye V(H) we havef(y)~ V(G). It is clear thatf(V(H)) = V(G), and therefore IGI = IHI. This gives that GR~ = X I \ H is
318
M. Lepovii'/ Discrete Mathematics 145 (1995) 315-320
isomorphic to Hs~ = X 2 \ G. Sincef(x) = x, we definitely have that graph G = GR, \ x is isomorphic to H = Hs~\X, which is a contradiction. This completes the proof. [] As an immediate consequence of the previous result, we get the next stagement. Proposition 5. Let GI, G 2 . . . . . G, (n >~ 2) be arbitrary connected strongly asymmetric graphs wihout common vertices. Then G~ u G 2 u . . . u G n is a strongly asymmetric graph if and only if for every i ~ j graph Gi is nonisomorphic to Gj (i,j = 1, 2 ..... n).
Proof. If G1 • G 2 L ; " " ~ G , is a strongly asymmetric graph, then Gg and Gi are obviously nonisomorphic graphs for any i ~ j (i,j = 1,2, ...,n). Now, assume that Gi and Gj are nonisomorphic graphs for any i # j. We shall prove that Ga u G2 L;... u G, is a strongly asymmetric graph by induction on n. This is obviously true for n - - 2 by Proposition 4. Assume that n ~> 3 and G 1 U G E U ... kd G k is a strongly asymmetric graph for any k < n. By this assumption we shall prove that G~ u G 2 u . . . ~ G, is a strongly asymmetric graph, too. On the contrary, assume that G = G~ U G E L ) . - . W G, is not strongly asymmetric. Then there exist two isomorphic graphs GR,Gsef~(G) with R g:S, where R = R i u R 2 w . . . u R . and S = S i u S 2 u . . . u S , ( R i , S i C_ V(GI), i = 1,2,.-.,n). Firstly, assume that R~ 4:0 and R~ = 0 for i = 2, 3..... n. Since GR and Gs are isomorphic graphs, then among the sets S~,$2 ..... S, exactly one must be nonempty - - assume Sio 4:0 and Si = 0 (i ~< n, i 4: io). Condition R # S gives that R1 4: S~o. It is clear that GR = (G~)R, U7=2 G~. If we assume that io = 1, then we get n
n
(G,)R, U Gi ~- (G1)s, U Gi. i=2
i=2
From the last relation it follows that (Gt)R, is isomorphic to (G1)s,. Since (GI)R,, (G1)s, Ef~(G1) and G~ is a strongly asymmetric graph we get a contradiction. Now, assume that io ~ 1. Without loss of generality we can suppose that i0 = 2. Whence we get
G1)R,UG2
Gi ~- G1u(G2)s~ i
U Gi. i=3
From the last relation we have ( G 1 ) R I U G 2 _ ~ G l w ( G 2 ) s 2 . Since G 2 and G~ are nonisomorphic graphs, we get that G2~(G2)s2, what is a contradiction since [(G2)s~[ = [G21 + 1. Consequently, if GR and Gs are two isomorphic graphs, then at least two sets Ri, Rj are nonempty, where i # j (i,j ..... n). Without loss of generality we can assume that R1 # 0 and R2 # 0 . Since R = Ug,~o Ri and R # S we shall also suppose that R1 ~ $1.
M. Lepovi~"/ Discrete Mathematics 145 (1995) 315-320
319
Put X1 = GR, X z = Gs and V(Xi) = V(G) u {x} (i = 1, 2), where x is the additional vertex of the graph Xi (i = 1,2). L e t f be a corresponding isomorphism from X~ to X2. Similarly to the previous proposition, we can conclude that f(x) = x. Let Yo be a fixed vertex from V(G2), We firstly assume that f(Yo)~ V(G2). Then for any vertex y e V(G2) we havef(y)E V(G2 ). Consequently, we get that Gg\ G2 ~- Gs\ G2. Since GR\ G2, Gs\ G2 ~-cff(Gl W G3, "-" UGn) and G~ w G3 w ' " k3 Gn is a strongly asymmetric graph by the induction hypothesis, we get a contradiction. Finally, assume that f ( y 0 ) e V(Gio) where i o ¢ 2. Then for any vertex y~V(Gz) we h a v e f ( y ) E V(Gio). This gives that Y1 = G~\G2 is isomorphic to Yz = Gs\Gio. Since f(x) = x, we can easily see that Y1\ x = G\ G2 is isomorphic to Y z \ x = G\ Gio. From this we have that Gio is isomorphic to G2, what is a contradiction. This completes the prof. [] Proposition 6. For every n >~6 there exists at least one connected strongly asymmetric graph of order n.
Proof. Let G be the graph displayed on Fig. 1. It is of order n = 6 and is a strongly asymmetric graph. Denote by O the graph having just one vertex. By Propositions 3 and 4, we get that G w O is a connected strongly asymmetric graph or order 7. If we assume that H is a connected strongly asymmetric graph of order k, then H w O is obviously a connected strongly asymmetric graph of order k + 1. Hence the induction in n completes the proof. Corollary 1. The class Sac of all connected strongly asymmetric graphs is infinite. Proposition 7. There exists at least one connected asymmetric graph which is not strongly asymmetric. A simple example with the least number of vertices is displayed in Fig. 2. Proof. It is easy to check that graph from Fig. 2 is asymmetric. Graphs X1 and displayed on Fig. 3 belong to the overset (~(G). Clearly, the mapping
6
5
8
1
2
3
3
?k 0 1
/
2
'X~ 4
Fig. 2
0
5
0 6
0 7
X 2
320
M. Lepovi? / Discrete Mathematics 145 (1995) 315-320 3
1
2
3
8
4
5
6
O 7
7
X, t
8
X2
Fig. 3
from V(X1) onto V(X2) induces an isomorphism from X1 to X2. Consequently, the graph from Fig. 2 is not strongly asymmetric. []
Corollary 2.
I f G is the graph displayed on Fig. 2, and 0 is a graph having just one vertex, then G u 0 is a connected asymmetric graph which is not strongly asymmetric.
Corollary 3.
For every n >>.7 there exists at least one connected asymmetric graph which is not strongly asymmetric.
References [1] D. Cvetkovi~, Theory of Graphs, Nau~na knjiga, Beograd (1981) (in Serbian). i-2] F. Harary, Graph Theory (Reading Mass., Boston, 1963).
Discrete Mathematics 145 (1995) 315- 320
Note On strongly asymmetric graphs Mirko Lepovid Ul. Tihomira Vuksanovii'a 32, 34000 Kragujevac, Yugoslavia
Received 19 January 1993; revised 9 November 1993
Abstract Let G be an arbitrary simple graph of order n. G is called strongly asymmetric if all induced overgraphs of G of order (n + 1) are nonisomorphic. In this paper we give some properties of such graphs and prove that the class 6ec of all connected strongly asymmetric graphs is infinite.
In this paper we consider only finite graphs having no loops or multiple edges. The vertex set of a graph G is denoted by V(G), and its order (number of vertices) by IGI. The edge set of G is denoted by E(G), and its number of edges by IE(G)b. Let S be any (possibly empty) subset of the vertex set V(G). Denote by Gs the graph obtained from the graph G by adding a new vertex x(x q~ V(G)), which is adjacent exactly to the vertices from S. Graph G is obviously an induced subgraph of the graph Gs, and Gs is an induced overgraph of G. Varying the set S ~ V(G) we get 2161 such graphs Gs. The family of all such graphs is denoted by Cg(G).We shall briefly call (~(G) the overset of G. We say that fg(G) is connected if every graph Hef#(G) except Go so is. It is easily seen that this happens if and only if the graph G is connected. We now introduce the following definition. Definition 1. Let G be an arbitrary graph of order n. Graph G is called strongly asymmetric if for any two sets $1,$2 c_ V(G) (St ~ $2), the graphs Gsl,Gs2 are nonisomorphic. Let F(G) be the corresponding group of automorphisms of a graph G. As is well known, if F(G) is trivial, then the graph G is called asymmetric. Next we prove that every strongly asymmetric graph is asymmetric. However, we will later see that there exist asymmetric graphs which are not strongly asymmetric. * This research was supported by Science Fund of Serbia, through the Mathematical Faculty of Belgrade. Elsevier Science B.V. SSDI 0 0 1 2 - 3 6 5 X ( 9 4 ) 0 0 0 4 5 - K
316
M. Lepovib / Discrete Mathematics 145 (1995) 315-320
Proposition 1. Every strongly asymmetric graph is also asymmetric. Proof. Let G be an arbitrary strongly asymmetric graph, and assume F(G) is the corresponding group of automorphisms. If we suppose that G is not an asymmetric graph, then there exists an automorphism feF(G) and at least two distinct vertices xl,x2~V(G) such that f ( x l ) = x2. Let Gi (i = 1,2) be the overgraph of G whose additional vertex ~ is adjacent just to the vertex x~. Since G is a strongly asymmetric graph, G1 is not isomorphic to G2. Define a mapping f * : V(G1)~ V(G2) as
f*(x) =f(x),
xeV(G); f*(.~) = 2.
It is clear that f * is an isomorphism of graph G1 to G2, wherefrom we get the statement. [] Let 6e be the class of all strongly asymmetric graphs. Besides, in this paper we consider the class 5ec ___5a of all connected strongly asymmetric graphs. The class 5a~ # 0. That statement is contained in the next proposition, as follows.
Proposition 2. There exists at least one connected strongly asymmetric graph. A simple example with the least number of vertices is displayed in Fig. 1, and it has 6 vertices. By a straightforward calculation, we can check that there is no connected strongly asymmetric graph with 2, 3, 4 or 5 vertices. Now, we prove that class Sac is infinite. First, we prove some auxiliary propositions. If G is an arbitrary graph, let (~ be the complementary graph of G, whose vertex set is also V(G) and two distinct vertices x, y ~ V(G) are adjacent in G if and only if x and y are not adjacent in G.
Proposition 3. If G is strongly asymmetric, then so is G. Proof. Let fC(G) = {GslS --- V(G) = V(G)} be the corresponding overset of the graph (7. It is sufficient to prove that for any sets $1, $2 ~- V(G) (St # $2), the graphs dsl, tTs~ are nonisomorphic. Let Ti = V(G)\S~ (i = 1, 2). Then obviously Tx # T2 and t~s, = Gr, (i = 1, 2). Since the graphs Grl, Gr2 are nonisomorphic, we immediately get that t~s~,ds2 are also nonisomorphic. []
3
A 0
£
~0
1
2
4
Fig. 1
0
0
5
6
M. Lepovi~/ Discrete Mathematics 145 (1995) 315 320
317
Let G and H be two connected graphs without common vertices. The union G w H is the graph whose vertex set is V(G)w V(H) and whose edge set is E(G)wE(H).
Proposition 4. Let G and H be two connected strongly asymmetric graphs without common vertices. Then G u H is a strongly asymmetric graph if and only if G and H are nonisomorphic. Proof. If G w H is a strongly asymmetric graph, then G and H are obviously nonisomorphic. Conversely, assume that G w H is not strongly asymmetric and G , H are nonisomorphic graphs. Then there exist at least two isomorphic graphs (GwH)R~Sl, (G w H ) R 2 ~ s ~ ( G w H), where Ri ~- V(G), Si ~_ V(H)(i = 1, 2) and R1 w $1 ~ R2 L)S2. If we suppose that R1 = 0, then it immediately follows that (GuH)RtuS~ = (G w H)sl = G u H s~ is disconnected, and consequently (G w H)R~s~ is disconnected, too. F r o m this we get that at least one of the sets R2, S 2 is empty one. If we assume that R2 = 0 we get G u Hs~ ~- G u Hs~, and therefore Hsl is isomorphic with Hs~. Since Hs~, Hs~e C~(H) (St v~ $2) and H is a strongly isomorphic graph, we get a contradiction. If R~ = 0 and $2 = 0, we get that
GwH& ~- GR2UH.
(1)
By assumption G and H are nonisomorphic, and by relation ( 1 ) w e get that G --- G R2, what is contradiction since IGR2[ = [G[ + 1. Consequently, (G u H)gl~sl and (G u H)R2~s: are two connected graphs. Put Xi = (GuH)R,~S,, and V(Xi)= V ( G u H ) w { x } (i = 1,2), where x is the additional vertex of the graph Xi(i = 1,2). L e t f b e an isomorphism from X 1 to X2. Since xe V(Xi) (i = 1, 2) is the only vertex adjacent to some vertices of G and some vertices of H, it follows that f(x) = x. Next we have R1 ¢ R2 or S1 :/: $2, and without loss of generality we will assume that R1 ~ R2. Let Yo be a fixed vertex from V(H). We firstly assume that f(yo)~ V(H). We shall prove that then f(y)~ V(H) for any vertex ye V(H). Indeed, assume that there exists a y~ V(H) such t h a t f ( y ) ~ V(G). Consider the case when y is adjacent to Yo. Then, we get that (f(Yo),f(Y))~ E(G w H), wherefrom we have that G u H is a connected graph, what is a contradiction. If y~ V(H) is not adjacent to Yo, then since H is a connected graph, there exist Yl,Y2 ..... yk~V(H) such that (Yo,Yl), (Y~,Y2), ...,(yk,y)eE(H). Having in mind the previous case, we get f(y~)E(H). Similarly we find that f(y2) ..... f(Yk),f(Y)~ V(H). Consequently, we get that GR1 = X I \ H is isomorphic to GR2 = X2\H. Since GR~, GR2~f~(G) and G is a strongly asymmetric graph, we get a contradiction. Finally, assume thatf(yo)~ V(G). Then for any vertex ye V(H) we havef(y)~ V(G). It is clear thatf(V(H)) = V(G), and therefore IGI = IHI. This gives that GR~ = X I \ H is
318
M. Lepovii'/ Discrete Mathematics 145 (1995) 315-320
isomorphic to Hs~ = X 2 \ G. Sincef(x) = x, we definitely have that graph G = GR, \ x is isomorphic to H = Hs~\X, which is a contradiction. This completes the proof. [] As an immediate consequence of the previous result, we get the next stagement. Proposition 5. Let GI, G 2 . . . . . G, (n >~ 2) be arbitrary connected strongly asymmetric graphs wihout common vertices. Then G~ u G 2 u . . . u G n is a strongly asymmetric graph if and only if for every i ~ j graph Gi is nonisomorphic to Gj (i,j = 1, 2 ..... n).
Proof. If G1 • G 2 L ; " " ~ G , is a strongly asymmetric graph, then Gg and Gi are obviously nonisomorphic graphs for any i ~ j (i,j = 1,2, ...,n). Now, assume that Gi and Gj are nonisomorphic graphs for any i # j. We shall prove that Ga u G2 L;... u G, is a strongly asymmetric graph by induction on n. This is obviously true for n - - 2 by Proposition 4. Assume that n ~> 3 and G 1 U G E U ... kd G k is a strongly asymmetric graph for any k < n. By this assumption we shall prove that G~ u G 2 u . . . ~ G, is a strongly asymmetric graph, too. On the contrary, assume that G = G~ U G E L ) . - . W G, is not strongly asymmetric. Then there exist two isomorphic graphs GR,Gsef~(G) with R g:S, where R = R i u R 2 w . . . u R . and S = S i u S 2 u . . . u S , ( R i , S i C_ V(GI), i = 1,2,.-.,n). Firstly, assume that R~ 4:0 and R~ = 0 for i = 2, 3..... n. Since GR and Gs are isomorphic graphs, then among the sets S~,$2 ..... S, exactly one must be nonempty - - assume Sio 4:0 and Si = 0 (i ~< n, i 4: io). Condition R # S gives that R1 4: S~o. It is clear that GR = (G~)R, U7=2 G~. If we assume that io = 1, then we get n
n
(G,)R, U Gi ~- (G1)s, U Gi. i=2
i=2
From the last relation it follows that (Gt)R, is isomorphic to (G1)s,. Since (GI)R,, (G1)s, Ef~(G1) and G~ is a strongly asymmetric graph we get a contradiction. Now, assume that io ~ 1. Without loss of generality we can suppose that i0 = 2. Whence we get
G1)R,UG2
Gi ~- G1u(G2)s~ i
U Gi. i=3
From the last relation we have ( G 1 ) R I U G 2 _ ~ G l w ( G 2 ) s 2 . Since G 2 and G~ are nonisomorphic graphs, we get that G2~(G2)s2, what is a contradiction since [(G2)s~[ = [G21 + 1. Consequently, if GR and Gs are two isomorphic graphs, then at least two sets Ri, Rj are nonempty, where i # j (i,j ..... n). Without loss of generality we can assume that R1 # 0 and R2 # 0 . Since R = Ug,~o Ri and R # S we shall also suppose that R1 ~ $1.
M. Lepovi~"/ Discrete Mathematics 145 (1995) 315-320
319
Put X1 = GR, X z = Gs and V(Xi) = V(G) u {x} (i = 1, 2), where x is the additional vertex of the graph Xi (i = 1,2). L e t f be a corresponding isomorphism from X~ to X2. Similarly to the previous proposition, we can conclude that f(x) = x. Let Yo be a fixed vertex from V(G2), We firstly assume that f(Yo)~ V(G2). Then for any vertex y e V(G2) we havef(y)E V(G2 ). Consequently, we get that Gg\ G2 ~- Gs\ G2. Since GR\ G2, Gs\ G2 ~-cff(Gl W G3, "-" UGn) and G~ w G3 w ' " k3 Gn is a strongly asymmetric graph by the induction hypothesis, we get a contradiction. Finally, assume that f ( y 0 ) e V(Gio) where i o ¢ 2. Then for any vertex y~V(Gz) we h a v e f ( y ) E V(Gio). This gives that Y1 = G~\G2 is isomorphic to Yz = Gs\Gio. Since f(x) = x, we can easily see that Y1\ x = G\ G2 is isomorphic to Y z \ x = G\ Gio. From this we have that Gio is isomorphic to G2, what is a contradiction. This completes the prof. [] Proposition 6. For every n >~6 there exists at least one connected strongly asymmetric graph of order n.
Proof. Let G be the graph displayed on Fig. 1. It is of order n = 6 and is a strongly asymmetric graph. Denote by O the graph having just one vertex. By Propositions 3 and 4, we get that G w O is a connected strongly asymmetric graph or order 7. If we assume that H is a connected strongly asymmetric graph of order k, then H w O is obviously a connected strongly asymmetric graph of order k + 1. Hence the induction in n completes the proof. Corollary 1. The class Sac of all connected strongly asymmetric graphs is infinite. Proposition 7. There exists at least one connected asymmetric graph which is not strongly asymmetric. A simple example with the least number of vertices is displayed in Fig. 2. Proof. It is easy to check that graph from Fig. 2 is asymmetric. Graphs X1 and displayed on Fig. 3 belong to the overset (~(G). Clearly, the mapping
6
5
8
1
2
3
3
?k 0 1
/
2
'X~ 4
Fig. 2
0
5
0 6
0 7
X 2
320
M. Lepovi? / Discrete Mathematics 145 (1995) 315-320 3
1
2
3
8
4
5
6
O 7
7
X, t
8
X2
Fig. 3
from V(X1) onto V(X2) induces an isomorphism from X1 to X2. Consequently, the graph from Fig. 2 is not strongly asymmetric. []
Corollary 2.
I f G is the graph displayed on Fig. 2, and 0 is a graph having just one vertex, then G u 0 is a connected asymmetric graph which is not strongly asymmetric.
Corollary 3.
For every n >>.7 there exists at least one connected asymmetric graph which is not strongly asymmetric.
References [1] D. Cvetkovi~, Theory of Graphs, Nau~na knjiga, Beograd (1981) (in Serbian). i-2] F. Harary, Graph Theory (Reading Mass., Boston, 1963).