On self-complementation

We prove that, with very few exceptions, every graph of order n, n = 0, 1 (mod 4) and size a t most n 1, is contained in a self-complementary graph of order n. We study a similar problem for digraphs. Throughout the paper, G and D will denote a finite graph and a finite digraph, respectively, without loops or multiple edges, with vertex-sets V(G) and V ( D ) , and edge-sets E(G) and E(D); define r (G) = IE(G)I, e(D) = (E(D)I. An edge of G joining x and y is denoted by xy, an edge of D from i to t by ( z , t ) , and a symmetric edge of D joining u and u by uu. Cj denotes a cycle of G of length i z= 3. G U H will refer to two vertex disjoint graphs G and H , and mG to m disjoint copies of G. If A C V(G), then G A is the subgraph of G induced byV(G) A . G is said to be a self-complementary graph (or S.C. graph) if it is isomorphic to its complement c, then, there exists a permutation (T of V(G), called S.C. permutation of G, such that xy is an edge of G if and only if a(x)u(y ) is an edge of (for simplicity, we use the notation ( ~ ( x y ) = u(x)cr(y)) . We say that G is contained in a graph G’ if there exists a subgraph in G’ isomorphic to G. The same is applicable to D . It is known [6,7,8] that if G is an S.C. graph of order n , then n = 0, 1 (mod 4), and its S.C. permutation has all its cycles of lengths being multiples of 4 (except one of length one if n is odd), and lengths of cycles of an S.C. permutation of an S . C . digraph D are even (except one of length one if the order of D is odd). Furthermore, if G is an S.C. On leave from University of Mining and Metallurgy, Krakow. Poland, Journal of Graph Theory, Vol. 8 (1985) 335-341 @ 1985 by John Wiley & Sons, Inc. CCC 0364-9024/85/03335-7$04.00 336 JOURNAL OF GRAPH THEORY graph with S.C. permutation cr! then .ry E E(G) if and only if u " ' ( s y ) E E(G) for uz even and (rnr E E ( c ) for M odd. In [3], there is proved that every S.C. graph contains a hamiltonian path and S.C. graphs with 2-factors have been characterized in [ S ] . Using similar arguments 1101, one can prove that every S.C. digraph contains a hamiltonian path (some edges can be oriented the wrong way) and that if an S.C. permutation of D has all the cycles of length at least 1, then L> contains a 2-factor. If u is any permutation whose cycles are even. except possibly one of length one, and if one of the cycles of cr has length at most 2, then there exist two digraphs L) and D' such that (J is an S . C . permutation of D and D ' , I ) has a 2-factor and L)' has no 2factor. In 141, it was proved that every graph of order IZ and size at most I ) is contained in its complement except for some exceptions (see also I I ,2,91). We need Lemmas I and 3 to prove that, with very few exceptions, every graph of order 11, 17 = 0, 1 (mod 4) and size at most 11 I is contained in an S . C . graph of order r z , and that every digraph of order n and size at most n is contained in an S.C. digraph of order 1 2 . Lemma 1. Let G be a graph of order n , n = 0, I (mod 4). Then, G is contained in an S.C. graph of order 17 if and only if there exists a permutation u of V(G) whose cycles have length equal to a multiple of 4 (except one of length one if n is odd) so that u'"' '(xyl E E ( c ) for every xy E HG). Proof. Let u be a permutation as defined in Lemma 1. We construct two sets of edges E,, and E, by the following algorithm: (i) For every s y E E(G) , assign u2"'(sy) to E,,, and assign c r " " + ' ( ~ y ) to E,., for all m 3 0. (ii) If there are vertices z , f E V ( C ) such that zt $Z E,, U E, , then assign (T""(zt) to Eh and u2"" ' (a) to E,. for all In 3 0. ( i i i ) If, after Eb and E,. have been enlarged by step ( i i ) . there is still another pair z , t E V(G) such that zt fj! Eh U E, . then repeat step ( i i ) , until all edges of the complete graph constructed on V(G) lie in Eh or E, . Clearly, the graph G , defined by V(G,) = U G ) , E ( G , ) = Eb u E, is complete, the graph G' defined by V(G') = V(G), E(G') = Eh is an S.C. graph and u is an S.C. permutation of G', whose complement is the graph G" defined by V(G") = V(G) and E ( C ' ) = E,. If G is contained in an S.C. graph, the existence of u is given by definition. I If the permutation (T of Lemma I exists, then u will be also called S.C. permutation of G. So, the S.C. permutations below will refer to cr of Lemma I and we denote by u(G) an arbitrary S.C. permutation (if it exists) of G. ON SELF-COMPLEMENTATION 337 Theorem 1. Let G be a graph of order n , I I = 0. I (mod 4) . If e(G) d n 1, then G is contained in an S.C. graph of order 1 1 , unless G is isomorphic to K , ,,,,, C3 U K , , C, U K, , , , . J I Z 2 5 ) . or C, U K , . Proof. We shall prove a stronger result. that is: G is contained in an S .C . graph whose S.C. permutation has all i t s cycles of length 4 (unless one of Length one if ti is odd), except if G is one of the forbidden graphs. The proof is by induction on n . The theorem is true for n = 4 and for n = 5. so, let I I 3 8 . I Z = 0, I (mod 4) and assume it is true for every graph of order p < IZ and size at most p 1, p = 0, I (mod 4). Consider a graph G of order n and size at most n 1 . We assume that G has precisely IZ I edges. We easily verify that K , , , , , , C3 U K , , C 3 U K , ,,,-, $ ( n 2 51, C, U K , are not contained in their complement, a fortiori in any S.C. graph. ‘Then. we assume that G is not isomorphic to any of these graphs. Property (P) and Lemma 2 given below are very useful to prove most of the cases. Property (P). G has four vertices .Y. p. :, t such that if G’ = G {.u,p.z,t}, then d(s ,G’) = d(t.G’) = 0 [resp. ti(?.,(;’) = tl(z.G’) = 01, and xz, x t , y t @ E(G) , and there exists an S.C. permutation of G’. (Observe that if 4 G ’ ) d IZ 6 then an S.C. permutation of G’ exists.) Lemma 2. If for x, y , z , f E V(G) Property (P) holds. then there is an S.C. permutation of G. Proof of Lemma 2 . We use Lemma 1 to show that for (T‘ = u(G’), (T = u’o(x ,y , t ,z ) is an S.C. permutation of G. Indeed, if d(.v,G’) = d(t,G’) = 0, then (~(xy) = m3(zr) = yt @ E ( G ) , a(zf) = a 3 ( x y ) = .rz @ E(G) , ( ~ ( y z ) = cr3(yz) = x t 6 E(G) , and for every E v(G’), ~ ( y i r ) = ta’(u) 6 E(G), a3(yrr) =X U ’ ~ ( L I ) @ E(G) , ( T ( Z I d ) = xa’ (u> @ E(G), (T3(zu) = ta”(u) 4 E(G). The argument is similar if d(y,G‘) = d(z,G‘) = 0. In particular, the components of G which are cycles will be assumed to have 3 or 4 vertices and we shall assume that there is at most one isolated vertex in G. Suppose first that G is connected. For P ( G ) = n 1, G is a tree, and since G # K , , , , ,, there are two endvertices p and 4 joined by a path ( p , x l , .... xk, 4) k 3 2 . So, either G’ { p , q , x , , . r k } admits an S.C. permutation and hence Property (P) holds for p , x , , x k , q or C’ = K, . , , -5 and Property (P) holds for PJ,, the vertex y of G‘ of maximum degree in G’ and an endneighbor of y. So assume G is not connected, and let TI , ..., T,, (LY 2 1) be the treecomponents of G , and H I , ..., Hfi ( p 2 1) the other components. I 338 JOURNAL OF GRAPH THEORY Case 1. G contains an isolated edge ah. Then every vertex of degree at least 3 [resp. at least 21 is joined to every vertex of degree at least I (except u,h) [resp. at least 21, otherwise G has Property (P) and Lemma 2 applies. Since n 3 8, there is a vertex .Y of degree at least 3 . Then d(.y,G) = n 3 k , where k is the number of isolated vertices ( k s 1 ) . G has Property (P) for the vertices a.b,x and every vertex t # n,b of degree at most 1 . Thus, in the following cases, we can assume that the number of vertices in 7; is different from 2 , for i = 1 , ..., a. Case 2. G contains an isolated vertex a . We consider the following subcases: Suhcase 2.1. d(.u,G) s 2 for every .r E V(G). Then a = 1 and H I = C, or C, for i = 1, ..., p. Subcase 2.I.a. n = 8 or n = 9, then G contains a cycle C = (xl, ..., x4.x,). Let yI, y z . y3 be three consecutive vertices of the second cycle C ‘ . Then u(G) = ~ o ( ~ ~ , x , , y , , y ~ ) o ( x ~ , x , , y , , ~ ~ ) , where % = P, and y , = a if n = 8 and %‘ = (a ) and y4 is the fourth vertex of C‘ if n = 9. Subcase 2.1.b. IZ 2 12 and G contains a cycle C = (x , , ..., x4.xI) of length 4. Consider three vertices yl, )I?, y , of a cycle C‘ # C of G and a vertex y 3 $Z V(C) U V(C‘) of degree at least 2. Choose w(G) = a(G’)o(.rl,y,,xz,y,)o(x4,x,, y 2 , y , ) . where G‘ = G {XI? .... -y4?yI,. ... ? y4}.