Bicyclic digraphs with extremal skew energy

Transcription

1 Electronic Journal of Linear Algebra Volume 3 Volume 3 (01) Article 01 Bicyclic digraphs with extremal skew energy Xiaoling Shen Yoaping Hou yphou@hunnu.edu.cn Chongyan Zhang Follow this and additional works at: Recommended Citation Shen, Xiaoling; Hou, Yoaping; and Zhang, Chongyan. (01), "Bicyclic digraphs with extremal skew energy", Electronic Journal of Linear Algebra, Volume 3. DOI: This Article is brought to you for free and open access by Wyoming Scholars Repository. It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository. For more information, please contact scholcom@uwyo.edu.

2 BICYCLIC DIGRAHS WITH EXTREMAL SKEW ENERGY XIAOLING SHEN, YAOING HOU, AND CHONGYAN ZHANG Abstract. Let G be a digraph and S( G) be the skew-adjacency matrix of G. The skew energy of G is the sum of the absolute values of eigenvalues of S( G). In this paper, the bicyclic digraphs with minimal and maximal skew energy are determined. Key words. Bicyclic digraph, Skew-adjacency matrix, Extremal skew energy. AMS subject classifications. 05C05, 05C50, 15A Introduction. Let G be a simple undirected graph of order n with vertex set V(G) = {1,...,n} and G be an orientation of G. The skew-adjacency matrix of G is the n n matrix S( G) = [s i,j ], s i,j = 1 and s j,i = 1 if i j is an arc of G, and s i,j = s j,i = 0 otherwise. Since S( G) is a real skew symmetric matrix, all eigenvalues {λ 1,..., λ n } of S( G) are pure imaginary numbers, the singular values of S( G) are just the absolute values { λ 1,..., λ n }. So the energy of S( G) defined as the sum of singular values of S( G) [1] is the sum of the absolute values of its eigenvalues. For convenience, the energy of S( G) is called skew energy of G ([1]), and denoted by E s ( G). Energy has close links to chemistry (see, for instance, [6]). Since the concept of the energy of simple undirected graphs was introduced by Gutman in [5], there has been lots of research papers on this topic. For a survey, we refer to Section 7 in [3] and references therein. Denote, as usual, the n-vertex path and cycle by n and C n, respectively. For the extremal energy of bicyclic graphs, let G(n) be the class of bicyclic graphs with n vertices and containing no disjoint odd cycles of lengths k and l with k + l = (mod ). Let Sn l be the graph obtained by connecting n l pendant vertices to a vertex of C l, Sn 3,3 be the graph formed by joining n pendant vertices to a vertex of degree three of the K e (see Fig. 1.1), and Sn, be the graph formed by joining n 5 pendant vertices to a vertex of degree three of the complete bipartite graph K,3. Let B n be the class of all bipartite bicyclic graphs of order n that are not the graph obtained from two cycles C a and C b (a,b 10 and a = b = (mod )) joined by an edge. Zhang and Zhou [1] showed that Sn 3,3 is the graph with Received by the editors on June 1, 011. Accepted for publication on December 18, 011. Handling Editor: Xingzhi Zhan. This project is supported by NSF of China. Department of Mathematics, Hunan Normal University, Changsha, Hunan 10081, China (xlshen003003@yahoo.com.cn, yphou@hunnu.edu.cn). 30

3 Bicyclic Digraphs With Extremal Skew Energy 31 minimal energy in G(n). In [10], Li et al. showed that 6,6 n is the graph with maximal energy in B n. Additional results on the energy of bicyclic graphs can be found in [, 7, 11, 13]. The skew energy was first introduced by C. Adiga et al. in [1]. Some properties of the skew energy of a digraph are given in [1]. A connected graph with n vertices and n+1 edges is called a bicyclic graph. In this paper, we are interested in studying the bicyclic graphs with extremal skew energy.. n S 3,3 n. 7 n 7 7, n 7 7 S 3,3 n, n Fig Graphs Sn 3,3, n, and their orientations. The rest of this paper is organized as follows. In Section, the bicyclic digraphs of each order n with minimal skew energy are determined. In Section 3, the bicyclic digraphs of each order n with maximal skew energy are determined.. Bicyclic graphs with minimal skew energy. Let G be a graph. A linear subgraph L of G is a disjoint union of some edges and some cycles in G ([]). A k-matching M in G is a disjoint union of k-edges. If k is the order of G, then a k- matching of G is called a perfect matching of G. The number of k-matchings of graph G is denoted by m(g,k). If C is an even cycle of G, then we say C is evenly oriented relative to an orientation G of G if it has an even number of edges oriented in the direction of the routing. Otherwise C is oddly oriented. We call a linear subgraph L of G evenly linear if L contains no cycle with odd length and denote by EL i (G) (or EL i for short) the set of all evenly linear subgraphs of G with i vertices. For a linear subgraph L EL i, denote by p e (L) (resp., p o (L)) the number of evenly (resp., oddly) oriented cycles in L relative to G. Denote the characteristic polynomial of S( G) by S ( G;x) = det(xi S( G)) = n b i x n i. i=0 Then b 0 = 1, b is the number of edges of G, all b i 0 and b i = 0 for all odd i.

4 3 X. Shen, Y. Hou, and C. Zhang We have the following results. Lemma.1. ([8]) Let G be an orientation of a graph G. Then b i ( G) = L EL i ( ) pe(l) po(l). Lemma.. ([8]) Let e = uv be an edge of G that is on no even cycle of G. Then (.1) S ( G;x) = S ( G e;x)+ S ( G u v;x). By equating the coefficients of polynomials in Eq. (.1), we have (.) b k ( G) = b k ( G e)+b k ( G u v). Furthermore, if e = uv is a pendant edge with pendant vertex v, then (.3) b k ( G) = b k ( G v)+b k ( G u v). For any orientation of a graph that does not contain any even cycle (in particular, a tree or a unicyclic non-bipartite graph), b k ( G) = m( G,k) by Lemma.1. In [9], the skew energy of G is expressed as the following integral formula: E S ( G) = 1 π 1 n t ln(1+ b k t k )dt. k=1 Thus E s ( G) is an increasing function of b k ( G), k = 0,1,..., n. Consequently, if G 1 and G are oriented graphs of G 1 and G, respectively, for which (.) b k ( G 1 ) b k ( G ) for all n k 0, then (.5) E s ( G 1 ) E s ( G ). Equality in (.5) is attained only if (.) is an equality for all n k 0. If the inequalities (.) hold for all k, then we write G 1 G or G G 1. If G 1 G, but not G G 1, then we write G 1 G. Let G be an orientation of a graph G. Let W be a subset of V(G) and W = V(G) \ W. The orientation G of G obtained from G by reversing the orientations

5 Bicyclic Digraphs With Extremal Skew Energy 33 of all arcs between W and W is said to be obtained from G by a switching with respect to W. Moreover, two orientations G and G of a graph G are said to be switching-equivalent if G can be obtained from G by a sequence of switchings. As noted in [1], since the skew adjacency matrices obtained by a switching are similar, their spectra and hence skew energies are equal. It is easy to verify that up to switching equivalence there are just two orientations of a cycle C: (1) Just one edge on the cycle has the opposite orientation to that of others, wecallit orientation+. ()All edgesonthecyclec havethe sameorientation, we denote this orientation by. So if a cycle is of even length and oddly oriented, then it is equivalent to the orientation +; if a cycle is of even length and evenly oriented, then it is equivalent to the orientation. The skew energy of a directed tree is the same as the energy of its underlying tree ([1]). So by switching equivalence, for a unicyclic digraph or bicyclic digraph, we only need to consider the orientations of cycles. Let C x, C y be two cycles in bicyclic graph G with t (t 0) common vertices. If t 1, then G contains exactly two cycles. If t, then G contains exactly three cycles. The third cycle is denoted by G z, where z = x+y t+. Without loss of generality, assume that x y z. For convenience, we denote by G + (resp., G ) the unicyclic graph on which the orientation of a cycle is of orientation + (resp., ), and denote by G the unicyclic graph on which the orientation of a cycle is of arbitrary orientation. If t 1, we denote by G a,b the bicyclic graph on which cycle C x is of orientation a and cycle C y is of orientation b, where a, b {+,, }. If t, we denote by G a,b,c the bicyclic graph on which C x is of orientation a, C y is of orientation b, C z is of orientation c, where a, b, c {+,, }. See examples in Fig... For the k-matching number of a graph G, we have the following. Lemma.3. Let e = uv be an edge of G. Then (i) m(g,k) = m(g e,k)+m(g u v,k 1). (ii) If G is a forest, then m(g,k) m( n,k), k 1. (iii) If H is a subgraph of G, then m(h,k) m(g,k), k 1. Moreover, if H is a proper subgraph of G, then the inequality is strict. We define m(g,0) = 1 and m(g,k) = 0 for k > n. In [9], the authors discussed the unicyclic digraph with extremal skew energy and established the following. Lemma.. ([9]) (1) Among all unicyclic digraphs on n vertices, S 3 n has the

6 3 X. Shen, Y. Hou, and C. Zhang t = 1, G +,+ t = 0, G +, t =, G +,+, Fig..1. Examples for orientation representations of bicyclic digraphs. minimal skew energy and S n has the second minimal skew energy for n 6; both S 3 5 and S 5 have the minimal skew energy, S 5 + has the second minimal skew energy for n = 5; C has the minimal skew energy, S 3 has the second minimal skew energy for n =. () Among all orientations of unicyclic graphs, n+ is the unique directed graph with maximal skew energy. For bicyclic digraphs, we have the following. Lemma.5. Let G be a bicyclic digraph of order n 8, G Sn 3,3. Then G (S 3,3 n ),,. roof. We prove the statement by induction on n. By Lemma.1, the characteristic polynomials of S((S 3,3 n ),, ), S( S 3 n ) and S( S n ) are: S ((S 3,3 n ),, ) = x n (x +(n+1)x +(n )), S ( S 3 n) = x n (x +nx +(n 3)), S ( S n ) = x n (x +nx +(n )). It suffices to prove that b ( G) > (n ) for G (S 3,3 n ),,. Let n = 8.

7 Bicyclic Digraphs With Extremal Skew Energy 35 Case 1.1. t 1. Subcase x = y =. Then we can choose an edge e = uv on some C such that G u v is connected. By Lemma.1, we have b (G, ) m(g,) = m(g e,)+m(g u v,1) m(g e,)+6 > m( 6,)+ ( 6 is a proper subgraph of G e) = 8 = b ((S 3,3 8 ),, ). Subcase Either x or y is. Without loss of generality, suppose that x =. Chose an edge e = uv on C y such that G u v has at least edges. By Lemma.1, b (G, ) m(g,) = m(g e,)+m(g u v,1) m(g e,)+ > m(s6,)+ (S 6 is a proper subgraph of G e) > 8 = b ((S 3,3 8 ),, ). Subcase Neither x nor y is. Then x = y = 3 or x = 3, y = 5. We can chose an edge e = uv on any cycle such that G u v contains at least 3 edges. By Lemma.1, we get b (G, ) = m(g,) = m(g e,)+m(g u v,1) m(g e,)+3 = b (G e)+3 (G e is a unicyclic graph without cycle of length ) > b ( S 3 8)+3 (by Lemma.) = 8 = b ((S 3,3 8 ),, ). Case 1.. t. Subcase Each cycle is of length. Then t = 3 and there are 3 vertices outside of those cycles, say v 1, v, v 3. Let v 1 be a pendant vertex of G and u 1 be the adjacent vertex of v 1, v be a pendant vertex of G v 1 and u be the adjacent vertex of v, v 3 be a pendant vertex of G v 1 v and u 3 be the adjacent vertex of v 3. By Eq. (.3), we have b (G, ) = b (G, v 1 )+b (G, u 1 v 1 )

8 36 X. Shen, Y. Hou, and C. Zhang b (G, v 1 )+3 b (G, v 1 v )+6 b (G, v 1 v v 3 )+9 > 8 = b ((S 3,3 8 ),, ). Subcase 1... There are two cycles of length in G, say, C x, C y, then t = and z = x+y t+ = 6. There are two vertices outside of G. Similar to the proof of subcase 1..1, we can obtain b (G, ) > b ((S 3,3 8 ),, ). Subcase There is just one cycle of length in G, say, C x. If C y = C 3, then t = and there are 3 vertices outside of G. Similar to the proof of subcase 1..1, we get b (G, ) > b ((S 3,3 8 ),, ). If y 5, we can chose an edge e = uv on C y such that m(g u v,1) 6. Then similar to the proof of subcase 1.1., we get that b (G, ) > b ((S 3,3 8 ),, ). Subcase 1... G contains no cycle of length. Similar to the proof of subcase 1.1.3, the result holds for n = 8. Suppose n > 8 and G (S 3,3 n ),, for any bicyclic digraph G of order n, n < n. Denote by p the number of pendant vertices in G. If p = 0, then G has no pendant vertex. Three cases are considered in the following. Case.1. t = 1. Let e = uv be an edge on C x and u is the common vertex of C x and C y. By Lemmas.1 and.3, and m( n,) = (n )(n 3), we have b (G, ) m(g,) = m(g e,)+m(g u v,1) = m( y n,)+m( x n x,1) = m( n,)+m( y n y,1)+n 8 = (n )(n 3) since (n )(n 3) = (n 6)(n+1)+ > 0 for n > 7. +n +n 8 > (n ) Case.. t. Suppose e = uv is an edge on C x and u is the common vertex of C x and C y. By Lemmas.1 and.3 b (G, ) m(g,) 6 = m(g e,)+m(g u v,1) 6 = m( y n,)+n 3 6

9 Bicyclic Digraphs With Extremal Skew Energy 37 = m( n,)+m( y n y,1)+n 9 = (n )(n 3) since (n )(n 3) 5 = (n 6)(n+1)+ > 0 for n > 7. +n +n 9 > (n ) Case.3. t = 0. Suppose that C x and C y are joined by a path of length a, n 8 a 0. Let e = uv be an edge on C x, where u is the vertex of degree 3. Similar to the proof of case.1, we obtain b (G, ) > (n ). Therefore, b (G, ) b ((Sn 3,3),, ) for p = 0. Let p 1 and v be a pendant vertex of G with corresponding unique edge uv. Since G, u v has at least 3 edges, by Eq. (.) and the induction hypothesis, b (G, ) = b (G, v)+b (G, u v) > b ((S 3,3 n 1 ),, )+3 = (n 1 )+3 > (n ) = b ((S 3,3 n ),, ). For n = 7, similar to the proof of Lemma.5 for n = 8, both (S, (Sn 3,3),, have the minimal skew energy. Since S ((S, n ),, ) = x n (x +(n+1)x +3(n 5)), 7 ),, and b ((S, 6 ),, ) = 3 and b ((S, 5 ),, ) = 0. In a similar way to the proof of Lemma.5 for n = 8, we can get that (S, 6 ),, has the minimal skew energy for n = 6 and (S, 5 ),, has the minimal skew energy for n = 5. By Lemma.5 and the above statements, we obtain the following. Theorem.6. Among all bicyclic digraphs of order n, (Sn 3,3),, has the minimal skew energy for n 8; both (S 3,3 7 ),, and (S, 7 ),, have the minimal skew energy for n = 7; (Sn,),, has the minimal skew energy for n = 5,6. 3. Bicyclic digraphs with maximal skew energy. For the path, by Lemmas.1 and.3, we can easily get the following statements. Lemma 3.1. Let F n be a forest of order n. Then F n n. Equality holds if and only if F n = n. Since the skew energy of a directed forest is the same as the energy of its underlying forest, by [6], we have the following.

10 38 X. Shen, Y. Hou, and C. Zhang Lemma 3.. n n n k n k k+1 n k 1 k 1 n k+1 3 n 3 1 n 1. Lemma 3.3. For any bicyclic graph G with t 1, G, G +,+. roof. If two cycles are of odd length, then by Lemma.1, for any orientation of G, b k ( G) = m(g,k), for all 0 k n. Thus G, = G +,+. If there is exactly one cycle of even length in G, say, C x, then b k (G, ) = m(g,k) m(g C x,k x ) b k(g +,+ ) = m(g,k)+m(g C x,k x ). If both x and y are even, then b k (G, ) = m(g,k) m(g C x,k x ) m(g C y,k y ) + m(g C x C y,k x+y ) b k (G ±, ) = m(g,k) ± m(g C x,k x ) m(g C y,k y ) m(g C x C y,k x+y ) b k (G +,+ ) = m(g,k) + m(g C x,k x )+m(g C y,k y ) + m(g C x C y,k x+y ). Lemma 3.. Let G be a bicyclic digraph of order n with t 1, G (, Then G (n,)+,+ for n 8. roof. We divide the proof into two cases. Case 1. There is at least one cycle of length odd, say, C x. n )+,+. (i) t = 1, we can choose an edge e = uv on C x such that u is the common vertex of two cycles. Obviously, G e is a unicyclic graph and G u v is a forest. By Eq. (.), Lemmas.3 and., we have b k ( G) = b k ( G e)+b k ( G u v) < b k ( n+ )+bk ( n ) (by Lemmas. and 3.) = m( n,k)+3m( n,k )+m( n,k 1) < m( n,k)+m( n,k 1)+5m( n,k )+m( n 8,k )

11 = b k ((, n ) +,+ ). Bicyclic Digraphs With Extremal Skew Energy 39 (ii) t = 0. We can choose an edge e = uv on C x such that u is a vertex in a path which connects C x and C y. Obviously, G e is a unicyclic graph and G u v is the disjoint union of a forest and a unicyclic graph. Claim 1. a n a b + n +, a. roof. By Lemmas. and 3., we have b k ( a n a +) b < b k ( a n a+ ) = m( a n a,k)+m( a n a,k ) < m( a n a,k)+m( a n a,k 1) +m( n 6,k ) < m( n,k)+m( n 6,k 1) +m( n 6,k ) = b k ( n + ). By Eq. (.), Lemmas.1,.3 and., we have b k ( G) = b k ( G e)+b k ( G u v) < b k ( n + )+bk ( n + ) (by Claim 1) = m( n,k)+m( n,k )+m( n,k 1) +m( n 8,k ) m( n,k)+m( n,k 1)+m( n,k ) m( n,k)+m( n,k 1)+m( n,k )+m( n 8,k ) = b k ((, n ) +,+ ). Case. Two cycles are of even lengths. By Lemma 3.3, we only need to consider G +,+. (1) t = 1. We can choose an edge e = uv in cycle C x, u is the common vertex of two cycles. By Lemma.1, we have b k (G +,+ ) = m(g,k)+m(g C x,k x )+m(g C y,k y )

12 350 X. Shen, Y. Hou, and C. Zhang m(g e,k)+m(g u v,k 1)+m( n,k ) +m(g C y,k y ) = b k ( G e + )+m(g u v,k 1)+m( n,k ) b k ( G e + )+m( n,k 1)+m( n,k )) < b k ( n+ )+m( n,k 1)+m( n,k ) = m( n,k)+m( n,k 1)+m( n,k ) m( n,k)+m( n,k 1)+m( n,k )+m( n 8,k ) = b k ((, n )+,+ ). (ii) t = 0. Choose an edge e = uv on the path which connects C x and C y. Assume that one of the two components of G e is of order j. By Eq. (.), we get b k (G +,+ ) = b k (G +,+ e)+b k (G +,+ u v) < b k ( + j n j + )+bk ( + j 1 n j 1+ ) ( or < b k ( + j n j + )+bk ( 3 n 5+ )) = b k ((, n ) +,+ ). Lemma 3.5. Let G be a bicyclic digraph of order n with t. Then G (, n ) +,+ for n 8. roof. We prove statement by dividing three cases. Case 1. x = y = z =. Then t = 3. If both C x and C y are oddly oriented, then C z must be evenly oriented. We can choose an edge e = uv such that G u v is disconnected. Without loss of generality, we assume that e is on C y. b k (G +,+, ) = m(g,k)+m(g C x,k )+m(g C y,k ) m(g C z,k ) m(g e,k)+m(g u v,k 1)+m(G C x,k ) +m(g C y,k ) b k ( G e)+m( n,k 1)+m( n,k ) < b k ( n+ )+m( n,k 1)+m( n,k ) = m( n,k)+m( n,k 1)+m( n,k ) m( n,k)+m( n,k 1)+m( n,k )

13 Bicyclic Digraphs With Extremal Skew Energy 351 +m( n 8,k ) = b k ((, n )+,+ ). If either C x or C y is oddly oriented, then C z must be oddly oriented. Similarly, we can prove that b k (G +,,+ ) < b k ((n, ) +,+ ) or b k (G,+,+ ) < b k ((n, ) +,+ ). If both C x and C y are evenly oriented, then C z is also evenly oriented. b k (G,, ) = m(g,k) m(g C x,k ) m(g C y,k ) m(g C z,k ) m(g,k)+m(g C x,k )+m(g C y,k ) m(g C z,k ) = b k (G +,,+ ) < b k ((, n )+,+ ). Case. x = y =, z. Then t = and z = 6. If both C x and C y are oddly oriented, then C z is oddly oriented. Since n 8, we can choose an edge e = uv such that G u v is disconnected and u is one of the common vertices between C x and C y. Without loss of generality, we suppose that e is on C y. Then both G C y and G C z are also disconnected. Note that G C x = G e C x, by Lemma.1, we get b k (G +,+,+ ) = m(g,k)+m(g C x,k )+m(g C y,k ) +m(g C z,k 3) m(g e,k)+m(g u v,k 1)+m(G e C x,k ) +m( n 6,k )+m( n 8,k 3) b k ( G e + )+m( n,k 1) +m( n 6,k )+m( n 8,k 3) < b k ( n+ )+m( n,k 1) +m( n 6,k )+m( n 8,k 3) = m( n,k)+m( n,k 1)+m( n,k ) +m( n 6,k )+m( n 8,k 3) m(n,k)+m( n,k 1)+m(n,k ) +m( n 8,k ) = b k ((, n )+,+ ). If either C x or C y is oddly oriented, then C z is evenly oriented. By Lemma

14 35 X. Shen, Y. Hou, and C. Zhang.1, b k (G +,, ) b k (G +,+,+ ) < b k ((n, ) +,+ ), or b k (G,+, ) b k (G +,+,+ ) < b k ((n,)+,+ ). If both C x and C y are evenly oriented, then C z is oddly oriented and b k (G,,+ ) b k (G +,+,+ ) < b k ((, n ) +,+ ). Case 3. There aren t two cycles with length. Since there is at least one cycle of even length in G, without loss of generality, we assume that C x is a cycle of minimal even length. Subcase 3.1. t is even. Subcase y isevenandbothc x andc y areoddlyoriented. Theny,z > x and C z is oddly oriented. Let e = uv be an edge on C y and u is the common vertex of C x and C y. Claim. m( n,k 1) m( n,k ) m( n l,k l). roof. By Lemma.3, we get m( n,k 1) = m( n 3,k 1)+m( n,k ) = m( n 3,k 1)+m( n 5,k )+m( n 6,k 3) l 1 = m( n (i+1),k i)+m( n l,k l). i=1 Claim follows immediately. By Lemma.1 and Claim, we get b k (G +,+,+ ) = m(g,k)+m(g C x,k )+m(g C y,k y ) +m(g C z,k z ) m(g e,k)+m(g u v,k 1)+m(G e C x,k ) +m( n 6,k 3) b k ( G e + )+m( n,k 1)+m( n 6,k 3) < b k ( n+ )+m(n,k 1)+m( n 8,k 3) +m( n 9,k ) = m( n,k)+m( n,k )+m( n,k 1)+m( n 5,k ) +m( n 8,k 3)+m( n 9,k ) m( n,k)+m( n,k 1)+m( n,k )

15 Bicyclic Digraphs With Extremal Skew Energy 353 +m( n 8,k 3)+m( n 8,k ) = b k ((, n ) +,+ ). If either C x or C y is oddly oriented, then C z is evenly oriented. By Lemma.1, b k (G +,, ) b k (G +,+,+ ) < b k ((n, ) +,+ ), or b k (G,+, ) b k (G +,+,+ ) < b k ((n,)+,+ ). If both C x and C y are evenly oriented, then C z is oddly oriented and b k (G,,+ ) b k (G +,+,+ ) < b k ((, n )+,+ ). Subcase y is odd. Then z is also odd. We can choose an edge e = uv on C y such that u is the common vertex and v is not the common vertex. By Lemma.1, we get b k (G,, ) b k (G +,, ) = b k ( G e + )+b k ( G u v) < b k ( n+ )+bk ( n ) = m( n,k)+m( n,k )+m( n,k 1) < m( n,k)+m( n,k 1)+m( n 5,k )+m( n,k ) m(n,k)+m( n,k 1)+m(n,k ) +m( n 8,k ) = b k ((, n )+,+ ). Subcase 3.. t is odd. Subcase y is even. Then y >. If both C x and C y are oddly oriented, then z is even and C z is evenly oriented. Let e = uv be an edge on C y and u is the common vertex between C x and C y. Then b k (G +,+, ) = m(g,k)+m(g C x,k )+m(g C y,k y ) m(g C z,k z ) m(g e,k)+m(g u v,k 1)+m(G C x,k ) +m( n 6,k 3) b k ( G e + )+m( n,k 1)+m( n 6,k 3) < b k ( n+ )+m(n,k 1)+m( n 6,k 3) = m( n,k)+m( n,k )+m( n,k 1)+m( n 5,k ) +m( n 6,k 3)

16 35 X. Shen, Y. Hou, and C. Zhang m(n,k)+m( n,k 1)+m(n,k ) +m( n 8,k ) = b k ((, n )+,+ ). If either C x or C y is oddly oriented, then C z is oddly oriented. Similar to the above proof, b k (G +,,+ ) < b k ((n,)+,+ ) or b k (G,+,+ ) < b k ((n,)+,+ ). If both C x and C y are evenly oriented, then C z is evenly oriented, so b k (G,, ) b k (G +,+, ) < b k ((, n ) +,+ ). Subcase 3... y is odd. Then z is odd too. Similar to the proof of subcase 3.1., we obtain b k (G,, ) b k (G +,, ) < b k ((, n ) +,+ ). Combining all those cases above, we complete the proof. By identifying two vertices of two cycles with length, we get a graph G, 7. For n = 6, 7, similar to the proofs of Lemmas 3., 3.5, we obtain that the following graphs have the maximal skew energy. Fig The maximal skew energy graph (, 6 ) +,+,+ for n = 6 and (G, 7 )+,+ for n = 7. By Lemmas 3. and 3.5, we obtain the following statement. Theorem 3.6. Among all bicyclic digraphs with order n 8, (n,)+,+ has the maximal skew energy; (G, 7 )+,+ has the the maximal skew energy for n = 7; (, 6 ) +,+,+ has the the maximal skew energy for n = 6. Acknowledgment. The authors would like to thank the referees for some very useful comments and valuable suggestions, which make a number of improvements on this paper.

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