Primitive Digraphs with the Largest Scrambling Index

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1 Primitive Digraph with the Larget Scrambling Index Mahmud Akelbek, Steve Kirkl 1 Department of Mathematic Statitic, Univerity of Regina, Regina, Sakatchewan, Canada S4S 0A Abtract The crambling index of a primitive digraph D i the mallet poitive integer k uch that for every pair of vertice u v, there i a vertex w uch that we can get to w from u v in D by directed walk of length k; it i denoted by k(d). In [1] we gave the upper bound on k(d) in term of the order the girth of a primitive digraph D. In thi paper, we characterize all the primitive digraph uch that the crambling index i equal to the upper bound. AMS claification: 05C0; 05C50 Key word: Scrambling index; Primitive digraph 1 Introduction There are numerou reult giving the upper bound on the econd larget modulu of eigenvalue of primitive tochatic matrice (ee [,5 8]). In [1], by uing Seneta [6] definition of coefficient of ergodicity, we have provided an attainable upper bound on the econd larget modulu of eigenvalue of a primitive matrix that make ue of the o-called crambling index (ee below). For vertice u, v w of a digraph D, if (u, w), (v, w) E(D), then vertex w i called a common out-neighbour of vertice u v. The crambling index of correponding author. addree: akelbek@math.uregina.ca (Mahmud Akelbek), kirkl@math.uregina.ca ( Steve Kirkl). 1 Reearch upported in part by a grant from the Natural Science Engineering Reearch Council of Canada under grant OGP Preprint ubmitted to Elevier April 008

2 a primitive digraph i the mallet poitive integer k uch that for every pair of vertice u v, there exit a vertex w uch that u k w v k w in D. The crambling index of D will be denoted by k(d). The main reult in [1] i the following. Theorem 1.1 [1] D be a primitive digraph with n vertice girth. Then k(d) K(n, ). (1) Equality hold if D = D,n gcd(n, ) = 1. Where D,n i a digraph a in Figure 1, K(n, ) = k(n, ) + n k(n, ) = ( 1 )n, when i odd, ( n 1 ), when i even. 1 n n Fig. 1. D,n In thi paper, we characterize all the primitive digraph D uch that k(d) = K(n, ). Some reult on crambling index For terminology notation ued here we follow [1] []. Let D = (V, E) denote a digraph (directed graph) with vertex et V = V (D), arc et E = E(D) order n. Loop are permitted but multiple arc are not. A u v walk in a digraph D i a equence of vertice u, u 1,..., u t, v V (D) a equence of arc (u, u 1 ), (u 1, u ),..., (u t, v) E(D), where the vertice

3 arc are not necearily ditinct. A cloed walk i a u v walk where u = v. A cycle i a cloed u v walk with ditinct vertice except for u = v. The notation u k v i ued to indicate that there i a u v walk of length k. The ditance from vertex u to vertex v in D, i the length of a hortet walk from u to v, denoted by d(u, v). A p-cycle i a cycle of length p, denoted C p. If the digraph D ha at leat one cycle, the length of a hortet cycle in D i called the girth of D, denoted (D). The number of arc entering (leaving) a vertex u i called the in-degree (out-degree) of u, denoted deg (u) (deg + (u)). A digraph D i called primitive if for ome poitive integer t there i a walk of length exactly t from each vertex u to each vertex v. If D i primitive, the mallet uch t i called the exponent of D, denoted by exp(d). A digraph D i primitive if only if it trongly connected the greatet common divior of all cycle length in D i equal to one []. For a poitive integer r, we define D r to be the digraph with the ame vertex et a D arc (u, v) if only if u r v in D. Conequently, the crambling index i the mallet poitive integer k uch that each pair of vertice ha a common out-neighbour in D k. For a vertice u, v V (D) (u v), we define k u,v (D) = min{k : u k w v k w, for ome w V (D)}. Then k(d) = max u,v V (D) {k u,v(d)}. Lemma.1 [1] Let p be poitive integer uch that gcd(p, ) = 1 p >. Then for each t, 1 t max{ 1, p/ }, the equation xp+y = t ha a unique integral olution (x, y) with x / y p/. Let D be a primitive digraph, let p be two different cycle length in D gcd(, p) = 1, where < p n. For u, v V (D), we can find a vertex w V (D) uch that there are directed walk from u to w v to w uch that both walk meet cycle of length p. Denote the length of thee directed walk by l(u, w) l(v, w). We ay that w i a double-cycle vertex of u v, we let l u,v = max{l(u, w), l(v, w)}. Lemma. [1] Let D be a primitive digraph, let p be two different cycle length in D. Suppoe that < p n gcd(, p) = 1. Then k u,v (D) min{ y, x p} + l u,v, () where (x, y) i the integer olution of the equation xp + y = r with minimum abolute value where l(u, w) l(v, w) r(mod).

4 Corollary. [1] Let D be a primitive digraph of order n with a Hamilton cycle, let the girth of D be, where 1 n 1 gcd(, n) = 1. If k(d) = K(n, ), then D contain a ubgraph iomorphic to D,n. Lemma.4 [1] Let D = D,n. Then for all vertice u v in D, l u,v (D) max{n, n }. Let r be the poitive integer that i defined a follow r n(mod), if i odd n i even, n (mod), if both n are odd. () Corollary.5 [1] Suppoe that gcd(, n) = 1,. Then for u, v V (D,n ), without lo of generality take u > v, k u,v (D,n ) = K(n, ) if only if u = n (1) v = n r t for ome t {0, 1,,, n r }, when i odd. () v = n, when i even. Lemma.6 [1] Let D be a primitive digraph with a Hamilton cycle let the girth of D be, where gcd(n, ) = 1, < n. Then either the cycle C i formed from conecutive vertice on the Hamilton cycle or there i another cycle of length p uch that gcd(, p) = q, where q when i even q when i odd. Lemma.7 [1] Let D be a primitive digraph with n vertice, uppoe that i the girth of D with. If there i another cycle of length p, < p n, uch that gcd(, p) = 1, then Furthermore, if p < n, then k(d) < K(n, ). k(d) K(n, ). (4) Let D be a primitive digraph L(D) = {, a 1,, a r } be the et of ditinct cycle length of D, where < a 1 < < a r. Lemma.8 [1] Let D be a primitive digraph with n vertice, be the girth of D with. Let L(D) = {, a 1,, a r }. If gcd(, a i ) 1 for each i = 1,,, r, Then k(d) < K(n, ). Corollary.9 [1] Let D be a primitive digraph of order n, be the girth of D with. If there i a cycle of length p, < p n, uch that 4

5 gcd(, p) < / or gcd(, p) / C C p, then k(d) < K(n, ). Characterization of primitive digraph with k(d) = K(n, ).1 Propertie of a primitive digraph D with k(d) = K(n, ) Let D be a primitive digraph with n vertice, be the girth of D, k(d) = K(n, ). Then by Lemma.7 Lemma.8 there i a cycle of length p, < p n, uch that gcd(, p) = 1 p = n. Since D contain a Hamilton cycle, then by Corollary. D contain D,n a a ubgraph. From the above, we conclude the following. Theorem.1 Let D be a primitive digraph with n vertice, let the girth of D be, uppoe that k(d) = K(n, ). Then (1) There i no cycle of length p, < p < n, uch that gcd(, p) = 1. () D contain D,n a a ubgraph gcd(, n) = 1. In the following we only conider primitive digraph that contain D,n a a ubgraph, we label the digraph D a in Figure 1. For D,n, by Corollary.5 we know all the pair of vertice u, v V (D,n ) uch that k u,v (D,n ) = K(n, ). Propoition. [4] The t-th power of a cycle of length p i the dijoint union of gcd(p, t) cycle of length p/ gcd(p, t). Definition. If the digraph D contain at leat two different cycle, then the ditance between two different cycle in D i defined a follow d(c, C ) = min{d(u, v) u C, v C }, where C C are different cycle in D. Lemma.4 Let D = D,n, gcd(n, ) = 1, let t be a poitive integer uch that t. Then (i) The digraph D t contain a Hamilton cycle t dijoint cycle of length /t. (ii) Every cycle of length /t i formed from /t conecutive vertice on the Hamilton cycle in D t. 5

6 Denote the t cycle of length /t in D t by H 1, H,, H t in order a in Figure, we ay that H i H (i+1)( mod t), where i = 1,,, t, are neighbour cycle in D t. We alo have the following: (iii) The ditance between two neighbour cycle of length /t in D t i either n or n + 1. t t Proof: (i) Since gcd(, n) = 1, then gcd(t, n) = 1. Therefore by Lemma., we know that D t contain a Hamilton cycle t dijoint cycle of length /t. (ii) For vertice i, 1 i t, we have i +pt C, 0 p 1. Alo we have t i t i + t t i + t t t i + ( t 1)t t i. Therefore every cycle of length /t i formed from /t conecutive vertice on the Hamilton cycle in D t. H t H 1 H t 1 H Fig.. D t (iii) There are two different type of directed path of length t in D,n. One type contain the arc 1, the other type doe not contain the arc 1. Oberving D t, we know that every arc in the Hamilton cycle in D t correpond to a directed path of length t in D,n that doe not contain the arc 1, all the other arc, we call them hortly -arc, correpond to directed path of length t in D,n that contain the arc 1. Alo notice that if u 1 u i an -arc, then 1 u 1 t (t 1) u. Let d(h i, H (i+1)( mod t) ) = q for ome i, then there exit a vertex u H i a vertex v H (i+1)( mod t) uch that d(u, v) = q in D t. From the digraph D t, we know that deg + (u) = deg (v) =. Hence u i the tarting vertex of an arc v i the ending vertex of an arc. Therefore 1 u t (t 1) v. 6

7 Since in D t, we have u q v, then in D,n we have u qt v thi directed walk doe not go through the arc 1. In D,n, the directed path from vertex u to vertex v without going through the arc 1 i of the form u l n n l v, where l 1, l t 1. Thu Hence n + 1 qt n (t 1) + (t 1), n + 1 qt n + (t 1) + t. n n q + 1. t t Therefore the ditance between any two neighbour cycle of length /t i n t or n t The cae i even Lemma.5 Let D be a primitive digraph that contain D,n a a ubgraph, where i the girth of D, gcd(n, ) = 1 i even. If D contain another cycle of length p, where p < n. Then k(d) < K(n, ). Proof. Let C p be the cycle of length p in the primitive digraph D. Cae 1: Suppoe gcd(, p) = r, with r <. Then by Corollary.9 we have k(d) < K(n, ). Cae : Suppoe gcd(, p) =. If C C p, we are alo done by Corollary.9. If C C p =, conider D. There are cycle of length cycle of length p. Let p = p. For u, v V (D ), l uv n. Hence k u,v (D ) ( 1 )p + n = p + n. Since p n, p n, we have k u,v (D) (n + p ) n + n < k(n, ) + n. Cae. gcd(, p) =. Since i even, then n i odd. We know there i only one pair of vertice u, v V (D,n ) uch that k u,v (D,n ) = k(n, ) + n, 7

8 they are vertex n n. Conider the digraph D. It i eay to ee that vertice n n are conecutive vertice on the Hamilton cycle in the digraph D, there are cycle of length cycle of length p repectively, where p = p p i odd (ince p = p ). Let p = t + 1 for ome nonnegative integer t. For vertex n, we can find a vertex w uch that the directed walk from vertex n to vertex w i a path through both cycle of length p, l(n, w) n p. Since in D, we have n 1 n. Then l(n, w) l(n, w) = 1 l(n, w) n p + 1. Therefore in the digraph D, we have n l(n,w)+t w n l(n,w)+p w. Thu k n,n (D ) n; hence k n,n (D) ( )n < k(n, ) + n. Cae 4. gcd(, p) =. Suppoe p = t, where 1 t < n. If t = 1, then p =. If the cycle C p i formed from vertice that are not conecutive on the Hamilton cycle, then by Lemma.6, there exit another cycle of length q uch that gcd(, q). For thi cae, from the previou reult we know that k n,n (D) < k(n, ) + n. If the cycle C p i formed by joining vertex i to vertex (i+ 1)(mod n), where i 1, then conider the ubgraph D p,n. Note that ince i 1, although p =, but C p C. Therefore D p,n D,n. In D p,n, the upper bound i attained for only one pair of vertice, they are vertex i 1 vertex (i+ )( mod n). Since i 1 n, we have k n,n (D p,n) < K(n, ). Therefore in the digraph D, we alo have k n,n (D) < k(n, ) + n. Now uppoe that t > 1, then < n. If C C p, there i at leat one vertex w belonging to the cycle C p uch that +1 w n 1. Otherwie the cycle C p only ha to contain vertice between vertex to vertex 1 n to n + 1. But there are only + uch vertice + < p. Hence for vertice n, we have l(n, w) < n. Then l(n, w) < n l(n, w) l(n ) =. In D,n, when n >, we get n n ( n 1 ) n n n. 8

9 When n <, we have n n ( n 1 ) Note that n 1 p = t. Hence n 1 n n +n ( 1)n. t let n 1 = t + t. Then ( n 1 ) = p + t, where n l(n,w) w p+t w l(n,w) n w. n w Therefore k n,n (D) l(n, w) + p + t < k(n, ) + n. If C C p =, for vertex n we can find a vertex w C p uch that l(n, w) n p. Then l(n, w) n p+ l(n, w) l(n, w) =. Since n 1 n 1 t, let n 1 t (mod t). For a nonnegative integer h we have n 1 = th + t. If t = 0, then ( n 1 ) = ht = hp, o n l(n,w) w hp w l(n,w) w n w. n Therefore k n,n (D) hp + l(n, w) < k(n, ) + n. If t 0, t > t > 0, we know that or equivalently Adding (t t ) on both ide, we get or Therefore we have n (n 1 ) =, (th + t ) n =. ht + t + (t t ) n = + (t t ), (h + 1)t ( n + (t t 1)) =. n l(n,w) w n+(t t 1) l(n,w) w n Then k n,n (D) n+(t t 1)+l(n, w) n+(t t 1)+n p = ( n 1 ) + n t < k(n, ) + n, a deired. w (h+1)p w. Theorem.6 Let D be a primitive digraph of order n girth, where i even. Then k(d) = K(n, ) if only if D = D,n gcd(n, ) = 1. 9

10 . The cae i odd Lemma.7 Let D be a primitive digraph that contain D,n a a ubgraph, where gcd(n, ) = 1, i odd. If D contain a cycle of length p with gcd(, p), then k(d) < K(n, ). Proof. Cae 1. gcd(, p) = l, l <. Then by Corollary.9 k(d) < k(n, ) + n. Cae. gcd(, p) =. If C C p, we are done by Corollary.9. If C C p =, conider D. There are cycle of length cycle of length p, let p = p. For u, v V (D ), we have l uv n. Hence Since p n p n k u,v (D ) ( 1 )p + n = p + n., we get k u,v (D) (n + p ) n + n < k(n, ) + n. Next we conider a primitive digraph D that contain D,n a a ubgraph, where gcd(, n) = 1 i odd, where the digraph D alo contain another cycle of length p with gcd(, p) =. Lemma.8 Let D be a primitive digraph that contain D,n a a ubgraph, where gcd(, n) = 1, i odd. Suppoe that the digraph D alo contain another cycle of length p with gcd(, p) =. If C C p, then k(d) < K(n, ). Proof. Suppoe that p = t that u i a vertex of D,n uch that k nu (D) = )n + n. ( 1 If u / C, then in the digraph D,n we have ( 1 )n n n u u m, where m i a poitive integer uch that m ( 1 )n = n u. If there i a vertex w uch that + 1 w u it belong to the cycle C p, then chooe w a the double-cycle vertex of u n. Then we have l(u, w) < u, l(n, w) < n l(n, w) l(u, w) = n u. Alo ince m > n > p p = t, then m = p + t for ome nonnegative integer t. Then 10

11 n l(n,w) w ( 1 )n w u l(u,w) w p+t w. Thu k n,u (D) ( 1 )n + l(n, w) < k(n, ) + n. Otherwie there i an arc from vertex j, u < j n, to vertex i, 1 i. Then we can get from vertex n to a vertex i on the cycle C in le than n tep. Therefore k n,u (D) < k(n, ) + n. Next conider u C. If p =, uppoe that the cycle C p i formed from conecutive vertice a in Figure. u 1 n v v+ Fig.. D,n {v v + } If v = u + 1, then l(n, w) < n l(u, w) = n. Therefore k n,u (D) < k(n, ) + n. If v u + 1, then conider the ubgraph D p,n. In D p,n, for ome vertex v we have k v 1,v (D p,n ) = K(n, ). Since v 1 u, n, then k n,u (D p,n ) < k(n, ) + n. Therefore k n,u (D) < k(n, ) + n. If the cycle C p i not formed from conecutive vertice, then by Lemma.6, there exit a cycle of length q uch that gcd(, q). In that cae, by Lemma.7, we have k(d) < k(n, ) + n. If p >, then take the firt vertex w on cycle C p from vertex n a the doublecycle vertex of u n. Since p, l(n, w) n. Since l(u, n) <, then l(u, w) < n. In the digraph D,n, there i a vertex u, u < u < n, uch that d(u, n) = d(n, u ) = n u, k n,u (D) = k(n, ) + n n n ( 1 )n 11

12 u u m, where m ( 1 )n = n u. Since m > n > p, then m = p + t for ome nonnegative integer t. In the digraph D we have n l(n,w) w p+t w u l(u,w) w ( 1 )n w, where l(u, w) l(n, w) = n u. Therefore k n,u (D) ( 1 )n + l(u, w) < )n + n. ( 1 Lemma.9 Let D be a primitive digraph that contain D,n a a ubgraph, uppoe that i odd,, that there i another cycle of length p uch that C C p = gcd(, p) =. If the cycle of length p i not formed from p conecutive vertice on the Hamilton cycle, then k(d) < K(n, ). Proof. Since the cycle of length p i not formed from p conecutive vertice on the Hamilton cycle, then there exit an arc from vertex i to vertex j, where + 1 i < j n j > i + 1. Then for any two vertice u, v V (D), we can get to vertice 1, C in le than n 1 tep. Therefore k(d) k(n, ) + n 1. The only remaining cae i that D i a digraph contructed from D,n by adding an arc from vertex u to vertex u + m 1, where i odd,, < u < n m + 1 m i a poitive integer uch that 1 m n u+1. Recall that in () we define the poitive integer r a follow r n (mod), if i odd, n i even, n (mod), if both n are odd. In both cae n r can be divided by. Let h = n r. (5) Note that in D,n, h+1 i the number of pair of vertice that attain the upper bound K(n, ). Lemma.10 Let D be a digraph contructed from D,n,, by adding an arc from vertex u to vertex u + m 1, where < u < n m + 1. Then k n,n r t (D) = K(n, ) if only if u = n r t + 1 n+h t 1 0(modm). 1

13 Proof. For the digraph D = D,n, the upper bound K(n, ) on the crambling index i attained by the pair of vertice n n r t, where 0 t n r. We only conider thoe pair of vertice. Suppoe that u = n r t + 1 for ome t. From the digraph we know that n r+t m n r t + m n r t n m n r t + m, n m (r + t m) = n r t = r + (h t), ince n = r + h. When n i even, ( n + h t) ( 1 )n = r + (h t). Suppoe m 1 q i the mallet nonnegative integer uch that ( n+h t + m 1 q) can be divided by p = m, where 0 q m 1. Then n r+t m n r t + m ( n+h t+m 1 q) n r t + m n r t n m n r t + m ( 1 )n+(m 1 q) n r t + m. Therefore k n,n r t (D) = ( 1 )n + n q. Since ( n+h t + m 1 q) can be divided by p = m, then n + h t 1 q(modm). Therefore if n+h t 1 0(modm), we have k n,n r t (D) = K(n, ). If n+h t 1 0(modm), then k n,n r t < K(n, ). Next we conider all other pair of vertice n u uch that k n,u (D,n ) = K(n, ). If u n r t + 1, let v = u + m 1. Conider the following three cae. Cae 1. n r t + 1 < u. We have n n v v 1

14 n r t n r t+n v v. In addition we have n r t + (n v) (n v) = n r t = r + (h t). Then we obtain n n v v ( n+h t+m 1 q) v n r t n r t+n v v ( 1 )n+(m 1 q) v. Therefore k n,n r t (D) = n r t + (n v) + ( 1 )n + (m 1 q) < n m + ( 1 )n + (m 1 q) = ( 1)n + n q k(n, ) + n. Cae. n r t > v. We have n n v v n r t n r t v v, n v (n r t v) = r + t. Alo Then ( n h + t) ( 1 )n = r + t. n n v v ( 1 )n+(m 1 q) v n r t n r t v v ( n h t+m 1 q) v. Therefore k n,n r t (D) = n v + ( 1 )n + (m 1 q) < n m + ( 1 )n + (m 1 q) = ( 1 )n + n q k(n, ) + n. Cae. u n r t v. Chooe v a the double-cycle vertex of n n r t. Then n n v v n r t n r t u+1 v. If n v > n r t u+1, ince n v (n r t u+1) = r+t (v u+1) = r + (t m) v > m, then k n,n r t (D) ( 1 )n + n v + (m 1 q) = ( 1 )n + n v + m q < k(n, ) + n. If n v < n r t u+1, then n r t u+1 (n v) = r t+v u+1 = r t + m = r + (m 1 t). Then ( 1 )n ( n t ) = r + (m 1 t) 14

15 for ome integer t. Therefore k n,n r t (D) ( 1 )n + n v + (m 1 q) = ( 1 )n + n v + m q < k(n, ) + n. Lemma.11 Let D be a digraph contructed from D,n ( ) by adding arc from vertex u i to vertex u i + m i 1, where u i >, m i 1, i = 1, u 1 u. Then k(d) < K(n, ). Proof. Let D i, i = 1,, be the ubgraph of D that contain D,n the cycle of length m i, then by Lemma.10, we know that there i at mot one pair of vertice, vertex n vertex u i 1, uch that k n,ui 1(D i ) = K(n, ). Since u 1 u, In the digraph D, we have k n,ui 1(D) < K(n, ). Concluding the above reult, we have the following theorem. Theorem.1 Let D be a primitive digraph of order n girth, where i odd. Then k(d) = K(n, ) if only if gcd(n, ) = 1 D = D,n or, D = D,n {n r t + 1 n r t + m} for ome m N ome t {1,,, n r 1} uch that n+h t 1 0(modm), where r h are a in () (5). Reference [1] M. Akelbek, S. Kirkl, Coefficient of ergodicity the crambling index, preprint. [] R.A. Brualdi, H.J. Ryer, Combinatorial Matrix Theory, Encyclopedia of Mathematic it Application 9, Cambridge Univerity Pre, Cambridge, [] D.J. Hartfiel, U.G. Rothblum, Convergence of inhomogeneou product of matrice coefficient of ergodicity, Linear Algebra Appl. 77 (1998), 1 9. [4] M. Kutz, The Angel Problem, Poitional Game, Digraph Root, Ph.D. thei, Freie Univeritat Berlin, 004. [5] U.G. Rothblum, C.P. Tan, Upper bound on the maximum modulu of ubdominant eigenvalue of nonnegative matrice, Linear Algebra Appl. 66 (1985), [6] E. Seneta, Coefficient of ergodicity: tructure application, Adv. Appl. Prob. 11 (1979),

16 [7] E. Seneta, Nonnegative Matrice Markov Chain, Springer-Verlag, New York, [8] C.P. Tan, Coefficient of ergodicity with repect to vector norm, J. Appl. Prob. 0 (198),

arxiv: v1 [math.mg] 25 Aug 2011

arxiv: v1 [math.mg] 25 Aug 2011 ABSORBING ANGLES, STEINER MINIMAL TREES, AND ANTIPODALITY HORST MARTINI, KONRAD J. SWANEPOEL, AND P. OLOFF DE WET arxiv:08.5046v [math.mg] 25 Aug 20 Abtract. We give a new proof that a tar {op i : i =,...,