THE SCRAMBLING INDEX OF PRIMITIVE DIGRAPHS CONSISTING OF A DIRECTED CYCLE AND A BI-DIRECTION PATH

  Bulletin of Mathematics

  ISSN Printed: 2087-5126; Online: 2355-8202 Vol. 06, No. 01 (2014), pp. 57–67. http://jurnal.bull-math.org

  

THE SCRAMBLING INDEX OF PRIMITIVE

DIGRAPHS CONSISTING OF A DIRECTED

CYCLE AND A BI-DIRECTION PATH

  

Zuhri and Saib Suwilo

Abstract.

  A strongly connected digraph is said tbe primitive provided there is a positive integer k such that for each pair of vertices u and v there are a walk from u to v and a walk from v to u of length k. The scrambling index of a primitive digraph D, denoted by k

  (D), is the smallest positive integer k such that for each pair of vertices u and v there is a vertex w with the property that that there are a walk from u to w and a walk for v to w of length k. We discuss the scrambling index of the class of primitive digraph on n vertices consisting of a directed cycle of odd length s and a bi directed path of length n − s. For such primitive digraph D we show that k (D) = 2s − 2 whenever s > n − s and k(D) = n − 1 whenever s ≤ n − s.

  

1. INTRODUCTION

Let D be a digraph. A directed walk of length m from a vertex u to a vertex v in D is a sequence of m arcs of the form u = v → v

  1 →

  v

  2

  → · · · → v m = v. For simplicity a directed walk of length m from a vertex u to a vertex v is denoted by u

  m

  −→ v. A u → v directed walk is

open whenever u 6= v and is closed whenever u = v. A u → v directed

path is a u → v directed walk without repeated vertices except possibly

u = v. A directed cycle is a closed directed path. The distance from a

  Received 12-10-2013, Accepted 15-12-2013. 2010 Mathematics Subject Classification

  : 05C20, 05C50 Key words and Phrases : Strongly connected digraph, primitive digraph, scrambling index. Zuhri & Saib Suwilo – Scrambling Index

vertex u to a vertex v, denoted by d(u, v), is the length of the shortest

u → v path. A digraph D is called symmetric provided that u → v is an

arc of D whenever v → u is an arc of D. By a bi-direction path or bi-path

connecting vertices u and v we mean a walk connecting u and v of the form

u = v → v → v → · · · → v m = v → v m − → v m − → · · · → v = u.

  1

  2

  1

  2 A digraph D is strongly connected provided that for each pair of ver-

  

tices u and v in D there are a u → v walk and a v → u walk. A strongly

connected digraph is primitive provided there exists a positive integer k

  k

  

such that for each pair of vertices u and v in D there are a u −→ v walk

  k

  

and a v −→ u walk. The smallest of such positive integer k is called the

exponent of D and is denoted by exp(D). Some result on the exponent of

primitive digraph can be found on [1, 2]. It is a well known result, see [1],

that a strongly connected digraph D is primitive if and only if the greatest

common divisor of the lengths of all directed cycles in D is 1.

  In 2009, Akelbek and Kirkland [3] introduced a new parameter of prim-

itive digraphs called scrambling index. The scrambling index of a primitive

digraph D, denoted by k(D), is the smallest positive integer k such that for

each pair of vertices u and v there exists a vertex w with the property that

  k k

  

there are a u −→ w walk and a v −→ w walk in D. Results on scrambling

index of primitive digraphs can be found on [4, 5, 6, 7, 8]. Liu and Huang [8]

discussed the scrambling index of symmetric digraphs. In particular, they

show that if D is a (n, s, n−s)-lollipop, that is a primitive symmetric digraph

on n vertices consisting of bi-direction cycle of odd length s and bi-direction

path of length n − s as shown in Figure 1, then k(D) = n − (s + 1)/2.

  

Figure 1: (n, s, n − s)-lollipop.

  In this paper we discuss the scrambling index of an (n, s, n−s)-cycle bi-

  n −s

  

path D , that is primitive digraphs on n vertices consisting of a directed

  s

  Zuhri & Saib Suwilo – Scrambling Index

  n −s

  

vertices of D as shown in Figure 2. In Section 2 we discuss a necessary

  s background on scrambling index. In Section 3, we present our result.

  

Figure 2: (n, s, n − s)-cycle bi-path.

2. NECESSARY BACKGROUND Let D be a primitive digraph and u and v be two distinct vertices in D.

  

The local scrambling index of u and v at the vertex w, denoted by k u,v (w),

  k

  

is defined to be the smallest positive integer k such that there are u −→ w

  k

  walk and v −→ w walk. That is

  k k

  k u,v (w) = min{k : there are u −→ w walk and v −→ w walk}. (1)

The local scrambling index of u and v in D, denoted by k (D), is defined

  u,v

  to be k u,v (D) = min {k u,v (w)}. (2)

  w ∈V

  

From the definition of k(D) and k u,v (D) we have for any pair of distinct

vertices u and v that k(D) ≥ k u,v (D). Furthermore, since D is strongly

  ′

  

connected for any positive integer ℓ ≥ k u,v (D) one can find a vertex w with

  ℓ ℓ ′ ′

  the property that there are u −→ w walk and v −→ w . This implies k(D) = max {k u,v (D)}. (3)

  6 u =v

  The following corollary presents primitivity of an (n, s, n − s)-cycle bi-path.

  n −s

  

Corollary 2.1 Let D be an (n, s, n − s)-cycle bi-path. If s is odd, then

  s n −s

  is primitive.

  Zuhri & Saib Suwilo – Scrambling Index

P roof. Notice that every bi-path v x → v x +1 → v x , for some s ≤ x ≤ n − 1,

¯

is a directed cycle of length 2. Since s is odd, the greatest common divisor

  n −s n −s of lengths of cycles in D is 1. Hence D is primitive. s s

3. MAIN RESULT

  n −s

  Let D be an (n, s, n − s)-cycle bi-path. In this section we discuss an

  s n −s

  

upper bound for k(D ). For this purpose we define C s to be the directed

  s

  

cycle of length s, C s : v → v → · · · → c s → v and P n −s be the bi-path of

  1

  2

  1

  

length 2(n − s), P n −s : v s → v s +1 → · · · → v n − → v n → v n − → v n − →

  1

  1

  1 · · · → v s → v s .

• 1

  Theorem 3.1 Let n be a positive integer with n ≥ 4 and s be an odd positive integer with s < n. Then ( if 2s − 2 s > n − s,

  −s n

  k(D ) =

  s n − 1 if s ≤ n − s. n −s

  

Proof. Since s is odd, by Corollary 2.1 D is a primitive digraph. We

  s split the proof into two cases, where s > n − s and s ≤ n − s.

  Case 1. s > n − s

  −s n

  

We first show that k(D ) ≥ 2s − 2. By (3), it suffices to show that there

  s n −s n −s

  

are vertices u and v in D such that k (D ) = 2s − 2. We show that

  u,v s s n −s

k v ,v (D ) = min {k v ,v (w)} = 2s − 2.

s

  1

  2

  1

  2

  w ∈V We consider four sub cases depending on the position of the vertex w.

  Sub Case

1.1. The vertex w = v . The length of any v → v walk is of

  1

  1

  1

  

the form s x + 2 y for some integers x ≥ 1 and y ≥ 0. The length of any

  1

  1

  1

  1

  v

  2 → v 1 walk is of the form (s − 1) + s x 2 + 2 y 2 for some integers x 2 , y 2 ≥ 0.

  

Notice also that there are v → v walk and v → v walk of the same length

  1

  1

  2

  1

  whenever s x

  1 + 2 y 1 = (s − 1) + s x 2 + 2 y 2 . Since this condition occurs the

  

first time when x = 1 and x = 1, we have 2y − 2y = s − 1. Therefore,

  1

  2

  1

  2 2y ≥ s − 1 and hence s x + 2 y ≥ 2s − 1. Thus k v ,v (v ) ≥ 2s − 1.

  1

  1

  1

  1

  1

  2 We next show that k v ,v (v ) ≤ 2s − 1. The walk that starts at v ,

  1

  1

  1

  2

moves to v s along the v → v s path, moves (s − 1)/2 times around the

  1

  

cycle v s → v s → v s , and finally moves to v using the v s → v arc is a

• 1

  1

  1

  

v → v walk of length 2s − 1. Notice also that the walk that starts at v ,

  1

  1

  2 Zuhri & Saib Suwilo – Scrambling Index v

  1 along the v 2 → v 1 path is a v 2 → v 1 walk of length 2s − 1. Therefore, 2s−1 2s−1 n −s

  

there are v −→ v walk and v −→ v walk in D . Hence by (1) we have

  1

  1

  2 1 s

  k v ,v (v

  1 ) ≤ 2s − 1. Hence we now conclude that

  1

  2 k v ,v (v ) = 2s − 1. (4)

  1

  1

2 Sub Case

  1.2. The vertex w = v j , for j = 2, 3, . . . , s − 1. The length of any v

  1 → v j walk is of the form j − 1 + s x 1 + 2 y 1 for some x 1 , y 1 ≥ 0 and the

length of any v → v j walk is of the form j−2+s x +2 y for some x , y ≥ 0.

  2

  2

  2

  2

  2 Notice also that there are v → v j walk and v → v j walk of the same length

  1

  2

  

whenever (j − 1) + s x + 2 y = (j − 2) + s x + 2 y for some x , x ≥ 1 and

  1

  1

  2

  2

  1

  2

  

y , y ≥ 0. Furthermore, (j − 1) + s x + 2 y = (j − 2) + s x + 2 y occurs

  1

  2

  1

  1

  2

  2

  

the first time when x = 1 and x = 2 or when x = 2 and x = 1. If x = 1

  1

  2

  1

  2

  1

  

and x = 2, then 2y − 2y = s − 1. Therefore, 2y ≥ (s − 1) and hence

  2

  1

  2

  1

(j − 1) + s x + 2 y ≥ 2s − 2 + j. If x = 2 and x = 1, then 2y − 2y = s + 1.

  1

  1

  1

  2

  2

  1 Therefore, 2y ≥ s + 1 and hence (j − 2) + s x + 2 y ≥ 2s − 1 + j. Thus we

  2

  2

  2 conclude that k v ,v (v j ) ≥ 2s − 2 + j for j = 2, 3, . . . , s − 1.

  1

2 We now show that for j = 2, 3, . . . , s − 1, k v ,v (v j ) ≤ 2s − 2 + j. The

  1

  2

walk that starts at v moves to v j along the v → v j path, then moves to

  1

  1

  

v s along the v j → v s path, then moves (s − 1)/2 times around the cycle

v s → v s → v s and back at v s , then finally moves to v j along the v s → v j

• 1

  

path, is a v → v j path of length 2s − 2 + j. Moreover, the walk that

  1

  

starts at v , then moves to v j along the v → v j path, and finally moves

  2

  2

  

two time around the cycle C s and back at v j , is a v → v j walk of length

  2 2s−2+j

  

2s − 2 + j. Therefore, for j = 2, 3, . . . , s − 1, there are v −→ v j walk anda

  1 2s−2+j v −→ v walk and hence by (1) we have k (v ) ≤ 2s − 2 + j. 2 j v ,v j

  1

  2 We now conclude that k v ,v (v j ) = 2s − 2 + j ≥ 2s (5)

  1

  2 for each j = 2, 3, . . . , s − 1.

  Sub Case

1.3. The vertex w = v . Notice that the length of any

  s −s n

  

v → v s walk in D is of the form (s − 1) + s x + 2 y for some integers

  1 s

  1

  1 n −s

  

x , y ≥ 0, and the length of any v → v walk in D is of the form

  1

  1 2 s s

  

(s − 2) + s x + 2 y for some integers x , y ≥ 0. Notice also that the

  2

  2

  2

  2

  expression (s − 1) + s x

  1 + 2 y 1 = (s − 2) + s x 2 + 2 y 2 can occur the first

  

time either when x = 0 and x = 1 or when x = 1 and x = 0. If x = 0

  1

  2

  1

  2

  1

  

and x = 1, then 2y − 2y = s − 1. This implies 2y ≥ s − 1 and hence

  2

  1

  2

  1

(s − 1) + s x + 2 y ≥ 2s − 2. If x = 1 and x = 0, then 2y − 2y = s + 1.

  1

  1

  1

  2

  2

  1 Zuhri & Saib Suwilo – Scrambling Index This implies 2y

  2 ≥ s + 1 and hence (s − 2) + s x 2 + 2 y 2 ≥ 2s − 1. Thus we conclude that k v ,v (v s ) ≥ 2s − 2.

  1

2 We now show that k v ,v (v s ) ≤ 2s−2. The walk that starts at v moves

  1

  1

  2 to v s along the v

  1 → v s path and finally moves (s − 1)/2 times around the

  

cycle v s → v s → v s is a v → v s walk of length 2s−2. The walk that starts

• 1

  1

  

at v moves to v s along the v → v s path and finally moves one time aound

  2

  2 2s−2

  

the cycle C s is a v → v s walk of length 2s−2. Therefore, there are v −→ v s

  2

  1 2s−2 n −s

  walk and v −→ v s walk in D . By (1) we have k v ,v (v ) ≤ 2s − 2.

  2 s

  2

  1

  2 Therefore, we now conclude that k v ,v (v s ) = 2s − 2. (6)

  1

2 Sub Case for

  1.4. The vertex w = v j j = s + 1, s + 2, . . . , n. Notice

that the length of any v → v walk is of the form (j − 1) + s x + 2 y

  1 j

  1

  1

  

for some integers x , y ≥ 0 and the length of any v → v j walk is of

  1

  1

  2

  

the form (j − 2) + s x + 2 y for some integers x , y ≥ 0. Notice that

  2

  2

  2

  2

  

(j − 1) + s x + 2 y = (j − 2) + s x + 2 y occurs the first time either

  1

  1

  2

  2

  

when x = 0 and x = 1 or when x = 1 and x = 0. If x = 0 and

  1

  2

  1

  2

  1

  

x = 1, then 2y − 2y = s − 1. This implies 2y ≥ (s − 1) and hence

  2

  1

  2

  1

  (j − 1) + s x

  1 + 2 y 1 ≥ s − 2 + j. If x 1 = 1 and x 2 = 0, then 2y 2 − 2y 1 = s + 1.

  

This implies 2y ≥ s + 1 and hence (j − 2) + s x + 2 y ≥ s − 1 + j. Thus

  2

  2

  2 we conclude that k v ,v (v j ) ≥ s − 2 + j for j = s + 1, s + 2, . . . , n.

  1

2 We now show that for j = s + 1, s + 2, . . . , n, k v ,v (v j ) ≤ s − 2 + j. The

  1

  2

walk that starts at v , then moves to v j along the v → v j path and finally

  1

  1

  

moves (s − 1)/2 times around the cycle v j → v j − → v j is a v → v j walk

  1

  1

  

of length s − 2 + j. The walk that starts at v , then moves one time around

  2

  

the cycle C s and back at v and finally moves to v j along the v → v j path

  2

  2

  

is a v → v walk of length s − 2 + j. Therefore, for j = s + 1, s + 2, . . . n,

  2 j − − s 2+j s 2+j n −s

  

there are v −→ v j walk and v −→ v j walk in D and hence by (1)

  1 2 s we have k v ,v (v j ) ≤ s − 2 + j.

  1

2 We now conclude that since j > s

  k v ,v (v j ) = s − 2 + j > 2s − 2 (7)

  1

  2 for each j = s + 1, s + 2, . . . , n.

  n −s

  From (4), (5), (6), (7), and (2) we now conclude that k v ,v (D ) =

  s

  1

  2 min ∈V {k (w)} = 2s − 2. Hence

  w v ,v

  1

  2

  n −s n −s

  k(D ) ≥ k v ,v (D ) = 2s − 2 (8)

  s s

  1

  

2 Zuhri & Saib Suwilo – Scrambling Index

  n −s

  We now show that k(D ) ≤ 2s − 2. We show that for each vertex

  s 2s−2 n −s

  v j , j = 1, 2, . . . , n, there is a v j −→ v s walk in D .

  s 2s−2

  Since v s lies on a cycle of length 2, there is a v s −→ v s . We consider

vertex v j 6= v s . Since s > n−s, for any vertex v j the distance d(v j , v s ) ≤ s−

1. If d(v j , v s ) ≡ 2s − 2 ( mod 2), then by using the cycle v s → v s → v s we

• 1 d (v j ,v s )

  2s−2

  

can extend the v j −→ v s path into a v j −→ v s walk. If d(v j , v s ) 6≡ ( mod

2), then d(v j , v s ) ≤ s − 2. Notice that the walk W v ,v that starts at v j

j s

  d (v ,v s )

  j

moves to v s along the v j −→ v s path and finally moves one time around

the cycle C is a v → v walk of length s + d(v , v ) ≡ 2s − 2 (mod 2).

  s j s j s 2s−2

  

Since s + d(v j , v s ) ≤ 2s − 2, we can extend the walk W v ,v into a v j −→ v s

j s

  2s−2

  

walk. Therefore, for each j = 1, 2, . . . , n, there is a v −→ v walk. This

  j s

  implies

  −s n

  k(D ) ≤ 2s − 2 (9)

  2 whenever s > n − s. n −s

  From (8) and (9), we now conclude that k(D ) = 2s − 2 whenever

  s s > n − s.

  Case 2. s ≤ n − s

  n −s

  

We first show that k(D ) ≥ n − 1. By (3) it suffices to show that there

  2 n −s n −s

  

are vertices u and v in D with the property that k u,v (D ) = n − 1.

  s s

  We will show that

  n −s

k v ,v (D ) = min {k v ,v (w)} = n − 1.

  n s n n−1 n−1

  w ∈V We consider four sub cases depending on the position of the vertex w.

  Sub Case for some

  2.1. The vertex w = v j j = 1, 2, . . . , s − 1. The length of any v n −

  1 → v j walk is of the form (n − 1 − s + j) + s x 1 + 2 y 1 for

  

some x , y ≥ 0. The length of any v n → v j walk is of the form (n − s + j) +

  1

  1

  

s x + 2 y for some x , y ≥ 0. Notice that (n − 1 − s + j) + s x + 2 y =

  2

  2

  2

  2

  1

  1

  

(n − s + j) + s x + 2 y occurs the first time either when x = 0 and x = 1

  2

  2

  1

  2

  

or when x = 1 and x = 0. If x = 0 and x = 1 we have 2y − 2y = s + 1

  1

  2

  1

  2

  1

  2

and hence 2y ≥ s + 1. This implies (n − 1 − s + j) + s x + 2 y ≥ n + j.

  1

  1

  1 If x = 1 and x = 0 we have 2y − 2y = s − 1 and hence 2y ≥ s − 1.

  1

  2

  2

  1

  2 This implies (n − s + j) + s x 2 + 2 y 2 ≥ n + j − 1. Thus we conclude that k v ,v (v j ) ≥ n − 1 + j.

  n n−1 We now show, for j = 1, 2, . . . , s − 1, that k v ,v (v j ) ≤ n − 1 + j. The n n−1 walk that starts at v n −

  1 moves to v j along the v n − 1 → v j path and finally Zuhri & Saib Suwilo – Scrambling Index moves one time around the cycle C s and back at v j is a v n −

  1 → v j of length

  

n − 1 + j. The walk that starts at v n moves (s − 1)/2 times around the

cycle v n → v n − → v n − and finally moves to v j along the v n → v j path is

  1

  1

  

a v n → v j walk of length n − 1 + j. Therefore, for j = 1, 2, . . . , s − 1, there

  n − n − 1+j 1+j n −s

  are v n −

  1 −→ v j walk and v n −→ v j walk in D and hence by (1) we s

  have k v ,v (v j ) ≤ n − 1 + j. n n−1

  We now conclude that k v ,v (v j ) = n − 1 + j ≥ n (10) n n−1 for each j = 1, 2, . . . , s − 1.

  Sub Case .

  2.2. The vertex w = v s The length of any v n → v s walk is

of the form n − s + s x + 2 y for some integers x , y ≥ 0. The length of

  1

  1

  1

  1

  

any v n − → v s walk is of the form (n − 1 − s) + s x + 2 y for some integers

  1

  2

  2

  

x , y ≥ 0. Notice that n − s + s x + 2 y = n − 1 − s + s x + 2 y occurs

  2

  2

  1

  1

  2

  2

the first time either when x = 0 and x = 1 or when x = 1 and x = 0.

  1

  2

  1

  2 If x 1 = 0 and x 2 = 1, then 2y 1 − 2y 2 = s − 1 and hence 2y 1 ≥ s − 1. This

  

implies n−s+s x +2 y ≥ n−1. If x = 1 and x = 0, then 2y −2y = s+1

  1

  1

  1

  2

  2

  1

  

and hence 2y ≥ s + 1. This implies n − s − 1 + s x + 2 y ≥ n. Therefore,

  2

  2

  2 we now conclude that k v ,v (v s ) ≥ n − 1.

  n n−1 We now show that k v ,v (v s ) ≤ n − 1. The walk that starts at v n , n n−1

  n −s

  

moves to v s along the v n −→ v s path and finally moves (s − 1)/2 times

around the cycle v → v → v is a v → v walk of length n − 1. The

  s s +1 s n s n −s−

  1

  

walk that starts at v n − moves to v s along the v n − −→ v s path and finally

  1

  1

moves one time around the cycle C s is a v n − → v s walk of length n − 1.

  1 n − n −

  1

  1 n −s

  

Therefore, there are v n − −→ v s walk and v n −→ v s walk in D and hence

  1 s by (1) we have k v ,v (v s ) ≤ n − 1.

  n n−1 We now conclude that k v ,v (v s ) = n − 1. (11) n n−1

  Sub Case 2.3. The vertex w = v for j = s + 1, s + 2, . . . , n − 1.

  j

  

The length of any v n → v j walk is of the form n − j + s x + 2 y for

  1

  1

  some integers x

  1 , y 1 ≥ 0. The length of any v n − 1 → v j walk is of the

  

form n − j − 1 + s x + 2 y for some integers x , y ≥ 0. We note that

  2

  2

  2

  2

  

n − j + s x + 2 y = n − j − 1 + s x + 2 y occurs the first time either

  1

  1

  2

  2

  

when x = 0 and x = 1 or when x = 1 and x = 0. If x = 1, then

  1

  2

  1

  2

  2

  

y ≥ j − s. This implies n − j − 1 + s x + 2 y ≥ n − 1 + j − s. If x = 1,

  2

  2

  2

  1

  

then y ≥ j − s. This implies n − j + s x + 2 y ≥ n + j − s. Therefore, for

  1

  1

  1 Zuhri & Saib Suwilo – Scrambling Index We now show, for j = s + 1, s + 2, . . . , n − 1, that k v ,v n (v j ) ≤ n−1

  n −j

  

n − 1 + j − s. The walk that starts at v n moves to v j along the v n −→ v j

path and finally moves j − (s + 1)/2 times around the cycle v j → v j → v j

• 1

  

is a v → v walk of length n − 1 + j − s. The walk that starts at v −

  n j n

  1 n − 1−s

  

moves to v s along the v n − −→ v s path, then moves one time around the

  1 j −s

  cycle C s and finally moves to v j along the v s −→ v j path is a v n −

  1 → v j

  

walk of length n − 1 + j − s. Therefore, for j = s + 1, s + 2, . . . , n − 1, there

  n − n − 1+j−s 1+j−s n −s

  

are v − −→ v walk and v −→ v walk in D and hence by (1)

  n 1 j n j s

  we have k v ,v (v j ) ≤ n − 1 + j − s. n n−1

  We now conclude that k v ,v (v j ) = n − 1 + j − s ≥ n (12) n n−1 for each j = s + 1, s + 2, . . . , n − 1.

  Sub Case

2.4. The vertex w = v . The length of any v → v walk

  n n n

  

is of the form s x + 2 y for some integers x , y ≥ 0. The length of any

  1

  1

  1

  1

  v n −

  1 → v n walk is of the form 1 + s x 2 + 2 y 2 for some integers x 2 , y 2 ≥ 0.

  

We note that s x + 2 y = 1 + s x + 2 y occurs the first time either when

  1

  1

  2

  2

  x

  1 = 0 and x 2 = 1 or when x 1 = 1 and x 2 = 0. If x 1 = 1, then y 1 ≥ n − s.

  

This implies s x + 2 y ≥ 2n − s. If x = 1, then y ≥ n − 1 − s. This implies

  1

  1

  2

  2

  

1 + s x + 2 y ≥ 2n − s − 1. Therefore, we conclude that k v ,v (v n ) ≥

  2

  2

  n n−1 2n − s − 1.

  We now show that k v ,v (v n ) ≤ 2n − s − 1. The walk that starts at n n−1

v n moves n − (s + 1)/2 times around the cycle v n → v n − → v n is a v n → v n

  1

  

walk of length 2n − s − 1. The walk that starts at v − moves to v along

  n 1 s n − 1−s

  

the v n − −→ v s path, then moves one time around the cycle C s and back

  1 n −s

  

at v s and finally moves to v n along the v s −→ v n is a v n − → v n walk of

  1 2n−s−1 2n−s−1

  length 2n − s − 1. Therefore, there are v n −

  1 −→ v n walk and v n −→ v n n −s

  walk in D and hence by (1) we have k v ,v (v n ) ≤ 2n − s − 1.

  s n

  n−1 We now conclude that k (v ) = 2n − s − 1 ≥ n − 1 + s ≥ n (13)

  v ,v n n

  n−1 since n − s ≥ s ≥ 1.

  n −s

  From (10), (11), (12), (13), and (2) we conclude that k v ,v (D ) = n s

  1 n − 1. This implies

  n −s n −s

  k(D ) ≥ k v ,v n (D ) = n − 1 (14)

  s s

  n−1 Zuhri & Saib Suwilo – Scrambling Index

  n −s

  We now show that k(D ) ≤ n − 1 whenever n − s ≥ s. We show

  s n − 1 n −s

  

that for each vertex v j , j = 1, 2, . . . , n there is a v j −→ v s walk in D .

  s

  

Suppose d(v j , v s ) ≡ n − 1 ( mod 2). Since v s lies on a cycle of length 2, then

  n −

  1

  

the v j → v s path of length d(v j , v s ) can be extended to a v j −→ v s walk. So

we assume that d(v , v ) 6≡ n − 1 (mod 2). Notice that the walk that starts

  j s

  

at v j moves to v s along the v j → v s path and finally moves one time around

the cycle C is a v → v walk of length s + d(v , v ) ≡ n − 1 (mod 2). If v

  s j s j s j

  

lies on the cycle C s , then s + d(v j , v s ) ≤ 2s − 1. Since s ≤ n − s, we now

have s + d(v j , v s ) ≤ 2s − 1 ≤ n − 1. If v j does not lie on C s that is v j lies on

P n −s , then d(v j , v s ) ≤ n − 1 − s < n − 1. Since v s lies on a cycle of length

2 and s + d(v j , v s ) ≡ n − 1 (mod 2), the v j → v s walk of length s + d(v j , v s )

  n −

  1

  

can be extended to a v j −→ v s walk. Therefore, for each j = 1, 2, . . . , n,

  n −

  1 n −s

  there is a v j −→ v s walk in D . Hence we conclude that

  s n −s

  k(D ) ≤ n − 1 (15)

  s whenever n − s ≥ s. n −s

  From (14) and (15) we conclude that k(D ) = n−1 whenever n−s ≥

  s s.

  As a consequence of Theorem 3.1 we have a general upper bound for

  n −s scrambling index of D . s

  

Corollary 3.2 Let n ≥ 4 be an integer and let s < n be an odd positive

integer. Then ( 2n − 6, if n is odd

  n −s

  k(D ) ≤

  s

  if 2n − 4, n is even.

  n −

2 Proof.

  By Theorem 3.1 k(D ) will be large whenever s > n − s. Notice

  s n −s

  

that if n is even, then s ≤ n − 2. This implies k(D ) = 2s − 2 ≤ 2n − 6.

  s −s n

  If n is even, then s ≤ n − 1. This implies k(D ) = 2s − 2 ≤ 2n − 4.

  s −s n

  We note that if n is odd, by Theorem 3.1, k(D ) = 2n−6 is achieved

  s n −s

  2

  

by the digraph D . If n is even, k(D ) = 2n − 4 is achieved by the

  n − s

  2

  1 digraph D .

  − n

  1 REFERENCES

  

1. R.A. Brualdi and H.J. Ryser . 1991. Combinatorial matrix theory, Cam-

  Zuhri & Saib Suwilo – Scrambling Index 2.

H. Schneider. 2003. Wielandt’s proof of the exponent inequality for primitive nonnegative matrices. Linear Algbera Appl., 353, 5–10.

  

3. M. Akelbek and S. Kirkland. 2009. Coefficient ergodicity and the scram-

bling index. Linear Algebra Appl., 430(4), 1111–1130.

  

4. M. Akelbek and S. Kirkland. 2009. Primitive digraphs with the largest

scrambling index. Linear Algebra Appl., 430(4), 1099–1110.

  

5. M. Akelbek, S. Fital, and J. Shen. 2009. A bound on the scrambling

index of a primitive matrix using Boolean rank. Linear Algebra Appl., 431(10), 1923–1931.

  

6. S. Chen and B. Liu . 2010. The scrambling index of symmetric primitive

matrices. Linear Algebra and Appl., 433(10), 1110–1126.

  

7. Y. Gao and Y. Shao. 2013. Scrambling index of primitive digraphs con-

sisting of exactly two cycles. Ars Combinatorica.

  8. B. Liu and Y. Huang. 2010. The scrambling index of primitive digraphs.

  Computer & Mathematics with Application , 60(3), 706-721.

  Zuhri : Department of Mathematics, University of Sumatera Utara Jl. Bioteknologi No 1 FMIPA USU, Medan 20155 Indonesia.

  E-mail: zuhri muin@yahoo.com Saib Suwilo

  : Department of Mathematics, University of Sumatera Utara Jl. Bioteknologi No 1 FMIPA USU, Medan 20155 Indonesia.

  E-mail: saibwilo@gmail.com