VERTEX EXPONENTS OF A CLASS OF TWO-COLORED DIGRAPH WITH ONE LOOP Siti Sahara, Saib Suwilo, Mardiningsih

  Bulletin of Mathematics Vol. 04, No. 02 (2012), pp. 105–122.

  VERTEX EXPONENTS OF A CLASS OF TWO-COLORED DIGRAPH WITH ONE LOOP _Abstract. Siti Sahara, Saib Suwilo, Mardiningsih

₍₂₎

  A two-colored digraph D is primitive provided there are nonnegative integers g and h such that for each pair of vertices u and v there exsist a (g, h)- walk from vertex u to vertex v. The smallest positive integer g

• h taken over all ₍₂₎ such nonnegative integers g and h is the exponent of a two-colored digraph D , _{(2) (2)} denoted by . The exponent of vertex v is exp(D ). Let v is a vertex of D ₍₂₎ the smallest positive integer g + h such that for every vertex u in D there is an

  (g, h)-walk from v to u, denoted by exp D (2) (v). This paper discuss the vertex ₍₂₎ exponents of primitive two-colored digraph S on n ≥ 3 vertices consisting of the ₂ cycle v → v n → v → · · · → v → v of length n and the red loop at v . For _{1 n−1 2 1 (2) 1} such two-colored digraph, if n-cycle in S has one blue arc and n − 1 red arcs, ₂ then its vertex exponents lies on [n − 2 + k, 2n − 3 + k] for all k = 1, 2, ..., n.

  Received 12-05-2012, Accepted 29-06-2012. 2010 Mathematics Subject Classification : 05C15, 05C20. Key words and phrases : Two-colored digraphs, primitive, exponent, vertex exponent.

  Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  1. INTRODUCTION (2)

  Fonarsini and Valcher [1] define a two-colored digraph D as a digraph whose each of it arcs is colored by either red or blue. A two-colored digraph (2)

  D is strongly connected provided for each pair of vertices u and v there is a (g, h)-walk from u to v, that is a walk consisting of g-red arcs and h-blue arcs. Let w be a walk. The number of red arcs in w denoted by r(w) and the number of blue arcs in w denoted by b(w) with length of w is r(w) l(w) = r(w) + b(w) and the vector is the composition of the walk b(w) w.

  (2) A two-colored digraph D is primitive provided there exist nonnega-

  (2) tive integers g and h such that for each pair of vertices u and v in D there is a (g, h)-walk from u to v [1]. The smallest positive integer g + h taken

  (2) over all nonnegative integers g and h is the exponent of D and denoted

  (2) by exp(D ) [2].

  (2) Let D be a strongly connected two-colored digraph and let C =

  (2) {C , C , ..., C } be a set all of cycles in D . Define a matrix cycle M to be

  1 2 q 2 by q as below r(C

  1 ) r(C 2 ) · · · r(C q ) M = , b(C ) b(C ) · · · b(C q )

  1

  2 where its ith column is the compositions of the ith cycle C i , i = 1, 2, · · · , q. (2)

  A two-colored digraph D is primitive if and only if the greatest common diviser of 2 by 2 of minors M is 1 [1].

  (2) Lemma 1.1 be a strongly connected at least one arc each colors.

  [1] Let D (2) (2)

  Let M be the matrix cycle of D . A two-colored digraph D is primitive if and only if the content of the matrix cycle M is 1.

  (2) Let D be a two-colored digraph on n vertices v , v , ..., v . Gao and

  1 2 n Shao [3] define a local concept of exponents of two-colored digraphs as fol-

  (2) lows. For any vertex v k in D , k = 1, 2, ..., n, the exponent of the vertex v k , denoted by exp (2) (v k ), is the smallest integer p +p such that for every

  1

  2 D (2) vertex v in D there is a (p , p )-walk from v k to v. It is customary to or-

  1

  2 (2) _{2 2} der the vertices v ,v ,...,v n of D such that exp (v ) ≤ exp (v ) ≤ · · · ≤ ₂

  1

  2 D

  1 D

  2 exp (v n ). Gao and Shao [3] discuss the vertex exponents for primitive two-

  D colored digraph of Wielandt type, that is a hamiltonian digraph consisting of

  Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  (2) Gao and Shao show that if the two-colored digraph W has only one blue

  2 arc v a → v a− , a = 2, 3, ..., n−1, then exp (2) (v k ) = n −2n+k−a+1. If the

  1 W (2) ₍₂₎

  2 two-colored digraph W has two blue arcs then exp (v k ) = n − 2n + k

  W

  2 or exp (2) (v ) = n − 2n + k + 1. Suwilo [4] discuss the vertex exponents k

  W (2) of two-colored digraph ministrong digraph D on n vertices consisting of two cycles of length n − 1 and n − 2 respectively. If the two-colored di-

  (2) (2) graph D has one blue arc, then the vertex exponents of D lies on

  2 2 (2) [n − 5n + 8, n − 3n + 1]. If the two-colored digraph D has two blue arcs,

  (2)

  2

  2 then the vertex exponents of D lies on [n − 4n + 4, n − n]. Suwilo and Syafrianty [5] discuss the vertex exponents of primitive two-colored digraph

  (2) D on n = 2m vertices with m ≥ 5 consisting of two cycles of length n − 1

  (2) and n − 3 respectively. For such that primitive two-colored digraph D ,

  3

  2

  3

  2 the vertex exponents lies on [(n − 5n + 4n + 4)/4, (n − 5n + 10n + 4)/4].

  Syahmarani and Suwilo [6] discuss the vertex exponents of two-colored di-

  2 graph whose underlying digraph is the digraph Hamiltonian L consisting n of the n-cycle v → v n → v n− → v n− → · · · → v → v and the arc

  1

  1

  2

  2

  1 v → v on n odd integer vertices. Syahmarani and Suwilo shows that

  1 n−

  2 (2)

  3

  2

  2 if exp(L n ) = (n − 2n + 1)/2, then the vertex exponents of L is lies on n

  (2)

  3

  2

  3

  2

  2 [(n − 2n − 3n + 4)/4, (n − 2n + 3n + 6)/4] and if exp(L n ) = 2n − 6n + 2,

  2

  2

  2 then the vertex exponents of L is lies on [n − 4n + 5, n − 2n − 1]. n

  This paper discuss the vertex exponents of a class of primitive two- (2) colored digraph S on n ≥ 3 vertices consisting of an n-cycle v → v n →

  1

  2 v → · · · → v → v and one loop at v . In section 2 we discuss primitive n−

  1

  2

  1

  1 (2) two-colored digraph S with one loop. In section 3 we discuss a way to

  2 set up a lower and an upper bound for vertex exponents of two-colored digraph consists of two cycles. In section 4 presents our main result on

  (2) vertex exponents of primitive two-colored digraph S .

  2

  2. TWO-COLORED DIGRAPH WITH ONE LOOP In this section we use the properties of the primitiveness of a two-colored di-

  (2) (2) graph in order to determine the color of each arc in S . Suppose that S

  2

  2 be a primitive two-colored digraph on n ≥ 3 vertices consisting of one loop at v and the cycle v → v n → v n− → · · · → v → v of length n. Since

  1

  1

  1

  2

  1 loop is also a cycle, assume that C 1 is a cycle of length 1 and C 2 is a cycle

  (2) of length n. Hence the cycle matrix of S can be expressed such of a form Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  1 − a n − b a b either M = or M = , for some integers

  1

  2 a b 1 − a n − b

  (2) 0 ≤ a ≤ 1 and 0 ≤ b ≤ n. Since S is primitive, then content of the cycle

  2 (2) matrix of S is 1, that is det(M ) = b−na = ±1 or det(M ) = na−b = ±1

  1

  2

  2 (2) if and only if a = 0 and b = 1. Clearly, the cycle matrix of S must either

  2 1 n − 1

  1 M 1 = or M 2 = . 1 1 n − 1

  Without lost of generality, we may assume that the cycle matrix of (2) 1 n − 1

  S is the matrix M = . Hence the cycle C of length 1 is a

  1

  2

  1 red loop and the cycle C of length n is a cycle consist of n − 1 red arcs and

  2 (2) exactly one blue arc. Furthermore, a primitive two-colored digraph S is

  2 classified on three types based on a blue arc and the red path whose exist in the cycle C as follows.

  2 (2)

  a. The primitive two-colored digraph S is of type A if the blue arc of

  2 C 2 is the arc v 2 → v 1 and the red path of length n − 1 of C 2 is the path v → v n → · · · → v → v .

  1

  3

  2 (2)

  b. The primitive two-colored digraph S is of type B if the blue arc of

  2 C is the arc v → v and the red path of length n − 1 is the path

  2 1 n v n → v n− → · · · → v → v .

  1

  2

  1 (2)

  c. The primitive two-colored digraph S is type of C if the blue arc of

  2 C is the arc v n−s → v n−s , s = 1, 2, ..., n − 2 and the red path of 2 +1 length n − 1 is the path v n−s → · · · → v → v n → · · · → v n−s .

  1 +1

  3. BOUNDS FOR VERTEX EXPONENTS This section we discuss a way to set up an upper and a lower bound for vertex exponents of a primitive two-colored digraph. Suwilo [4] gives one technique in order to determine an upper and a lower bound for vertex ex- ponents of primitive two-colored digraph consisting of two cycles. Now we start with the lower bound as the following lemma.

  (2) Lemma 3.1 Let D be a primitive two-colored digraph consisting two r(C ) b(C )

  1

  2 cycles with cycle matrix . Let be any vertex

  M = v k Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  (2) in and suppose there is an to vertex in

  D (g, h)-walk from vertex v k v j g u (2) with for some nonnegative integers

  D = M u and v. Then h v u r(p ) k,j

  −

  1 for some path from to

  ≥ M p v k v j (k,j) v b(p k,j )

  Proof. For some j = 1, 2, ..., n, Let p be a path from vertex v to vertex k,j k

  (2) v j . Since D consisting of two cycles and every walk can be decomposed into path and a cycle, we have g x r(p )

  1 k,j

• = M (1) h x b(p k,j )

  2 (2) for some x , x ≥ 0. Since D is primitive, M is an invertible matrix.

  1

  2 g u

  By considering = M and equation (1) we have the following h v equation u x r(p k,j )

  1

• M = M v x b(p )

  2 k,j x u r(p k,j )

  1 M = M − x 2 v b(p k,j ) x u r(p k,j )

  1 −

  1 = − M ≥ 0 x

  2 v b(p k,j ) u r(p ) k,j

  −

  1 hence ≥ M and the lemma holds. v b(p k,j ) By considering of Lemma 3.1 we have the following theorem.

  (2) Theorem 3.2 be a primitive two-colored digraph consisting

  [6] Let D (2) of two cycles C and C . Let v be a vertex in D . For some vertex v i and

  1 2 k (2) in define v j D u = b(C )r(p k,j ) − r(C )b(p k,j ) and v = r(C )b(p k,j ) −

  2

  2

  1 g u b(C )r(p ). Then ≥ M + and hence exp (2) (v ) ≥ l(C )u

  1 k,j k

  1 D h v l(C )v .

  2 Proof.

  Suppose that the vertex exponents of v k is holded by an (g, h)- g u walk with = M for some nonnegative integers u and v and we

  Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  have the following equation u r(p k,j ) b(C )r(p k,j ) − r(C )b(p k,j ) −

  2

  2

  1 ≥ M = (2) v b(p k,j ) r(C )b(p k,j ) − b(C )r(p k,j )

  1

  1 for some path p from vertex v to vertex v j . If for some vertex v j , j = k,j k

  1, 2, ..., n we have the value b(C )r(p k,j ) − r(C )b(p k,j ) ≥ 0, then we define

  2

  2 u = b(C )r(p ) − r(C )b(p ) ≥ 0 (3)

  2 k,j 2 k,j and if for some vertex v i , i = 1, 2, ..., n we have the value r(C )b(p k,i ) −

  1 b(C )r(p ) ≥ 0, then we define

  1 k,i v = r(C )b(p k,i ) − b(C )r(p k,i ) ≥ 0. (4)

  1

  1 Thus u ≥ u and v ≥ v . By Lemma (3.1) we have g u u

  = M ≥ M (5) h v v and hence exp (2) (v ) = g + h ≥ (r(C ) + b(C ))u + (r(C ) + b(C ))v = k

  1

  1

  2

  2 D l(C )u + l(C )v .

  1

  2 In the next Proposition, we describe an upper bound of vertex expo- nents of any vertex in two-colored digraph in term of a specified vertex. We define d(v , v) to be the distance from v to v, that is the sortest walk from k k v k to v.

  (2) Proposition 3.3 [4] Suppose that D is a primitive two-colored digraph on

  (2) with exponent (2) n vertices. Let v be a vertex in D exp (v). Then for each

  D (2) vertex v , k = 1, 2, ..., n in D we have exp (2) (v ) ≤ exp (2) (v) + d(v , v). k k k

  D D Proof. For each k = 1, 2, ..., n, let p be a (r(p ), b(p ))-path from v to k,v k,v k,v k ₍₂₎ vertex v with length d(v k , v). Since the vertex exponent of v is exp (v),

  D there is an (g, h)-walk with length exp (2) (v) = g +h from v to every vertex k

  D (2) v j , j = 1, 2, ..., n. This implies that for each vertex v k in D , there is an (g + r(p k,v ), h + b(p k,v ))-walk from vertex v k to each vertex v j , namely the walk that starts at v k , moves to v along the path (r(p k,v ), b(p k,v ))-path and then moves to v j by using an (g, h)-walk from v to v j . Now, we conclude exp (2) (v ) ≤ exp (2) (v) + d(v , v) k k

  D D

  Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  Proposition 3.4 gives in order to determine an upper bound for vertex exponents of primitive two-colored digraph contains of two cycles.

  (2) Proposition 3.4 [4] Let D be a two-colored digraph consisting of two

  (2) cycles and . Let be a vertex of that belongs to cycles and

  C C v k D C

  1

  2

  1 C . If for each i = 1, 2, ..., n and for some positive integers s and t, there is

  2 a path from to such that the system p k,i v k v i r(p k,i ) g

  M x + = (6) b(p ) h k,i has nonnegative integer solution, then (2) exp (v k ) ≤ g + h.

  D T Proof.

  Assume that the solution of equation (6) is x = (x , x ) . Since

  1

  2 (2)

  D is primitive, M is an invertible matrix and hence x and x cannot be

  1

  2 both zero. Since v k is belongs to both cycles, then there are three possible cases.

  If x > 0 and x > 0, then the walk that starts at v k to v i , moves x

  1

  2

  1 times around the cycles C and moves x times around the cycle C and

  1

  2

  2 back at v k , then moves to v i along the path p k,i is an (g, h)-walk from v k to v i . If x

  1 = 0 and x 2 > 0, then the walk that starts at v k to v i , moves x

  2 times around the cycles C and back at v k , then moves to v i along the path

  2 p k,i is an (g, h)-walk from v k to v i . Similarly if x > 0 and x = 0, then the

  1

  2 walk starts at v to v i , moves x times around the cycles C and back at k

  1

  1 v k , then moves to v i along to the path p k,i is an (g, h)-walk from v k to v i .

  Therefore for every vertex v , i = 1, 2, ..., n, there is an (g, h)-walk from v to i ₍₂₎ k v i . The definition of exponent of vertex v k implies that exp (v k ) ≤ g + h.

  D

  4. MAIN RESULTS (2)

  Let S be a primitive two-colored digraph such as discussed in section 2

  2 (2) and let v be a fixed vertex in S . Lemma (3.1) asserted that the vertex k

  2 (2) exponents of S is depends heavily on how large the number u in equation

  2 (3) and the number v in equation (4). Note that the number u will be large when the path p contains as many red arcs and as few blue arcs as k,j possible. Likewise the number v will be large when the path p k,j contains as many blue arcs and as few red arcs as possible. Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  (2) Lemma 4.1 Let S be a primitive two-colored digraph of type A, then ₍₂₎

  2 exp (v k ) = n − 2 + k for all k = 1, 2, ..., n. S _{2 (2)}

  Proof . We next show that exp (v k ) = n−2+k for all k = 1, 2, ..., n. Since S ₂ the red path of length n − 1 is the path v → v n → v → · · · → v → v ,

  1 (n−1)

  3

  2 ∗ ∗ set x = v 1 and y = v 2 . We split the proof into three cases.

  Case : k = 1

  We first show that the lower bound for vertex exponents of primitive two- (2) ₍₂₎ ∗ colored digraph S is exp (v k ) ≥ n − 2 + k. To do that, taking y = v 2 ,

  2 S ₂ there is a unique path p k from vertex v to vertex v which is (k+n−2, 0)- v ,v k ₂

  2 path. Using this path and the definition of u in equation (3) we have _{∗ ∗} u = b(C )r(p k,y ) − r(C )b(p k,y ) = 1(k + n − 2) − (n − 1)0 = n − 2 + k. (7)

  2

  2 ∗

  Taking x = v , there are two path p k from vertex v to vertex v , they 1 v ,v k ₁

  1 are (1, 0)-path and (k + n − 2, 1)-path. Using the (k + n − 2, 1)-path p v k ,v ₁ and the definition of v in equation (4) we have _{∗ ∗} v = r(C )b(p k,x ) − b(C )r(p k,x ) = 1(1) − 0(k + n − 2) = 1. (8)

  1

  1 Using the path (1, 0)−path and the definition of v in equation (4) we have _{∗ ∗} v = r(C )b(p ) − b(C )r(p ) = 1(0) − 0(1) = 0. (9)

  1 k,x 1 k,x Since the lower bound achieved if the value of v is small, from equations (8) and (9) we have v = 0. By lemma (3.1), equations (7) and (9) we have g u n − 2 + k

  ≥ M = (10) h v hence ₍₂₎ exp (v k ) ≥ n − 2 + k (11)

  S ₂ for k = 1.

  By considering of equation (10), we next show exp (2) (v ) ≤ n − 2 + k k

  S ₂ for k = 1 by showing that the system x r(p ) n − 1

  1 1,1

• M = (12)

Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  has a nonnegative integer solution which is x 1 ≥ 0 and x 2 ≥ 0.

  (2) Since S is primitive, M is an invertible and the solution of the system

  2 (12) is the integer vector x n − 1 − r(p ) − b(p )(1 − n)

  1 1,1 1,1 = x −b(p )

  2 1,1 Path p from v to v we can choose (1, 0)-path with r(p ) = 1 and

  1,1

  1 1 1,1 b(p 1,1 ) = 0, we have x 1 = n − 2 > 0 and x 2 = 0. Since the system (12) has nonnegative integer solution and vertex v lies on both cycles, Proposisi

  1 (3.4) guarantees exp (2) (v k ) ≤ n − 2 + k. (13)

  S ₂ Using equations (11) and (13) we have exp (2) (v ) = n − 2 + k (14) k

  S ₂ for k = 1.

  For the following cases, we next show that the lower bound for vertex (2) ₍₂₎ exponents of primitive two colored digraph S is exp (v k ) ≥ n − 2 + k

  2 S ₂ for all k = 2, 3, ..., n.

  Case : k = 2

  ∗ Taking y = v , we see that path p v k ,v from v k to v is (k − 3 + n, 1)-

  2 ²

  2 path. Using this path and the definiton of u in equation (3) we have _{∗ ∗} u = b(C )r(p k,y ) − r(C )b(p k,y ) = 1(k − 3 + n) − (n − 1)1 = k − 2. (15)

  2

  2 ∗

  Next taking x = v 1 , there is a (k − 2, 1)-path p v k ,v from v k to v ₁ 1 . Using this path and equation (4) we have _{∗ ∗} v = r(C

  1 )b(p k,x ) − b(C 1 )r(p k,x ) = 1(1) − 0(k − 2) = 1. (16) By Lemma (3.1), equations (15) and (16) we conclude that g u n − 3 + k

  ≥ M = . (17) Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  Therefore exp (2) (v k ) ≥ n − 2 + k (18) S ₂ for k = 2.

  Case : 3 ≤ k ≤ n (2)

  ∗ Taking y = v , for each vertex v in S , path p from v to v is

  2 k k,v k ₂

  2

  2 (k − 2, 0)-path. Using this path and equation (3) we have _{∗ ∗} u = b(C )r(p ) − r(C )b(p ) = 1(k − 2) − (n − 1)0 = k − 2. (19)

  2 k,y 2 k,y (2)

  ∗ Next taking x = v , for each vertex v in S , path p from v to v is

  1 k k,v k ₁

  1

  2 (k − 2, 1)-path. Using this path and equation (4) we have _{∗ ∗} v = r(C )b(p ) − b(C )r(p ) = 1(1) − 0(k − 2) = 1. (20)

  1 k,x 1 k,x By Lemma (3.1), equations (19) and (20) we conclude that g u 1 n − 1 k − 2 n − 3 + k

  ≥ M = = . (21) h v

  1

  1

  1 Hence ₍₂₎ exp (v k ) ≥ n − 2 + k (22) S ₂ for all k = 3, 4, ..., n.

  Now from equations (18) and (22) we can conclude that ₍₂₎ exp (v k ) ≥ n − 2 + k (23) S ₂ for all k = 2, 3, ..., n.

  We next show the upper bound for vertex exponents of primitive two (2) colored digraph S is exp (2) (v k ) ≤ n − 2 + k for all k = 2, 3, ..., n. we

  2 ₍₂₎ S ₂ first show that exp (v ) = n and then using Propositon (3.3) in order to

  2 S ₂ (2) determine the upper bound for vertex exponents of S for all k = 2, 3, ..., n.

  2 ₍₂₎ From equation (23) we know that for k = 2, exp (v 2 ) ≥ n. Thus, it

  S

  Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph ₍₂₎

  remains to show that exp (v 2 ) ≤ n. By considering equation (21) we S ₂ show that for each i = 1, 2, ..., n, there is a path from vertex v 2 to vertex v i consisting of n − 1 red arcs and one blue arc. It is suffices to show that the system x r(p ) n − 1

  1 2,i M + = (24) x

  2 b(p 2,i )

  1 has a nonnegative integer solution which is x ≥ 0 and x ≥ 0.

  1

  2 Let path p be a path from vertex v to v , i = 1, 2, ..., n. Since 2,i 2 i

  (2) −

  1 S is primitive, the cycle matrix M is an invertible matrix with M =

  2 1 1 − n and the solution of the system (24) is

  1 x (n − 1)b(p ) − r(p ) 1 2,i 2,i

  = x 1 − b(p ) 2 2,i

  If i = 1, path p from v to v i is (0, 1)-path with r(p ) = 0 and 2,i 2 2,i b(p ) = 1, we have x = n − 1 > 0 and x = 0. Notice that for any v ,

  2,i

  1 2 i i = 2, 3, ..., n, there is a path p from v to v i with 1 ≤ r(p ) ≤ n − 1 and

  2,i 2 2,i b(p ) = 1, hence x ≥ 0 and x = 0. These fact implies that the system

  2,i

  1

  2 (24) has a nonnegative integer solution and Propositon (3.4) guarantees _{(2) (2) (2)} that exp (v

  2 ) ≤ n. Since exp (v 2 ) ≥ n and exp (v 2 ) ≤ n, we have S S S _{2 2 2} exp (2) (v ) = n. For every k = 2, 3, ..., n, we know that the sortest walk from

  2 S _{2 (2)} v k to v is d(v k , v ) = k − 2, Proposition (3.3) implies that exp (v k ) ≤

  2

  2 ₍₂₎ S ₂ exp (v 2 ) + d(v k , v 2 ), hence

  S _{2 (2)} exp (v k ) ≤ n − 2 + k (25)

  S ₂ for all k = 2, 3, ..., n.

  Combining of equations (23) and (25) we can conclude that vertex ex- (2) ₍₂₎ ponent of S is exp (v k ) = n − 2 + k for all k = 2, 3, ..., n.

  2 S ₂ From cases 1, 2 and 3 we have exp (2) (v ) = n − 2 + k for all k = k S ₂ 1, 2, ..., n.

  (2) Lemma 4.2 Let be a primitive two-colored digraph of type B, then

  ₍₂₎ Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph exp (v k ) = n − 1 + k for all k = 1, 2, ..., n.

  S ₂ Proof.

  We next show that vertex exponents of primitive two-colored di- (2) ₍₂₎ graph S is exp (v k ) = n − 1 + k for all k = 1, 2, ..., n. Since the red path

  2 S ₂ ∗ of length n − 1 is the path v n → v n− → v n− → · · · → v → v , set x = v n

  1

  2

  2

  1 ∗ and y = v . We first show that exp (2) (v ) ≥ n − 1 + k for all k = 1, 2, ..., n.

  1 k S ₂ We split the proof into two cases when k = 1 and when 2 ≤ k ≤ n.

  Case : k = 1 ∗

  Taking y = v 1 , there are two path p v k ,v from v k to v ₁ 1 . They are (1, 0)-path and (k + n − 2, 1)-path. Using (k + n − 2, 1)-path p v k ,v and equation (3) we ₁ have _{∗ ∗} u = b(C )r(p k,y ) − r(C )b(p k,y ) = 1(k + n − 2) − (n − 1)1 = k − 1. (26)

  2

  2 Using (1, 0)-path and equation (3) we also have _{∗ ∗} u = b(C )r(p k,y ) − r(C )b(p k,y ) = 1(1) − (n − 1)0 = 1. (27)

  2

  2 Since the lower bound achieved when the value of u is small, from equations (26) and (27) choose u = k − 1.

  ∗ Next taking x = v n , path p v k ,v n from v to v n is (0, 1)-path. Using k this path and equation (4) we have _{∗ ∗} v = r(C )b(p k,x ) − b(C )r(p k,x ) = 1(1) − 0(0) = 1. (28)

  1

  1 By Lemma (3.1), equations (26) and (28) we have g u 1 n − 1 k − 1 n − 2 + k ≥ M = = . h v

  1

  1

  1 Therefore, we can conclude that exp (2) (v k ) = g + h ≥ n − 1 + k (29) S ₂ for k = 1.

  Case

  Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  (2) ∗

  Taking y = v , for every vertex v k in S , there is a unique path from

  1

  2 v k to v which is a (k − 1, 0)-path. Using this path and equation (3) we have

  1 _{∗ ∗} u = b(C )r(p k,y ) − r(C )b(p k,y ) = 1(k − 1) − 1(0) = k − 1. (30)

  2

  2 (2)

  ∗ Next taking x = v n , for every vertex v k in S , there is a unique path from

  2 v to v which is a (k − 1, 1)-path. Using this path and equation (3) we have k n _{∗ ∗} v = r(C )b(p k,x ) − b(C )r(p k,x ) = 1(1) − 0(k − 1) = 1. (31)

  1

  1 By Lemma (3.1), equations (30) and (31) we have g u 1 n − 1 k − 1 n − 2 + k

  ≥ M = = (32) h v

  1

  1

  1 hence exp (2) (v ) = g + h ≥ n − 1 + k (33) k

  S ₂ for all k = 2, 3, ..., n.

  Combining equations (29) and (33) we can conclude that exp (2) (v ) ≥ n − 1 + k (34) k

  S ₂ for all k = 1, 2, ..., n.

  (2) We next show that the upper bound for vertex exponents of S is

  2 exp (2) (v ) ≤ n − 1 + k for all k = 1, 2, ..., n. First we must show that k

  S _{(2) 2} exp (v ) = n and then using Prosition (3.3) in order to determine the

  1 S ₂ upper bound of other vertices. From equation (34) we see that for k = 1, we have exp (2) (v ) ≥ n. Thus, it remains to show that exp (2) (v ) ≤ n.

  1

  1 S ₂ S ₂ By considering equation (32) we show that for each vertex v i , i = 1, 2, ..., n, there is a path from v to v consisting of n − 1 red arcs and one blue arc.

  1 i It is suffices to show that the system x r(p ) n − 1

  1 1,i

• M = (35) x b(p )

  1 2 1,i has a nonnegative integer solution which is x , x ≥ 0.

  1

  2

  Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  Let p 1,i be a path from v 1 to v i . Notice that since M is an invertible matrix, equation (35) has the solution as below x (n − 1)b(p ) − r(p )

  1 1,i 1,i = x 1 − b(p )

  2 1,i For any vertex v i , i = 1, 2, ..., n, there exist a path p from v to

  1,i

  1 v with 0 ≤ r(p ) ≤ n − 1 and b(p ) = 1 which is imply x ≥ 0 and i 1,i 1,i

  1 x = 0. Therefore, we can conclude that for each vertex v i , i = 1, 2, ..., n,

  2 there is a path p from v to v hence the system of equation (35) has a

  1,i 1 i nonnegative integer solution. Since vertex v lies on both cycles C and _{(2) (2)}

  1

  1 C 2 , Propotiton (3.4) guarantees that exp (v 1 ) ≤ n. Since exp (v 1 ) ≥ S S _{2 2} n and exp (2) (v ) ≤ n, exp (2) (v ) = n. Proposition (3.3) implies that

  1

  1 _{(2) (2)} S S _{2 2} exp (v k ) ≤ exp (v ) + d(v k , v ), hence we have

  1

  1 S S _{2 2 (2)} exp (v k ) ≤ n − 1 + k (36)

  S ₂ for all k = 1, 2, ..., n.

  Finally, combining equations (34) and (36) we can conclude that the (2) ₍₂₎ vertex exponent of S is exp (v k ) = n − 1 + k for all k = 1, 2, ..., n.

  2 S ₂ (2)

  Lemma 4.3 Let be a primitive two-colored digraph of type C, then S

  2 exp (2) (v ) = n − 1 + k + s for all k = 1, 2, ..., n and s = 1, 2, ..., n − 2. k

  S ₂ Proof. We next show that exp (2) (v ) = n − 1 + k + s for k = 1, 2, ..., n k S ₂ and s = 1, 2, ..., n − 2. Since the red path of length n − 1 is the path

  ∗ v n−s → · · · → v → v n → · · · → v n−s , set x = v n−s as intial vertex and

  1 +1 ∗

  ∗ y = v n− 1+s as terminal vertex and then using path from v k to y to find

  ∗ the value of u and also using path from v k to v x to find the value of v . ₍₂₎ We first show that exp (v k ) ≥ n − 1 + k + s for all k = 1, 2, ..., n and

  S ₂ s = 1, 2, ..., n − 2. We split the proof into two cases.

  Case : 1 ≤ k ≤ n − s

  (2) ∗

  Taking y = v n−s , for every vertex v k in S , path from v k to v n−s

• 1
• 1

  2 is (k − 1 + s, 0)-path. Using this path and equation (3) we have _{∗ ∗} u = b(C )r(p k,y ) − r(C )b(p k,y ) = 1(k − 1 + s) − (n − 1)0 = k + s − 1.(37)

  2

  2

  Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  (2) ∗

  Next taking x = v , for every vertex v in S , path from v to v is n−s k k n−s

  2 (k − 1 + s, 1)-path. Using this path and equation (4) we have _{∗ ∗} v = r(C )b(p ) − b(C )r(p ) = 1(1) − 0(k − 1 + s) = 1. (38)

  1 k,x 1 k,x By Lemma (3.1), equations (37) and (38) we have g u 1 n − 1 k + s − 1 n − 2 + k + s

  ≥ M = = . (39) h v

  1

  1

  1 Therefore, ₍₂₎ exp (v k ) = g + h ≥ n − 1 + k + s (40) S ₂ for all k = 1, 2, ..., n − s and s = 1, 2, ..., n − 2.

  Case : n − s + 1 ≤ k ≤ n

  ∗ Similiarly with the first case, taking y = v n−s , for every vertex v k in

• 1 (2)

  S , path p v k ,v from v k to v n−s is (k + s − n − 1, 0)-path. Using this

• 1 2 n−s+1

  path and equation (3) we have _{∗ ∗} u = b(C )r(p ) − r(C )b(p ) = 1(k + s − n − 1) − (n − 1)0 = k + s − n − 1.(41) 2 k,y 2 k,y

  (2) ∗

  Next taking x = v n−s , for every vertex v in S , path p v k ,v from v to k k _n−s

  2 v n−s is (k + s − n − 1, 1)-path. Using this path and equation (4) we have _{∗ ∗} v = r(C )b(p ) − b(C )r(p ) = 1(1) − 0(k + s − n − 1) = 1. (42)

  1 k,x 1 k,x By Lemma (3.1), equations (41) and (42) we have g u 1 n − 1 k + s − n − 1 k + s − 2

  ≥ M = = (43) h v ₍₂₎

  1

  1

  1 hence exp (v k ) = g + h ≥ k + s − 1.

  S ₂ We note that for (k + s − n − 1, 1)-path from vertex v to vertex v , k n−s the system x k + s − n − 1 k + s − 2

  1 = + M

  Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  has a nonnegative integer solution x 1 = n − 1 > 0 and x 2 = 0. Proposition (3.4) asserted that if x = n − 1 > 0 and x = 0, then walk start at v k

  1

  2 to v n−s , moves n − 1 times around the cycle C . This implies there is no

  1 k + s − 2 walk from vertex v to vertex v with composition , hence k n−s

  1 exp (2) (v ) > k + s − 1. k

  S ₂ Notice that the sortest walk from vertex v k to vertex v n−s with at least k + s − 2 red arcs and 1 blue arc is the walk starts at v , moves to v and k

  1 then moves n − 1 times around the cycle C and then back at vertex v k ,

  1 finaly moves to vertex v n−s belongs to the path p k,n−s is a path from vertex n − 3 + k + s v to v n−s . The compositon of this walk is . Thus we now k

  2 have exp (2) (v ) ≥ n − 1 + k + s (44) k

  S ₂ for all n − s + 1 ≤ k ≤ n and s = 1, 2, ..., n − 2.

  Combining equations (40) and (44) we can conclude that exp (2) (v ) ≥ n − 1 + k + s (45) k

  S ₂ for all k = 1, 2, ..., n and s = 1, 2, ..., n − 2 ₍₂₎

  Next we show exp (v k ) ≤ n − 1 + k + s for all k = 1, 2, ..., n and s = S _{2 (2)}

  1, 2, ..., n − 2. We first show exp (v ) = n + s and then using Proposition

  1 S ₂ (3.3) in order to determine the vertex exponents of other vertices. From ₍₂₎ equation (45) we know that for k = 1, exp (v ) ≥ n + s. Thus, its

  1 ₍₂₎ S ₂ remains to show exp (v 1 ) ≤ n + s. By considering equation (39) for each

  S ₂ i = 1, 2, ..., n, we show there is a path from v to v i consisting of n − 1 + s

  1 red arcs and one blue arc. It is suffices to show that the system x r(p ) n − 1 + s

  1 1,i M + = (46) x

  2 b(p 1,i )

  1 has a nonnegative integer solution which is x , x ≥ 0.

  1

  2 (2)

  Let p be a path from v to v i , i = 1, 2, ..., n. Since S is primitive, 1,i

  1

  2 M is an invertivle matrix. Hence the solution of the system (46) is below x s − r(p ) − (1 − n)b(p )

  1 1,i 1,i =

  Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  For any vertex v i , i = 1, 2, ..., n − s, there exist a path p 1,i from v

  1 to v i with s ≤ r(p ) ≤ n − 1 and b(p ) = 1 which implies x > 0 and

  1,i 1,i

  1 x = 0. Moreover, if i = n − s + 1, n − s + 2, ..., n, there exist a path p

  2 1,i from v to v i with 1 ≤ r(p ) ≤ s and b(p ) = 0 hence x ≥ 0 and x = 1.

  1 1,i 1,i

  1

  2 Therefore, we can conclude that for each i = 1, 2, ..., n, there is a path p 1,i from v to v hence the system (46) has a nonnegative integer solution. 1 i

  Since the vertex v lies on both cycles, Proposition (3.4) guarantees that

  1 exp (2) (v ) ≤ n + s. Since exp (2) (v ) ≥ n + s and exp (2) (v ) ≤ n + s, we

  1

  1

  1 S S S _{2 2 2} have exp (2) (v ) = n + s. We see that for every k = 1, 2, ..., n, the sortest

  1 S ₂ walk from v k to v is d(v k , v ) = k − 1. Proposition (3.3) guarantees that

  1

  1 exp (2) (v ) ≤ exp (2) (v ) + d(v , v ), hence k 1 k

  1 S S _{2 2} exp (2) (v k ) ≤ n − 1 + k + s (47)

  S ₂ for all k = 1, 2, ..., n and s = 1, 2, ..., n − 2.

  By equations (45) and (47) we conclude that the vertex exponents of (2) primitve two-colored digraph S is exp (2) (v ) = n − 1 + k + s for all k

  2 S ₂ k = 1, 2, ..., n and s = 1, 2, ..., n − 2.

  (2) Teorema 4.4 Let S be a primitive two-colored digraph on n ≥ 3 vertices

  2 (2) consisting one red loop at v and the n-cycle. If n-cycle in S contains ex-

  1

  2 actly one blue arc and (2) n−1 red arcs, then n−2+k ≤ exp (v k ) ≤ 2n−3+k

  S ₂ for all k = 1, 2, ..., n.

  Proof. By Lemma (4.1) through Lemma (4.3) we have that for each k = 1, 2, ..., n, the smallest vertex exponents of primitive two-colored di- (2) graph S is achieved by Lemma (4.1), that is exp (2) (v ) = n − 2 + k k

  2 S ₂ (2) and the biggest vertex exponents of primitive two-colored digraph S is

  2 achieved by Lemma (4.3) for s = n − 2, that is exp (2) (v ) = 2n − 3 + k. k

  S ₂ (2)

  This implies for any k = 1, 2, ..., n, the vertex exponents of S is n−2+k ≤ ₍₂₎ 2 exp (v k ) ≤ 2n − 3 + k.

  S ₂

  5. CONCLUSION In this paper we study the vertex exponents of two-colored digraph consist-

  (2)

  Siti Sahara et al. - Vertex Exponent of Two-Colored Digraph

  way that Suwilo [4] gives in setting up a lower and an upper bounds for ver- tex exponents of two-colored digraph consist of two cycles, we have that if (2) n-cycle in S has only one blue arc, then each of its vertices has exponents

  2 that lies on the interval [n − 2 + k, 2n − 3 + k].

  References

  [1] Fornasini, E. and Valcher, M. E. Directed graphs, 2D state models, and characteristic polynomials of irreducible matrix pair. Linear Algebra Appl . 263, 275-310, (1997). [2] Shader, B.L. and Suwilo, S. Exponents of nonnegative matrix pairs.

  Linear Algebra Appl . 363, 275-293, (2003). [3] Gao, Y and Shao, Y. Generalized exponents of primitive two-colored digraphs. Linear Algebra Appl. 430, 1550-1565, (2009).

  [4] Suwilo, S. Vertex exponents of two-colored primitive extremal min- istrong digraph. Global Journal of Technology and Optimization. Vol 2 No 2, 166-174, (2011). [5] Suwilo, S and Syafrianty, A. Vertex exponents of a class of two colored digraphs with even number of vertices. East-West J. of Mathematics.

  Special volume, 1-15, (2012). [6] Syahmarani, A and Suwilo, S. Vertex exponents of a class of two-colored hamiltonian digraphs. J.Indones.Math.Soc. Vol.18, No.1, 1-19, (2012).

  : Department of Mathematics, Faculty of Mathematics and Natural Sciences, University of Sumatera Utara, Medan 20155, Indonesia

  E-mail: sahararosseros@gmail.com

  : Department of Mathematics, Faculty of Mathematics and Natural Sciences, University of Sumatera Utara, Medan 20155, Indonesia

  E-mail: saib@usu.ac.id

  MARDININGSIH : Department of Mathematics, Faculty of Mathematics and Nat- ural Sciences, University of Sumatera Utara, Medan 20155, Indonesia

  E-mail: mardiningsih@usu.ac.id