Primitive Zero-Symmetric Sign Pattern Matrices with Zero Diagonal Attaining the Maximum Base

Abstract

The base set of primitive zero-symmetric sign pattern matrices with zero diagonal is {1,2,…,2𝑛−1}. In this paper, the primitive zero-symmetric sign pattern matrices with zero diagonal attaining the maximal base 2𝑛−1 are characterized.

1. Introduction

A sign pattern matrix (or sign pattern) 𝐴 is a matrix whose entries are from the set {1,−1,0}. Notice that for a square sign pattern matrix 𝐴, in the computation of (the signs of) the entries of the power 𝐴^𝑘, an ambiguous sign may arise when a positive sign is added to a negative sign. So a new symbol # was introduced in [1] to denote such an ambiguous sign. The powers of a square sign pattern have been investigated to some extent, see, for example, [1–12]. In [1], the set Γ={1,−1,0,#} is defined as the generalized sign set and the matrices with entries in the set Γ are called generalized sign pattern matrices, and the addition and multiplication involving the symbol # are defined as follows: ()()(),({0.−1+1=1+−1=#;𝑎+#=#∀𝑎∈Γ0⋅#=#⋅=0;𝑏⋅#=#⋅𝑏=#∀𝑏∈Γ⧵}) (1.1)

From now on, we assume that all the matrix operations considered in this paper are operations of the matrices over the set Γ.

Definition 1.1. 𝐴square generalized sign pattern matrix 𝐴 is called powerful if each power of 𝐴 contains no # entry.
In [1], Li et al. introduced the concepts of base and period for (powerful) sign pattern matrices which are the generalizations of the concepts of “index of convergence” and period for square nonnegative matrices. These concepts are extended from (powerful) sign pattern matrices to (square) generalized sign pattern matrices by You et al. in [12] as follows.

Definition 1.2 (see [12]). Let 𝐴 be a square generalized sign pattern matrix of order 𝑛 and 𝐴,𝐴²,𝐴³,… the sequence of powers of 𝐴. (Since there are only 4^𝑛2 different generalized sign patterns of order 𝑛, there must be repetitions in the sequence.) Suppose 𝐴^𝑙 is the first power that is repeated in the sequence, that is, 𝑙 is the least positive integer such that 𝐴^𝑙=𝐴^𝑙+𝑝 holds for some positive integer 𝑝. Then 𝑙 is called the generalized base (or simply base) of 𝐴, and is denoted by 𝑙(𝐴). The least positive integer 𝑝 such that 𝐴^𝑙=𝐴^𝑙+𝑝 holds for 𝑙=𝑙(𝐴) is called the generalized period (or simply period) of 𝐴 and is denoted by 𝑝(𝐴).
For a sign pattern matrix 𝐴, we use |𝐴| to denote the (0,1) matrix obtained from 𝐴 by replacing each nonzero entry by 1.
A nonnegative square matrix 𝐴 is primitive if some power 𝐴^𝑘>0. The least such as 𝑘 is called the primitive exponent (or simply exponent) of 𝐴, denoted by exp(𝐴). For convenience, a square sign pattern matrix 𝐴 is called primitive if |𝐴| is primitive, and in this case we define exp(𝐴)=exp(|𝐴|).

Definition 1.3 (see [3]). A square sign pattern matrix 𝐴=(𝑎_𝑖𝑗)_𝑛×𝑛 is called zero-pattern symmetric (abbreviated zero-symmetric, or simply ZS) if |𝐴| is symmetric.
It is well known that graph-theoretical methods are often useful in the study of the powers of square matrices, so we now introduce some graph-theoretical concepts.
Let 𝐷 be a digraph with vertex set 𝑉 and arc set 𝐸 (which permits loops but no multiple arcs). By assigning a sign of 1 or −1 to each arc of the digraph 𝐷, we obtain a signed digraph 𝑆. By a walk 𝑊 in the digraph 𝐷 (or the signed digraph 𝑆), we mean a sequence of vertices (𝑣₀,𝑣₁,𝑣₂,…,𝑣_𝑘) such that 𝑒_𝑖=(𝑣_𝑖−1,𝑣_𝑖) is an arc of 𝐷 for 𝑖=1,…,𝑘. The number 𝑘 is called the length of the walk 𝑊, denoted by 𝑙(𝑊). If the vertices 𝑣₀,𝑣₁,…,𝑣_𝑘−1 are distinct, the walk 𝑊 is called a path, if 𝑣_𝑘=𝑣₀, the path 𝑊 is called a cycle in 𝐷 (and in 𝑆). The sign of the walk 𝑊 in  𝑆, denoted by sgn(𝑊), is defined to be ∏^𝑘_𝑖=1sgn(𝑒_𝑖) where sgn(𝑒_𝑖) is the sign of the arc 𝑒_𝑖.

Let 𝐴=(𝑎_𝑖𝑗) be a sign pattern matrix of order 𝑛. Then the associated digraph 𝐷(𝐴) of 𝐴 is defined to be the digraph with vertex set 𝑉={1,2,…,𝑛} and arc set 𝐸={(𝑣_𝑖,𝑣_𝑗)∣𝑎_𝑖𝑗≠0}. The associated signed digraph 𝑆(𝐴) is obtained from 𝐷(𝐴) by assigning the sign of 𝑎_𝑖𝑗 to each arc (𝑣_𝑖,𝑣_𝑗) in 𝐷(𝐴).

Definition 1.4 (see [10]). Let 𝐷 be a digraph (permitting loops but no multiple arcs). Digraph 𝐷 is called primitive if there is a positive integer 𝑘 such that for all ordered pairs of vertices 𝑣_𝑖 and 𝑣_𝑗 (not necessarily distinct) in 𝐷, there exists a walk of length 𝑘 from 𝑣_𝑖 to 𝑣_𝑗. The least such 𝑘 is called the primitive exponent of 𝐷, denoted by exp(𝐷).
It is well known that a digraph 𝐷 is primitive if and only if 𝐷 is strongly connected and the greatest common divisor (gcd for short) of the lengths of all the cycles of 𝐷 is 1 (see [13]).

Definition 1.5 (see [3]). Let 𝑆 be a signed digraph of order 𝑛. Then there is a sign pattern matrix 𝐴 of order 𝑛 whose associated signed digraph 𝑆(𝐴) is 𝑆. We say that 𝑆 is powerful if 𝐴 is powerful. Also we define 𝑙(𝑆)=𝑙(𝐴).
We say that a sign pattern matrix 𝐴=(𝑎_𝑖𝑗) has zero diagonal if 𝑎_𝑖𝑖=0 for all 𝑖. A digraph 𝐷 with vertex set {𝑣₁,…,𝑣_𝑛} and arc set 𝐸 is called symmetric provided that (𝑣_𝑖,𝑣_𝑗)∈𝐸 iff (𝑣_𝑗,𝑣_𝑖)∈𝐸 for all 𝑖,𝑗. It is clear that a sign pattern matrix 𝐴 is ZS iff its associated digraph 𝐷(𝐴) is symmetric. For simplicity, we represent a symmetric (signed) digraph by its underlying graph.
The base set of primitive ZS sign pattern matrices and the base set of primitive ZS sign pattern matrices with zero diagonal are given, respectively, in [3, 10]. In [2], Cheng and Liu characterized the primitive ZS sign pattern matrices with the maximum base.
In this paper, we characterize the primitive sign pattern matrices with zero diagonal attaining the maximum base. Our main result is given in the following theorem.

Theorem 1.6. Let 𝐴=(𝑎_𝑖𝑗) be an 𝑛×𝑛 primitive zero-symmetric sign pattern matrix with zero diagonal. Then 𝑙(𝐴)≤2𝑛−1 (1.2) and the equality holds if and only if 𝐴 is nonpowerful and skew symmetric, namely, 𝑎_𝑖𝑗=−𝑎_𝑗𝑖 for all 1≤𝑖≤𝑗≤𝑛, and the associated digraph 𝐷(𝐴) is isomorphic to 𝐺 (see Figure 1).

Figure 1 

The graph  𝐺. (All 2-cycles in 𝐺 are negative, 𝑟 is odd, 3≤𝑟≤𝑛).

The proof of Theorem 1.6 will be given in Section 3.

2. Preliminary Results

In this section, we introduce some theorems, definitions, and lemmas which we need to use in the proof of our main result in Section 3.

In [1], Li et al. showed that if an irreducible sign pattern matrix 𝐴 is powerful, then 𝑙(𝐴)=𝑙(|𝐴|). That is to say the study of the base for a primitive powerful sign pattern matrix is essentially the study of the base (i.e., exponent) for primitive (0,1) matrices. Therefore, for a primitive powerful ZS sign pattern matrix with zero diagonal, Theorem 2.1 gives the base.

Theorem 2.1 (see [8]). Let 𝐴 be an 𝑛×𝑛 primitive symmetric (0,1) matrix with zero diagonal. Then exp(𝐴)≤2𝑛−4 and the primitive exponent set of 𝑛×𝑛 primitive symmetric (0,1) matrix with zero diagonal is {2,3,…,2𝑛−4}⧵𝐷, where 𝐷 is the set of odd numbers in {𝑛−2,𝑛−1,…,2𝑛−5}.

Theorem 2.2 (see [10]). Let 𝐴 be an 𝑛×𝑛 primitive ZS sign pattern matrix with zero diagonal. Then 𝑙(𝐴)≤2𝑛−1.

By Theorems 2.1 and 2.2, the sign pattern matrices with zero diagonal attaining this upper bound 𝑙(𝐴)=2𝑛−1 must be nonpowerful. So it remains to consider nonpowerful sign pattern matrices with zero diagonal.

Definition 2.3. Two walks 𝑊₁ and 𝑊₂ in a signed digraph are called a pair of SSSD walks, if they have the same initial vertex, same terminal vertex, and same length, but they have different signs.

Lemma 2.4 (see [8]). Let 𝐷 be a symmetric digraph. Then 𝐷 is primitive if and only if 𝐷 is strongly connected and there exists an odd cycle in 𝐷.

Lemma 2.5 (see [1, 12]). If 𝑆 is a primitive signed digraph, then  𝑆  is nonpowerful if and only if 𝑆 contains a pair of cycles 𝐶₁ and 𝐶₂, with lengths 𝑝₁ and 𝑝₂, respectively, satisfying one of the following conditions:(𝐴₁) 𝑝₁ is odd and 𝑝₂ is even and sgn(𝐶₂)=−1;(𝐴₂) both 𝑝₁ and 𝑝₂ are odd and sgn(𝐶₁)=−sgn(𝐶₂).

Lemma 2.6 (see [12]). Let 𝑆 be a primitive, nonpowerful signed digraph. Then we have the following.(1)There is an integer 𝑘 such that there exists a pair of SSSD walks of length 𝑘 from each vertex 𝑥 to each vertex 𝑦 in 𝑆.(2)If there exists a pair of SSSD walks of length 𝑘 from each vertex 𝑥 to each vertex 𝑦, then there also exists a pair of SSSD walks of length 𝑘+1 from each vertex 𝑢 to each vertex 𝑣 in 𝑆.(3)The minimal such 𝑘 (as in (1)) is just 𝑙(𝑆), the base of 𝑆.

Lemma 2.7 (see [7]). Suppose that an 𝑛×𝑛 sign pattern matrix 𝐴=(𝑎_𝑖𝑗) is skew symmetric. Let 𝑆(𝐴) be the associated signed digraph of 𝐴. Let 𝑟 be an odd integer with 3≤𝑟≤𝑛 (𝑛≥3). Then 𝑙(𝑆(𝐴))=2𝑛−1 if and only if 𝑆(𝐴) is isomorphic to 𝐺 (see Figure 1).

3. Main Results

For an undirected walk 𝑊 of graph 𝐺 and two vertices 𝑥, 𝑦 on 𝑊, we denote by 𝑄_𝑊(𝑥→𝑦) a shortest path from 𝑥 to 𝑦 on 𝑊 and by 𝑄(𝑥→𝑦) a shortest path from 𝑥 to 𝑦 on 𝐺. For a cycle 𝐶 of 𝐺, if 𝑥 and 𝑦 are two (not necessarily distinct) vertices on 𝐶 and 𝑃 is a path from 𝑥 to 𝑦 along 𝐶, then 𝐶⧵𝑃 denotes the path or cycle from 𝑥 to 𝑦 along 𝐶 obtained by deleting the edges of 𝑃.

Lemma 3.1. Let 𝐴 be an 𝑛×𝑛 primitive nonpowerful ZS sign pattern matrix with zero diagonal. If all the 2 cycles in 𝑆(𝐴) are positive, then 𝑙(𝐴)≤2𝑛−2.

Proof. Since 𝐴 is primitive, it follows from Lemma 2.4 that 𝑆(𝐴) is strongly connected and there is an odd cycle 𝐶^′ in 𝑆(𝐴) such that 𝑙(𝐶^′)=𝑙^′. Since 𝐴 has zero diagonal, there are no loops in 𝑆(𝐴) and so 𝑙^′≥3. Without loss of generality, we assume that 𝐶^′ is an odd cycle with the least length in 𝑆(𝐴).
Case  1. There exists at least one negative even cycle in 𝑆(𝐴).
Let 𝐶 be a negative even cycle in 𝑆(𝐴). Without loss of generality, we assume that 𝐶 is a negative even cycle with the least length in 𝑆(𝐴). Since all 2 cycles in 𝑆(𝐴) are positive, 𝑙(𝐶)=𝑙≥4.
Subcase  1.1. 𝐶 and 𝐶^′ have no common vertices.
Let 𝑃 be the shortest path from 𝐶 to 𝐶^′. Suppose 𝑃 intersects 𝐶 at vertex 𝑢 and intersects 𝐶^′ at vertex 𝑣 and there are 𝑘 vertices on 𝑃 where 𝑘≥2. Let 𝐺₀=𝐶∪𝑃∪𝐶^′. Let 𝑥 and 𝑦 be any two (not necessarily distinct) vertices in 𝑆(𝐴). Suppose that 𝑃₁ is the shortest path from 𝑥 to 𝐺₀ and intersects 𝐺₀ at vertex 𝑥^′ and 𝑃₂ is the shortest path from 𝑦 to 𝐺₀ and intersects 𝐺₀ at vertex 𝑦^′. Then 0≤𝑙(𝑃₁),  𝑙(𝑃₂)≤𝑛−𝑙−𝑙^′−𝑘+2, where 𝑙≥4 and 𝑙^′≥3. By the proof of Case  1 of Lemma  4.5 in [3], there exists a pair of SSSD walks from 𝑥 to 𝑦 of length 2𝑛−2. Therefore, by Lemma 2.6, 𝑙(𝐴)≤2𝑛−2.
Subcase  1.2. 𝐶 and 𝐶^′ have at least one common vertex.
Let 𝐺₁=𝐶∪𝐶^′. Let 𝑥 and 𝑦 be any two (not necessarily distinct) vertices in 𝑆(𝐴). Suppose that 𝑃₁ is the shortest path from 𝑥 to 𝐺₁ and intersects 𝐺₁ at vertex 𝑥^′ and 𝑃₂ is the shortest path from 𝑦 to 𝐺₁ and intersects 𝐺₁ at vertex 𝑦^′. Denote 𝐶∩𝐶^′ by 𝑅. Assume 𝑅 has 𝑚 vertices, where 1≤𝑚≤min(𝑙,𝑙^′), then 0≤𝑙𝑃₁,𝑙𝑃₂≤𝑛−𝑙−𝑙^′+𝑚. (3.1) If 𝑚≥2, then 𝑅 is a path with vertex set 𝑉(𝑅)={𝑣₁,𝑣₂,…,𝑣_𝑚} and edge set 𝐸(𝑅)={(𝑣₁,𝑣₂),(𝑣₂,𝑣₃),…,(𝑣_𝑚−1,𝑣_𝑚)}. In this case, if 𝑚=min(𝑙,𝑙^′)=𝑙, then there exists another odd cycle with length 𝑙^′−𝑙+2<𝑙^′, which contradicts our assumption that 𝐶^′ is an odd cycle with the least length in 𝑆(𝐴). Therefore, 𝑚<𝑙 and if 𝑚=min(𝑙,𝑙^′), then 𝑚=𝑙^′.
Subcase  1.2.1. 𝑥^′=𝑦^′ and 1≤𝑚≤min(𝑙,𝑙^′).
Subcase  1.2.1.1. 𝑥^′=𝑦^′∈𝐶⧵𝑅. See Figure 2(a).
We consider two subcases: the subcase 1≤𝑚<min(𝑙,𝑙^′) and the subcase 𝑚=min(𝑙,𝑙^′)=𝑙^′.
First, we consider the subcase 1≤𝑚<min(𝑙,𝑙^′). Without loss of generality, we assume that 𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁))≤𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣_𝑚)). Let 𝑤=𝑙(𝑃₁)+𝑙(𝐶)+𝑙(𝑃₂). If 𝑤 is even, set 𝑊^′=𝑃₁+𝐶+𝑃₂. (3.2) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣₁+𝑄_{𝐶′⧵𝑅}𝑣₁→𝑣_𝑚+𝑄_𝐶⧵𝑅𝑣_𝑚→𝑦^′+𝑃₂. (3.3) Let 𝑊₁=𝑊^′𝑊+𝐶,𝑤iseven,_′′𝑊+𝐶,otherwise,²=𝑊^′+𝑙2𝐶₂𝑊,𝑤iseven,_′′+𝑙2𝐶₂,otherwise, (3.4) where 𝐶₂ is a positive 2-cycle that contains vertex 𝑥. Since 𝐶^′ is odd, 𝑙(𝑄_{𝐶′⧵𝑅}(𝑣₁→𝑣_𝑚)) and 𝑙(𝑄_𝑅(𝑣₁→𝑣_𝑚)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is even, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑚+2𝑙=2𝑛−2𝑙^′+2𝑚. (3.5) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑚+(𝑙−𝑚+1)+𝑙^′−𝑚+1+𝑙=2𝑛−𝑙^′+2. (3.6) Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Then, we consider the subcase 𝑚=min(𝑙,𝑙^′)=𝑙^′. Without loss of generality, we assume that 𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁))≤𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣_𝑚)). Let 𝑤=𝑙𝑃₁+2𝑙𝑄_𝐶⧵𝑅𝑥^′→𝑣₁+𝑙𝑃₂. (3.7) If 𝑤 is even, set 𝑊^′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣₁+𝑄_𝐶⧵𝑅𝑣₁→𝑥^′+𝑃₂. (3.8) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣₁+𝑄_{𝐶′⧵𝑅}𝑣₁→𝑣_𝑚+𝑄_𝐶⧵𝑅𝑣_𝑚→𝑥^′+𝑃₂. (3.9) Let 𝑊₁=𝑊^′𝑊+𝐶,𝑤iseven,_′′𝑊+𝐶,otherwise,²=𝑊^′+𝑙2𝐶₂𝑊,𝑤iseven,_′′+𝑙2𝐶₂,otherwise, (3.10) where 𝐶₂ is a positive 2-cycle that contains vertex 𝑥. Since 𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁)+𝑄_{𝐶′⧵𝑅}(𝑣₁→𝑣_𝑚)+𝑄_𝐶⧵𝑅(𝑣_𝑚→𝑥^′) is an odd cycle, 𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁)) and 𝑙(𝑄_{𝐶′⧵𝑅}(𝑣₁→𝑣_𝑚)+𝑄_𝐶⧵𝑅(𝑣_𝑚→𝑥^′)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is even, 𝑙𝑊₁=𝑙𝑊₂≤2(𝑛−𝑙)+𝑙−𝑙^′+1+𝑙=2𝑛−𝑙^′+1. (3.11) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2(𝑛−𝑙)+𝑙−𝑙^′+2+𝑙=2𝑛−𝑙^′+2. (3.12) Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  1.2.1.2. 𝑥^′=𝑦^′∈𝐶^′⧵𝑅. See Figure 2(b).
The proof for this subcase is similar to that of the Subcase 1.2.1.1 and is omitted.
Subcase  1.2.1.3. 𝑥^′=𝑦^′∈𝑅. See Figure 2(c).
Let 𝑤=𝑙(𝑃₁)+𝑙(𝑃₂). If 𝑤 is even, set 𝑊^′=𝑃₁+𝑃₂. (3.13) Otherwise, set 𝑊^′′=𝑃₁+𝑃₂+𝐶^′. (3.14) Let 𝑊₁=𝑊^′𝑊+𝐶,𝑤iseven,_′′𝑊+𝐶,otherwise,²=𝑊^′+𝑙2𝐶₂𝑊,𝑤iseven,_′′+𝑙2𝐶₂,otherwise, (3.15) where 𝐶₂ is a positive 2-cycle that contains vertex 𝑥. Since 𝑙^′ is odd, we see that both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is even, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑚+𝑙=2𝑛−𝑙−2𝑙^′+2𝑚. (3.16) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑚+𝑙^′+𝑙=2𝑛−𝑙−𝑙^′+2𝑚. (3.17) Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  1.2.2. 𝑥^′≠𝑦^′ and 2≤𝑚≤min(𝑙,𝑙^′).
Subcase  1.2.2.1. 𝑥^′∈𝐶⧵𝑅 and 𝑦^′∈𝐶⧵𝑅. See Figure 3(a).
Without loss of generality, we assume that 𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁))≤𝑙(𝑄_𝐶⧵𝑅(𝑦^′→𝑣₁)). Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁))+𝑙(𝑄_𝑅(𝑣₁→𝑣_𝑚))+𝑙(𝑄_𝐶⧵𝑅(𝑣_𝑚→𝑦^′))+𝑙(𝑃₂). If 𝑤 is even, set 𝑊^′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣₁+𝑄_𝑅𝑣₁→𝑣_𝑚+𝑄_𝐶⧵𝑅𝑣_𝑚→𝑦^′+𝑃₂. (3.18) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣₁+𝑄_{𝐶′⧵𝑅}𝑣₁→𝑣_𝑚+𝑄_𝐶⧵𝑅𝑣_𝑚→𝑦^′+𝑃₂. (3.19) Let 𝑊₁=𝑊^′𝑊+𝐶,𝑤iseven,_′′𝑊+𝐶,otherwise,²=𝑊^′+𝑙2𝐶₂𝑊,𝑤iseven,_′′+𝑙2𝐶₂,otherwise, (3.20) where 𝐶₂ is a positive 2-cycle that contains vertex 𝑥. Since 𝐶^′ is odd, 𝑙(𝑄_{𝐶′⧵𝑅}(𝑣₁→𝑣_𝑚)) and 𝑙(𝑄_𝑅(𝑣₁→𝑣_𝑚)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Noting that 𝑊₃=𝑃₁+𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁)+𝑄_𝑅(𝑣₁→𝑣_𝑚)+𝑄_𝐶⧵𝑅(𝑣_𝑚→𝑦^′) and 𝑊₄=𝑃₁+𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁)+𝑄_{𝐶′⧵𝑅}(𝑣₁→𝑣_𝑚)+𝑄_𝐶⧵𝑅(𝑣_𝑚→𝑦^′) are paths of 𝑆(𝐴), we have 𝑙(𝑊₃),  𝑙(𝑊₄)≤𝑛−1. Then 𝑙𝑊₁=𝑙𝑊₂≤(𝑛−1)+𝑛−𝑙−𝑙^′+𝑚+𝑙=2𝑛−𝑙^′+𝑚−1. (3.21) Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  1.2.2.2. 𝑥^′∈𝐶⧵𝑅 and 𝑦^′∈𝑅. See Figure 3(b).
Without loss of generality, we assume that 𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁))≤𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣_𝑚)). Then we consider two subcases: the subcase 𝑥^′∈𝐶⧵𝑅, 𝑦^′∈𝑅 and 𝑦^′≠𝑣₁ and the subcase 𝑥^′∈𝐶⧵𝑅 and 𝑦^′=𝑣₁.
First, we consider the subcase 𝑥^′∈𝐶⧵𝑅, 𝑦^′∈𝑅 and 𝑦^′≠𝑣₁. Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁))+𝑙(𝑄_𝑅(𝑣₁→𝑦^′))+𝑙(𝑃₂). If 𝑤 is even, set 𝑊^′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣₁+𝑄_𝑅𝑣₁→𝑦^′+𝑃₂. (3.22) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣₁+𝑄_{𝐶′⧵𝑅}𝑣₁→𝑣_𝑚+𝑄_𝑅𝑣_𝑚→𝑦^′+𝑃₂. (3.23) Let 𝑊₁=𝑊^′𝑊+𝐶,𝑤iseven,_′′𝑊+𝐶,otherwise,²=𝑊^′+𝑙2𝐶₂𝑊,𝑤iseven,_′′+𝑙2𝐶₂,otherwise, (3.24) where 𝐶₂ is a positive 2-cycle that contains vertex 𝑥. Since 𝐶^′ is odd, 𝑙(𝑄_𝑅(𝑣₁→𝑦^′)) and 𝑙(𝑄_{𝐶′⧵𝑅}(𝑣₁→𝑣_𝑚))+𝑙(𝑄_𝑅(𝑣_𝑚→𝑦^′)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Noting that 𝑊₃=𝑃₁+𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁)+𝑄_𝑅(𝑣₁→𝑦^′) and 𝑊₄=𝑃₁+𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁)+𝑄_{𝐶′⧵𝑅}(𝑣₁→𝑣_𝑚)+𝑄_𝑅(𝑣_𝑚→𝑦^′) are two paths of 𝑆(𝐴), then we have 𝑙(𝑊₃),  𝑙(𝑊₄)≤𝑛−1. Then 𝑙𝑊₁=𝑙𝑊₂≤(𝑛−1)+𝑛−𝑙−𝑙^′+𝑚+𝑙=2𝑛−𝑙^′+𝑚−1. (3.25) Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Then, we consider the subcase 𝑥^′∈𝐶⧵𝑅 and 𝑦^′=𝑣₁. Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣_𝑚))+𝑙(𝑄_𝑅(𝑣_𝑚→𝑣₁(𝑦^′)))+𝑙(𝑃₂). If 𝑤 is even, set 𝑊^′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣_𝑚+𝑄_𝑅𝑣_𝑚→𝑣₁𝑦^′+𝑃₂. (3.26) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣_𝑚+𝑄_{𝐶′⧵𝑅}𝑣_𝑚→𝑣₁𝑦^′+𝑃₂. (3.27) Let 𝑊₁=𝑊^′𝑊+𝐶,𝑤iseven,_′′𝑊+𝐶,otherwise,²=𝑊^′+𝑙2𝐶₂𝑊,𝑤iseven,_′′+𝑙2𝐶₂,otherwise, (3.28) where 𝐶₂ is a positive 2-cycle that contains vertex 𝑥. Since 𝐶^′ is odd, 𝑙(𝑄_𝑅(𝑣_𝑚→𝑣₁(𝑦^′))) and 𝑙(𝑄_{𝐶′⧵𝑅}(𝑣_𝑚→𝑣₁(𝑦^′))) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Noting that 𝑊₃=𝑃₁+𝑄_𝐶⧵𝑅(𝑥^′→𝑣_𝑚)+𝑄_𝑅(𝑣_𝑚→𝑣₁(𝑦^′)) and 𝑊₄=𝑃₁+𝑄_𝐶⧵𝑅(𝑥^′→𝑣_𝑚)+𝑄_{𝐶′⧵𝑅}(𝑣_𝑚→𝑣₁(𝑦^′)) are two paths of 𝑆(𝐴), we have 𝑙(𝑊₃),  𝑙(𝑊₄)≤𝑛−1. Then 𝑙𝑊₁=𝑙𝑊₂≤(𝑛−1)+𝑛−𝑙−𝑙^′+𝑚+𝑙=2𝑛−𝑙^′+𝑚−1. (3.29) Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  1.2.2.3. 𝑥^′∈𝐶⧵𝑅 and 𝑦^′∈𝐶^′⧵𝑅. See Figure 3(c).
The proof for this subcase is similar to that of the Subcase 1.2.2.1 and is omitted.
Subcase  1.2.2.4. 𝑥^′∈𝑅 and 𝑦^′∈𝑅. See Figure 3(d).
Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶′(𝑥^′→𝑦^′))+𝑙(𝑃₂). If 𝑤 is even, set 𝑊^′=𝑃₁+𝑄_𝐶′𝑥^′→𝑦^′+𝑃₂. (3.30) Otherwise, set 𝑊^′′=𝑃₁+𝐶^′⧵𝑄_𝐶′𝑥^′→𝑦^′+𝑃₂. (3.31) Let 𝑊₁=𝑊^′𝑊+𝐶,𝑤iseven,_′′𝑊+𝐶,otherwise,²=𝑊^′+𝑙2𝐶₂𝑊,𝑤iseven,_′′+𝑙2𝐶₂,otherwise, (3.32) where 𝐶₂ is a positive 2-cycle that contains vertex 𝑥. Since 𝐶^′ is odd, 𝑙(𝑄_𝐶′(𝑥^′→𝑦^′)) and 𝑙(𝐶^′⧵𝑄_𝐶′(𝑥^′→𝑦^′)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then if 𝑤 is even, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑙+𝑚^′−12+𝑙=2𝑛+2𝑚−𝑙−3𝑙^′2−12. (3.33) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑚+𝑙^′−1+𝑙=2𝑛−𝑙−𝑙^′+2𝑚−1. (3.34) Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  1.2.2.5. 𝑥^′∈𝑅 and 𝑦^′∈𝐶^′⧵𝑅. See Figure 3(e).
The proof for this subcase is similar to that of Subcase 1.2.2.4, so we omit it.
Subcase  1.2.2.6. 𝑥^′∈𝐶^′⧵𝑅 and 𝑦^′∈𝐶^′⧵𝑅. See Figure 3(f).
Without loss of generality, we assume that 𝑙(𝑄_{𝐶′⧵𝑅}(𝑥^′→𝑣₁))≤𝑙(𝑄_{𝐶′⧵𝑅}(𝑦^′→𝑣₁)). Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_{𝐶′⧵𝑅}(𝑥^′→𝑣₁))+𝑙(𝑄_𝑅(𝑣₁→𝑣_𝑚))+𝑙(𝑄_{𝐶′⧵𝑅}(𝑣_𝑚→𝑦^′))+𝑙(𝑃₂). If 𝑤 is even, set 𝑊^′=𝑃₁+𝑄_{𝐶′⧵𝑅}𝑥^′→𝑣₁+𝑄_𝑅𝑣₁→𝑣_𝑚+𝑄_{𝐶′⧵𝑅}𝑣_𝑚→𝑦^′+𝑃₂. (3.35) Otherwise, set 𝑊^′′=𝑃₁+𝑄_{𝐶′⧵𝑅}𝑥^′→𝑦^′+𝑄_{𝐶′⧵𝑅}𝑦^′→𝑣_𝑚+𝑄_{𝐶′⧵𝑅}𝑣_𝑚→𝑦^′+𝑃₂. (3.36) Let 𝑊₁=𝑊^′𝑊+𝐶,𝑤iseven,_′′𝑊+𝐶,otherwise,²=𝑊^′+𝑙2𝐶₂𝑊,𝑤iseven,_′′+𝑙2𝐶₂,otherwise, (3.37) where 𝐶₂ is a positive 2-cycle that contains vertex 𝑥. Since 𝐶^′ is odd, 𝑙(𝑄_{𝐶′⧵𝑅}(𝑥^′→𝑦^′)) and 𝑙(𝑄_{𝐶′⧵𝑅}(𝑥^′→𝑣₁))+𝑙(𝑄_𝑅(𝑣₁→𝑣_𝑚))+𝑙(𝑄_{𝐶′⧵𝑅}(𝑣_𝑚→𝑦^′)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Noting that 𝑊₃=𝑃₁+𝑄_{𝐶′⧵𝑅}(𝑥^′→𝑣₁)+𝑄_𝑅(𝑣₁→𝑣_𝑚)+𝑄_{𝐶′⧵𝑅}(𝑣_𝑚→𝑦^′) and 𝑊₄=𝑃₁+𝑄_{𝐶′⧵𝑅}(𝑥^′→𝑦^′)+𝑄_{𝐶′⧵𝑅}(𝑦^′→𝑣_𝑚) are two paths of 𝑆(𝐴), then we have 𝑙(𝑊₃)≤𝑛−1 and 𝑙(𝑊₄)≤𝑛−𝑚. Then, if 𝑤 is even, 𝑙𝑊₁=𝑙𝑊₂≤(𝑛−1)+𝑛−𝑙−𝑙^′+𝑚+𝑙=2𝑛−𝑙^′+𝑚−1. (3.38) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤𝑛−𝑙−𝑙^′+𝑚+(𝑛−𝑚)+𝑙^′−𝑚−1+𝑙=2𝑛−𝑚−1. (3.39) Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  1.2.3. 𝑚=1 and 𝑥^′≠𝑦^′. We assume that 𝑉(𝐶)∩𝑉(𝐶^′)={𝑢}.
Subcase  1.2.3.1. 𝑥^′∈𝐶 and 𝑦^′∈𝐶. See Figure 4(a).
Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶(𝑥^′→𝑢))+𝑙(𝑄_𝐶(𝑢→𝑦^′))+𝑙(𝑃₂). If 𝑤 is even, set 𝑊^′=𝑃₁+𝑄_𝐶𝑥^′→𝑢+𝑄_𝐶𝑢→𝑦^′+𝑃₂. (3.40) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝐶𝑥^′→𝑢+𝑄_𝐶𝑢→𝑦^′+𝑃₂+𝐶^′. (3.41) Let 𝑊₁=𝑊^′𝑊+𝐶,𝑤iseven,_′′𝑊+𝐶,otherwise,²=𝑊^′+𝑙2𝐶₂𝑊,𝑤iseven,_′′+𝑙2𝐶₂,otherwise, (3.42) where 𝐶₂ is a positive 2-cycle that contains vertex 𝑥. Since 𝑙^′ is odd, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is even, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+1+(𝑙−1)+𝑙=2𝑛−2𝑙^′+1. (3.43) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+1+(𝑙−1)+𝑙^′+𝑙=2𝑛−𝑙^′+1. (3.44)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  1.2.3.2. 𝑥^′∈𝐶, 𝑦^′∈𝐶^′, and 𝑥^′≠𝑦^′≠𝑢. See Figure 4(b).
Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶(𝑥^′→𝑢))+𝑙(𝑄_𝐶′(𝑢→𝑦^′))+𝑙(𝑃₂). If 𝑤 is even, set 𝑊^′=𝑃₁+𝑄_𝐶𝑥^′→𝑢+𝑄_𝐶′𝑢→𝑦^′+𝑃₂. (3.45) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝐶𝑥^′→𝑢+𝐶^′⧵𝑄_𝐶′𝑢→𝑦^′+𝑃₂. (3.46) Let 𝑊₁=𝑊^′𝑊+𝐶,𝑤iseven,_′′𝑊+𝐶,otherwise,²=𝑊^′+𝑙2𝐶₂𝑊,𝑤iseven,_′′+𝑙2𝐶₂,otherwise, (3.47) where 𝐶₂ is a positive 2-cycle that contains vertex 𝑥. Since 𝐶^′ is odd, 𝑙(𝐶^′⧵𝑄_𝐶′(𝑢→𝑦^′)) and 𝑙(𝑄_𝐶′(𝑢→𝑦^′)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Noting that 𝑊₃=𝑃₁+𝑄_𝐶(𝑥^′→𝑢)+𝑄_𝐶′(𝑢→𝑦^′) and 𝑊₄=𝑃₁+𝑄_𝐶(𝑥^′→𝑢)+𝐶^′⧵𝑄_𝐶′(𝑢→𝑦^′) are two paths of 𝑆(𝐴), we have 𝑙(𝑊₃),  𝑙(𝑊₄)≤𝑛−1. Then, 𝑙𝑊₁=𝑙𝑊₂≤(𝑛−1)+𝑛−𝑙−𝑙^′+1+𝑙=2𝑛−𝑙^′. (3.48)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  1.2.3.3. 𝑥^′∈𝐶^′ and 𝑦^′∈𝐶^′. See Figure 4(c).
The proof for this subcase is similar to that of the Subcase 1.2.3.1 and is omitted.
Thus in each of the above subcases, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2. Therefore, we get 𝑙(𝐴)≤2𝑛−2 by Lemma 2.6.
Case  2. There exists no negative even cycle in 𝑆(𝐴).
Since 𝐴 is primitive, nonpowerful and there exist no negative even cycles in 𝑆(𝐴), it follows from Lemma 2.5 that there exist two odd cycles 𝐶 and 𝐶^′ with different signs in 𝑆(𝐴). We assume that 𝑙(𝐶)=𝑙 and 𝑙(𝐶^′)=𝑙^′. Since the trace of 𝐴 is zero, there are no loops in 𝑆(𝐴) and so 𝑙,  𝑙^′≥3. Without loss of generality, we assume 𝑙≥𝑙^′≥3.
Subcase  2.1. 𝐶 and 𝐶^′ have no common vertices.
Let 𝑃 be the shortest path from 𝐶 to 𝐶^′. Suppose 𝑃 intersects 𝐶 at vertex 𝑢 and intersects 𝐶^′ at vertex 𝑣 and there are 𝑘 vertices on 𝑃 where 𝑘≥2. Let 𝐺₂=𝐶∪𝑃∪𝐶^′. Let 𝑥 and 𝑦 be two arbitrary (not necessarily distinct) vertices in 𝑆(𝐴). Suppose that 𝑃₁ is the shortest path from 𝑥 to 𝐺₂ and intersects 𝐺₂ at vertex 𝑥^′ and 𝑃₂ is the shortest path from 𝑦 to 𝐺₂ and intersects 𝐺₂ at vertex 𝑦^′. Then 0≤𝑙𝑃₁,𝑙𝑃₂≤𝑛−𝑙−𝑙^′−𝑘+2. (3.49) Since 𝐶 and 𝐶^′ have diffident signs, if there exists an even walk 𝑊 with length 𝑙(𝑊)≤2𝑛−2−𝑙, then 𝑊₁=𝑊+𝐶 and 𝑊₂=𝑊+((𝑙−𝑙^′)/2)𝐶₂+𝐶^′ have the same length 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2 and different signs. As the proof of Subcase  1.1, we can construct an even walk 𝑊 with length 𝑙(𝑊)≤2𝑛−2−𝑙. So there exists a pair of SSSD walks from 𝑥 to 𝑦 of length 2𝑛−2. It remains to consider the following subcase.
Subcase  2.2. 𝐶 and 𝐶^′ have at least one common vertex.
Let 𝐺₃=𝐶∪𝐶^′. Let 𝑥 and 𝑦 be two arbitrary (not necessarily distinct) vertices in 𝑆(𝐴). Suppose that 𝑃₁ is the shortest path from 𝑥 to 𝐺₃ and intersects 𝐺₃ at vertex 𝑥^′ and 𝑃₂ is the shortest path from 𝑦 to 𝐺₃ and intersects 𝐺₃ at vertex 𝑦^′. Denote 𝐶∩𝐶^′ by 𝑅. Assume 𝑅 has 𝑚 vertices, where 1≤𝑚≤min(𝑙₁,𝑙₂), then 0≤𝑙𝑃₁,𝑙𝑃₂≤𝑛−𝑙₁−𝑙₂+𝑚. (3.50) If 𝑚>1, then 𝑅 is a path with vertex set 𝑉(𝑅)={𝑣₁,𝑣₂,…,𝑣_𝑚} and edge set 𝐸(𝑅)={(𝑣₁,𝑣₂),(𝑣₂,𝑣₃),…,(𝑣_𝑚−1,𝑣_𝑚)}. If 𝑚=𝑙^′, since there exists no negative even cycles in 𝑆(𝐴), then 𝐶 and 𝐶^′ have the same sign, a contradiction. Therefore, 1≤𝑚<𝑙^′.
Subcase  2.2.1. 𝑥^′∈𝐶⧵𝑅 and 𝑦^′∈𝐶⧵𝑅. See Figure 3(a).
Without loss of generality, we assume that 𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁))≤𝑙(𝑄_𝐶⧵𝑅(𝑦^′→𝑣₁)). Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁))+𝑙(𝑄_𝑅(𝑣₁→𝑣_𝑚))+𝑙(𝑄_𝐶⧵𝑅(𝑣_𝑚→𝑦^′))+𝑙(𝑃₂). If 𝑤 is odd, set 𝑊^′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣₁+𝑄_𝑅𝑣₁→𝑣_𝑚+𝑄_𝐶⧵𝑅𝑣_𝑚→𝑦^′+𝑃₂. (3.51) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣₁+𝑄_{𝐶′⧵𝑅}𝑣₁→𝑣_𝑚+𝑄_𝐶⧵𝑅𝑣_𝑚→𝑦^′+𝑃₂. (3.52) Let 𝑊₁=𝑊^′𝑊+𝐶,𝑤isodd,_′′𝑊+𝐶,otherwise,²=𝑊^′+𝐶^′+𝑙−𝑙^′2𝐶₂𝑊,𝑤isodd,_′′+𝐶^′+𝑙−𝑙^′2𝐶₂,otherwise, (3.53) where 𝐶₂ is a positive 2-cycle that contains 𝑥. Since 𝐶^′ is odd, 𝑙(𝑄_𝑅(𝑣₁→𝑣_𝑚)) and 𝑙(𝑄_{𝐶′⧵𝑅}(𝑣₁→𝑣_𝑚)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑚+𝑙+𝑙=2𝑛−2𝑙^′+2𝑚. (3.54) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑚+(𝑙−𝑚+1)+𝑙^′−𝑚+1+𝑙=2𝑛−𝑙^′+2. (3.55)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  2.2.2. 𝑥^′∈𝐶⧵𝑅 and 𝑦^′∈𝑅. See Figure 3(b).
Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶(𝑥^′→𝑦^′))+𝑙(𝑃₂). If 𝑤 is odd, set 𝑊^′=𝑃₁+𝑄_𝐶𝑥^′→𝑦^′+𝑃₂. (3.56) Otherwise, set 𝑊^′′=𝑃₁+𝐶⧵𝑄_𝐶𝑥^′→𝑦^′+𝑃₂. (3.57) Let 𝑊₁=𝑊^′𝑊+𝐶,𝑤isodd,_′′𝑊+𝐶,otherwise,²=𝑊^′+𝐶^′+𝑙−𝑙^′2𝐶₂𝑊,𝑤isodd,_′′+𝐶^′+𝑙−𝑙^′2𝐶₂,otherwise, (3.58) where 𝐶₂ is a positive 2-cycle that contains 𝑥. Since 𝐶 is odd, 𝐶⧵𝑄_𝐶(𝑥^′→𝑦^′) and 𝑄_𝐶(𝑥^′→𝑦^′) have different parity, Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑚+𝑙−12+𝑙=2𝑛−2𝑙^′+2𝑚−𝑙+12. (3.59) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑚+(𝑙−1)+𝑙=2𝑛−2𝑙^′+2𝑚−1. (3.60) Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  2.2.3. 𝑥^′∈𝐶⧵𝑅 and 𝑦^′∈𝐶^′⧵𝑅. See Figure 3(c).
Without loss of generality, we assume that 𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁))≤𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣_𝑚)). Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁))+𝑙(𝑄_{𝐶′⧵𝑅}(𝑣₁→𝑦^′))+𝑙(𝑃₂). If 𝑤 is odd, set 𝑊^′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣₁+𝑄_{𝐶′⧵𝑅}𝑣₁→𝑦^′+𝑃₂. (3.61) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣₁+𝑄_𝑅𝑣₁→𝑣_𝑚+𝑄_{𝐶′⧵𝑅}𝑣_𝑚→𝑦^′+𝑃₂. (3.62) Let 𝑊₁=𝑊^′𝑊+𝐶,𝑤isodd,_′′𝑊+𝐶,otherwise,²=𝑊^′+𝐶^′+𝑙−𝑙^′2𝐶₂𝑊,𝑤isodd,_′′+𝐶^′+𝑙−𝑙^′2𝐶₂,otherwise, (3.63) where 𝐶₂ is a positive 2-cycle that contains 𝑥. Since 𝐶^′ is odd, 𝑙(𝑄_{𝐶′⧵𝑅}(𝑣₁→𝑦^′)) and 𝑙(𝑄_𝑅(𝑣₁→𝑣_𝑚))+𝑙(𝑄_{𝐶′⧵𝑅}(𝑣_𝑚→𝑦^′)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Noting that 𝑊₃=𝑃₁+𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁)+𝑄_{𝐶′⧵𝑅}(𝑣₁→𝑦^′) and 𝑊₄=𝑃₁+𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁)+𝑄_𝑅(𝑣₁→𝑣_𝑚)+𝑄_{𝐶′⧵𝑅}(𝑣_𝑚→𝑦^′) are paths of 𝑆(𝐴), then we have 𝑙(𝑊₃),  𝑙(𝑊₄)≤𝑛−1. Therefore, 𝑙𝑊₁=𝑙𝑊₂≤(𝑛−1)+𝑛−𝑙−𝑙^′+𝑚+𝑙=2𝑛−𝑙^′+𝑚−1. (3.64) Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  2.2.4. 𝑥^′∈𝑅 and 𝑦^′∈𝑅. See Figure 3(d).
The proof for this subcase is similar to that of Subcase 2.2.2, so we omit it.
Subcase  2.2.5. 𝑥^′∈𝑅 and 𝑦^′∈𝐶^′⧵𝑅. See Figure 3(e).
The proof for this subcase is similar to that of Subcase 2.2.2, so we omit it.
Subcase  2.2.6. 𝑥^′∈𝐶^′⧵𝑅 and 𝑦^′∈𝐶^′⧵𝑅. See Figure 3(f).
The proof for this subcase is similar to that of Subcase 2.2.1, so we omit it.
From all the above subcases, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2. Therefore, we get 𝑙(𝐴)≤2𝑛−2 by Lemma 2.6.

(a)

(b)

(c)

Figure 2 

Illustrate for Subcase 1.2.1 of Lemma 3.1.

(a)

(b)

(c)

(d)

(e)

(f)

Figure 3 

Illustration for Subcases  1.2.2 and 2.2 of Lemma 3.1.

(a)

(b)

(c)

Figure 4 

Illustrate for Subcase 1.2.3 of Lemma 3.1.

Lemma 3.2. Let 𝐴 be an 𝑛×𝑛 primitive nonpowerful ZS sign pattern matrix with zero diagonal. If there exist a negative 2-cycle and a positive 2-cycle in 𝑆(𝐴), then 𝑙(𝐴)≤2𝑛−2.

Proof. Since 𝐴 is primitive, it follows from Lemma 2.4 that 𝑆(𝐴) is strongly connected and there is an odd cycle 𝐶 in 𝑆(𝐴) with length 𝑙(𝐶)=𝑙. Since 𝐴 has zero diagonal, there are no loops in 𝑆(𝐴) and so 𝑙≥3. Without loss of generality, we assume that 𝐶 is an odd cycle with the least length in 𝑆(𝐴). Since 𝐴 is ZS and 𝑆(𝐴) contain a positive 2-cycle and a negative 2-cycle, there exists a positive 2-cycle 𝐶^′₂ and a negative 2-cycle 𝐶₂^′′ such that 𝑉(𝐶^′₂)∩𝑉(𝐶₂^′′)≠∅. Let 𝑢∈𝑉(𝐶^′₂)∩𝑉(𝐶₂^′′).
Let 𝑃 be the shortest path from 𝑢 to 𝐶. Let 𝑥 and 𝑦 be two arbitrary (not necessarily distinct) vertices in 𝑆(𝐴). Suppose there are 𝑘 vertices on 𝑃 and 𝑃 intersects 𝐶 at 𝑣, where 𝑘≥1. Suppose 𝑃₁ is the shortest path from 𝑥 to 𝑃∪𝐶 and 𝑃₁ intersects 𝑃∪𝐶 at 𝑥^′ and 𝑃₂ is the shortest path from 𝑦 to 𝑃∪𝐶 and 𝑃₂ intersects 𝑃∪𝐶 at 𝑦^′ where 0≤𝑙(𝑃₁),  𝑙(𝑃₂)≤𝑛−𝑙−𝑘+1. We consider the following three cases.
Case  1. 𝑥^′∈𝑃 and 𝑦^′∈𝑃.
By the Subcase 2.1 of Lemma  4.2 in [3], there exists a pair of SSSD walks from 𝑥 to 𝑦 of length 2𝑛−2.
Case  2. Only one of 𝑥^′ and 𝑦^′ belongs to 𝑃.
By the Subcase2.2 of Lemma  4.2 in [3], there exists a pair of SSSD walks from 𝑥 to 𝑦 of length 2𝑛−2.
It remains to consider the following case.
Case  3. 𝑥^′∉𝑃 and 𝑦^′∉𝑃. See Figure 5.
Subcase  3.1. 𝑥^′=𝑦^′.
Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶(𝑥^′→𝑣))+2𝑙(𝑃)+𝑙(𝑄_𝐶(𝑣→𝑦^′))+𝑙(𝑃₂). If 𝑤 is even, set 𝑊^′=𝑃₁+𝑄_𝐶𝑥^′→𝑣+2𝑃+𝑄_𝐶𝑣→𝑦^′+𝑃₂. (3.65) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝐶𝑥^′→𝑣+2𝑃+𝐶⧵𝑄_𝐶𝑣→𝑦^′+𝑃₂. (3.66) Let 𝑊₁=𝑊^′+𝐶^′₂𝑊,𝑤iseven,_′′+𝐶^′₂𝑊,otherwise,²=𝑊^′+𝐶₂^′′𝑊,𝑤iseven,_′′+𝐶₂^′′,otherwise. (3.67) Since 𝐶 is odd, 𝑙(𝑄_𝐶(𝑣→𝑥^′)) and 𝑙(𝐶⧵𝑄_𝐶(𝑣→𝑥^′)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤2(𝑛−𝑙−𝑘+1)+2(𝑘−1)+(𝑙−1)+2=2𝑛−𝑙+1. (3.68) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2(𝑛−𝑙−𝑘+1)+2(𝑘−1)+𝑙+2=2𝑛−𝑙+2. (3.69)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  3.2. 𝑥^′≠𝑦^′.
Without loss of generality, we assume that 𝑙(𝑄_𝐶(𝑥^′→𝑣))≤𝑙(𝑄_𝐶(𝑦^′→𝑣)). Let 𝑃₃=𝑄_𝐶(𝑥^′→𝑣) and 𝑃₄=𝑄_𝐶⧵𝑃3(𝑣→𝑦^′). Let 𝑤=𝑙(𝑃₁)+𝑙(𝑃₃)+𝑙(𝑃₄)+2𝑙(𝑃)+𝑙(𝑃₂), If 𝑤 is odd, set 𝑊^′=𝑃₁+𝑄_𝐶𝑥^′→𝑣+2𝑃+𝑃₄+𝑃₂. (3.70) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝐶𝑥^′→𝑣+2𝑃+𝑄_𝐶𝑣→𝑥^′+𝑄_{𝐶⧵{𝑃3+𝑃4}}𝑥^′→𝑦^′+𝑃₂. (3.71) Let 𝑊₁=𝑊^′+𝐶^′₂𝑊,𝑤iseven,_′′+𝐶^′₂𝑊,otherwise,²=𝑊^′+𝐶₂^′′𝑊,𝑤iseven,_′′+𝐶₂^′′,otherwise. (3.72) Since 𝐶 is odd, 𝑙(𝑃₃)+𝑙(𝑃₄) and 𝑙(𝑄_{𝐶⧵{𝑃3+𝑃4}}(𝑥^′→𝑦^′)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤2(𝑛−𝑙−𝑘+1)+2(𝑘−1)+(𝑙−1)+2=2𝑛−𝑙+1. (3.73) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2(𝑛−𝑙−𝑘+1)+2(𝑘−1)+(𝑙−1)+(𝑙−2)+2=2𝑛−1. (3.74)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Therefore, 𝑙(𝐴)≤2𝑛−2 by Lemma 2.6.

Figure 5 

Illustration for Case 3 of Lemma 3.2.

Lemma 3.3. Let 𝐴 be an 𝑛×𝑛 primitive nonpowerful ZS sign pattern matrix with zero diagonal. If 𝑙(𝐴)=2𝑛−1, then all 2-cycles in 𝑆(𝐴) are negative.

Proof. Assume that 𝑙(𝐴)=2𝑛−1. By Lemma 3.1, 𝑆(𝐴) has at least one negative 2-cycle. It then follows from Lemma 3.2 that all 2-cycles in 𝑆(𝐴) are negative.

Lemma 3.4. Let 𝐴 be an 𝑛×𝑛 primitive nonpowerful ZS sign pattern matrix with zero diagonal. Suppose that 𝑙(𝐴)=2𝑛−1. Then there exists an odd cycle 𝐶 in 𝑆(𝐴) and 𝐶 is the only cycle of length at least 3 in 𝑆(𝐴).

Proof. Since 𝐴 is primitive, it follows from Lemma 2.4 that 𝑆(𝐴) is strongly connected and there is an odd cycle 𝐶=(𝑣₁,𝑣₂,…,𝑣_𝑙) in 𝑆(𝐴) with length 𝑙(𝐶)=𝑙. Since 𝐴 has zero diagonal, there is no loop in 𝑆(𝐴) and so 𝑙≥3. Consider the following two directed cycles: 𝐶¹=𝑣₁,𝑣₂,…,𝑣_𝑙,𝑣₁,𝐶²=𝑣₁,𝑣_𝑙,…,𝑣₂,𝑣₁. (3.75) Since there exists no positive 2-cycle in 𝑆(𝐴) by Lemma 3.3, the arcs (𝑣_𝑖,𝑣_𝑖+1) and (𝑣_𝑖+1,𝑣_𝑖) have different signs. Thus sgn(𝐶¹)=−sgn(𝐶²) by the fact that 𝑙 is odd. Without loss of generality, we assume that 𝐶 is an odd cycle with the least length in 𝑆(𝐴).
If 𝑙(𝐴)=2𝑛−1, suppose there exists another cycle 𝐶^′ with length 𝑙(𝐶^′)=𝑙^′≥3, we consider the following two cases.
Case  1. 𝐶 and 𝐶^′ have no common vertices.
Let 𝑃 be the shortest path from 𝐶 to 𝐶^′. Suppose 𝑃 intersects 𝐶 at vertex 𝑢 and intersects 𝐶^′ at vertex 𝑣 and there are 𝑘 vertices on 𝑃 where 𝑘≥2. Let 𝐺₄=𝐶∪𝑃∪𝐶^′. Let 𝑥 and 𝑦 be two arbitrary (not necessarily distinct) vertices in 𝑆(𝐴). Suppose that 𝑃₁ is the shortest path from 𝑥 to 𝐺₄ and intersects 𝐺₄ at vertex 𝑥^′ and 𝑃₂ is the shortest path from 𝑦 to 𝐺₄ and intersects 𝐺₄ at vertex 𝑦^′. Then 0≤𝑙𝑃₁,𝑙𝑃₂≤𝑛−𝑙−𝑙^′−𝑘+2. (3.76) We consider the following six subcases.
Subcase  1.1. 𝑥^′∈𝐶 and 𝑦^′∈𝐶. See Figure 6(a).
Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶(𝑥^′→𝑦^′))+𝑙(𝑃₂). If 𝑤 is odd, set 𝑊^′=𝑃₁+𝑄_𝐶𝑥^′→𝑦^′+𝑃₂. (3.77) Otherwise, set 𝑊^′′=𝑃₁+𝐶⧵𝑄_𝐶𝑥^′→𝑦^′+𝑃₂. (3.78) Let 𝑊₁=𝑊^′+𝐶¹𝑊,𝑤isodd,_′′+𝐶¹𝑊,otherwise,²=𝑊^′+𝐶²𝑊,𝑤isodd,_′′+𝐶²,otherwise. (3.79) Since 𝐶 is odd, 𝑙(𝑄_𝐶(𝑥^′→𝑦^′)) and 𝑙(𝐶⧵𝑄_𝐶(𝑥^′→𝑦^′)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′−𝑘+2+𝑙−12+𝑙=2𝑛−2𝑙^′𝑙−2𝑘−2+72. (3.80) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′−𝑘+2+𝑙+𝑙=2𝑛−2𝑙^′−2𝑘+4. (3.81)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  1.2. 𝑥^′∈𝐶 and 𝑦^′∈𝑃. See Figure 6(b).
Let 𝑊=𝑙(𝑃₁)+𝑙(𝑄_𝐶(𝑥^′→𝑢))+𝑙(𝑄_𝑃(𝑢→𝑦^′))+𝑙(𝑃₂). If 𝑤 is odd, set 𝑊^′=𝑃₁+𝑄_𝐶𝑥^′→𝑢+𝑄_𝑃𝑢→𝑦^′+𝑃₂. (3.82) Otherwise, set 𝑊^′=𝑃₁+𝐶⧵𝑄_𝐶𝑥^′→𝑢+𝑄_𝑃𝑢→𝑦^′+𝑃₂. (3.83) Let 𝑊₁=𝑊^′+𝐶¹𝑊,𝑤isodd,_′′+𝐶¹𝑊,otherwise,²=𝑊^′+𝐶²𝑊,𝑤isodd,_′′+𝐶²,otherwise. (3.84) Since 𝐶 is odd, 𝑙(𝑄_𝐶(𝑥^′→𝑢)) and 𝑙(𝐶⧵𝑄_𝐶(𝑥^′→𝑢)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′−𝑘+2+𝑙−12+(𝑘−1)+𝑙=2𝑛−2𝑙^′𝑙−𝑘−2+52. (3.85) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′−𝑘+2+(𝑘−1)+𝑙+𝑙=2𝑛−2𝑙^′−𝑘+3. (3.86)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  1.3. 𝑥^′∈𝐶 and 𝑦^′∈𝐶^′. See Figure 6(c).
Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶(𝑥^′→𝑢))+𝑙(𝑃)+𝑙(𝑄_𝐶′(𝑣→𝑦^′))+𝑙(𝑃₂). If 𝑤 is odd, set 𝑊^′=𝑃₁+𝑄_𝐶𝑥^′→𝑢+𝑃+𝑄_𝐶′𝑣→𝑦^′+𝑃₂. (3.87) Otherwise, set 𝑊^′′=𝑃₁+𝐶⧵𝑄_𝐶𝑥^′→𝑢+𝑃+𝑄_𝐶′𝑣→𝑦^′+𝑃₂. (3.88) Let 𝑊₁=𝑊^′+𝐶¹𝑊,𝑤isodd,_′′+𝐶¹𝑊,otherwise,²=𝑊^′+𝐶²𝑊,𝑤isodd,_′′+𝐶²,otherwise. (3.89) Since 𝐶 is odd, 𝑙(𝑄_𝐶(𝑥^′→𝑢)) and 𝑙(𝐶⧵𝑄_𝐶(𝑥^′→𝑢)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′−𝑘+2+𝑙−12+()+𝑙𝑘−1^′2𝑙+𝑙=2𝑛−𝑘−2−3𝑙^′2+52. (3.90) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′()+𝑙−𝑘+2+𝑙+𝑘−1^′2+𝑙=2𝑛−𝑘−3𝑙^′2+3. (3.91)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  1.4. 𝑥^′∈𝑃 and 𝑦^′∈𝑃. See Figure 6(d).
Without loss of generality, we assume that 𝑙(𝑄_𝑃(𝑥^′→𝑢))≤𝑙(𝑄_𝑃(𝑦^′→𝑢)). Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝑃(𝑥^′→𝑢))+𝑙(𝑃)+𝑙(𝑄_𝑃(𝑣→𝑦^′))+𝑙(𝑃₂). If 𝑤 is odd, set 𝑊^′=𝑃₁+𝑄_𝑃𝑥^′→𝑢+𝑃+𝑄_𝑃𝑣→𝑦^′+𝑃₂. (3.92) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝑃𝑥^′→𝑢+𝑃+𝑄_𝑃𝑣→𝑦^′+𝑃₂+𝐶. (3.93) Let 𝑊₁=𝑊^′+𝐶¹𝑊,𝑤isodd,_′′+𝐶¹𝑊,otherwise,²=𝑊^′+𝐶²𝑊,𝑤isodd,_′′+𝐶²,otherwise. (3.94) Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′−𝑘+2+2(𝑘−1)+𝑙=2𝑛−𝑙−2𝑙^′+2. (3.95) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′−𝑘+2+2(𝑘−1)+𝑙+𝑙=2𝑛−2𝑙^′+2. (3.96)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  1.5. 𝑥^′∈𝑃 and 𝑦^′∈𝐶^′. See Figure 6(e).
Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝑃(𝑥^′→𝑢))+𝑙(𝑃)+𝑙(𝑄_𝐶′(𝑣→𝑦^′))+𝑙(𝑃₂), if 𝑤 is odd, set 𝑊^′=𝑃₁+𝑄_𝑃𝑥^′→𝑢+𝑃+𝑄_𝐶′𝑣→𝑦^′+𝑃₂. (3.97) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝑃𝑥^′→𝑢+𝑃+𝑄_𝐶′𝑣→𝑦^′+𝑃₂+𝐶. (3.98) Let 𝑊₁=𝑊^′+𝐶¹𝑊,𝑤isodd,_′′+𝐶¹𝑊,otherwise,²=𝑊^′+𝐶²𝑊,𝑤isodd,_′′+𝐶²,otherwise. (3.99) Since 𝐶 is odd, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′()+𝑙−𝑘+2+2𝑘−1^′2+𝑙=2𝑛−𝑙−3𝑙^′2+2. (3.100) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′()+𝑙−𝑘+2+2𝑘−1^′2+𝑙+𝑙=2𝑛−3𝑙^′2+2. (3.101)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  1.6. 𝑥^′∈𝐶^′ and 𝑦^′∈𝐶^′. See Figure 6(f).
Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶′(𝑥^′→𝑣))+𝑙(𝑄_𝐶′(𝑣→𝑦^′))+2𝑙(𝑃)+𝑙(𝑃₂). If 𝑤 is odd, set 𝑊^′=𝑃₁+𝑄_𝐶′𝑥^′→𝑣+2𝑃+𝑄_𝐶′𝑣→𝑦^′+𝑃₂. (3.102) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝐶′𝑥^′→𝑣+2𝑃+𝑄_𝐶′𝑣→𝑦^′+𝑃₂+𝐶. (3.103) Let 𝑊₁=𝑊^′+𝐶¹𝑊,𝑤isodd,_′′+𝐶¹𝑊,otherwise,²=𝑊^′+𝐶²𝑊,𝑤isodd,_′′+𝐶²,otherwise. (3.104) Since 𝐶 is odd, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′−𝑘+2+2(𝑘−1)+𝑙^′+𝑙=2𝑛−𝑙−𝑙^′+2. (3.105) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′−𝑘+2+2(𝑘−1)+𝑙^′+𝑙+𝑙=2𝑛−𝑙^′+2. (3.106)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Therefore, 𝑙(𝐴)≤2𝑛−2 by Lemma 2.6, a contradiction.
Case  2. 𝐶 and 𝐶^′ have at least one common vertex.
Let 𝐺₅=𝐶∪𝐶^′. Let 𝑥 and 𝑦 be two arbitrary (not necessarily distinct) vertices in 𝑆(𝐴). Suppose that 𝑃₁ is the shortest path from 𝑥 to 𝐺₅ and intersects 𝐺₅ at vertex 𝑥^′ and 𝑃₂ is the shortest path from 𝑦 to 𝐺₅ and intersects 𝐺₅ at vertex 𝑦^′. Denote 𝐶∩𝐶^′ by 𝑅. Assume 𝑅 has 𝑚 vertices, where 1≤𝑚≤min(𝑙₁,𝑙₂), then 0≤𝑙𝑃₁,𝑙𝑃₂≤𝑛−𝑙₁−𝑙₂+𝑚. (3.107) If 𝑚>1, then 𝑅 is a path with vertex set 𝑉(𝑅)={𝑣₁,𝑣₂,…,𝑣_𝑚} and edge set 𝐸(𝑅)={(𝑣₁,𝑣₂),(𝑣₂,𝑣₃),…,(𝑣_𝑚−1,𝑣_𝑚)}. If 𝑙^′ is odd, 𝑚=min(𝑙,𝑙^′)=𝑙 by the minimality of 𝐶. If 𝑙^′ is even, suppose min(𝑙,𝑙^′)=𝑙^′, then there exists an odd cycle 𝐶^′ with 𝑙(𝐶^′)<𝑙 which contradicts our assumption that 𝐶 is an odd cycle with the least length in 𝑆(𝐴). Therefore, if 𝑚=min(𝑙,𝑙^′), then 𝑚=𝑙<𝑙^′. We consider the following three subcases.
Subcase  2.1. 𝑥^′∈𝐶 and 𝑦^′∈𝐶.
Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶(𝑥^′→𝑦^′))+𝑙(𝑃₂). If 𝑤 is odd, set 𝑊^′=𝑃₁+𝑄_𝐶𝑥^′→𝑦^′+𝑃₂. (3.108) Otherwise, set 𝑊^′′=𝑃₁+𝐶⧵𝑄_𝐶𝑥^′→𝑦^′+𝑃₂. (3.109) Let 𝑊₁=𝑊^′+𝐶¹𝑊,𝑤isodd,_′′+𝐶¹𝑊,otherwise,²=𝑊^′+𝐶²𝑊,𝑤isodd,_′′+𝐶²,otherwise. (3.110) Since 𝐶 is odd, 𝑙(𝑄_𝐶(𝑥^′→𝑦^′)) and 𝑙(𝐶⧵𝑄_𝐶(𝑥^′→𝑦^′)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑚+𝑙−12+𝑙=2𝑛−2𝑙^′𝑙+2𝑚−2−12. (3.111) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑚+𝑙+𝑙=2𝑛−2𝑙^′+2𝑚. (3.112)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  2.2. 𝑥^′∈𝐶⧵𝑅 and 𝑦^′∈𝐶^′⧵𝑅.
Without loss of generality, we assume that 𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁))≤𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣_𝑚)). Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁))+𝑙(𝑄_{𝐶′⧵𝑅}(𝑣₁→𝑦^′))+𝑙(𝑃₂). If 𝑤 odd, set 𝑊^′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣₁+𝑄_{𝐶′⧵𝑅}𝑣₁→𝑦^′+𝑃₂. (3.113) Otherwise, set 𝑊^′′=𝑃₁+𝑄_𝐶⧵𝑅𝑥^′→𝑣_𝑚+𝑄_𝑅𝑣_𝑚→𝑣₁+𝑄_{𝐶′⧵𝑅}𝑣₁→𝑦^′+𝑃₂. (3.114) Let 𝑊₁=𝑊^′+𝐶¹𝑊,𝑤isodd,_′′+𝐶¹𝑊,otherwise,²=𝑊^′+𝐶²𝑊,𝑤isodd,_′′+𝐶²,otherwise. (3.115) Since 𝐶 is odd, 𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣₁)) and 𝑙(𝑄_𝐶⧵𝑅(𝑥^′→𝑣_𝑚))+𝑙(𝑄_𝑅(𝑣_𝑚→𝑣₁)) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even.
Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑚+(𝑙−𝑚+1)+𝑙^′−𝑚+1+𝑙=2𝑛−𝑙^′+2. (3.116) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2𝑛−𝑙−𝑙^′+𝑚+(𝑙−𝑚)+𝑙^′−1+𝑙=2𝑛−𝑙^′+𝑚−1. (3.117)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Subcase  2.3. 𝑥^′∈𝐶^′⧵𝑅 and 𝑦^′∈𝐶^′⧵𝑅.
The verification for this subcase is similar to that of the Subcase 2.2 and is omitted.
Therefore, 𝑙(𝐴)≤2𝑛−2 by Lemma 2.6, a contradiction.

(a)

(b)

(c)

(d)

(e)

(f)

Figure 6 

Illustration for Lemma 3.4.

Lemma 3.5. Let 𝐴 be an 𝑛×𝑛 primitive nonpowerful ZS sign pattern matrix with zero diagonal. If 𝑙(𝐴)=2𝑛−1, then 𝑆(𝐴) is isomorphic to 𝐺 (see Figure 1).

Proof. Let 𝐴 be an 𝑛×𝑛 primitive nonpowerful ZS sign pattern matrix with zero diagonal. Since 𝐴 is primitive, it follows from Lemma 2.4 that 𝑆(𝐴) is strongly connected and there is an odd cycle 𝐶=(𝑣₁,𝑣₂,…,𝑣_𝑙) in 𝑆(𝐴) with length 𝑙(𝐶)=𝑙. Since 𝐴 has zero diagonal, there is no loop in 𝑆(𝐴) and so 3≤𝑙≤𝑛. Without loss of generality, we assume that 𝐶 is an odd cycle with the least length in 𝑆(𝐴). If 𝑙(𝐴)=2𝑛−1, then by Lemma 3.3, there exists no positive 2-cycle in 𝑆(𝐴). Then 𝑆(𝐴) is isomorphic to 𝐺 by Lemma 2.7. We will give anther proof of the theorem.
Denote the vertex set of 𝑆(𝐴) by 𝑉 and the vertex set of 𝐶 by 𝑉^′. By Lemmas 3.4 and 3.3, the cycle 𝐶 is the only cycle of length at least 3 in 𝑆(𝐴) and there exists no positive 2-cycle in 𝑆(𝐴). Consider the two directed cycles 𝐶¹=(𝑣₁,𝑣₂,…,𝑣_𝑙,𝑣₁) and 𝐶²=(𝑣₁,𝑣_𝑙,…,𝑣₂,𝑣₁). Since there exists no positive 2-cycle in 𝑆(𝐴), the arcs (𝑣_𝑖,𝑣_𝑖+1) and (𝑣_𝑖+1,𝑣_𝑖) have different signs. Thus sgn(𝐶¹)=−sgn(𝐶²) by the fact that 𝑙 is odd. Let 𝑥 and 𝑦 be two arbitrary (not necessarily distinct) vertices in 𝑆(𝐴). Suppose that 𝑃₁ is the shortest path from 𝑥 to 𝐶 and intersects 𝐶 at vertex 𝑥^′ and 𝑃₂ is the shortest path from 𝑦 to 𝐶 and intersects 𝐶 at vertex 𝑦^′. If 𝑙=𝑛, 𝑆(𝐴) is isomorphic to an odd cycle of length 𝑛. If 𝑙<𝑛, it is enough to show that there exists a vertex 𝑢∉𝑉^′ in 𝑆(𝐴) such that 𝑙(𝑃)=𝑛−𝑙 where 𝑃 is the shortest path from 𝑢 to 𝐶 and 𝑃 intersects 𝐶 at 𝑣. Suppose not, then 𝑙𝑃₁,𝑙𝑃₂≤𝑛−𝑙−1. (3.118) Now we consider the following three cases.
Case  1. 𝑥∈𝑉⧵𝑉^′ and 𝑦∈𝑉⧵𝑉^′.
Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶(𝑥^′→𝑦^′))+𝑙(𝑃₂)). If 𝑤 is odd, set 𝑊^′′=𝑃₁+𝑄_𝐶𝑥^′→𝑦^′+𝑃₂. (3.119) Otherwise, set 𝑊^′=𝑃₁+𝐶⧵𝑄_𝐶𝑥^′→𝑦^′+𝑃₂. (3.120) Let 𝑊₁=𝑊^′+𝐶¹𝑊,𝑤isodd,_′′+𝐶¹𝑊,otherwise,²=𝑊^′+𝐶²𝑊,𝑤isodd,_′′+𝐶²,otherwise. (3.121) Since 𝐶 is odd, 𝑄_𝐶(𝑥^′→𝑦^′) and 𝐶⧵𝑄_𝐶(𝑥^′→𝑦^′) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤2(𝑛−𝑙−1)+𝑙−12𝑙+𝑙=2𝑛−2−52. (3.122) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤2(𝑛−𝑙−1)+𝑙+𝑙=2𝑛−2. (3.123)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Case  2. 𝑥∈𝑉⧵𝑉^′ and 𝑦∈𝑉.
In this case, 𝑦^′=𝑦. Let 𝑤=𝑙(𝑃₁)+𝑙(𝑄_𝐶(𝑥^′→𝑦)). If 𝑤 is odd, set 𝑊^′′=𝑃₁+𝑄_𝐶𝑥^′→𝑦. (3.124) Otherwise, set 𝑊^′=𝑃₁+𝐶⧵𝑄_𝐶𝑥^′→𝑦. (3.125) Let 𝑊₁=𝑊^′+𝐶¹𝑊,𝑤isodd,_′′+𝐶¹𝑊,otherwise,²=𝑊^′+𝐶²𝑊,𝑤isodd,_′′+𝐶²,otherwise. (3.126) Since 𝐶 is odd, 𝑄_𝐶(𝑥^′→𝑦) and 𝐶⧵𝑄_𝐶(𝑥^′→𝑦) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤(𝑛−𝑙−1)+𝑙−12𝑙+𝑙=𝑛+2−32. (3.127) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤(𝑛−𝑙−1)+𝑙+𝑙=𝑛+𝑙−1. (3.128)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
Case  3. 𝑥∈𝑉 and 𝑦∈𝑉.
In this case, 𝑥^′=𝑥 and 𝑦^′=𝑦. Let 𝑤=𝑄_𝐶(𝑥→𝑦). If 𝑤 is odd, set 𝑊^′′=𝑄_𝐶𝑥→𝑦. (3.129) Otherwise, set 𝑊^′=𝐶⧵𝑄_𝐶𝑥→𝑦. (3.130) Let 𝑊₁=𝑊^′+𝐶¹𝑊,𝑤isodd,_′′+𝐶¹𝑊,otherwise,²=𝑊^′+𝐶²𝑊,𝑤isodd,_′′+𝐶²,otherwise. (3.131) Since 𝐶 is odd, 𝑄_𝐶(𝑥→𝑦) and 𝐶⧵𝑄_𝐶(𝑥→𝑦) have different parity. Therefore, both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even. Then, if 𝑤 is odd, 𝑙𝑊₁=𝑙𝑊₂≤𝑙−12+𝑙=3𝑙2−12. (3.132) Otherwise, 𝑙𝑊₁=𝑙𝑊₂≤𝑙+𝑙=2𝑙. (3.133)
Since both 𝑙(𝑊₁) and 𝑙(𝑊₂) are even, it follows that 𝑙(𝑊₁)=𝑙(𝑊₂)≤2𝑛−2. We see that the pair 𝑊₁, 𝑊₂ is a pair of SSSD walks with even length. Therefore, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2.
From all the above cases, there exists a pair of SSSD walks from 𝑥 to 𝑦 with length 2𝑛−2, therefore, 𝑙(𝐴)≤2𝑛−2 by Lemma 2.6, a contradiction.

Proof of Theorem 1.6. We get 𝑙(𝐴)≤2𝑛−1 from Theorem 2.2. Suppose that 𝐴 is nonpowerful and skew symmetric and 𝑆(𝐴) is isomorphic to 𝐺. Then 𝑙(𝐴)=2𝑛−1 by Lemma 2.7.
Conversely, suppose that 𝑙(𝐴)=2𝑛−1. We need to prove that 𝐴 is nonpowerful and skew symmetric and 𝑆(𝐴) is isomorphic to 𝐺. In [1], Li et al. showed that if an irreducible sign pattern matrix 𝐴 is powerful, then 𝑙(𝐴)=𝑙(|𝐴|). Therefore, if 𝐴 is powerful, then 𝑙(𝐴)≤2𝑛−4 by Theorem 2.1, which contradicts 𝑙(𝐴)=2𝑛−1. Hence 𝐴 is nonpowerful. Consequently, 𝐴 is skew symmetric by Lemma 3.3 and 𝑆(𝐴) is isomorphic to 𝐺 by Lemma 3.5.

Acknowledgments

This paper is supported by NSFC (61170311), Chinese Universities Specialized Research Fund for the Doctoral Program (20110185110020), and Sichuan Province Sci. & Tech. Research Project (12ZC1802).