Nowhere-Zero Flows on Signed Wheels and Signed Fans

Abstract

A signed graph $(G, \sigma )$ is flow-admissible if there exists an orientation $\tau$ and a positive integer k such that $(G, \sigma )$ admits a nowhere-zero k-flow. Bouchet (J Combin Theory Ser B 34:279–292, 1983) conjectured that every flow-admissible signed graph has a nowhere-zero 6-flow. In this paper, we show that each flow-admissible signed wheel admits a nowhere-zero 4-flow if and only if G is not the specified graph. Moreover, there are infinitely many signed wheels which do not admit a nowhere-zero 3-flow. We also prove each flow-admissible signed fan admits a nowhere-zero 4-flow. Similarly, there are infinitely many signed fans which do not admit a nowhere-zero 3-flow.

1 Introduction

Graphs in this paper are finite and simple. We follow the notation and terminology in [1] except otherwise stated. An n-cycle is a cycle on n vertices. For $k\ge 2$, a wheel $W_k$ is the graph obtained from a k-cycle by adding a new vertex, called the center of the wheel, which is adjacent to every vertex of the k-cycle. An n -path is a path of length n. For $k\ge 2$, an Fan $F_k$ is the graph obtained from a k-path by adding a new vertex which is adjacent to every vertex of the k-path.

In a graph G, an edge $e=xy$ is viewed as two half-edges: one half-edge incident with x, and the other incident with y. Denote by E(G) the set of all edges of G, and by H(G) the set of all half-edges of G. For $h\in H(G)$, let $e_h$ be the edge containing h, let $v_h$ be the vertex incident with h and let $h'$ be the other half-edge of $e_h$. For a vertex $v, H_G(v)$ (or shortly H(v)) is the set of half-edges incident with v, and $E_G(v)$ [or shortly E(v)] is the set of edges incident with v.

Suppose that G is a graph and $\sigma {:}\;E(G)\rightarrow \{1,-1\}$ is a mapping. Then the pair $(G,\sigma )$ is called a signed graph. An edge e is called a positive edge (or a negative edge) if $\sigma (e)=1$ [or $\sigma (e)=-1$]. Given a signed graph $(G,\sigma )$, an orientation of $(G,\sigma )$ is a mapping $\tau {:}\;H(G)\rightarrow \{1,-1\}$ such that for each edge e, if $h, h'$ are the two half-edges of e, then $\tau (h)\tau (h^\prime )=-\sigma (e)$. We view $\tau$ as an assignment of directions to the half-edges of G. If $\tau (h)=1$, then the half-edge h is oriented away from $v_h$; if $\tau (h)=-1$, then the half-edge h is oriented towards $v_h$. The pair $(G,\tau )$ is called a bidirected graph. A negative edge is called an in-edge if both half-edges are oriented from the end-vertices, and out-edge otherwise. The signed graph $(G,\sigma )$ is called the underlying graph of $(G,\tau )$, and the mapping $\sigma$ is called the signature of $\tau$. For a signed graph, denote by n(G) and n(v) the number of negative edges in G and the number of negative edges incident with v, respectively.

Let $(G,\tau )$ be a bidirected graph. For a mapping $f{:}\;E\rightarrow R$, the boundary of f is the map $\partial f{:}\;V(G)\rightarrow R$ defined as $\partial f(v)=\sum _{h\in H(v)}\tau (h)f(e_h)$ for each vertex v. If $\partial f=0$ (called Kirchhoff’s law), then f is a flow in $(G,\tau )$. If f is an integer flow in $(G,\tau )$ and $1\le |f(e)|\le k-1$ for each edge e, then f is a nowhere-zero k-flow in $(G,\tau )$.

For a bidirected graph $(G,\tau )$, as mentioned in [10], the existence or non-existence of a nowhere-zero k-flow is determined by the signature of $(G,\tau )$. Thus, if an orientation $(G, \tau )$ of $(G, \sigma )$ admits a nowhere-zero k-flow, then every orientation of $(G, \sigma )$ admits a nowhere-zero k-flow. A signed graph $(G, \sigma )$ is flow-admissible if there exist an orientation $\tau$ and some positive integer k such that $(G, \sigma )$ admits a nowhere-zero k-flow. The flow number of a flow-admissible signed graph G is the smallest integer k such that G admits a nowhere-zero k-flow.

Clearly, if a graph G admits a nowhere-zero k-flow, then it also admits a nowhere-zero h-flow for each $h\ge k$. Let A be an abelian group with identity 0. A map $f: E(G)\rightarrow A{\setminus }\{0\}$ is a nowhere-zero A-flow if $\partial f: V(G)\rightarrow A$ with $\partial f(v)\equiv 0$ for each vertex v.

The problem of nowhere-zero k-flows in signed graphs was first studied by Bouchet [2], who proposed the following conjecture.

Conjecture 1.1

Every flow-admissible signed graph admits a nowhere-zero 6-flow.

Bouchet [2] proved that every flow-admissible signed graph admits a nowhere-zero 216-flow, which was improved to a nowhere-zero 30-flow by Zýka [16]. Khelladi [5] proved that every 4-connected flow-admissible signed graph admits a nowhere-zero 18-flow, which was improved to 4-flow by Raspaud and Zhu [10]. Xu and Zhang [15] proved that Bouchet’s conjecture holds for 6-edge-connected flow-admissible signed graphs. Wu et al. [14] proved that every 8-edge-connected flow-admissible signed graph admits a nowhere-zero 3-flow. Wei et al. [13] proved that every 3-edge-connected flow-admissible signed graph admits a nowhere-zero 15-flow.

However, a few classes of signed graphs whose flow numbers were determined are known. Such numbers of signed complete graphs and signed complete bipartite graphs were determined by Màčajová and Rollová [8]; the numbers of signed cubic graphs by Máčajová and Škoviera [9], and the number of signed regular graphs by Schubert and Steffen [11]. On the other hand, wheels played special role in investigating the existence of nowhere-zero 3-flows and $Z_3$-connectivity of a graph [3, 6, 7]. Motivated by the above observations, we present the following results in this paper.

Theorem 1.2

Let G be a flow-admissible signed wheel. Then G admits a nowhere-zero 4-flow if and only if G is not the bidirected graph $(W_5, \sigma ')$ in Fig. 1, which admits a nowhere-zero 5-flow. Moreover, for each $n\ge 3$, there exists a mapping $\sigma : E(W_n)\rightarrow \{1,-1\}$ such that the signed wheel $G=(W_n,\sigma )$ has no nowhere-zero 3-flow.

Fig. 1

The bidirected graph $(W_5, \sigma ^\prime )$

Theorem 1.3

Each flow-admissible signed fan admits a nowhere-zero 4-flow. Moreover, for each $n\ge 5$, there exists a mapping $\sigma {:}\;E(F_n)\rightarrow \{1,-1\}$ such that the signed fan $G=(F_n,\sigma )$ has no nowhere-zero 3-flow.

Remark (1) There exist infinitely many signed wheels which admit a nowhere-zero 3-flow in the sense that for even signed wheels. Let G be the signed wheel $W_{2k}$ with vertex set $V=\{u,v_1, v_2, \ldots , v_{2k}\}$, where u is the center and $v_1, \ldots , v_{2k}$ are on the outer cycle, and two negative edges $v_1v_2, v_3v_4$. We assume, without loss of generality, that $v_1v_2$ is an out-edge. We orient the edges of G in such a way that all the three half-edges incident with v are oriented into it if $v \in \{v_1, v_2, v_5, v_7,\ldots , v_{2k-1}\}$, and oriented from it if $v\in \{v_3, v_4, v_6, v_8,\ldots , v_{2k}\}$. For the center u, the number of all the edges coming into u is equal to the number of all the edges coming from u. This shows that G has a modulo 3-orientation. By Theorem 1.5 in [15], G admits a nowhere-zero 3-flow.

For an odd signed wheel $W_{2k+1}$ with two negative edges $uv_1, v_2v_3$, one can similarly show that it has a modulo 3-orientation and hence it admits a nowhere-zero 3-flow.

(2) Tutte in [12] proved that a graph admits a nowhere-zero k-flow if and only if it admits a nowhere-zero modular k-flow for any positive integer $k\ge 3$. One naturally asks whether the signed graph version of this Tutte’ theorem holds. Xu and Zhang in [15] confirmed this problem for $k=3$. It is easy to see that the signed graph $(W_5, \sigma ')$ in Fig. 1 admits a nowhere-zero modular 4-flow but no nowhere-zero 4-flow. This tells us that there is no signed graph version of the Tutte’s result.

In each figure in this paper, we illustrate negative edges by using thick lines while positive edges by usual lines. The number of arrows on a half-edge represents the flow value on an edge as well as the orientation. Since both half-edges of a positive edge are oriented consistently, we use the orientation as well as the flow value for each positive edge only once.

We organize this paper as follows. In Sect. 2, we state some known results that will be used in the proofs of our theorems. We will prove Theorems 1.2 and 1.3 in Sects. 3 and  4, respectively.

2 Preliminaries

A cycle in this paper is a connected 2-regular graph. A cycle C in a signed graph is balanced if the number of negative edges in C is even and unbalanced otherwise. An unbalanced graph is a signed graph containing an unbalanced cycle. A signed graph which is not unbalanced is a balanced graph.

Let $(G, \sigma )$ be a signed graph. For $v\in V(G)$, a switching at v defines a graph $(G, \sigma ')$ with $\sigma '(e)=-\sigma (e)$ if $e\in E(v)$ and $\sigma '(e)=\sigma (e)$ otherwise. Two signed graphs G and H are switching equivalent if they can be obtained from each other by a series of switchings. Assume that $(G, \sigma ')$ is obtained from $(G, \sigma )$ by a switch at a vertex v and $(G, \tau )$ is an orientation of $(G, \sigma )$. Let $\tau ^\prime : H(G)\rightarrow \{1, -1\}$ such that $\tau ^\prime (h)=-\tau (h)$ if $h\in H(v)$ and $\tau ^\prime (h)=\tau (h)$ otherwise. Then $\tau ^\prime$ is an orientation of $(G, \sigma ')$. Moreover, if f is a flow in $(G, \tau )$, then f is also a flow in $(G, \tau ^\prime )$.

Some results on the existence of nowhere-zero flows on signed graphs and graphs are summarized as follows [10, 15].

Lemma 2.1

Let G be a bidirected graph, $E_0$ be a subset of E, and $G_{E_0}$ be the bidirected graph obtained from G by reversing the orientation of each edge in $E_0$. Then G admits a nowhere-zero k-flow if and only if $G_{E_0}$ admits a nowhere-zero k-flow.

Lemma 2.2

• (1) A signed graph is balanced if and only if it is equivalent to a graph with all positive edges.

• (2) Let G be a 2-edge-connected signed graph. Then G is not flow-admissible if and only if it is switching equivalent to a signed graph with exactly one negative edge.

The following lemma is due to Lai [6].

Lemma 2.3

If n is even, then $W_n$ admits a nowhere-zero 3-flow; if n is odd, then $W_n$ admits a nowhere-zero 4-flow.

For each orientation D of $(G,\sigma )$, let $d^+_D(v)$ [resp. $d^-_D(v)$] denote the number of half-arcs leaving (resp. entering) a vertex v. We drop the subscript D if there is no confusion. An orientation D of $(G,\sigma )$ is called a modulo 3-orientation if $d^+_D(v)\equiv d^-_D(v)$ (mod 3) for all $v\in V(G)$. As shown by Tutte [12], a graph G admits a modulo 3-orientation if and only if it has a nowhere-zero 3-flow; Xu and Zhang [15] extended this equivalence to signed graphs. Let $Z_k$ be the cyclic group of order k.

Theorem 2.4

Let G be a 2-edge-connected signed graph. Then G admits a nowhere-zero 3-flow if and only if G admits a modulo 3-orientation (or, a nowhere-zero $Z_3$-flow).

3 Signed Wheel

In this section, for a wheel $W_n$, we always assume that $V(W_n)=\{u, v_1, \ldots , v_n\}$, where u is the center, $C=v_1\ldots v_nv_1$ is a cycle, and the subscripts of vertices take modulo n. In order to prove Theorem 1.2, we state some lemmas first.

Let $(C^\prime , \sigma )$ be an unbalanced cycle, that is, $(C^\prime , \sigma )$ contains an odd number of negative edges. It is easy to see that there always exist $X\subseteq V(C)$ such that after switching at X, the resulting graph contains exactly one negative edge. By Lemma 2.2(1), we have the following statement.

Lemma 3.1

Each signed wheel $G=(W_n,\sigma )$ is switching equivalent to one with at most one negative edge on $C=v_1\ldots v_nv_1$.

In the rest of this section, by Lemma 3.1, we may assume that $G=(W_n,\sigma )$ has at most one negative edge on the cycle C. By vertex switchings, we further assume that $n(v)\le \frac{d(v)}{2}$ for any $v\in V(G)$. It follows that $n(G)\le \frac{n}{2}+1$.

Lemma 3.2

Let $G=(W_n,\sigma )$ be a flow-admissible signed wheel on $n\le 4$ vertices. Then G admits a nowhere-zero 4-flow.

Proof

Recall that $n(v)\le \frac{d(v)}{2}$ and $n(G)\le \frac{n}{2}+1$. If $n=3$, then $n(G)=0$ or 2 by Lemma 2.2(2). If $n(G)=0$, then G is balanced and hence admits a nowhere-zero 4-flow by Lemmas 2.2(1) and 2.3. Thus, $n(G)=2$. By symmetry, G is the bidirected graph $(W_3,\sigma _1)$ in Fig. 2, which admits a nowhere-zero 4-flow as depicted in Fig. 2.

Fig. 2

Flow-admissible and unbalanced signed wheels on $n\le 4$

Let $n=4$. If G is balanced, then it admits a nowhere-zero 4-flow by Lemmas 2.2(1) and  2.3. Thus, assume that G is unbalanced. Then $2\le n(G)\le 3$. If $n(G)=2$, then G is switching equivalent to one of $(W_4, \sigma _1), (W_4, \sigma _2),$ and $(W_4, \sigma _3)$ in Fig. 2, each of which admits a nowhere-zero 4-flow. If $n(G)=3$, then two of these three edges forms a 2-path passing the center and the third is an independent edge. In this case, by vertex switchings, the graph is equivalent to the signed wheel $(W_4, \sigma _2)$, which admits a nowhere-zero 4-flow in Fig. 2. $\square$

Lemma 3.3

The bidirected graph $(W_5,\sigma ')$ in Fig. 1 does not admit a nowhere-zero 4-flow and indeed admits a nowhere-zero 5-flow.

Proof

We first show that $(W_5,\sigma ')$ admits no nowhere-zero 4-flow. Suppose otherwise that $(W_5, \sigma ')$ admits a nowhere-zero 4-flow f. We use notation and the orientation of the edges of $(W_5, \sigma ')$ as shown in Fig. 1. Define $f(v_1v_5)=x$ and $f(uv_i)=x_i$ where $i=1, 2, 3, 5$. It follows that $f(v_jv_{j+1})=x+\sum ^j_{t=1} x_t$ for $j=1,2,3$ and $f(v_4v_5)=x+x_5$. For vertex u, by the Kirchhoff’s law, $f(uv_1)+f(uv_3)+f(uv_4)+f(uv_5)=f(uv_2)$. For vertex $v_4$, by the Kirchhoff’s law, $f(uv_4)=f(v_5v_4)+f(v_3v_4)$. It follows that $x+x_1+x_3+x_5=0$ and $f(uv_4)=x+x_2$. Note that $x, x_i\in \{-3, -2, -1, 1, 2, 3\}$ and $0<|f(v_1v_2)|=| x+x_1|\le 3$ and $0<|f(v_5v_1|=|x+x_5|\le 3$. By symmetry, we may assume that $x\in \{1, 2, 3\}$. It is not hard to check that all the possible tuples for $(x, x_1, x_3, x_5)$ are $(1,-3,1,1), (1,-2,-1,2), (1,-2,3,-2), (1,1,1,-3), (1,1,-3,1), (1,2,-1,-2), (2,-3,2,-1)$, $(2,-1,2,-3), (2,-1,-2,1), (2,1,-2,-1), (3,-2,1,-2), (3,-1,-1,-1)$. In what follows, we will deduce a contradiction in each case. If $(x, x_1, x_3, x_5)=(1,-3,1,1)$, then $f(v_2v_3)=x_2-2, f(v_3v_4)=x_2-1,$ and $f(uv_4)=x_2+1$. It is easy to see that there exists no $x_2\in \{-3, -2, -1, 1, 2, 3\}$ such that $x_2-2, x_2-1, x_2+1\in \{-3, -2, -1, 1, 2, 3\}$, contrary to that f is a nowhere-zero 4-flow of $(W_5,\sigma ')$. For other cases, by a similar argument one can obtain a contradiction.

Thus, $(W_5,\sigma ')$ has no nowhere-zero 4-flow. Meanwhile, $(W_5,\sigma ')$ admits a nowhere-zero 5-flow, which is presented in Fig. 3. $\square$

Fig. 3

Flow-admissible and unbalanced signed wheels on $n=5$

Lemma 3.4

Let $G=(W_5,\sigma )$ be a flow-admissible signed wheel. Then G admits a nowhere-zero 4-flow if and only if G is not $(W_5,\sigma ')$ shown in Fig. 3.

Proof

If G is balanced, then G is equivalent to a graph with all positive edges by Lemma 2.2(1). By Lemma 2.3, G admits a nowhere-zero 4-flow. Thus, we may assume that G is unbalanced. Note that $2\le n(G)\le 3$. We first show that there are five non-equivalent unbalanced signed wheels on five vertices, which are depicted in Fig. 3.

We now consider the mutual positions of negative edges. Recall that at most one negative edge is on the cycle C. We first assume that $n(G)=2$. It follows that there are either both negative edges incident with the center or one incident with the center and the other on the cycle. In the former, G is $(W_5, \sigma _1)$ or $(W_5, \sigma _2)$ shown in Fig. 3. In the latter, G is $(W_5,\sigma _3)$ or $(W_5,\sigma _4)$ shown in Fig. 3.

Now we assume that $n(G)=3$. Since at most one negative edge is on the cycle C and the number of negative edges incident with the center is at most two, two negative edges are incident with the center and the other is on the cycle. If the two negative edges incident with the center are two edges of a 3-cycle, then G is switching equivalent to $(W_5, \sigma _4)$. Thus, the two negative edges incident with the center are not two edges of a 3-cycle, the G is $(W_5,\sigma _4)$ in Fig. 3 by vertex switching.

Since $(W_5,\sigma _i)$ for $1\le i\le 4$ in Fig. 3 admits a nowhere-zero 4-flow and by Lemma 3.3, we conclude that $(W_5,\sigma )$ admits a nowhere-zero 4-flow if and only if $(W_5,\sigma ^\prime )$. $\square$

Lemma 3.5

Assume that $n\ge 4$ and there exist two adjacent vertices $v_j$ and $v_{j+1}$ such that each edge in $E(v_j)\cup E(v_{j+1})$ is positive, where the subscripts take modulo n. If $G'=(G{\setminus } \{v_{j}\})\cup \{v_{j-1}v_{j+1}\}$ admits a nowhere-zero 4-flow, then so does G.

Proof

Let $f^\prime$ be a nowhere-zero 4-flow of $G'$ and $f^\prime (v_{j-1}v_{j+1})=a, f^\prime (uv_{j+1})=b$. We assume, without loss of generality, that in $G', v_{j-1}v_{j+1}$ is oriented from $v_{j-1}$ to $v_{j+1}, v_{j+1}v_{j+2}$ is from $v_{j+1}$ to $v_{j+2}, and uv_{j+1}$ is from u to $v_{j+1}$. By Lemma 2.1, we may assume that $uv_j$ is from u to $v_j, v_{j-1}v_{j}$ is from $v_{j-1}$ to $v_{j}$, and $v_jv_{j+1}$ is from $v_j$ to $v_{j+1}$ in G. We can now define a nowhere-zero 4-flow f on G from $f'$.

Define $f(e)=f^\prime (e)$ if $e\not \in E(v_j)\cup \{uv_{j+1}\}$ and let $f(v_{j-1}v_j)=a, f(uv_j)=x$, $f(v_jv_{j+1})=a+x, f(uv_{j+1})=b-x$. Since $f^\prime (v_{j+1}v_{j+2})=a+b$, at most one of a and b equals 3 (or $-3$). We claim that there exists $x\in \{\pm 1, \pm 2, \pm 3\}$ such that $a+x, b-x \in \{\pm 1, \pm 2, \pm 3\}$. Suppose otherwise that for any $x\in \{\pm 1, \pm 2, \pm 3\}$ such that $a+x\notin \{\pm 1, \pm 2, \pm 3\}$ or $b-x\notin \{\pm 1, \pm 2, \pm 3\}$. If $a+x=0$ and $b-x=0$, then $a+b=0$, a contradiction. By symmetry, we may assume that $a+x=0$ or $|a+x|\ge 4$. In the former case, $a+x=0, b-x\not =0, a+b\in \{\pm 1, \pm 2, \pm 3\}$. If $b-x\in \{\pm 1, \pm 2, +3\}$, then let $f(uv_j)=x+1$, $f(v_jv_{j+1})=a+x+1, f(uv_{j+1})=b-x-1$. If $b-x=-3$, then let $f(uv_j)=x-1$, $f(v_jv_{j+1})=a+x-1, f(uv_{j+1})=b-x+1$. In each case, we obtain a contradiction. In the latter case, we assume, without loss of generality, that $a+x\ge 4$. Since $x\in \{\pm 1, \pm 2, \pm 3\}, a\ge 1$, and $x\ge 1$. If $a=3$ (or $b=-3$), then by the Kirchhoff’s law $b\not =3$ (or $a\not =3$). Let $f(uv_j)=x=-1$, $f(v_jv_{j+1})=a-1, f(uv_{j+1})=b+1$. Thus, assume that $a\not =3$ and $b\not =-3$. In this case, $f(uv_j)=x=1$, $f(v_jv_{j+1})=a+1, f(uv_{j+1})=b-1$. In each case, we obtain a contradiction. $\square$

Lemma 3.6

Assume that $n\ge 7$ and $P=v_{i-1}v_iv_{i+1}v_{i+2}v_{i+3}v_{i+4}$ is a segment on C. Further assume that each edge on P is positive, edges $uv_i$ and $uv_{i+2}$ are negative, and edges $uv_{i+1}$ and $uv_{i+3}$ are positive. If $G'=(G{\setminus }\{v_i, v_{i+1}, v_{i+2},$ $v_{i+3}\})\cup \{v_{i-1}v_{i+4}\}$ admits a nowhere-zero 4-flow, then so does G.

Proof

Let $f^\prime$ be a nowhere-zero 4-flow of $G'$ and $f^\prime (v_{i-1}v_{i+4})=a$. We assume, without loss of generality, that in $G', v_{i-1}v_{i+4}$ is positive and oriented from $v_{i-1}$ to $v_{i+4}$. By Lemma 2.1, we may assume that $uv_i$ is an out-edge, $uv_{i+2}$ is an in-edge, $uv_{i+1}$ is from $v_{i+1}$ to $u, uv_{i+3}$ is from u to $v_{i+3}$, and $v_jv_{j+1}$ is from $v_j$ to $v_{j+1}$ for $i-1\le j\le i+3$ in G. We can now define a nowhere-zero 4-flow f on G from $f'$.

Define $f(e)=f^\prime (e)$ if $e\not \in \cup ^{i+3}_{j=i} E(v_j)$ and $f(v_{i-1}v_{i})=f(v_{i+3}v_{i+4})=a$. Let $f(uv_i)=f(uv_{i+2})=x$, $f(uv_{i+1})=f(uv_{i+3})=y, f(v_iv_{i+1})=a+x, f(v_{i+1}v_{i+2})=a+x-y$, and $f(v_{i+2}v_{i+3})=a-y$. If $\mid a\mid =1$, let $x=y=2$. If $\mid a\mid =2$, let $x=y=1$. If $a=-3$, let $x=2$ and $y=-2$. If $a=3$, let $x=-2$ and $y=2$. Then f is as desired. $\square$

Lemma 3.7

None of $(W_4, \sigma _2)$ in Fig. 2 and $(W_5,\sigma _1)$ in Fig. 3 admits nowhere-zero 3-flow.

Proof

We only prove that $(W_4, \sigma _2)$ does not admit a nowhere-zero 3-flow. The proof for the case that $(W_5,\sigma _1)$ admits no nowhere-zero 3-flow is similar. Suppose to the contrary that $(W_4, \sigma _2)$ admits a nowhere-zero 3-flow. Note that both $v_2v_3$ and $uv_4$ are negative edges. We assume, without loss of generality, that $uv_4$ is an out-edge. By Lemma 2.4, $v_1v_4$ is oriented from $v_1$ to $v_4$ and $v_3v_4$ is oriented from $v_3$ to $v_4$. By Lemma 2.4 again, all edges incident with $v_1$ are oriented from $v_1$, all edges incident with $v_2$ are oriented into $v_2$, and all edges incident with $v_3$ are oriented into $v_3$. This leads that $v_3v_4$ cannot be oriented, a contradiction. $\square$

Lemma 3.8

Let $G=(W_n,\sigma )$ be a flow-admissible signed wheel on $n\ge 3$ vertices. Then G does not admit a nowhere-zero 4-flow if and only if G is $(W_5,\sigma ^\prime )$ in Fig. 1.

Proof

If G is $(W_5, \sigma ')$ in Fig. 1, then G does not admit a nowhere-zero 4-flow by Lemma 3.3.

Conversely, we proceed our proof by induction on n. If $n\le 5$, our theorem follows by Lemmas 3.2 and 3.4. Thus, assume that $G=(W_n,\sigma )$ is a flow-admissible signed wheel on $n\ge 6$ vertices.

If G is balanced, then our conclusion follows Lemmas 2.2(1) and  2.3. Thus, we assume that G is unbalanced. Recall that C contains at most one negative edge and $n(v)\le \frac{d(v)}{2}$ for $v\in V(G)$. It follows that $2\le n(G)\le \frac{n}{2}+1$.

Claim 1 If there exist two adjacent vertices $v_i$ and $v_{i+1}$ such that each edge in $E(v_i)\cup E(v_{i+1})$ is positive, then G admits a nowhere-zero 4-flow.

Proof of Claim 1

Let $G'=(G{\setminus }\{v_i\})\cup \{v_{i-1}v_{i+1}\}$. In this case, $|V(G')|=|V(G)|-1$ and $n(G')=n(G)$. It follows our assumption and Lemma 2.2(2) that $G'$ is flow-admissible. Applying induction hypothesis to $G'$, $G'$ admits a nowhere-zero 4-flow and so does G by Lemma 3.5. This proves Claim 1.

By Claim 1, we may assume that every pair of vertices $v_i$ and $v_{i+1}$ has at least one vertex incident with a negative edge for $i\in \{1, 2, \ldots , n\}$.

Claim 2 If the cycle C has no negative edge, then G admits a nowhere-zero 4-flow.

Proof of Claim 2

In this case, $n(G)=n(u)\le \frac{n}{2}$. By Claim 1, for each edge $v_iv_{i+1}\in E(C)$, at least one of $\{ uv_i, uv_{i+1}\}$ is negative. On the other hand, if both $uv_i$ and $uv_{i+1}$ are negative, then switching at u, we obtain that $uv_i, uv_{i+1}$ are positive. It follows Claim 1 that G admits a nowhere-zero 4-flow. Thus, we may assume that for any pair of $uv_i$ and $uv_{i+1}$, where $1\le i\le n$, exactly one of them is negative. This implies that n must be even and $n(G)=n(u)=\frac{n}{2}$. We assume, without loss of generality, that each of $\{uv_1, uv_3,\ldots , uv_{n-1}\}$ is negative. If $n\ge 8$, let $G'=(G{\setminus } \{v_1, v_{2}, v_{3},v_{4}\})\cup \{v_{5}v_{n}\}$. By Lemma 3.2 and applying induction hypothesis to $G', G'$ admits a nowhere-zero 4-flow and so does G by Lemma 3.6. Let $n=6$. Assume that each negative edge is an out-edge, $uv_{i}$ is oriented from u to $v_{i}$, where $i\in \{2, 4, 6\}$, and $v_{j}v_{j+1}$ is from $v_{j}$ to $v_{j+1}$, where $j=1,2,\ldots ,6$ and the subscripts of v take modulo 6. Define $f(uv_{1})=f(uv_{2})=f(uv_{4})=f(uv_{5})=1$, $f(uv_{3})=f(uv_{6})=-2$ and the values of f on C are assigned in order 2, 3, 1, 2, 3, and 1 from $v_1v_2$. Thus, f is as desired. This proves Claim 2.

We are ready to complete the proof of Lemma 3.8. By Lemma 3.1 and Claim 2, we may assume that the cycle C contains exactly one negative edge.

We assume, without loss of generality, that $v_1v_2$ is the unique negative edge on C. Since $n(v)\le \frac{d(v)}{2}$ for $v\in V(G), uv_1$ and $uv_2$ are positive. If $uv_n$ is positive, then we obtain that each edge in $E(v_n)\cup E(v_1)$ is positive by switching at $v_2$. By Claim 1, G admits a nowhere-zero 4-flow. Thus, we may assume that $uv_n$ is negative. By symmetry, we also assume that $uv_3$ is negative. If $uv_4$ is negative, then we obtain that each edge in $E(v_3)\cup E(v_4)$ is positive by switching at u and hence G admits a nowhere-zero 4-flow by Claim 1. Thus, assume that $uv_4$ is positive. This implies that $uv_i$ is negative if $i\ge 3$ and i is odd while $uv_j$ is positive if $j\ge 4$ and j is even. It follows that n is odd and $n\ge 7$. Define $G'=(G{\setminus } \{v_3, v_{4}, v_{5}, v_{6}\})\cup \{v_{2}v_{7}\}$. If $n=7$, then $|V(G')|=3$ and $G'$ admits a nowhere-zero 4-flow by Lemma 3.2. Thus, assume that $n\ge 9$. Applying the induction hypothesis to $G', G'$ admits a nowhere-zero 4-flow and so does G by Lemma 3.6. $\square$

Proof of Theorem 1.2

By Lemma 3.8, it is sufficient to show that for $n\ge 3$, there exists a mapping $\sigma {:}\;E(W_n)\rightarrow \{1,-1\}$ such that the signed wheel $G=(W_n,\sigma )$ has no nowhere-zero 3-flow.

For $n=3$, let $\sigma (e)=1$ for every edge $e\in E(W_n)$. Then $(W_3,\sigma )$ is balanced and hence it has no nowhere-zero 3-flow by Lemma 2.3. For $4\le n\le 5$, the theorem follows from Lemma 3.7. Thus, we may assume that $n\ge 6$.

Assume first that n is even and each of $\{v_1v_2, v_4v_5, uv_3\}$ is negative. We assume, without loss of generality, that $v_1v_2$ is an in-edge. Suppose to the contrary that G admits a nowhere-zero 3-flow. By Theorem 2.4, G has a modulo 3-orientation. Note that for each vertex of degree 3, all the three edges incident with it are oriented either out or into it. Since $v_1v_2$ is an in-edge, it forces that $v_iv_{i+1}$ is from $v_i$ to $v_{i+1}$ for $i\in \{2,5,7,\ldots , n-1\}, v_iv_{i+1}$ is from $v_{i+1}$ to $v_i$ for $i\in \{3,6,8,\ldots , n\}$, $uv_i$ is from $v_i$ to u for $i\in \{1,2,4, 5,7,\ldots , n-1\}, uv_i$ is from u to $v_i$ for $i\in \{6,8,\ldots , n\}$, where the subscripts of v take modulo n.

On the other hand, observe the center u. The number of all the edges coming into u is equal to the number of all the edges coming from u plus 4, which implies $d^+(u)-d^-(u)\not =0$ (mod 3), contrary to Theorem 2.4.

Thus, we assume that n is odd. In this case, assume that each of $\{v_1v_2, uv_3, uv_5\}$ is negative. We assume, without loss of generality, that $v_1v_2$ is an in-edge. By the argument above, for each vertex on C, all edges incident with it are from or into it. Thus, $v_iv_{i+1}$ is from $v_i$ to $v_{i+1}$ for $i\in \{2,4,6,\ldots , n-1\}, v_iv_{i+1}$ is from $v_{i+1}$ to $v_i$ for $i\in \{3,5,7,\ldots , n-2\}$, and $v_1v_n$ is oriented from $v_1$ to $v_n, uv_i$ is from $v_i$ to u for $i\in \{1,2,4, 6, 8,\ldots , n-1\}, uv_i$ is from u to $v_i$ for $i\in \{7,9,\ldots , n\}$, where the subscripts of v take modulo n. On the other hand, The number of all the edges coming into u is equals to the number of all the edges coming from u plus 5, which implies that $d^+(u)-d^-(u)\not =0$ (mod 3), contrary to Theorem 2.4.$\square$

4 Proof of Theorem 1.3

In this section, for an fan $F_n$, where $n\ge 2$, we assume that $V(F_n)=\{u, v_1, \ldots , v_n\}$, $P=v_1\ldots v_n$ is a path, and u is adjacent to $v_i$, where $1\le i\le n$.

Lemma 4.1

$G=(F_n,\sigma )$ has no negative edge on P by vertex switchings.

Proof

If not, let $v_iv_{i+1}$, where $1\le i\le n-1$, be the first negative edge on P from $v_1$ to $v_{n}$. By vertex switchings on $v_i, v_{i-1}, \ldots , v_1$ in order, there is no negative edge in $\{v_1v_2, v_2v_3, \ldots , v_{i}v_{i+1}\}$, contrary to the choice of $v_iv_{i+1}$. $\square$

By Lemma 4.1, in the rest of this section, we assume that $G=(F_n,\sigma )$ has no negative edge on the path P. By vertex switchings, we may assume that $n(u)\le \frac{d(u)}{2}$. It follows that $n(G)=n(u)\le \frac{d(u)}{2}$.

Lemma 4.2

Assume that $n\ge 5$ and there exist two adjacent positive edges $uv_j$ and $uv_{j+1}$ where $1\le j\le n-1$.

• (1) Let $G'=G{\setminus } \{v_{1}\}$ if $j=1$.

• (2) Let $G'=G{\setminus } \{v_{n}\}$ if $j=n-1$.

• (3) Let $G'=(G{\setminus } \{v_{j}\})\cup \{v_{j-1}v_{j+1}\}$ if $2\le j\le n-2$.

If $G'$ admits a nowhere-zero 4-flow, then so does G.

Proof

(1) Let $f'$ be a nowhere-zero 4-flow of $G'$ and $f'(uv_{2})=f'(v_{2}v_{3})=a$. We assume, without loss of generality, that in $G', v_{2}v_{3}$ is oriented from $v_{2}$ to $v_{3}, uv_{2}$ is from u to $v_{2}$. We further assume that $uv_1$ is from u to $v_1$ and $v_{1}v_{2}$ is from $v_{1}$ to $v_{2}$ in G. We can now define a nowhere-zero 4-flow f on G from $f'$.

Define $f(e)=f'(e)$ if $e\not \in E(v_1)\cup \{uv_{2}\}$ and let $f(uv_1)=f(v_{1}v_2)=x, f(uv_{2})=a-x$. If $a\in \{-2, -1, 2, 3\}$, let $x=1$. If $a\in \{-3, 1\}$, let $x=-1$. Then f is as desired.

The proof of (2) is quite similar to (1). The proof of (3) is similar with that of Lemma 3.5. $\square$

With the same argument in the proof of Lemma 3.6, we obtain next lemma.

Lemma 4.3

Assume that $n\ge 6$ and there exist four edges $uv_i, uv_{i+1}, uv_{i+2}$, and $uv_{i+3}$ of E(u), where $2\le i\le n-4$, such that $uv_i$ and $uv_{i+1}$ are negative while $uv_{i+1}$ and $uv_{i+3}$ are positive. If $G'=(G{\setminus } \{v_i, v_{i+1}, v_{i+2},$ $v_{i+3}\})\cup \{v_{i-1}v_{i+4}\}$ admits a nowhere-zero 4-flow, then so does G.

Lemma 4.4

If $G=(F_n,\sigma )$ is a flow-admissible signed fan on $n\le 5$ vertices, then G admits a nowhere-zero 4-flow.

Proof

Recall that $n(G)=n(u)\le \frac{n}{2}$. Since G is flow-admissible, either $n(G)=0$ or $n(G)\ge 2$ by Lemma 2.2(2). If $n(G)=0$, then it is easy to verify that G admits a nowhere-zero 4-flow. If $n(G)\ge 2$, then $4\le n\le 5$ and $n(G)=2$ since $n(G)\le \frac{n}{2}$. Since $n(G)=2$, at most one of $uv_1$ and $uv_2$ is negative for otherwise, we obtain a signed fan contains exactly one negative edge $v_2v_3$ by switching at $v_1, v_2$, contrary to that $G=(F_n,\sigma )$ is flow-admissible. By symmetry, at most one of $uv_{n-1}$ and $uv_n$ is negative. First, let $n=4$. By above argument, exactly one of $uv_1$ and $uv_2$ ($uv_3$ and $uv_4$) is negative. Thus, G is equivalent to one of $(F_4, \sigma _1), (F_4, \sigma _2)$ in Fig. 4, each of which admits a nowhere-zero 4-flow. Thus, let $n=5$. It is easy to see that G contains two positive edges $uv_i$ and $uv_{i+1}$ except $(F_5, \sigma ^\prime )$ in Fig. 4. If $i=1$, define $G^\prime =G\backslash \{v_{1}\}$. If $i=4$, define $G^\prime =G\backslash \{v_{5}\}$. If $2\le i\le 3$, define $G^\prime =(G\backslash \{v_{i}\})\cup \{v_{i-1}v_{i+1}\}$. Then $G'$ is a signed fan with two negative edges. By Lemma 2.2(2), $G'$ is flow-admissible. On the other hand, $G'$ is a $(F_4, \sigma )$ and hence $G^\prime$ admits a nowhere-zero 4-flow by above argument and so does G by Lemma 4.2. $\square$

Fig. 4

Three unbalanced flow-admissible signed fans

Lemma 4.5

None of $(F_5, \sigma ')$ and $(F_6, \sigma ')$ admits nowhere-zero 3-flow in Fig. 5.

Fig. 5

Two signed fans that admit no nowhere-zero 3-flow

Proof

We only prove that $(F_5, \sigma ')$ does not admit a nowhere-zero 3-flow. The proof for the case that $(F_6,\sigma ')$ is similar. Suppose otherwise that $(F_5, \sigma ')$ admits a nowhere-zero 3-flow. We assume, without loss of generality, that the negative edge $uv_2$ is an out-edge. By Theorem 2.4, $(F_5, \sigma ')$ has a modulo 3-orientation. Thus, all the edges incident with $v_2$ are oriented into $v_2$, all the edges incident with $v_3$ are oriented from $v_3$, all the edges incident with $v_4$ are into $v_4$, edge $uv_1$ is oriented from u to $v_1$, and edge $uv_5$ is from u to $v_5$. In this case, $d^+(u)=2$ and $d^-(u)=3$, which implies that $d^+(u)\not = d^-(u)$ (mod 3), contrary to Theorem 2.4. $\square$

Lemma 4.6

For $n\ge 2$, let $G=(F_n,\sigma )$ be a flow-admissible signed fan on $n\ge 2$ vertices. Then G admits a nowhere-zero 4-flow.

Proof

We proceed our proof by induction on n. If $n\le 5$, our theorem follows by Lemma 4.4. Thus, assume that $n\ge 6$.

We first assume that G is balanced. By Lemma 2.2(1), G is equivalent to a graph with all positive edges. It is easy to see that G admits a nowhere-zero 3-flow. Thus, we assume that G is unbalanced. By Lemma 2.2(2), G has at least two negative edges. We establish the following claims. $\square$

Claim 1 If there exist two adjacent vertices $v_i$ and $v_{i+1}$ such that $uv_i$ and $uv_{i+1}$ are positive, where $1\le i\le n-1$, then G admits a nowhere-zero 4-flow.

Proof of Claim 1

Recall that $n(G)=n(u)\le \frac{d(u)}{2}$. First, assume that $i=1$. By Lemma 2.2(2) and induction hypothesis, $G^\prime =G{\setminus } \{v_1\}$ admits a nowhere-zero 4-flow and so does G by Lemma 4.2(1). By symmetry, if $i=n-1, G$ admits a nowhere-zero 4-flow. Thus, assume that $2\le i\le n-2$. Let $G'=(G{\setminus }\{v_i\})\cup \{v_{i-1}v_{i+1}\}$. It is clear that $2\le n(G')=n(G)=n(u)\le \frac{d_G(u)}{2}$. By Lemma 2.2(2), $G'$ is flow-admissible. By induction hypothesis, $G'$ admits a nowhere-zero 4-flow and so does G by Lemma 4.2(3). This proves Claim 1.

By Claim 1, we may assume that for each pair of $\{ uv_i, uv_{i+1}\}$, where $1\le i\le n-1$, at least one of $uv_i, uv_{i+1}$ is negative. If both of them are negative, then by switching at u, we obtain that $uv_i, uv_{i+1}$ are positive. It follows by Claim 1 that G admits a nowhere-zero 4-flow. Thus, we may assume that for every pair of two edges $uv_i$ and $uv_{i+1}$, exactly one of them is negative. This implies that n must be even and $n(G)=\frac{n}{2}$. Thus, we prove next claim.

Claim 2 If n is even and $n(G)=\frac{n}{2}$, then G admits a nowhere-zero 4-flow.

Proof of Claim 2

By symmetry, we may assume that each of $\{uv_2, uv_4, \ldots , uv_{n}\}$ is negative.

If $n\ge 8$, let $G^\prime =(G{\setminus } \{v_2, v_{3}, v_{4}, v_{5}\})\cup \{v_{1}v_{6}\}$. It is clear that $n(G')=\frac{n}{2}-2\ge 2$. Thus, $G^\prime$ is flow-admissible by Lemma 2.2(2). By induction hypothesis, $G^\prime$ admits a nowhere-zero 4-flow and so does G by Lemma 4.3. Let $n=6$. By Lemma 2.1, we assume that each negative edge is an out-edge, $uv_{i}$ is oriented from u to $v_{i}$, where $i\in \{1, 3, 5\}$, and $v_{j}v_{j+1}$ is from $v_{j}$ to $v_{j+1}$, where $j=1,2,\ldots ,5$. Define $f(uv_{1})=f(uv_{2})=f(uv_{3})=f(uv_{6})=1$, $f(uv_{4})=f(uv_{5})=-2$ and the values of f on P are assigned in order 1, 2, 3, 1, $-1$ from $v_1v_2$. Thus, f is as desired. This proves Claim 2. $\square$

Proof of Theorem 1.3

By Lemma 4.6, it is sufficient to show that for $n\ge 5$, there exists a mapping $\sigma {:}\;E(F_n)\rightarrow \{-1, 1\}$ such that the signed fan $(F_n, \sigma )$ has no nowhere-zero 3-flow. If $n\le 6$, then our theorem follows from Lemma 4.5. Thus, we may assume that $n\ge 7$.

Suppose to the contrary that G admits a nowhere-zero 3-flow. Assume that G has two negative edges $uv_3$ and $uv_5$ where $uv_3$ is an in-edge. By Theorem 2.4, G has a modulo 3-orientation. $\square$

Note that for each vertex of degree 3, all the three edges incident with it are oriented either out or into it. Note that $uv_3$ is an in-edge. If n is even, then $uv_i$ is from u to $v_i$ for $i\in \{1,2,4,6,8,10,\ldots , n-2\}$. $uv_i$ is from $v_i$ to u for $i\in \{7,9, \ldots , n-1, n\}$. If n is odd, then $uv_i$ is from u to $v_i$ for $i\in \{1,2,4,6,8,10,\ldots , n-1, n\}$. $uv_i$ is from $v_i$ to u for $i\in \{7,9, \ldots , n-2\}$.

On the other hand, observe the vertex u. If n is even, then the number of all the edges coming from u is equal to the number of all the edges coming into u plus 4, which implies $d^+(u)-d^-(u)\not =0$ (mod 3), contrary to Theorem 2.4.

Thus, we may assume that n is odd. Then the number of all the edges coming from u is equal to the number of all the edges coming into u plus 7, which implies that $d^+(u)-d^-(u)\not =0$ (mod 3), contrary to Theorem 2.4. $\square$