Proof of the Caccetta-Haggkvist conjecture for digraphs with small independence number

TL;DR

This paper proves the Caccetta-H"aggkvist conjecture for digraphs with small independence number, extending previous results to a broader class of graphs with independence number up to (k+1)/2.

Contribution

It generalizes the proof of the conjecture to digraphs with independence number at most (k+1)/2, beyond the previously known case of independence number two.

Findings

Confirmed the conjecture for digraphs with independence number ≤ (k+1)/2.

Extended the class of graphs for which the conjecture holds.

Provided a new bound relating independence number and cycle length.

Abstract

For a digraph $G$ and $v \in V(G)$, let $\delta^+(v)$ be the number of out-neighbors of $v$ in $G$. The Caccetta-H\"{a}ggkvist conjecture states that for all $k \ge 1$, if $G$ is a digraph with $n = |V(G)|$ such that $\delta^+(v) \ge n/k$ for all $v \in V(G)$, then G contains a directed cycle of length at most $k$. In [2], N. Lichiardopol proved that this conjecture is true for digraphs with independence number equal to two. In this paper, we generalize that result, proving that the conjecture is true for digraphs with independence number at most $(k+1)/2$.

Equations

$$|V(G)\setminus S_p|\ge (p-1)d+2d+1=(2\alpha (G)-2)d+1$$

Full text

1 Introduction and definitions

For the rest of the paper, we use the words cycle and path to refer to a directed cycle and directed path, respectively, and every graph considered is a digraph. Furthermore, every digraph $G$ is simple, meaning it has no loops or parallel edges. Let the girth $g(G)$ of a digraph $G$ be the length of its shortest cycle, and for a vertex $v\in V(G)$, let $\delta^{+}(v)$ denote the number of out-neighbors of $v$ in $G$. Let $\Delta^{+}(G)=\displaystyle\min_{v\in V(G)}\delta^{+}(v)$ be the minimum out-degree of a vertex in $G$. For vertices $u,v\in V(G)$, let the distance $d(u,v)$ from $u$ to $v$ be the length of the shortest path from $u$ to $v$ (define this to be zero if $u=v$). For $v\in V(G)$ and $i\geq 1$, let $N_{i}^{+}(v)$ be the set of vertices $u$ with $d(v,u)=i$, and let $N_{i}^{-}(v)$ be the set of vertices $u$ with $d(u,v)=i$. For a digraph $G$, call a set of vertices $H\subset V(G)$ independent if there are no edges between any two vertices of $H$. Let the independence number $\alpha(G)$ of a digraph $G$ be the size of the largest independent set $H\subset V(G)$. For disjoint sets $S_{1},S_{2}\subset V(G)$, say that $S_{1}$ is stable with $S_{2}$ if there are no edges between a vertex in $S_{1}$ and a vertex in $S_{2}$.

We begin with the following simple observation.

Lemma 1.1

Suppose that $G$ is a digraph containing a cycle; then $g(G)\leq 2\alpha(G)+1$.

Proof.  Let $C$ be a cycle of $G$ with minimum length, and suppose $C$ has at least $2\alpha(G)+2$ vertices. Then there exists a subset $S\subset V(C)$ of size $\alpha(G)+1$ such that no pair of vertices of $S$ are adjacent in $C$. Then there is an edge in $G$ between some pair of vertices in $S$, which gives a shorter cycle in $G$, a contradiction. This proves Lemma 1.1.

The next lemma immediately follows from Lemma 1.1, and is used repeatedly throughout the paper.

Lemma 1.2

Suppose $G$ is a digraph with $g(G)\geq 2\alpha(G)$, and that $H\subset G$ is a subgraph of $G$ with $\alpha(H)\leq\alpha(G)-1$. Then $H$ is acyclic.

Proof.  If $H$ contains a cycle, then Lemma 1.1 shows that $H$ contains a cycle of length at most $2\alpha(G)-1$, which is a contradiction. This proves Lemma 1.2.

In this paper, we deal with the following formulation of the Caccetta-Häggkvist conjecture, which was introduced in [References]:

Conjecture 1.1** (Caccetta-Haggkvist)**

For $d\geq 1$, $k\geq 1$, if $G$ is a digraph with $n=|V(G)|\leq kd$ and $\Delta^{+}(G)\geq d$, then $g(G)\leq k$.

For $k=1$ and $k=2$ it follows that the digraph is not simple, a contradiction. So, to prove Conjecture 1.1, we can assume $k\geq 3$.

Now, Lemma 1.1 gives that Conjecture 1.1 is true for $\alpha(G)\leq(k-1)/2$. In this paper, we prove that Conjecture 1.1 is true for $\alpha(G)\leq(k+1)/2$.

2 Main Results

We need the following two lemmas.

Lemma 2.1

Suppose that $G$ is an acyclic digraph; then for all $v\in V(G)$, there exists a path of length at most $2\alpha(G)-1$ to a vertex of out-degree zero in $G$.

Proof.  Since $G$ is acyclic, there exists a path from $v$ to a vertex of out-degree zero in $G$. Let $P=(v,v_{2},\cdots,v_{k})$ be a shortest such path. Then $P$ is induced, so if $k\geq 2\alpha(G)+1$ then $\{v,v_{3},\cdots,v_{k}\}\subset V(G)$ is an independent set of size at least $\alpha(G)+1$, which is a contradiction. Thus $P$ has length at most $2\alpha(G)-1$, as desired. This proves Lemma 2.1.

Lemma 2.2

Let $G$ be an simple digraph with minimum out-degree $d\geq 1$, $\alpha(G)\geq 3$, and $g(G)\geq 2\alpha(G)$. Set $p=2\alpha(G)-3$, and suppose $v\in V(G)$ is a vertex with $\delta^{+}(v)=d$. For odd $1\leq i\leq p$, let $S_{i}$ be the subgraph of $G$ induced by the vertex set $V(G)\setminus(N_{1}^{+}(v)\cup\{v\}\cup\left(\bigcup_{j=1}^{i}N_{j}^{-}(v)\right)$. Then, for each odd $1\leq i\leq p$, there exists a unique vertex $v_{i}\in S_{i}$ such that $N_{1}^{+}(v_{i})\subset N_{i}^{-}(v)$. Furthermore, $|V(G)\setminus S_{p}|\geq(2\alpha(G)-2)d+1$.

Proof.  Every $w\in N_{1}^{+}(v)$ has $N_{1}^{+}(w)\subset N_{1}^{+}(v)\cup S_{p}$, since otherwise we obtain a cycle of length at most $2\alpha(G)-1$, a contradiction. Since $|N_{1}^{+}(v)|=d$, it follows that $V(S_{i})\neq\emptyset$ for odd $1\leq i\leq p$.

Now, for odd $1\leq i\leq p$, we iteratively choose $v_{i}\in S_{i}$ such that $N_{1}^{+}(v_{i})\subset N_{i}^{-}(v)$. Let $\{v_{1},v_{3},\cdots,v_{i-2}\}$ be vertices such that $N_{1}^{+}(v_{j})\subset N_{j}^{-}(v)$ for all odd $1\leq j\leq i-2$ (if $i=1$, this set of vertices is empty). The set $T=\{v,v_{1},v_{3},\cdots,v_{i-2}\}$ (if $i=1$, then $T=\{v\}$) is stable with $S_{i}$, so $\alpha(S_{i})\leq\alpha(G)-(i+1)/2$, and thus $S_{i}$ is acyclic by Lemma 1.1. Thus there exists $v_{i}\in S_{i}$ with out-degree zero in $S_{i}$.

Now, we claim that $N_{1}^{+}(v_{i})\subset N_{i}^{-}(v)$. If not, then $v_{i}$ has an edge to a vertex $w_{1}\in N_{1}^{+}(v)$, which has an edge to a vertex $w_{2}\in S_{i}$. Let $H$ be the subgraph of $S_{i}$ induced by the set of vertices with no edge to $v_{i}$. We may assume $w_{2}\in H$. We have that $\{v,v_{1},\cdots,v_{i}\}$ is stable with $H$, so $\alpha(H)\leq\alpha(G)-(i+3)/2$. Then by Lemma $2.1$ there exists a path $(w_{2}\cdots w_{j})$ of length at most $2\alpha(G)-i-4$ from $w_{2}$ to a vertex $w_{j}\in H$ with out-degree zero in $H$. If $w_{j}$ has out-degree in $S_{i}$ equal to zero, then since $|N_{1}^{+}(v)|=d$, it follows that $w_{j}$ has an out-neighbor in $N_{i}^{-}(v)$ and we obtain a cycle of length at most $2\alpha(G)-1$, a contradiction. If instead $w_{j}$ has an out-neighbor to $w_{j+1}\in S_{i}\setminus H$, then we again obtain a cycle of length at most $2\alpha(G)-1$, a contradiction. It follows that $v_{i}$ has $N_{1}^{+}(v_{i})\subset N_{i}^{-}(v)$ for odd $1\leq i\leq p$, as claimed.

Now, for odd $1\leq i\leq p$, let $V_{i}$ be the set of vertices $u\in S_{i}$ such that $u$ has out-degree zero in $S_{i}$. Let $H=\{v\}\cup V_{1}\cup V_{3}\cup\cdots\cup V_{p}$. For $v_{i}\in V_{i}$, since $N_{1}^{+}(v_{i})\subset N_{i}^{-}(v)$, it follows that $H$ is an independent set, so $|V(H)|\leq\alpha(G)$. We also know that the $V_{i}$ are nonempty, so $|V(H)|\geq\alpha(G)$. Thus $|V_{i}|=1$ for all odd $1\leq i\leq p$. This proves the first part of the lemma, namely that for each odd $1\leq i\leq p$ there exists a unique vertex $v_{i}\in S_{i}$ with $N_{1}^{+}(v_{i})\subset N_{i}^{-}(v)$. For the remainder of the proof, let $\{v_{1},v_{3},\cdots,v_{p}\}$ be those unique vertices.

For odd $3\leq i\leq p$, define $X_{i}=N_{1}^{+}(v_{i})\subset N_{i}^{-}(v)$, and let $T_{i}=N_{i}^{-}(v)\cup N_{i-1}^{-}(v)\setminus X_{i}$. $\{v\}$ is stable with $X_{i}$, so by Lemma 1.2, $X_{i}$ is acyclic and contains a vertex $u_{i}\in V(X_{i})$ with out-degree zero in $X_{i}$. We claim that $N_{1}^{+}(u_{i})\subset T_{i}$, and consequently $|T_{i}|\geq d$. If not, then there exists a path of length at most 2 from $u_{i}$ to a vertex $w_{2}\in S_{i}$. Since $\{v,v_{1},\cdots v_{i-2}\}$ is stable with $S_{i}$, $\alpha(S_{i})\leq\alpha(G)-(i+1)/2$. Lemma 2.1 gives a path from $w_{2}$ to $v_{i}$ of length at most $2\alpha(G)-5$. These two paths together form a cycle in $G$ of length at most $2\alpha(G)-2$, which is a contradiction.

Thus, for odd $3\leq i\leq p$, we have $|X_{i}|+|T_{i}|\geq 2d$. Also, $N_{1}^{+}(v_{i})\subset N_{1}^{-}(v)$ gives $|N_{1}^{-}(v)|\geq d$, and by the definition of $v$ we have $|N_{1}^{+}(v)|=d$. Together with the vertex $v$, these give:

[TABLE]

as desired. This proves Lemma 2.2.

Lemma 2.2 is used to prove the following two theorems.

Theorem 2.1

Suppose that $G$ is a digraph with minimum out-degree $d\geq 1$ and $n=|V(G)|\leq 2\alpha(G)d$; then $g(G)\leq 2\alpha(G)$.

Proof.  As mentioned above, it suffices to consider simple digraphs $G$ with $\alpha(G)\geq 2$. The case $\alpha(G)=2$ is proved in [References], so we may further assume that $\alpha(G)\geq 3$. Now, for the sake of contradiction, suppose that $g(G)\geq 2\alpha(G)+1$. Then Lemma 2.2 implies that $|V(G)\setminus S_{p}|\geq(2\alpha(G)-2)d+1$, which together with $|V(G)|\leq 2\alpha(G)d$ gives $|S_{p}|\leq 2d-1$. $H=\{v,v_{1},v_{3},\cdots,v_{p-2}\}$ is stable with $S_{p}$, so $\alpha(S_{p})=1$ and $S_{p}$ is a transitive tournament. Let $(w_{1}\cdots w_{r})$ be the unique Hamiltonian path of the transitive tournament $S_{p}$.

Now, $J=\{v_{1},v_{3},\cdots,v_{p}\}$ is stable with $N_{1}^{+}(v)$, so $N_{1}^{+}(v)$ is a transitive tournament. Let its unique Hamiltonian path be $(u_{1}\cdots u_{d})$. $N_{1}^{+}(u_{d})\subset S_{p}$, so there is an out-neighbor $w_{k}$ of $u_{d}$ with $k\geq d$. It follows that $w_{k}$ has an edge to a vertex not in $S_{p}$. An edge from $w_{k}$ to $v$ or to $w^{\prime}\in N_{i}^{-}(v)$ for some $1\leq i\leq p$ yields a cycle of length at most $2\alpha(G)$, a contradiction. If, instead, $w_{k}$ has an edge to $u^{\prime}\in N_{1}^{+}(v)$, then $u^{\prime}$ has an edge to $u_{d}$, and we obtain a cycle of length at most three, a contradiction. This proves Theorem 2.1.

Theorem 2.2

Suppose $G$ is a digraph with minimum out-degree $d\geq 1$ and $n=|V(G)|\leq(2\alpha(G)-1)d$; then $g(G)\leq 2\alpha(G)-1$.

Proof.  As mentioned above, it suffices to consider simple digraphs $G$ with $\alpha(G)\geq 2$. The case $\alpha(G)=2$ is proved in [References], so we further assume that $\alpha(G)\geq 3$. For the sake of contradiction, suppose $g(G)\geq 2\alpha(G)$. Lemma 2.2 gives a set of vertices $\{v_{i}\}$ indexed by odd $1\leq i\leq p$ such that $N_{1}^{+}(v_{i})\subset N_{i}^{-}(v)$. $v_{1}$ is stable with $N_{1}^{+}(v)$ (otherwise we obtain a cycle of length four), so Lemma 1.2 gives that $N_{1}^{+}(v)$ is acyclic. So, there exists $u\in N_{1}^{+}(v)$ with out-degree zero in $N_{1}^{+}(v)$. If $u$ has an edge to any vertex not in $S_{p}$, we obtain a cycle of length at most $2\alpha(G)-1$, a contradiction. Thus, we must have $N_{1}^{+}(u)\subset S_{p}$ and it follows that $|S_{p}|\geq d$.

But Lemma 2.2 also gives that $|V(G)\setminus S_{p}|\geq(2\alpha(G)-2)d+1$, which together with $|S_{p}|\geq d$ implies that $|V(G)|\geq(2\alpha(G)-1)d+1$, contradicting the assumption that $|V(G)|\leq(2\alpha(G)-1)d$. This proves Theorem 2.2.

Theorem 2.3

Conjecture 1.1 is true for digraphs $G$ with $\alpha(G)\leq(k+1)/2$.

Proof.  Theorem 2.1 and Theorem 2.2 together with Lemma 1.1 give the desired result. This proves Theorem 2.3.