On a conjecture of Bondy and Vince

TL;DR

This paper confirms Bondy and Vince's long-standing conjecture that graphs with a bounded number of low-degree vertices typically contain two cycles with lengths differing by one or two, except for finitely many counterexamples.

Contribution

The paper proves the conjecture for all nonnegative integers k, extending previous partial results and resolving a 20-year-old open problem.

Findings

Confirmed the conjecture for all k

Identified only finitely many counterexamples

Extended previous partial results

Abstract

Twenty years ago Bondy and Vince conjectured that for any nonnegative integer $k$, except finitely many counterexamples, every graph with $k$ vertices of degree less than three contains two cycles whose lengths differ by one or two. The case $k\leq 2$ was proved by Bondy and Vince, which resolved an earlier conjecture of Erd\H{o}s et. al.. In this paper we confirm this conjecture for all $k$.

Equations

$$f(k)\ge f(k-1)+7k+f(3)\text{mbox}holdsforallk\ge 4.$$

$$\sum _{i=1}^t|D_i|\le \sum _{i=1}^{t}2|\mathcal{B}(G_1)\cap V(C_i)|\le 2|\mathcal{B}(G_1)|\le 2k.$$

$$\sum _{i=t+1}^{m-1}|D_i|\le \sum _{i=t+1}^{m-1}2|\mathcal{B}(G_{t+1})\cap V(C_i)|\le 2|\mathcal{B}(G_{t+1})|\le 2k.$$

Full text

The study on cycles has been extensive in the literature and various problems on the existence of cycles with specified lengths were investigated decades ago (e.g., Erdős [2, 3]). One such question, proposed by Erdős et. al. (see the discussion in [1]), asked whether every graph with minimum degree at least three contains two cycles whose lengths differ by one or two. This was answered affirmatively by Bondy and Vince [1] in the following stronger theorem: with the exception of $K_{1}$ and $K_{2}$, every graph having at most two vertices of degree less than three contains two cycles of lengths differing by one or two. They further conjectured the following generalization. (All graphs referred here are simple.)

Conjecture. (Bondy and Vince [1]) Let $k$ be any nonnegative integer. With finitely many exceptions, every graph having at most $k$ vertices of degree less than three has two cycles whose lengths differ by one or two.

The authors of [1] remarked that they can also prove for $k=3$ (with twelve exceptions) and the case $k=4,5$ likely can be solved using their analysis. We mention that the theorem in [1] was generalized in another direction by Fan [4]. For related topics, results and open problems, we refer interested readers to [6, 7, 8, 10, 11] and their references.

In this paper, we confirm the above conjecture of Bondy and Vince by the following.

Theorem. Every graph, having at most $k$ vertices of degree less than three and at least $5k^{2}$ vertices, contains two cycles whose lengths differ by one or two.

This also can be viewed as a resilience type answer to the above question of Erdős et. al.. Let $G$ be an $n$-vertex graph with minimum degree at least three. Then one can derive that by deleting any $\sqrt{n}/5$ edges from $G$, the remaining graph still contains two cycles of lengths differing by one or two. Also by repeating the following procedure: first apply this theorem to find a pair of two cycles of lengths differing by one or two and then delete two edges to destroy these two cycles, one can in fact find $\Omega(\sqrt{n})$ such pairs of cycles in $G$.

For the proof, we will need a lemma of Bondy and Vince [1] (the proof of which uses an argument based on Thomassen and Toft [9]). Let $C$ be a cycle in a graph $G$. A bridge of $C$ is either a chord of $C$ or a subgraph of $G$ obtained from a component $B$ in $G-V(C)$ by adding all edges between $B$ and $C$. We call vertices of the bridge not in $C$ internal.

Lemma. ([1]) Let $G$ be a 2-connected graph, not a cycle, and let $C$ be an induced cycle in $G$ some bridge $B$ of which has as many internal vertices as possible. Then either $B$ is the only bridge of $C$, or else that $B$ is a bridge containing exactly two vertices $u,v$ in $V(C)$ and every other bridge of $C$ is a path from $u$ to $v$.

We are ready to present the proof of our result.

Proof of Theorem. Throughout this proof, let $\mathcal{B}(G)$ denote the set of all vertices with degree at most two in a graph $G$, and we say a pair of cycles is good if their lengths differ by one or two. Let $f(1)=f(2)=3,f(3)=14,f(4)=56,f(5)=116$ and $f(k)=5k^{2}$ for $k\geq 6$ so that $\{f(k)-f(k-1)\}$ is strictly increasing and

[TABLE]

We will prove by induction on $k$ that every graph $G$ with $|\mathcal{B}(G)|\leq k$ and at least $f(k)$ vertices contains a good pair of cycles. The case $k\leq 2$ follows by the aforementioned theorem of Bondy and Vince [1], so we may assume that $k\geq 3$ and the statement holds for any integer $\ell<k$.

We begin by showing some useful facts on graphs $H$ with $|\mathcal{B}(H)|=k$. For a subgraph $F$ of $H$, by $H-F$ we denote the graph obtained from $H$ by deleting all vertices of $F$.

Suppose that there exists a cut-vertex $u$ in $H$. Let $B_{1}$ be a component of $H-\{u\}$ and $B_{2}=H-\{u\}-B_{1}$. For $i\in[2]$, let $b_{i}=|\mathcal{B}(H)\cap B_{i}|$. It is clear that $k-1\leq b_{1}+b_{2}\leq k$. We also have $b_{i}\geq 2$. Indeed, otherwise say some $b_{i}\leq 1$, then $H_{i}=H[B_{i}\cup\{u\}]$ is a graph with at least three vertices (since $\delta(H)\geq 2$) and at most two vertices of degree less than three; by the theorem of Bondy and Vince, it contains a good pair of cycles (so does $H$), a contradiction. This already proves for $k=3$ (as it is impossible to have $b_{1}+b_{2}\geq 4$). Now we consider $k\geq 4$. Since $H_{i}$ contains no good pair of cycles and $|\mathcal{B}(H_{i})|\leq b_{i}+1\leq k-1$, by induction we have $f(b_{1}+1)+f(b_{2}+1)>|V(H_{1})|+|V(H_{2})|>|V(H)|\geq f(k-1)+f(3)=\max_{2\leq s\leq k/2}\{f(s+1)+f(k-s+1)\}$333Here the equality follows from the fact that $f(k)-f(k-1)$ is increasing., again a contradiction. This finishes the proof. ∎

A cycle $C$ in a graph $H$ is called feasible if it is induced and $H-C$ is connected.

For $F,F^{\prime}\subseteq H$, let $N_{F}(F^{\prime})$ be the set of vertices in $F$ adjacent to some vertex in $F^{\prime}$.

Suppose there exist two consecutive vertices in $C$, say $u_{1},u_{2}$, belonging to $A$. As noted above there is no $u_{i},u_{i+s+1}\in A$ for any $i$. So $u_{s},u_{s+1},u_{s+2},u_{s+3}$ are not in $A$ and the pair $u_{i},u_{i+s+1}$ for any $3\leq i\leq s-1$ has at most one vertex in $A$, which together imply that $|A|\leq s-1$, a contradiction. This shows that no two consecutive vertices of $C$ can be in $A$ and thus the vertices of $C$ alternate between $A$ and $\mathcal{B}(H)\cap V(C)$. Finally let us note that $r$ is divisible by four, as otherwise $s$ is odd, so if $u_{i}\in A$ then $u_{i+s+1}\in A$, a contradiction. ∎

Consider $H-D$. We have that $|V(H-D)|$ is at least $f(2)$ for $k=3$ (by Claim 3, $|D|\leq|C|=4$) and is at least $f(k-1)+f(3)$ for $k\geq 4$, and $\mathcal{B}(H-D)=(\mathcal{B}(H)-\mathcal{B}(H)\cap V(C))\cup A^{\prime}$. So $|\mathcal{B}(H-D)|=k-s+|A^{\prime}|$. First we see $|A^{\prime}|\geq s$; as otherwise $|\mathcal{B}(H-D)|\leq k-1$ and by induction $H-D$ has a good pair of cycles. If $V(C)\backslash D=\emptyset$, then there are exactly $|A|=s$ edges between $C$ and $H-C$, so the deletion of $D=V(C)$ will result in at most $s$ new vertices of degree at most two, that is $|A^{\prime}|\leq s$. Therefore in this case, $|A^{\prime}|=s=|A|$ and $A\cup A^{\prime}$ induces a matching of size $s$ in $H$. Now suppose $V(C)\backslash D=\{u\}$. There are exactly $s-1$ edges between $D$ and $H-C$, so the deletion of $D$ will result in at most $s-1$ new vertices of degree at most two in $H-C$, which, plus possibly $u\in A^{\prime}$ (if and only if $d_{H}(u)=4$), show that $|A^{\prime}|\leq s$. Therefore, again we have $|A^{\prime}|=s=|A|$; in fact $d_{H}(u)=4$, $A\cap A^{\prime}=\{u\}$ and $A\cup A^{\prime}-\{u\}$ induces a matching of size $s-1$ in $H$.

The above analysis also demonstrates that every vertex in $A^{\prime}$ has degree two in $H-D$ and thus $\delta(H-D)\geq 2$. By Claim 1, $H-D$ is 2-connected. This finishes the proof. ∎

Let $G_{0}$ be a graph with $|\mathcal{B}(G_{0})|=k$ and at least $f(k)$ vertices. Suppose for a contradiction that $G_{0}$ has no good pair of cycles.

We will define a sequence of subgraphs $G_{0}\supseteq G_{1}\supseteq...\supseteq G_{m}$. First we show there exists a 2-connected $G_{1}\subseteq G_{0}$ with $|\mathcal{B}(G_{1})|=k$ and $|V(G_{1})|\geq f(k)-k$. Let $S$ be the set of vertices of degree at most one in $G_{0}$ and let $G_{1}=G_{0}-S$. Then $|\mathcal{B}(G_{1})|\leq|\mathcal{B}(G_{0})|-|S|+|S^{\prime}|$ where $S^{\prime}$ denotes the set of vertices in $G_{1}$ adjacent to $S$ (so clearly $|S^{\prime}|\leq|S|$). We must have $|\mathcal{B}(G_{1})|=k$; otherwise, $|\mathcal{B}(G_{1})|\leq k-1$ and $|V(G_{1})|\geq f(k)-k\geq f(k-1)$, implying a good pair of cycles in $G_{1}\subseteq G_{0}$. This shows that $S\cup S^{\prime}$ induces a matching of size $|S|$ in $G_{0}$ and every vertex in $S^{\prime}$ is in $\mathcal{B}(G_{1})\backslash\mathcal{B}(G_{0})$. Therefore $\delta(G_{1})\geq 2$. As $|V(G_{1})|\geq f(k)-k\geq f(k-1)+f(3)$ for $k\geq 4$, by Claim 1 we conclude that $G_{1}$ is 2-connected.

Now suppose we have defined $G_{i}$ for some $i\geq 1$. If $G_{i}$ is 2-connected with $|\mathcal{B}(G_{i})|=k$ and of order at least $f(2)+4$ for $k=3$ and at least $f(k-1)+f(3)+2k$ for $k\geq 4$, then

• •

let $C_{i}$ be a feasible cycle in $G_{i}$ (by Claim 2), with the preference to be a four-cycle,

• •

let $A_{i}=N_{C_{i}}(G_{i}-C_{i})$, and further

• •

let $D_{i}=\{v\in V(C_{i})|\leavevmode\nobreak\ d_{G_{i}}(v)\leq 3\}$, $G_{i+1}=G_{i}-D_{i}$ and $A^{\prime}_{i}=\mathcal{B}(G_{i+1})\setminus\mathcal{B}(G_{i})$.

Otherwise we terminate. Let $G_{0}\supseteq G_{1}\supseteq...\supseteq G_{m}$ be the sequence of subgraphs we obtain as above. Then we can apply Claims 3 and 4 for $G_{i},C_{i},A_{i}$ for each $1\leq i\leq m-1$. In particular, we see that $G_{i}$ for all $1\leq i\leq m$ is 2-connected with $|\mathcal{B}(G_{i})|=k$. So the reason we terminate at $G_{m}$ is because the order of $G_{m}$ is at most $f(2)+3$ for $k=3$ and at most $f(k-1)+f(3)+2k-1$ for $k\geq 4$. By (1), this implies that $\sum_{i=1}^{m-1}|D_{i}|=|V(G_{1})|-|V(G_{m})|$ is at least $5$ for $k=3$ and at least $(f(k)-k)-(f(k-1)+f(3)+2k-1)>4k$ for $k\geq 4$.

In what follows, we will investigate properties on the cycles $C_{1},...,C_{m-1}$, which eventually will lead a contradiction to the above lower bound of $\sum_{i=1}^{m-1}|D_{i}|$.

We first show that if $A_{i}^{\prime}\not\subseteq\mathcal{B}(G_{i+1})\cap V(C_{i+1})$, then $C_{i+1}$ is a feasible cycle in $G_{i}$. Clearly we have $A_{i}^{\prime}\cap A_{i+1}=\emptyset$ (as any vertex in $A_{i}^{\prime}$ has degree two in $G_{i+1}$, while vertices in $A_{i+1}$ have degree at least three in $G_{i+1}$). Note that $V(C_{i+1})$ is partitioned into $\mathcal{B}(G_{i+1})\cap V(C_{i+1})$ and $A_{i+1}$. So $A_{i}^{\prime}\not\subseteq\mathcal{B}(G_{i+1})\cap V(C_{i+1})$ implies that $A_{i}^{\prime}$ intersects with $G_{i+1}-C_{i+1}$. By Claim 4, we see that any vertex in $A_{i}^{\prime}$ is adjacent to $D_{i}$, so $G_{i+1}-C_{i+1}$ is adjacent to $D_{i}$. Therefore, $G_{i}-C_{i+1}$ is connected and thus $C_{i+1}$ is feasible in $G_{i}$.

We now assert that there exists some $t$ such that $|C_{1}|=...=|C_{t}|=4$ and $|C_{i}|\geq 8$ for each $t+1\leq i\leq m-1$. By Claim 3 each $|C_{i}|$ is divisible by four, so it suffices to prove that if $|C_{i}|\geq 8$, then $|C_{i+1}|\geq 8$. Suppose this is not true. Then $|C_{i+1}|=4<|C_{i}|$, implying that $|A_{i}^{\prime}|=|C_{i}|/2>|C_{i+1}|/2=|\mathcal{B}(G_{i+1})\cap V(C_{i+1})|$. So $C_{i+1}$ is a feasible cycle (of length four) in $G_{i}$. But this contradicts our preference for choosing $C_{i}$ in $G_{i}$, finishing the proof.

Next we claim that for each $1\leq i\leq t-1$, $C_{i+1}$ is feasible in $G_{i}$. We have $|C_{i}|=|C_{i+1}|=4$. So $|A_{i}^{\prime}|=2=|\mathcal{B}(G_{i+1})\cap V(C_{i+1})|$. If $A_{i}^{\prime}=\mathcal{B}(G_{i+1})\cap V(C_{i+1})$, then this will give two possible configurations between $C_{i}$ and $C_{i+1}$ and in each case we can find a good pair of cycles easily (of lengths 4 and 5, or 4 and 6). So we may assume $A_{i}^{\prime}\neq\mathcal{B}(G_{i+1})\cap V(C_{i+1})$. Therefore, $C_{i+1}$ is a feasible cycle in $G_{i}$.

We also claim that for each $i\geq t+1$, $C_{i+1}$ is feasible in $G_{i}$. Let $C_{i}=u_{0}u_{1}...u_{s-1}u_{0}$ where $s\geq 8$. So $|A_{i}|=|A_{i}^{\prime}|\geq 4$. By Claims 3 and 4, we may assume that $u_{0},u_{2},u_{4}\in A_{i}-V(C_{i})\backslash D_{i}$ and $x,y,z\in A_{i}^{\prime}-V(C_{i})\backslash D_{i}$ with $u_{0}x,u_{2}y,u_{4}z\in E(G_{i})$. We may also assume that $x,y,z\in A_{i}^{\prime}\subseteq\mathcal{B}(G_{i+1})\cap V(C_{i+1})$ (as otherwise, $C_{i+1}$ is feasible in $G_{i}$). Let $P$ denote one of the two segments of $C_{i+1}$ between $x$ and $y$ with $|E(P)|\geq|C_{i+1}|/2$. Clearly $|E(P)|\geq 4$ is even. If $|E(P)|\equiv 2$ mod $4$, then $C^{\prime}=P\cup xu_{0}u_{1}u_{2}y$ forms a cycle in $G_{i}$ whose length is $2$ mod $4$ and thus is at least 10. Then there exist two vertices $a,b\in V(P)\cap A_{i+1}$ which divide the cycle $C^{\prime}$ into two segments of lengths differing by two. Also there exists a path with endpoints $a,b$ and all internal vertices in $G_{i+1}-C_{i+1}$, which is internally disjoint from $C^{\prime}$. Putting the above together, we find a good pair of cycles in $G_{i}$. So $|E(P)|\equiv$ 0 mod $4$. Similarly any segment of $C_{i+1}$ between $y$ and $z$ has length [math] mod $4$. Therefore any segment (say $Q$) of $C_{i+1}$ between $x$ and $z$ has length [math] mod $4$ as well. Then $Q\cup xu_{0}u_{1}u_{2}u_{3}u_{4}z$ forms a cycle of length $2$ mod $4$. If $|E(Q)|=4$, then $C_{i}$ and $(C_{i}-\{u_{1},u_{2},u_{3}\})\cup u_{0}x\cup Q\cup zu_{4}$ form a good pair of cycles (whose lengths differ by two). So $|E(Q)|\geq 8$. Again there exist two vertices $a,b\in V(Q)\cap A_{i+1}$ dividing the cycle $Q\cup xu_{0}u_{1}u_{2}u_{3}u_{4}z$ into two segments of lengths differing by two. By similar arguments as above this enables us to find a good pair of cycles in $G_{i}$, proving the claim.

Finally we are completing the proof using the above two claims. We point out that for each $1\leq i\leq t-1$, $C_{i+1}$ and $C_{i}$ play the same role in $G_{i}$. Repeatedly applying this, one can conclude that in fact all four-cycles $C_{1},C_{2},...,C_{t}$ are feasible in $G_{1}$. By Claims 3 and 4, each $V(C_{i})\backslash A_{i}=\mathcal{B}(G_{1})\cap V(C_{i})$ provides $|C_{i}|/2\geq|D_{i}|/2$ distinct vertices in $\mathcal{B}(G_{1})$. So

[TABLE]

Similarly, $C_{t+1},C_{t+2},...,C_{m-1}$ are feasible in $G_{t+1}$ and using this fact, we can derive that

[TABLE]

Putting the above together, we have $\sum_{i=1}^{m-1}|D_{i}|\leq 4k$, but we have seen that $\sum_{i=1}^{m-1}|D_{i}|>4k$ for $k\geq 4$. This contradiction finishes the proof for the case $k\geq 4$. For $k=3$, by Claim 3, all feasible cycles must be of length four, so $t=m-1$. And we obtained that $\sum_{i=1}^{t}|D_{i}|\geq 5$. This tells that there are at least two four-cycles $C_{1},C_{2}$, each providing two vertices in $\mathcal{B}(G_{1})$. But it is impossible as $3=|\mathcal{B}(G_{1})|\geq 4$. The proof now is completed.

We didn’t make lots of efforts to optimise the constant in the proof as we tend to believe that the quadratic bound $O(k^{2})$ can be further improved (perhaps, to a linear term $O(k)$).

To conclude this paper we would like to mention a conjecture of [7], which seems to be a natural generalization for the question of Erdős et. al. and has implications (if true) for other related problems: For any nonnegative integer $k$, every graph with minimum degree at least $k+1$ contains $k$ cycles $C_{1},...,C_{k}$ with $|C_{i+1}|=|C_{i}|+d$ for $d\in\{1,2\}$.

Acknowledgement. We would like to thank the referees for their careful reading and helpful suggestions.

Note added in proof. After this paper was submitted for publication, a proof to the problem mentioned in the last paragraph was announced in [5].