# Prove that connected graph $G$ with $\delta (G)\geq 3$ contains a cycle $C$ such that $G-E(C)$ is still connected.

In a connected simple graph every vertex has a degree at least $$3$$. Prove that the graph contains a cycle such that the graph remains connected when the edges of this cycle are deleted.

I've tried this: Pull out of $$G$$ a cycle $$C$$ with smallest perimeter $$p$$.

Claim: No two nonconsecutive vertices in the circle $$C$$ are adjacent (else we could make smaller cycle then we already have).

Let $$C_1,C_2,...,C_k$$ be all components in $$G-C$$ and I want to prove $$k=1$$.

If $$n_i = |V(C_i)|$$ and $$e_i= |E(C_i)|$$, then we have $$e_i = {3n_i\over 2}-b_i$$ where $$b_i$$ is a number of edges between $$C_i$$ and $$C$$. Because $$G$$ is connected we have $$b_i\geq 1$$ for each $$i$$. Because of the claim we have: $$b_1+b_2+\cdots +b_k = p$$

I have no idea what to do now.

Edit: As Misha Lavrov pointed in a comment, this attemt is not correct.

• You're assuming that the graph is $3$-regular, while we're only given that it has minimum degree $3$. Also, taking a minimum-length cycle is not always sufficient. This graph, for example, has $10$ cycles of length $3$, only $8$ of which leave the graph connected when removed. You could even use this method to find an example where none of the minimum-length cycles work. – Misha Lavrov Feb 3 at 15:57
• Yes but, can we say that is true if $G$ is 3 regular? @MishaLavrov – Aqua Feb 4 at 15:48
• Not sure. The one bit of partial progress that's easy to point out is that this holds for graphs with at least $2n-1$ edges: set aside a spanning tree to ensure connectedness, and you still have $n$ edges left, which contain a cycle. – Misha Lavrov Feb 4 at 17:01
• Is there a counterexample for non-simple graphs? – bof Feb 5 at 2:33
• I don't know how to solve the problem, so I don't know if this is a helpful hint or not. If the statement is true, then it remains true under the weaker assumption, that there is at most one vertex of degree less than $3$. In trying to prove this more general statement by induction, it may help to know that a minimal counterexample must be $2$-connected. – bof Feb 5 at 2:37

Suppose $$G$$ to be a counterexample with a minimal number of edges.

$$G$$ has no isthmus $$vw$$.

Otherwise, contract edge $$vw$$ and consider a cycle $$C$$ of the reduced graph whose removal does not disconnect the reduced graph. Then $$C$$ is entirely within one of the components of $$G$$ produced by deleting $$vw$$ and so deleting it from $$G$$ does not disconnect $$G$$ and there is no counterexample after all.

No two adjacent points $$v,w$$ of $$G$$ can both have degree at least $$4$$.

Otherwise simply remove edge $$vw$$.

$$G$$ has no triangle $$u,v,w$$.

Otherwise, removing edges $$uv,vw,wu$$ must disconnect the graph. Without loss of generality we can assume that $$u$$ is then in a separate component from $$v$$ and $$w$$.

If $$\rho (u)=3$$, then let $$x\in N(u)/\{v,w\}$$. Then $$xu$$ is an isthmus.

So $$\rho (u)>3$$ and therefore $$\rho (v)=\rho (w)=3$$.

If $$(N(v)\bigcap N(w))/\{u\}=\{ y\}$$, then contracting all edges joining vertices in $$\{u,v,w,y\}$$ produces a simple connected graph with all vertices of degree at least $$3$$ and we can proceed using minimality as in the 'isthmus' proof.

Otherwise, we can assume $$N(v)=\{ u,w,y\}$$ and $$N(w)=\{ u,v,z\}$$ with $$y\ne z$$. Then contracting all edges joining vertices in $$\{u,v,w\}$$ produces a simple connected graph with all vertices of degree at least $$3$$ and we can again proceed using minimality.

G is 3-regular

Otherwise let $$N(u)=\{x,y,v\}$$ with $$\rho (v)>3$$. Reduce $$G$$ by removing $$u$$ and adding edge $$xy$$. The new graph has a cycle $$C$$ whose removal does not disconnect the graph. If $$xy$$ is in this cycle then replace it by $$xuy$$. Removing the possibly amended cycle $$C$$ does not disconnect $$G$$.

Proof

Let $$N(u)=\{v,a,b\}$$ and $$N(v)=\{u,c,d\}$$ . Delete $$u$$ and $$v$$ and add in edges $$ab$$ and $$cd$$. The new graph has a cycle $$C$$ whose removal does not disconnect the graph.

If neither $$ab$$ nor $$cd$$ is in $$C$$ then removing $$C$$ does not disconnect $$G$$.

If, say, $$ab$$ but not $$cd$$ is in $$C$$ then replace it in the cycle by $$aub$$. Removing the possibly amended cycle $$C$$ does not disconnect $$G$$ and again there is no counterexample.

Finally suppose both $$ab$$ and $$cd$$ are in $$C$$. Part of $$C$$ is a path from $$a$$ to say $$c$$ (or to $$d$$) which does not include $$ab$$ or $$cd$$. Combining this path with $$auvc$$ then gives a cycle which can be removed without disconnecting $$C$$.

• One question here though: If say $uv$ is in $C$ but $vw$ is not in $C$, then how do you know $G \setminus E(C')$ is connected, where $C' \doteq C -\{uv\}+\{uxv\}$? What if edge $wv$ is a cut-edge in the new graph minus the edges in $C$? – Mike Feb 6 at 19:57
• By assumption the removal of C does not disconnect and the only change we have then made is to replace $uv$ by $ux,xv$. The edge $vw$ is not in $C'$ and so is not going to be deleted. – S. Dolan Feb 6 at 20:06
• Yes I see the first part of that. But the edge $vw$ is not in $G$ either though. – Mike Feb 6 at 20:08
• I see the point you're making now. The fact that $xw$ cannot be an isthmus might help circumvent the issue but it needs some thought. Thanks for pointing that out -I'll get back to you on that. – S. Dolan Feb 6 at 20:39
• I only am aware of this because I got stuck on it too! Meanwhile I do admire how cleanly you make your points above, I am still working on being able to write that crisply. – Mike Feb 6 at 20:43

NOTE (2/6/2021): This answer was posted today and is a redo from the one posted on Wednesday. I came up with this independently of S. Dolan.

Proof SKETCH: Let $$G$$ be the graph on $$n$$ vertices where every vertex has degree at least 3. We assume that every simple graph of minimum degree 3 and on at most $$n-1$$ vertices, has a cycle such that the graph remains connected even after removing the edges of the cycle.

Case A: $$G$$ has at least one triangle $$T$$. Let $$T$$ be a triangle in $$G$$. Let us write $$T=y_1y_2y_3$$.

Subcase A.1: $$G \setminus E(T)$$ has 2 components with $$y_1 \in C_1$$ and $$y_2,y_3 \in C_2$$.

A.1.1 If there is only one vertex $$w_2 \in C_2$$ that is adjacent to either $$y_2$$ or $$y_3$$, AND there is only one vertex $$w_1 \in C_1$$:

A.1.1.1 If $$w_2$$ has 2 neighbors in $$C_2 \setminus \{y_2,y_3\}$$ then add the edge $$w_1w_2$$ and remove vertices $$y_1,y_2,y_3$$ Then the resulting graph $$G'_{n-1}$$ is simple and has minimum degree 3. Also, a cycle $$C'_{n-1}$$ in $$G'_{n-1}$$ so that $$G'_{n-1} \setminus E(C'_{n-1})$$ is connected, translates naturally to a cycle $$C_{n}$$ in $$G$$ so that $$G\setminus E(C_n)$$ is connected.

A.1.1.2 If $$w_2$$ has only one neighbor $$w'_2$$ in $$C_2 \setminus \{y_2,y_3\}$$ then add the edge $$w_1w'_2$$ and remove vertices $$y_1,y_2,y_3,w_2$$. Then the resulting graph $$G'_{n-1}$$ is simple and has minimum degree 3. Also, a cycle $$C'_{n-1}$$ in $$G'_{n-1}$$ so that $$G'_{n-1} \setminus E(C'_{n-1})$$ is connected, translates naturally to a cycle $$C_{n}$$ in $$G$$ so that $$G\setminus E(C_n)$$ is connected.

A.1.2 If there is either more than one vertex in $$C_2$$ adjacent to either $$y_2$$ or $$y_3$$ or there is more than one vertex in $$C_1$$ adjacent to $$w_1$$ collapse $$T$$ into a vertex $$v_T$$ and add the edges $$v_Tw$$; $$w \not \in T$$; $$w$$ adjacent in $$G$$ to a vertex in $$T$$. Then the resulting graph $$G'_{n-1}$$ is simple and has minimum degree 3. Also, a cycle $$C'_{n-1}$$ in $$G'_{n-1}$$ so that $$G'_{n-1} \setminus E(C'_{n-1})$$ is connected, translates naturally to a cycle $$C_{n}$$ in $$G$$ so that $$G\setminus E(C_n)$$ is connected.

Subcase A.2: $$G \setminus E(T)$$ has 3 components. Collapse $$T$$ into a vertex $$v_T$$ and add the edges $$v_Tw$$; $$w \not \in T$$; $$w$$ adjacent in $$G$$ to a vertex in $$T$$. Call the resulting graph $$G'_{n-1}$$. Then the resulting graph $$G'_{n-1}$$ is simple and has minimum degree 3. Also, a cycle $$C'_{n-1}$$ in $$G'_{n-1}$$ so that $$G'_{n-1} \setminus E(C'_{n-1})$$ is connected, translates naturally to a cycle $$C_{n}$$ in $$G$$ so that $$G\setminus E(C_n)$$ is connected.

Subcase A.3: $$G \setminus E(T)$$ has one component. Then we would be done as here we have a cycle namely $$T$$ so that $$G \setminus E(T)$$ is connected.

Case B: $$G$$ has no triangles. If $$G$$ has no triangles, then pick a vertex $$v_n$$ and write $$G_{n-1} \doteq G \setminus \{v_n\}$$, and write $$u_1,u_2,\ldots, u_k$$ as the neighbors of $$v_n$$ in $$G$$. Add to $$G_{n-1}$$ a matching $$M$$ of $$\lfloor \frac{k}{2} \rfloor$$ edges between the $$u_i$$s.

(C) If there remains [at most] 1 vertex $$u_3$$ of degree 2 in the resulting graph, then contract $$u_3$$ to an edge $$e_{u_3}$$ and call the resulting graph $$G'_{n-1}$$. As $$G$$ is triangle free the resulting $$G'_{n-1}$$ is simple. Let $$C'_{n-1}$$ be a cycle so that $$G'_{n-1}\setminus E(C'_{n-1})$$ is connected.

We consider 2 possibilities:

Subcase B.1: $$C'_{n-1}$$ contains 2 or more arcs in $$M$$. Then let $$u_1$$ and $$u_2$$ be such that (i) $$u_1$$ and $$u_2$$ are incident to distinct edges in $$M \cap C'$$, and (ii) there is a path $$P'_{n-1}$$ from $$u_1$$ to $$u_2$$ in $$C'_{n-1}$$ that contains no arcs in $$M$$ and does not contain $$e_{u_3}$$. One can observe that there exists such a $$P'_{n-1}$$. Then let $$C_n=vu_1P'_{n-1}u_2v$$. Then $$G \setminus E(C_n)$$ is connected. [Indeed let us suppose that there is a vertex $$u_3$$ as in (C) above. Then let $$u$$ and $$w$$ be any 2 vertices in $$V(G) \setminus \{u_3,v_n\}$$, and let $$P'_{uw}$$ be a path in $$G'_{n-1} \setminus E(C'_{n-1})$$. Then if $$P'_{uw}$$ contains any arcs $$u_ju_{j+1}$$ in $$M$$ or $$e_{u_3}$$ then as $$M$$ is a matching $$\{u_j,u_{j+1}\} \cap \{u_1,u_2\}$$ is empty. So replace $$u_ju_{j+1}$$ with $$u_jv_nu_{j+1}$$ and $$e_{u_3}$$ with $$xu_3y$$ where $$xy \doteq e_{u_3}$$ and so the resulting walk $$W_{uw}$$ is in $$G \setminus E(C_n)$$ so $$u$$ and $$w$$ are connected to each other in $$G \setminus E(C_n)$$ for each $$u,w \in V(G) \setminus \{u_3,v_n\}$$. If $$P'_{uw}$$ does not contain any arc in $$M$$ or $$e_{u_3}$$ then $$P'_{uw}$$ is in $$G \setminus E(C_n)$$ and so $$u$$ and $$w$$ are connected to each other for each in $$G \setminus E(C_n)$$ for each $$u,w \in V(G) \setminus \{u_3,v_n\}$$. But then as $$e_{u_3} \not \in P'_{n-1}$$ it follows that $$xe_{u_3}y$$ is in $$G \setminus E(C_n)$$ so $$u_3$$ is in the same component as
every other vertex in $$V(G) \setminus \{u_3,v_n\}$$, and as $$v_n$$ has degree 3, there is an edge in $$G \setminus E(C_n)$$ between $$v_n$$ and another vertex, so $$v_n$$ is in this component as well. And so $$G \setminus E(C_n)$$ indeed has only 1 component. ]

Subcase B.2: $$C'_{n-1}$$ contains exactly 1 arc $$e=u_1u_2$$ in $$M$$ and $$e_{u_3}$$ as in (C) above exists and is in $$C'_{n-1}$$ as well. Then write $$e_{u_3} \doteq xy$$; $$x$$ and $$y$$ the endpoints of $$e_{u_3}$$ in $$G'_{n-1}$$; so that the path $$P'_{n-1}=xx_2x_3 \ldots x_lu_1$$ in $$C'_{n-1}$$ contains neither $$u_2$$ nor $$y$$. Then let $$P_n=u_3xx_2x_3 \ldots x_lu_1$$ and let $$C_n=v_nu_3xx_2 \ldots x_lu_1v_n$$. Then $$G \setminus E(C_n)$$ is connected. [Indeed let $$u$$ and $$w$$ be any 2 vertices in $$V(G) \setminus \{u_3,v_n\}$$, and let $$P'_{uw}$$ be a path in $$G'_{n-1} \setminus E(C'_{n-1})$$. Then if $$P'_{uw}$$ contains any arcs $$u_ju_{j+1}$$ in $$M$$ then $$\{u_j,u_{j+1}\} \cap \{u_1,u_2\}$$ is empty [as $$u_1u_2$$ is in $$C'_{n-1}$$ so it cannot be in $$P'_{uw}$$ and also $$u_1u_2 \in M$$ a matching], so replace with $$u_jv_nu_{j-1}$$ and so the resulting walk $$W'_{uw}$$ is in $$G \setminus E(C_n)$$ so $$u$$ and $$w$$ are connected to each other in $$G \setminus E(C_n)$$ for all $$V(G) \setminus \{u_3,v_n\}$$. But then the edge $$v_nu_2$$ is in $$G \setminus E(C_n)$$, and so is the edge $$u_3y$$.]

The remaining cases are easier and won't be covered here.

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