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Consider vertex $v$ in graph $G$. Let $v$ have at least three adjacent vertices with degrees of two. Prove that $G$ is not a Hamiltonian graph.

Proof

Suppose $G$ is Hamiltonian.

Let $x$, $y$ and $z$ be the vertex adjacent to $v$.

We know that $G$ contains at least 4 vertices: $v$, $x$, $y$ and $z$.

$G$ must contain at least one more vertex $w$ because otherwise the sum of all degrees ($2+2+2+3=9$) would not equal twice the number of edges ($2\cdot 4=8$).

Now, there cannot be an edge between any of the edges $x$, $y$ and $z$, because that would create a cycle, e.g. $vxyv$. If we tried to construct a Hamiltonian cycle, there would be no way to "exit" without passing through the same vertex twice and whatever we try to do (i.e add more vertices), this problem still occurs.

Thus $G$ cannot be Hamiltonian.

Another approach I tried

Suppose $G$ is Hamiltonian with $n$ vertices.

Let $x$, $y$ and $z$ be the vertex adjacent to $v$.

From Dirac's theorem it follows that each vertex must have a degree of $\frac{n}{2}$ and therefore for all pairs of adjacent vertices the sum of their degrees must equal at least $n$.

$G$ must contain at least one more vertex $w$ because otherwise the sum of all degrees ($2+2+2+3=9$) would not equal twice the number of edges ($2\cdot 4=8$).

So $n\geq5$ and therefore there can be no edges between any of the pairs $x$, $y$ and $z$, because $2+2<5$.

...

Now there should be a way to reach a contradiction here but I don't really know how.

Any help or tips are much appreciated!

RobPratt
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t-y
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  • You only considered the case where you have a cycle $vxyv$. What if there is none ? Try to find how many edges adjacent to $v$ must belong to the hamiltonian path. – caduk May 11 '22 at 09:08
  • Isn’t the case exactly the same, no matter what vertex (x, y or z) I choose?

    If there is no cycle, that would contradict the fact that x, y and z have degrees of two? They all are adjacent to v and then some other vertex (i.e not x, y or z). If we add a vertex (w in this case), it must be connected to some other vertex, since otherwise we would not be able to construct a Hamiltonian cycle?

     – t-y May 11 '22 at 09:49
  • Then you would get a longer cycle than just $vxyv$, which could have some vertices with degree greater than 2, so there would be some way to "exit" the cycle... – caduk May 11 '22 at 12:24

1 Answers1

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If $\ x,y,z\ $ are adjacent to $\ v\ $ and have degree two, then each of them is adjacent to exactly one other vertex, $\ x\ $ to $\ x'\ne v\ $, $\ y\ $ to $\ y'\ne v\ $ and $\ z\ $ to $\ z'\ne v\ $, say. Now each vertex of a Hamiltonian cycle must be incident on exactly two of its edges, and since any Hamiltonian cycle of $\ G\ $ must pass through $\ x\ $, $\ y\ $ and $\ z\ $, it must therefore include both of the edges incident on each of them. In particular, it must include all three of the edges $\ \{x,v\}\ $, $\ \{y,v\}\ $ and $\ \{z,v\}\ $. But this would require $\ v\ $ to be incident on at least three edges of the cycle, which is impossible.

lonza leggiera
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