Assume a graph $(V,E)$ with $n$ vertices and degree sequence $d_1 \geq d_2 \geq \dots \geq d_n$.

The purpose of this question is to understand when the graph join $G^{(k)}$ (defined as the union of $k$ copies of $G$ with all the edges between vertices that belong to different copies of $G$ present) is Hamiltonian.

$G^{(k)}$ has $kn$ vertices with degree sequence:

$$(d_1 + (k-1)n, \dots ,d_1 + (k-1)n, d_2 + (k-1)n, \dots, d_2 + (k-1)n, \dots, d_n + (k-1)n, \dots, d_n + (k-1)n)$$ (each $d_i$ is repeated $k$ times).

Using Dirac's theorem, if

$$ d_i + (k-1)n \geq \frac{kn}{2}, \, i =1,2,\dots,n \iff d_i \geq \frac{n(2-k)}{2}, \, i =1,2,\dots,n $$

then $G^{(k)}$ is Hamiltonian. But $k \geq 2$, so the inequality is always true.

There's probably a mistake somewhere, 'cause this would imply that every graph join is Hamiltonian.


Isn't it Hamiltonian? Let $v_{i,j}$ be the copy of vertex $i$ in copy $j$ of $G$. Then we have the path$$ v_{1,1},v_{1,2},\dots, v_{1,k},v_{2,1},v_{2,2},\dots,v_{2,k}\dots,v_{n,k},v_{1,1}$$ We must have either $n>1$ or $k>2$ so that $G^{(k)}$ has at least $3$ vertices for Dirac's theorem to apply. In the case where $n=1$ and $k=2$, $G^(2)$ has a single edge, so is not Hamiltonian.


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