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The following theorem is well known.

Theorem 1. If $G$ is a graph containing a set $S \subset V(G)$ such that $G-S$ has more than $|S|$ components, then $G$ is not Hamiltonian.

We know the converse of Theorem 1 is false. We can find some examples. But can we find examples with any connectivity that show it wrong?

Let $c(G)$ denote the number of components of a graph $G$. A graph $G$ is tough if the number of $c(G-S) \le |S|$ for all $S \subset V(G)$.

That is to say, we aim to find non-Hamiltonian tough graphs with connectivity $k$ where $k$ can be arbitrarily large.

For $k=2$, we can find the following graph. But for $k\ge 3$, can we continue?

enter image description here

licheng
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    For $k=3$, the Petersen graph. – Misha Lavrov Jul 12 '23 at 12:28
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    For $k = 1$ we have $P_2$. (Yes, it is very special case.) – Smylic Jul 12 '23 at 16:24
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    Even for $k = 0$ we have $2K_2$, why not? – Smylic Jul 12 '23 at 16:27
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    @Smylic Very good, you are correct. I have made the corrections in the post. I am more concerned about whether an example can be constructed for any general $k$. In fact, if there isn't, perhaps for specific connectivity, this condition would then become a necessary and sufficient condition. I would rather believe that these examples can be constructed. – licheng Jul 13 '23 at 02:56

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Consider a graph $H \cong K_{\ell} + (\ell - 1)K_1$ for $\ell \ge 3$ that has $\ell$ universal vertices and $(\ell - 1)$ vertices with no neighbors other than these universal vertices. Now replace every vertex of your example $G$ by a copy of $H$, and every edge by $\ell^2$ edges between the universal vertices of the corresponding copies of $H$. Now we have a graph $G'$ with $8(2\ell - 1)$ vertices and $11 \ell^2 + 12\ell(\ell - 1)$ edges.

The graph $G'$ is $\ell$-connected, because $H$ is $\ell$-connected and $\ell$ vertices of each copy of $H$ are adjacent to at least $\ell$ vertices of other copies (and yes, graph $G$ is connected).

The graph $G'$ is non-Hamiltonian. Suppose the opposite, that there is a Hamiltonian cycle $C'$. Then vertices of each copy of $H$ make exactly one contiguous subsequence of cyclic sequence of vertices corresponding to $C'$. This comes from two facts: 1) each of $(\ell - 1)$ non-universal vertices of the copy of $H$ is surrounded by two universal vertices of the same copy; 2) the number of universal vertices is greater than the number of non-universal vertices exactly by $1$. Thus we can contract vertices of each copy of $H$ into one, corresponding to a vertex of $G$. Note that adjacent vertices of different copies of $H$ correspond to adjacent vertices of $G$. Therefore the Hamiltonian cycle $C'$ of graph $G'$ turns to a Hamiltonian cycle $C$ of the graph $G$. This controversy proves the initial statement.

And the last (but not the least), $c(G' - S') \le |S'|$ for any $S' \subset V(G')$. If less than $\ell$ vertices are deleted from a copy of $H$ then it remains connected itself and connected to the same other copies of $H$. Removing $\ell$ universal vertices of a copy of $H$ produces $\ell - 1$ isolated vertices and has (in sense of components) the same effect on the remaining graph as removing a vertex of $G$ from $G$. So every $S' \subset V(G')$ corresponds to some $S \subset V(G)$. Therefore $$\begin{aligned}c(G' - S') &\le (\ell - 1) \cdot \left\lfloor\frac{|S'|}{\ell}\right\rfloor + c(G - S)\\ &\le (\ell - 1) \cdot \left\lfloor\frac{|S'|}{\ell}\right\rfloor + |S|\\ &\le (\ell - 1) \cdot \left\lfloor\frac{|S'|}{\ell}\right\rfloor + \left\lfloor\frac{|S'|}{\ell}\right\rfloor\\ &= \ell \cdot \left\lfloor\frac{|S'|}{\ell}\right\rfloor \le |S'|.\end{aligned}$$

Smylic
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