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A minimum-spanning-tree (MST) path is always $V-1$ edges and a Hamiltonian Cycle (HC) is always $V$ edges.

Because the HC has an extra edge we could say that in general, the weight of every Hamiltonian cycle of a connected graph will be more.

If we take an edge of every Hamiltonian cycle we will find an MST and by that, the MST weight will be always lower than HC.

  • There is a way to prove this where MST tree not necessary belongs to the HC?
  • There is more definitive proof for this?
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  • $\begingroup$ Possibly a typo: "Because the HC has an extra edge we could say that in general, the weight of every Hamiltonian cycle of a connected graph will be less." Did you mean 'more' instead of 'less' ? $\endgroup$ – lox Jun 9 at 17:33
  • $\begingroup$ Thanks for the correction! edited the question. $\endgroup$ – user2668676 Jun 9 at 17:51
  • $\begingroup$ Hint: if you remove any edge from a Hamiltonian cycle, you get a spanning tree. $\endgroup$ – Yuval Filmus Jun 9 at 20:53
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Suppose:

  • $G$ is some graph with non-zero, positive weights.

  • The minimum spanning tree of $G$ is denoted by $MST$

  • $H_C$ is a hamiltonian cycle in $G$.

  • $H_P$ is a hamiltonian path in $G$, given by removing any one edge of $H_C$.

Since all the edges are positive and non-zero, we immediately get that:

$$w(H_C) > w(H_P)$$

Also, $H_P$ is a subgraph that includes all vertices of $G$ (and connects them), so $H_P$ in fact spans $G$.

Hence, by definition of minimum spanning tree:

$$w(H_P) \geq w(MST)$$

It follows that $$w(H_C) > w(H_P) \geq w(MST)$$


In case we allow zero weights, the inequality holds, it simply will not be strictly greater (for the same reasons noted in the answer). That is: $$w(H_C) \geq w(H_P) \geq w(MST)$$

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