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A minimum spanning tree of a finite connected undirected weighted graph minimizes the sum of edge weights over all spanning trees. Kruskal's algorithm repeatedly accepts a least-weight edge joining different current connected components. The cut-exchange argument preserves an optimum containing every accepted edge, including when weights tie.

For nonnegative edge costs, an MST gives a lower bound on a travelling salesman tour: removing one tour edge leaves a spanning path whose cost is no greater than the tour cost, and the minimum spanning tree costs no more than that path.

Nonnegativity is needed for this comparison. A triangle with all three edge weights equal to minus one has tour cost $-3$ and minimum spanning tree cost $-2$, so the latter is not a lower bound. The MST optimization problem itself permits negative edge weights.

A travelling salesman tour is a cyclic visiting order that visits each required vertex exactly once before returning to its start. In a complete graph, it is a Hamiltonian cycle. For an ordering $\pi$ of $n$ vertices and costs $c_{ij}$, its length is ${\sum }_{k=0}^{n-1}{c}_{\pi (k),\pi ((k+1)\mathrm{mod}n)}$.

In a symmetric problem, rotations and reversal represent the same undirected tour, giving $(n-1)!/2$ tours for $n\ge 3$. Thus four labeled locations have three distinct tours under this convention. Asymmetric costs generally distinguish a tour from its reversal. A closed walk allowed to revisit vertices is a different feasible object unless the problem explicitly permits that change.