Metric travelling salesman problem

This is the travelling salesman problem on a complete graph with nonnegative symmetric distances satisfying the triangle inequality. Repeated visits can be shortcut without increasing cost. Assigning distance one to edges of an input graph and two to its nonedges creates a metric instance with a tour of cost at most the number of vertices exactly when the graph has a Hamiltonian cycle. Thus the metric problem remains NP-hard.

Double every edge of a minimum spanning tree to obtain a connected even-degree multigraph, traverse an Euler circuit, and shortcut repeated vertices. The triangle inequality prevents any shortcut from increasing cost. Deleting an edge of an optimal tour gives a spanning tree, so the minimum spanning tree has cost at most the optimal tour. The returned tour therefore costs at most twice optimum and is computable in polynomial time. In the relative-error convention this is an error-one approximation.

Compute a minimum spanning tree, then a minimum-weight perfect matching on its odd-degree vertices. Their multiset union is connected with even degrees, so it has an Euler circuit. Shortcut repeated vertices to obtain a metric tour. The tree costs at most the optimal tour, and shortcutting the optimal tour to its odd-degree subset gives a cyclic order whose alternating matchings show that the added matching costs at most half the optimum. The result is a polynomial-time $3/2$-approximation algorithm for symmetric metric instances.