Spanning closed walks and TSP in 3-connected planar graphs

• Authors: Ken-ichi Kawarabayashi;Kenta Ozeki
• Venue: Proceedings of the twenty-third annual ACM-SIAM symposium on Discrete Algorithms
• Year: 2012

We consider the following problem which is motivated by two different contexts independently, namely graph theory and combinatorial optimization. Given a 3-connected planar graph G with n vertices, is there a spanning closed walk W with at most 4n/3 edges? In graph theory, the above question is motivated by the famous hamiltonian result by Tutte in 1956 which says that every 4-connected planar graph is hamiltonian (a simpler proof is given by Thomassen in 1983). What happens if we relax the 4-connectivity? There is a 3-connected planar graph that is not hamiltonian, but how about a spanning close walk (which is exactly a traveling salesman tour. Sometimes such a walk is called hamiltonian walk)? How many edges are necessary to cover all the vertices of a 3-connected planar graph by a closed walk? This is exactly the above question. In combinatorial optimization, the famous traveling salesman problem in metric graphs is one of most fundamental NP-hard optimization problems. In spite of a vast amount of research several important questions remain open. In particular, the best known upper bound is not believed to be best possible. A promising direction to improve this approximation guarantee, has long been to understand the power of a linear program known as the Held-Karp relaxation [11]. On the one hand, the best lower bound on its integrality gap (for the symmetric case) is 4/3 and indeed the famous (so called 4/3-)conjecture said that this lower bound would be tight [10]. Goemans pointed out that there is a planar graph that achieves this bound. So he brought attention to the above question, i.e, the famous 4/3-conjecture is always true for 3-connected planar graphs. We prove the above problem in the following strong form; Given a circuit graph (which is obtained from a 3-connected planar graph by deleting one vertex) with n vertices, there is a spanning closed walk with at most 4(n − 1)/3 edges such that each edge is used at most twice. Moreover, our proof is constructive (and purely combinatorial) in a sense that there is an O(n2) algorithm to construct, given a 3-connected planar graph, such a walk. We shall construct an example that shows that the bound 4(n − 1)/3 is essentially tight. We also point out that 2-connected planar graphs may not have such a walk, as K2,n−2 shows.