  A long-standing conjecture for the traveling salesman problem (TSP) states
that the integrality gap of the standard linear programming relaxation of the
TSP is at most 4/3. Despite significant efforts, the conjecture remains open.
  We consider the half-integral case, in which the LP has solution values in
$\{0, 1/2, 1\}$. Such instances have been conjectured to be the most difficult
instances for the overall four-thirds conjecture. Karlin, Klein, and Oveis
Gharan, in a breakthrough result, were able to show that in the half-integral
case, the integrality gap is at most 1.49993. This result led to the first
significant progress on the overall conjecture in decades; the same authors
showed the integrality gap is at most $1.5- 10^{-36}$ in the non-half-integral
case. For the half-integral case, the current best-known ratio is 1.4983, a
result by Gupta et al.
  With the improvements on the 3/2 bound remaining very incremental even in the
half-integral case, we turn the question around and look for a large class of
half-integral instances for which we can prove that the 4/3 conjecture is
correct.
  The previous works on the half-integral case perform induction on a hierarchy
of critical tight sets in the support graph of the LP solution, in which some
of the sets correspond to "cycle cuts" and the others to "degree cuts". We show
that if all the sets in the hierarchy correspond to cycle cuts, then we can
find a distribution of tours whose expected cost is at most 4/3 times the value
of the half-integral LP solution; sampling from the distribution gives us a
randomized 4/3-approximation algorithm. We note that the known bad cases for
the integrality gap have a gap of 4/3 and have a half-integral LP solution in
which all the critical tight sets in the hierarchy are cycle cuts; thus our
result is tight.

Data Structures and Algorithms;Optimization and Control