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.

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