Comment on ‘the strong NP-completeness of 3-PARTITION problem with B⩾k^m’ by Zhongyi Jiang, Fangfang Chen, Chunqing Wu

The note of Jiang and colleagues is concerned with the following problem.

3-PARTITION: Given 3m+1 positive integer numbers x₁, …, x_3m and B such that ∑_j=1^3mx _j =mB and B/4< x _j < B/2 for j=1, …, 3m, is there a partition of the set A={1, …, 3m} into m disjoint subsets A _l such that for l=1, …, m?

The main result of the authors is Theorem 1 stating that 3-PARTITION is NP-hard in the strong sense even if B⩾k^m, where k⩾2 is a given integer. This result is incorrect. In the proof, Jiang and colleagues use a series of transformations of 3-PARTITION, each time obtaining a new problem by resetting B:=kB and x _j :=kx _j , j=1, …, 3m. In the first new problem, B and x _j are multiples of k; in the second new problem, they are multiples of k² and so on. Each new problem is a special case of 3-PARTITION. It remains NP-hard in the strong sense until B and x _j are multiples of k^c, where c is a given positive integer. However, passing from a problem with B and x _j being multiples of k^c to a problem with B and x _j being multiples of k^m cannot be done by using multiplier k, and their proof fails. This passage needs multiplier k^m−c, but then the reduction is not pseudopolynomial.

Furthermore, if the logic of the authors is correct, then 𝒫=𝒩𝒫. Indeed, substituting k and m by m and 3m, respectively, does not change the logic of the proof of Theorem 1. Then the statement of this theorem would be ‘3-PARTITION is NP-hard in the strong sense even if B⩾m^3m’. However, 3-PARTITION can be solved in O(m^3m) time by full enumeration. This means that the strongly NP-hard problem is solvable in O(B) time, which is pseudopolynomial, and 𝒫=𝒩𝒫.