Precedence Scheduling with Unit Execution Time is Equivalent to Parametrized Biclique

Abstract

We consider the following scheduling problem. Given m machines and n jobs with unit execution times and a precedence relation between the jobs, the problem is to assign each job to one of the machines. The objective is to find a schedule that minimizes the makespan i.e. the length of the schedule. We reduce $$3\text {-CNF-SAT}$$ to this problem and obtain a new lower bound for the running time of $$2^{o\sqrt{n\log n}}$$ assuming the Expontential Time Hypothesis $$\text {ETH}$$. This improves the previous lower bound of $$2^{o\sqrt{n}}$$ also due to the $$\text {ETH}$$ and a reduction by Ullman [13] or, alternatively, a reduction from the k-Clique problem by Lenstra and Rinnooy Kan [10]. For the corresponding decision problem of whether there is a schedule with target makespan $$\mathbf{T }=3$$ or not, we further show the equivalence to a classical graph problem, the parametrized Biclique problem. The equivalence also holds for the same scheduling problem with the additional restriction that no job has both a predecessor and a successor. By this we show that an improved lower bound for the running time for the Biclique problem will lead to an improved lower bound for the running time for our scheduling problem and vice versa. Moreover a transfered lower bound for the running time from the Biclique problem would also hold for the running time of approximation algorithms with ratio better than $$\frac{4}{3}\text {OPT}$$. That is, if for example there was no algorithm solving Biclique in $$2^{on}$$ and U was the set of vertices in the Biclique problem, then there would be no approximation algorithm finding a solution for the introduced scheduling problem with $$\varTheta |U|$$ jobs, that finds a solution with a target makespan smaller than $$\frac{4}{3}$$ times the optimal makespan in time $$2^{on}$$.