From Approximate to Optimal Solutions: A Case Study of Number Partitioning

Given a set of numbers, the two-way partitioning problem is to divide them into two subsets, so that the sum of the numbers in each subset are as nearly equal as possible. The problem is NP-complete, and is contained in many scheduling applications. Based on a polynomial-time heuristic due to Karmarkar and Karp, we present a new algorithm, called Complete Karmarkar Karp (CKK), that optimally solves the general number-partitioning problem. CKK significantly outperforms the best previously-known algorithms for this problem. By restricting the numbers to twelve significant digits, we can optimally solve two-way partitioning problems of arbitrary size in practice. CKK first returns the Karmarkar-Karp solution, then continues to find better solutions as time allows. Almost five orders of magnitude improvement in solution quality is obtained within a minute of running time. Rather than building a single solution one element at a time, CKK constructs subsolutions, and combines them in all possible ways. CKK is directly applicable to the 0/1 knapsack problem, since it can be reduced to number partitioning. This general approach may also be applicable to other NP-hard problems as well. 1 Introduction and Overview Consider the following very simple scheduling problem. We are given two identical machines, a set of jobs, and the time required to process each job on either machine. Assign each job to one of the machines, in order to complete all the jobs in the shortest elapsed time. In other words, divide the job processing times into two subsets, so that the sum of the times in each subset are as nearly equal as possible. This is the two-way number partitioning problem, and is NP-complete[l]. The generalization to K-way partitioning with K machines is straightforward , with the cost function being the difference between the largest and smallest subset sums. This basic problem is likely to occur as a subproblem in many practical scheduling applications. For example, consider the set of numbers (4, 5, 6, 7, 8). If we divide it into the two subsets (7,8) and (4,5,6), the sum of each subset is 15, and the difference of the subset sums is zero. In addition to being optimal, this is also a perfect partition. Note that if the sum of all the numbers is odd, a perfect partition will have a subset difference of one. We first present previous work on this problem, including a polynomial-time greedy heuristic, and an exponential-time algorithm …