From approximate to exact integer programming

Abstract

Approximate integer programming is the following: For a given convex body \(K \subseteq {\mathbb {R}}^n\), either determine whether \(K \cap {\mathbb {Z}}^n\) is empty, or find an integer point in the convex body \(2\cdot (K - c) +c\) which is K, scaled by 2 from its center of gravity c. Approximate integer programming can be solved in time \(2^{O(n)}\) while the fastest known methods for exact integer programming run in time \(2^{O(n)} \cdot n^n\). So far, there are no efficient methods for integer programming known that are based on approximate integer programming. Our main contribution are two such methods, each yielding novel complexity results. First, we show that an integer point \(x^* \in (K \cap {\mathbb {Z}}^n)\) can be found in time \(2^{O(n)}\), provided that the remainders of each component \(x_i^* \mod \ell \) for some arbitrarily fixed \(\ell \ge 5(n+1)\) of \(x^*\) are given. The algorithm is based on a cutting-plane technique, iteratively halving the volume of the feasible set. The cutting planes are determined via approximate integer programming. Enumeration of the possible remainders gives a \(2^{O(n)}n^n\) algorithm for general integer programming. This matches the current best bound of an algorithm by Dadush (Integer programming, lattice algorithms, and deterministic, vol. Estimation. Georgia Institute of Technology, Atlanta, 2012) that is considerably more involved. Our algorithm also relies on a new asymmetric approximate Carathéodory theorem that might be of interest on its own. Our second method concerns integer programming problems in equation-standard form \(Ax = b, 0 \le x \le u, \, x \in {\mathbb {Z}}^n\). Such a problem can be reduced to the solution of \(\prod _i O(\log u_i +1)\) approximate integer programming problems. This implies, for example that knapsack or subset-sum problems with polynomial variable range \(0 \le x_i \le p(n)\) can be solved in time \((\log n)^{O(n)}\). For these problems, the best running time so far was \(n^n \cdot 2^{O(n)}\).