A polynomial relaxation-type algorithm for linear programming

The paper proposes a polynomial algorithm for solving systems of linear inequalities. The algorithm uses a polynomial relaxation-type procedure which either finds a solution for Ax = b,0 ≤ x ≤ 1, or decides that the system has no integer solutions. 1 Polynomial algorithm The polynomial algorithm described in [2] either solves a linear system Ax = b,0 ≤ x ≤ 1, or decides that it has no 0,1-solutions. Here, 0 and 1 denote all-zero and all-one vectors, respectively. The core of the algorithm is a divide-and-conquer procedure which may be considered as a generalization of the relaxation method (see [1] and [5]). This short paper describes a way of converting the mentioned polynomial algorithm, which is further called the LP algorithm, into a polynomial algorithm for solving linear systems. This approach substantially differs from well known polynomial algorithms like the ellipsoid method by Khachiyan [4] and the projective algorithm by Karmarkar [3]. 1 Consider a linear system Ax = b,0 ≤ x ≤ u, (1.1) where all coefficients are integer. Without loss of generality we assume that uj > 0 for all j = 1, . . . , n. Let us introduce additional variables ξ and x̄ = (x̄1, . . . , x̄n) T and consider the system Ax− bξ = b, x̄+ x− uξ = u, 0 ≤ x̄ ≤ (1 + T )u, 0 ≤ x ≤ (1 + T )u, 0 ≤ ξ ≤ T , (1.2) where T is n! times absolute value of the product of all components of A, b, and u. The size of T is polynomially bounded. Lemma 1.1 The system (1.2) has an integer solution if and only if (1.1) is feasible. Proof. Indeed, if (x∗, x̄∗, ξ∗) is an integer solution of (1.2), then 1 1+ξ∗ x∗ is feasible for (1.1). Now let us assume that the system (1.1) is feasible. Then (1.1) defines a bounded polyhedron. Consider a vertex x of this polyhedron. Since all coefficients of the system are integer, Cramer’s theorem implies that x is representable as x = ( x1 x′′ 1 , . . . , x ′ n x′′ n )T where |xj| and |x′′ j | are integers bounded by T. Let ξ∗ = | ∏n l=1 x ′′ l | − 1. The ξ∗ is a nonnegative integer bounded by T . Let us set x∗ := (1 + ξ∗)x# and x̄∗ := (1 + ξ∗)u− x∗. We have Ax∗ = (1 + ξ∗)b because Ax = b. The vector x∗ is nonnegative because ξ∗ ≥ 0 and x ≥ 0. Moreover, it is integer because (1 + ξ)|xj | = | n ∏ l=1 x′′ l | · |xj| |x′′ j | = | n ∏ l 6=j x′′ l | · |xj|.