NP-Hardness Results for Some Linear and Quadratic ... - CiteSeerX

NP-Hardness Results for Some Linear and Quadratic Problems∗ Jiˇr´ı Rohn†

Abstract Several problems concerning norms, linear inequalities, linear equations, linear programming and quadratic programming are proved to be NP-hard.

Key words. Norm, linear inequalities, linear equations, linear programming, quadratic programming, NP-hardness

1

Introduction

The first part of this report (sections 2 to 5) was originally made as a transcript of transparencies of seminar talks1 . Improvements and consequences found shortly after the transcription had been completed were added as Appendices 1 to 4. In this rather incoherent form, the main result is Theorem 2, supported by Proposition 2 (already known in a slightly different setting). Among other consequences, it is shown that computing kAk∞,1 within accuracy 12 is NP-hard (Corollary 9), which in turn implies that the same is true for computing the maximal value of a convex quadratic program (Corollary 11) and for one of the two bounds on the optimal value of a linear program with inexact right-hand side (Corollary 12). Another result (Corollary 3) shows that checking sensitivity of a system of linear equations is an NP-hard problem. ˇ This work was supported in part by the Czech Republic Grant Agency under grant GACR 201/95/1484 † Institute of Computer Science, Academy of Sciences of the Czech Republic, Prague ([email protected]), and Faculty of Mathematics and Physics, Charles University, Prague, Czech Republic ([email protected]) 1 held in Prague, November 1994, and in Leipzig, December 1994 ∗

1

2

M C-matrices

The following concept will be used as a basic tool throughout this report: Definition A real symmetric n × n matrix A = (aij ) is called an M C-matrix2 if it is of the form ( =n if i = j aij ∈ {0, −1} if i 6= j (i, j = 1, . . . , n). Proposition 1 If A is an M C–matrix, then A−1 is nonnegative and symmetric positive definite. Proof. By definition, A is of the form A = nI − A0 = n(I − where A0 ≥ 0 and k n1 A0 k∞ ≤

n−1 n

< 1, hence

−1

A

1 A0 ) n

∞ 1X 1 = ( A0 )j ≥ 0. n 0 n

A is symmetric by definition; it is positive definite since for x 6= 0, xT Ax ≥ nkxk22 −

X

|xi xj | = (n + 1)kxk22 − kxk21 ≥ kxk22 > 0.

i6=j

Hence A−1 is also symmetric and positive definite.

2

The next result is due to Poljak and Rohn [8] (given there in a slightly different formulation without using the concept of an M C-matrix). We add the proof for completeness. Proposition 2 The following decision problem is NP-complete: Instance. An M C-matrix A and a positive integer L. Question. Is z T Az ≥ L for some z ∈ {−1, 1}n ? Proof. Let (N, E) be a graph with N = {1, . . . , n}. Let A = (aij ) be given by aij = 2

   n

if i = j 0 if {i, j} ∈ / E, i 6= j   −1 if {i, j} ∈ E, i = 6 j

from ”maximum cut”; explained in the proof of Proposition 2

2

then A is an M C–matrix. For S ⊆ N , define a cut by c(S) = Card{{i, j} ∈ E; exactly one of i, j is in S}. If z is given by

(

1 if k ∈ S −1 if k ∈ /S

zk = then

1 c(S) = (z T Az + 2Card(E) − n2 ), 4 hence c(S) ≥ L if and only if z T Az ≥ 4L − 2Card(E) + n2 . Since the problem ”c(S) ≥ L” (maximum cut in a graph) is NP-complete (Garey and Johnson [1]), the current problem is NP-hard. It is obviously in the class NP, since a guessed solution z can be verified in polynomial time; hence it is NP-complete. 2

3

The result

Theorem 1 below forms a common basis for several NP-hardness results listed in the next section. Proposition 3 Let A be an M C–matrix and L a positive integer. Then z T Az ≥ L holds for some z ∈ {−1, 1}n if and only if the system −e ≤ LA−1 x ≤ e has a solution satisfying kxk1 ≥ 1 (where e = (1, 1, . . . , 1)T and kxk1 =

P i

|x|i ).

Proof. ⇒: Let z T Az ≥ L. Put x=

Az z T Az 3

then

¯ ¯ ¯ Lz ¯ ¯ ≤ |z| = e ¯ T

|LA−1 x| = ¯¯

z Az

and

eT |Az| z T Az = = 1. z T Az z T Az ⇐: If |LA−1 x| ≤ e and kxk1 ≥ 1, then for z given by zi = 1 if xi ≥ 0 and zi = −1 otherwise we have kxk1 =

L ≤ Lkxk1 = Lz T x = Lz T AA−1 x ≤ |z T A|e = z T Az. 2 Theorem 1 The following decision problem is NP–complete: Instance. A nonnegative symmetric positive definite rational matrix A. Question. Does the system −e ≤ Ax ≤ e (where e = (1, 1, . . . , 1)T ) have a solution satisfying kxk1 ≥ 1 ? Proof. According to Propositions 2 and 3, the NP-complete problem ”z T Az ≥ L” can be polynomially reduced to this one (if A is an M C–matrix, then LA−1 is nonnegative symmetric positive definite), hence the current problem is NP-hard. If the problem has a solution, then it also has a rational solution of the form x=

Az z T Az

(proof of Proposition 3) which can be checked in polynomial time; thus the problem belongs to the class NP, hence it is NP-complete. 2

4

Corollaries

The following five corollaries are direct consequences of Theorem 1. The instances are always assumed to be rational without further notice. Corollary 1 The following problem is NP-hard: Instance. A ∈ Rm×n , b ∈ Rm , m ≥ 2n, L positive integer. Question. Does each solution of the system Ax ≤ b satisfy kxk1 < L ? 4

Corollary 2 The following problem is NP-hard: Instance. A, B ∈ Rn×n , b ∈ Rn . Question. Does the system Ax + B|x| ≤ b have a solution? Corollary 3 The following problem is NP-hard: Instance. A nonnegative symmetric positive definite A ∈ Rn×n , b ∈ Rn , δ > 0, ² > 0; denote x = A−1 b. Question. Does the solution of each Ax0 = b0 with kb0 − bk∞ < δ satisfy kx0 − xk1