2015
DOI: 10.1016/j.orl.2015.01.008
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On the lattice programming gap of the group problems

Abstract: Abstract. Given a full-dimensional lattice Λ ⊂ Z k and a cost vector l ∈ Q k >0 , we are concerned with the family of the group problemsThe lattice programming gap gap(Λ, l) is the largest value of the minima in (0.1) as r varies over Z k . We show that computing the lattice programming gap is NP-hard when k is a part of input. We also obtain lower and upper bounds for gap(Λ, l) in terms of l and the determinant of Λ. Introduction and statement of resultsConsider the integer programming problemGomory [11] defi… Show more

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Cited by 3 publications
(8 citation statements)
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“…Here b ranges over all integers such that (5) is feasible and bounded. Notice that computing Gap(c, a) when n is a part of input is NP-hard (see Aliev [1] and Eisenbrand et al [9]). For any fixed n, the integer programming gap can be computed in polynomial time due to results of Hoşten and Sturmfels [16] (see also Eisenbrand and Shmonin [10]).…”
Section: Introductionmentioning
confidence: 99%
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“…Here b ranges over all integers such that (5) is feasible and bounded. Notice that computing Gap(c, a) when n is a part of input is NP-hard (see Aliev [1] and Eisenbrand et al [9]). For any fixed n, the integer programming gap can be computed in polynomial time due to results of Hoşten and Sturmfels [16] (see also Eisenbrand and Shmonin [10]).…”
Section: Introductionmentioning
confidence: 99%
“…Here b ranges over all integers such that (5) is feasible and bounded. Notice that computing Gap(c, a) when n is a part of input is NP-hard (see Aliev [1] and Eisenbrand et al [9]).…”
Section: Introductionmentioning
confidence: 99%
“…Appendix: Proof of Theorem 5We will denote for A ∈ Q(T ) the index of a maximum coordinate by i(A) and we set c A = −e i(A) . The tuples (A, c A ) are generic and in view of Theorem 1.8 we findGap c A (A) ≥ ρ n−1 a n T n−1 T 1−1/(n−1) ≤ 2 n T n−1/(n−1) …”
mentioning
confidence: 65%
“…Note that for n = 2 we have Gap(Λ A , l) = l 1 (|A τ | − 1) and thus (4.4) implies (1.8). For n > 2, the bound (1.8) immediately follows from (4.4) and Theorem 1.2(i) in [1].…”
Section: Proof Of Theoremmentioning
confidence: 85%
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