2017
DOI: 10.1016/j.tcs.2016.11.034
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Parameterized complexity of Strip Packing and Minimum Volume Packing

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Cited by 4 publications
(4 citation statements)
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“…In this problem, we are given a set of n rectangles S, and positive integers k, W ∈ N, and the objective is to decide whether all the rectangles in S can be packed in a rectangle (called a strip) of dimensions k × W . In [5], it was shown that if the maximum of the dimensions of the input rectangles, denoted by ℓ, is fixed, then the problem is FPT by k. Thus, the question whether the problem is FPT parameterized by k + ℓ remained open. By a straightforward reduction, we resolve this question as a corollary of our theorem.…”
Section: Our Contribution and Main Proof Ideasmentioning
confidence: 99%
“…In this problem, we are given a set of n rectangles S, and positive integers k, W ∈ N, and the objective is to decide whether all the rectangles in S can be packed in a rectangle (called a strip) of dimensions k × W . In [5], it was shown that if the maximum of the dimensions of the input rectangles, denoted by ℓ, is fixed, then the problem is FPT by k. Thus, the question whether the problem is FPT parameterized by k + ℓ remained open. By a straightforward reduction, we resolve this question as a corollary of our theorem.…”
Section: Our Contribution and Main Proof Ideasmentioning
confidence: 99%
“…In this problem, we are given a set of n rectangles S, and positive integers k, W ∈ N, and the objective is to decide whether all the rectangles in S can be packed in a rectangle (called a strip) of dimensions k × W . In [5], it was shown that if the maximum of the dimensions of the input rectangles, denoted by , is fixed (i.e., a constant independent of the input), then the problem is FPT parameterized by k. Specifically, running time of 8 k n O( 2 ) W was attained, which is not FPT with respect to k + . Thus, the question whether the problem is FPT parameterized by k + remained open.…”
Section: Our Contribution and Main Proof Ideasmentioning
confidence: 99%
“…Assume that f (N i,j (2)) = (12(i − 1) + 9, 8(j − 1) + 5). Observe that N i,j (2) has four neighbors, N i,j (1), N i,j (3), N i,j (4) and N i,j (5), therefore one of them is embedded at (12(i − 1) + 10, 8(j − 1) + 5). This is a contradiction, since from Lemma 4.12 we get that f ((12(i − 1) + 10, 8(j − 1) + 5)) = (12(i − 1) + 10, 8(j − 1) + 5) and f is an injection.…”
Section: Lemma 411 Let G Be a Grid Graph And Let F Be A Grid Graph Em...mentioning
confidence: 99%
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