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Register a new userPath Puzzles: Discrete Tomography with a Path Constraint is Hard
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Added by
Jeffrey Bosboom
Publication date
2018
fields
Informatics Engineering
and research's language is
English
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We prove that path puzzles with complete row and column information--or equivalently, 2D orthogonal discrete tomography with Hamiltonicity constraint--are strongly NP-complete, ASP-complete, and #P-complete. Along the way, we newly establish ASP-completeness and #P-completeness for 3-Dimensional Matching and Numerical 3-Dimensional Matching.
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We study the complexity of symmetric assembly puzzles: given a collection of simple polygons, can we translate, rotate, and possibly flip them so that their interior-disjoint union is line symmetric? On the negative side, we show that the problem is strongly NP-complete even if the pieces are all polyominos. On the positive side, we show that the problem can be solved in polynomial time if the number of pieces is a fixed constant.
In this paper, we show that deciding rigid foldability of a given crease pattern using all creases is weakly NP-hard by a reduction from Partition, and that deciding rigid foldability with optional creases is strongly NP-hard by a reduction from 1-in-3 SAT. Unlike flat foldability of origami or flexibility of other kinematic linkages, whose complexity originates in the complexity of the layer ordering and possible self-intersection of the material, rigid foldability from a planar state is hard even though there is no potential self-intersection. In fact, the complexity comes from the combinatorial behavior of the different possible rigid folding configurations at each vertex. The results underpin the fact that it is harder to fold from an unfolded sheet of paper than to unfold a folded state back to a plane, frequently encountered problem when realizing folding-based systems such as self-folding matter and reconfigurable robots.
We introduce a computational origami problem which we call the segment folding problem: given a set of $n$ line-segments in the plane the aim is to make creases along all segments in the minimum number of folding steps. Note that a folding might alter the relative position between the segments, and a segment could split into two. We show that it is NP-hard to determine whether $n$ line segments can be folded in $n$ simple folding operations.
Given a set P of n points in the plane, a unit-disk graph G_{r}(P) with respect to a radius r is an undirected graph whose vertex set is P such that an edge connects two points p, q in P if the Euclidean distance between p and q is at most r. The length of any path in G_r(P) is the number of edges of the path. Given a value lambda>0 and two points s and t of P, we consider the following reverse shortest path problem: finding the smallest r such that the shortest path length between s and t in G_r(P) is at most lambda. It was known previously that the problem can be solved in O(n^{4/3} log^3 n) time. In this paper, we present an algorithm of O(lfloor lambda rfloor cdot n log n) time and another algorithm of O(n^{5/4} log^2 n) time.
Deciding whether a family of disjoint line segments in the plane can be linked into a simple polygon (or a simple polygonal chain) by adding segments between their endpoints is NP-hard.
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