منابع مشابه
Packing anchored rectangles
Let S be a set of n points in the unit square [0, 1], one of which is the origin. We construct n pairwise interior-disjoint axis-aligned empty rectangles such that the lower left corner of each rectangle is a point in S, and the rectangles jointly cover at least a positive constant area (about 0.09). This is a first step towards the solution of a longstanding conjecture that the rectangles in s...
متن کاملPacking Boundary-Anchored Rectangles
In this paper, we study the boundary-anchored rectangle packing problem in which we are given a set P of points on the boundary of an axis-aligned squareQ. The goal is to find a set of disjoint axis-aligned rectangles in Q such that each rectangle is anchored at some point in P , each point in P is used to anchor at most one rectangle, and the total area of the rectangles is maximized. We show ...
متن کاملOn Packing Almost Half of a Square with Anchored Rectangles: A Constructive Approach
In this paper, we consider the following geometric puzzle whose origin was traced to Allan Freedman [13, 22] in the 1960s by Dumitrescu and Tóth [10]. The puzzle has been popularized of late by Peter Winkler [23]. Let Pn be a set of n points, including the origin, in the unit square U = [0, 1] . The problem is to construct n axis-parallel and mutually disjoint rectangles inside U such that the ...
متن کاملPacking Rectangles into 2OPT Bins Using Rotations
We consider the problem of packing rectangles into bins that are unit squares, where the goal is to minimize the number of bins used. All rectangles can be rotated by 90 degrees and have to be packed nonoverlapping and orthogonal, i.e., axis-parallel. We present an algorithm for this problem with an absolute worst-case ratio of 2, which is optimal provided P 6= NP.
متن کاملOn Partitioning and Packing Products with Rectangles
In [1] we introduced and studied for product hypergraphs Hn = Qni=1Hi , where Hi = (Vi; Ei) , the minimal size (Hn) of a partition of Vn =Qni=1 Vi into sets that are elements of En = Qni=1 Ei . The main result was that (Hn) = n Y i=1 (Hi); (1) if the Hi's are graphs with all loops included. A key step in the proof concerns the special case of complete graphs. Here we show that (1) also holds wh...
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ژورنال
عنوان ژورنال: Combinatorica
سال: 2015
ISSN: 0209-9683,1439-6912
DOI: 10.1007/s00493-015-3006-1