Search results
Results from the WOW.Com Content Network
In computational geometry, the largest empty rectangle problem, [2] maximal empty rectangle problem [3] or maximum empty rectangle problem, [4] is the problem of finding a rectangle of maximal size to be placed among obstacles in the plane. There are a number of variants of the problem, depending on the particularities of this generic ...
Here, a d-dimensional rectangular range is defined to be a Cartesian product of d intervals of real numbers, which is a subset of R d. The problem is named after Victor Klee , who gave an algorithm for computing the length of a union of intervals (the case d = 1) which was later shown to be optimally efficient in the sense of computational ...
The big rectangle has width m and length T + 3m. Every solution to the 3-partition instance induces a packing of the rectangles into m subsets such that the total length in each subset is exactly T, so they exactly fit into the big rectangle. Conversely, in any packing of the big rectangle, there must be no "holes", so the rectangles must not ...
Second, compare the solution gained by the first step to the best solution which uses a small number of sets. Third, return the best out of all examined solutions. This algorithm achieves an approximation ratio of 1 − 1 / e − o ( 1 ) {\displaystyle 1-1/e-o(1)} .
Khandhawit, Pagonakis & Sriswasdi (2013) used a min-max strategy for area of a convex set containing a segment, a triangle and a rectangle to show a lower bound of 0.232239 for a convex cover. In the 1970s, John Wetzel conjectured that a 30° circular sector of unit radius is a cover with area π / 12 ≈ 0.2618 {\displaystyle \pi /12\approx 0. ...
The area thus obtained is referred to as the sofa constant. The exact value of the sofa constant is an open problem. The leading solution, by Joseph L. Gerver, has a value of approximately 2.2195. In November 2024, Jineon Baek posted an arXiv preprint claiming that Gerver's value is optimal, which if true, would solve the moving sofa problem. [2]
Because the "sweep" of the area under the involute is bounded by a tangent line (see diagram and derivation below) which is not the boundary (¯) between overlapping areas, the decomposition of the problem results in four computable areas: a half circle whose radius is the tether length (A 1); the area "swept" by the tether over an angle of 2 ...
An a × b rectangle can be packed with 1 × n strips if and only if n divides a or n divides b. [ 15 ] [ 16 ] de Bruijn's theorem : A box can be packed with a harmonic brick a × a b × a b c if the box has dimensions a p × a b q × a b c r for some natural numbers p , q , r (i.e., the box is a multiple of the brick.) [ 15 ]