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A variant of this algorithm that works for matrices of arbitrary shapes and is faster in practice [7] splits matrices in two instead of four submatrices, as follows. [9] Splitting a matrix now means dividing it into two parts of equal size, or as close to equal sizes as possible in the case of odd dimensions.
The fair polygon partitioning problem [20] is to partition a (convex) polygon into (convex) pieces with an equal perimeter and equal area (this is a special case of fair cake-cutting). Any convex polygon can be easily cut into any number n of convex pieces with an area of exactly 1/n. However, ensuring that the pieces have both equal area and ...
A container, usually a two- or three-dimensional convex region, possibly of infinite size. Multiple containers may be given depending on the problem. A set of objects, some or all of which must be packed into one or more containers. The set may contain different objects with their sizes specified, or a single object of a fixed dimension that ...
A simple lower bound is obtained by dividing the total amount of product by the size of each master roll. The total product required is 1380 x 22 + 1520 x 25 + ... + 2200 x 20 = 407160 mm. Each master roll is 5600 mm, requiring a minimum of 72.7 rolls, which means 73 rolls or more are required.
Fifteen equal circles packed within the smallest possible square. Only four equilateral triangles are formed by adjacent circles. Packing circles in simple bounded shapes is a common type of problem in recreational mathematics. The influence of the container walls is important, and hexagonal packing is generally not optimal for small numbers of ...
The basic idea behind a Riemann sum is to "break-up" the domain via a partition into pieces, multiply the "size" of each piece by some value the function takes on that piece, and sum all these products. This can be generalized to allow Riemann sums for functions over domains of more than one dimension.
Now there are two "good" cases - cases in which we can use these pieces to attain a weighted-proportional division respecting the different entitlements: Thre are several combinations of the pieces that give each their due share. Case 1: A subset of the pieces whose sum is 5 is produced if George marks the 3 piece and two of the three 1-pieces.
However, there are three distinct ways of partitioning a square into three similar rectangles: [1] [2] The trivial solution given by three congruent rectangles with aspect ratio 3:1. The solution in which two of the three rectangles are congruent and the third one has twice the side length of the other two, where the rectangles have aspect ...