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Relative to a reference layer with positioning A, two more positionings B and C are possible. Every sequence of A, B, and C without immediate repetition of the same one is possible and gives an equally dense packing for spheres of a given radius. The most regular ones are FCC = ABC ABC ABC... (every third layer is the same) HCP = AB AB AB AB...
Square packing in a square is the problem of determining the maximum number of unit squares (squares of side length one) that can be packed inside a larger square of side length . If a {\displaystyle a} is an integer , the answer is a 2 , {\displaystyle a^{2},} but the precise – or even asymptotic – amount of unfilled space for an arbitrary ...
For example, it is possible to pack 147 rectangles of size (137,95) in a rectangle of size (1600,1230). Packing different rectangles in a rectangle : The problem of packing multiple rectangles of varying widths and heights in an enclosing rectangle of minimum area (but with no boundaries on the enclosing rectangle's width or height) has an ...
In one dimension it is packing line segments into a linear universe. [10] In dimensions higher than three, the densest lattice packings of hyperspheres are known up to 8 dimensions. [11] Very little is known about irregular hypersphere packings; it is possible that in some dimensions the densest packing may be irregular.
A compact binary circle packing with the most similarly sized circles possible. [7] It is also the densest possible packing of discs with this size ratio (ratio of 0.6375559772 with packing fraction (area density) of 0.910683). [8] There are also a range of problems which permit the sizes of the circles to be non-uniform.
It is also possible to tell most C and C++ compilers to "pack" the members of a structure to a certain level of alignment, e.g. "pack(2)" means align data members larger than a byte to a two-byte boundary so that any padding members are at most one byte long.
Circle packing in a circle is a two-dimensional packing problem with the objective of packing unit circles into the smallest possible larger circle. Table of solutions, 1 ≤ n ≤ 20 [ edit ]
At the beginning of the solution, a simple method can be used to determine as many boxes as possible. This method uses conjunctions of possible places for each block of boxes. For example, in a row of ten cells with only one clue of 8, the bound block consisting of 8 boxes could spread from the right border, leaving two spaces to the left;