hypercube
a simple capture of acomplex optical sculpture made of mirrors and thin neon lights...
A hypercube of dimension n has 2n "sides" (a 1-dimensional line has 2 end points; a 2-dimensional square has 4 sides or edges; a 3-dimensional cube has 6 2-dimensional faces; a 4-dimensional tesseract has 8 cells). The number of vertices (points) of a hypercube is 2n (a cube has 23 vertices, for instance).
A simple formula to calculate the number of "n-2"-faces in an n-dimensional hypercube is: 2n2 − 2n
The number of m-dimensional hypercubes (just referred to as m-cube from here on) on the boundary of an n-cube is
, where and n! denotes the factorial of n.
For example, the boundary of a 4-cube (n=4) contains 8 cubes (3-cubes), 24 squares (2-cubes), 32 lines (1-cubes) and 16 vertices (0-cubes).
This identity can be proved by combinatorial arguments; each of the 2n vertices defines a vertex in a m-dimensional boundrary. There are ways of choosing which lines ("sides") that defines the subspace that the boundrary is in. But, each side is counted 2m times since it has that many vertices, we need to divide with this number. Hence the identity above.
hypercube
a simple capture of acomplex optical sculpture made of mirrors and thin neon lights...
A hypercube of dimension n has 2n "sides" (a 1-dimensional line has 2 end points; a 2-dimensional square has 4 sides or edges; a 3-dimensional cube has 6 2-dimensional faces; a 4-dimensional tesseract has 8 cells). The number of vertices (points) of a hypercube is 2n (a cube has 23 vertices, for instance).
A simple formula to calculate the number of "n-2"-faces in an n-dimensional hypercube is: 2n2 − 2n
The number of m-dimensional hypercubes (just referred to as m-cube from here on) on the boundary of an n-cube is
, where and n! denotes the factorial of n.
For example, the boundary of a 4-cube (n=4) contains 8 cubes (3-cubes), 24 squares (2-cubes), 32 lines (1-cubes) and 16 vertices (0-cubes).
This identity can be proved by combinatorial arguments; each of the 2n vertices defines a vertex in a m-dimensional boundrary. There are ways of choosing which lines ("sides") that defines the subspace that the boundrary is in. But, each side is counted 2m times since it has that many vertices, we need to divide with this number. Hence the identity above.