Pascal’s triangle and hypercubes

A 3D projection of an tesseract performing an ...

A 3D projection of an tesseract performing an isoclinic rotation. (Photo credit: Wikipedia)

This blog entry poses a genuine question from me and, I hope, a puzzle that might interest many. It is inspired by an entry in the “Short Dictionary of Heuristic” found in How to Solve it: A New Aspect of Mathematical Method.

First, a quick description of Pascal’s Triangle.

This is the triangle of integers – with 1 at the apex – where each number is the sum of the numbers immediately above it – hence it begins (apologies for the slightly unorthodox presentation – a consequence of’s insistence on suppressing leading spaces on lines):

1 1
1 2 1
1 3 3 1
1 4 6 4 1

…and so on.

As is so often the case with maths, this is more than just a pleasing pattern. The numbers also, correspond, for instance, to the coefficients of the expansion of the binomial (x+a)^n eg:

(x+a)^0 = 1
(x+a)^1 = x+a
(x+a)^2 = x^2 + 2ax + a^2
(x+a)^3 = x^3 +3ax^2 +3a^2x + a^3 and so on…

But it is the relationship of the triangle to geometry I am interested in.

The simplest (the only) null dimensional shape: a point, has 1 vertex: 1
The simplest 1 dimensional shape – a line: has 2 vertices (null dimensions) and 1 one-dimensional shape: 2 1
The simplest 2 dimensional shape – a triangle: 3 vertices, 3 one dimensional elements (bounding lines) and 1 two-dimensional space: 3 3 1
The simplest 3 dimensional shape – a tetrahedron: 4 vertices, 6 bounding lines, 4 two-dimensional elements (faces) and contains 1 three dimensionsal space: 4 6 4 1

So we can see Pascal’s triangle emerge there (and thus propose by induction that the simplest four dimensional space has 5 vertices, 10 bounding lines, 10 faces, contains 5 three dimensional spaces and 1 four dimensional space). This induction is correct – the shape is the pentachoron.

What puzzles me is whether we can draw any analogies for other shapes.

So for the families of squares and cubes:

We could have the same first two members – a point, 1
A line: 2, 1
Then a square: 4 vertices, 4 bounding lines, 1 space
A cube: 8 vertices, 12 bounding lines, 6 faces, 1 space
A hypercube (tesseract): 16 vertices, 32 bounding lines, 24 faces, 8 three dimensional spaces (cells) and 1 four dimensional space.

So we have a triangle like this:

2 1
4 4 1
8 12 6 1
16 32 24 8 1

My question is: is there some fundamental relation between the rows of this triangle and there is for Pascal’s triangle? I cannot see one, but there looks like there should be one!

3 thoughts on “Pascal’s triangle and hypercubes

  1. Pingback: I’ve turned the Curse of Dimensionality into the Blessing of Layering Complex Data |

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