
Re: combinatorics in three dimensions: what is being enumerated?
Posted:
Apr 4, 2013 3:19 AM


On 04/03/2013 08:06 PM, fom wrote: > On 4/3/2013 5:26 AM, David Bernier wrote: >> I'm counting something. What could it be? >> >> 000 >> 001 >> 010 >> 011 >> 100 >> 101 >> 110 >> 111 >> ============ >> Subtotal: 8 >> >> >> 00* >> 0*0 >> *00 >> 0*1 >> *01 >> 01* >> *10 >> *11 >> 10* >> 1*0 >> 1*1 >> 11* >> ============= >> Subtotal: 12 [...]
>> Clue given in the ROT13 substitution cipher: >> >> pbhagvat gur snprf bs ulcrephorf eniv inxvy >> >> >> >> David Bernier > > > I don't really care what you think you are > enumerating since it is uninterpreted and > I am not going to decode the cipher. > > 8 octants > > 4 quadrants for each of 3 hyperplanes > > 6 rays > > 1 point > > > I imagine that if I sat down and wrote > it all out, I could give it a coherent > labeling with respect to your "don't care" > states. > > Aw heck. Took the time. Yes, it works. > > Were you thinking of something different? [...]
I understand your enumeration of octants, quadrants, rays and point (the origin, presumably).
Yes, the enumeration problem I was working on is "counting faces in a hypercube", in a 1996 book by Ravi Vakil.
The name of the book is "A Mathematical Mosaic: Patterns & Problem Solving"
There are illustrations that go with his description. The problem is introduced in the Section on combinatorics, with the headline: "Counting the Faces of Hypercubes" on page 45. The object is to count the total number (combined) of zerocells, onecells, twocells and threecells in a regular hexaedron or cube. Just to be clear, zerocells are vertices, onecells are edges, twocells are faces and threecells is the cube itself.
The claim is that a hypercube in n dimensions has 3^n cells, counting 0cells (vertices), 1cells, ... up to ncells (the hypercube itself).
In my cryptic notation, 1*1 represents the edge joining the vertex (1,0,1) to the vertex (1, 1, 1) in the cube with vertices (0,0,0) , (0,0,1), (0, 1,0), (1,0,0), (0,1,1), (1, 1,0), (1,0,1) and (1,1,1) in R^3.
By "face" in Ravi Vakil's heading "Counting Faces in a Hypercube" is meant (for the cube, regular polyhedron) either a vertex, edge, face or the hexaedron itself.
The hexaedron has 8 vertices, 12 edges, 6 faces and 1 3dimensional "cell" .
8+12+6+1 = 27 = 3^3.
A square has: 4 vertices 4 edges 1 square face ===== 9 "cells" of dimensions 0, 1 or 3.
The proof given by Ravi Vakil on page 47 uses mathematical induction, plus the idea that a 4dimensional hypercube is formed by moving a cube from "height" = 0 in the 4th dimension to "height" = 1 in the 4th dimension.
Each cell of the ordinary cube (as above) gives rise while moving "up" in the 4th dimension to exactly three cells in or on the hypercube.
As the ordinary cube has 27 cells, the 4D hypercube has 81 cells.
David Bernier
 Jesus is an Anarchist.  J.R.

