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Topic: Lesson Plan: Cube to Hexapent Mapping
Replies: 1   Last Post: Mar 21, 2006 7:02 PM

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Kirby Urner

Posts: 4,709
Registered: 12/6/04
Lesson Plan: Cube to Hexapent Mapping
Posted: Mar 20, 2006 5:01 AM

I got a lot of this thinking from another math teacher here in Portland, Glenn Stockton, and adapted it for a simpler pre-college context.

Start with a 4x4x4 cube. You only need 3 digits running from 0 to 3, i.e. 000 - 333 with a base 4 place value, to identify all the cells.

For example, the "bottom floor" could be:

000 001 002 003
010 011 012 013
020 021 022 023
030 031 032 033

and the top floor:

300 301 302 303
310 311 312 313
320 321 322 323
330 331 332 333

Students could provide the intermediate two floors as an exercise.

So that's a 4x4x4 cube, as I said. Our goal is to map this to two concentric hexapents, one inside the other, each hexapent having 32 cells. 2*32 = 4*4*4, so there's a one-to-one mapping possible (this is the kind of stuff Glenn is into).

What's a hexapent? The simplest 32 cell version is the trunctated icosahedron or buckyball (C60), with 12 pentagonal cells and 20 hexagonal cells. Yes, the soccer ball.

A soccer ball has 32 polygonal regions. Two concentric soccer balls (faces aligned per Glenn's Z-axis) would have our 64 = 4x4x4 cells.

In my curriculum, we want students to learn (10*f*f + 2) as a sequence generator. This is the number of balls in an expanding cuboctahedral shell or, equivalently, an icosahedral shell. Lower-case f stands for frequency.[1]

V = 10*f*f + 2
N = V - 2
F = 2*N
E = 3*N
V + F = E + 2

Above are basic equations for the classic geodesic sphere, derived from an icosahedral shell of V balls (hubs). F is the number of triangular facets (faces, windows), E the number of edges (struts). The last line is Euler's Law and holds true for any uncored (unbored) polyhedron.[2]

The first hexapent (soccer ball, buckyball) appears with f = 3, i.e. we start out with 92 Vertices, then "bash out" groups of 6 and 5 triangular windows, converting the whole skeleton to nothing but 12 pentagons (at what had been the 12 vertices of an icosahedron) and the rest hexagons (20 of them).

The next bigger hexapent occurs when f = 6, then 9 and so on, multiples of 3. F = H*6 + P*5, where H = number of hexagons and P = number of pentagons = constant 12. You can make H a function of f: H = (20*f*f - 60)/6 (simplify), e.g. where f=3, (180-60)/6 = 20. Note that you need positive integer f to be divisible by 3 for H to be a whole number (an excercise in analyzing prime factors).

Although this is a very geometric lesson, there's algebraic content as well, as one may think of several ways to map the boxes in our 4x4x4 cube to the 64 cells of the two concentric hexapents. This suggests an "arrow diagram" mapping a set in a domain to a range within a codomain.

This "one-to-one" mapping between the 4x4x4 cube and the 2 concentric hexapents will help drive/reinforce our discussion of related concepts e.g. "one-to-many" and "many-to-one" -- concepts important in database work, and in the theory of functions, with its surjective, injective and bijective relationships. How the teacher wants to fit these in, or leave them out, depends on the surrounding curriculum (a network of anchoring topics).

Note that you might prefer base 2 over base 4, in which case the 64 cells in question may be addressed as:
000000, 000001, 000010, 000011... 111111. (111111 = 32+16+8+4+2+1 = 63). Those with a leading digit 0 could be on the inner hexapent, those with a leading 1 on the outer hexapent.

Obviously we're into teaching about multiple bases here. That's a big part of our computer-friendly Silicon Forest curriculum.[3]

4D Solutions


[1] (10*f*f + 2): http://www.research.att.com/~njas/sequences/A005901

[2] Re Euler's V + F = E + 2, see Cromwell's Polyhedra:

[3] Silicon Forest:


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