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From: leeper@mtgzz.UUCP (m.r.leeper)
Newsgroups: net.math
Subject: Re: Euler's(?) formula
Message-ID: <848@mtgzz.UUCP>
Date: Mon, 24-Jun-85 02:34:04 EDT
Article-I.D.: mtgzz.848
Posted: Mon Jun 24 02:34:04 1985
Date-Received: Wed, 19-Jun-85 04:32:42 EDT
References: <1832@ut-ngp.UTEXAS>
Organization: AT&T Information Systems Labs, Middletown NJ
Lines: 46

This is not a solution, but a comment on extending Euler's formula.
You state it that F + V = E + 2.

Consider a vertex to be a zero dimensional boundary.
Consider an edge to be a one dimensional boundary.
etc.

For polyhedra in any number of dimensions:
The sum of the number of even-dimensional boundaries always exceeds the
odd-dimensional boundaries by 1.

Take a common cube which has 8 vertices, 12 edges, etc.

Dim	E	O
0	8
1		12
2	6
3		1
sum	14  -   13  =  1


Now join two tetrahedra at one vertex:

Dim	E	O
0	7
1		12
2	8
3		2
sum	15  -   14  =  1

Now try a variation, a two dimensional tear-drop shape has one vertex,
one edge, and encloses one two dimensional space.  2 - 1 = 1.

The generalizes the Euler formula.  The Euler formula assumes that
there is one three-dimensional area enclosed and nothing above three
dimensions.  I think this points to an induction proof of itself and
Euler.  Build up to the an arbitrary figure.  Start with a point.  To
turn it into a line segment add an edge missing one vertex.  That adds
one even-dimenional boundary, a vertex, and one odd dimensional
boundary, the difference remains one.  If you merge two vertices you
lose a vertex and gain two-dimensional space, this again leaves the
difference the same.


				Mark Leeper
				...ihnp4!mtgzz!leeper