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# An Electrician’s (or a Plumber’s) Proof of Euler’s Polyhedral Formula

Euler’s famous polyhedral formula,

$V-E+F=2,\tag1$

relates the numbers of vertices, edges, and faces of a polyhedron “without holes,” i.e., one that is sphere-like and in three dimensions. If the polyhedron has one hole, as in Figure 2, then we subtract $$2$$ from the right-hand side of the formula, which becomes $$V-E+F=0$$. The same thing happens for each additional hole (or, putting it differently, “handle on the sphere”). I will describe an argument based on electric circuits, leading to Euler’s formula. I learned this beautiful idea from Peter Lax, and I therefore lay no claim to originality, except for any errors.

Figure 1. V, E, and F are the numbers of vertices, edges, and faces of a polyhedron.
Imagine our polyhedron as a wire frame, the edges being conducting wires, each of resistance 1 ohm, welded together at the vertices. Let us connect two vertices (chosen arbitrarily) to a battery, adjusting the voltage so as to drive the current of exactly $$1$$ ampere.  Now, nature will pick a specific value for each edge’s current. In doing so, she obeys Kirchhoff’s laws: the currents satisfy some equations that determine the currents. Let us take it for granted that

$\textrm{the number of unknown currents} = \textrm{the number of independent equations}.$

This sentence is already Euler’s formula in disguise! Indeed, the left-hand side is $$E$$, one unknown current per wire. For the right-hand side, Kirchhoff’s laws state the following:

(i) The sum of currents entering each vertex is zero, giving $$V$$ equations. But one of these equations is redundant, since it results from adding up all the others (I leave out the simple verification), yielding $$V-1$$ equations.

(ii) The sum of voltage drops around each face is zero. This gives $$F$$ equations, one of which is the sum of the remaining ones and thus redundant, for the total of $$F-1$$ equations.

Figure 2. A polyhedron with a hole; two non-contractible circuits are indicated.
Summarizing,

$E=(V-1)+(F-1),$

which amounts to $$(1)$$. For a polyhedron with a hole, as illustrated in Figure 2, we must add two more equations, expressing the fact that the voltage drop over each of two non-contractible circuits is zero, resulting in $$E=(V-1)+(F-1)+2$$, or

$V-E+F=0.$

Admittedly the proof is not rigorous as given, since, for instance, I did not eliminate the possibility of more redundant equations.

Although this proof would have sounded strange in Euler’s pre-electricity days, it could be reformulated in plumber’s terms by treating the polyhedron as a network of tubes (with porous blockages playing the role of resistors), currents as the mass flow per second, and voltages as pressures.