Recipe

Solve Euler's formula for an unknown count

Use when: you know any two of a connected plane graph's vertex count, edge count, and face count, and need the third

Goal: the missing count, plus confirmation the graph's connectedness hypothesis actually holds

Taught in Euler's formula for planar graphs

1. 1

   Euler's formula v-e+f=2 only holds for a connected plane graph. A disconnected graph with c components lands on v-e+f=1+c instead, so check connectedness first or the rest of this recipe silently gives the wrong answer.

   Example

   Two separate triangles (c=2) give v-e+f=1+2=3, not 2 -- a warning sign that the graph is disconnected, not that the formula is wrong.

   Why: skipping this check is the single most common way to misapply Euler's formula, since the arithmetic still "works" for the wrong target value.

2. 2

   List whichever two of v, e, f the problem gives you. If the graph is drawn and crossing-free, f can be counted directly off the drawing (don't forget the unbounded outer face); v and e are read off directly either way.

   Example

   A plane embedding has 6 vertices and 10 edges; the face count f is unknown.

   Why: Euler's formula relates exactly three quantities, so having two of them pins down the third with no further information needed.

3. 3

   Rearrange the formula for whichever of v, e, f is unknown, then substitute the two known values.

   Example

   f = 2-v+e = 2-6+10 = 6.

   • f unknown — f = 2-v+e
   • v unknown — v = 2+e-f
   • e unknown — e = v+f-2

   Why: this is just Euler's formula solved for a different variable -- no new theorem, only algebra.