Euler’s Formula for Polyhedra Calculator (V − E + F = 2: find vertices, edges or faces)

Understanding the Euler's polyhedron formula calculator — concept and principle

This calculator uses Euler's formula V − E + F = 2 to find the missing number of vertices, edges or faces of a polyhedron, or to check all three. It handles solids with holes through them, tests whether the counts are possible at all, and says whether they match a Platonic solid. Use it for geometry homework and 3D modelling checks.

For any polyhedron without holes, vertices minus edges plus faces is always 2.

How it is calculated

V − E + F = 2 − 2g, where g is the number of holes through the solid (g = 0 for an ordinary polyhedron)

For g = 0: E = V + F − 2; V = 2 + E − F; F = 2 + E − V

Possibility check: 2E ≥ 3F and 2E ≥ 3V

The last row compares the counts with the five Platonic solids; matching counts alone do not prove the shape.

The principle

Every edge joins two vertices and borders two faces. If you flatten a polyhedron without holes onto a plane and remove faces or collapse edges one at a time, V − E + F never changes, and it ends as 2. Each hole through the solid lowers it by 2. The inequalities follow from counting edge ends: every face has at least 3 edges and every vertex meets at least 3, while each edge is counted twice.

Worked example

Default inputs: solve for edges, V = 8, F = 6, no holes.

E = V + F − 2 = 8 + 6 − 2 = 12, so E = 12.
Check: V − E + F = 8 − 12 + 6 = 2, matching the expected 2 (2 − 2 × 0).
Formula satisfied: yes — the counts fit.
2E = 24 ≥ 3F = 18 and ≥ 3V = 24: both hold.
Same counts as a Platonic solid: cube (regular hexahedron).

Each point is the calculator re-run with only “Vertices V” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Vertices V	Result
2	E = 6
4	E = 8
6	E = 10
8	E = 12
10	E = 14
12	E = 16
16	E = 20

How to read the result

If V − E + F is not 2 for a solid without holes, a count is wrong, which is a useful check for hand counts and 3D models. Different shapes can share the same counts; a triangular-faced shape and a box can both have 8 vertices, so the Platonic match is only a hint.

Common mistakes

Counting each edge twice, once per face.
Applying V − E + F = 2 to a solid with a tunnel through it.
Forgetting hidden faces or vertices at the back of a drawing.
Assuming any counts that satisfy the formula form a real polyhedron.

Keep in mind: The formula needs a connected solid with flat faces whose surface has no self-intersections. Unusual shapes need careful counting.

Frequently asked questions

What is Euler's formula for polyhedra?

V − E + F = 2 for any convex polyhedron, or any polyhedron without holes.

How many edges does a cube have?

12, since 8 − 12 + 6 = 2.

What changes for a doughnut-shaped solid?

With one hole, V − E + F = 0.

Sources: E. W. Weisstein, "Polyhedral Formula", Wolfram MathWorld

Checked 9 Nov 2026. Plus Magazine (University of Cambridge), “Plus Advent Calendar Door #14: Euler’s polyhedron formula”, 14 Dec 2013: V − E + F = 2 for all polyhedra except those with holes running through them; for shapes with holes the Euler characteristic V − E + F equals 2 minus 2 times the number of holes; cube 8 − 12 + 6 = 2. Edge-count inequalities derived here by counting edge–face and edge–vertex incidences.

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