What this calculator does
Euler's formula says that for any convex polyhedron, vertices minus edges plus faces always equals 2. A cube has 8 vertices, 12 edges and 6 faces, and 8 − 12 + 6 is 2. A tetrahedron has 4, 6 and 4, and that comes to 2 as well. The shape can be regular or thoroughly irregular and the relationship still holds.
Because the three counts are locked together, any one of them follows from the other two. That is what this calculator does. It is worth knowing the limits, though: the value of 2 applies to convex polyhedra and to anything topologically equivalent to a sphere. A solid with a hole through it gives 0 instead, and each further hole subtracts another 2.
The formula
The formula V − E + F = 2 is rearranged for whichever quantity is being solved for: V = 2 − F + E, E = V + F − 2, or F = 2 − V + E. The two values you supply are substituted in and the third is returned. The check line then adds all three back together, and it should come to 2 for any valid convex polyhedron.
| Term | Meaning |
|---|---|
| V | Vertices, the corner points where edges meet. |
| E | Edges, the line segments where two faces meet. |
| F | Faces, the flat polygons making up the surface. The count includes every face, with no outside or base treated differently. |
| Euler characteristic | The value of V − E + F, which is 2 for a sphere-like surface and drops by 2 for every hole through the solid. |
The inputs explained
| Field | What to enter |
|---|---|
| Solve for | Which of the three counts you want calculated. The other two are read from the fields below. |
| Vertices (V) | The number of vertices. Ignored when vertices are the quantity being solved for. |
| Edges (E) | The number of edges. Ignored when edges are the quantity being solved for. |
| Faces (F) | The number of faces. Ignored when faces are the quantity being solved for. |
When to use it
Filling in a count you did not measure
Edges are the most tedious of the three to count on a physical model or a drawing, and they are also the ones the formula recovers most easily from the vertices and faces, which are far quicker to tally.
Checking a model or a mesh
Running the three counts from a 3D model through the formula catches structural problems quickly. A result other than 2 means either the surface has holes or the mesh has faults in it, and both are worth knowing about early.
Working through a geometry exercise
Polyhedron problems routinely give two counts and ask for the third, or give all three and ask whether such a solid can exist. The check line answers the second kind directly.
Worked examples
Every figure in the tables below is produced by this page’s own calculator at build time, so the numbers and the tool always agree. Select any row to load that scenario.
Can any of the three counts be recovered from the other two?
The same solid with each of the three quantities solved for in turn, holding the other two fixed.
| Solving for | Value returned | Check: V − E + F |
|---|---|---|
| Vertices | 20 | 2 (should equal 2 for a convex polyhedron) |
| Edges | 30 | 2 (should equal 2 for a convex polyhedron) |
| Faces | 12 | 2 (should equal 2 for a convex polyhedron) |
How do vertices and edges trade off at a fixed face count?
The face count held at 8 while the edge count rises, solving for vertices each time.
| Edges | Vertices returned | Check: V − E + F |
|---|---|---|
| 12 | 6 | 2 (should equal 2 for a convex polyhedron) |
| 14 | 8 | 2 (should equal 2 for a convex polyhedron) |
| 16 | 10 | 2 (should equal 2 for a convex polyhedron) |
| 18 | 12 | 2 (should equal 2 for a convex polyhedron) |
Questions
Does the formula work for every solid?
For every convex polyhedron, and more generally for any surface that could be deformed into a sphere without tearing. It fails for solids with holes: a shape like a picture frame gives 0 rather than 2, and each further hole subtracts another 2 from the result.
What are the counts for the five Platonic solids?
The tetrahedron has 4 vertices, 6 edges and 4 faces. The cube has 8, 12 and 6, and the octahedron has 6, 12 and 8, the same numbers swapped. The dodecahedron has 20, 30 and 12, and the icosahedron has 12, 30 and 20, again a swap. Every one of them satisfies V − E + F = 2.
If my three counts satisfy the formula, does the solid exist?
Not necessarily. The formula is a requirement every convex polyhedron meets, not a guarantee that any triple meeting it can be built. Further conditions apply, such as every face needing at least three edges and every vertex joining at least three.
Why is the Euler characteristic 2 rather than some other number?
It falls out of the surface being sphere-like. Flattening a polyhedron into a planar graph and then removing faces one at a time leaves the quantity V − E + F unchanged throughout, and the simplest remaining figure gives 2. The number is a property of the shape of the surface, not of the particular polyhedron on it.
For the surface area and volume of a rectangular solid, see the cuboid calculator. For the volume of a range of standard shapes, see the volume of common shapes calculator.