Euler's Polyhedron Formula Calculator (V − E + F)

One formula that ties a solid together

Count the corners, edges, and flat faces of any convex polyhedron, and a small miracle happens every time: vertices minus edges plus faces always equals 2.

V − E + F = 2

A cube has 8 vertices, 12 edges, and 6 faces: 8 − 12 + 6 = 2. A soccer ball (a truncated icosahedron) has 60 vertices, 90 edges, and 32 faces: 60 − 90 + 32 = 2. It does not matter how lopsided or intricate the solid is, as long as it is convex and has no holes. Leonhard Euler noticed this in 1758, and it became one of the first theorems of topology. This calculator uses it two ways: enter any two of the three counts and it solves for the third, or enter all three and it checks whether your shape is a valid convex polyhedron.

The Euler characteristic

The number 2 is not magic. It is the Euler characteristic χ of a sphere, and a convex polyhedron is really just a sphere with flat panels. Deform a cube into a ball and the counts still describe the same surface, so the total holds. Change the surface and the number changes with it.

When it is not 2

Punch a hole through the solid and you get something shaped like a doughnut, a torus, whose Euler characteristic is 0. Two holes give −2, three give −4. The pattern is χ = 2 − 2g, where g is the genus, the number of holes. So if you tally V, E, and F on a model and get anything other than 2, either the mesh has an error or the shape is not topologically a sphere. That single check is why 3D modelers and finite-element engineers lean on this formula to catch broken meshes.

Euler is necessary, but it is not enough

Getting χ = 2 does not prove a solid exists. Take V = 3, E = 4, F = 3: it satisfies the formula perfectly, and no polyhedron on earth has three corners. Two more counting rules close that gap.

Every face is bounded by at least three edges, and every edge borders exactly two faces. Add up edges face by face and you count each one twice, so 2E ≥ 3F.

Every vertex has at least three edges meeting at it, and every edge ends at exactly two vertices. Same argument, so 2E ≥ 3V.

Run those on the impostor above: 2E = 8, but 3F = 9 and 3V = 9. Both fail, and the shape is ruled out. Run them on a cube: 2E = 24 against 3F = 18 and 3V = 24. Both hold, with the second exactly tight because every corner of a cube has precisely three edges.

The calculator applies all three tests, so it will tell you when counts pass Euler and still cannot be built.

A quick reference

The five Platonic solids all obey it: tetrahedron (4, 6, 4), cube (8, 12, 6), octahedron (6, 12, 8), dodecahedron (20, 30, 12), and icosahedron (12, 30, 20). Every one returns 2. The tetrahedron is the smallest polyhedron there is; nothing with fewer than 4 vertices, 6 edges or 4 faces exists.

The same relationship governs connected planar graphs, where the faces count the regions the graph carves the plane into, including the unbounded outer region.