Introduction
Euler's rule says that for a solid without holes, the number of vertices minus the number of edges plus the number of faces is 2. A rule like that is worth testing by hand, on solids you can picture, including one built to break it.
This reading works through four solids, counting each part with care, and shows how the result changes when a hole runs through the solid.
The Problem
You have four solids. Each is a polyhedron: a solid bounded by flat polygon faces that meet along straight edges and at corners called vertices.
- A tetrahedron: a pyramid with a triangular base, so four triangular faces in all.
- A cube.
- An octahedron: two square-based pyramids glued base to base, giving eight triangular faces.
- A "picture frame": a square frame built from four rectangular beams, with a square hole through the middle.
For each solid, count the vertices V, the edges E, and the faces F, and compute V - E + F. Then explain the pattern. The quantity V - E + F is called the Euler characteristic: vertices minus edges plus faces, a number that tracks how many holes a solid has. In 1813 Simon Lhuilier showed that a solid with g holes has Euler characteristic 2 - 2g (O'Connor and Robertson 1996c).
Working It Through
Step 1: Count the tetrahedron
A tetrahedron has 4 corners, so V = 4. Every pair of corners is joined by an edge. There are 6 pairs among 4 corners, so E = 6. It has 4 triangular faces, so F = 4.
V - E + F = 4 - 6 + 4 = 2
You can check the edge count another way. Each of the 4 triangles has 3 sides, which makes 12 sides in total. Every edge is shared by exactly 2 faces, so each edge was counted twice: 12 \div 2 = 6.
Step 2: Count the cube
A cube has 4 corners on top and 4 on the bottom, so V = 8. It has 4 edges around the top, 4 around the bottom, and 4 vertical edges joining them, so E = 12. It has 6 square faces, so F = 6.
V - E + F = 8 - 12 + 6 = 2
The shared-edge check works here too: 6 squares with 4 sides each give 24 sides, and 24 \div 2 = 12.
Step 3: Count the octahedron
Picture the octahedron as a square "equator" with one point above it and one below. That's 4 vertices around the middle, plus a top and a bottom, so V = 6. There are 4 edges around the square, 4 edges running up to the top point, and 4 running down to the bottom point, so E = 12. The upper pyramid has 4 triangular faces and the lower one has 4, so F = 8. The square where the pyramids were glued is inside the solid, so it isn't a face.
V - E + F = 6 - 12 + 8 = 2
Check: 8 triangles with 3 sides each give 24 sides, and 24 \div 2 = 12.
Step 4: Count the picture frame
Figure: A square picture frame made of four beams with mitered corners, drawn at an angle so its 16 vertices are visible: 4 outer and 4 inner corners on the top face, and the same 8 on the bottom face. Top and bottom are each divided into 4 trapezoids, and the outer and inner walls are each 4 rectangles. Alt text: A thick square frame with a square hole in the middle, its top face split into four trapezoids by lines running from each inner corner to the matching outer corner.
Look at the top of the frame. It's a square ring, with 4 outer corners and 4 inner corners. The bottom is the same. So V = 8 + 8 = 16.
Each beam's top face is a trapezoid, a four-sided shape with two parallel sides of different lengths, because the beams meet at angled corners. The top of the frame therefore has 4 outer edges, 4 inner edges, and 4 short edges joining each inner corner to its outer corner: 12 edges. The bottom has another 12. Then there are vertical edges: 4 at the outer corners and 4 at the inner corners, 8 in all. So E = 12 + 12 + 8 = 32.
For the faces, count 4 trapezoids on top, 4 on the bottom, 4 rectangles on the outer walls, and 4 rectangles on the inner walls, around the hole. So F = 16.
V - E + F = 16 - 32 + 16 = 0
Check: all 16 faces have 4 sides, giving 64 sides in all, and 64 \div 2 = 32. The answer isn't a slip. The frame really does give 0.
Step 5: Interpret the results
Three solids gave 2 and one gave 0. The difference isn't size or angle; you could make the cube enormous or the octahedron lopsided and the counts wouldn't change. The difference is the hole. The tetrahedron, cube, and octahedron have no hole through them, so g = 0 and 2 - 2g = 2. The picture frame has one hole, so g = 1 and 2 - 2g = 0.
The frame also explains why the count measures something real. You could build the frame from eight beams instead of four, or split each wall into smaller rectangles, and the counts of faces and edges would change. The total would still be 0, because each extra edge you add comes with an extra face or an extra vertex that cancels it (Richeson 2008).
Key Considerations
Count each edge once. Every edge of a polyhedron sits between exactly two faces. If you count edges face by face, you'll count each one twice. The shared-edge check, adding up the sides of all faces and dividing by 2, catches this mistake.
The common mistake: double-counting, or mislabeling the frame's faces. The most frequent error is counting the edges face by face and forgetting to divide by 2, which gives 64 edges for the frame instead of 32. A second error is to treat the frame's top and bottom as squares. The top of the frame is a square ring, and a square ring isn't a single polygon face: a face must be a polygon with no hole in it. Splitting the ring into four trapezoids is what makes it a legitimate set of faces. If you count the ring as one face, you get the wrong number of faces and edges, and the total comes out wrong.
What the formula assumes. Euler's own argument assumed the solid was convex, so that the straight line between any two of its points stays inside it (O'Connor and Robertson 1996c). The picture frame isn't convex: a line from one side of the hole to the other passes through empty space. The rule 2 - 2g is the corrected version that covers solids like it.
Summary
Counting carefully, with each edge counted once, gives 2 for every solid without a hole and 0 for the frame with one hole, exactly as 2 - 2g predicts.
| Solid | V | E | F | V - E + F | Holes |
|---|
| Tetrahedron | 4 | 6 | 4 | 2 | 0 |
| Cube | 8 | 12 | 6 | 2 | 0 |
| Octahedron | 6 | 12 | 8 | 2 | 0 |
| Picture frame | 16 | 32 | 16 | 0 | 1 |
- Every edge count was checked by adding the sides of all faces and dividing by 2.
- The three solids without holes give 2; the frame, with one hole, gives 0.
- Both results match Lhuilier's rule V - E + F = 2 - 2g, with g the number of holes.
References
- O'Connor, J. J., and E. F. Robertson. 1996c. "A History of Topology." MacTutor History of Mathematics Archive, University of St Andrews. Last updated May 1996.
- Richeson, David S. 2008. Euler's Gem: The Polyhedron Formula and the Birth of Topology. Princeton, NJ: Princeton University Press.
- Free: None
- Free companion: O'Connor and Robertson 1996c