This results in just 64 / 4 = 16 vertices on a hypercube. But four cubes meet at each vertex, so we have in fact overcounted by a factor of 4. With 8 vertices on each of 8 cubes, we have 64 vertices all together. This implies that there are in fact 96 / 3 = 32 edges on a hypercube.įinally, we count the vertices. But three cubes meet at each edge, so we’ve overcounted by a factor of 3. Since 8 cubes contribute 12 edges each, there are 96 edges in total. But since the cubes meet square-to-square, we’ve overcounted by a factor of 2, so that there are just 48 / 2 = 24 squares on a hypercube. With 8 cubes, we have 8 x 6 = 48 squares. Let’s use this analogy to check that we’ve counted correctly. This pattern continues into higher dimensions as well. By analogy, on the hypercube, we should have 2 squares merging, 3 edges, and 4 vertices. On the cube, we observed that 2 edges merged, then 3 vertices. On the square, we saw that 2 vertices merged. We note that the “8” comes from the sequence 2, 4, 6, 8,… for 2 vertices on a segment, 4 edges on a square, 6 squares on a cube, etc. Now begin with the eight cubes on a hypercube. This means we’ve overcounted by a factor of 3, so there are 24 / 3 = 8 vertices on a cube.

So there are 6 x 4 = 24 vertices on the 6 squares - but when we join the squares together, three vertices merge to one. This means there are 24 / 2 = 12 edges on the cube.įor vertices, we note that three squares meet at each vertex. But as with the vertices on the square, when we join two squares at their edges, two edges merge, so we’ve overcounted by a factor of 2. To count edges, we see that each square on the cube has 4 edges, for a total of 6 x 4 = 24 edges. What happens when we look at the cube in the same way? Let’s start with the six squares. Thus, there are 8 / 2 = 4 vertices on a square (as we know). But when we join the edges at vertices, two vertices merge, so we’ve overcounted by a factor of 2. Now to count vertices, we note that each edge has two vertices - for 4 x 2 = 8 vertices.

We’ll start with one dimension less than the object we’re examining - later we’ll use the squares on a cube, and the cubes on a hypercube. Let’s start with the number of edges on a square this time. So we begin by looking at the square and cube in a different way, and thinking (again) by analogy from there. This suggests that maybe we were right all along…. Mainly, we thought by analogy - but how could we be sure our intuition was correct? Today, we look at another say to make these same counts, and obtain the same results. Last week we counted vertices, edges, faces, and cells on a hypercube.

Number of edges in a hypercube