考虑旋转等价性时，两种颜色给四面体边染色的不同方案数

Hey there! Let's break down how to count the distinct ways to color a tetrahedron's edges with two colors (like purple/orange), where any two colorings that can be rotated into each other count as the same. We'll use Burnside's Lemma here—it's the standard tool for counting symmetric equivalence classes, and it works by averaging the number of colorings fixed by each symmetry operation.

First, let's list out all the rotational symmetries of a tetrahedron (we only consider proper rotations, no reflections—since the problem specifies rotations):

• There are 12 total rotational symmetry operations, split into 3 categories:

1. Identity rotation (1 operation) #

This is just doing nothing—every coloring stays the same under this operation. A tetrahedron has 6 edges, each with 2 color choices, so the number of fixed colorings here is:
2^6 = 64

2. 120°/240° rotations about vertex-to-face-center axes (8 operations) #

Each vertex has an axis going straight to the center of the opposite triangular face. For each axis, there are two non-identity rotations: 120° and 240°. With 4 vertices, that's 4 × 2 = 8 total operations.

For these rotations, the edges get grouped into two disjoint 3-cycles: one cycle includes the three edges connected to the vertex, the other includes the three edges of the opposite face. For a coloring to be fixed under this rotation, all edges in each cycle must be the same color. Each cycle has 2 color options, so each operation fixes 2^2 = 4 colorings.

Total fixed colorings for this category: 8 × 4 = 32

3. 180° rotations about midpoint-to-midpoint axes of opposite edges (3 operations) #

A tetrahedron has 3 pairs of opposite edges (edges that don't share a vertex). For each pair, there's a rotation axis going through the midpoints of the two edges, rotating 180°. That's 3 operations total.

Under this rotation, the edges split into two 1-cycles (the two opposite edges themselves, which stay in place) and two 2-cycles (pairs of edges that swap places). For a coloring to be fixed, each cycle's edges must match: the two fixed edges can be any color, and each swapped pair must be the same color. That gives 2^4 = 16 fixed colorings per operation.

Total fixed colorings for this category: 3 × 16 = 48

Final Calculation with Burnside's Lemma #

Now we add up all the fixed colorings and divide by the total number of rotational operations (12) to get the number of distinct equivalence classes:
(64 + 32 + 48) / 12 = 144 / 12 = 12

So the total number of distinct colorings is 12.

内容的提问来源于stack exchange，提问作者Shuryu Kisuke