Sperner Triangle Boundary Parity

worked-example

Let's consider a mesh with denominator n=2 and favorite corner 1, with tie priority 120. We assign barycentric coordinates [2-i-j, i, j] to each vertex (i, j) and assign colors according to the rules:

Mesh n=2, coordinates: (0,0): barycentric [2,0,0], colors: score(0)=2, label=0 (1,0): barycentric [1,1,0], colors: score(0)=1, score(1)=1*2=2, label=1 (2,0): barycentric [0,2,0], colors: score(1)=2*2=4, label=1 (0,1): barycentric [1,0,1], colors: score(0)=1, score(2)=1, tie priority 120 -> label=2 (1,1): barycentric [0,1,1], colors: score(1)=1*2=2, score(2)=1, label=1 (0,2): barycentric [0,0,2], colors: score(2)=2, label=2 Vertices labels: (0,0)=0, (1,0)=1, (2,0)=1 (0,1)=2, (1,1)=1 (0,2)=2 Triangles: 1: (0,0),(1,0),(0,1) labels (0,1,2) -> fully labeled! 2: (1,0),(1,1),(0,1) labels (1,1,2) 3: (1,0),(2,0),(1,1) labels (1,1,1) 4: (0,1),(1,1),(0,2) labels (2,1,2) Result: First fully labeled triangle is #1. Total fully labeled is 1.

experiment

In the interactive module at Sperner Triangle Boundary Parity, you can verify the combinatorial certificate on different grids:

1. Change the Mesh denominator to control the resolution.
2. Adjust the Favored corner to bias colors.
3. Set the Tie priority to resolve ties differently.
4. Observe that the number of fully labeled triangles is always odd.

limitations

This exact counting method relies on discrete barycentric integer coordinates where i,j ≥ 0, i+j ≤ n. The rule that no opposite face uses a forbidden label ensures boundary parity. This is a finite combinatorial certificate rather than a numerical fixed-point solver, so its scope is bound to the simplicial structure rather than an arbitrary domain or mapping.

references

For more background on Sperner's Lemma and fixed-point theorems, see:

• MIT 18.304 Seminar in Discrete Mathematics: Sperner's Lemma - outlines the boundary parity edge-counting argument.