ConceptComplete

Euler Formula and Applications

Euler's formula nm+f=2n - m + f = 2 for planar graphs provides powerful constraints on graph structure, leading to edge bounds and planarity tests.

TheoremEuler's Formula for Planar Graphs

For any connected planar graph with nn vertices, mm edges, and ff faces: nm+f=2n - m + f = 2

For disconnected graphs with kk components: nm+f=k+1n - m + f = k + 1.

ProofProof by Induction

Induct on mm. Base: m=0m=0 gives nn isolated vertices, f=1f=1 outer face, so n0+1=n+1=k+1n - 0 + 1 = n+1 = k+1 ✓.

Inductive step: Remove an edge ee. If ee is a bridge, it reduces components by 1 and faces stay same: (n)(m1)+f=k(n) - (m-1) + f = k becomes nm+f=k+1n - m + f = k+1 after restoring ee. If ee is not a bridge, it merges two faces: (n)(m1)+(f1)=2(n) - (m-1) + (f-1) = 2 gives nm+f=2n - m + f = 2 ✓.

TheoremEdge Bounds for Planar Graphs

For a connected planar graph with n3n \geq 3 vertices and mm edges:

  1. m3n6m \leq 3n - 6 (general bound)
  2. If no triangles: m2n4m \leq 2n - 4 (bipartite bound)

These bounds immediately prove K5K_5 and K3,3K_{3,3} are non-planar.

Example$K_5$ is Non-Planar

K5K_5 has n=5n=5 vertices and m=10m=10 edges. If planar, m3n6=9m \leq 3n - 6 = 9. But 10>910 > 9, contradiction. Thus K5K_5 cannot be planar.

Similarly, K3,3K_{3,3} is bipartite with n=6n=6, m=9m=9. Bipartite bound gives m2(6)4=8<9m \leq 2(6) - 4 = 8 < 9, so K3,3K_{3,3} is non-planar.

DefinitionAverage Face Degree

In a planar graph, each face ii has degree did_i (number of edges bounding it). Then: i=1fdi=2m\sum_{i=1}^f d_i = 2m This double-counts edges: each edge bounds two faces (or one face twice if a bridge).

Remark

Euler's formula extends to surfaces: on a surface of genus gg (sphere has g=0g=0, torus g=1g=1), the formula becomes: nm+f=22gn - m + f = 2 - 2g This Euler characteristic is a topological invariant, connecting graph theory to algebraic topology.