Graph Coloring and Planarity

6.1 Vertex Coloring

Bounds

• $\chi(G) \leq \Delta(G) + 1$ (greedy coloring bound)
• $\chi(G) \geq \omega(G)$ (clique number)
• Every planar graph has $\chi(G) \leq 4$ (Four Color Theorem)

6.2 Planar Graphs

Euler's Formula

Non-Planar Graphs

• $K_5$ (complete graph on 5 vertices) — non-planar
• $K_{3,3}$ (complete bipartite) — non-planar Kuratowski's Theorem: A graph is non-planar iff it contains a subdivision of $K_5$ or $K_{3,3}$.

✅ Practice Questions

Solution

$\chi(K_n) = n$ — every vertex must have a distinct color since all pairs are adjacent. Q2: Is $K_4$ planar? Verify Euler's formula. Solution

Yes, $K_4$ can be drawn as a triangle with a vertex inside. $V=4, E=6, F=4$. $4-6+4=2$. ✓ Q3: Show $K_{3,3}$ is non-planar. Solution

If $K_{3,3}$ were planar, by Euler: $E \leq 3V - 6 = 3(6)-6 = 12$. But $K_{3,3}$ has $3 \times 3 = 9$ edges, which satisfies this inequality. Need stronger argument: In a bipartite planar graph, $E \leq 2V - 4$. $9 \leq 2(6)-4 = 8$ is false. So $K_{3,3}$ is non-planar. Join Discord PreviousGraph Theory BasicsNextGraph Coloring Apps