Unit 7: Planar Graphs, Euler's Formula, Kuratowski's Theorem & Duality
Comprehensive theory of graph planarity and embeddings: planar embeddings, Jordan curve theorem in planarity, faces/regions, Euler's polyhedral formula V - E + F = 2, maximum edge bounds for planar graphs (E <= 3V - 6), non-planarity of K_5 and K_3,3, Kuratowski's theorem, Wagner's minors, geometric and combinatorial duals, and Whitney's theorem on planar duality.
§7.1 Planar Embeddings, Jordan Curves & Face Partition
1. Planar Graphs and Planar Embeddings
A graph is an abstract topological entity. When drawn in a continuous geometric space, edges may or may not cross.
Definition 7.1 (Planar Graph and Plane Graph):
- A graph $G$ is called planar if it can be drawn in the Euclidean plane $\mathbb{R}^2$ (or on the 2-sphere $S^2$) such that no two edges intersect, except at their common endpoints.
- A specific non-crossing drawing of a planar graph in $\mathbb{R}^2$ is called a planar embedding (or a plane graph).
2. Faces and Regions of a Plane Graph
A planar embedding partitions the plane $\mathbb{R}^2 \setminus G$ into topologically connected open domains called faces (or regions).
Definition 7.2 (Faces and the Outer Face): Let $G$ be a plane graph.
- The connected components of $\mathbb{R}^2 \setminus G$ are the faces of $G$.
- Exactly one face is unbounded, called the exterior face (or infinite face).
- All other faces are bounded, called interior faces.
- The boundary of a face $f$ consists of the vertices and edges that enclose $f$.
- The degree of a face $f$, denoted $\deg(f)$, is the number of edges along its boundary (with cut-edges counted twice).
Theorem 7.1 (Handshaking Lemma for Faces): In any plane graph $G$ with edge set $E$ and face set $F$:
Proof:
Every edge either separates two distinct faces $f_1 \ne f_2$ (contributing 1 to $\deg(f_1)$ and 1 to $\deg(f_2)$), or is a bridge surrounded on both sides by the same face $f$ (contributing 2 to $\deg(f)$). In either case, each edge contributes exactly 2 to the sum of face degrees. $\blacksquare$
§7.2 Euler's Polyhedral Formula $V - E + F = 2$ & Corollaries
1. Euler's Formula for Connected Plane Graphs
Published by Leonhard Euler in 1758 for convex 3D polyhedra, this formula forms the bedrock of topological graph theory:
Theorem 7.2 (Euler's Formula): Let $G$ be a connected plane graph with $V$ vertices, $E$ edges, and $F$ faces. Then:
Complete Formal Proof (by Induction on Edges):
We prove the formula by induction on the number of edges $E$ for any connected plane graph on $V$ vertices:
- Base Case (Trees):
If $G$ contains no cycles, then $G$ is a tree. By Theorem 3.1, $E = V - 1$. Since a tree contains no cycles, it does not divide the plane into interior faces; it has only the single unbounded exterior face: $F = 1$. Substituting these values:
The base case holds with exact equality!
- Inductive Step:
Assume the formula $V - E + F = 2$ holds for all connected plane graphs with strictly fewer than $E$ edges. Let $G$ be a connected plane graph with $E$ edges that is not a tree. Since $G$ is not a tree, $G$ must contain at least one simple cycle $C$. Choose an edge $e$ lying on the cycle $C$. By the Jordan Curve Theorem, the cycle $C$ separates the plane into an interior domain and an exterior domain. Therefore, the edge $e$ lies on the boundary of two distinct faces $f_1 \ne f_2$. Now, remove the edge $e$ from $G$, forming the graph $G' = G - e$. Notice:
- Since $e$ lay on a cycle, $G'$ remains connected.
- The vertex count is unchanged: $V' = V$.
- The edge count decreases by 1: $E' = E - 1$.
- The two faces $f_1$ and $f_2$ merge into a single unified face in $G'$, so the face count decreases by 1: $F' = F - 1$.
By the induction hypothesis applied to $G'$:
Substituting $V' = V$, $E' = E - 1$, and $F' = F - 1$:
Therefore, $V - E + F = 2$ holds for all connected plane graphs. $\blacksquare$
Corollary 7.2.1 (Disconnected Planar Graphs): If a planar graph $G$ has $\omega(G) = k$ connected components:
§7.3 Edge Upper Bounds & Non-Planarity of $K_5$ and $K_{3,3}$
1. Maximum Edges in Simple Planar Graphs
In a simple plane graph, every face boundary must contain at least 3 edges (no self-loops or multiple edges):
Theorem 7.3 (Planar Edge Upper Bound): In any simple planar graph $G$ with $V \ge 3$ vertices and $E$ edges:
If $G$ is triangle-free (contains no $C_3$, such as a bipartite graph):
Complete Proof:
1. General Bound ($E \le 3V - 6$):
By the Face Handshaking Lemma (Theorem 7.1) and $\deg(f) \ge 3$:
Substitute this into Euler's formula $V - E + F = 2$:
2. Triangle-Free Bound ($E \le 2V - 4$):
If $G$ contains no triangles, then every face boundary must have at least 4 edges: $\deg(f) \ge 4$.
Substitute into Euler's formula:
2. The Canonical Non-Planar Graphs: $K_5$ and $K_{3,3}$
Theorem 7.4 (Non-Planarity of $K_5$ and $K_{3,3}$):
- The complete graph $K_5$ is non-planar.
- The complete bipartite graph $K_{3,3}$ (the Utility Graph) is non-planar.
Proof of 1 ($K_5$):
For $K_5$, vertex count $V = 5$, edge count $E = \binom{5}{2} = 10$. If $K_5$ were planar, it would satisfy Theorem 7.3:
$10 \le 9$ is a contradiction! Therefore, $K_5$ is non-planar. $\blacksquare$
Proof of 2 ($K_{3,3}$):
For $K_{3,3}$, vertex count $V = 3 + 3 = 6$, edge count $E = 3 \times 3 = 9$. $K_{3,3}$ is bipartite, so it contains no odd cycles; in particular, it is triangle-free! If $K_{3,3}$ were planar, it would satisfy the triangle-free bound:
$9 \le 8$ is a contradiction! Therefore, $K_{3,3}$ is non-planar. $\blacksquare$
§7.4 Kuratowski's Theorem, Subdivisions & Wagner's Minors
1. Graph Subdivisions and Homeomorphism
Definition 7.3 (Elementary Subdivision): An elementary subdivision of an edge $e = \{u, v\}$ is the operation of replacing $e$ with two edges $\{u, w\}$ and $\{w, v\}$ by inserting a new vertex $w$ of degree 2. Two graphs $G_1$ and $G_2$ are homeomorphic (or topological minors) if both can be obtained from the same graph by a sequence of elementary subdivisions.
A graph $H$ is a subdivision of $K$ if $H$ can be obtained from $K$ by subdividing zero or more edges. Notice that subdividing an edge does not change planarity: $G$ is planar if and only if any subdivision of $G$ is planar.
2. Kuratowski's Theorem
In 1930, Polish mathematician Kazimierz Kuratowski established the definitive characterization of planarity:
Theorem 7.5 (Kuratowski's Theorem, 1930): A graph $G$ is planar if and only if it contains no subgraph that is a subdivision of $K_5$ or $K_{3,3}$.
3. Wagner's Theorem and Graph Minors
In 1937, Klaus Wagner formulated an alternative characterization based on minor operations:
- Minor Operations: Edge deletion, vertex deletion, and edge contraction (merging the two endpoints of an edge).
- A graph $H$ is a minor of $G$ if $H$ can be obtained from $G$ by a finite sequence of deletions and contractions.
Theorem 7.6 (Wagner's Theorem, 1937): A graph $G$ is planar if and only if it contains neither $K_5$ nor $K_{3,3}$ as a minor.
§7.5 Geometric Duals, Combinatorial Duality & Whitney's Theorem
1. Construction of the Geometric Dual $G^*$
Let $G$ be a plane graph with vertex set $V$, edge set $E$, and face set $F$.
Definition 7.4 (Geometric Dual $G^$): The geometric dual $G^ = (V^, E^)$ is constructed as follows:
- For each face $f_i \in F(G)$, place a corresponding dual vertex $v_i^ \in V^$.
- For each edge $e \in E(G)$ that separates face $f_i$ and face $f_j$, draw a dual edge $e^ \in E^$ connecting $v_i^$ and $v_j^$, crossing $e$ exactly once.
(If $e$ is surrounded on both sides by the same face $f_i$, $e^$ is a self-loop on $v_i^$).
Dual Parameters:
If $G$ is connected, then $(G^)^ \cong G$ (the dual of the dual is isomorphic to the primal graph).
2. Whitney's Planar Duality Theorem
Theorem 7.7 (Whitney's Duality Theorem, 1932): A graph $G$ is planar if and only if it possesses a combinatorial dual: That is, there exists a graph $G^$ and an edge bijection $\phi: E(G) \to E(G^)$ such that for every subset $S \subseteq E(G)$:
Step-by-step rigorous derivations with unskipped proofs, categorized into Foundational Concepts, Advanced Structural Analysis, and Honors / Proof Challenge tiers.
- Verify Euler's formula $V - E + F = 2$ for the 1-skeletons of all five Platonic solids: (a) Regular Tetrahedron, (b) Cube (Hexahedron), (c) Regular Octahedron, (d) Regular Dodecahedron, (e) Regular Icosahedron. 2. A connected planar graph has 10 vertices and 15 edges. Determine the exact number of faces in any planar embedding of $G$.
Part 1: Verification of Euler's Formula on the Platonic Solids
(a) Regular Tetrahedron:
- $V = 4$, $E = 6$, $F = 4$ (all faces triangular).
- $V - E + F = 4 - 6 + 4 = 2$. Verified!
(b) Cube (Hexahedron):
- $V = 8$, $E = 12$, $F = 6$ (all faces quadrilateral).
- $V - E + F = 8 - 12 + 6 = 2$. Verified!
(c) Regular Octahedron:
- $V = 6$, $E = 12$, $F = 8$ (all faces triangular).
- $V - E + F = 6 - 12 + 8 = 2$. Verified!
(d) Regular Dodecahedron:
- $V = 20$, $E = 30$, $F = 12$ (all faces pentagonal).
- $V - E + F = 20 - 30 + 12 = 2$. Verified!
(e) Regular Icosahedron:
- $V = 12$, $E = 30$, $F = 20$ (all faces triangular).
- $V - E + F = 12 - 30 + 20 = 2$. Verified!
Notice the beautiful duality:
- Tetrahedron is self-dual ($V=4, F=4$).
- Cube ($V=8, F=6$) is dual to Octahedron ($V=6, F=8$).
- Dodecahedron ($V=20, F=12$) is dual to Icosahedron ($V=12, F=20$).
Part 2: Face Count for $V = 10, E = 15$
Given $V = 10$ and $E = 15$ in a connected planar graph. By Euler's Formula (Theorem 7.2):
Substitute values:
Any planar embedding of $G$ will contain exactly 7 faces (6 interior faces and 1 exterior face). $\blacksquare$
- Prove that every simple planar graph must contain at least one vertex of degree at most 5: $\delta(G) \le 5$. 2. The Petersen graph has 10 vertices and 15 edges, and has girth 5 (its shortest cycle has length 5). Use generalized Euler face bounds to prove rigorously that the Petersen graph is non-planar.
Part 1: Proof that $\delta(G) \le 5$ in every Simple Planar Graph
Let $G = (V, E)$ be a simple planar graph with $V = n$ vertices and $E = m$ edges.
- If $n \le 6$, then $\delta(G) \le n - 1 \le 5$ trivially.
- Assume $n \ge 7$.
Assume for contradiction that $\delta(G) \ge 6$. Then every vertex in $G$ has degree at least 6:
Summing over all $n$ vertices and applying Euler's Handshaking Lemma:
However, by Theorem 7.3, every simple planar graph must satisfy:
Combining both inequalities:
This is a direct mathematical impossibility! Therefore, our assumption that every vertex has degree $\ge 6$ was false. Hence, every simple planar graph must contain at least one vertex with degree $\le 5$:
Part 2: Non-Planarity of the Petersen Graph via Girth Bound
Let $G$ be the Petersen graph:
- Vertices: $V = 10$.
- Edges: $E = 15$ (3-regular graph on 10 vertices).
- Girth: $g(G) = 5$ (contains no 3-cycles or 4-cycles; shortest cycle is $C_5$).
Suppose for contradiction that the Petersen graph were planar. In any planar embedding of $G$, the boundary of every face $f$ must be a cycle. Because the shortest cycle in $G$ has length 5, every face boundary must contain at least 5 edges:
By the Face Handshaking Lemma (Theorem 7.1):
Now substitute into Euler's Formula $V - E + F = 2$:
Rearranging for $E$:
Now evaluate this bound for the Petersen graph ($V = 10$):
Since $E = 15$, we have:
This is an absolute contradiction! Therefore, the Petersen graph cannot be embedded in the plane without edge crossings. Hence the Petersen graph is non-planar. $\blacksquare$
- Prove that any graph $G$ containing a subgraph homeomorphic to $K_5$ or $K_{3,3}$ is non-planar. 2. Demonstrate that the Petersen graph contains a minor isomorphic to $K_5$, and identify the explicit edge contraction sequence.
Part 1: Proof that Kuratowski Subgraphs are Non-Planar
Let $G$ be a graph containing a subgraph $H \subseteq G$ such that $H$ is a subdivision of $K \in \{K_5, K_{3,3}\}$.
- In Theorem 7.4, we proved from first principles that $K_5$ and $K_{3,3}$ are non-planar.
- Consider an elementary subdivision of an edge $e = \{u, v\}$ into $\{u, w\}$ and $\{w, v\}$.
If a graph $H$ has a planar embedding, placing vertex $w$ on the curve representing edge $e$ preserves the planar embedding. Conversely, if $H$ with vertex $w$ has a planar embedding, erasing the degree-2 vertex $w$ and joining the two curve segments preserves planarity. Therefore, a graph is planar if and only if any subdivision of the graph is planar.
- Because $K_5$ and $K_{3,3}$ are non-planar, every subdivision of $K_5$ and every subdivision of $K_{3,3}$ is non-planar.
- If $G$ contains a non-planar subgraph $H$, then $G$ cannot be planar (a planar embedding of $G$ would immediately restrict to a planar embedding of $H$).
Therefore, any graph containing a subdivision of $K_5$ or $K_{3,3}$ is non-planar. $\blacksquare$
Part 2: $K_5$ Minor in the Petersen Graph
The Petersen graph consists of:
- Outer 5-cycle: $V_{\text{out}} = \{u_0, u_1, u_2, u_3, u_4\}$ with edges $\{u_i, u_{i+1 \pmod 5}\}$.
- Inner 5-star: $V_{\text{in}} = \{v_0, v_1, v_2, v_3, v_4\}$ with edges $\{v_i, v_{i+2 \pmod 5}\}$.
- Five spoke edges connecting corresponding vertices: $E_{\text{spokes}} = \{ \{u_i, v_i\} : i = 0, \dots, 4 \}$.
Edge Contraction Sequence: Contract each of the 5 spoke edges $\{u_0, v_0\}, \{u_1, v_1\}, \{u_2, v_2\}, \{u_3, v_3\}, \{u_4, v_4\}$:
- For each $i \in \{0, 1, 2, 3, 4\}$, contracting $\{u_i, v_i\}$ merges $u_i$ and $v_i$ into a single composite vertex $w_i$.
- What edges exist between the composite vertices $w_i$?
- The outer cycle edges $\{u_i, u_{i+1}\}$ become edges between $w_i$ and $w_{i+1 \pmod 5}$.
- The inner star edges $\{v_i, v_{i+2}\}$ become edges between $w_i$ and $w_{i+2 \pmod 5}$.
- Together, for every pair of indices $i \ne j$, either $|i - j| \equiv 1 \pmod 5$ (an outer edge) or $|i - j| \equiv 2 \pmod 5$ (an inner edge)!
- Every pair of distinct vertices $\{w_i, w_j\}$ ($0 \le i < j \le 4$) is connected by an edge!
The contracted graph has 5 vertices $\{w_0, w_1, w_2, w_3, w_4\}$, and every pair is connected by an edge. This is precisely the complete graph $K_5$!
By Wagner's Theorem (Theorem 7.6), because the Petersen graph contains $K_5$ as a minor, the Petersen graph is non-planar. $\blacksquare$