14.2 Graph Theory: Trees, Cycles, Paths, and Planarity

14.2 Graph Theory: Trees, Cycles, Paths, and Planarity

Graph theory on the GRE Mathematics Subject Test evaluates vertex degrees, connectivity, Eulerian and Hamiltonian conditions, tree invariants, and planar graph embeddings. Fluency with Euler's formula, the Handshaking Lemma, and topological criteria is essential for solving discrete mathematics problems accurately.

---

Simple Graphs and the Handshaking Lemma

A simple graph $G = (V, E)$ has vertices $V$ and undirected edges $E \subseteq \binom{V}{2}$ without loops or parallel edges. The degree $\deg(v)$ is the number of edges incident to $v$.

Handshaking Lemma

For any finite undirected graph $G = (V, E)$: $\sum_{v \in V} \deg(v) = 2|E|$ Because the sum of vertex degrees is even, every graph contains an even number of vertices of odd degree.

Degree Sequences and Havel-Hakimi Theorem

A non-increasing sequence $d_1 \ge d_2 \ge \dots \ge d_n$ of non-negative integers is graphic if it is the degree sequence of a simple graph. The Havel-Hakimi theorem states that $d_1, \dots, d_n$ is graphic if and only if the sequence formed by deleting $d_1$ and subtracting 1 from the next $d_1$ terms is graphic.

---

Bipartite Graphs and Odd Cycles

A graph $G$ is bipartite if $V = V_1 \cup V_2$ with $V_1 \cap V_2 = \emptyset$ such that every edge joins $V_1$ and $V_2$.

• Odd-cycle characterization: A graph $G$ is bipartite if and only if $G$ contains no odd cycles.
• Chromatic Number: Every bipartite graph with $|E| \ge 1$ has chromatic number $\chi(G) = 2$.
• Complete Bipartite Graph $K_{m, n}$: Has $|V| = m + n$, $|E| = mn$, and is Hamiltonian if and only if $m = n \ge 2$.

---

Trees and Spanning Trees

A tree $T$ is an undirected graph satisfying any of these equivalent definitions:

1. $T$ is connected and acyclic.
2. $T$ is connected with $|E| = |V| - 1$.
3. $T$ is acyclic with $|E| = |V| - 1$.
4. Any two vertices are connected by a unique simple path.
5. $T$ is minimally connected (removing any edge disconnects it).
6. $T$ is maximally acyclic (adding any edge creates a unique cycle).

Leaves and Degree Distribution

Every finite tree with $|V| \ge 2$ contains at least two leaves ($\deg(v) = 1$). If internal vertices have degrees $d_1, \dots, d_k \ge 2$, the leaf count $\ell$ satisfies $\ell = 2 + \sum_{i=1}^k (d_i - 2)$.

Cayley's Formula and Spanning Trees

• Cayley's Formula: The number of distinct labeled trees on $n$ vertices is $n^{n-2}$.
• Matrix Tree Theorem (Kirchhoff): The number of spanning trees of $G$ equals any cofactor of the Laplacian matrix $L = D - A$, where $D$ is the diagonal degree matrix and $A$ is the adjacency matrix.

---

Eulerian and Hamiltonian Graphs

Euler's Criterion

• Connected $G$ has an Eulerian circuit $\iff$ every vertex has even degree.
• Connected $G$ has an Eulerian trail (not a circuit) $\iff$ exactly 2 vertices have odd degree.

Hamiltonian Sufficient Conditions

• Dirac's Theorem: If simple $G$ has $n \ge 3$ vertices and $\deg(v) \ge n/2$ for all $v \in V$, then $G$ is Hamiltonian.
• Ore's Theorem: If simple $G$ has $n \ge 3$ vertices and $\deg(u) + \deg(v) \ge n$ for every non-adjacent pair $u, v$, then $G$ is Hamiltonian.
• Note: Dirac and Ore conditions are sufficient but not necessary: cycle $C_n$ is Hamiltonian with all degrees 2.

---

Planar Graphs and Kuratowski's Theorem

A graph is planar if it can be drawn in $\mathbb{R}^2$ without intersecting edges. A planar embedding partitions the plane into $F$ faces.

Euler's Formula

For any connected planar graph: $V - E + F = 2$ For a planar graph with $k$ connected components, $V - E + F = 1 + k$.

Planarity Edge Bounds

Since each face is bounded by at least 3 edges ($2E \ge 3F$), substituting into Euler's formula gives: $E \le 3V - 6 \quad (\text{for } V \ge 3)$ If $G$ is triangle-free (girth $\ge 4$, including all bipartite planar graphs), then $2E \ge 4F$, yielding: $E \le 2V - 4 \quad (\text{for } V \ge 4)$

Non-Planar Prototypes and Kuratowski's Theorem

• $K_5$ is non-planar: $V = 5, E = 10 > 3(5) - 6 = 9$.
• $K_{3,3}$ is non-planar: Bipartite ($V = 6, E = 9 > 2(6) - 4 = 8$).
• Kuratowski's Theorem: $G$ is planar $\iff G$ contains no subgraph homeomorphic to $K_5$ or $K_{3,3}$ (Wagner's variant: no $K_5$ or $K_{3,3}$ graph minor).
• Four Color Theorem: Every planar graph is 4-vertex-colorable ($\chi(G) \le 4$).

---

Step-by-Step Worked Problem

Problem: A connected planar graph has 10 vertices and partitions the plane into 8 faces. Can this graph be bipartite?

Solution:

1. Apply Euler's formula: For connected planar graphs, $V - E + F = 2$. $10 - E + 8 = 2 \implies E = 16$
2. Evaluate the bipartite planar bound: Bipartite graphs contain no odd cycles (girth $\ge 4$). Every face is bounded by at least 4 edges, so $2E \ge 4F$, which implies $F \le E/2$.
3. Check maximum edge bound: $E \le 2V - 4 = 2(10) - 4 = 16$ The graph has $E = 16$, which satisfies the bound with equality.
4. Constructive verification: The complete bipartite graph $K_{2, 8}$ has $V = 10$ vertices and $E = 2 \times 8 = 16$ edges. It is bipartite and planar.
5. Conclusion: Yes, such a graph can be bipartite.

---

GRE Exam Traps & Pitfalls

Trap 1: Confusing Eulerian Parity with Hamiltonian Conditions Eulerian circuits depend solely on local vertex degree parity (all degrees even). Hamiltonian cycles depend on global cycle structure; degree criteria (Dirac/Ore) are sufficient, never necessary.

Trap 2: Using $E \le 3V - 6$ on Bipartite Graphs The bound $E \le 3V - 6$ fails to prove non-planarity for bipartite graphs like $K_{3,3}$ ($9 \le 12$ holds). For triangle-free graphs, always use the tighter bound $E \le 2V - 4$.

Trap 3: Labeled vs. Unlabeled Trees in Cayley's Formula Cayley's formula $n^{n-2}$ counts labeled trees on $n$ distinct vertices. Unlabeled non-isomorphic trees are vastly fewer and lack a simple power formula.

Trap 4: Neglecting Component Counts in Euler's Formula $V - E + F = 2$ requires connectedness. If a planar graph has $k$ components, use $V - E + F = 1 + k$.