Back to Combinatorics

Chapter 15: Fundamentals of Graph Theory

October 11, 2026 Wasil Zafar 33 min read

The graph structures and counting theorems combinatorics needs

Table of Contents

  1. Why do labeled trees have such a simple counting formula?
  2. 15.1 Graphs, subgraphs, degree sequences
  3. 15.2 Trees, spanning trees, Cayley's formula
  4. 15.3 Connectivity, Eulerian and Hamiltonian structures
  5. 15.4 Planar graphs and Euler's formula
  6. 15.5 Graph coloring and chromatic polynomials
  7. 15.6 Matrix-tree theorem
  8. 15.7 Menger's theorem and connectivity dualities
  9. 15.8 Tutte polynomials and deletion–contraction
  10. 15.9 Graph minors and forbidden-subgraph characterizations
  11. Compare tree counts, colorings, and deletion–contraction recurrences

Why do labeled trees have such a simple counting formula?

Graph theory and combinatorics are taught together because a graph is fundamentally a combinatorial structure. Combinatorics is the study of discrete structures, counting, and arrangements, while graph theory studies relationships between discrete pairs of objects.

1. Graphs Are Set-Theoretic Objects

Formally, a graph $G = (V, E)$ consists of a set of vertices $V$ and a set of edges $E$, where each edge is a 2-element subset of $V$. Because edges are subsets selected from $\binom{n}{2}$ candidate pairs, analyzing graphs relies directly on combinatorial subset selection and binomial coefficients.

2. Core Questions Are Combinatorial
  • Enumerative (Counting): How many labeled trees exist on $n$ vertices? (Cayley's formula: $n^{n-2}$). How many proper vertex colorings exist? (Chromatic polynomials).
  • Existence & Extremal: How many edges force a cycle or a complete clique? (Turán's Theorem).
  • Ramsey Theory: How large must a graph be to guarantee complete order or fixed substructures?
3. Shared Proof Techniques
  • Pigeonhole Principle: Proves that any simple graph with $n \ge 2$ vertices contains at least two vertices with identical degrees.
  • Double Counting: Summing vertex degrees yields the Handshaking Lemma: $\sum_{v \in V} d_G(v) = 2m$.
  • Inclusion–Exclusion: Used to compute chromatic polynomials and count matchings.
  • Probabilistic Method: Erdős's technique for proving graph existence via random spaces.
4. Computer Science Applications

Combinatorics governs the search space and time complexity of graph algorithms. Network optimization (minimum spanning trees, shortest paths, max flows) and intractable routing puzzles (Traveling Salesperson Problem exploring $(n-1)!/2$ Hamiltonian tours) rely directly on these counting foundations.

The Foundation of Graph Theory in Combinatorics A conceptual diagram showing how vertex sets and edge subsets form graphs, leading to key combinatorial pillars: enumeration, extremal bounds, structural dualities, and network algorithms. Graph G = (V, E) Subsets E ⊆ ⎛V⎞ / 2 Enumeration Cayley's formula: nⁿ⁻² Prüfer codes & trees Extremal Bounds Handshaking lemma Turán's theorem & degrees Structural Dualities Max flow = Min cut Menger's connectivity Invariants Chromatic polynomials Tutte polynomial T(x,y)

Suppose four named devices must be connected by an undirected network that has a route between every pair and contains no redundant cycles. Which links can we choose, and how many different networks are possible? If every pair of devices can be linked, there are $16$ valid answers. For $n$ named devices, the answer is the remarkably compact $n^{n-2}$.

The labels matter: a network centered at device $1$ differs structural-wise from one centered at device $2$. Restrictions on available links also matter: the formula counts trees in the complete graph $K_n$, not spanning trees in arbitrary restricted networks. We will explain this count by encoding a tree as a sequence, then connect tree counting to degree constraints, connectivity, colorings, and deletion–contraction.

This chapter develops the graph theory needed for combinatorial proofs and enumeration. Review Combinatorial Thinking, Induction and Recursive Reasoning, and Matchings. Linear algebra is needed only for the optional matrix-tree section. For a dedicated structural treatment, use Carnegie Mellon's Graph Theory course notes; for graph representations and full algorithm implementations, use Princeton's graph algorithms materials. math.cmu.edu

Motivating Problem

Why do labeled trees have such a simple counting formula?

Inputs. Use vertices $\{1,2,3,4\}$ and any subset of the six possible links. Require connectivity and no cycle; these networks are called trees.

Two distinct topological shapes exist for $n=4$: the star (edges $\{12,13,14\}$) and the path (edges $\{12,23,34\}$). Here $ij$ abbreviates the unordered edge $\{i,j\}$.

ShapeLabeling countReason
Star$4$Choose its central vertex ($4$ choices); the other three vertices become leaves.
Path$4!/2 = 12$Order the four labels along a line ($4! = 24$); reversing the order yields identical undirected edges, so divide by $2$.

These are the only tree shapes on four vertices, as the degree argument in Section 15.2 will confirm. Thus there are $4 + 12 = 16 = 4^{4-2}$ labeled trees.

Tree Shapes on Four Labeled Vertices Left: The Star shape with center 1 and leaves 2, 3, 4 (1 of 4 labelings). Right: The Path shape 1-2-3-4 (1 of 12 labelings). Star Shape (4 Distinct Labelings) 1 2 3 4 Center choices: {1}, {2}, {3}, or {4} Path Shape (12 Distinct Labelings) 1 2 3 4 Ordered paths: 4! / 2 symmetry = 12

Question to revisit. Can every tree on $n$ labels be encoded by exactly $n-2$ entries, each chosen independently from those $n$ labels? For the star centered at $1$, repeatedly remove the smallest leaf and record its neighbor: remove $2$, record $1$; remove $3$, record $1$. Its Prüfer code is $(1,1)$. We will prove that this procedure is reversible.

15.1 Graphs, subgraphs, degree sequences

Concept & Definition. Unless stated otherwise, a graph is finite, undirected, and simple: $G=(V,E)$, where $V$ is a vertex set and $E$ is a set of unordered pairs of distinct vertices. There are no self-loops (edges from a vertex to itself) or parallel edges (multiple edges joining the same pair). Write $n=|V|$ for the number of vertices and $m=|E|$ for the number of edges. Vertices joined by an edge are adjacent; the degree $d_G(v)$ is the number of edges incident with vertex $v$.

Subgraphs & Containment Types. Subgraphs capture local structural views of a larger network:

  • A general subgraph $H=(V',E')$ satisfies $V'\subseteq V$ and $E'\subseteq E$, where all edge endpoints in $E'$ lie in $V'$.
  • An induced subgraph $G[S]$, for a subset $S\subseteq V$, keeps $S$ as its vertex set and retains all edges of $G$ whose endpoints both lie in $S$. You cannot cherry-pick edges; if two vertices in $S$ were joined in $G$, their edge must exist in $G[S]$.
  • A spanning subgraph keeps all vertices ($V'=V$), while potentially deleting edges ($E'\subseteq E$).

Our primary running example throughout this chapter is a square with one diagonal:

$$ V(G)=\{A,B,C,D\},\qquad E(G)=\{AB,BC,CD,DA,AC\}. $$

Subgraphs of the Running Graph Comparison of the full running graph G (Square ABCD with diagonal AC), an induced subgraph G[{A,B,C}], and a spanning subgraph C4. Full Graph G A B C D Degrees: (3, 2, 3, 2), m=5 Induced G[{A,B,C}] A B C D Retains all edges between A,B,C Spanning C₄ (Delete AC) A B C D Keeps all V={A,B,C,D}, omits AC

Its degrees in order $(A,B,C,D)$ are $(3,2,3,2)$. The induced subgraph $G[\{A,B,C\}]$ is a triangle $K_3$; the spanning subgraph with edges $AB,BC,CD,DA$ is a four-cycle $C_4$. The familiar complete graph $K_n$ contains all $\binom{n}{2}$ possible edges. A complete bipartite graph $K_{a,b}$ consists of two disjoint vertex partitions of sizes $a$ and $b$, containing all $ab$ possible cross-edges and no internal edges.

Handshake Lemma (Double Counting Fundamentals). Counting vertex–edge incidences in two ways yields:

$$ \sum_{v\in V}d_G(v)=2m. $$

Why it works: Every edge has exactly two endpoints, so placing a token at every endpoint-edge incidence puts two tokens on every edge. Consequently, the sum of all degrees is always an even number. A fundamental corollary is that the number of vertices of odd degree in any graph is always even. In our running example, $d(A)+d(B)+d(C)+d(D) = 3+2+3+2 = 10 = 2(5)$, confirming $m=5$.

Degree Sequences & Havel–Hakimi Theorem. A degree sequence lists the vertex degrees of a graph, typically in nonincreasing order $(d_1, d_2, \ldots, d_n)$. A sequence is graphical if there exists a simple graph realizing those exact degrees.

Necessary conditions for a sequence to be graphical are: $0\le d_i\le n-1$ and an even sum. However, these are not sufficient. For instance, $(3,3,1,1)$ has an even sum ($8$) and respects $d_i \le 3$. But two vertices of degree $3$ in a 4-vertex graph must connect to every other vertex, forcing the remaining two vertices to have degree at least $2$. Thus $(3,3,1,1)$ cannot be realized by any simple graph.

The definitive test for graphical sequences is the Havel–Hakimi reduction algorithm:

Havel–Hakimi Reduction Algorithm:
  1. Sort the integer sequence in nonincreasing order: $(d_1, d_2, \ldots, d_n)$.
  2. Remove the first element $d_1$, and subtract $1$ from each of the next $d_1$ entries.
  3. Re-sort the remaining sequence in nonincreasing order.
  4. Repeat steps 1–3 until either:
    • A negative entry appears, or there are insufficient remaining elements $\implies$ NOT graphical.
    • An all-zero sequence is obtained $\implies$ IS graphical.
Havel-Hakimi Reduction Trace Trace of Havel-Hakimi reduction starting from (3,3,2,2), subtracting 1 from 3 elements to get (2,1,1), then (0,0). Step 1: Initial (3, 3, 2, 2) Delete 3, subtract 1 from next 3 Step 2: Reduced (2, 1, 1) Delete 2, subtract 1 from next 2 Step 3: Base Case (0, 0) ✓ Graphical sequence proven!

Why the reduction is valid. A realization of the reduced sequence extends directly by adding a vertex adjacent to those $d_1$ selected high-degree vertices. For the converse, start with any realization and call its maximum-degree vertex $v$ ($d(v)=d_1$). Suppose a desired high-degree neighbor $a$ is absent while an undesired lower-degree neighbor $b$ is present. Since $d(a)\ge d(b)$ and $b$ meets $v$ while $a$ does not, there must exist a vertex $w\ne v,a,b$ adjacent to $a$ but not $b$. Swapping edges $\{vb, aw\}$ with $\{va, bw\}$ preserves every vertex degree, introduces no loops or parallel edges, and increases the overlap between $v$'s neighbors and the desired high-degree set. Repeating this swap makes all desired adjacencies available; deleting $v$ yields the reduced graphical realization.

Exercise. Is $(2,2,2,2)$ graphical? Is its realization necessarily a tree? Hint: draw a square. Solution: The four-cycle $C_4$ realizes $(2,2,2,2)$, so it is graphical. However, $C_4$ contains a cycle, so it is not a tree. Degree constraints alone do not establish acyclicity.

15.2 Trees, spanning trees, Cayley's formula

Fundamentals & Definitions.

  • A path is a sequence of distinct connected vertices.
  • A graph is connected if a path exists between every pair of vertices.
  • A cycle is a closed path with at least three distinct vertices.
  • A tree is a connected graph containing no cycles. A forest is an acyclic graph (a disjoint union of trees).
  • A leaf is a vertex of degree $1$. Every tree on $n \ge 2$ vertices has at least two leaves.

Key Structural Theorem: A tree on $n$ vertices has exactly $n-1$ edges. Proof by Leaf Induction: Find a longest path in $T$; its endpoint $v$ must be a leaf (otherwise the path could be extended, or an extra edge would close a cycle). Removing leaf $v$ and its incident edge yields a smaller tree $T'$ on $n-1$ vertices. By induction, $T'$ has $n-2$ edges, so $T$ has $(n-2) + 1 = n-1$ edges.

Summing over components, a forest with $n$ vertices and $c$ components has $n-c$ edges. A connected graph contains a spanning tree: repeatedly delete edges from cycles until no cycle remains while preserving connectivity. Consequently, any connected graph on $n$ vertices with exactly $n-1$ edges is a tree.

Cayley's Formula. For $n\ge2$, the number of distinct labeled trees on the set $[n]=\{1,2,\ldots,n\}$ is:

$$ n^{n-2}. $$

Proof via Prüfer Sequences (Encoding & Decoding Bijection). Heinz Prüfer established a 1-to-1 correspondence between labeled trees on $[n]$ and sequences of length $n-2$ with entries from $[n]$.

Prüfer Encoding (Tree $\to$ Sequence)
  1. Locate the smallest labeled leaf $\ell$ in current tree $T$.
  2. Record the label of $\ell$'s unique neighbor $a_i$ in the code.
  3. Remove leaf $\ell$ and its edge from $T$.
  4. Repeat until only $2$ vertices remain. Output sequence $(a_1, a_2, \ldots, a_{n-2})$.
Prüfer Decoding (Sequence $\to$ Tree)
  1. Start with sequence $S$ and available label set $L = \{1, 2, \ldots, n\}$.
  2. Find the smallest label $\ell \in L$ that does not appear in remaining sequence $S$.
  3. Add edge $\{\ell, s_1\}$ to the tree, where $s_1$ is the first element of $S$.
  4. Remove $\ell$ from $L$ and drop $s_1$ from $S$. Repeat until $S$ is empty.
  5. Join the final two labels remaining in $L$.

Degree–Occurrence Invariant: In a Prüfer code, vertex $i$ appears exactly $d_i - 1$ times, where $d_i$ is its degree in the tree. This immediate consequence refines Cayley's count for prescribed positive degree vectors $(d_1, \ldots, d_n)$ via multinomial coefficients:

$$ \frac{(n-2)!}{\prod_{i=1}^n(d_i-1)!}. $$

Prüfer Sequence Decoding Workshop Visual trace decoding sequence (4,4,2) on vertices {1,2,3,4,5} to build tree with edges 1-4, 3-4, 2-4, 2-5. Decoding Prüfer Code (4, 4, 2) on Vertices {1, 2, 3, 4, 5} Step 1 Code: (4, 4, 2) | L: {1..5} Min absent from code: 1 Add Edge {1, 4} Remaining Code: (4, 2) Remaining L: {2, 3, 4, 5} Step 2 Code: (4, 2) | L: {2..5} Min absent from code: 3 Add Edge {3, 4} Remaining Code: (2) Remaining L: {2, 4, 5} Step 3 Code: (2) | L: {2, 4, 5} Min absent from code: 4 Add Edge {4, 2} Remaining Code: () Remaining L: {2, 5} Final Step Join final pair {2, 5} 1 3 4 2 5 Tree Degrees: (1,2,1,3,1)
Proof Workshop

Trees, Spanning Trees, Cayley's Formula: Worked Example

Decode $(4,4,2)$ on label set $[5]$.

Remaining SequenceAvailable LabelsSmallest Absent LabelEdge Added
$(4,4,2)$$1,2,3,4,5$$1$$\{1,4\}$
$(4,2)$$2,3,4,5$$3$$\{3,4\}$
$(2)$$2,4,5$$4$$\{2,4\}$
Empty$2,5$Final pair$\{2,5\}$

The decoded tree has edges $\{14,34,24,25\}$ and degree vector $(1,2,1,3,1)$. Re-encoding removes leaves $1,3,4$ in order, regenerating sequence $(4,4,2)$.

15.3 Connectivity, Eulerian and Hamiltonian structures

Route Hierarchy & Fundamentals.

  • A walk is an alternating sequence of vertices and edges $v_0, e_1, v_1, e_2, \ldots, v_k$ (repeating vertices and edges allowed).
  • A trail is a walk with no repeated edges.
  • A path is a walk with no repeated vertices (and thus no repeated edges).
  • A component is a maximal connected subgraph.
  • A bridge is an edge whose deletion increases the component count. An edge is a bridge if and only if it lies on no cycle.
  • A cut vertex is a vertex whose deletion increases the component count.

Eulerian Structures (Traversing Every Edge). An Euler circuit is a closed trail using every edge of $G$ exactly once.

  • Theorem (Euler, 1736): A connected graph contains an Euler circuit if and only if every vertex has an even degree.
  • A connected graph contains an open Euler trail (starting and ending at distinct vertices) if and only if exactly two vertices have odd degree (which serve as the start and end points).

Hamiltonian Structures (Visiting Every Vertex). A Hamiltonian cycle is a closed path visiting every vertex of $G$ exactly once before returning to the start.

  • Unlike Eulerian circuits (which depend locally on degree parity), determining whether a graph contains a Hamiltonian cycle is NP-complete.
  • Necessary conditions: A Hamiltonian simple graph on $n \ge 3$ vertices must have minimum degree $\delta(G) \ge 2$ and can contain no cut vertices.
Euler circuits and Hamiltonian cycles impose different conditions Left: triangles OAB and OCD share only O. O is a cut vertex of degree 4. Has Euler circuit, but no Hamiltonian cycle. Right: square ABCD with diagonal AC. Has Hamiltonian cycle along perimeter, but no Euler circuit due to odd degrees at A and C. Two Triangles Sharing Vertex O Square with Diagonal AC OAB CD AB CD deg 2deg 2 deg 2deg 2 deg 4 (Cut Vertex) deg 3deg 2 deg 2deg 3 Euler circuit: YES O → A → B → O → C → D → O Hamiltonian cycle: NO (O is a cut vertex) Hamiltonian cycle: YES A → B → C → D → A (Boundary) Euler circuit: NO (A, C have odd degrees)

Exercise. Classify the star $K_{1,3}$: connected, Euler circuit, open Euler trail, Hamiltonian cycle? Hint: its degrees are $3,1,1,1$. Answer: Connected, but none of the three requested tours exists. Four odd-degree vertices rule out both Eulerian tours; degree-$1$ leaf vertices rule out a Hamiltonian cycle.

15.4 Planar graphs and Euler's formula

Concept & Definitions. A graph is planar if it can be drawn in the two-dimensional plane such that no two edges cross each other. A plane graph is a specific planar embedding of such a graph. Drawing a plane graph partitions the plane into contiguous geometric regions called faces, including one infinite unbounded outer face.

Euler's Formula for Planar Graphs. For any connected plane graph with $n$ vertices, $m$ edges, and $f$ faces:

$$ n - m + f = 2. $$

Proof Mechanism by Edge Restoration: Begin with a spanning tree of the plane graph. It has $n$ vertices, $n-1$ edges, and exactly $1$ face (the exterior region). Checking Euler's formula gives $n - (n-1) + 1 = 2$. Now restore the remaining $m - (n-1)$ non-tree edges one by one in their plane positions. Each added edge connects two vertices within an existing face, splitting that face into two independent regions. Thus, each edge restoration increases both $m$ and $f$ by $1$, leaving $n - m + f = 2$ invariant.

Euler's Planar Face Formula and Non-Planar Graphs Left: Plane drawing of square plus diagonal showing 3 faces (F1, F2, F3). Right: K5 and K3,3 non-planar structures. Plane Graph (Square + Diagonal) Face F₁ Face F₂ Outer Face F₃ A B C D n=4, m=5, f=3 ⇒ 4 - 5 + 3 = 2 ✓ Planar Edge Bounds & Canonical Obstructions Complete Graph K₅ n = 5, m = 10 Bound: m ≤ 3n - 6 10 ≤ 3(5) - 6 = 9 ✗ Non-Planar! Bipartite K₃,₃ n = 6, m = 9 (No C₃) Bound: m ≤ 2n - 4 9 ≤ 2(6) - 4 = 8 ✗ Non-Planar!

Planar Edge Bounds. Summing boundary lengths across all face walks yields $\sum \text{len}(F_i) = 2m$. In a simple connected planar graph with $n \ge 3$, every face boundary requires at least $3$ edges, so $3f \le 2m$. Substituting $f \le \frac{2}{3}m$ into Euler's formula yields the maximum edge bound:

$$ m \le 3n - 6. $$

If $G$ is also triangle-free (no $C_3$, such as bipartite graphs), every face boundary requires at least $4$ edges ($4f \le 2m$), giving the tighter bound:

$$ m \le 2n - 4. $$

15.5 Graph coloring and chromatic polynomials

Concept & Fundamentals. A proper vertex coloring assigns a color to each vertex such that no two adjacent vertices share the same color. The chromatic number $\chi(G)$ is the minimum number of colors required to properly color $G$. The chromatic polynomial $P_G(q)$ is a counting function that outputs the exact number of proper vertex colorings of $G$ using a palette of $q$ available colors.

Null Graph ($n$ isolated)

$P_G(q) = q^n$

Every vertex chooses any of $q$ colors independently.
Trees on $n$ Vertices

$P_T(q) = q(q-1)^{n-1}$

Root vertex has $q$ choices; every child avoids its parent's color ($q-1$).
Complete Graph $K_n$

$P_{K_n}(q) = q(q-1)\cdots(q-n+1)$

Every vertex requires a distinct color choice.

Deletion–Contraction Recurrence for Colorings. For any edge $e = uv$ in $G$:

$$ P_G(q) = P_{G-e}(q) - P_{G/e}(q). $$

Why the minus sign? Consider proper $q$-colorings of $G-e$ (where constraint edge $e$ is removed). In $G-e$, endpoints $u$ and $v$ can either receive different colors or the same color:

  • Colorings of $G-e$ where $c(u) \ne c(v)$ are valid proper colorings of $G$.
  • Colorings of $G-e$ where $c(u) = c(v)$ correspond to proper colorings of contracted graph $G/e$ (where $u$ and $v$ are merged).

Thus, $P_{G-e}(q) = P_G(q) + P_{G/e}(q)$, which rearranging yields $P_G(q) = P_{G-e}(q) - P_{G/e}(q)$.

Deletion-Contraction Recurrence for Graph Coloring Visual breakdown of P_G(q) = P_{G-e}(q) - P_{G/e}(q) on triangle K3. Graph G (K₃) u v w Target Edge e = vw = Deletion G - e (Path P₃) u v w P(q) = q(q-1)² − Contraction G / e u vw P(q) = q(q-1)

Worked Example (Square + Diagonal Graph). In $G$, color non-adjacent vertices $A,C$ first in $q(q-1)$ ways (forcing $c(A) \ne c(C)$). Vertices $B$ and $D$ are each adjacent to both $A$ and $C$, so each must choose from the remaining $q-2$ available colors. Thus:

$$ P_G(q) = q(q-1)(q-2)^2. $$

For $q=3$ colors, $P_G(3) = 3(2)(1)^2 = 6$ proper three-colorings exist, confirming $\chi(G)=3$.

15.6 Matrix-tree theorem

Optional Extension: Algebraic Enumeration. For a loopless multigraph $G$ with adjacency matrix $A$ and diagonal degree matrix $D = \mathrm{diag}(d(v_1), \ldots, d(v_n))$, the Graph Laplacian Matrix is defined as:

$$ L = D - A. $$

Properties of $L$: Diagonal entries $L_{ii} = d(v_i)$; off-diagonal entries $L_{ij} = -m_{ij}$ (where $m_{ij}$ is the number of edges between $v_i$ and $v_j$). Every row and column sum of $L$ is identically $0$.

Graph Laplacian Matrix and Determinant Cofactor Construction of Laplacian L = D - A for square with diagonal, showing deletion of row/column D to compute det(L^(D)) = 8. Laplacian L = D − A for Running Graph G Full 4×4 Laplacian L ⎡ 3 -1 -1 -1 ⎤ ⎢ -1 2 -1 0 ⎥ ⎢ -1 -1 3 -1 ⎥ ⎣ -1 0 -1 2 ⎦ ← Delete Row/Col 4 (D) Reduced Laplacian L^{(D)} det ⎡ 3 -1 -1 ⎤ ⎢ -1 2 -1 ⎥ = 8 ⎣ -1 -1 3 ⎦ Spanning Trees τ(G) = 8 ✓

Matrix-Tree Theorem (Kirchhoff, 1847). Delete any arbitrary row $r$ and column $r$ from $L$ to obtain reduced matrix $L^{(r)}$. The number of spanning trees $\tau(G)$ is given by:

$$ \tau(G) = \det L^{(r)}. $$

15.7 Menger's theorem and connectivity dualities

Concept & Duality. Connectivity evaluates network robustness. For non-adjacent target vertices $s$ and $t$:

  • An $s$–$t$ vertex separator $S \subseteq V \setminus \{s,t\}$ is a set of vertices whose deletion disconnects $s$ and $t$.
  • Paths are internally vertex-disjoint if they share no common vertices other than $s$ and $t$.

Menger's Theorem (Vertex Form). For non-adjacent vertices $s, t$:

$$ \max \{\text{internally vertex-disjoint } s\text{–}t \text{ paths}\} = \min \{|S| : S \text{ separates } s, t\}. $$

Menger's Theorem Duality and Network Flow Reduction Left: 2 vertex-disjoint paths B-A-D and B-C-D separated by {A,C}. Right: Vertex splitting transformation for max-flow min-cut duality. Disjoint Paths vs Separator Set B D A C Separator S = {A, C} (|S| = 2) Network Flow Transformation (Vertex Splitting) v_in cap = 1 v_out Reduces Menger's Theorem directly to Ford-Fulkerson Max-Flow Min-Cut Duality!

Global Connectivity Hierarchy. For connected simple graphs with $n \ge 2$:

$$ \kappa(G) \le \lambda(G) \le \delta(G), $$

where $\kappa(G)$ is vertex connectivity, $\lambda(G)$ is edge connectivity, and $\delta(G)$ is minimum degree.

15.8 Tutte polynomials and deletion–contraction

Optional Extension: The Master Unifying Invariant. For a multigraph $G=(V,E)$ and edge subset $A \subseteq E$, let $c(A)$ denote the number of connected components of spanning subgraph $(V,A)$. Define the rank function $r(A) = n - c(A)$.

The Tutte polynomial $T_G(x,y)$ is defined as:

$$ T_G(x,y)=\sum_{A\subseteq E} (x-1)^{r(E)-r(A)}(y-1)^{|A|-r(A)}. $$

Its deletion–contraction recurrence rules are:

$$ T_G(x,y)=\begin{cases} T_{G-e}(x,y)+T_{G/e}(x,y),& e \text{ is an ordinary edge (neither loop nor bridge)},\\ x T_{G/e}(x,y),& e \text{ is a bridge},\\ y T_{G-e}(x,y),& e \text{ is a loop}. \end{cases} $$

Tutte Polynomial Reduction Tree for K3 Reduction tree for K3 yielding Tutte polynomial x^2 + x + y. Tutte Polynomial Deletion–Contraction Tree: T_{K₃}(x,y) = x² + x + y T(K₃) Deletion G - e (Path P₃) Bridges ⇒ x² Contraction G / e (C₂) Bridge + Loop ⇒ x + y

Universal Specializations:

  • $T_G(1,1) = \tau(G)$ (Number of Spanning Trees).
  • $T_G(2,1) = 2^{|E|-n+c}$ (Number of Spanning Subgraphs).
  • Chromatic Polynomial: $P_G(q) = (-1)^{r(E)} q^{c(E)} T_G(1-q, 0)$.

15.9 Graph minors and forbidden-subgraph characterizations

Optional Extension: Structural Containment Hierarchy.

Containment TypeOperations AllowedKey Properties
SubgraphDelete vertices, delete edges.Preserves maximum degree $\Delta(G)$.
Induced SubgraphDelete vertices (retains all induced edges).Preserves cliques and independent sets.
SubdivisionReplace edges with internally vertex-disjoint paths.Topological containment.
Graph MinorDelete vertices, delete edges, contract edges.Universal structural containment.
Hierarchy of Graph Containment Operations Visual comparison showing Subgraph -> Subdivision -> Minor via edge contraction. 1. Subgraph Delete Vertices / Edges H ⊆ G 2. Topological Minor Subdivide Edges into Paths Subdivision TS(H) ⊆ G 3. Graph Minor Contract Edges e = uv H ≤_minor G

Classical Planarity Characterizations:

  • Kuratowski's Theorem (1930): A finite graph is planar if and only if it contains no subgraph that is a subdivision of $K_5$ or $K_{3,3}$.
  • Wagner's Theorem (1937): A finite graph is planar if and only if it contains neither $K_5$ nor $K_{3,3}$ as a minor.

Robertson–Seymour Graph Minor Theorem (2004): In any infinite sequence of finite undirected graphs, some graph is a minor of a later one. Consequently, every minor-closed property of finite graphs is characterizable by a finite set of forbidden minors.

Compare tree counts, colorings, and deletion–contraction recurrences

Return to running graph $G$ ($V=\{A,B,C,D\}$, $E=\{AB,BC,CD,DA,AC\}$). Spanning trees count connected backbone topologies under addition, whereas proper colorings count conflict-free state assignments under subtraction.

Comparison of Tree and Chromatic Deletion-Contraction Trees Dual tree showing spanning trees adding counts vs chromatic polynomials subtracting counts. Dual Recurrence Tree on Target Edge e = AC Graph G (Square + AC) Deletion G - e (C₄) Spanning Trees: τ(C₄) = 4 Chromatic: P(3) = 18 Contraction G / e Spanning Trees: τ(G/e) = 4 Chromatic: P(3) = 12
Investigation

Compare tree counts, colorings, and deletion–contraction recurrences

1. Warm-up. Find the degrees of $G$, $G-e$, and $G/e$.

Solution: $G$ has degrees $(3,2,3,2)$ with degree sum $10$ ($m=5$). $G-e = C_4$ has degrees $(2,2,2,2)$ with sum $8$ ($m=4$). $G/e$ merges $A,C$ into $X$, yielding degrees $(4,2,2)$ in order $(X,B,D)$ with sum $8$ ($m=4$ parallel edges retained).

2. Spanning Tree Recurrence. Prove $\tau(G) = \tau(G-e) + \tau(G/e)$ and compute all counts.

Solution: Partition spanning trees by whether they contain edge $e=AC$:

  • Trees avoiding $AC$ are spanning trees of $G-e = C_4$, giving $\tau(C_4) = 4$.
  • Trees containing $AC$ correspond bijectively to spanning trees of $G/e$, giving $2 \times 2 = 4$ choices.

Thus, $\tau(G) = 4 + 4 = 8$. Of the $\binom{5}{3} = 10$ three-edge subsets, exactly two form triangles ($\{AB,BC,AC\}$ and $\{AC,CD,DA\}$), leaving $10 - 2 = 8$ valid spanning trees.

3. Chromatic Recurrence. Calculate $P_G(q) = P_{G-e}(q) - P_{G/e}(q)$ and evaluate at $q=3$.

Solution: $P_{G-e}(q) = (q-1)^4 + (q-1)$ for $C_4$. $P_{G/e}(q) = q(q-1)^2$. Subtraction gives:

$$ P_G(q) = q(q-1)(q-2)^2. $$

At $q=3$: $P_{G-e}(3) = 18$, $P_{G/e}(3) = 12$, yielding $P_G(3) = 18 - 12 = 6$.

4. Exhaustive Enumeration Checksum Table.

Quantity$G-e$$G/e$ (parallel edges)Graph $G$Recurrence Check
Spanning Trees $\tau$$4$$4$$8$$4 + 4 = 8$ ✓
3-Colorings $P(3)$$18$$12$$6$$18 - 12 = 6$ ✓

Next in the Series

In Chapter 16: Directed Graphs and Network Flows, we'll convert feasibility, assignments, and connectivity into integral flow networks.