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NotesMath AI HLTopic 3.14Introduction to graph theory
Back to Math AI HL Topics
3.14.12 min read

Introduction to graph theory

IB Mathematics: Applications and Interpretation • Unit 3

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Contents

  • Vertices, edges, degree & the handshake lemma
  • Naming graphs: simple, connected, complete, bipartite, trees
A graph is dots joined by lines: Picture a map of 5 towns with roads between some of them. Draw each town as a dot and each road as a line joining two dots — that picture is a graph.

The dots are vertices (one is a vertex), the lines are edges. A graph only records which towns are connected, not how far apart they are drawn — so the same graph can be sketched in many ways.

The degree of a vertex is simply how many edge-ends touch it — how many roads leave that town. A town with 3 roads has degree 3.
A loop (an edge from a vertex back to itself) adds 2 to the degree.
The handshake lemma — why degrees add to twice the edges: Every edge has two ends, and each end adds 1 to the degree of the vertex it touches.

So if you add up the degrees of all vertices, you count every edge exactly twice (once for each end).

That gives the handshake lemma: the sum of all degrees equals 2 × (number of edges). It's a fast check on any graph — and it forces the number of odd-degree vertices to be even.
E = number of edges. So the total degree is always even.

IB-style question — degrees of a road network

A small road network links 5 towns A, B, C, D, E. The roads are: A–B, A–C, A–D, B–C, C–D, C–E.

Write the degree of each town, and use the handshake lemma to check the number of roads.

Step by step

  1. Count the roads meeting each town. A touches B, C, D → 3 roads.
  2. B touches A, C → 2. C touches A, B, D, E → 4. D touches A, C → 2. E touches C → 1.
  3. Add all degrees.
  4. Handshake lemma: total degree = 2E, so E = 12 ÷ 2 = 6 — which matches the six roads listed.

Final answer

Degrees 3, 2, 4, 2, 1; total 12 = 2 × 6, confirming there are 6 roads.

Special families of graphs: The exam expects you to recognise and use these names:

Simple — no loops (an edge from a vertex to itself) and no repeated edges between the same two vertices. Most real networks are simple.

Connected — you can travel from any vertex to any other by following edges (no isolated piece).

Complete — every pair of vertices is joined by an edge. The complete graph on n vertices is written Kₙ.

Bipartite — the vertices split into two groups, and edges only go between the groups, never within a group (e.g. students on one side, clubs on the other).

Tree — a connected graph with no cycles. A tree on n vertices has exactly n − 1 edges — the fewest edges that still keep everything connected.
In Kₙ every vertex has degree n − 1.

IB-style question — a complete social network

Six members of a project team are each directly messaging every other member, modelled by the complete graph K₆.

(a) How many direct message links are there? (b) State the degree of each member.

Step by step

  1. (a) In Kₙ every pair is joined, so count pairs: edges = n(n − 1)/2.
  2. (b) In a complete graph each vertex joins to all n − 1 others.
  3. Check with the handshake lemma: 6 vertices × degree 5 = 30 = 2 × 15. ✓

Final answer

(a) 15 links. (b) Every member has degree 5.

IB-style question — is it a tree?

A company's reporting structure is connected, has 9 staff (vertices) and 8 reporting links (edges), with no cycles.

Explain whether this is a tree, and what would happen to connectivity if one link were removed.

Step by step

  1. A tree is connected with no cycles — both hold here.
  2. Check the edge count: a tree on n vertices has n − 1 edges.
  3. A tree has the fewest edges that keep it connected, so removing any single edge disconnects it.

Final answer

Yes — it is a tree (connected, no cycles, 8 = 9 − 1 edges). Removing any link would break it into two disconnected groups.

IB Exam Questions on Introduction to graph theory

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Define

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AO1
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AO2
Explain

Give reasons WHY — cause and effect within Introduction to graph theory.

AO3
Evaluate

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AO3
Discuss

Present arguments FOR and AGAINST with a balanced conclusion.

AO3

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