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Chapter 38. Euler's Formula

Chapter 38. Euler’s Formula

Euler’s formula is the basic counting law for connected plane graphs. It relates three quantities: the number of vertices, the number of edges, and the number of faces. If a connected planar graph is drawn without crossings, then

ve+f=2. v - e + f = 2.

Here vv is the number of vertices, ee is the number of edges, and ff is the number of faces, including the unbounded outer face. This formula holds for every finite connected plane graph.

38.1 Statement of the Formula

Let GG be a connected plane graph. Write

v=V(G),e=E(G),f=number of faces of G. v = |V(G)|, \qquad e = |E(G)|, \qquad f = \text{number of faces of }G.

Then Euler’s formula states that

ve+f=2. v - e + f = 2.

The number

ve+f v - e + f

is called the Euler characteristic of the plane embedding. For connected plane graphs, its value is always 22. The drawing may change, the shapes of the faces may change, and the outer face may change, but the value remains the same.

38.2 First Examples

A single vertex has

v=1,e=0,f=1. v = 1,\qquad e = 0,\qquad f = 1.

Hence

ve+f=10+1=2. v - e + f = 1 - 0 + 1 = 2.

A triangle has

v=3,e=3,f=2. v = 3,\qquad e = 3,\qquad f = 2.

The two faces are the bounded triangular face and the unbounded outer face. Therefore

ve+f=33+2=2. v - e + f = 3 - 3 + 2 = 2.

A square with one diagonal has

v=4,e=5,f=3. v = 4,\qquad e = 5,\qquad f = 3.

The diagonal divides the inside of the square into two triangular faces, and there is also the outer face. Thus

ve+f=45+3=2. v - e + f = 4 - 5 + 3 = 2.

These examples show the pattern. Adding an edge may increase the number of faces. Adding a vertex may increase the number of edges. The alternating sum remains fixed.

38.3 Trees as the Base Case

A tree with vv vertices has

e=v1. e = v - 1.

A tree has no cycle, so a plane drawing of a tree encloses no bounded face. Therefore it has only one face, the outer face:

f=1. f = 1.

Substituting into Euler’s expression gives

ve+f=v(v1)+1=2. v - e + f = v - (v-1) + 1 = 2.

Thus Euler’s formula holds for every tree.

This case is important because every connected graph contains a spanning tree. A connected plane graph can be reduced to a tree by deleting suitable edges from cycles.

38.4 Proof by Deleting Cycle Edges

Let GG be a connected plane graph.

If GG is a tree, the formula has already been proved. Suppose GG is not a tree. Then GG contains at least one cycle.

Choose an edge aa that lies on a cycle. Removing aa does not disconnect the graph, because the rest of the cycle still gives a path between the endpoints of aa. Hence the new graph remains connected.

What happens to vv, ee, and ff?

The number of vertices is unchanged:

v=v. v' = v.

The number of edges decreases by one:

e=e1. e' = e - 1.

The removed edge separated two faces. After the edge is deleted, those two faces merge into one. Therefore

f=f1. f' = f - 1.

Now compute:

ve+f=v(e1)+(f1)=ve+f. v' - e' + f' = v - (e-1) + (f-1) = v - e + f.

Thus deleting a cycle edge preserves the value of ve+fv - e + f.

Repeat this process until no cycle remains. The resulting graph is a tree. Since the expression ve+fv - e + f did not change during the deletion process, and since it equals 22 for the final tree, it also equals 22 for the original graph.

Therefore every connected plane graph satisfies

ve+f=2. v - e + f = 2.

This is the standard edge-deletion proof of Euler’s formula.

38.5 Why the Outer Face Counts

The outer face must be counted. Without it, the formula would fail.

For a triangle, the bounded region inside the triangle is one face. If only bounded faces were counted, then

v=3,e=3,f=1, v = 3,\qquad e = 3,\qquad f = 1,

and

ve+f=1. v - e + f = 1.

Counting the outer face gives f=2f = 2, and the formula becomes

33+2=2. 3 - 3 + 2 = 2.

The outer face is not an exception. It is a genuine face of the embedding. It is unbounded, but it is still one of the connected regions left by the drawing.

38.6 Face Degrees

The degree of a face is the number of edge appearances along its boundary. An edge may appear twice on the same face boundary, as happens with a bridge.

For a connected plane graph, the sum of all face degrees is

2e. 2e.

Each edge has two sides. In ordinary cases, those two sides border two different faces. If the edge is a bridge, both sides border the same face. In either case, the edge contributes two appearances to the total face-boundary count.

Thus

faces Fdeg(F)=2e. \sum_{\text{faces } F} \deg(F) = 2e.

This identity is the face version of the handshaking lemma.

38.7 The Edge Bound for Simple Planar Graphs

Euler’s formula implies that simple planar graphs cannot have too many edges.

Let GG be a simple connected planar graph with v3v \geq 3. Since GG is simple, every face has degree at least 33. Therefore

3f2e. 3f \leq 2e.

Euler’s formula gives

f=2v+e. f = 2 - v + e.

Substitute this into the inequality:

3(2v+e)2e. 3(2 - v + e) \leq 2e.

Expanding,

63v+3e2e. 6 - 3v + 3e \leq 2e.

Hence

e3v6. e \leq 3v - 6.

So every simple planar graph with at least three vertices satisfies

e3v6. e \leq 3v - 6.

This inequality is one of the most useful consequences of Euler’s formula. It gives an immediate obstruction to planarity.

38.8 Application to K5K_5

The complete graph K5K_5 has

v=5 v = 5

and

e=(52)=10. e = \binom{5}{2} = 10.

If K5K_5 were planar, the edge bound would give

e3v6. e \leq 3v - 6.

For v=5v = 5,

3v6=156=9. 3v - 6 = 15 - 6 = 9.

But K5K_5 has 1010 edges. Since

10>9, 10 > 9,

K5K_5 is not planar.

This proof avoids any attempt to draw all possible pictures of K5K_5. The contradiction comes from counting alone.

38.9 Bipartite Planar Graphs

A stronger edge bound holds for bipartite planar graphs.

Let GG be a simple connected bipartite planar graph with v3v \geq 3. A bipartite graph has no odd cycle. In particular, it has no triangle. Therefore every face has degree at least 44.

Thus

4f2e. 4f \leq 2e.

Equivalently,

2fe. 2f \leq e.

Using Euler’s formula,

f=2v+e. f = 2 - v + e.

Substitute:

2(2v+e)e. 2(2 - v + e) \leq e.

Then

42v+2ee. 4 - 2v + 2e \leq e.

Therefore

e2v4. e \leq 2v - 4.

So every simple connected bipartite planar graph with at least three vertices satisfies

e2v4. e \leq 2v - 4.

38.10 Application to K3,3K_{3,3}

The complete bipartite graph K3,3K_{3,3} has two parts of size 33. It has

v=6 v = 6

and

e=33=9. e = 3 \cdot 3 = 9.

If K3,3K_{3,3} were planar, the bipartite planar edge bound would give

e2v4. e \leq 2v - 4.

For v=6v = 6,

2v4=124=8. 2v - 4 = 12 - 4 = 8.

But K3,3K_{3,3} has 99 edges. Since

9>8, 9 > 8,

K3,3K_{3,3} is not planar.

Together, K5K_5 and K3,3K_{3,3} are the two classical minimal obstructions that appear in Kuratowski’s theorem.

38.11 Disconnected Plane Graphs

Euler’s formula in the form

ve+f=2 v - e + f = 2

assumes that the graph is connected.

If a plane graph has cc connected components, then the formula becomes

ve+f=c+1. v - e + f = c + 1.

For example, two disjoint triangles have

v=6,e=6. v = 6,\qquad e = 6.

The drawing has three faces: the inside of the first triangle, the inside of the second triangle, and the common outer face. Thus

f=3. f = 3.

Hence

ve+f=66+3=3. v - e + f = 6 - 6 + 3 = 3.

Since c=2c = 2,

c+1=3. c + 1 = 3.

The formula agrees.

38.12 Euler’s Formula and Polyhedra

Euler’s formula first appears naturally in the study of polyhedra. A convex polyhedron has vertices, edges, and faces, and these satisfy

VE+F=2. V - E + F = 2.

A convex polyhedron can be projected onto the plane by choosing one face as the outer face. The resulting drawing is a connected plane graph. This explains why the same formula governs both convex polyhedra and connected planar graphs.

For a cube,

V=8,E=12,F=6. V = 8,\qquad E = 12,\qquad F = 6.

Then

VE+F=812+6=2. V - E + F = 8 - 12 + 6 = 2.

For a tetrahedron,

V=4,E=6,F=4. V = 4,\qquad E = 6,\qquad F = 4.

Then

VE+F=46+4=2. V - E + F = 4 - 6 + 4 = 2.

The graph-theoretic version and the polyhedral version are expressions of the same topological invariant.

38.13 Limits of the Formula

Euler’s formula gives necessary conditions for planarity, but it does not give sufficient conditions.

If a graph violates

e3v6, e \leq 3v - 6,

then it is nonplanar. But if a graph satisfies this inequality, it may still be nonplanar.

For example, K3,3K_{3,3} has

v=6,e=9. v = 6,\qquad e = 9.

The general planar bound gives

3v6=12. 3v - 6 = 12.

Since 9129 \leq 12, the general bound alone does not rule out planarity. The bipartite bound is needed.

Even that is still not a complete planarity test. Complete characterizations require deeper theorems, such as Kuratowski’s theorem or Wagner’s theorem.

38.14 Summary

Euler’s formula states that every connected plane graph satisfies

ve+f=2. v - e + f = 2.

It can be proved by reducing a connected plane graph to a tree through deletion of cycle edges. Each such deletion lowers both the edge count and the face count by one, leaving ve+fv - e + f unchanged.

The formula has several important consequences:

ResultFormula
Connected plane graphve+f=2v - e + f = 2
Disconnected plane graph with cc componentsve+f=c+1v - e + f = c + 1
Simple planar graph, v3v \geq 3e3v6e \leq 3v - 6
Simple bipartite planar graph, v3v \geq 3e2v4e \leq 2v - 4
Sum of face degreesdeg(F)=2e\sum \deg(F) = 2e

Euler’s formula is a counting theorem, but its force is structural. It shows that planar graphs are sparse, explains the nonplanarity of K5K_5 and K3,3K_{3,3}, and connects graph theory with the topology of surfaces.