Lemma 5.1
PROOFS IN CHAPTER 5
X
deg(v) = 2|E|.
v∈V
Proof
Each edge has two ends.
Proposition 5.2 Let G1 = (V1 , E1 ) and G2 = (V2 , E2 ) be graphs and suppose that f : V1 → V2 is an
isomorphism. Then deg(f (v)) = deg(v), for each v ∈ V1 .
Proof
Let v have degree n in G1 , and w1 , w2 , . . . , wn be the vertices in G1 adjacent to v. Then f (w1 ), . . . , f (wn )
are the vertices in G2 adjacent to f (w). Hence f (w) has degree n in G2 .
Lemma 5.3 Let G be a connected graph and suppose that e is an edge of G which occurs in a circuit.
Then the graph H, say, obtained from G by deleting the edge e, is also connected.
Proof
Let u, w be any two vertices of G. By assumption they are joined by a wander in G, say u = v0 , v1 , . . . , vn =
w. If e is not one of the edges in this wander, then u is joined to w in H by the same wander.
However if e does occur in the wander in G from u to w we can proceed as follows. Since e is contained
in a circuit of G there is a path in G, not involving e, from the first vertex of e to the second. We can
use this path to replace each occurrence of e in our wander from u to w and thereby obtain a wander in
H from u to w.
The vertices of H are the same as the vertices of G. Hence we have shown that every pair of vertices
u, w in H is joined by a wander in H. In other words, H is connected.
Lemma 5.4 Suppose that G is a graph without circuits. Then G is connected if and only if G has the
property that the addition of any new edge (between existing vertices) creates a circuit.
Proof
Suppose G is connected.
Then if u, v are any pair of vertices of G they are joined by a path. So if {u, v} is not already an edge,
adding this edge creates a circuit.
Suppose adding any new edge between existing vertices creates a circuit.
Then given any pair u, v of vertices such that {u, v} is not an edge of G, by assumption adding {u, v}
to G creates a circuit, so u must already have been joined to v by a path. And given any pair u, v of
vertices such that {u, v} is an edge of G, the vertex u is joined to the vertex v by that edge. Thus every
pair u, v of vertices of G is connected by a path. So G is connected.
Lemma 5.5 A tree T which has ≥ 2 vertices contains a terminal vertex.
Proof
Let v0 , v1 , . . . , vr be a path of longest length in T , subject to the condition that vr 6= v0 . (There is a
longest such path, though not necessarily just one, since by definition such a path can only go through
each vertex once.)
Then vr cannot be joined by a single edge to any of v0 , . . . , vr−2 since this would mean T would contain
a circuit. And vr cannot be joined by a single edge to any other vertex of T except vr−1 since this would
mean we could find a longer path in T . Hence vr is joined only to vr−1 by an edge. Hence vr is a terminal
vertex.
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Proposition 5.6 A tree with (precisely) n ≥ 2 vertices has (precisely) n − 1 edges.
Proof
Induction on n.
Case n = 2
A tree with exactly two vertices has a single edge.
Assume a tree with k vertices has k − 1 edges, and let T be a tree with k + 1 vertices.
Choose a terminal vertex v of T (by 5.5). Then v is joined to the rest of T by a single edge e. Remove
v and e from T and call the remaining graph S. Since T is connected and has no circuits, and removing
a ‘twig’ neither disconnects a graph nor creates circuits, the graph S is also a tree. But S has k vertices
so it has k − 1 edges by the inductive hypothesis. Hence T which has the edges of S together with the
extra edge e, altogether has k edges.
The statement of the Proposition follows by the principle of induction.
Lemma 5.7 If G has n vertices, n − 1 edges and is connected then G is a tree.
Proof
If G is not a tree then it must have circuits (since by assumption G is connected). Now by the top-down
method we can remove edges from G until we have a spanning tree T . But T has n − 1 edges (by 5.5),
so we cannot have removed any. Hence G = T .
Theorem 5.9 Let G be a connected, costed graph. Then any spanning tree T for G constructed by
Kruskal’s algorithm is an economy tree for G.
Proof
Using Kruskal’s algorithm, choose a spanning tree T with edges e1 , e2 , . . . , , er , chosen in that order. T
will remain unchanged for the whole of the proof.
Let S be any spanning tree of G. Define the discrepancy of S to be the number of edges of T not in S.
We proceed by induction on the discrepancy, n, of S. Our statement P (n) is that every spanning tree S
of discrepancy n has c(T ) ≤ c(S).
Case n = 0. Then S = T so P (0) is true.
Assume P (k) is true and let S be a graph of discrepancy k + 1.
Now let ei be the first of the edges e1 , e2 , . . . , er of T that does not lie in S. Adding ei to S creates a
circuit, and this circuit must contain some edge f not in T (since T contains no circuits).
Let R be the graph obtained by adding ei to S and then removing f . Note that R is a tree (being a
connected graph with one more vertex than edges) and that it has discrepancy one less than that of S.
Hence by the inductive hypothesis c(T ) ≤ c(R).
But c(ei ) ≤ c(f ) (else we would have chosen f rather than ei when building the tree T by Kruskal’s
algorithm). Hence c(R) ≤ c(S).
Putting together the two inequalities gives c(T ) ≤ c(S) and completes the proof of the inductive step.
The statement of the theorem now follows by the principle of induction.
Proposition 5.10 The graph K3,3 is not planar and the graph K5 is not planar.
Proof follows below from that of 5.11 (Euler’s formula)
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Theorem 5.11 (Euler’s Formula) For a connected planar graph, |V | − |E| + |F | = 2, where |F | is the
number of faces.
Proof
Sketch proof only. Induction on the number of faces. If G has just one face then it has no circuits and
there it is a tree. Then |V | = n, |E| = n − 1 and |F |=1. Now assume the result for every connected
planar graph with k faces let G be a planar graph with k + 1 faces. Remove any single edge from G that
lies in the boundary of two different faces. This reduces |E| by one and also reduces |F | by one (since
two face are amalgamated into one). So the total |V | − |E| + |F | is unchanged. But the new graph has
k faces, so it has |V | − |E| + |F | = 2 by the inductive hypothesis. Hence G has |V | − |E| + |F | = 2, and
we have proved the induction step.
Proof of Proposition 5.10 Suppose K5 is planar.
The graph K5 has 5 vertices and 10 edges, so if it is planar it has 7 faces (by Euler’s formula). Each face
has at least 3 edges and each edge lies in at most two faces, so in order to have 7 faces it must have at
least (7 × 3)/2 edges. But it actually has 10 edges. Contradiction.
Suppose K3,3 is planar.
K3,3 has 6 vertices and 9 edges. So to be planar it would have to have 5 faces (by Euler’s formula). But
each face of K3,3 would have to have at least 4 edges around its boundary (since houses and utilities
alternate). As each edge can be in at most two faces this means we need a total of (5 × 4)/2 = 10 edges.
But K3,3 has only 9 edges. So it cannot be planar.
Theorem 5.12 (Kuratowski’s Theorem (1930)) A graph is planar if and only if it contains no graph
which is a subdivision of K5 or of K3,3 .
Proof ‘Only if’ is obvious from Proposition 5.10. But ‘if’ is hard, and we shall not prove it in this course.
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