1.2. PATHS, CYCLES, AND TRAILS - People.stat.sfu.ca

27 Chapter 1: Fundamental Concepts Section 1.2: Paths, Cycles, and Trails 28
u ∈ S and v /∈ S. Now there is an x, v-path formed by extending an x, upath
along the edge uv. This contradicts the choice of S, so in fact S is all
of V (G). Since there are paths from x to all other vertices, the transitivity
of the connection relation implies that G is connected.
1.2.23. a) If a connected simple graph G is not a complete graph, then every
vertex of G belongs to some induced subgraph isomorphic to P3. Let v be a
vertex of G. If the neighborhood of v is not a clique, then v has a pair x, y
of nonadjacent neighbors; {x, v, y} induces P3. If the neighborhood of v is a
clique, then since G is not complete there is some vertex y outside the set S
consisting of v and its neighbors. Since G is connected, there is some edge
between a neighbor w of v and a vertex x that is not a neighbor of v. Now
the set {v, w, x} induces P3, since x is not a neighbor of v.
One can also use cases according to whether v is adjacent to all other
vertices or not. The two cases are similar to those above.
b) When a connected simple graph G is not a complete graph, G may
have edges that belong to no induced subgraph isomorphic to P3. In the
graph below, e lies in no such subgraph.
•
e
•
• •
1.2.24. If a simple graph with no isolated vertices has no induced subgraph
with exactly two edges, then it is a complete graph. Let G be such a graph.
If G is disconnected, then edges from two components yield four vertices
that induce a subgraph with two edges. If G is connected and not complete,
then G has nonadjacent vertices x and y. Let Q be a shortest x, y-path; it
has length at least 2. Any three successive vertices on Q induce P3, with
two edges.
Alternatively, one can use proof by contradiction. If G is not complete,
then G has two nonadjacent vertices. Considering several cases (common
neighbor or not, etc.) always yields an induced subgraph with two edges.
1.2.25. Inductive proof that every graph G with no odd cycles is bipartite.
Proof 1 (induction on e(G)). Basis step (e(G) = 0): Every graph with
no edges is bipartite, using any two sets covering V (G).
Induction step (e(G) > 0): Discarding an edge e introduces no odd
cycles. Thus the induction hypothesis implies that G − e is bipartite.
If e is a cut-edge, then combining bipartitions of the components of
G − e so that the endpoints of e are in opposite sets produces a bipartition
of G. If e is not a cut-edge of G, then let u and v be its endpoints, and let
X, Y be a bipartition of G − e. Adding e completes a cycle with a u, v-path
in G − e; by hypothesis, this cycle has even length. This forces u and v to be
in opposite sets in the bipartition X, Y . Hence the bipartition X, Y of G − e
is also a bipartition of G.
Proof 2 (induction on n(G)). Basis step (n(G) = 1): A graph with one
vertex and no odd cycles has no loop and hence no edge and is bipartite.
Induction step (n(G) > 1): When we discard a vertex v, we introduce
no odd cycles. Thus the induction hypothesis implies that G −v is bipartite.
Let G1, . . . , Gk be the components of G − v; each has a bipartition. If v has
neighbors u, w in both parts of the bipartition of Gi, then the edges uv and
vw and a shortest u, w-path in Gi form a cycle of odd length. Hence we can
specify the bipartition Xi, Yi of Gi so that Xi contains all neighbors of v in
Gi. We now have a bipartition of G by letting X = Xi and Y = {v}∪( Yi).
1.2.26. A graph G is bipartite if and only if for every subgraph H of G,
there is an independent set containing at least half of the vertices of H.
Every bipartite graph has a vertex partition into two independent sets, one
of which must contain at least half the vertices (though it need not be a
maximum independent set). Since every subgraph of a bipartite graph is
bipartite, the argument applies to all subgraphs of a bipartite graph, and
the condition is necessary.
For the converse, suppose that G is not bipartite. By the characterization
of bipartite graphs, G contains an odd cycle H. This subgraph H
has no independent set containing at least half its vertices, because every
set consisting of at least half the vertices in an odd cycle must have two
consecutive vertices on the cycle.
1.2.27. The “transposition graph” on permutations of [n] is bipartite. The
partite sets are determined by the parity of the number of pairs i, j such
that i < j and ai > aj (these are called inversions). We claim that each
transposition changes the parity of the number of inversions, and therefore
each edge in the graph joins vertices with opposite parity. Thus the
permutations with an even number of inversions form an independent set,
as do those with an odd number of inversions. This is a bipartition, and
thus the graph is bipartite.
Consider the transposition that interchanges the elements in position
r and position s, with r < s. No pairs involving elements that are before r
or after s have their order changed. If r < k < s, then interchanging ar and
as changes the order of ar and ak, and also it changes the order of ak and as.
Thus for each such k the number of inversions changes twice and retains
the same parity. This describes all changes in order except for the switch
of ar and as itself. Thus the total number of changes is odd, and the parity
of the number of inversions changes.