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Pro5 10

A connected graph G with size at least 2 is nonseparable if and only if any two adjacent edges lie on a common cycle. The proof demonstrates that if G is nonseparable, then adjacent edges can be part of a cycle, and conversely, if adjacent edges are on a common cycle, then no vertex can be a cut-vertex, ensuring G remains nonseparable. Thus, the conditions are equivalent.

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Anmol Murti
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43 views1 page

Pro5 10

A connected graph G with size at least 2 is nonseparable if and only if any two adjacent edges lie on a common cycle. The proof demonstrates that if G is nonseparable, then adjacent edges can be part of a cycle, and conversely, if adjacent edges are on a common cycle, then no vertex can be a cut-vertex, ensuring G remains nonseparable. Thus, the conditions are equivalent.

Uploaded by

Anmol Murti
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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5.

10 Prove that a connected graph G of size at least 2 is


nonseparable if, and only if any two adjacent edges of G lie on a
common cycle of G.

Proof: Evidently, if G has size at least 2, then G must have


order at least 3.

⇒: Suppose that G is nonseparable and let e = uv and f = vw be


arbitrary adjacent edges of G. Then, since the order of G must be
at least 3, Theorem 5.7 implies there is a cycle C containing u and
w. The cycle C contains two distinct u - w paths P and P ′ . At
least one of them does not contain v. Say for definiteness it is
P. Then P followed by the path w, v , u is a cycle in G containing
e and f.

⇐: Suppose any two adjacent edges of G lie on a common cycle.


Let v be an arbitrary vertex of G. Since G is connected,
deg(v) ≥ 1. If deg(v) = 1, then v is not a cut-vertex. Thus, we
may assume deg(v) ≥ 2. Then (1) either all vertices of G adjacent
to v lie in the same component of G - v, in which case G - v is
connected and v is not a cut-vertex, or (2) there must be distinct
neighboring vertices u and w adjacent to v with u and w lying in
different components of G - v, in which case G - v is separated and
v is a cut-vertex.
We shall show the second possibility cannot happen. Let u and
w be any two neighbors of v. Then e = uv and f = vw are adjacent
edges in G. By hypothesis there is a cycle C in G containing e and
f. Observe that C cannot contain any other of the edges of G that
may be incident with v. It follows that by removing v and the
edges e and f from C, we obtain a u - w path P in G that is also a
path in G - v. Thus, it is not possible to have u separated from
w in G - v.
Thus, with the present hypotheses, the second possibility
above cannot happen, and v cannot be a cut-vertex of G. Since v
was arbitrary, G must be nonseparable.//

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