Prove that any two connected graphs of order n with degree sequence (2,2, …, 2) are isomorphic.

Name: Problem 10, Chapter 11: Introductory Combinatorics
Uploaded: 2020-08-06T01:16:47-08:00
Duration: 10 min 8 s
Channel: Chris Trentman
Description: Prove that any two connected graphs of order n with degree sequence (2,2, …, 2) are isomorphic.

Prove that any two connected graphs of order n with degree...

Key Concepts

Connected Graph

A connected graph is one in which there exists a path between every pair of vertices. This concept is crucial because connectivity guarantees that the graph is a single cohesive structure rather than a collection of disjoint parts, which is a key property when considering isomorphism between graphs.

Degree Sequence

The degree sequence of a graph is a list of its vertex degrees. For a graph with degree sequence (2,2,…,2), each vertex has exactly two neighbors. This uniformity strongly constrains the graph's structure, often leading to special types such as cycles in the context of connected graphs.

Graph Isomorphism

Graph isomorphism is the concept that two graphs are structurally the same if there is a bijective mapping between their vertex sets that preserves adjacency. Demonstrating isomorphism means showing that, despite possible differences in the appearance or labeling of the vertices, the two graphs have an identical connectivity pattern.

Cycle Graph

A cycle graph is a connected graph where each vertex has degree 2, forming a single closed loop. Recognizing that any connected graph with the degree sequence (2,2,…,2) is essentially a cycle graph is key to proving their isomorphism, as all cycle graphs of the same order are isomorphic.

Transcript

00:01 We're asked to prove that the property of being bipartite is invariant.

00:18 So suppose that the graph g is bipartite and h is another graph, and we have that g is isomorphic to h.

01:00 We want to show that h is bipartite.

01:07 So since they're isomorphic, that means there is function f from v1 to v2, such that f is 1 to 1 and onto, and we have that in the expanded sense of the isomorphism of graphs, an edge uv lies in e1, if and only if the edge f of u, you f of v lies in e2, and since g is bipartite, it follows that we can partition vertex set v1 into the disjoint union of v2 and v3, or sorry, not v2, v3 and v4.

02:47 And we have that because the graph is bipartite, follows that uvv is an edge for graph g, if and we have that if u belongs to v3 and so if uv is an edge in g, and it follows that u is an element of v3 and v4, or we have that u is an element of v4, or we have that u is an element of v4, well, i'll keep it this way.

04:02 So it follows that the graph contains only edges between a vertex of v3 and a vertex of v4.

04:09 Now, define v5 to be the set of all vertices f of v in the graph h such that v3.

04:33 So this is the image of v3 essentially.

04:38 And define v6 to be the set of all f of v, such that v lies in v4.

04:49 So this is the image of v4 in the graph h, you could say, that is the vertices.

05:01 So we have that f is one to one.

05:09 It follows that v5 and v6 are disjoint for otherwise if we had some...