Graph theory and simple circuit help

brad sue
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Hi,
I have this problem I don't understand I have been on it for days now:

Show that the existence of a simple circuit of length k, where k is a positive integer greater than 2, is an isomorphism invariant.

please can someone help me with it?
Thank you
B.
 
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Er, are you talking about graph theory?

What is the definition of

C is a simple circuit of length k?​

So what happens to C if you apply an isomorphism?
 
Yes, it is graph theory.
A simple circuit is a path starting to a point and end to the same point, passing through each edge once.
So what happens to C if you apply an isomorphism?
I really don't have any idea!
 
Okay, what is a path? We need to get to a formal statement, so that we can start proving things... so I'll cut to the chase. Using your definition of circuit (something feels slightly off, but I think it's minor, and I'll let you worry about it), a circuit of length k is:

(1) An alternating sequence of vertices and edges:
vertex_0, edge_1, vertex_1, edge_2, ..., edge_k, vertex_k

(2) The i-th edge is an edge between the (i-1)-th vertex and the i-th vertex.

(3) Every edge in your graph occurs exactly once in the sequence of edges.

(4) vertex_0 = vertex_k


So, can you write down a formal statement of what you're trying to prove?
 
MMMM!
Not sure
What do you want me to understand is that 2 graphs are isomorphic if they both have a simple circuit of same length ( besides same # of vertices, edges and degree for vertices)??
Not sure Perhaps I an tired!
 
What do you want me to understand is that 2 graphs are isomorphic if they both have a simple circuit of same length
Nope; that's backwards.

Show that the existence of a simple circuit of length k, where k is a positive integer greater than 2, is an isomorphism invariant.​

This wants you to show that if two graphs are isomorphic, then one has such a circuit iff the other does.
 
Man! I still don't find it.
Please, I need more help with this problem.
 
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