Prove G contains a cycle of length at least k+1

  • Thread starter cxc001
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  • #1
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This is a graph theory related question.

Let G be a simple graph with min. degree k, where k>=2. Prove that G contains a cycle of length at least k+1.

Am I suppose to use induction to prove G has a path length at least k first, then try to prove that G has a cycle of length at least k+1? Or should I go directly use induction to prove G contains a cycle of length at least k+1???
 

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  • #2
fzero
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Consider one of the vertices, [tex]\nu_k[/tex] of degree [tex]k[/tex]. Let the set of vertices which are connected to this one be [tex]\{ \nu_i^{(k)} \}[/tex]. What is the minimum path length of a cycle [tex]C^{(k)}[/tex] that connects all of the [tex]\nu_i^{(k)}[/tex]? Can you see a way to construct a cycle [tex]C'[/tex] of degree [tex]\text{deg}\, C^{(k)}+1[/tex]?
 
  • #3
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Tell me if I'm not on the right track.

Use induction on k.

Pick a path P of maximum length, and suppose vertex vi is a vertex on this path, which has degree at least k, with a set of adjacent vertices {w1,w2,…,wj}, the adjacent vertex set must be on the path.
The minimum path length of a cycle Ci that connected all of vertex set {w1,w2,…,wj } is k.
Then extend the path to a cycle by adding the edge wjvi, so the resulting cycle has length at least k+1.
We're done! Does this induction make sense?
 
  • #4
Office_Shredder
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with a set of adjacent vertices {w1,w2,…,wj}, the adjacent vertex set must be on the path.

This is not likely to be true
 
  • #5
fzero
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Your path P is too arbitrary and it's unlikely that you can prove very much given the constraints. However, it should be possible to argue that a simple graph contains a cycle that actually contains the vertex set wi and consider the path length of a closely related cycle that includes v.
 

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