Find Two Equal Length Paths in a Tree Graph

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Discussion Overview

The discussion revolves around the problem of proving that in any simple graph with more than two vertices, two equal paths of maximal length must intersect. The focus includes the exploration of extremal arguments and the implications of path intersections in tree graphs.

Discussion Character

  • Exploratory
  • Homework-related

Main Points Raised

  • Some participants propose using an extremal argument to demonstrate that two equal length maximum paths in a tree must intersect, suggesting that if they do not intersect, the graph would not remain connected.
  • Others question the clarity of the problem statement, noting that the title appears to be a fragment and does not fully articulate the problem being addressed.
  • A later reply clarifies the intended problem statement, emphasizing that it should assert the necessity of intersection for maximal length paths in a simple graph.
  • One participant references a similar homework problem discussed previously, indicating that the reasoning about common vertices in maximal paths aligns with established ideas.

Areas of Agreement / Disagreement

Participants express some agreement on the need for paths to intersect in connected graphs, but there is uncertainty regarding the clarity of the problem statement and the specific details of the proof being sought.

Contextual Notes

The discussion highlights limitations in the initial problem statement and the need for clearer definitions, as well as the potential for misunderstanding the requirements of the proof.

Superyoshiom
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In order to try and sovle this problem, my idea here was to use an extremal argument, with two equal length maximum paths on a tree. I said that no matter where the two graphs started, they'd end up having to cross paths since if there were two distinct paths with equal lengths the graph would no longer be connected; the paths still needed to intersect.
 
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Superyoshiom said:
In order to try and sovle this problem, my idea here was to use an extremal argument, with two equal length maximum paths on a tree. I said that no matter where the two graphs started, they'd end up having to cross paths since if there were two distinct paths with equal lengths the graph would no longer be connected; the paths still needed to intersect.
What exactly are you trying to prove?

The title says "Prove that in any simple graph with more than two vertices, two equal"
which is a fragment that doesn't state the full problem... I don't see the problem stated in the body here either.
 
StoneTemplePython said:
What exactly are you trying to prove?

The title says "Prove that in any simple graph with more than two vertices, two equal"
which is a fragment that doesn't state the full problem... I don't see the problem stated in the body here either.
I'm sorry, the title cut off for some reason, it should say: "Prove that in any simple graph with more than two vertices, two equal paths of maximal length must intersect."
 
Superyoshiom said:
I'm sorry, the title cut off for some reason, it should say: "Prove that in any simple graph with more than two vertices, two equal paths of maximal length must intersect."
Got it. This was, basically, a homework problem in the calculus forums last week. And your sketch is right -- either there are common vertices in these maximal length paths or the graph isn't connected
 
Superyoshiom said:
I'm sorry, the title cut off for some reason, it should say: "Prove that in any simple graph with more than two vertices, two equal paths of maximal length must intersect."
I've updated your thread title for you.
 

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