Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that a graph G contains no cycles IF AND ONLY IF the intersection of any two intersecting paths is also a path in G.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

HINT: If there are paths $P_1$ and $P_2$ whose intersection $H$ is not a path, there must be vertices $u,v\in H$ such that $H$ does not contain a path from $u$ to $v$. On the other hand, you know that $P_1$ does contain a path from $u$ to $v$, and $P_2$ contains a path from $v$ to $u$, so ... ?

Showing that if $G$ has a cycle, then it has two paths whose intersection is not a path is very straightforward; the hint for the other direction should be of some help here as well.

share|improve this answer
    
Why does P2 necessarily contain a path from v to u? –  Heisenberg Sep 30 '12 at 0:21
    
@Heisenberg: Any path necessarily contains a subpath between any two of its vertices. E.g., the path $v_0v_1\dots v_n$ contains the directed subpaths $v_kv_{k+1}\dots v_m$ and $v_mv_{m-1}\dots v_k$ between $v_k$ and $v_m$, where $0\le k\le m\le n$. –  Brian M. Scott Sep 30 '12 at 0:25
    
So I'm assuming the only way that P1 and P2 can have paths from u to v and v to u respectively is if H is a cycle containing u and v? –  Heisenberg Sep 30 '12 at 0:30
    
@Heisenberg: That's right: you can paste together a path from $u$ to $v$ in $P_1$ and a path from $v$ to $u$ in $P_2$ to get a cycle. –  Brian M. Scott Sep 30 '12 at 0:33
    
Thank you so much! –  Heisenberg Sep 30 '12 at 0:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.