Unique path in a connected graph
$begingroup$
An undirected connected graph with $n - 1$ edges has only one unique path between any 2 vertices. Is this true. If so, how.
combinatorics graph-theory
edited Dec 16 '18 at 14:43
Maria Mazur
46.6k1260119
asked Dec 16 '18 at 13:45
RobustPath_91RobustPath_91
82
$endgroup$
add a comment |
$begingroup$
An undirected connected graph with $n - 1$ edges has only one unique path between any 2 vertices. Is this true. If so, how.
combinatorics graph-theory
edited Dec 16 '18 at 14:43
Maria Mazur
46.6k1260119
asked Dec 16 '18 at 13:45
RobustPath_91RobustPath_91
82
$endgroup$
add a comment |
$begingroup$
An undirected connected graph with $n - 1$ edges has only one unique path between any 2 vertices. Is this true. If so, how.
combinatorics graph-theory
edited Dec 16 '18 at 14:43
Maria Mazur
46.6k1260119
asked Dec 16 '18 at 13:45
RobustPath_91RobustPath_91
82
$endgroup$
An undirected connected graph with $n - 1$ edges has only one unique path between any 2 vertices. Is this true. If so, how.
combinatorics graph-theory
combinatorics graph-theory
edited Dec 16 '18 at 14:43
Maria Mazur
46.6k1260119
asked Dec 16 '18 at 13:45
RobustPath_91RobustPath_91
82
edited Dec 16 '18 at 14:43
Maria Mazur
46.6k1260119
asked Dec 16 '18 at 13:45
RobustPath_91RobustPath_91
82
edited Dec 16 '18 at 14:43
Maria Mazur
46.6k1260119
edited Dec 16 '18 at 14:43
Maria Mazur
46.6k1260119
edited Dec 16 '18 at 14:43
Maria Mazur
46.6k1260119
46.6k1260119
asked Dec 16 '18 at 13:45
RobustPath_91RobustPath_91
82
asked Dec 16 '18 at 13:45
RobustPath_91RobustPath_91
82
asked Dec 16 '18 at 13:45
RobustPath_91RobustPath_91
82
82
add a comment |
add a comment |
1 Answer
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$begingroup$
This graph is a tree and so it does not contain a cycle. So if there are two points $A,B$ with two paths between them $$P_1 : Ato ...to B$$ and $$P_2 : Ato ...to B$$ then path $P_2^{-1}circ P_1$ would be a cycle wich starts and ends with $A$.
answered Dec 16 '18 at 13:52
Maria MazurMaria Mazur
46.6k1260119
$endgroup$
$begingroup$
Technically the walk you give is not a cycle, as vertices may be repeated. It must contain a cycle however.
$endgroup$
– Bob Krueger
Dec 17 '18 at 4:27
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
This graph is a tree and so it does not contain a cycle. So if there are two points $A,B$ with two paths between them $$P_1 : Ato ...to B$$ and $$P_2 : Ato ...to B$$ then path $P_2^{-1}circ P_1$ would be a cycle wich starts and ends with $A$.
answered Dec 16 '18 at 13:52
Maria MazurMaria Mazur
46.6k1260119
$endgroup$
$begingroup$
Technically the walk you give is not a cycle, as vertices may be repeated. It must contain a cycle however.
$endgroup$
– Bob Krueger
Dec 17 '18 at 4:27
add a comment |
$begingroup$
This graph is a tree and so it does not contain a cycle. So if there are two points $A,B$ with two paths between them $$P_1 : Ato ...to B$$ and $$P_2 : Ato ...to B$$ then path $P_2^{-1}circ P_1$ would be a cycle wich starts and ends with $A$.
answered Dec 16 '18 at 13:52
Maria MazurMaria Mazur
46.6k1260119
$endgroup$
$begingroup$
Technically the walk you give is not a cycle, as vertices may be repeated. It must contain a cycle however.
$endgroup$
– Bob Krueger
Dec 17 '18 at 4:27
add a comment |
$begingroup$
This graph is a tree and so it does not contain a cycle. So if there are two points $A,B$ with two paths between them $$P_1 : Ato ...to B$$ and $$P_2 : Ato ...to B$$ then path $P_2^{-1}circ P_1$ would be a cycle wich starts and ends with $A$.
answered Dec 16 '18 at 13:52
Maria MazurMaria Mazur
46.6k1260119
$endgroup$
This graph is a tree and so it does not contain a cycle. So if there are two points $A,B$ with two paths between them $$P_1 : Ato ...to B$$ and $$P_2 : Ato ...to B$$ then path $P_2^{-1}circ P_1$ would be a cycle wich starts and ends with $A$.
answered Dec 16 '18 at 13:52
Maria MazurMaria Mazur
46.6k1260119
edited Dec 16 '18 at 14:41
answered Dec 16 '18 at 13:52
Maria MazurMaria Mazur
46.6k1260119
answered Dec 16 '18 at 13:52
Maria MazurMaria Mazur
46.6k1260119
answered Dec 16 '18 at 13:52
Maria MazurMaria Mazur
46.6k1260119
46.6k1260119
$begingroup$
Technically the walk you give is not a cycle, as vertices may be repeated. It must contain a cycle however.
$endgroup$
– Bob Krueger
Dec 17 '18 at 4:27
add a comment |
$begingroup$
Technically the walk you give is not a cycle, as vertices may be repeated. It must contain a cycle however.
$endgroup$
– Bob Krueger
Dec 17 '18 at 4:27
$begingroup$
Technically the walk you give is not a cycle, as vertices may be repeated. It must contain a cycle however.
$endgroup$
– Bob Krueger
Dec 17 '18 at 4:27
$begingroup$
Technically the walk you give is not a cycle, as vertices may be repeated. It must contain a cycle however.
$endgroup$
– Bob Krueger
Dec 17 '18 at 4:27
add a comment |
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