A cycle in an undirected graph












0












$begingroup$


A cycle is a simple path of length at least $1$ which begins and ends at the same vertex.



In an undirected graph, a cycle must be of length at least $3$.



Could you explain me why that stands??










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$endgroup$












  • $begingroup$
    You must be considering undirected simple graphs: Undirected graphs with no (self) loops or parallel edges.
    $endgroup$
    – user183763
    Nov 19 '14 at 12:03
















0












$begingroup$


A cycle is a simple path of length at least $1$ which begins and ends at the same vertex.



In an undirected graph, a cycle must be of length at least $3$.



Could you explain me why that stands??










share|cite|improve this question









$endgroup$












  • $begingroup$
    You must be considering undirected simple graphs: Undirected graphs with no (self) loops or parallel edges.
    $endgroup$
    – user183763
    Nov 19 '14 at 12:03














0












0








0





$begingroup$


A cycle is a simple path of length at least $1$ which begins and ends at the same vertex.



In an undirected graph, a cycle must be of length at least $3$.



Could you explain me why that stands??










share|cite|improve this question









$endgroup$




A cycle is a simple path of length at least $1$ which begins and ends at the same vertex.



In an undirected graph, a cycle must be of length at least $3$.



Could you explain me why that stands??







discrete-mathematics graph-theory






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share|cite|improve this question











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share|cite|improve this question










asked Nov 19 '14 at 11:48









Mary StarMary Star

3,02982268




3,02982268












  • $begingroup$
    You must be considering undirected simple graphs: Undirected graphs with no (self) loops or parallel edges.
    $endgroup$
    – user183763
    Nov 19 '14 at 12:03


















  • $begingroup$
    You must be considering undirected simple graphs: Undirected graphs with no (self) loops or parallel edges.
    $endgroup$
    – user183763
    Nov 19 '14 at 12:03
















$begingroup$
You must be considering undirected simple graphs: Undirected graphs with no (self) loops or parallel edges.
$endgroup$
– user183763
Nov 19 '14 at 12:03




$begingroup$
You must be considering undirected simple graphs: Undirected graphs with no (self) loops or parallel edges.
$endgroup$
– user183763
Nov 19 '14 at 12:03










1 Answer
1






active

oldest

votes


















2












$begingroup$

In an undirected simple graph, there are no self loops (which are cycles of length 1) or parallel edges (which are cycles of length 2). Thus all cycles must be of length at least 3.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    And a simple path can't use the same edge twice, so $A$-to-$B$-to-$A$ doesn't count as a cycle of length 2.
    $endgroup$
    – Gerry Myerson
    Nov 19 '14 at 12:10










  • $begingroup$
    A path is simple if all edges and all vertices on the path, except possibly the first and last vertices, are distinct, right?? Could you explain me further why in an undirected simple graph there is no self loops or parallel edges??
    $endgroup$
    – Mary Star
    Nov 19 '14 at 12:13








  • 2




    $begingroup$
    @MaryStar That is what "simple graph" means. Most people assume that when you say "graph" you really mean "simple graph". If parallel edges and self-loops are allowed, it would be called a "multigraph" instead.
    $endgroup$
    – Arthur
    Nov 19 '14 at 12:16













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1 Answer
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2












$begingroup$

In an undirected simple graph, there are no self loops (which are cycles of length 1) or parallel edges (which are cycles of length 2). Thus all cycles must be of length at least 3.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    And a simple path can't use the same edge twice, so $A$-to-$B$-to-$A$ doesn't count as a cycle of length 2.
    $endgroup$
    – Gerry Myerson
    Nov 19 '14 at 12:10










  • $begingroup$
    A path is simple if all edges and all vertices on the path, except possibly the first and last vertices, are distinct, right?? Could you explain me further why in an undirected simple graph there is no self loops or parallel edges??
    $endgroup$
    – Mary Star
    Nov 19 '14 at 12:13








  • 2




    $begingroup$
    @MaryStar That is what "simple graph" means. Most people assume that when you say "graph" you really mean "simple graph". If parallel edges and self-loops are allowed, it would be called a "multigraph" instead.
    $endgroup$
    – Arthur
    Nov 19 '14 at 12:16


















2












$begingroup$

In an undirected simple graph, there are no self loops (which are cycles of length 1) or parallel edges (which are cycles of length 2). Thus all cycles must be of length at least 3.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    And a simple path can't use the same edge twice, so $A$-to-$B$-to-$A$ doesn't count as a cycle of length 2.
    $endgroup$
    – Gerry Myerson
    Nov 19 '14 at 12:10










  • $begingroup$
    A path is simple if all edges and all vertices on the path, except possibly the first and last vertices, are distinct, right?? Could you explain me further why in an undirected simple graph there is no self loops or parallel edges??
    $endgroup$
    – Mary Star
    Nov 19 '14 at 12:13








  • 2




    $begingroup$
    @MaryStar That is what "simple graph" means. Most people assume that when you say "graph" you really mean "simple graph". If parallel edges and self-loops are allowed, it would be called a "multigraph" instead.
    $endgroup$
    – Arthur
    Nov 19 '14 at 12:16
















2












2








2





$begingroup$

In an undirected simple graph, there are no self loops (which are cycles of length 1) or parallel edges (which are cycles of length 2). Thus all cycles must be of length at least 3.






share|cite|improve this answer









$endgroup$



In an undirected simple graph, there are no self loops (which are cycles of length 1) or parallel edges (which are cycles of length 2). Thus all cycles must be of length at least 3.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 19 '14 at 12:03







user183763



















  • $begingroup$
    And a simple path can't use the same edge twice, so $A$-to-$B$-to-$A$ doesn't count as a cycle of length 2.
    $endgroup$
    – Gerry Myerson
    Nov 19 '14 at 12:10










  • $begingroup$
    A path is simple if all edges and all vertices on the path, except possibly the first and last vertices, are distinct, right?? Could you explain me further why in an undirected simple graph there is no self loops or parallel edges??
    $endgroup$
    – Mary Star
    Nov 19 '14 at 12:13








  • 2




    $begingroup$
    @MaryStar That is what "simple graph" means. Most people assume that when you say "graph" you really mean "simple graph". If parallel edges and self-loops are allowed, it would be called a "multigraph" instead.
    $endgroup$
    – Arthur
    Nov 19 '14 at 12:16




















  • $begingroup$
    And a simple path can't use the same edge twice, so $A$-to-$B$-to-$A$ doesn't count as a cycle of length 2.
    $endgroup$
    – Gerry Myerson
    Nov 19 '14 at 12:10










  • $begingroup$
    A path is simple if all edges and all vertices on the path, except possibly the first and last vertices, are distinct, right?? Could you explain me further why in an undirected simple graph there is no self loops or parallel edges??
    $endgroup$
    – Mary Star
    Nov 19 '14 at 12:13








  • 2




    $begingroup$
    @MaryStar That is what "simple graph" means. Most people assume that when you say "graph" you really mean "simple graph". If parallel edges and self-loops are allowed, it would be called a "multigraph" instead.
    $endgroup$
    – Arthur
    Nov 19 '14 at 12:16


















$begingroup$
And a simple path can't use the same edge twice, so $A$-to-$B$-to-$A$ doesn't count as a cycle of length 2.
$endgroup$
– Gerry Myerson
Nov 19 '14 at 12:10




$begingroup$
And a simple path can't use the same edge twice, so $A$-to-$B$-to-$A$ doesn't count as a cycle of length 2.
$endgroup$
– Gerry Myerson
Nov 19 '14 at 12:10












$begingroup$
A path is simple if all edges and all vertices on the path, except possibly the first and last vertices, are distinct, right?? Could you explain me further why in an undirected simple graph there is no self loops or parallel edges??
$endgroup$
– Mary Star
Nov 19 '14 at 12:13






$begingroup$
A path is simple if all edges and all vertices on the path, except possibly the first and last vertices, are distinct, right?? Could you explain me further why in an undirected simple graph there is no self loops or parallel edges??
$endgroup$
– Mary Star
Nov 19 '14 at 12:13






2




2




$begingroup$
@MaryStar That is what "simple graph" means. Most people assume that when you say "graph" you really mean "simple graph". If parallel edges and self-loops are allowed, it would be called a "multigraph" instead.
$endgroup$
– Arthur
Nov 19 '14 at 12:16






$begingroup$
@MaryStar That is what "simple graph" means. Most people assume that when you say "graph" you really mean "simple graph". If parallel edges and self-loops are allowed, it would be called a "multigraph" instead.
$endgroup$
– Arthur
Nov 19 '14 at 12:16




















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