Having No Directed Cycles Guarantees a Vertex of Zero Outdegree
Is it true that a directed graph with a finite number of vertices and with no directed cycles has at least one vertex whose out-degree is zero?
Here is my idea:
Suppose there is no vertex with out-degree zero, this implies there is an outgoing edge from each vertex. Suppose there are $n$ vertices then there must be $n$ edges since out-degree is greater than or equal to 1, but we can't have this since it creates a directed cycle. Hence, there must be a vertex whose out-degree is zero.
Is this correct? Can you provide a mathematical proof or counterexample?
graph-theory directed-graphs
edited Nov 26 at 1:51
Austin Mohr
20k35097
asked Nov 26 at 0:57
thetraveller
1515
add a comment |
Is it true that a directed graph with a finite number of vertices and with no directed cycles has at least one vertex whose out-degree is zero?
Here is my idea:
Suppose there is no vertex with out-degree zero, this implies there is an outgoing edge from each vertex. Suppose there are $n$ vertices then there must be $n$ edges since out-degree is greater than or equal to 1, but we can't have this since it creates a directed cycle. Hence, there must be a vertex whose out-degree is zero.
Is this correct? Can you provide a mathematical proof or counterexample?
graph-theory directed-graphs
edited Nov 26 at 1:51
Austin Mohr
20k35097
asked Nov 26 at 0:57
thetraveller
1515
add a comment |
Is it true that a directed graph with a finite number of vertices and with no directed cycles has at least one vertex whose out-degree is zero?
Here is my idea:
Suppose there is no vertex with out-degree zero, this implies there is an outgoing edge from each vertex. Suppose there are $n$ vertices then there must be $n$ edges since out-degree is greater than or equal to 1, but we can't have this since it creates a directed cycle. Hence, there must be a vertex whose out-degree is zero.
Is this correct? Can you provide a mathematical proof or counterexample?
graph-theory directed-graphs
edited Nov 26 at 1:51
Austin Mohr
20k35097
asked Nov 26 at 0:57
thetraveller
1515
Is it true that a directed graph with a finite number of vertices and with no directed cycles has at least one vertex whose out-degree is zero?
Here is my idea:
Suppose there is no vertex with out-degree zero, this implies there is an outgoing edge from each vertex. Suppose there are $n$ vertices then there must be $n$ edges since out-degree is greater than or equal to 1, but we can't have this since it creates a directed cycle. Hence, there must be a vertex whose out-degree is zero.
Is this correct? Can you provide a mathematical proof or counterexample?
graph-theory directed-graphs
graph-theory directed-graphs
edited Nov 26 at 1:51
Austin Mohr
20k35097
asked Nov 26 at 0:57
thetraveller
1515
edited Nov 26 at 1:51
Austin Mohr
20k35097
asked Nov 26 at 0:57
thetraveller
1515
edited Nov 26 at 1:51
Austin Mohr
20k35097
edited Nov 26 at 1:51
Austin Mohr
20k35097
edited Nov 26 at 1:51
Austin Mohr
20k35097
20k35097
asked Nov 26 at 0:57
thetraveller
1515
asked Nov 26 at 0:57
thetraveller
1515
asked Nov 26 at 0:57
thetraveller
1515
1515
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You basically have the right idea, but I would organize it as a contraposition.
Statement: Let $G$ be a directed graph with a finite number of vertices. If $G$ has no directed cycles, then it has at least one vertex with outdegree zero.
Contrapositive: Let $G$ be a directed graph with a finite number of vertices. If $G$ has no vertex with outdegree zero, then it has a directed cycle.
Proof (sketch): Start at any vertex. Since it's outdegree is not zero, you can travel along an edge. Your new position is also a vertex with nonzero outdegree, so you can travel along an edge. Keep travelling in this way (it's always possible, since every vertex has nonzero outdegree). Since $G$ is finite, you must eventually return to a vertex you've previously visited, which indicates the presence of a directed cycle.
answered Nov 26 at 1:46
Austin Mohr
20k35097
thanks for the detailed proof
– thetraveller
Nov 26 at 3:07
add a comment |
Take the longest directed path in $G$. The end vertex of the path has the desired property.
answered Nov 26 at 2:37
hbm
957156
How do you know there is a longest directed path?
– Henning Makholm
Nov 26 at 2:43
1
It is a finite graph.
– hbm
Nov 26 at 5:15
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You basically have the right idea, but I would organize it as a contraposition.
Statement: Let $G$ be a directed graph with a finite number of vertices. If $G$ has no directed cycles, then it has at least one vertex with outdegree zero.
Contrapositive: Let $G$ be a directed graph with a finite number of vertices. If $G$ has no vertex with outdegree zero, then it has a directed cycle.
Proof (sketch): Start at any vertex. Since it's outdegree is not zero, you can travel along an edge. Your new position is also a vertex with nonzero outdegree, so you can travel along an edge. Keep travelling in this way (it's always possible, since every vertex has nonzero outdegree). Since $G$ is finite, you must eventually return to a vertex you've previously visited, which indicates the presence of a directed cycle.
answered Nov 26 at 1:46
Austin Mohr
20k35097
thanks for the detailed proof
– thetraveller
Nov 26 at 3:07
add a comment |
You basically have the right idea, but I would organize it as a contraposition.
Statement: Let $G$ be a directed graph with a finite number of vertices. If $G$ has no directed cycles, then it has at least one vertex with outdegree zero.
Contrapositive: Let $G$ be a directed graph with a finite number of vertices. If $G$ has no vertex with outdegree zero, then it has a directed cycle.
Proof (sketch): Start at any vertex. Since it's outdegree is not zero, you can travel along an edge. Your new position is also a vertex with nonzero outdegree, so you can travel along an edge. Keep travelling in this way (it's always possible, since every vertex has nonzero outdegree). Since $G$ is finite, you must eventually return to a vertex you've previously visited, which indicates the presence of a directed cycle.
answered Nov 26 at 1:46
Austin Mohr
20k35097
thanks for the detailed proof
– thetraveller
Nov 26 at 3:07
add a comment |
You basically have the right idea, but I would organize it as a contraposition.
Statement: Let $G$ be a directed graph with a finite number of vertices. If $G$ has no directed cycles, then it has at least one vertex with outdegree zero.
Contrapositive: Let $G$ be a directed graph with a finite number of vertices. If $G$ has no vertex with outdegree zero, then it has a directed cycle.
Proof (sketch): Start at any vertex. Since it's outdegree is not zero, you can travel along an edge. Your new position is also a vertex with nonzero outdegree, so you can travel along an edge. Keep travelling in this way (it's always possible, since every vertex has nonzero outdegree). Since $G$ is finite, you must eventually return to a vertex you've previously visited, which indicates the presence of a directed cycle.
answered Nov 26 at 1:46
Austin Mohr
20k35097
You basically have the right idea, but I would organize it as a contraposition.
Statement: Let $G$ be a directed graph with a finite number of vertices. If $G$ has no directed cycles, then it has at least one vertex with outdegree zero.
Contrapositive: Let $G$ be a directed graph with a finite number of vertices. If $G$ has no vertex with outdegree zero, then it has a directed cycle.
Proof (sketch): Start at any vertex. Since it's outdegree is not zero, you can travel along an edge. Your new position is also a vertex with nonzero outdegree, so you can travel along an edge. Keep travelling in this way (it's always possible, since every vertex has nonzero outdegree). Since $G$ is finite, you must eventually return to a vertex you've previously visited, which indicates the presence of a directed cycle.
answered Nov 26 at 1:46
Austin Mohr
20k35097
answered Nov 26 at 1:46
Austin Mohr
20k35097
answered Nov 26 at 1:46
Austin Mohr
20k35097
answered Nov 26 at 1:46
Austin Mohr
20k35097
20k35097
thanks for the detailed proof
– thetraveller
Nov 26 at 3:07
add a comment |
thanks for the detailed proof
– thetraveller
Nov 26 at 3:07
thanks for the detailed proof
– thetraveller
Nov 26 at 3:07
thanks for the detailed proof
– thetraveller
Nov 26 at 3:07
add a comment |
Take the longest directed path in $G$. The end vertex of the path has the desired property.
answered Nov 26 at 2:37
hbm
957156
How do you know there is a longest directed path?
– Henning Makholm
Nov 26 at 2:43
1
It is a finite graph.
– hbm
Nov 26 at 5:15
add a comment |
Take the longest directed path in $G$. The end vertex of the path has the desired property.
answered Nov 26 at 2:37
hbm
957156
How do you know there is a longest directed path?
– Henning Makholm
Nov 26 at 2:43
1
It is a finite graph.
– hbm
Nov 26 at 5:15
add a comment |
Take the longest directed path in $G$. The end vertex of the path has the desired property.
answered Nov 26 at 2:37
hbm
957156
Take the longest directed path in $G$. The end vertex of the path has the desired property.
answered Nov 26 at 2:37
hbm
957156
answered Nov 26 at 2:37
hbm
957156
answered Nov 26 at 2:37
hbm
957156
answered Nov 26 at 2:37
hbm
957156
957156
How do you know there is a longest directed path?
– Henning Makholm
Nov 26 at 2:43
1
It is a finite graph.
– hbm
Nov 26 at 5:15
add a comment |
How do you know there is a longest directed path?
– Henning Makholm
Nov 26 at 2:43
1
It is a finite graph.
– hbm
Nov 26 at 5:15
How do you know there is a longest directed path?
– Henning Makholm
Nov 26 at 2:43
How do you know there is a longest directed path?
– Henning Makholm
Nov 26 at 2:43
1
1
It is a finite graph.
– hbm
Nov 26 at 5:15
It is a finite graph.
– hbm
Nov 26 at 5:15
add a comment |
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