Graph with same number of edges and vertices is a loop?Graph Theory - How can I calculate the number of vertices and edges, if given this exampleCounting number of vertices given a graphProve that every connected undirected graph with n vertices has at least n-1 edges.Eulerian graph with odd/even vertices/edgesProve that every graph has two vertices that are endpoints of the same number of edgesNumber of vertices of a complete graph with $n$ edges40 Vertices And A Connected Graph, Minimum Number Of Edges?A directed complete graph with equal number of incoming and outgoing edgesvertices and edges on a cycleGraph Theory - Number of edges and vertices
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Graph with same number of edges and vertices is a loop?
Graph Theory - How can I calculate the number of vertices and edges, if given this exampleCounting number of vertices given a graphProve that every connected undirected graph with n vertices has at least n-1 edges.Eulerian graph with odd/even vertices/edgesProve that every graph has two vertices that are endpoints of the same number of edgesNumber of vertices of a complete graph with $n$ edges40 Vertices And A Connected Graph, Minimum Number Of Edges?A directed complete graph with equal number of incoming and outgoing edgesvertices and edges on a cycleGraph Theory - Number of edges and vertices
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Suppose I have a graph which has the same number of edges as the number of vertices, and suppose also that each vertex has at least two edges coming out of it. Does it follow by these conditions that it is a loop?
graph-theory
asked 8 hours ago
Johnny T.Johnny T.
6091516
$endgroup$
add a comment |
$begingroup$
Suppose I have a graph which has the same number of edges as the number of vertices, and suppose also that each vertex has at least two edges coming out of it. Does it follow by these conditions that it is a loop?
graph-theory
asked 8 hours ago
Johnny T.Johnny T.
6091516
$endgroup$
$begingroup$
Is the graph connected?
$endgroup$
– Fabio Somenzi
8 hours ago
add a comment |
$begingroup$
Suppose I have a graph which has the same number of edges as the number of vertices, and suppose also that each vertex has at least two edges coming out of it. Does it follow by these conditions that it is a loop?
graph-theory
asked 8 hours ago
Johnny T.Johnny T.
6091516
$endgroup$
Suppose I have a graph which has the same number of edges as the number of vertices, and suppose also that each vertex has at least two edges coming out of it. Does it follow by these conditions that it is a loop?
graph-theory
graph-theory
asked 8 hours ago
Johnny T.Johnny T.
6091516
asked 8 hours ago
Johnny T.Johnny T.
6091516
asked 8 hours ago
Johnny T.Johnny T.
6091516
asked 8 hours ago
Johnny T.Johnny T.
6091516
asked 8 hours ago
Johnny T.Johnny T.
6091516
6091516
$begingroup$
Is the graph connected?
$endgroup$
– Fabio Somenzi
8 hours ago
add a comment |
$begingroup$
Is the graph connected?
$endgroup$
– Fabio Somenzi
8 hours ago
$begingroup$
Is the graph connected?
$endgroup$
– Fabio Somenzi
8 hours ago
$begingroup$
Is the graph connected?
$endgroup$
– Fabio Somenzi
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First, you mean cycle graph, not loop (which is an edge from a vertex to itself).
Second, the answer is "yes", but only if the graph is both finite and connected, and each vertex has exactly two edges coming out of it.
If the graph is not finite, the answer is "no" -- consider the integers, with each integer $n$ connected to both $n-1$ and $n+1$.
If the graph is not connected, the answer is "no", as the graph may be several disjoint cycles.
If a vertex has more than two edges coming out of it, then it need not be a cycle. For example, consider $K_4$, the complete graph on 4 vertices.
Note: if the graph is finite, and the number of vertices equals the number of edges, and each vertex has at least two edges incident to it, then each vertex must have exactly two edges incident to it, by an edge-counting argument.
Here is the edge-counting argument. Consider each edge as split into two half-edges. Each half-edge is incident to just one vertex, and two half-edges combine to form one edge. Now, suppose that each vertex has at least two edges incident to it. By way of contradiction, suppose that some vertex has more than two edges incident to it. Let's go from vertex to vertex counting how many half-edges it has. Each vertex has at least two, and some vertex has MORE than two, so the total number of half-edges is strictly greater than twice the number of vertices. Hence, the number of edges is strictly greater than the number of vertices. But we assumed that they were equal, so we have a contradiction.
answered 8 hours ago
vadim123vadim123
76.9k899193
$endgroup$
$begingroup$
Could you possibly elaborate on the edge-counting argument part? I would like to know how I can deduce this. Thank you.
$endgroup$
– Johnny T.
8 hours ago
add a comment |
$begingroup$
To add to vadim23's answer: It should also be noted that [relaxing away from the condition that every vertex has degree exactly 2] there are connected graphs with precisely the same number of edges as vertices, that is NOT a cycle.
Here is an example: Let $G$ be the graph on $1,2,3,4$ where the edges are $12,23,31,14$ i.e., a 3-cycle plus an additional edge between the 4th vertex and a vertex on the 3-cycle. [Note that $G$ has 4 edges, and 4 vertices. However, vertex 4 has degree 1 and vertex 1 has degree 3.]
Can you generalize this? Can you construct, for general $n$, a graph on $n$ vertices and $n$ edges that is not a cycle?
answered 8 hours ago
MikeMike
5,471614
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
First, you mean cycle graph, not loop (which is an edge from a vertex to itself).
Second, the answer is "yes", but only if the graph is both finite and connected, and each vertex has exactly two edges coming out of it.
If the graph is not finite, the answer is "no" -- consider the integers, with each integer $n$ connected to both $n-1$ and $n+1$.
If the graph is not connected, the answer is "no", as the graph may be several disjoint cycles.
If a vertex has more than two edges coming out of it, then it need not be a cycle. For example, consider $K_4$, the complete graph on 4 vertices.
Note: if the graph is finite, and the number of vertices equals the number of edges, and each vertex has at least two edges incident to it, then each vertex must have exactly two edges incident to it, by an edge-counting argument.
Here is the edge-counting argument. Consider each edge as split into two half-edges. Each half-edge is incident to just one vertex, and two half-edges combine to form one edge. Now, suppose that each vertex has at least two edges incident to it. By way of contradiction, suppose that some vertex has more than two edges incident to it. Let's go from vertex to vertex counting how many half-edges it has. Each vertex has at least two, and some vertex has MORE than two, so the total number of half-edges is strictly greater than twice the number of vertices. Hence, the number of edges is strictly greater than the number of vertices. But we assumed that they were equal, so we have a contradiction.
answered 8 hours ago
vadim123vadim123
76.9k899193
$endgroup$
$begingroup$
Could you possibly elaborate on the edge-counting argument part? I would like to know how I can deduce this. Thank you.
$endgroup$
– Johnny T.
8 hours ago
add a comment |
$begingroup$
First, you mean cycle graph, not loop (which is an edge from a vertex to itself).
Second, the answer is "yes", but only if the graph is both finite and connected, and each vertex has exactly two edges coming out of it.
If the graph is not finite, the answer is "no" -- consider the integers, with each integer $n$ connected to both $n-1$ and $n+1$.
If the graph is not connected, the answer is "no", as the graph may be several disjoint cycles.
If a vertex has more than two edges coming out of it, then it need not be a cycle. For example, consider $K_4$, the complete graph on 4 vertices.
Note: if the graph is finite, and the number of vertices equals the number of edges, and each vertex has at least two edges incident to it, then each vertex must have exactly two edges incident to it, by an edge-counting argument.
Here is the edge-counting argument. Consider each edge as split into two half-edges. Each half-edge is incident to just one vertex, and two half-edges combine to form one edge. Now, suppose that each vertex has at least two edges incident to it. By way of contradiction, suppose that some vertex has more than two edges incident to it. Let's go from vertex to vertex counting how many half-edges it has. Each vertex has at least two, and some vertex has MORE than two, so the total number of half-edges is strictly greater than twice the number of vertices. Hence, the number of edges is strictly greater than the number of vertices. But we assumed that they were equal, so we have a contradiction.
answered 8 hours ago
vadim123vadim123
76.9k899193
$endgroup$
$begingroup$
Could you possibly elaborate on the edge-counting argument part? I would like to know how I can deduce this. Thank you.
$endgroup$
– Johnny T.
8 hours ago
add a comment |
$begingroup$
First, you mean cycle graph, not loop (which is an edge from a vertex to itself).
Second, the answer is "yes", but only if the graph is both finite and connected, and each vertex has exactly two edges coming out of it.
If the graph is not finite, the answer is "no" -- consider the integers, with each integer $n$ connected to both $n-1$ and $n+1$.
If the graph is not connected, the answer is "no", as the graph may be several disjoint cycles.
If a vertex has more than two edges coming out of it, then it need not be a cycle. For example, consider $K_4$, the complete graph on 4 vertices.
Note: if the graph is finite, and the number of vertices equals the number of edges, and each vertex has at least two edges incident to it, then each vertex must have exactly two edges incident to it, by an edge-counting argument.
Here is the edge-counting argument. Consider each edge as split into two half-edges. Each half-edge is incident to just one vertex, and two half-edges combine to form one edge. Now, suppose that each vertex has at least two edges incident to it. By way of contradiction, suppose that some vertex has more than two edges incident to it. Let's go from vertex to vertex counting how many half-edges it has. Each vertex has at least two, and some vertex has MORE than two, so the total number of half-edges is strictly greater than twice the number of vertices. Hence, the number of edges is strictly greater than the number of vertices. But we assumed that they were equal, so we have a contradiction.
answered 8 hours ago
vadim123vadim123
76.9k899193
$endgroup$
First, you mean cycle graph, not loop (which is an edge from a vertex to itself).
Second, the answer is "yes", but only if the graph is both finite and connected, and each vertex has exactly two edges coming out of it.
If the graph is not finite, the answer is "no" -- consider the integers, with each integer $n$ connected to both $n-1$ and $n+1$.
If the graph is not connected, the answer is "no", as the graph may be several disjoint cycles.
If a vertex has more than two edges coming out of it, then it need not be a cycle. For example, consider $K_4$, the complete graph on 4 vertices.
Note: if the graph is finite, and the number of vertices equals the number of edges, and each vertex has at least two edges incident to it, then each vertex must have exactly two edges incident to it, by an edge-counting argument.
Here is the edge-counting argument. Consider each edge as split into two half-edges. Each half-edge is incident to just one vertex, and two half-edges combine to form one edge. Now, suppose that each vertex has at least two edges incident to it. By way of contradiction, suppose that some vertex has more than two edges incident to it. Let's go from vertex to vertex counting how many half-edges it has. Each vertex has at least two, and some vertex has MORE than two, so the total number of half-edges is strictly greater than twice the number of vertices. Hence, the number of edges is strictly greater than the number of vertices. But we assumed that they were equal, so we have a contradiction.
answered 8 hours ago
vadim123vadim123
76.9k899193
edited 5 hours ago
answered 8 hours ago
vadim123vadim123
76.9k899193
answered 8 hours ago
vadim123vadim123
76.9k899193
answered 8 hours ago
vadim123vadim123
76.9k899193
76.9k899193
$begingroup$
Could you possibly elaborate on the edge-counting argument part? I would like to know how I can deduce this. Thank you.
$endgroup$
– Johnny T.
8 hours ago
add a comment |
$begingroup$
Could you possibly elaborate on the edge-counting argument part? I would like to know how I can deduce this. Thank you.
$endgroup$
– Johnny T.
8 hours ago
$begingroup$
Could you possibly elaborate on the edge-counting argument part? I would like to know how I can deduce this. Thank you.
$endgroup$
– Johnny T.
8 hours ago
$begingroup$
Could you possibly elaborate on the edge-counting argument part? I would like to know how I can deduce this. Thank you.
$endgroup$
– Johnny T.
8 hours ago
add a comment |
$begingroup$
To add to vadim23's answer: It should also be noted that [relaxing away from the condition that every vertex has degree exactly 2] there are connected graphs with precisely the same number of edges as vertices, that is NOT a cycle.
Here is an example: Let $G$ be the graph on $1,2,3,4$ where the edges are $12,23,31,14$ i.e., a 3-cycle plus an additional edge between the 4th vertex and a vertex on the 3-cycle. [Note that $G$ has 4 edges, and 4 vertices. However, vertex 4 has degree 1 and vertex 1 has degree 3.]
Can you generalize this? Can you construct, for general $n$, a graph on $n$ vertices and $n$ edges that is not a cycle?
answered 8 hours ago
MikeMike
5,471614
$endgroup$
add a comment |
$begingroup$
To add to vadim23's answer: It should also be noted that [relaxing away from the condition that every vertex has degree exactly 2] there are connected graphs with precisely the same number of edges as vertices, that is NOT a cycle.
Here is an example: Let $G$ be the graph on $1,2,3,4$ where the edges are $12,23,31,14$ i.e., a 3-cycle plus an additional edge between the 4th vertex and a vertex on the 3-cycle. [Note that $G$ has 4 edges, and 4 vertices. However, vertex 4 has degree 1 and vertex 1 has degree 3.]
Can you generalize this? Can you construct, for general $n$, a graph on $n$ vertices and $n$ edges that is not a cycle?
answered 8 hours ago
MikeMike
5,471614
$endgroup$
add a comment |
$begingroup$
To add to vadim23's answer: It should also be noted that [relaxing away from the condition that every vertex has degree exactly 2] there are connected graphs with precisely the same number of edges as vertices, that is NOT a cycle.
Here is an example: Let $G$ be the graph on $1,2,3,4$ where the edges are $12,23,31,14$ i.e., a 3-cycle plus an additional edge between the 4th vertex and a vertex on the 3-cycle. [Note that $G$ has 4 edges, and 4 vertices. However, vertex 4 has degree 1 and vertex 1 has degree 3.]
Can you generalize this? Can you construct, for general $n$, a graph on $n$ vertices and $n$ edges that is not a cycle?
answered 8 hours ago
MikeMike
5,471614
$endgroup$
To add to vadim23's answer: It should also be noted that [relaxing away from the condition that every vertex has degree exactly 2] there are connected graphs with precisely the same number of edges as vertices, that is NOT a cycle.
Here is an example: Let $G$ be the graph on $1,2,3,4$ where the edges are $12,23,31,14$ i.e., a 3-cycle plus an additional edge between the 4th vertex and a vertex on the 3-cycle. [Note that $G$ has 4 edges, and 4 vertices. However, vertex 4 has degree 1 and vertex 1 has degree 3.]
Can you generalize this? Can you construct, for general $n$, a graph on $n$ vertices and $n$ edges that is not a cycle?
answered 8 hours ago
MikeMike
5,471614
edited 8 hours ago
answered 8 hours ago
MikeMike
5,471614
answered 8 hours ago
MikeMike
5,471614
answered 8 hours ago
MikeMike
5,471614
5,471614
add a comment |
add a comment |
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$begingroup$
Is the graph connected?
$endgroup$
– Fabio Somenzi
8 hours ago