Equivalent Definition of a Tree Graph Theory: Restriction for Being Connected?
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I was looking at some equivalent definitions of a tree in graph theory and one is the following where we let $G$ be an undirected graph:
$G$ is a tree iff $G$ is connected and has $n-1$ edges where $n$ denotes the number of vertices in $G$.
My question is why is there the restriction that $G$ is connected here? Isn't it true that $G$ is a tree iff $|E(G)|=|V(G)|-1$? I thought $|E(G)|=|V(G)|-1$ implies that $G$ is connected.
graph-theory trees
asked Dec 5 at 2:05
W. G.
5431416
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up vote
0
down vote
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I was looking at some equivalent definitions of a tree in graph theory and one is the following where we let $G$ be an undirected graph:
$G$ is a tree iff $G$ is connected and has $n-1$ edges where $n$ denotes the number of vertices in $G$.
My question is why is there the restriction that $G$ is connected here? Isn't it true that $G$ is a tree iff $|E(G)|=|V(G)|-1$? I thought $|E(G)|=|V(G)|-1$ implies that $G$ is connected.
graph-theory trees
asked Dec 5 at 2:05
W. G.
5431416
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was looking at some equivalent definitions of a tree in graph theory and one is the following where we let $G$ be an undirected graph:
$G$ is a tree iff $G$ is connected and has $n-1$ edges where $n$ denotes the number of vertices in $G$.
My question is why is there the restriction that $G$ is connected here? Isn't it true that $G$ is a tree iff $|E(G)|=|V(G)|-1$? I thought $|E(G)|=|V(G)|-1$ implies that $G$ is connected.
graph-theory trees
asked Dec 5 at 2:05
W. G.
5431416
I was looking at some equivalent definitions of a tree in graph theory and one is the following where we let $G$ be an undirected graph:
$G$ is a tree iff $G$ is connected and has $n-1$ edges where $n$ denotes the number of vertices in $G$.
My question is why is there the restriction that $G$ is connected here? Isn't it true that $G$ is a tree iff $|E(G)|=|V(G)|-1$? I thought $|E(G)|=|V(G)|-1$ implies that $G$ is connected.
graph-theory trees
graph-theory trees
asked Dec 5 at 2:05
W. G.
5431416
asked Dec 5 at 2:05
W. G.
5431416
edited Dec 5 at 2:13
asked Dec 5 at 2:05
W. G.
5431416
asked Dec 5 at 2:05
W. G.
5431416
asked Dec 5 at 2:05
W. G.
5431416
5431416
add a comment |
add a comment |
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You mean $|E|=|V|-1$, not $|V|=|E|-1$.
You can have a graph with four vertices, three of which are conected with three edges but isolated from the remaining one. This graph clearly complies with $|E|=|V|-1$, but is not connected.
answered Dec 5 at 2:12
Rócherz
2,7012721
Sorry about the $|E|=|V|-1$ thing. I get what you mean. Do you know if it would be true for graphs without isolated vertices?
– W. G.
Dec 5 at 2:16
I'm sure it does in that case. I appreciate your help!
– W. G.
Dec 5 at 2:32
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You mean $|E|=|V|-1$, not $|V|=|E|-1$.
You can have a graph with four vertices, three of which are conected with three edges but isolated from the remaining one. This graph clearly complies with $|E|=|V|-1$, but is not connected.
answered Dec 5 at 2:12
Rócherz
2,7012721
Sorry about the $|E|=|V|-1$ thing. I get what you mean. Do you know if it would be true for graphs without isolated vertices?
– W. G.
Dec 5 at 2:16
I'm sure it does in that case. I appreciate your help!
– W. G.
Dec 5 at 2:32
add a comment |
up vote
2
down vote
accepted
You mean $|E|=|V|-1$, not $|V|=|E|-1$.
You can have a graph with four vertices, three of which are conected with three edges but isolated from the remaining one. This graph clearly complies with $|E|=|V|-1$, but is not connected.
answered Dec 5 at 2:12
Rócherz
2,7012721
Sorry about the $|E|=|V|-1$ thing. I get what you mean. Do you know if it would be true for graphs without isolated vertices?
– W. G.
Dec 5 at 2:16
I'm sure it does in that case. I appreciate your help!
– W. G.
Dec 5 at 2:32
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You mean $|E|=|V|-1$, not $|V|=|E|-1$.
You can have a graph with four vertices, three of which are conected with three edges but isolated from the remaining one. This graph clearly complies with $|E|=|V|-1$, but is not connected.
answered Dec 5 at 2:12
Rócherz
2,7012721
You mean $|E|=|V|-1$, not $|V|=|E|-1$.
You can have a graph with four vertices, three of which are conected with three edges but isolated from the remaining one. This graph clearly complies with $|E|=|V|-1$, but is not connected.
answered Dec 5 at 2:12
Rócherz
2,7012721
answered Dec 5 at 2:12
Rócherz
2,7012721
answered Dec 5 at 2:12
Rócherz
2,7012721
answered Dec 5 at 2:12
Rócherz
2,7012721
2,7012721
Sorry about the $|E|=|V|-1$ thing. I get what you mean. Do you know if it would be true for graphs without isolated vertices?
– W. G.
Dec 5 at 2:16
I'm sure it does in that case. I appreciate your help!
– W. G.
Dec 5 at 2:32
add a comment |
Sorry about the $|E|=|V|-1$ thing. I get what you mean. Do you know if it would be true for graphs without isolated vertices?
– W. G.
Dec 5 at 2:16
I'm sure it does in that case. I appreciate your help!
– W. G.
Dec 5 at 2:32
Sorry about the $|E|=|V|-1$ thing. I get what you mean. Do you know if it would be true for graphs without isolated vertices?
– W. G.
Dec 5 at 2:16
Sorry about the $|E|=|V|-1$ thing. I get what you mean. Do you know if it would be true for graphs without isolated vertices?
– W. G.
Dec 5 at 2:16
I'm sure it does in that case. I appreciate your help!
– W. G.
Dec 5 at 2:32
I'm sure it does in that case. I appreciate your help!
– W. G.
Dec 5 at 2:32
add a comment |
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