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 '18 at 2:05
W. G.W. G.
5641516
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add a comment |
$begingroup$
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 '18 at 2:05
W. G.W. G.
5641516
$endgroup$
add a comment |
$begingroup$
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 '18 at 2:05
W. G.W. G.
5641516
$endgroup$
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 '18 at 2:05
W. G.W. G.
5641516
asked Dec 5 '18 at 2:05
W. G.W. G.
5641516
edited Dec 5 '18 at 2:13
W. G.
asked Dec 5 '18 at 2:05
W. G.W. G.
5641516
asked Dec 5 '18 at 2:05
W. G.W. G.
5641516
asked Dec 5 '18 at 2:05
W. G.W. G.
5641516
5641516
add a comment |
add a comment |
1 Answer
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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 '18 at 2:12
RócherzRócherz
2,7762721
$endgroup$
$begingroup$
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?
$endgroup$
– W. G.
Dec 5 '18 at 2:16
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I'm sure it does in that case. I appreciate your help!
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– W. G.
Dec 5 '18 at 2:32
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
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 '18 at 2:12
RócherzRócherz
2,7762721
$endgroup$
$begingroup$
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?
$endgroup$
– W. G.
Dec 5 '18 at 2:16
$begingroup$
I'm sure it does in that case. I appreciate your help!
$endgroup$
– W. G.
Dec 5 '18 at 2:32
add a comment |
$begingroup$
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 '18 at 2:12
RócherzRócherz
2,7762721
$endgroup$
$begingroup$
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?
$endgroup$
– W. G.
Dec 5 '18 at 2:16
$begingroup$
I'm sure it does in that case. I appreciate your help!
$endgroup$
– W. G.
Dec 5 '18 at 2:32
add a comment |
$begingroup$
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 '18 at 2:12
RócherzRócherz
2,7762721
$endgroup$
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 '18 at 2:12
RócherzRócherz
2,7762721
answered Dec 5 '18 at 2:12
RócherzRócherz
2,7762721
answered Dec 5 '18 at 2:12
RócherzRócherz
2,7762721
answered Dec 5 '18 at 2:12
RócherzRócherz
2,7762721
2,7762721
$begingroup$
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?
$endgroup$
– W. G.
Dec 5 '18 at 2:16
$begingroup$
I'm sure it does in that case. I appreciate your help!
$endgroup$
– W. G.
Dec 5 '18 at 2:32
add a comment |
$begingroup$
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?
$endgroup$
– W. G.
Dec 5 '18 at 2:16
$begingroup$
I'm sure it does in that case. I appreciate your help!
$endgroup$
– W. G.
Dec 5 '18 at 2:32
$begingroup$
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?
$endgroup$
– W. G.
Dec 5 '18 at 2:16
$begingroup$
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?
$endgroup$
– W. G.
Dec 5 '18 at 2:16
$begingroup$
I'm sure it does in that case. I appreciate your help!
$endgroup$
– W. G.
Dec 5 '18 at 2:32
$begingroup$
I'm sure it does in that case. I appreciate your help!
$endgroup$
– W. G.
Dec 5 '18 at 2:32
add a comment |
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