Which of the following is true in a tree: $E=O(V^2)$ or $E=O(V)$?
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I am doing a computer science assignment.
In the assignment, I was asked to find an algorithm with time complexity $O(|E|)$. The input is a connected acyclic graph (a free tree). I have come up with an algorithm costs $O(V^2)$.
I am thinking of whether the time complexity of my algorithm satisfy the requirement or not. Because I know that in a complete graph, $E = O(V^2)$. However, in this question, I am dealing with a tree with $V-1$ edges. So the question is which of the following is true in a tree: $E=O(V^2)$ or $E=O(V)$?
graph-theory
edited Nov 23 at 4:32
user302797
19.7k92252
asked Nov 22 at 12:59
Billy Cheung
61
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I am doing a computer science assignment.
In the assignment, I was asked to find an algorithm with time complexity $O(|E|)$. The input is a connected acyclic graph (a free tree). I have come up with an algorithm costs $O(V^2)$.
I am thinking of whether the time complexity of my algorithm satisfy the requirement or not. Because I know that in a complete graph, $E = O(V^2)$. However, in this question, I am dealing with a tree with $V-1$ edges. So the question is which of the following is true in a tree: $E=O(V^2)$ or $E=O(V)$?
graph-theory
edited Nov 23 at 4:32
user302797
19.7k92252
asked Nov 22 at 12:59
Billy Cheung
61
That's the wrong question, because $O(V) subset O(V^2)$. It could be fixed by using $Theta$ instead of $O$.
– Peter Taylor
Nov 23 at 6:25
add a comment |
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0
down vote
favorite
up vote
0
down vote
favorite
I am doing a computer science assignment.
In the assignment, I was asked to find an algorithm with time complexity $O(|E|)$. The input is a connected acyclic graph (a free tree). I have come up with an algorithm costs $O(V^2)$.
I am thinking of whether the time complexity of my algorithm satisfy the requirement or not. Because I know that in a complete graph, $E = O(V^2)$. However, in this question, I am dealing with a tree with $V-1$ edges. So the question is which of the following is true in a tree: $E=O(V^2)$ or $E=O(V)$?
graph-theory
edited Nov 23 at 4:32
user302797
19.7k92252
asked Nov 22 at 12:59
Billy Cheung
61
I am doing a computer science assignment.
In the assignment, I was asked to find an algorithm with time complexity $O(|E|)$. The input is a connected acyclic graph (a free tree). I have come up with an algorithm costs $O(V^2)$.
I am thinking of whether the time complexity of my algorithm satisfy the requirement or not. Because I know that in a complete graph, $E = O(V^2)$. However, in this question, I am dealing with a tree with $V-1$ edges. So the question is which of the following is true in a tree: $E=O(V^2)$ or $E=O(V)$?
graph-theory
graph-theory
edited Nov 23 at 4:32
user302797
19.7k92252
asked Nov 22 at 12:59
Billy Cheung
61
edited Nov 23 at 4:32
user302797
19.7k92252
asked Nov 22 at 12:59
Billy Cheung
61
edited Nov 23 at 4:32
user302797
19.7k92252
edited Nov 23 at 4:32
user302797
19.7k92252
edited Nov 23 at 4:32
user302797
19.7k92252
19.7k92252
asked Nov 22 at 12:59
Billy Cheung
61
asked Nov 22 at 12:59
Billy Cheung
61
asked Nov 22 at 12:59
Billy Cheung
61
61
That's the wrong question, because $O(V) subset O(V^2)$. It could be fixed by using $Theta$ instead of $O$.
– Peter Taylor
Nov 23 at 6:25
add a comment |
That's the wrong question, because $O(V) subset O(V^2)$. It could be fixed by using $Theta$ instead of $O$.
– Peter Taylor
Nov 23 at 6:25
That's the wrong question, because $O(V) subset O(V^2)$. It could be fixed by using $Theta$ instead of $O$.
– Peter Taylor
Nov 23 at 6:25
That's the wrong question, because $O(V) subset O(V^2)$. It could be fixed by using $Theta$ instead of $O$.
– Peter Taylor
Nov 23 at 6:25
add a comment |
1 Answer
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It is not. As you said $|E| = |V| - 1$ here. Hence, if your algorithm is in $O(|V|^2)$ it is not satisfied $O(|E|)$ here.
answered Nov 22 at 14:27
OmG
1,949617
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1 Answer
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It is not. As you said $|E| = |V| - 1$ here. Hence, if your algorithm is in $O(|V|^2)$ it is not satisfied $O(|E|)$ here.
answered Nov 22 at 14:27
OmG
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up vote
0
down vote
It is not. As you said $|E| = |V| - 1$ here. Hence, if your algorithm is in $O(|V|^2)$ it is not satisfied $O(|E|)$ here.
answered Nov 22 at 14:27
OmG
1,949617
add a comment |
up vote
0
down vote
up vote
0
down vote
It is not. As you said $|E| = |V| - 1$ here. Hence, if your algorithm is in $O(|V|^2)$ it is not satisfied $O(|E|)$ here.
answered Nov 22 at 14:27
OmG
1,949617
It is not. As you said $|E| = |V| - 1$ here. Hence, if your algorithm is in $O(|V|^2)$ it is not satisfied $O(|E|)$ here.
answered Nov 22 at 14:27
OmG
1,949617
answered Nov 22 at 14:27
OmG
1,949617
answered Nov 22 at 14:27
OmG
1,949617
answered Nov 22 at 14:27
OmG
1,949617
1,949617
add a comment |
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That's the wrong question, because $O(V) subset O(V^2)$. It could be fixed by using $Theta$ instead of $O$.
– Peter Taylor
Nov 23 at 6:25