Doubt regarding the variable by which time complexity is measured
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In order to assert that a given algorithm for graphs runs in polynomial time, must the variable in the big-O function that represents the run time (denoted henceforth as $O(f(n))$) be the number of vertices? Or can one say that a graph algorithm runs in polynomial time if the variable $n$ is something else (like the number of edges in a graph).
Edit: Could an algorithm that is $O(N)$ (where $N$ is the number of trees in the graph) also be considered as being in $P$?
graph-theory asymptotics computational-complexity computability
asked Dec 29 '18 at 12:25
Aryaman GuptaAryaman Gupta
356
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add a comment |
$begingroup$
In order to assert that a given algorithm for graphs runs in polynomial time, must the variable in the big-O function that represents the run time (denoted henceforth as $O(f(n))$) be the number of vertices? Or can one say that a graph algorithm runs in polynomial time if the variable $n$ is something else (like the number of edges in a graph).
Edit: Could an algorithm that is $O(N)$ (where $N$ is the number of trees in the graph) also be considered as being in $P$?
graph-theory asymptotics computational-complexity computability
asked Dec 29 '18 at 12:25
Aryaman GuptaAryaman Gupta
356
$endgroup$
1
$begingroup$
The variable could be either of them or a subset of them. It will depend on the case for which time complexity is being determined.
$endgroup$
– toric_actions
Dec 29 '18 at 12:29
1
$begingroup$
$Nodesleq 2Edges$ and $Edgesleq Nodes^2$ so a polynomial result for one implies a polynomial result for the other. (Assuming all nodes touch at least one edge)
$endgroup$
– Michael
Dec 29 '18 at 15:49
$begingroup$
If an algorithm is $O(N)$ (Where $N$) is the number of trees in a graph. then is that algorithm still in the complexity class $P$?
$endgroup$
– Aryaman Gupta
Dec 29 '18 at 17:34
add a comment |
$begingroup$
In order to assert that a given algorithm for graphs runs in polynomial time, must the variable in the big-O function that represents the run time (denoted henceforth as $O(f(n))$) be the number of vertices? Or can one say that a graph algorithm runs in polynomial time if the variable $n$ is something else (like the number of edges in a graph).
Edit: Could an algorithm that is $O(N)$ (where $N$ is the number of trees in the graph) also be considered as being in $P$?
graph-theory asymptotics computational-complexity computability
asked Dec 29 '18 at 12:25
Aryaman GuptaAryaman Gupta
356
$endgroup$
In order to assert that a given algorithm for graphs runs in polynomial time, must the variable in the big-O function that represents the run time (denoted henceforth as $O(f(n))$) be the number of vertices? Or can one say that a graph algorithm runs in polynomial time if the variable $n$ is something else (like the number of edges in a graph).
Edit: Could an algorithm that is $O(N)$ (where $N$ is the number of trees in the graph) also be considered as being in $P$?
graph-theory asymptotics computational-complexity computability
graph-theory asymptotics computational-complexity computability
asked Dec 29 '18 at 12:25
Aryaman GuptaAryaman Gupta
356
asked Dec 29 '18 at 12:25
Aryaman GuptaAryaman Gupta
356
edited Dec 29 '18 at 17:31
Aryaman Gupta
asked Dec 29 '18 at 12:25
Aryaman GuptaAryaman Gupta
356
asked Dec 29 '18 at 12:25
Aryaman GuptaAryaman Gupta
356
asked Dec 29 '18 at 12:25
Aryaman GuptaAryaman Gupta
356
356
1
$begingroup$
The variable could be either of them or a subset of them. It will depend on the case for which time complexity is being determined.
$endgroup$
– toric_actions
Dec 29 '18 at 12:29
1
$begingroup$
$Nodesleq 2Edges$ and $Edgesleq Nodes^2$ so a polynomial result for one implies a polynomial result for the other. (Assuming all nodes touch at least one edge)
$endgroup$
– Michael
Dec 29 '18 at 15:49
$begingroup$
If an algorithm is $O(N)$ (Where $N$) is the number of trees in a graph. then is that algorithm still in the complexity class $P$?
$endgroup$
– Aryaman Gupta
Dec 29 '18 at 17:34
add a comment |
1
$begingroup$
The variable could be either of them or a subset of them. It will depend on the case for which time complexity is being determined.
$endgroup$
– toric_actions
Dec 29 '18 at 12:29
1
$begingroup$
$Nodesleq 2Edges$ and $Edgesleq Nodes^2$ so a polynomial result for one implies a polynomial result for the other. (Assuming all nodes touch at least one edge)
$endgroup$
– Michael
Dec 29 '18 at 15:49
$begingroup$
If an algorithm is $O(N)$ (Where $N$) is the number of trees in a graph. then is that algorithm still in the complexity class $P$?
$endgroup$
– Aryaman Gupta
Dec 29 '18 at 17:34
1
1
$begingroup$
The variable could be either of them or a subset of them. It will depend on the case for which time complexity is being determined.
$endgroup$
– toric_actions
Dec 29 '18 at 12:29
$begingroup$
The variable could be either of them or a subset of them. It will depend on the case for which time complexity is being determined.
$endgroup$
– toric_actions
Dec 29 '18 at 12:29
1
1
$begingroup$
$Nodesleq 2Edges$ and $Edgesleq Nodes^2$ so a polynomial result for one implies a polynomial result for the other. (Assuming all nodes touch at least one edge)
$endgroup$
– Michael
Dec 29 '18 at 15:49
$begingroup$
$Nodesleq 2Edges$ and $Edgesleq Nodes^2$ so a polynomial result for one implies a polynomial result for the other. (Assuming all nodes touch at least one edge)
$endgroup$
– Michael
Dec 29 '18 at 15:49
$begingroup$
If an algorithm is $O(N)$ (Where $N$) is the number of trees in a graph. then is that algorithm still in the complexity class $P$?
$endgroup$
– Aryaman Gupta
Dec 29 '18 at 17:34
$begingroup$
If an algorithm is $O(N)$ (Where $N$) is the number of trees in a graph. then is that algorithm still in the complexity class $P$?
$endgroup$
– Aryaman Gupta
Dec 29 '18 at 17:34
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Informally: An algorithm is polynomial time if the algorithm runs in time polynomial in the size of the input. Suppose the input for an algorithm is a graph $G$. Then letting $n$ be the number of vertices and $m$ the number of edges, if the algorithm runs in time polynomial in either $n$ or $m$, then that algorithm is considered a polynomial-time algorithm, since the size of $G$ is $n+m$.
An algorithm that runs in time say linear in the number of trees $G$ has [which may be exponential in the size of $G$] is not in $P$. Why? The graph $G$ itself encodes all the information needed for the algorithm and so could be the input. So the algorithm would not run in time polynomial in the size of $G$.
answered Dec 30 '18 at 2:38
MikeMike
4,171412
$endgroup$
$begingroup$
@HenningMakholm sure. But the OP asked about graphs explicitly, and generalizing, it is whether an algorithm runs in time polynomial in the size of the input.
$endgroup$
– Mike
Dec 30 '18 at 2:44
1
$begingroup$
Ah, I think you're right, actually. Never mind.
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:46
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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votes
$begingroup$
Informally: An algorithm is polynomial time if the algorithm runs in time polynomial in the size of the input. Suppose the input for an algorithm is a graph $G$. Then letting $n$ be the number of vertices and $m$ the number of edges, if the algorithm runs in time polynomial in either $n$ or $m$, then that algorithm is considered a polynomial-time algorithm, since the size of $G$ is $n+m$.
An algorithm that runs in time say linear in the number of trees $G$ has [which may be exponential in the size of $G$] is not in $P$. Why? The graph $G$ itself encodes all the information needed for the algorithm and so could be the input. So the algorithm would not run in time polynomial in the size of $G$.
answered Dec 30 '18 at 2:38
MikeMike
4,171412
$endgroup$
$begingroup$
@HenningMakholm sure. But the OP asked about graphs explicitly, and generalizing, it is whether an algorithm runs in time polynomial in the size of the input.
$endgroup$
– Mike
Dec 30 '18 at 2:44
1
$begingroup$
Ah, I think you're right, actually. Never mind.
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:46
add a comment |
$begingroup$
Informally: An algorithm is polynomial time if the algorithm runs in time polynomial in the size of the input. Suppose the input for an algorithm is a graph $G$. Then letting $n$ be the number of vertices and $m$ the number of edges, if the algorithm runs in time polynomial in either $n$ or $m$, then that algorithm is considered a polynomial-time algorithm, since the size of $G$ is $n+m$.
An algorithm that runs in time say linear in the number of trees $G$ has [which may be exponential in the size of $G$] is not in $P$. Why? The graph $G$ itself encodes all the information needed for the algorithm and so could be the input. So the algorithm would not run in time polynomial in the size of $G$.
answered Dec 30 '18 at 2:38
MikeMike
4,171412
$endgroup$
$begingroup$
@HenningMakholm sure. But the OP asked about graphs explicitly, and generalizing, it is whether an algorithm runs in time polynomial in the size of the input.
$endgroup$
– Mike
Dec 30 '18 at 2:44
1
$begingroup$
Ah, I think you're right, actually. Never mind.
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:46
add a comment |
$begingroup$
Informally: An algorithm is polynomial time if the algorithm runs in time polynomial in the size of the input. Suppose the input for an algorithm is a graph $G$. Then letting $n$ be the number of vertices and $m$ the number of edges, if the algorithm runs in time polynomial in either $n$ or $m$, then that algorithm is considered a polynomial-time algorithm, since the size of $G$ is $n+m$.
An algorithm that runs in time say linear in the number of trees $G$ has [which may be exponential in the size of $G$] is not in $P$. Why? The graph $G$ itself encodes all the information needed for the algorithm and so could be the input. So the algorithm would not run in time polynomial in the size of $G$.
answered Dec 30 '18 at 2:38
MikeMike
4,171412
$endgroup$
Informally: An algorithm is polynomial time if the algorithm runs in time polynomial in the size of the input. Suppose the input for an algorithm is a graph $G$. Then letting $n$ be the number of vertices and $m$ the number of edges, if the algorithm runs in time polynomial in either $n$ or $m$, then that algorithm is considered a polynomial-time algorithm, since the size of $G$ is $n+m$.
An algorithm that runs in time say linear in the number of trees $G$ has [which may be exponential in the size of $G$] is not in $P$. Why? The graph $G$ itself encodes all the information needed for the algorithm and so could be the input. So the algorithm would not run in time polynomial in the size of $G$.
answered Dec 30 '18 at 2:38
MikeMike
4,171412
answered Dec 30 '18 at 2:38
MikeMike
4,171412
answered Dec 30 '18 at 2:38
MikeMike
4,171412
answered Dec 30 '18 at 2:38
MikeMike
4,171412
4,171412
$begingroup$
@HenningMakholm sure. But the OP asked about graphs explicitly, and generalizing, it is whether an algorithm runs in time polynomial in the size of the input.
$endgroup$
– Mike
Dec 30 '18 at 2:44
1
$begingroup$
Ah, I think you're right, actually. Never mind.
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:46
add a comment |
$begingroup$
@HenningMakholm sure. But the OP asked about graphs explicitly, and generalizing, it is whether an algorithm runs in time polynomial in the size of the input.
$endgroup$
– Mike
Dec 30 '18 at 2:44
1
$begingroup$
Ah, I think you're right, actually. Never mind.
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:46
$begingroup$
@HenningMakholm sure. But the OP asked about graphs explicitly, and generalizing, it is whether an algorithm runs in time polynomial in the size of the input.
$endgroup$
– Mike
Dec 30 '18 at 2:44
$begingroup$
@HenningMakholm sure. But the OP asked about graphs explicitly, and generalizing, it is whether an algorithm runs in time polynomial in the size of the input.
$endgroup$
– Mike
Dec 30 '18 at 2:44
1
1
$begingroup$
Ah, I think you're right, actually. Never mind.
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:46
$begingroup$
Ah, I think you're right, actually. Never mind.
$endgroup$
– Henning Makholm
Dec 30 '18 at 2:46
add a comment |
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$begingroup$
The variable could be either of them or a subset of them. It will depend on the case for which time complexity is being determined.
$endgroup$
– toric_actions
Dec 29 '18 at 12:29
1
$begingroup$
$Nodesleq 2Edges$ and $Edgesleq Nodes^2$ so a polynomial result for one implies a polynomial result for the other. (Assuming all nodes touch at least one edge)
$endgroup$
– Michael
Dec 29 '18 at 15:49
$begingroup$
If an algorithm is $O(N)$ (Where $N$) is the number of trees in a graph. then is that algorithm still in the complexity class $P$?
$endgroup$
– Aryaman Gupta
Dec 29 '18 at 17:34