Doubt regarding the variable by which time complexity is measured












0












$begingroup$


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







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$endgroup$








  • 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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 17:31







Aryaman Gupta

















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














  • 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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $G$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @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











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1 Answer
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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 $G$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @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
















1












$begingroup$

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 $G$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @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














1












1








1





$begingroup$

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 $G$.






share|cite|improve this answer









$endgroup$



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 $G$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    @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


















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