The time complexity of the n-ary cartesian product over n sets
$begingroup$
Recall that the Cartesian product $Atimes A$ is defined as the set $lbrace (x,y):xin A ,y in A rbrace$ . Thus, if for example, $A=lbrace 1,2,3 rbrace$,$A times A=lbrace(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)rbrace$, then the time complexity is $O(n^2 log n)$, where n refers to the number of members of the set (I presume). But suppose I would like to evaluate $A times A times A$ or any finite number of sets in $A times A times A$... What would the time complexity be then?
computational-complexity
edited Jun 30 '15 at 22:57
mich95
6,98211126
asked Jun 30 '15 at 22:01
Carl HillCarl Hill
62
$endgroup$
add a comment |
$begingroup$
Recall that the Cartesian product $Atimes A$ is defined as the set $lbrace (x,y):xin A ,y in A rbrace$ . Thus, if for example, $A=lbrace 1,2,3 rbrace$,$A times A=lbrace(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)rbrace$, then the time complexity is $O(n^2 log n)$, where n refers to the number of members of the set (I presume). But suppose I would like to evaluate $A times A times A$ or any finite number of sets in $A times A times A$... What would the time complexity be then?
computational-complexity
edited Jun 30 '15 at 22:57
mich95
6,98211126
asked Jun 30 '15 at 22:01
Carl HillCarl Hill
62
$endgroup$
add a comment |
$begingroup$
Recall that the Cartesian product $Atimes A$ is defined as the set $lbrace (x,y):xin A ,y in A rbrace$ . Thus, if for example, $A=lbrace 1,2,3 rbrace$,$A times A=lbrace(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)rbrace$, then the time complexity is $O(n^2 log n)$, where n refers to the number of members of the set (I presume). But suppose I would like to evaluate $A times A times A$ or any finite number of sets in $A times A times A$... What would the time complexity be then?
computational-complexity
edited Jun 30 '15 at 22:57
mich95
6,98211126
asked Jun 30 '15 at 22:01
Carl HillCarl Hill
62
$endgroup$
Recall that the Cartesian product $Atimes A$ is defined as the set $lbrace (x,y):xin A ,y in A rbrace$ . Thus, if for example, $A=lbrace 1,2,3 rbrace$,$A times A=lbrace(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)rbrace$, then the time complexity is $O(n^2 log n)$, where n refers to the number of members of the set (I presume). But suppose I would like to evaluate $A times A times A$ or any finite number of sets in $A times A times A$... What would the time complexity be then?
computational-complexity
computational-complexity
edited Jun 30 '15 at 22:57
mich95
6,98211126
asked Jun 30 '15 at 22:01
Carl HillCarl Hill
62
edited Jun 30 '15 at 22:57
mich95
6,98211126
asked Jun 30 '15 at 22:01
Carl HillCarl Hill
62
edited Jun 30 '15 at 22:57
mich95
6,98211126
edited Jun 30 '15 at 22:57
mich95
6,98211126
edited Jun 30 '15 at 22:57
mich95
6,98211126
6,98211126
asked Jun 30 '15 at 22:01
Carl HillCarl Hill
62
asked Jun 30 '15 at 22:01
Carl HillCarl Hill
62
asked Jun 30 '15 at 22:01
Carl HillCarl Hill
62
62
add a comment |
add a comment |
1 Answer
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$begingroup$
I assume the task is generating a string representation of $A^k$.
If $A$ has $n$ elements, we have to list $n^k$ tuples with $k$ components each.
However the size of the numbers themselves grows with growing $n$, needing about $log_{10}(n) + 1$ digits per component.
So that will give an output of roughly $n^k k log_{10} n$ digits.
answered Jun 30 '15 at 22:45
mvwmvw
31.5k22252
$endgroup$
$begingroup$
Again, thank you for the response. My only other question is if this is considered "fast" to compute. Is it faster or slower than polynomial time?
$endgroup$
– Carl Hill
Jul 4 '15 at 1:42
$begingroup$
The number of digits seems to be $O(n^{k+1})$ so needing polynomial time. Yes that is fast in quotes.
$endgroup$
– mvw
Jul 4 '15 at 6:48
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
I assume the task is generating a string representation of $A^k$.
If $A$ has $n$ elements, we have to list $n^k$ tuples with $k$ components each.
However the size of the numbers themselves grows with growing $n$, needing about $log_{10}(n) + 1$ digits per component.
So that will give an output of roughly $n^k k log_{10} n$ digits.
answered Jun 30 '15 at 22:45
mvwmvw
31.5k22252
$endgroup$
$begingroup$
Again, thank you for the response. My only other question is if this is considered "fast" to compute. Is it faster or slower than polynomial time?
$endgroup$
– Carl Hill
Jul 4 '15 at 1:42
$begingroup$
The number of digits seems to be $O(n^{k+1})$ so needing polynomial time. Yes that is fast in quotes.
$endgroup$
– mvw
Jul 4 '15 at 6:48
add a comment |
$begingroup$
I assume the task is generating a string representation of $A^k$.
If $A$ has $n$ elements, we have to list $n^k$ tuples with $k$ components each.
However the size of the numbers themselves grows with growing $n$, needing about $log_{10}(n) + 1$ digits per component.
So that will give an output of roughly $n^k k log_{10} n$ digits.
answered Jun 30 '15 at 22:45
mvwmvw
31.5k22252
$endgroup$
$begingroup$
Again, thank you for the response. My only other question is if this is considered "fast" to compute. Is it faster or slower than polynomial time?
$endgroup$
– Carl Hill
Jul 4 '15 at 1:42
$begingroup$
The number of digits seems to be $O(n^{k+1})$ so needing polynomial time. Yes that is fast in quotes.
$endgroup$
– mvw
Jul 4 '15 at 6:48
add a comment |
$begingroup$
I assume the task is generating a string representation of $A^k$.
If $A$ has $n$ elements, we have to list $n^k$ tuples with $k$ components each.
However the size of the numbers themselves grows with growing $n$, needing about $log_{10}(n) + 1$ digits per component.
So that will give an output of roughly $n^k k log_{10} n$ digits.
answered Jun 30 '15 at 22:45
mvwmvw
31.5k22252
$endgroup$
I assume the task is generating a string representation of $A^k$.
If $A$ has $n$ elements, we have to list $n^k$ tuples with $k$ components each.
However the size of the numbers themselves grows with growing $n$, needing about $log_{10}(n) + 1$ digits per component.
So that will give an output of roughly $n^k k log_{10} n$ digits.
answered Jun 30 '15 at 22:45
mvwmvw
31.5k22252
answered Jun 30 '15 at 22:45
mvwmvw
31.5k22252
answered Jun 30 '15 at 22:45
mvwmvw
31.5k22252
answered Jun 30 '15 at 22:45
mvwmvw
31.5k22252
31.5k22252
$begingroup$
Again, thank you for the response. My only other question is if this is considered "fast" to compute. Is it faster or slower than polynomial time?
$endgroup$
– Carl Hill
Jul 4 '15 at 1:42
$begingroup$
The number of digits seems to be $O(n^{k+1})$ so needing polynomial time. Yes that is fast in quotes.
$endgroup$
– mvw
Jul 4 '15 at 6:48
add a comment |
$begingroup$
Again, thank you for the response. My only other question is if this is considered "fast" to compute. Is it faster or slower than polynomial time?
$endgroup$
– Carl Hill
Jul 4 '15 at 1:42
$begingroup$
The number of digits seems to be $O(n^{k+1})$ so needing polynomial time. Yes that is fast in quotes.
$endgroup$
– mvw
Jul 4 '15 at 6:48
$begingroup$
Again, thank you for the response. My only other question is if this is considered "fast" to compute. Is it faster or slower than polynomial time?
$endgroup$
– Carl Hill
Jul 4 '15 at 1:42
$begingroup$
Again, thank you for the response. My only other question is if this is considered "fast" to compute. Is it faster or slower than polynomial time?
$endgroup$
– Carl Hill
Jul 4 '15 at 1:42
$begingroup$
The number of digits seems to be $O(n^{k+1})$ so needing polynomial time. Yes that is fast in quotes.
$endgroup$
– mvw
Jul 4 '15 at 6:48
$begingroup$
The number of digits seems to be $O(n^{k+1})$ so needing polynomial time. Yes that is fast in quotes.
$endgroup$
– mvw
Jul 4 '15 at 6:48
add a comment |
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