Can the asymptotic expression $mathcal{O(log(n) cdot log(m))}$ be simplified?
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I constructed an algorithm $A$ with input $(n,m)$ and I found that it has runtime
$mathcal{O(log(n) cdot log(m))}$. I was asking myself if this expression can be simplified somehow, but I could not find a way. Do you see a possiblity here?
asymptotics
asked Nov 17 at 11:20
3nondatur
362111
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up vote
1
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I constructed an algorithm $A$ with input $(n,m)$ and I found that it has runtime
$mathcal{O(log(n) cdot log(m))}$. I was asking myself if this expression can be simplified somehow, but I could not find a way. Do you see a possiblity here?
asymptotics
asked Nov 17 at 11:20
3nondatur
362111
1
What could be simpler? You use the two input values $n,m$ once each, you apply a simple standard function to each, you combine the two by the simple operation of multiplication
– Hagen von Eitzen
Nov 17 at 11:21
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I constructed an algorithm $A$ with input $(n,m)$ and I found that it has runtime
$mathcal{O(log(n) cdot log(m))}$. I was asking myself if this expression can be simplified somehow, but I could not find a way. Do you see a possiblity here?
asymptotics
asked Nov 17 at 11:20
3nondatur
362111
I constructed an algorithm $A$ with input $(n,m)$ and I found that it has runtime
$mathcal{O(log(n) cdot log(m))}$. I was asking myself if this expression can be simplified somehow, but I could not find a way. Do you see a possiblity here?
asymptotics
asymptotics
asked Nov 17 at 11:20
3nondatur
362111
asked Nov 17 at 11:20
3nondatur
362111
asked Nov 17 at 11:20
3nondatur
362111
asked Nov 17 at 11:20
3nondatur
362111
asked Nov 17 at 11:20
3nondatur
362111
362111
1
What could be simpler? You use the two input values $n,m$ once each, you apply a simple standard function to each, you combine the two by the simple operation of multiplication
– Hagen von Eitzen
Nov 17 at 11:21
add a comment |
1
What could be simpler? You use the two input values $n,m$ once each, you apply a simple standard function to each, you combine the two by the simple operation of multiplication
– Hagen von Eitzen
Nov 17 at 11:21
1
1
What could be simpler? You use the two input values $n,m$ once each, you apply a simple standard function to each, you combine the two by the simple operation of multiplication
– Hagen von Eitzen
Nov 17 at 11:21
What could be simpler? You use the two input values $n,m$ once each, you apply a simple standard function to each, you combine the two by the simple operation of multiplication
– Hagen von Eitzen
Nov 17 at 11:21
add a comment |
1 Answer
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1
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The only thing I can think of is
$$mathcal O(log(n^{log m})) $$
But this is silly and not really any simpler. Go with what you have.
answered Nov 17 at 11:23
Matt Samuel
36.2k63463
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The only thing I can think of is
$$mathcal O(log(n^{log m})) $$
But this is silly and not really any simpler. Go with what you have.
answered Nov 17 at 11:23
Matt Samuel
36.2k63463
add a comment |
up vote
1
down vote
accepted
The only thing I can think of is
$$mathcal O(log(n^{log m})) $$
But this is silly and not really any simpler. Go with what you have.
answered Nov 17 at 11:23
Matt Samuel
36.2k63463
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The only thing I can think of is
$$mathcal O(log(n^{log m})) $$
But this is silly and not really any simpler. Go with what you have.
answered Nov 17 at 11:23
Matt Samuel
36.2k63463
The only thing I can think of is
$$mathcal O(log(n^{log m})) $$
But this is silly and not really any simpler. Go with what you have.
answered Nov 17 at 11:23
Matt Samuel
36.2k63463
answered Nov 17 at 11:23
Matt Samuel
36.2k63463
answered Nov 17 at 11:23
Matt Samuel
36.2k63463
answered Nov 17 at 11:23
Matt Samuel
36.2k63463
36.2k63463
add a comment |
add a comment |
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What could be simpler? You use the two input values $n,m$ once each, you apply a simple standard function to each, you combine the two by the simple operation of multiplication
– Hagen von Eitzen
Nov 17 at 11:21