Mixing
What will be a more meaningful way to represent a O$frac{n-8}{2} choose frac{n}{4}-2$ Time Complexity
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I completed the Time Analysis for an algorithm and wanted to check if i can represent its TC in a more readable way.
The best I have at the moment is O$frac{n-8}{2} choose frac{n}{4}-2$
What could be are 'easier' way to represent this TC?
discrete-mathematics algorithms
asked Jan 2 at 18:20
John SeppardJohn Seppard
52
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add a comment |
$begingroup$
I completed the Time Analysis for an algorithm and wanted to check if i can represent its TC in a more readable way.
The best I have at the moment is O$frac{n-8}{2} choose frac{n}{4}-2$
What could be are 'easier' way to represent this TC?
discrete-mathematics algorithms
asked Jan 2 at 18:20
John SeppardJohn Seppard
52
$endgroup$
add a comment |
$begingroup$
I completed the Time Analysis for an algorithm and wanted to check if i can represent its TC in a more readable way.
The best I have at the moment is O$frac{n-8}{2} choose frac{n}{4}-2$
What could be are 'easier' way to represent this TC?
discrete-mathematics algorithms
asked Jan 2 at 18:20
John SeppardJohn Seppard
52
$endgroup$
I completed the Time Analysis for an algorithm and wanted to check if i can represent its TC in a more readable way.
The best I have at the moment is O$frac{n-8}{2} choose frac{n}{4}-2$
What could be are 'easier' way to represent this TC?
discrete-mathematics algorithms
discrete-mathematics algorithms
asked Jan 2 at 18:20
John SeppardJohn Seppard
52
asked Jan 2 at 18:20
John SeppardJohn Seppard
52
asked Jan 2 at 18:20
John SeppardJohn Seppard
52
asked Jan 2 at 18:20
John SeppardJohn Seppard
52
asked Jan 2 at 18:20
John SeppardJohn Seppard
52
52
add a comment |
add a comment |
1 Answer
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This
$${{frac{n-8}{2}} choose {frac{n}{4}-2}} quad in quad theta(2^{frac{n}{2}} left(sqrt{n}right)^{-1})$$
[Informally, the "$-frac{8}{2}$" and the "$-2$" don't change the asymptotics]
answered Jan 2 at 18:24
MikeMike
3,333311
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
This
$${{frac{n-8}{2}} choose {frac{n}{4}-2}} quad in quad theta(2^{frac{n}{2}} left(sqrt{n}right)^{-1})$$
[Informally, the "$-frac{8}{2}$" and the "$-2$" don't change the asymptotics]
answered Jan 2 at 18:24
MikeMike
3,333311
$endgroup$
add a comment |
$begingroup$
This
$${{frac{n-8}{2}} choose {frac{n}{4}-2}} quad in quad theta(2^{frac{n}{2}} left(sqrt{n}right)^{-1})$$
[Informally, the "$-frac{8}{2}$" and the "$-2$" don't change the asymptotics]
answered Jan 2 at 18:24
MikeMike
3,333311
$endgroup$
add a comment |
$begingroup$
This
$${{frac{n-8}{2}} choose {frac{n}{4}-2}} quad in quad theta(2^{frac{n}{2}} left(sqrt{n}right)^{-1})$$
[Informally, the "$-frac{8}{2}$" and the "$-2$" don't change the asymptotics]
answered Jan 2 at 18:24
MikeMike
3,333311
$endgroup$
This
$${{frac{n-8}{2}} choose {frac{n}{4}-2}} quad in quad theta(2^{frac{n}{2}} left(sqrt{n}right)^{-1})$$
[Informally, the "$-frac{8}{2}$" and the "$-2$" don't change the asymptotics]
answered Jan 2 at 18:24
MikeMike
3,333311
edited Jan 2 at 18:30
answered Jan 2 at 18:24
MikeMike
3,333311
answered Jan 2 at 18:24
MikeMike
3,333311
answered Jan 2 at 18:24
MikeMike
3,333311
3,333311
add a comment |
add a comment |
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