Running Times

asa.hoshi · **Joined:** Thu, 11 May 2006 11:24:10 UTC **Posts:** 332

Hi all.

I'm a bit stuck on this question.

So an algorithm A has a running time in O(n^2)

and in $\Omega (n log_2 n)$ .

State whether and why it is consistent or inconsistent with the claim mentioned above for the following statements.

a) in $O(n^{\frac{3}{2}})$
b) in O(n)

c) in $\Theta(n^2)$
d) in $\Theta(n^3)$

my answers were...
a) no. because in $O(n^{\frac{3}{2}})$ , O(n^2)

grows faster
b) no. same as above.
c) yes. does grow at the same rate as O(n^2)

, and is faster than $\Omega (n log_2 n)$ .
d) no. does not grow at $\Theta(n^3)$ . it is faster.

and i on the right track?

Thanks.

Shadow · **Posted:** Thu, 10 Mar 2011 14:28:34 UTC

asa.hoshi wrote:

Hi all.

I'm a bit stuck on this question.

So an algorithm A has a running time in O(n^2)

and in $\Omega (n log_2 n)$ .

State whether and why it is consistent or inconsistent with the claim mentioned above for the following statements.

a) in $O(n^{\frac{3}{2}})$
b) in O(n)

c) in $\Theta(n^2)$
d) in $\Theta(n^3)$

my answers were...
a) no. because in $O(n^{\frac{3}{2}})$ , O(n^2)

grows faster
b) no. same as above.
c) yes. does grow at the same rate as O(n^2)

, and is faster than $\Omega (n log_2 n)$ .
d) no. does not grow at $\Theta(n^3)$ . it is faster.

and i on the right track?

Thanks.

For the last one, what is "it"? If you mean the algorithm, then no, you're wrong, the algorithm does not grow faster it IS faster, which means the complexity is slower, i.e. shorter run-time. Be careful with how you use "faster" when talking about asymptotics.

asa.hoshi · **Joined:** Thu, 11 May 2006 11:24:10 UTC **Posts:** 332

i think i meant to say that n^3 grows faster then n^2 for q.d). And Cos they dont grow at the same speed, ans is no. Then is that right?

Shadow · **Posted:** Thu, 10 Mar 2011 22:19:37 UTC

asa.hoshi wrote:

i think i meant to say that n^3 grows faster then n^2 for q.d). And Cos they dont grow at the same speed, ans is no. Then is that right?

In particular it's because n^3

isn't between the two bounds (it's not O(n^2)

)

asa.hoshi · **Joined:** Thu, 11 May 2006 11:24:10 UTC **Posts:** 332

Shadow wrote:

asa.hoshi wrote:

i think i meant to say that n^3 grows faster then n^2 for q.d). And Cos they dont grow at the same speed, ans is no. Then is that right?

In particular it's because n^3

isn't between the two bounds (it's not O(n^2)

)

i think im confusing myself here.

but for a) in $O(n^{\frac{3}{2}})$ .

Okay, so i know that comparing $n^{\frac{3}{2}}$ and n^2

. Obviously, n^2

grows faster. And $n^{\frac{3}{2}}$ is faster than nlog_2(n)

.

So, having said that is it consistent with the fact that the alrogithm whose running time is in O(n^2)

and $\Omega (n log_2 n)$ ? Because it is within the bounds of both running times. Does that make sense?

Or... is it because, the question says, no faster than $n^{\frac{3}{2}}$ , but the algorithm can run faster than $n^{\frac{3}{2}}$ and up to n^2

? So answer is inconsistent.

Shadow · **Posted:** Mon, 14 Mar 2011 17:47:11 UTC

asa.hoshi wrote:

Shadow wrote:

asa.hoshi wrote:

i think i meant to say that n^3 grows faster then n^2 for q.d). And Cos they dont grow at the same speed, ans is no. Then is that right?

In particular it's because n^3

isn't between the two bounds (it's not O(n^2)

)

i think im confusing myself here.

but for a) in $O(n^{\frac{3}{2}})$ .

Okay, so i know that comparing $n^{\frac{3}{2}}$ and n^2

. Obviously, n^2

grows faster. And $n^{\frac{3}{2}}$ is faster than nlog_2(n)

.

So, having said that is it consistent with the fact that the alrogithm whose running time is in O(n^2)

and $\Omega (n log_2 n)$ ? Because it is within the bounds of both running times. Does that make sense?

Or... is it because, the question says, no faster than $n^{\frac{3}{2}}$ , but the algorithm can run faster than $n^{\frac{3}{2}}$ and up to n^2

? So answer is inconsistent.

The first conclusion is correct. You should recall that the big O means something akin to $\le$ and $\Omega$ means something akin to $\ge$

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