- #1

nrobidoux

- 25

- 0

## Homework Statement

A computer can perform 10

^{10}ops/s. Assume 1 op per 1 input. Given the following algorithmic complexities how many inputs can be performed in an hour.

*n*^{2}*n*^{3}- 100
*n*^{2} *n*log*n*- 2
^{n} - 2
^{2n}

## Homework Equations

## The Attempt at a Solution

[/B]

*C*= 1·10

^{10}ops/s · 3.6·10

^{2}s/hr

*C = 3.6*·10

^{12}ops/hr

Set each complexity equal to

*C*.

I know this is almost basic algebra but I haven't done this in decades. I think I'm good in the first 3... it's the last 3 I'm a bit confused on. Mostly

*n*log

*n*. Did some googling of the rules but I'm still confused there, and I've never seen a log of a log...

----------------------------------------------

*n*

^{2}

*= C*

n

n

^{(2 · 1/2)}=

*C*

^{1/2}

*n*=

*C*

^{1/2}

----------------------------------------------

The next is basically the same except it's cubic:

*n*=

*C*

^{1/3}

----------------------------------------------

100

*n*

^{2}=

*C*

n

n

^{2}

*= C*/100

*n*

^{(2 · 1/2)}

*=*(

*C*/100)

^{1/2}

*n =*(

*C*/100)

^{1/2}

----------------------------------------------

*n*log

_{b}(

*n*) =

*C*

... (more Googling)...

... (more Googling)...

I have no clue... I found another rule. On the surface I think I applied them correctly but I don't know how to get

*n*by itself. I got:

*b*=

^{C}*n*and

^{n}*b*=

^{C/n}*n*... of course if I really looked at those two...

----------------------------------------------

2

^{n}=

*C*

n= log

n

_{2}(

*C*)

----------------------------------------------

2

^{2n}=

*C*

2

^{n}= log

_{2}(

*C*)

*n*= log

_{2}( log

_{2}(

*C*))