CS61B

34.2 Math Problems Out of Nowhere

A Math Problem

Is N!\in \Omega((\frac{N}{2})^\frac{N}{2})Prove your answer. [Recall that ∈ Ω can be informally be interpreted to mean >=. In other words, does factorial grow at least as quickly as (\frac{N}{2})^\frac{N}{2}?

10!

10 * 9 * 8 * 7 * 6 * … * 1

5 * 5 * 5 * 5 * 5

$N!>(\frac{N}{2})^\frac{N}{2}$ for large N, therefore $N!\in \Omega((\frac{N}{2})^\frac{N}{2})$

Another Math Problem

Given: N!>(\frac{N}{2})^\frac{N}{2}, which we used to prove our answer to the previous problem. Show that log(N!) \in \Omega(N * logN). [Recall: log means an unspecified base]

We have that $N!>(\frac{N}{2})^\frac{N}{2}$

Taking the log of both sides, we have that $log(N!)>log((\frac{N}{2})^\frac{N}{2})$ .
Bringing down the exponent we have that $log(N!)>\frac{N}{2}\*log((\frac{N}{2}))$ .
Discarding the unnecessary constant, we have $log(N!)>\Omega(N\*log(\frac{N}{2}))$ .
From there, we have that $log(N!)>\Omega(N\*log(N))$ . [since $log(\frac{N}{2})$ is the same thing asymptotically as $log(N)$ ]

In other words, $log(N!)$ grows at least as quickly as $N\*log(N)$ .

Last Math Problem

In the previous problem, we showed that log(N!) \in \Omega(N * logN). Now show that N*logN \in \Omega(log(N!)).

Proof:

$log(N!)=log(N)+log(N-1)+...+log(1)$
$N\*logN=log(N)+log(N)+...+log(N)$
Therefore $N\*logN \in \Omega(log(N!))$

Omega and Theta

Given N*logN \in \Omega(log(N!)) and log(N!) \in \Omega(N * logN). Which of the following can we say?

A. N*logN \in \Theta(log(N!))
B. log(N!) \in \Theta(N * logN)
C. Both A and B
D. Neither

Answer: C. Both A and B

Summary

We’ve shown that $N\*logN \in \Theta(log(N!))$ .

In other words, these two functions grow at the same rate asymptotically.

As for why we did this, we will see in a little while…