## Contents

# Rhea Section for ECE 608 Professor Ghafoor, Spring 2009

If you create a page that belongs to this course, please write

[[Category:ECE608Spring2009ghafoor]]

at the top of the page. You may also add any other category you feel is appropriate (e.g., "homework", "Fourier", etc.).

## What's New

- Discussion about the availability of hw solutions online
**- New discussion!** - Are you for or against using plus or minus grades?

## TA

Hamza Bin Sohail Office Hours: Tuesday & Thursday 4:30-5:30PM in EE306

### Course Website

http://cobweb.ecn.purdue.edu/~ee608/

### Newsgroup

On the news.purdue.edu server: purdue.class.ece608

One way to access: SSH to a server at Purdue (ie expert.ics.purdue.edu) and type "lynx news.purdue.edu/purdue.class.ece608"

On Ubuntu, you can use the "Pan" newsreader.

- "sudo apt-get install pan"
- Set "news.purdue.edu" as the server. You do not need to enter login information.
- Type "purdue.class.ece608" in the box in the upper-left of the screen.
- After a delay (half a minute?) the newsgroup will appear in the left pane. You can right click the group to "subscribe".

### Reviewed Algorithms

### Discussions

Area to post questions, set up study groups, etc.

## 4-4

Attempted solution for 4-4 part (d): $ T(n) = 3T(n/3+5)+n/2 $

We use the iteration method. Start with recursion tree:

- Root node: $ \frac{n}{2} $
- First level: $ 3\frac{\frac{n}{3}+5}{2} $
- Second level: $ 9\frac{\frac{n}{3}+5}{4} $
- i
^{th}level: $ \left(\frac{3}{2}\right)^i\left(\frac{n}{3}+5\right) $

*Note:* above is not correct

$ \left(\frac{n}{3}+5\right)\sum{\left(\frac{3}{2}\right)^i} $

## Randomization (and prob 5.3-3)

As we've seen, Permute-With-All does not produce a uniform random permutation. Curious to see how it affects the distribution of permutations? Here's an experiment with the erroneous randomization algorithm: False randomize

In the solution given for this problem, their argument is that $ n^n $ is not a multiple of $ n! $ for $ n\ge 3 $. Intuitively, $ n! $ for $ n\ge 3 $ includes more than one prime factor, namely 2 and 3, whereas $ n^n $ of course has only one.

## Prob 5.4-5

Is there an intuitive explanation as to why the nested summations in the solution evaluate to a "combination"?