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In the solution given for this problem, their argument is that <math>n^n</math> is not a multiple of <math>n!</math> for <math>n\ge 3</math>. Intuitively, <math>n!</math> for <math>n\ge 3</math> includes more than one prime factor, namely 2 and 3, whereas <math>n^n</math> of course has only one. | In the solution given for this problem, their argument is that <math>n^n</math> is not a multiple of <math>n!</math> for <math>n\ge 3</math>. Intuitively, <math>n!</math> for <math>n\ge 3</math> includes more than one prime factor, namely 2 and 3, whereas <math>n^n</math> of course has only one. | ||
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| + | |||
| + | |||
| + | == Prob 5.4-5 == | ||
| + | Is there an intuitive explanation as to why the nested summations in the solution evaluate to a "combination"? | ||
Revision as of 21:27, 14 February 2009
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.).
ECE 608 professor Ghafoor Spring 2009
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} $
- ith 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"?
