Assignment 5 -- due Wednesday, February 21
at the beginning of lab
CS 4521 Spring Semester, 2007
15 Points
Topics: Quicksort, randomized quicksort, and their run times
The assignment:
consists of two parts:
In the first part, you are asked to implement quicksort,
randomized quicksort, and heapsort,
and compare their running times with merge sort;.
in the second part, you are asked to prove a fact about the run time
of quicksort.
Part 1: Comparing the run times of sorting routines (10 points)
Implement the following three sorting routines:
quicksort (page 146), randomized-quicksort (page 154),
and heapsort (page 136) --
actually, you can implement "min-heapsort" that sorts in decreasing
order, which is easy to do with the minHeapify() routine of the previous
assignment.
Note that you may have to implement accessors and mutators, since the
compiler might not allow direct access to member data.
After you have coded all three of the sort routines, load the array
A
with the integers 1, 2, ...,
n
in order, i.e.,
A[i] = i.
This should produce worst-case performance for quicksort; the run times
of the other two sorts should by independent of the order of the input.
Make timing runs for each of the sorts for the largest value of
n
that you used for merge sort in Assignment 2
and record the times for each of the four sorts:
merge sort (from Assignment 2) and the three new ones.
Compare the results -- which is fastest? slowest?
Discussion: for randomized-quicksort,
you can use the C library routine
rand()
to obtain (pseudo-) random number (int) in the range 0 to
RAND_MAX (for example,
RANDOM(p,r) could simply return
p + rand() % (r - p + 1)
-- or, better: replace Line 1 of
RANDOMIZED-PARTITION()
with
i <- p + rand() % (r - p + 1) ).
Note:
rand()
&
RAND_MAX
are
declared in
<stdlib.h>.
What To Hand In:
The results of your timing runs and the
code for each of your new sort routines.
Part 2: Proving a fact about quicksort's run time (5 points)
- (4 points) Exercise 7.4-2, page 159.
Similarly to Section 7.4.1, we want to solve the recurrence:
T(n) = min ( T(q) + T(n - q - 1) ) + Θ(n)
where the minimum is taken over q in [0,n-1].
The problem suggests that T(n) ≥ cnlg(n), which can be shown to be
true by the substitution method for a suitable value of c.
Using the Fact: f(q) = q lg q + (a-q) lg (a-q) attains its
minimum at q = a/2 (and substituting n-1 for a), we have
T(n) ≥ c(n-1) lg((n-1)/2) + Θ(n)
= cn lg(n-1) - c( lower order terms in n ) + Θ(n)
≥ cn lg(n/2) - c( lower order terms in n ) + Θ(n) (for what n?)
= cn lg n - c( new lower order terms in n ) + Θ(n)
≥ cn lg n
for a suitable value of c (how should it be chosen?).
You can assume the Θ(n) term can
be replaced by c1n for some c1 > 0.
Don't forget to prove the Fact.
Page URL: http://www.d.umn.edu
/~ddunham/cs4521s07/assignments/a5/assignment.html
Page Author: Doug Dunham
Last Modified: Saturday, 17-Feb-2007 17:18:57 CST
Comments to: ddunham@d.umn.edu