**
Assignment 2 -- due Monday, September 20
at the ***beginning* of lab

CS 4521 Fall Semester, 2010

25 Points

**
Topics: Growth of Functions, Comparison of Sorts
**

**
The assignment:
**

consists of two parts: in the **first part**, you are asked to solve problems
from the text; in the **second part**, you are asked to implement insertion
sort and merge sort and run them to find the "cross-over" point where
merge sort becomes faster than insertion sort.

### Part 1: Growth of Functions (10 points)

Do the following Exercises from the text:
- (5 points) Exercise 3.1-4, page 53. Prove both of your answers.
- (5 points) Exercise 3.1-5, page 53. Hint: prove
the "if" and the "only if" parts separately.

### Part 2: Comparing run times for sorting (15 points)

The run time of merge sort is independent of the sorting of its input
(or at least it
should be). On the other hand the run time of insertion sort depends on
the input: in the best case, when the array is already sorted, the run
time is Θ(n); in the worst case -- the array is reverse-sorted --
the run time is Θ(n^{2});
in the average case when each new number is
inserted about half way down the sorted part of the array, the run time
is also Θ(n^{2}),
but the constant multiplier is about 1/2 of what it is for the worst case.
There are two subparts to this part: first determine the "cross-over"
value of n where merge sort first beats insertion sort; in the second
subpart, verify what I said above that insertion sort is twice as fast
in the average case as it is in the worst case.
#### Subpart A: Comparison of average case run times for
insertion and merge sort (12 points)

Implement insertion sort and merge sort as two routines.
Have the user enter
**n**,
the number of integers to be sorted.
Rather than generating
**n**
random numbers with a random number generator,
have your program generate the "average case" situation described in
the lab: put the numbers 1, 2, 3, ..., n/2 in that order into
`A`
at every other index, and put the numbers
n, n-1, n-2, ..., n/2 + 1 in that order into the remaining indices.
You may assume
**n**
is even or odd if one or the other is easier.
To time your `insert`
and `merge` programs,
you can use the `clock()`
function in `time.h`
(so you must
`#include <time.h>`
at the top of your program).
The function
`clock()`
returns a value of type
`clock_t`
which contains the number of milliseconds since
the program started - *unfortunately*
rounded to the nearest 100th of a
second (see below for a fix to this problem).
Thus to get timings of sections of code, you could do something like:

clock_t tstart, tend, ttotal ;
tstart = clock() ;
/* code to be timed */
tend = clock() ;
ttotal = tend - tstart ;
printf( "time = %u\n", ttotal ) ;

Note that the type
`clock_t`
is an unsigned (probably long)
integer. I'm sure this can also be done in C++ (and maybe
`cout << ttotal << endl ;`
would work, but I'm not sure of the details -- please let me know of any
subtle details you encounter if you use C++).
But there is a problem with getting accurate timings.
Most small programs, such as insertion sort
and merge sort run so fast that the times reported are very small --
maybe less than 100th of a second even when sorting 100 numbers.
To cure this problem and magnify the results,
put your array intialization and call to your sort routine in a
`for`
loop that executes `numIterations`
( = 100) times, and then divide your total time by
`numIterations`
(you may even have to make
`numIterations`
= 1000, or 10000, ...).

Also, for merge sort, to avoid allocating and de-allocating memory on
each call to the
`merge()`
subroutine, you can make the temporary arrays
`L`
and
`R`
global (or pass their names, i.e. pointers, as parameters to
`merge()`
and
`merge-sort()`).

Finally, run your two programs, timing them, for different values of
`n`,
until you find the "cross-over" value of
`n`
for which merge sort first beats insertion sort.
You might try to do a sort of "binary search" for that value -- i.e. try
`n`
= 100, and if insertion wins, try
`n`
= 1000. Assuming that merge wins at
`n`
= 1000, try
`n`
= 500, etc.
I think that the cross-over value is somewhere between 10 and 1000.
Note that different people might get different answers, due to different
encodings of the algorithms.

**Pseudocode:**

Generate average-case data in the array OrigData[]
(1) Time the following loop to see how much time copying takes
For i = 1 to NumIterations
copy OrigData[] to A[] // requires an inner loop
(2) Time insertion sort:
For i = 1 to NumIterations
copy OrigData[] to A[]
InsertionSort( A, 1, n )
(3) Time merge sort:
For i = 1 to NumIterations
copy OrigData[] to A[]
MergeSort( A, 1, n )

To get the actual times for insertion sort and merge sort, subtract the
copying time (1) from the times of segments (2) and (3).
#### Subpart B: The worst/average case factor of two (3 points)

Add the following code to your program:
Generate worst-case data in the array OrigData[]
(4) Time insertion sort:
For i = 1 to NumIterations
copy OrigData[] to A[]
InsertionSort( A, 1, n )

To generate worst-case data, fill the array
` OrigData[]` with the numbers
n, n-1, n-2, ... 2, 1. Then time a run of your program for the
largest value of n that you used in Subpart 1 -- the run time for (4)
should be roughly twice as long as the run time for (2) (after subtracting the
copying time from both). Report the two run times and the
worst case / average case ratio.
** What To Hand In For Part 2: **
The results of your timing runs and the
code for each of your sort routines.
For subpart A, show timing runs for n = "cross-over" value, and
for n-10 and n+10.
For subpart B, report the two run times and the
worst case / average case ratio.

**Page URL: **http://www.d.umn.edu
/~ddunham/cs4521f10/assignments/a2/assignment.html

**Page Author: **Doug Dunham

**Last Modified: **Tuesday, 19-Oct-2010 15:56:25 CDT

**Comments to: **ddunham@d.umn.edu