CS 660: Intro

CS 660: Combinatorial Algorithms
Review of Mathematical Analysis of Algorithms

[To Lecture Notes Index]
San Diego State University -- This page last updated 8/29/95

Contents of Intro Lecture

References

Introduction To Algorithms, Corman, Leiserson,Rivest, Chapters 1-4

Mathematical Analysis of Algorithms

Model of Computing

If analysis of algorithms is the answer, what is the question?

Given two or more algorithms for the same task, which is better?

Under which condition is bubble sort better than insertion sort?

What computing resources does an algorithm require?

How long will it take bubble sort to sort a list of N items?

Goal of mathematical analysis is a function of the resources required of an algorithm

On what computer?

What is a Computer? Random-access machine (RAM)

Single processor
Instructions executed sequentially
Each operation requires the same amount of time

Single cost vs. Lg(N) cost

Time required for basic operation?

3 + 6: 1234!

Insertion Sort A[0] = - infinity
for K = 2 to N do
begin

J = K;
Key = A[J];
while Key < A[J-1] do
begin

A[J] = A[J-1];
J = J - 1;

end while;
A[J] = Key;

end for;
Complexity

Resources required by the algorithm as a function of the input size

Worst-case Analysis

Complexity of an algorithm based on worst input of each size

Average-case Analysis

Complexity of an algorithm averaged over all inputs of each size

Insertion Sort

	Comparisons	Element moves
worst case	(N+1)N/2 - 1	(N-1)N/2
average case	(N+1)N/4 - 1/2	(N-1)N/4

Timing Analysis

Timing in C on Rohan

main()
{

int k, iterations;
for (iterations = 0; iterations < 50; iterations++)
{ start(); /* start the timer */
for (k = 0; k < 2000000; k++) /* do some work */ k = k;
stop(); /* stop the timer */

printf("Time taken: %ld\n", report());
};

}
Result on Rohan
Time Frequency Occurred
30 2
31 2
32 9
33 10
34 11
35 9
36 5
37 1
39 1 Source for Timing C Code on Rohan
#include <stdio.h>
#include <sys/times.h>
#include <limits.h>

static struct tms _start; /* Stores the starting time*/
static struct tms _stop; /* Stores the ending time*/

int start()
{
times(&_start);
}

int stop()
{
times(&_stop);
}

unsigned long report()
{
return _stop.tms_utime - _start.tms_utime;
}

main()
{
int k, iterations;

for (iterations = 0; iterations < 50; iterations++)
{
start();: /* start the timer */

for (k = 0; k < 2000000; k++): /* do some work */
k = k;

stop();: /* stop the timer */

printf("Time taken: %ld\n", report());
};

}

Handling Measurement Errors [1]

Repeat a measurement n times
Let the measurements be labeled

Let

and

The confidence interval for the true measurement is[2]:

The value of t determine the probability the measurement is in the interval

When n >= 50
Probability value of t

50%	80%	90%	95%	99%
0.67	1.28	1.64	1.96	2.58

In Example

, s = 3.15, selecting t = 1.96 we get

95% confidence interval is (32.83, 34.57) Student t table - When n < 50

n	90%	95%	99%
1	3.078	6.314	31.821
2	1.886	2.920	6.965
3	1.638	2.353	4.541
4	1.533	2.132	3.747
5	1.476	2.015	3.365

6	1.440	1.943	3.143
7	1.415	1.895	2.998
8	1.397	1.860	2.896
9	1.383	1.833	2.821
10	1.372	1.812	2.764

20	1.325	1.725	2.528
30	1.310	1.697	2.457
40	1.303	1.684	2.423

Estimating Complexity from Timing Results

Fun with Functions
Let f(n) = 3n*n + 4n + 5 and g(n) = 3n*n

Fact: g(n) is an approximation of f(n)

Notation: f(n) = g(n) +

n	f(n)	g(n)	% error
1	12	3	75.00%
10	345	300	13.04%
20	1285	1200	6.61%
30	2825	2700	4.42%
40	4965	4800	3.32%
50	7705	7500	2.66%
60	11045	10800	2.22%
70	14985	14700	1.90%
80	19525	19200	1.66%
90	24665	24300	1.48%
100	30405	30000	1.33%
200	120805	120000	0.67%
300	271205	270000	0.44%