dr deepak b phatakcs101/2011.2/lectures/lecture_slides/201… · dr. deepak b phatak 3. iit bombay....

CS 101Computer Programming and utilization

Dr Deepak B PhatakSubrao Nilekani Chair Professor

Department of CSE, Kanwal Rekhi BuildingIIT Bombay

Session 6, Iterative SolutionsFriday, August 12, 2011

[Some ideas: courtesy Prof Sohoni, and Prof Ranade]

IIT BOMBAY

Session 6, Iterative solutions

Dr. Deepak B Phatak 2

IIT BOMBAY

Overview

• Iterative solutions using a count– The ‘for’ statement in C++

• Estimating the value of loge a

• Hemachandra numbers

• Announcements– Labs next week



IIT BOMBAY

Finding sum of numbers

• Find the sum of N numbers– Value of N is given as input

• Values of each individual number given as input



IIT BOMBAY

Repeating the iteration-block a given number of times

• Execute a block of statements a fixed number of times, say N

• We need a ‘counting mechanism’ , which will

• start with a prescribed initial value, say, count = 1

• If count is <= N, continue iterative execution,

else jump out of the iterative block

• add 1 to the count every time, after we execute the block



IIT BOMBAY

Finding sum of N numbers

cin >> N;count = 1;while( count < = N ){

count = count + 1;}

return 0;}


Block of statements to be repeated


IIT BOMBAY

Finding sum of N numbers#include<iostream>using namespace std;int main(){

int a, sum = 0, N; int count;cout<< “Give number of values to be added ”; cin >> N;count = 1;while( count < = N){

cout<<"enter the next number:"; cin>>a;sum = sum +a;count = count + 1;

} cout << “Sum is “ << sum << endl;return 0;

}



IIT BOMBAY

Solutions which use such counting mechanism

• Several computational problems need such mechanism– Adding a fixed number of terms of any sequence– Calculating average marks for a given number of students– Calculating N!

• N! = 1 x 2 x 3 x 4 x 5 ….. x N• No input value needed, we can generate each ‘term’



IIT BOMBAY

Finding value of N!#include<iostream>using namespace std;int main(){

int a, FactN = 1, N; int count;cout<< “Give number whose factorial is needed:”; cin >> N;count = 1;while(count < = N){

FactN = FactN * count;count = count + 1;

} cout << “Factorial is ” << FactN << endl;return 0;

}



IIT BOMBAY

Repetitive execution by counting

• C++ provides an instruction to implement such iteration control– ‘for’ instruction

• This instruction permits us to define these three actions at a single place in our program– Initial value of a count, – condition to continue execution of the iterative block

(to check if count is within the prescribed limit)– Increment the count after each iteration

• These actions are defined at the beginning of the iterative block



IIT BOMBAY

Repetitive execution using for

for ( count = 1; count <=N; count ++){

// statements forming the iterative-block

}• It initializes the count before beginning the first iteration• At the beginning of each iteration, it checks condition

– continue execution if the condition is true, – else jump out, to exit the ‘for’ block

• It increments the count immediately after each iteration



IIT BOMBAY

Program for adding N numbers using for statement

int main(){int number, N, sum, count;sum = 0;cout << “How many numbers do you wish to add ” ; cin >> N;for (count =1; count <=N; count ++){

cout<<“\nenter next number to be added: ”; cin >> number;sum = sum + number;

} cout << endl << “sum is ” << sum << endl;return 0;

}



IIT BOMBAY

Finding maximum of N given numbers

int main() { int count, number, N, max; cout << “How many numbers?: ”; cin >> N;cout << “give first number: ”; cin >> number; max = number;for (count =2; count <=N; count ++){

cout << endl << “give next number: ”; cin >> number;if (number > max) { max = number;}

}cout << endl << “Maximum of all given numbers is: ” << max;return 0;

}



IIT BOMBAY

Sum of N natural numbers

int main(){int i, N, Sum = 0;cin >> Nfor(i=1; i <= N; i=i+1){

sum = sum + i;}cout << “Sum of” << N << “natural numbers is ”;cout << sum << “endl”;return 0;

}



IIT BOMBAYFactorial of a given number n

n! = 1 * 2 * 3 * 4 * … * n

int main() {int nfactorial, n, i;cout << “Give value of n “; cin >> n;nfactorial = 1;for (i =2; i <= n; i++){

nfactorial = nfactorial * i;};cout << “Factorial of “ << n << “ is” << nfactorial << endl;return 0;

}



IIT BOMBAYGeneral format and semantics of the ‘for’ statement

for( init-group; condition; increment-group ){iterative-block

}1. Execute “init-group”. An assignment statement. “loop initialization”2. Evaluate “condition” . “loop test”.

if true, continue with step 3;

if false, execution of ‘for’ statement ends, control jumps to step 6

3. Execute “iterative-block”. “loop body”

4. Execute “increment-group”. “loop increment”

5. Go back to step 2.

6. Execution continues with the next statement following the ‘for’ blockSession 6, Iterative solutions


IIT BOMBAY

Variations in specifications of ‘for’ loop

for (i = 7; i < N ; i = i + 2){ …}

for (x = 1.5; x < = 2.8; x = x + 0.0001){…}

for (x= 1.5, count =1; count <=N; x +=0.0001, count++){…}

for (i = 0, j=N; i<j; i++, j--){…}

for (Number = 0; Number <= 20;){…}

for (;;){…}



IIT BOMBAY

Computing Natural Logarithm

• Natural Logarithm of a number a is defined as

• A computer cannot integrate– We must use arithmetic operations to estimate integral value– Estimate area under the curve f(x) = 1/x from 1 to a.

• We will approximate the area by many small rectangles.– Each having a fixed width w– Sum of the areas = approximate value of the integral

1/ x dx1

a



IIT BOMBAY

Integral

1 a

calculate the area under the curvebetween 1 and a.

x

y

y = 1/x,



IIT BOMBAY

Riemann Integral

1 a

calculate Sum of areas of allrectangles between 1 and a.



IIT BOMBAY

Riemann Integral

1 a

Each rectangle has a fixed width wWhat is the height?



IIT BOMBAYRiemann Integral using n rectangles

If there are n rectangles of equal width ww = (a-1) / n

1 a

w




1 ax coordinate of bottom left corner of i th rectangle

ith rectangle: x = 1 + (i-1)w

x coordinate of bottom left corner of these rectangles:1st rectangle: x= 1; 2nd rectangle: x = 1 + w;

3rd rectangle: x = 1 + 2w;. . .




1 a

Height h of i th Rectangle

h = 1 / x

= 1/(1 + (i-1)*w)




1 a

Area of i th Rectangle = w * h

= w * (1 / x )

= w * (1/(1 + (i-1)*w)



IIT BOMBAY

How many rectangles?

• More the merrier! Say 1000.• Width of all rectangles taken together = a - 1.• Width w of each rectangle= (a - 1)/1000• x coordinate of bottom left corner of ith rectangle

x = 1 + (i-1) * w.• Height h of ith rectangle

1/x = 1/(1+(i-1)*w) • Area of this rectangle is

width x height= w * (1/(1+ (i-1) *w))



IIT BOMBAY

Program to compute logarithm of a given value

int main(){int i;float a, area=0, w;cout << “Give number whose logarithm is to be found ”; cin >> a;w = (a-1)/1000.0;for(i=1; i <= 1000; i=i+1){

area = area + w*(1/(1+(i-1)*w));}cout << “Log of ” << a << “is ” << area;return 0;

}



IIT BOMBAY

Introduction to Hemachandra’s Problem

• Suppose I have to build a wall of length 4 feet. I have bricks which are 2 feet long, and also 1 foot long. In how many ways I can lay the bricks, so that I fill the 4 feet?

• Possibilities: 1, 1, 1, 12, 1, 11, 2, 11, 1, 22, 2

• In how many ways can this be done, if the length of wall is– 3 feet?– 5 feet?



IIT BOMBAY

Hemachandra’s Problem …

• Possibilities, if length of wall is 3, 4, or 5 feet:

3 feet wall 4 feet wall 5 Feet wall1, 1, 1 1, 1, 1, 1 1, 1, 1, 1, 12, 1 2, 1, 1 2, 1, 1, 11, 2 1, 2, 1 1, 2, 1, 1

1, 1, 2 1, 1, 2, 12, 2 2, 2, 1

1, 1, 1, 22, 1, 21, 2, 2



IIT BOMBAY


• Possibilities, if length of wall is 3, 4, or 5 feet: 3 feet wall 4 feet wall 5 Feet wall1, 1, 1 1, 1, 1, 1 1, 1, 1, 1, 12, 1 2, 1, 1 2, 1, 1, 11, 2 1, 2, 1 1, 2, 1, 1

1, 1, 2 1, 1, 2, 12, 2 2, 2, 1

1, 1, 1, 22, 1, 21, 2, 2

• Possibilities for 5 feet wall =– Possibilities for 4 feet wall + possibilities for 3 feet wall



IIT BOMBAY

Hemachandra’s Actual Problem

• Poetic meter is made of short syllables and long syllables. : Long syllable (l) = 2 beats. Short syllable (s) = 1 beat,

• Example of a poetic meter

ya kun den du tu sha r ha r dha va laL L L S S L S L S S S L

ya shubh r vas tra vru taL L S L L S L

• Suppose I am designing a poetic meter with 8 beats. – How many ways are there of filling 8 beats?



IIT BOMBAY

Hemachandra’s Solution

• “By the method of Pingala, it is enough to observe that the last beat is long or short”

• Pingala: mathematician/poet from 200 B.C.

• Hemachandra is giving credit to someone who lived more than 1000 years before him!!

• Copy if necessary, and if permitted,

• But always give credit



IIT BOMBAY

Hemachandra’s solution contd.

• An 8 beat pattern can be made by– Take any 7 beat pattern and add a Short beatOr – Take any 6 Beat pattern and add a Long beat

• Thus the total number of possible 8 beat patterns is simply= Number of 7 bit patterns + number of 6 beat patterns

|8 beat patterns| = |7 beat patterns| + |6 beat patterns|



IIT BOMBAY

Algebraically..

• Hn = number of patterns with n beats

H8 = H7 + H6

In general Hn = Hn-1 + Hn-2

• Does this help us to compute H8?

We need to know H7, H6, for which we need H5, ...



IIT BOMBAY

Algorithm Idea

• H1 = number of patterns with 1 beat = 1 {S}• H2 = Number with 2 beats = 2 {SS, L}• H3 = H2 + H1 = 2 + 1 = 3 {SSS, SL, LS}• H4 = H3 + H2 = 3 + 2 = 5 • H5 = H4 + H3 = 5 + 3 = 8• H6 = H5 + H4 = 8 + 5 = 13• H7 = H6 + H5 = 13 + 8 = 21 • H8 = H7 + H6 = 21 + 13 = 34 ...



IIT BOMBAY

Program to compute Hn

int n; cin >> n; // which Hemachandra number to computeint hprev = 1, hcurrent = 2, hnext;for (int i=3; i <= n; i++){

hnext = hprev + hcurrent;// prepare for next iteration, by shifting values of objectshprev = hcurrent; hcurrent = hnext;

}cout << hnext;



IIT BOMBAY

On Hemachandra Numbers

• Mathematics from poetry!

• This interesting series models many real-life things– Number of petals in many flowers.

• Ratio of consecutive terms tends to a limit.– Golden Ratio

• These are more commonly known as Fibonacci numbers!!

• Hemachandra lived before Fibonacci.– Pingal lived several centuries earlier



IIT BOMBAY

Lab Announcements

• From next week, students attending a lab will be organized into groups of 12 to 16 students. Each group will have two teams of 6 to 8 students each, and will be mentored by 2 Teaching Assistants. (One TA per team)

• Students are required to seat as per the earmarked locations to facilitate this arrangement

• Make up Lab will be held on Saturday in OSL– From 14:00 to 16:00 Hrs

• If there are more students, they can come to another make-up lab to be held on Tuesday from 18:00 to 20:00– Only about 40 students can be accommodated, since the lab

is primarily meant for a special group of students



IIT BOMBAY

Announcement for special Lectures

• Special Lectures will be held on Saturday, 13 August 2011 in Room o SIC 301, Kanwal Rekhi Building, 3rd floor

• 14:00 to 16:00. Hindi lecture – by Deepak B Phatak

• 16:00 to 18:00 Discussion of programming problems– By Nagesh Rau Karmali



IIT BOMBAY

Lecture schedule, Quizzes and Exams


• Complete course schedule be available on moodle and on the course home page on 15th

• The first quiz will not be held the scheduled date, 2nd Sept 11– It will be held on Friday, 26 August 11, during regular class

hours in this lecture hall itself

dr deepak b phatakcs101/2011.2/lectures/lecture_slides/201… · dr. deepak b phatak 3. iit bombay....

Documents