cs 101 iit bombay computer programming and utilizationcs101/2009.2/lectures/...cs 101 - lecture 5...

CS 101

Computer Programming and Utilization

Dr Deepak B Phatak

Subrao Nilekani Chair Professor

Department of CSE, Kanwal Rekhi Building

IIT Bombay

Lecture 5, More Numerical computing

(Slides courtesy Prof Ranade)

IIT BOMBAY

Dr. Deepak B Phatak 2

IIT BOMBAY

CS 101 - Lecture 5 More numerical Computing

Overview

• Review of CPP Dumbo

• Iterative Numerical computations

•Factorial of a given integer

•Hemachandra Numbers

•Finding roots of an equation

• Course and Lab organization


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C++ program structure

#include <iostream>

using namespace std;

int main(){

---

--- statements/instructions;

---

return(0);

}

• We will learn later what each of these lines mean. Right now, this is a mandatory structure


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C++ Data types

• We have seen numerical and string data

7.45, 12.3E-12, “Ranade”, “\n”,

• Constant values are written like this by us

C++ stores these in an internal format

• Numerical values have an associated type

int, long, float, double

• Memory location names, called variables,

must be predefined with associated data type

int count; float a, b, sum; double largevalue;


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C++ statements

• Each instruction of our program can ordinarily convey any one of the 3 actions

•Assignment, input, output

• Assignement

sum = a + b;

area = (x-1.0)*(1/(x-1.0);

m = m + x/2.5

•The last one is to be seen as “reassignment”


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C++ statements ...

• “increment” is special form of reassignment

count = count + 1;

or count += 1;

or, more simply, ++count or count++

• Input

cin >> x >> maxvalue;

• Output

cout << “Value of y is “ << y << “\n”;


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Conditional execution

• Program instructions are normally executed in

the given sequence

• C++ can examine conditions, and based on the

result, can execute conditions out of sequence

if (condition)

{ statement group 1};

else

{ statement group 2};


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Conditional execution ...

• Based on the age of a passenger, the ticket

cost is different. Let us say, it is Rs 25.50 for

an adult, and Rs 12.75 for a child (< 12 Years)

cin >> age;

ticket = 12.75;

ticket = 25.50;

cout << “Rs. “ << ticket << endl;


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Conditional execution ...

• Suppose we alter the sequence of assigning a value to ticket

cin >> age;

ticket = 25.50;

ticket = 12.75;


• The assignment to variable ticket will always be the last assigned value irrespective of age!


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Conditional execution …

• Use of the if – else statements will permit us to correctly evaluate the ticket

cin >> age;

if (age > 12)

{ticket = 25.50};

else

{ticket = 12.75};


• What if elders are charged Rs 20?


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If – else if ladder

cin >> age;

if (age > 60){

ticket = 20.00;}

else if (age > 12){

ticket = 25.50;}

else {

ticket = 12.75;}



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Repetitive actions

int nfactorial, n, i

cin >> n

nafactorial = 1;

for (i =1; i <= n, i++){

nfactorial = nfactorial * i;

};

cout << “factorial ” << n << “ is ” << nfactorial;


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Complete program for n!

#include <iostream>

using namespace std;

int main() {

int nfactorial, n, i;

cout << “ Give value of n” << endl;

cin >> n;


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Complete program for n! …

nafactorial = 1;

for (i =1; i <= n, i++){

nfactorial = nfactorial * i;

};

cout <<“factorial ”<<n<< “is ”<<nfactorial<< endl;

Return(0);

}


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Hemachandra’s Problem (12th century AD)

Suppose I have to build a wall of length 8 feet. I have bricks 2 feet long and also 1 foot long. In how many ways I can lay the bricks so that I fill the 8 feet?

Possibilities:

2,2,2,2;

1,1,1,1,1,1,1,1;

2,2,2,1,1

....


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Hemachandra’s Actual Problem

Suppose I am designing a poetic meter with 8 beats. The meter is made of short syllables and long syllables.

Short syllable (s) = 1 beat,

Long syllable (l) = 2 beats.

How many ways are there of filling 8 beats?

Example of a poetic meterya kun den du tu sha r ha r dha va la ya shubh r vas tra vru ta

l l l s s l s l s s s l l l s l l s s


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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 500 A.D.

Hemachandra is giving credit to someone who

lived hundreds of years before him!!

Copy if necessary and if permitted,

but always give credit


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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, ...


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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 ...


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Program to compute Hn

int n; cin >> n; // which number to compute

int hprev = 1, hcurrent = 2;

for(int i=3; i <= n; i++){

hnext = hprev + hcurrent;

// prepare for next iteration

hprev = hcurrent;

hcurrent = hnext; }

cout << hnext;


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Code is tricky!

Need a comment:

/* At the begining of an iteration

hcurrent = Hi-1 write ith term if you like.

hprev = Hi-2

where i is the value of variable i */

Can you prove this?

Will mathematical induction help?

Proving this is enough

-- hnext = hprev + hcurrent –

hence correct answer will be generated.


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Proof by induction

Base case: At the beginning of the first iteration is this true? Yes, i will have value 3, and hprev = 1 = H1, hcurrent = 2 = H2

Suppose it is true at some later iteration, when i has value v >= 3. By induction hypothesis, hprev and hcurrent have values Hv-1, Hv-2 respectively

The first statement hnext = hprev + hcurrent makes hnext = Hv-1 + Hv-2 = Hv. After this the statement hprev = hcurrent makes hprev = Hv-1. The next statement hcurrent = hnext makes hcurrent=Hv.

In the next iteration i will have value v+1. But hprev,hcurrent will have exactly the right values!


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On Hemachandra Numbers

Mathematics from poetry!

Series is very interesting.

- Number of petals in many flowers.

- Ratio of consecutive terms tends to a limit.

What are these numbers more commonly known as?

Fibonacci numbers!!

Hemachandra lived before Fibonacci.


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Newton Raphson method

Method to find the root of f(x), i.e. x s.t. f(x)=0.

Method works if:

f(x) and f '(x) can be easily calculated.

A good initial guess is available.

Example: To find square root of k.

use f(x) = x2 - k. f’ (x) = 2x.

f(x), f’ (x) can be calculated easily.

2,3 arithmetic operations

Initial guess x0 = 1 always works! can be proved.


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How to get better xi+1 given xi

f(x)

xi

xi+1

Point A =(xi,0) known.

A

B

C

f’ (xi) = AB/AC = f(xi)/(xi - xi+1) xi+1

= (xi- f(xi)/f’ (xi))

Calculate f(xi).

Point B=(xi,f(x

i)) known

Approximate f by tangent

C= intercept on x axis

C=(xi+1

,0)


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Square root of k

xi+1

= (xi- f(xi)/f’ (xi))

f(x) = x2 - k, f’ (x) = 2x

xi+1 = xi - (xi2 - k)/(2xi) = (xi + k/xi)/2

Starting with x0=1, we compute x1, then x2, then...

We can get as close to sqrt(k) as required.

Proof not part of the course.


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Code

float k;

cin >> k;

float xi=1; // Initial guess. Known to work.

for(int i=0; i < 10; i++){

// 10 iterations

xi = (xi + k/xi)/2;

}

cout << xi;


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Another way

float xi, k; cin >> k;

for( xi = 1 ;

// Initial guess. Known to work.

xi*xi – k > 0.001 || k - xi*xi > 0.001 ;

// until error in the square is at most 0.001

xi = (xi + k/xi)/2);

cout << xi;


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Yet Another way

float k; cin >> k;

float xi=1;

while(xi*xi – k > 0.001 || k - xi*xi > 0.001){

xi = (xi + k/xi)/2 ;

}

cout << xi;

}


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While statement

while (condition) { loop body}

check condition, then execute loop body if true.

Repeat.

If loop body is a single statement, then need not

use { }. Always putting braces is recommended;

if you insert a statement, you may forget to put

them, so do it at the beginning.

True for other statements also: for/repeat/if.


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For vs. while

If there is a “control” variable with initial value,

update rule, and whose value distinctly defines

each loop iteration, use for.

If loop executes fixed number of times, use for.


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Homework

Write a program to calculate the cube root

using Newton Raphson method.

Check how many iterations are needed to get

good answers. Should be very few.


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evaluation approach for CS101

Quizzes 10%

Assignments 10%

Course Project 30%

Mid-semester exam 20%

End Semester Exam 30%

Minimum passing grade (DD) at >= 40% marks


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Consulting support

Monday 13:30 to 17:30

Tuesday 13:30 to 17:30

TAs will be available to help you with

your queries regarding lectures/labs and

with programming problems

Additional hours for lab access on both

these days

//possibly also Thursday 13:30 to 17:30

// will be announced tomorrow

cs 101 iit bombay computer programming and utilizationcs101/2009.2/lectures/...cs 101 - lecture 5...

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