CS 101 Computer Programming and Utilization. Lecture 5, More Numerical computing (Slides courtesy Prof Ranade)

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1 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)

2 Overview Review of CPP Dumbo Iterative Numerical computations Factorial of a given integer Hemachandra Numbers Finding roots of an equation Course and Lab organization 2

3 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 3

4 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; 4

5 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 5

6 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 ; 6

7 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}; 7

8 Conditional execution... Based on the age of a passenger, the ticket cost is different. Let us say, it is Rs for an adult, and Rs for a child (< 12 Years) cin >> age; ticket = 12.75; ticket = 25.50; cout << Rs. << ticket << endl; 8

9 Conditional execution... Suppose we alter the sequence of assigning a value to ticket cin >> age; ticket = 25.50; ticket = 12.75; cout << Rs. << ticket << endl; The assignment to variable ticket will always be the last assigned value irrespective of age! 9

10 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}; cout << Rs. << ticket << endl; What if elders are charged Rs 20? 10

11 If else if ladder cin >> age; if (age > 60){ ticket = 20.00;} else if (age > 12){ ticket = 25.50;} else { ticket = 12.75;} cout << Rs. << ticket << endl; 11

12 Repetitive actions int nfactorial, n, i cin >> n nafactorial = 1; for (i =1; i <= n, i++){ nfactorial = nfactorial * i; }; cout << factorial << n << is << nfactorial; 12

13 Complete program for n! #include <iostream> using namespace std; int main() { int nfactorial, n, i; cout << Give value of n << endl; cin >> n; 13

14 Complete program for n! nafactorial = 1; for (i =1; i <= n, i++){ nfactorial = nfactorial * i; }; cout << factorial <<n<< is <<nfactorial<< endl; Return(0); } 14

15 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,

16 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 meter ya 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 16

17 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 17

18 Hemachandra s solution contd. S : Class of 8 beat patterns with short last beat. L : Class of 8 beat patterns with long last beat. Each 8 beat pattern is in class L or class S S = all 7 beat patterns + short beat appended. L = all 6 beat patterns + long beat appended class S = Number of patterns with 7 beats class L = Number of patterns with 6 beats 8 beat patterns = class S + class L = 7 beat patterns + 6 beat patterns 18

19 Algebraically.. H n = number of patterns with n beats H 8 = H 7 + H 6 In general H n = H n-1 + H n-2 Does this help us to compute H 8? We need to know H 7, H 6, for which we need H 5,... 19

20 Algorithm Idea H 1 = number of patterns with 1 beat = 1 {S} H 2 = Number with 2 beats = 2 {SS, L} H 3 = H 2 + H 1 = = 3 {SSS, SL, LS} H 4 = H 3 + H 2 = = 5 H 5 = H 4 + H 3 = = 8 H 6 = H 5 + H 4 = = 13 H 7 = H 6 + H 5 = = 21 H 8 = H 7 + H 6 = =

21 Program to compute H n 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; 21

22 Code is tricky! Need a comment: /* At the begining of an iteration hcurrent = H i-1 write ith term if you like. hprev = H i-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. 22

23 Proof by induction Base case: At the beginning of the first iteration is this true? Yes, i will have value 3, and hprev = 1 = H 1, hcurrent = 2 = H 2 Suppose it is true at some later iteration, when i has value v >= 3. By induction hypothesis, hprev and hcurrent have values H v-1, H v-2 respectively The first statement hnext = hprev + hcurrent makes hnext = H v-1 + H v-2 = H v. After this the statement hprev = hcurrent makes hprev = H v-1. The next statement hcurrent = hnext makes hcurrent=h v. In the next iteration i will have value v+1. But hprev,hcurrent will have exactly the right values! 23

24 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. 24

25 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) = x 2 - k. f (x) = 2x. f(x), f (x) can be calculated easily. 2,3 arithmetic operations Initial guess x 0 = 1 always works! can be proved. 25

26 How to get better x i+1 given x i Point A =(x i,0) known. B Calculate f(x i ). Point B=(x i,f(x i )) known f(x) C x i+1 A x i Approximate f by tangent C= intercept on x axis C=(x i+1,0) f (x i ) = AB/AC = f(x i )/(x i - x i+1 ) x i+1 = (x i - f(x i )/f (x i )) 26

27 Square root of k x i+1 = (x i - f(x i )/f (x i )) f(x) = x 2 - k, f (x) = 2x x i+1 = x i - (x i2 - k)/(2x i ) = (x i + k/x i )/2 Starting with x 0 =1, we compute x 1, then x 2, then... We can get as close to sqrt(k) as required. Proof not part of the course. 27

28 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; 28

29 Another way float xi, k; cin >> k; for( xi = 1 ; // Initial guess. Known to work. xi*xi k > k - xi*xi > ; // until error in the square is at most xi = (xi + k/xi)/2); cout << xi; 29

30 Yet Another way float k; cin >> k; float xi=1; while(xi*xi k > k - xi*xi > 0.001){ xi = (xi + k/xi)/2 ; } cout << xi; } 30

31 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. 31

32 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. 32

33 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. 33

34 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 34

35 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 35