1. Two-week ISTE workshop on Effective teaching/learning of computer programming Dr Deepak B Phatak Subrao Nilekani Chair Professor Department of CSE, Kanwal Rekhi Building IIT Bombay Lecture 3, Fundamental programming concepts (Iterative algorithms) Tuesday 29 June 2010

2. OVERVIEW  Need for repetitive execution of instructions Problem of finding maximum of given numbers  Instructions for specifying iteration with ‘for’ With ‘While’  Problem of estimating value of log x  General problem of finding roots of a polynomial

3. Revisiting problem to find Maximum  We saw that if we have to find maximum of 5 numbers we could write: int a, b, c, d, e, max; cin >>a >>b >>c >>d >e ; max = a; if (b > max) {max = b}; if (c > max) {max = c}; if (d > max) {max = d}; if (e > max) {max = e}; cout<<“Max is”<< max << "\n";  What do we do if the number of numbers is very large

4. A possible strategy  We notice a repetitive pattern in our program, which is if (number > max) { max = number}  If we could execute this instruction repeatedly, every time with a different value of number, we will get the desired result. To ensure that this comparison is made with a new value every time, we consider repeating a modifying pattern cin >> number; if (number > max) { max = number}

5. Further analysis of the repetitive strategy  If we repeat this block five times, we will find the maximum of five numbers. But we need an initial value for max.  If we assume all the given numbers are positive, then we could assign an arbitrary negative initial value to max  Expanding our strategy in words, we wish to say something like: max = -999; // repeat the following block 5 times {cin >> number; if (number > max) { max = number} }

6. Repetitive execution  We need some kind of a ‘counting mechanism’ which will add 1 to the count every time we execute the block, and will stop repeated execution of the block once the count reaches 5  The ‘ for’ instruction provides such mechanism max = -999; for (count =1, count <=5; count ++) {cin >> number; if (number > max) { max = number} }

7. Repetitive execution  The artificial initial value for max in not a good idea (What if we are given negative numbers also).  A better thing would be to read the first given number separately, assign it to max, and then repeat the block only 4 times cin >> number; max = number; for (count =1, count <=4; count ++) {cin >> number; if (number > max) { max = number} }

8. Finding maximum of N given numbers // Program 4 (prog4.cpp) #include using namespace std; // To find maximum of N given numbers int main() { int count, number, N, max; cin >> N; cin >> number; max = number; for (count =1, count <=N-1; count ++){ cin >> number; if (number > max) { max = number} }

9. General format and semantics of the ‘for’ statement for( xxx; yyy; zzz ) { www; } 0. Execute “xxx;”. Must be an assignment statement. “loop initialization” 1. Evaluate condition yyy. “loop test”. If true, continue with step 2; if false, execution of ‘for’ statement ends. 2. Execute www. “loop body” 3. Execute zzz. “loop increment” 4. Go back and repeat from 1.

10. Sum of N natural numbers ---- int main(){ int i, N, Sum = 0; cin >> N for(i=1; i < N; i=i+1){ sum = sum + i; } cout << “Sum of” << N < “numbers is”; cout << sum << “\n”; return 0; }

11. Computing Natural Logarithm  Natural Logarithm of a number a is defined as  A computer cannot integrate Must use arithmetic operations.  Estimate area under the curve f(x) = 1/x from 1 to a.  Area approximated by small rectangles. 1/ x dx 1 a

12. Riemann Integral 1 a calculate Sum of areas of all rectangles between 1 and a.

13. Riemann Integral using n rectangles Width w of i th Rectangle (same for all rectangles)‏ 1 a w = (a-1)/n)‏

14. Riemann Integral using n rectangles 1 a x coordinate of i th Rectangle x = 1 + (i-1)w

15. Riemann Integral using n rectangles 1 a Height h of i th Rectangle h = 1 / x = 1/(1 + (i-1)w)‏

16. How many rectangles?  More the merrier! Say 1000.  Total width of rectangles = a - 1.  Width w of each rectangle= (a - 1)/1000  x coordinate of left side of i th rectangle x = 1 + (i-1)w.  Height of i th rectangle 1/x = 1/(1+(i-1)w)‏

17. Program to compute logarithm of a given value int main(){ int i; float x, area=0, w; cin >> x; w = (x-1)/1000.0; for(i=1; i <= 1000; i=i+1){ area = area + w*(1/(1+(i-1)*w)); } cout << “Log of ” << x << “is ” << area; return 0; }

18. Factorial of a number int nfactorial, n, i; cin >> n; nfactorial = 1; for (i =1; i <= n, i++){ nfactorial = nfactorial * i; }; cout << nfactorial;

19. Hemachandra’s Problem (12 th 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....

20. 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 l l l s s l s l s s s l ya shubh r vas tra vru ta l l s l l s s

21. 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

22. 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|

23. 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,...

24. 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 = 2 + 1 = 3 {SSS, SL, LS}  H 4 = H 3 + H 2 = 3 + 2 = 5  H 5 = H 4 + H 3 = 5 + 3 = 8  H 6 = H 5 + H 4 = 8 + 5 = 13  H 7 = H 6 + H 5 = 13 + 8 = 21  H 8 = H 7 + H 6 = 21 + 13 = 34...

25. Program to compute H n int n; cin >> n; // which number to compute int hprev = 1, hcurrent = 2, hnext; For (int i=3; i <= n; i++){ hnext = hprev + hcurrent; // prepare for next iteration hprev = hcurrent; hcurrent = hnext; } cout << hnext;

26. 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.

27. Newton Raphson method  Method to find the root of f(x), i.e. x such that f(x)=0.  Method works if: f(x) and f '(x) can be easily calculated. and 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. only 2 or 3 arithmetic operations needed Initial guess x 0 = 1 always works! can be proved.

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

29. 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 i 2 - 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.

30. Program segment 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;

31. 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;

32. 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;

33. While statement while (condition) { loop body};  check condition, if true then execute loop body. 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.

34. for and 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.

35. THANK YOU