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Functions : Recursion

Functions : Recursion. Lecture 12 4.2.2002. Recursion. a + (a+1) + (a+2) + ... + b = a + ((a+1) + (a+2) + ... + b) sum(a,b) = a + sum(a+1,b) Base case ? factorial (n) = n*factorial(n-1) Base case ?. Recursion. int factorial (int n) { int temp; if (n<=1) return 1;

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Functions : Recursion

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  1. Functions : Recursion Lecture 12 4.2.2002 Sudeshna Sarkar, IIT Kharagpur

  2. Recursion a + (a+1) + (a+2) + ... + b = a + ((a+1) + (a+2) + ... + b) sum(a,b) = a + sum(a+1,b) Base case ? factorial (n) = n*factorial(n-1) Base case ? Sudeshna Sarkar, IIT Kharagpur

  3. Recursion int factorial (int n) { int temp; if (n<=1) return 1; temp=factorial(n-1); return n*temp; } int sum (int a, int b) { int temp; if (a>b) return 0; temp=sum(a+1,b); return a+temp; } Sudeshna Sarkar, IIT Kharagpur

  4. Recursion 1 4 sum(a,b) { temp=sum(2,4); return 1+temp; } main () { . . . x = sum(1,4); . . . } int sum (int a, int b) { int temp; if (a>b) return 0; temp=sum(a+1,b); return a+temp; } sum(2,4) { temp=sum(3,4); return 2+temp; } sum(3,4) { temp=sum(4,4); return 3+temp; } sum(4,4) { temp=sum(5,4); return 4+temp; } sum(5,4) { return 0; } Sudeshna Sarkar, IIT Kharagpur

  5. Recursion sum(1,4) { temp=sum(2,4); return 1+temp; } main () { . . . x = sum(1,5); . . . } int sum (int a, int b) { int temp; if (a>b) return 0; temp=sum(a+1,b); return a+temp; } sum(2,4) { temp=sum(3,4); return 2+temp; } sum(3,4) { temp=sum(4,4); return 3+temp; } sum(4,4) { temp=0; return 4; } Sudeshna Sarkar, IIT Kharagpur

  6. Recursion sum(1,4) { temp=sum(2,4); return 1+temp; } main () { . . . x = sum(1,5); . . . } int sum (int a, int b) { int temp; if (a>b) return 0; temp=sum(a+1,b); return a+temp; } sum(2,4) { temp=sum(3,4); return 2+temp; } sum(3,4) { temp=4; return 3+4; } Sudeshna Sarkar, IIT Kharagpur

  7. Recursion sum(1,4) { temp=sum(2,4); return 1+temp; } main () { . . . x = sum(1,5); . . . } int sum (int a, int b) { int temp; if (a<=b) return 0; temp=sum(a+1,b); return a+temp; } sum(2,4) { temp=7; return 2+7; } Sudeshna Sarkar, IIT Kharagpur

  8. Recursion sum(1,4) { temp=9; return 1+9; } main () { . . . x = sum(1,5); . . . } int sum (int a, int b) { int temp; if (a<=b) return 0; temp=sum(a+1,b); return a+temp; } Sudeshna Sarkar, IIT Kharagpur

  9. void foo () { char ch; scanf (“%c”, &ch); if (ch==‘\n’) return; foo (); printf (“%c”, &ch); } What will the program output on the following input ? PDS \n Sudeshna Sarkar, IIT Kharagpur

  10. int sumSquares (int m, int n) { if (m<n) /* Recursion */ return sumSquares(m, n-1) + n*n; else /* Base Case */ return n*n ; } Example int sumSquares (int m, int n) { if (m<n) /* Recursion */ return m*m + sumSquares(m+1, n); else /* Base Case */ return m*m ; } Sudeshna Sarkar, IIT Kharagpur

  11. Example int sumSquares (int m, int n) { int middle ; if (m==n) return m*m; else { middle = (m+n)/2; return sumSquares(m,middle) + sumSquares(middle+1,n) ; } 5 . . . 7 8 . . . 10 n m mid mid+1 Sudeshna Sarkar, IIT Kharagpur

  12. Call Tree sumSquares(5,10) sumSquares(5,10) sumSquares(5,10) sumSquares(8,10) sumSquares(5,7) sumSquares(10,10) sumSquares(8,9) sumSquares(7,7) sumSquares(5,6) sumSquares(9,9) sumSquares(6,6) sumSquares(8,8) sumSquares(5,5) Sudeshna Sarkar, IIT Kharagpur

  13. Annotated Call Tree 355 sumSquares(5,10) sumSquares(5,10) 245 110 sumSquares(5,10) sumSquares(8,10) sumSquares(5,7) 100 49 145 61 sumSquares(10,10) sumSquares(8,9) sumSquares(7,7) sumSquares(5,6) 36 81 25 64 sumSquares(9,9) sumSquares(6,6) sumSquares(8,8) sumSquares(5,5) 49 100 36 81 25 64 Sudeshna Sarkar, IIT Kharagpur

  14. Trace sumSq(5,10) = (sumSq(5,7) + sumSq(8,10)) = (sumSq(5,6) + (sumSq(7,7)) + (sumSq(8,9) + sumSq(10,10)) = ((sumSq(5,5) + sumSq(6,6)) + sumSq(7,7)) + ((sumSq(8,8) + sumSq(9,9)) + sumSq(10,10)) = ((25 + 36) + 49) + ((64 + 81) + 100) = (61 + 49) + (145 + 100) = (110 + 245) = 355 Sudeshna Sarkar, IIT Kharagpur

  15. Recursion : The general idea • Recursive programs are programs that call themselves • to compute the solution to a subproblem having these properties : 1. the subproblem is smaller than the overall problem or, simpler in the sense that it is closer to the final solution 2. the subproblem can be solved directly (as a base case) or recursively by making a recursive call. 3. the subproblem’s solution can be combined with solutions to other subproblems to obtain the solution to the overall problem. Sudeshna Sarkar, IIT Kharagpur

  16. Think recursively • Break a big problem into smaller subproblems of the same kind, that can be combined back into the overall solution : Divide and Conquer Sudeshna Sarkar, IIT Kharagpur

  17. Exercise 1. Write a recursive function that computes xn, called power (x, n), where x is a floating point number and n is a non-negative integer. 2. Write an improved recursive version of power(x,n) that works by breaking n down into halves, squaring power(x, n/2), and multiplying by x again if n is odd. 3. To calculate the square root of x (a +ve real) by Newton’s method, we start with an initial approximation a=x/2. If |x-aa| epsilon, we stop with the result a.Otherwise a is replaced with the next approximation, (a+x/a)/2. Write a recursive method, sqrt(x) to compute square root of x by Newton’s method. Sudeshna Sarkar, IIT Kharagpur

  18. Common Pitfalls • Infinite Regress : a base case is never encountered • a base case that never gets called : fact (0) • running out of resources :each time a function is called, some space is allocated to store the activation record. With too many nested calls, there may be a problem of space. int fact (int n) { if (n==1) return 1; return n*fact(n-1); } Sudeshna Sarkar, IIT Kharagpur

  19. Tower of Hanoi A B C Sudeshna Sarkar, IIT Kharagpur

  20. Tower of Hanoi A B C Sudeshna Sarkar, IIT Kharagpur

  21. Tower of Hanoi A B C Sudeshna Sarkar, IIT Kharagpur

  22. Tower of Hanoi A B C Sudeshna Sarkar, IIT Kharagpur

  23. void towers (int n, char from, char to, char aux) { if (n==1) { printf (“Disk 1 : %c -> &c \n”, from, to) ; return ; } towers (n-1, from, aux, to) ; printf (“Disk %d : %c  %c\n”, n, from, to) ; towers (n-1, aux, to, from) ; } Sudeshna Sarkar, IIT Kharagpur

  24. Disk 1 : A -> C Disk 2 : A -> B Disk 1 : C -> B Disk 3 : A -> C Disk 1 : B -> A Disk 2 : B -> C Disk 1 : A -> C Disk 4 : A -> B Disk 1 : C -> B Disk 2 : C -> A Disk 1 : B -> A Disk 3 : C -> B Disk 1 : A -> C Disk 2 : A -> B Disk 1 : C -> B towers (4, ‘A’, ‘B’, ‘C’) ; Sudeshna Sarkar, IIT Kharagpur

  25. Recursion may be expensive ! • Fibonacci Sequence: • fib (n) = n if n is 0 or 1 • fib (n) = fib (n-2) + fib(n-1) if n>= 2. int fib (int n) { if (n==0 or n==1) return 1; return fib(n-2) + fib(n-1) ; } Sudeshna Sarkar, IIT Kharagpur

  26. Call Tree 5 fib (5) 2 3 fib (3) fib (4) 1 2 1 1 fib (1) fib (2) fib (2) fib (3) 1 1 1 0 1 1 0 fib (0) fib (1) fib (1) fib (2) fib (0) fib (1) 1 1 1 1 0 0 0 fib (1) fib (0) 1 0 Sudeshna Sarkar, IIT Kharagpur

  27. Iterative fibonacci computation int fib (int n) { if (n <= 1) return n; lofib = 0 ; hifib = 1 ; for (i=2; i<=n; i++) { x = lofib ; lofib = hifib ; hifib = x + lofib; } return hifib ; } i = 2 3 4 5 6 7 x = 0 1 1 2 3 5 lofib= 0 1 1 2 3 5 8 hifib= 1 1 2 3 5 8 13 Sudeshna Sarkar, IIT Kharagpur

  28. Merge Sort • To sort an array of N elements, • 1. Divide the array into two halves. Sort each half. • 2. Combine the two sorted subarrays into a single sorted array • Base Case ? Sudeshna Sarkar, IIT Kharagpur

  29. Binary Search revisited • To search if key occurs in an array A (size N) sorted in ascending order: • mid = N/2; • if (key == N/2) item has been found • if (key < N/2) search for the key in A[0..mid-1] • if (key > N/2) search for key in A[mid+1..N] • Base Case ? Sudeshna Sarkar, IIT Kharagpur

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