1. CS212: Data Structures and Algorithms
Lecture # 7
Recursion

2. Outline Towards Recursion Recursion Recursive Programming
Rules of Recursion
Exercises

3. Towards Recursion Factorial Function n != 1 if n == 0
Given a +ive integer n, n factorial is defined as the product of all integers between 1 and n, including n.
n != if n == 0
n != n*(n-1) * (n-2) * * 1 if n > 0
Algorithm that accepts an integer n and returns the value of n!
prod = 1;
for(x = n; x > 0; x--)
prod*= x;
return(prod);
This algorithm is an iterative algorithm

4. Towards Recursion
Multiplying n by the product of all integers from n-1 to 1 yields the product of all integers from n to 1.
For any n > 0, we see that n! equals n * (n-1)!
n! = 1 if n == 0
n! = n(n-1)! if n > 0
Is this a recursive definition?
A definition which defines an object in terms of a simpler case of itself, is called a recursive definition

5. Towards Recursion 1. 5! = 5 * 4! 2. - - - - - - - 4! = 4 * 3!
1. 5! = 5 * 4!
! = 4 * 3!
! = 3 * 2!
! = 2 * 1!
! = 1 * 0!
! = 1
Solve from line 6 to 1.
6´ ! = 1
5´ ! = 1*0! = 1 * 1 = 1
4´ ! = 2*1! = 2 * 1 = 2
3´ ! = 3*2! = 3 * 2 = 6
2´ ! = 4*3! = 4 * 6 = 24
1´ ! = 5*4! = 5 *24 = 120

6. Towards Recursion Multiplication of Natural Number
The product a * b, where a and b are positive integers, may be defined as
a * b = a if b ==1
a * b = a * (b-1) + a if b > 1
e.g.
6 * 3 = 6 * 2 + 6
= 6 * =
= 18

7. Towards Recursion Fibonnacci Sequence fib(0) = 0 fib(1) = 1
0, 1, 1, 2, 3, 5, 8, 13, 21, 34,
Each element is the sum of the two preceding elements with
fib(0) = 0
fib(1) = 1
fib(n) = n if n ==0 or n ==1
fib(n) = fib(n-2) + fib(n-1) if n >= 2
e.g.
Compute fib(5)

8. Towards Recursion
Fib(5)
Fib(3)
Fib(1)
Fib(2)
Fib(0)
Fib(4)

9. Towards Recursion All these problems have a recursive definition
A definition which defines an object in terms of a simpler case of itself, is called a recursive definition

10. Recursion Every recursive definition has two parts,
A Recursive part &
A Non-Recursive part, also called as Base Case
e.g. in the recursive definition of n!
n! = if n == 0 Base Case
n! = n*(n-1) * (n-2) *…* 1 if n > 0 Recursive Part

11. Recursion
The recursive part of the n! definition is used several times, terminating with the non-recursive part
1. 5! = 5 * 4!
! = 4 * 3!
! = 3 * 2!
! = 2 * 1!
! = 1 * 0!
! = 1
Solve from line 6 to 1.
6´ ! = 1
5´ ! = 1*0! = 1 * 1 = 1
4´ ! = 2*1! = 2 * 1 = 2
3´ ! = 3*2! = 3 * 2 = 6
2´ ! = 4*3! = 4 * 6 = 24
1´ ! = 5*4! = 5 *24 = 120

12. Recursion All recursive definitions have to have a base case
If they didn't, there would be no way to terminate the recursive path
Such a definition would cause infinite recursion
This problem is similar to an infinite loop

13. Recursion
How can we translate recursive definition into a computer program?
Recursion is a fundamental programming technique that can provide a solution to the problems that have a recursive definition

14. Recursive Programming
A method/function which can invoke itself, if set up that way, is called a recursive method/function
The code of a recursive method/function must be structured to handle both the base case and the recursive case

15. Recursive Programming
Each call to the method sets up a new execution environment, with new parameters and local variables
As always, when the method completes, control returns to the method that invoked it (which may be an earlier invocation of itself)

16. Recursive Programming
Consider again the recursive definition of the factorial of a positive integer.
n! = 1 if n == 0 Base Case
n! = n*(n-1) * (n-2) * …* 1 if n > 0 Recursive Part
We can write the recursive function using this definition as
int factorial(int n) {
if(n == 0) //Base Case
return 1;
else
return n * factorial(n-1); //Recursion }

17. Recursive Programming
main
factorial(3)
factorial(2)
factorial(1)
2*factorial(1)
3 * factorial(2)
return 1
return 6
1*factorial(0)
factorial(0)
return 2

18. Recursive Programming
Multiplication of natural numbers:
The product a * b, where a and b are positive integers, may be defined as
a * b = a if b ==1
a * b = a * (b-1) + a if b > 1
int multiplication(int a, int b) {
if(b == 1)
return a;
else if(b > 1)
return multiplication(a, b-1) + a; }

19. Recursive Programming
Fibonnacci Sequence
fib(n) = n if n ==0 or n ==1
fib(n) = fib(n-2) + fib(n-1) if n >= 2
int fib(int n) {
if(n == 0 || n == 1)
return n;
else
return fib(n-2) + fib(n-1); }

20. Hanoi Tower Problem Statement
Given N discs (of strictly decreasing size) stacked on one needle and two empty needles, it is required to stack all the discs onto a second needle in decreasing order of size
The third needle may be used as temporary storage
The movement of the discs is restricted by the following rules:
Only one disc may be moved at a time
A disc may be moved from any needle to any other
At no time may a larger disc rest upon a smaller disc

21. Hanoi Tower Needle A Needle B Needle C (start of (interme- (completion
problem) diate) of problem)
Solution Steps:
1. Move N –1 discs from A to B
2. Move disc N from A to C
3. Move N –1 discs from B to C

22. Hanoi Tower
void tower(char source, char destination, char intermediate, int n) {
if(n==1) {
cout<<"Move disk 1 from needle "<<source<<" to needle "<<destination<<endl; }
else {
tower(source, intermediate, destination, n-1);
cout<<"Move disk "<<n<<" from needle "<<source<<" to needle "<<destination<<endl;
tower(intermediate, destination, source, n-1);

23. Hanoi Tower Move disk 3 from A to C
tower(‘A’, ‘C’, ‘B’, 3); //A is source, C is destination
//B is intermediate
Move disk 1 from A to C
Move disk 2 from A to B
Move disk 1 from C to B
Move disk 3 from A to C
Move disk 1 from B to A
Move disk 2 from B to C

24. Rules of Recursion Base Case Making Progress
There must be a way out for a recursive algorithm
It must not generate an infinite sequence of calls on itself
Without a non-recursive exit, no recursive function can ever be computed
Making Progress
For the cases that are to be solved recursively, the recursive call must always be to a case that makes progress toward a base case

25. Rules of Recursion Example 1- - - - - - int Bad(int N) 2- - - - - - {
if(N == 0)
return 0;
else
return Bad (N / 3+1) + N - 1; }

26. Rules of Recursion Design Rule
Assume that all the recursive calls work
Never duplicate work by solving the same instance of a problem in separate recursive calls
Example: Fibonacci Numbers

27. Exercises Example void print_out(int N) { if(N < 10) cout<<N;
A +ive integer to be printed
Assume that only I/O routines available will take a single digit number and output it to the terminal
Write a recursive algorithm to print the number digit by digit
void print_out(int N) {
if(N < 10)
cout<<N;
else
print_out(N/10);
cout<<(N%10); }

28. Exercises
The problem of computing the sum of all the numbers between 1 and any positive integer N can be recursively defined as:
i = 1 N
N-1
N-2
= N +
= N + (N-1) +
= etc.

29. Exercises int sum(int n) { if(n==1) return n; else
return n + sum(n-1); }

30. Exercises
Ackerman's function is defined recursively on a non negative integers as follows
A(m, n) = n if m = = 0
A(m, n) = A(m-1, 1) if m != 0, n = = 0
A(m, n) = A(m-1, A(m, n-1)) if m != 0, n != 0

31. Exercises int ackerman(int m, int n) { if(m==0) return n+1;
if(m!=0&&n==0)
return ackerman(m-1, 1);
if(m!=0&&n!=0)
return ackerman(m-1, ackerman(m,n-1)); }

32. Exercises
The Greatest Common Divisor (GCD) of two positive integers x and y is defined as
GCD(x, y) = y if (y <= x && x % y ==0)
GCD(x, y) = GCD(y, x) if (x < y)
GCD(x, y) = GCD(y, x % y) otherwise

33. Exercises int GCD(int x,int y) { if (y <= x && x % y ==0) return y;
else if (x < y)
return GCD(y,x);
else
return GCD(y, x % y); }