Recursion

Kasun Ranga Wijeweera

(Email: krw19870829@gmail.com)

Department of Statistics and Computer Science

Faculty of Science

University of Peradeniya

What is Recursion? • A fundamental concept in computer science and mathematics• A recursive program is one that calls itself• There must be a termination condition

Recurrences • Recursive definitions of functions are quite common in

mathematics• The simplest type, involving integer arguments are called

recurrence relations • Most familiar such a function is the factorial function, defined

by the formula

N ! = N * (N - 1) !, for N >=1 with 0 ! = 1

Recurrences • The corresponding simple recursive program is as follows

int factorial (int N)

{

if (N == 0) return 1;

return N * factorial (N - 1);

}

• Problem:

factorial (-1) leads to an infinite loop

Recurrences • A second well-known recurrence relation is the one that defines the

Fibonacci numbers

F (N) = F (N - 1) + F (N - 2), for N >= 2 with F (0) = F (1) =1

• This defines the sequence

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610,…

Recurrences • The corresponding recursive program is as follows

int fibonacci (int N)

{

if (N <= 1) return 1;

return fibonacci (N - 1) + fibonacci (N - 2);

}

• The running time of this program is exponential

Recurrences • To compute F (N) in linear time

#define max 25

int fibonacci (int N)

{

int i, F[max];

F[0] = 1; F[1] = 1;

for(i = 2; i <= max; i++)

F[i] = F[i - 1] + F[i - 2];

return F[N];

}

Divide and Conquer • Most of the recursive programs use two recursive calls, each

operating on about half the input• This is so called “divide and conquer” paradigm• Used to achieve significant economics• They normally do not reduce to trivial loops like the factorial

program• They normally do not lead to excessive re-computing as Fibonacci

program, because the input is divided without overlap

Divide and Conquer: Example • Let us consider the task of drawing the markings for each inch on a

ruler • There is a mark at the 1/2 point• Slightly shorter marks at 1/4 intervals• Still shorter marks at 1/8 intervals, etc

Divide and Conquer: Example • Suppose the desired resolution is 1/(2^n)• Put a mark at every point between 0 and (2^n), end points not

included• A procedure mark (x, h) is used to mark h units high at x

position• The middle mark should be n units high• The marks in the left and right halves should be n-1 units high

Divide and Conquer: Example • The relevant “divide and conquer” recursive program is as follows

rule (int a, int r, int h)

{

int m = (a + r) / 2;

if (h > 0)

{

mark (m, h);

rule (a, m, h-1);

rule (m, r, h-1);

}

}

Any Questions?

Thank You!