Recursive

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Recursive Definitions

•In a recursive definition, an object is defined in terms of itself.

•We can recursively define sequences, functions and sets.

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Recursively Defined Sequences•Example:

The sequence {an} of powers of 2 is given byan = 2n for n = 0, 1, 2, … .

•The same sequence can also be defined recursively:

a0 = 1

an+1 = 2an for n = 0, 1, 2, …

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Recursively Defined Functions

We can use the following method to define a function with the natural numbers as its domain:

1. Specify the value of the function at zero.

2. Give a rule for finding its value at any integer from its values at smaller integers.

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Constructing Recursion

o To construct a recursive algorithm you have to

find out:

1.Recursive step

2.Base step

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Recursively Defined Functions•Example:

f(0) = 3

f(n + 1) = 2f(n) + 3

•f(0) = 3•f(1) = 2f(0) + 3 = 23 + 3 = 9•f(2) = 2f(1) + 3 = 29 + 3 = 21•f(3) = 2f(2) + 3 = 221 + 3 = 45•f(4) = 2f(3) + 3 = 245 + 3 = 93

Example : Give an inductive definition of the factorial

function F(n) = n!, Then compute F(5).

Solution:

(1) Specify the initial value of factorial function: F(0)=1.

(2) Rule for finding F(n + 1) from F(n):

F(n+1) = (n + 1) F(n)

F(5) = 5 • F(4)

=5 • 4 • F(3)

=5 • 4 • 3 • F(2)

=5 • 4 • 3 • 2 • F(1)

=5 • 4 • 3 • 2 • 1 • F(0)

= 5 • 4 • 3 • 2 • 1 • 1=120

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Recursively Defined Functions•A famous example: The Fibonacci numbers

f(0) = 0, f(1) = 1f(n) = f(n – 1) + f(n - 2)

•f(0) = 0•f(1) = 1•f(2) = f(1) + f(0) = 1 + 0 = 1•f(3) = f(2) + f(1) = 1 + 1 = 2•f(4) = f(3) + f(2) = 2 + 1 = 3•f(5) = f(4) + f(3) = 3 + 2 = 5•f(6) = f(5) + f(4) = 5 + 3 = 8

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Recursive Algorithms

Example II: Recursive Fibonacci Algorithm

procedure fibo(n: nonnegative integer)

if n = 0 then fibo(0) := 0

else if n = 1 then fibo(1) := 1

else fibo(n) := fibo(n – 1) + fibo(n – 2)

n 0 1 2 3 4 5 6 7 8 9 10 11

Fib(n) 0 1 1 2 3 5 8 13 21 34 55 89

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Recursive Algorithms

•Recursive Fibonacci Evaluation:

f(4)

f(3)

f(2)

f(1) f(0)

f(1)

f(2)

f(1) f(0)

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General Algorithm

if (stopping condition) then

solve simple problem (base)

else

use recursion to solve smaller problem

combine solutions from smaller problem

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Convert from decimal to binaryThis method converts an integer number to its binary equivalent.

Base step: dec2bin(n) = n if n is 0 or 1

Recursive step: dec2bin(n) = dec2bin (n/2) , (n mod 2)

Algorithm dec2bin(n):

If n < 2 Print nelse dec2bin(n / 2) Print n mod 2

Example (Convert from decimal to binary)

n=12• dec2bin(12) = dec2bin (6) print (12 mod 2) 0• dec2bin(6) = dec2bin (3) , print (6mod 2) 0• dec2bin(3) = dec2bin (3/2) , print (3 mod 2) 1• dec2bin(3/2) = 1

1 1 0 0

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Example (Convert from decimal to binary)

n=17• dec2bin(17) = dec2bin (17/2) print (17 mod 2) 1• dec2bin(8) = dec2bin (8/2) , print (8 mod 2) 0• dec2bin(4) = dec2bin (4/2) , print (4 mod 2) 0• dec2bin(2) = dec2bin (2/2) , print (2 mod 2) 0• Dec2bin(1) = 1

1 0 0 0 1

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Example (Convert from decimal to binary)

n=90• Dec2bin(90) = dec2bin (90/2) print (90 mod 2) 0• dec2bin(45) = dec2bin (45/2) print (45 mod 2) 1• dec2bin(22) = dec2bin (22/2) print (22 mod 2) 0• dec2bin(11) = dec2bin (11/2) print (11 mod 2) 1• dec2bin(5) = dec2bin (5/2) print (5 mod 2) 1• dec2bin(2) = dec2bin (2/2) , print (2 mod 2) 0• Dec2bin(1) = 1

1 0 1 1 0 115

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class Method { public static void dec2bin( int n){

if ( n < 2 ) System.out.print( n );

else { dec2bin( n / 2 ); System.out.print( n % 2 ); } }}class Dec2Bin{

public static void main(String [] arg){ int i=10;

dec2bin(i);}

}

Output:1010

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Example : The Sum of the First N Positive Integers

• Definition:sum(n) = n + (n-1) + (n-2) + … + 1 for any integer n > 0

• Recursive relation;sum(n) = n + [(n-1) + (n-2) + … + 1]

= n + sum(n-1)Looks so nice, but how about n == 1?sum(1) = 1 + sum(0), but the argument to sum( ) must be positive

• Final recursive definition:sum(n) = 1 if n = 1 (Base case)

= n + sum(n-1) if n > 1 (Recursive call)

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Recursive Definition of sum(n)int sum(int n){ if (n == 1) return 1; else return n + sum(n-1);}

n = 3

sum(n-1)? =

return?

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Box trace of sum(3)

n = 3A: sum(n-1)?=

return?

n = 2A: sum(n-1)?=

return?

n = 1

return 1

n = 3A: sum(n-1)? =

return?

n = 2A: sum(n-1)?=return?

n = 1

return 1

Each box corresponds to a function’s activation record or stack.

cout << sum(3);

1

33

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Basic Recursions

Home Work

Give a recursive algorithm for computing the greatest common divisor of two nonnegative integers a and b with a < b.

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Computing GCD of A and B

• Basis for recursionGCD (a, 0) = a (base case)

GCD (a, b) = GCD (b, a mod b) (recursion)

• Recursive methodGcd(a , b)

{

if (b != 0)

return gcd(b, a % b); // recursion

return a; // base case

}

Computing GCD of 33 and 55

GCD (a, 0) = a (base case)

GCD (a, b) = GCD (b, a mod b) (recursion)

• GCD (33,55) = GCD (55 , 33%55) = GCD (55 , 33)

• GCD (55,33) = GCD (33, 55%33) = GCD (33 , 22)

• GCD (22,33) = GCD (22, 33%22) = GCD (22 , 11)

• GCD (11,22) = GCD (11, 22%11) = GCD (11 , 0)

• GCD (11 , 0) = 11

• GCD(33,55) = 11

Trace of recursive method: gcd

gcd(33,55)

gcd(55,33)

gcd(33,22)

gcd(22,11)

gcd(11,0)

return 11

return 11

return 11

return 11

return 11

Callerint x = gcd(33,55);

call

call

call

call

call