CS 202 Computer Science II Lab Fall 2009 October 29.

CS 202Computer Science II Lab

Fall 2009 October 29

Today Topics

• Makefile• Recursion

Makefile

main.out: main.o stack$(VAL).o g++ -o main.out main.o stack$(VAL).o

main.o: main.cpp stack$(VAL).hg++ -c main.cpp

Stack$(VAL).o: stack$(VAL).cpp stack$(VAL).hg++ -c stack$(VAL).cpp

$ make VAL=1

$ make VAL=2

Recursion

• Recursion– A method of specifying a process by means of itself.

• Recursive Function– A way of defining a function using recursion.

• ExampleF(x) = 30x + 7 Direct Implementation

F(x) = F(x-1) - 30 Recursion Implementation

Recursion

• Base Case of a Recursively Defined Function– A value or values for which the recursive function is

concretely defined.

x F(x)

0 7

1 37

2 67

3 97

F(x) = 30x + 7

Base case: F(0) = 7

Recursion

• Recursive Algorithm: – An algorithm that calls itself with smaller and

smaller subsets of the original input, doing a little bit or work each time, until it reaches the base case.

int F(int x){     if (x=0)              return 7;     else              return F(x-1) + 30;}

int F(int x){     return (x= =0)  ?  7 : F(x-1)+30;}

Recursion Example

• factorial(n) n! = nx(n-1)x(n-2)x…x2x1 Direct Imp.

n! = nx(n-1)! Recursive Imp.

Base case: If n = 0 then factorial(n) = 1

int factorial(int n){     if (n>0)              return n*factorial(n-1);     else              return 1;}

int factorial(int n){     return (n) ? n*factorial(n-1) : 1; }

Recursion Example• Fibonacci numbers

• A sequence of numbers named after Leonardo of Pisa, known as Fibonacci.

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

If n = 0 then fib(n) = 1

If n = 1 then fib(n) = 1

If n > 1 then fib(n) = fib(n - 1) + fib(n - 2)

Recursion Example• Fibonacci numbers

int fib(int n){        if (n==0 or n==1)        {                return 1;        }        else        {                return fib(n-1)+fib(n-2);        }}

Recursion Example

• The greatest common divisor (GCD) of two integers m and n

int tryDivisor(int m, int n, int g){

if ( ( (m % g) == 0 ) && ( (n % g) == 0 ) )

return g;

else

return tryDivisor(m, n, g - 1);

}

int gcd(int m, int n) {

return tryDivisor(m, n, n); //use n as our first guess

}

gcd(6, 4)

tryDivisor(6, 4, 4)

tryDivisor(6, 4, 3)

tryDivisor(6, 4, 2)

=> 2

GCD

int gcd(int m, int n) {

if(m == n)

return m;

elseif (m > n)

return gcd(m-n, n);

else

return gcd(m, n-m);

}

gcd(468, 24)

gcd(444, 24)

gcd(420, 24)

...

gcd(36, 24)

gcd(12, 24) (Now n is bigger)

gcd(12, 12) (Same)

=> 12

Extremely Smart way to compute GCD

int gcd(int m, int n)

{

if ( (m % n) == 0 )

return n;

else

return gcd(n, m % n);

}

gcd(468, 24)

gcd(24, 12)

=> 12

gcd(135, 19)

gcd(19, 2)

gcd(2, 1)

=> 1

Questions?