Calculating the Fibonacci function definition of n items listed below:
int fib(int n){
if(n==0)
return 1;
else if(n==1)
return 2;
else
return fib(n-1)+fib(n-2);
}
Will fib (1) -fib (9) the number of times each function call fib
answer:
For small amounts of data for n
f (1) call 1
f (2) call 3 times
f (3) call 5 times
f (4) calls 9 times
Found a pattern: open start from n = 3, the number of calls for the first two plus one
That is: f (3) = f (2) + f (1) + 1 = 5
f(4) =f(3) +f(4) +1=9
f(5) =f(4) +f(3) +1=15
f(6) =f(5) +f(4) +1=25
f(7) =f(6) +f(5) +1=41
f(8) =f(7) +f(6) +1=67
f(9) =f(8) +f(7) +1=109