fibonacci series - recursive summation

Question

Ok, I initially wrote a simple code to return the Fibonacci number from the series based on the user input..

n=5 will produce 3..

static int fibonacci(in         


        

Answer 1

Your modification to your fibonacci program does indeed work to calculate sums. However, the way you have used recursion is inefficient. One way to deal with this is with a "dynamic programming" approach, where calculated values are cached so that they can be re-used by the second recursive call. However, the n-th Fibonacci number can be forward calculated from the base. A recursive implementation of this would be:

public static int fib_r (int a, int b, int n) {
    if (n == 1) return a;
    if (n == 2) return b;
    return fib_r(b, a+b, n-1);
}

public static int fib (int n) {
    return fib_r(0, 1, (n > 0) ? n : 1);
}

The corresponding code for the sum would be:

public static int sumfib_r (int a, int b, int n) {
    if (n == 1) return a;
    if (n == 2) return b;
    return sumfib_r(b, a+b+1, n-1);
}

public static int sumfib (int n) {
    return sumfib_r(0, 1, (n > 0) ? n : 1);
}

Tail recursion will often be changed by the compiler/interpreter into a simple loop as part of tail call removal.

You asked:

I still couldn't figure out how the summation of the series works if I add 1. Can someone please explain??

This question is really about understanding the algorithm, which I would suppose is topical on SO. But, math is required to describe why the algorithm works. So, this is really a math question. There is a well known theorem regarding the sum of Fibonacci numbers. If F[i] is the i-th Fibonacci number, and S[n] is the sum of the first n Fibonacci numbers, then the theorem above states:

    S[n] = F[n+2] - 1

So, if we consider that by definition of S[n+2],

S[n+2] = S[n+1] + F[n+2]

Then, substituting S[n] + 1 for F[n+2]:

S[n+2] = S[n+1] + S[n] + 1

Which you should recognize is your "add 1 modified" fibonacci function.

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Below is a proof by induction that your program computes the sum that I provided in my original answer. Let F represent your fibonacci function, and let S represent your "add 1 modified" fibonacci function.

F[1] = 0
F[2] = 1
F[i] = F[i-1] + F[i-2] for i > 1

S[1] = 0
S[2] = 1
S[i] = S[i-1] + S[i-2] + 1 for i > 1

Then, you want a proof that for k > 0:

         k
       .---  
S[k] =  >   F[i]
       `---
       i = 1

Note that the above summation is true if and only if:

S[1] = F[1]
S[k] = F[k] + S[k-1] for k > 1

The proof is fairly straight forward. The base cases are trivially true.

S[1] = F[1] = 0
S[2] = F[2] + F[1] = 1
S[3] = S[2] + S[1] + 1 = F[3] + F[2] + F[1] = 2

The induction step is: Given that for some k > 2, S[j+1] = F[j+1] + S[j] for 0 < j < k+1, prove that the equality holds true if j = k+1, that is: S[k+2] = F[k+2] + S[k+1].

    S[k+2] = S[k+1] + S[k] + 1
=>  S[k+2] = (F[k+1] + S[k]) + (F[k] + S[k-1]) + 1
=>  S[k+2] = (F[k+1] + F[k]) + (S[k] + S[k-1] + 1)
=>  S[k+2] = F[k+2] + S[k+1]

This completes the proof.