## Wednesday, June 8, 2016

### Sum of Fibonacci Numbers - GeeksforGeeks

Sum of Fibonacci Numbers - GeeksforGeeks
Given a number positive number n, find value of f0 + f1 + f2 + …. + fn where fi indicates i'th Fibonacci number. Remember that f0 = 0, f1 = 1, f2 = 1, f3 = 2, f4 = 3, f5 = 5, …

The idea is to find relationship between the sum of Fibonacci numbers and n’th Fibonacci number.
F(i) refers to the i’th Fibonacci number.
S(i) refers to sum of Fibonacci numbers till F(i),
```We can rewrite the relation F(n+1) = F(n) + F(n-1) as below
F(n-1)    = F(n+1)  -  F(n)

Similarly,
F(n-2)    = F(n)    -  F(n-1)
.       .           .
.       .          .
.       .          .
F(0)      = F(2)    -  F(1)
------------------------------- ```
Adding all the equations, on left side, we have
F(0) + F(1) + … F(n-1) which is S(n-1).
Therefore,
S(n-1) = F(n+1) – F(1)
S(n-1) = F(n+1) – 1
S(n) = F(n+2) – 1 —-(1)
In order to find S(n), simply calculate the (n+2)’th Fibonacci number and subtract 1 from the result.
F(n) can be evaluated in O(log n) time using either method 5 or method 6
`// Create an array for memoization`
`int` `f[MAX] = {0};`

`// Returns n'th Fibonacci number using table f[]`
`int` `fib(``int` `n)`
`{`
`    ``// Base cases`
`    ``if` `(n == 0)`
`        ``return` `0;`
`    ``if` `(n == 1 || n == 2)`
`        ``return` `(f[n] = 1);`

`    ``// If fib(n) is already computed`
`    ``if` `(f[n])`
`        ``return` `f[n];`

`    ``int` `k = (n & 1)? (n+1)/2 : n/2;`

`    ``// Applying above formula [Note value n&1 is 1`
`    ``// if n is odd, else 0.`
`    ``f[n] = (n & 1)? (fib(k)*fib(k) + fib(k-1)*fib(k-1))`
`           ``: (2*fib(k-1) + fib(k))*fib(k);`

`    ``return` `f[n];`
`}`

`// Computes value of first Fibonacci numbers`
`int` `calculateSum(``int` `n)`
`{`
`    ``return` `fib(n+2) - 1;`
`}`
Method 6 (O(Log n) Time)
Below is one more interesting recurrence formula that can be used to find n’th Fibonacci Number in O(Log n) time.
```If n is even then k = n/2:
F(n) = [2*F(k-1) + F(k)]*F(k)

If n is odd then k = (n + 1)/2
F(n) = F(k)*F(k) + F(k-1)*F(k-1)```
How does this formula work?
The formula can be derived from above matrix equation.
Taking determinant on both sides, we get
(-1)n = Fn+1Fn-1 – Fn2
Moreover, since AnAm = An+m for any square matrix A, the following identities can be derived (they are obtained form two different coefficients of the matrix product)
FmFn + Fm-1Fn-1 = Fm+n-1
By putting n = n+1,
FmFn+1 + Fm-1Fn = Fm+n
Putting m = n
F2n-1 = Fn2 + Fn-12
F2n = (Fn-1 + Fn+1)Fn = (2Fn-1 + Fn)Fn (Source: Wiki)
To get the formula to be proved, we simply need to do following
If n is even, we can put k = n/2
If n is odd, we can put k = (n+1)/2
`// Create an array for memoization`
`int` `f[MAX] = {0};`

`// Returns n'th fuibonacci number using table f[]`
`int` `fib(``int` `n)`
`{`
`    ``// Base cases`
`    ``if` `(n == 0)`
`        ``return` `0;`
`    ``if` `(n == 1 || n == 2)`
`        ``return` `(f[n] = 1);`

`    ``// If fib(n) is already computed`
`    ``if` `(f[n])`
`        ``return` `f[n];`

`    ``int` `k = (n & 1)? (n+1)/2 : n/2;`

`    ``// Applyting above formula [Note value n&1 is 1`
`    ``// if n is odd, else 0.`
`    ``f[n] = (n & 1)? (fib(k)*fib(k) + fib(k-1)*fib(k-1))`
`           ``: (2*fib(k-1) + fib(k))*fib(k);`

`    ``return` `f[n];`
`}`

DP:
`   ``static` `int` `fib(``int` `n)`
`    ``{`
`        ``/* Declare an array to store Fibonacci numbers. */`
`    ``int` `f[] = ``new` `int``[n+``1``];`
`    ``int` `i;`
`     `
`    ``/* 0th and 1st number of the series are 0 and 1*/`
`    ``f[``0``] = ``0``;`
`    ``f[``1``] = ``1``;`
`    `
`    ``for` `(i = ``2``; i <= n; i++)`
`    ``{`
`       ``/* Add the previous 2 numbers in the series`
`         ``and store it */`
`        ``f[i] = f[i-``1``] + f[i-``2``];`
`    ``}`
`     `
`    ``return` `f[n];`
`    ``}`
`int` `fib(``int` `n)`
`{`
`  ``int` `a = 0, b = 1, c, i;`
`  ``if``( n == 0)`
`    ``return` `a;`
`  ``for` `(i = 2; i <= n; i++)`
`  ``{`
`     ``c = a + b;`
`     ``a = b;`
`     ``b = c;`
`  ``}`
`  ``return` `b;`
`}`
Method 4 ( Using power of the matrix {{1,1},{1,0}} )
This another O(n) which relies on the fact that if we n times multiply the matrix M = {{1,1},{1,0}} to itself (in other words calculate power(M, n )), then we get the (n+1)th Fibonacci number as the element at row and column (0, 0) in the resultant matrix.
The matrix representation gives the following closed expression for the Fibonacci numbers:
`    ``static` `int` `fib(``int` `n)`
`    ``{`
`    ``int` `F[][] = ``new` `int``[][]{{``1``,``1``},{``1``,``0``}};`
`    ``if` `(n == ``0``)`
`        ``return` `0``;`
`    ``power(F, n-``1``);`
`    `
`       ``return` `F[``0``][``0``];`
`    ``}`
`     `
`     ``/* Helper function that multiplies 2 matrices F and M of size 2*2, and`
`     ``puts the multiplication result back to F[][] */`
`    ``static` `void` `multiply(``int` `F[][], ``int` `M[][])`
`    ``{`
`    ``int` `x =  F[``0``][``0``]*M[``0``][``0``] + F[``0``][``1``]*M[``1``][``0``];`
`    ``int` `y =  F[``0``][``0``]*M[``0``][``1``] + F[``0``][``1``]*M[``1``][``1``];`
`    ``int` `z =  F[``1``][``0``]*M[``0``][``0``] + F[``1``][``1``]*M[``1``][``0``];`
`    ``int` `w =  F[``1``][``0``]*M[``0``][``1``] + F[``1``][``1``]*M[``1``][``1``];`
`     `
`    ``F[``0``][``0``] = x;`
`    ``F[``0``][``1``] = y;`
`    ``F[``1``][``0``] = z;`
`    ``F[``1``][``1``] = w;`
`    ``}`

`    ``/* Helper function that calculates F[][] raise to the power n and puts the`
`    ``result in F[][]`
`    ``Note that this function is designed only for fib() and won't work as general`
`    ``power function */`
`    ``static` `void` `power(``int` `F[][], ``int` `n)`
`    ``{`
`    ``int` `i;`
`    ``int` `M[][] = ``new` `int``[][]{{``1``,``1``},{``1``,``0``}};`
`    `
`    ``// n - 1 times multiply the matrix to {{1,0},{0,1}}`
`    ``for` `(i = ``2``; i <= n; i++)`
`        ``multiply(F, M);`
`    ``}`
The method 4 can be optimized to work in O(Logn) time complexity. We can do recursive multiplication to get power(M, n) in the prevous method (Similar to the optimization done in this post)
`    ``static` `int` `fib(``int` `n)`
`    ``{`
`    ``int` `F[][] = ``new` `int``[][]{{``1``,``1``},{``1``,``0``}};`
`    ``if` `(n == ``0``)`
`        ``return` `0``;`
`    ``power(F, n-``1``);`
`     `
`    ``return` `F[``0``][``0``];`
`    ``}`
`     `
`    ``static` `void` `multiply(``int` `F[][], ``int` `M[][])`
`    ``{`
`    ``int` `x =  F[``0``][``0``]*M[``0``][``0``] + F[``0``][``1``]*M[``1``][``0``];`
`    ``int` `y =  F[``0``][``0``]*M[``0``][``1``] + F[``0``][``1``]*M[``1``][``1``];`
`    ``int` `z =  F[``1``][``0``]*M[``0``][``0``] + F[``1``][``1``]*M[``1``][``0``];`
`    ``int` `w =  F[``1``][``0``]*M[``0``][``1``] + F[``1``][``1``]*M[``1``][``1``];`
`    `
`    ``F[``0``][``0``] = x;`
`    ``F[``0``][``1``] = y;`
`    ``F[``1``][``0``] = z;`
`    ``F[``1``][``1``] = w;`
`    ``}`
`     `
`    ``/* Optimized version of power() in method 4 */`
`    ``static` `void` `power(``int` `F[][], ``int` `n)`
`    ``{`
`    ``if``( n == ``0` `|| n == ``1``)`
`      ``return``;`
`    ``int` `M[][] = ``new` `int``[][]{{``1``,``1``},{``1``,``0``}};`
`     `
`    ``power(F, n/``2``);`
`    ``multiply(F, F);`
`     `
`    ``if` `(n%``2` `!= ``0``)`
`       ``multiply(F, M);`
`    ``}`

Brute Force approach is pretty straight forward, find all the Fibonacci numbers till f(n-1) and then add them up.
`int` `calculateSum(``int` `n)`
`{`
`    ``if` `(n <= 0)`
`       ``return` `0;`

`    ``int` `fibo[n+1];`
`    ``fibo[0] = 0, fibo[1] = 1;`

`    ``// Initialize result`
`    ``int` `sum = fibo[0] + fibo[1];`

`    ``// We are using zero indexing`
`    ``//so the nth fibonacci number is`
`    ``//n-1 th fibonacci number`
`    ``for` `(``int` `i=2; i<=n; i++)`
`    ``{`
`        ``fibo[i] = fibo[i-1]+fibo[i-2];`
`        ``sum += fibo[i];`
`    ``}`

`    ``return` `sum;`
`}`
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