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.
DP:
Brute Force approach is pretty straight forward, find all the Fibonacci numbers till f(n-1) and then add them up.
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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),
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).
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)
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 memoizationint 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 numbersint 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.
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.
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)
(-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
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
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 memoizationint 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];} 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.
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;}