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# C++ Program For Fibonacci Numbers

The Fibonacci numbers are the numbers in the following integer sequence.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……..

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation

Fn = Fn-1 + Fn-2

with seed values

F0 = 0 and F1 = 1. Given a number n, print n-th Fibonacci Number.

Examples:

Input  : n = 2
Output : 1

Input  : n = 9
Output : 34
Recommended Practice

Write a function int fib(int n) that returns Fn. For example, if n = 0, then fib() should return 0. If n = 1, then it should return 1. For n > 1, it should return Fn-1 + Fn-2

For n = 9
Output:34

The following are different methods to get the nth Fibonacci number.

Method 1 (Use recursion)
A simple method that is a direct recursive implementation mathematical recurrence relation is given above.

## C++

 // Fibonacci Series using Recursion  #include  using namespace std;     int fib(int n)  {      if (n <= 1)          return n;      return fib(n - 1) + fib(n - 2);  }     int main()  {      int n = 9;      cout << fib(n);      getchar();      return 0;  }

Output

34

Time Complexity: Exponential, as every function calls two other functions.

If the original recursion tree were to be implemented then this would have been the tree but now for n times the recursion function is called

Original tree for recursion

                          fib(5)
/                \
fib(4)                fib(3)
/        \              /       \
fib(3)      fib(2)         fib(2)   fib(1)
/    \       /    \        /      \
fib(2)   fib(1)  fib(1) fib(0) fib(1) fib(0)
/     \
fib(1) fib(0)

Optimized tree for recursion for code above

fib(5)

fib(4)

fib(3)

fib(2)

fib(1)

Extra Space: O(n) if we consider the function call stack size, otherwise O(1).

Method 2: (Use Dynamic Programming)
We can avoid the repeated work done in method 1 by storing the Fibonacci numbers calculated so far.

## C++

 // C++ program for Fibonacci Series    // using Dynamic Programming   #include  using namespace std;     class GFG{         public:  int fib(int n)  {             // Declare an array to store       // Fibonacci numbers.      // 1 extra to handle       // case, n = 0       int f[n + 2];       int i;         // 0th and 1st number of the       // series are 0 and 1      f = 0;      f = 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];      }  };     // Driver code  int main ()  {      GFG g;      int n = 9;             cout << g.fib(n);      return 0;  }

Output

34

Time complexity: O(n) for given n

Auxiliary space: O(n)

Method 3: (Space Optimized Method 2)
We can optimize the space used in method 2 by storing the previous two numbers only because that is all we need to get the next Fibonacci number in series.

## C++

 // Fibonacci Series using Space Optimized Method  #include  using namespace std;     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;  }     // Driver code  int main()  {      int n = 9;             cout << fib(n);      return 0;  }

Output

34

Time Complexity: O(n)
Extra Space: O(1)

Method 4: Using power of the matrix {{1, 1}, {1, 0}}
This is another O(n) that 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: ## C++

 #include  using namespace std;     // Helper function that multiplies 2   // matrices F and M of size 2*2, and  // puts the multiplication result   // back to F[][]   void multiply(int F, int M);     // 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   void power(int F, int n);     int fib(int n)  {      int F = { { 1, 1 }, { 1, 0 } };             if (n == 0)          return 0;                 power(F, n - 1);             return F;  }     void multiply(int F, int M)  {      int x = F * M +               F * M;      int y = F * M +               F * M;      int z = F * M +               F * M;      int w = F * M +               F * M;             F = x;      F = y;      F = z;      F = w;  }     void power(int F, int n)  {      int i;      int M = { { 1, 1 }, { 1, 0 } };             // n - 1 times multiply the       // matrix to {{1,0},{0,1}}      for(i = 2; i <= n; i++)          multiply(F, M);  }     // Driver code  int main()  {      int n = 9;             cout << " " <<  fib(n);             return 0;  }

Output

 34

Time Complexity: O(n)
Extra Space: O(1)

Method 5: (Optimized Method 4)
Method 4 can be optimized to work in O(Logn) time complexity. We can do recursive multiplication to get power(M, n) in the previous method (Similar to the optimization done in this post)

## C++

 // Fibonacci Series using Optimized Method   #include  using namespace std;     void multiply(int F, int M);  void power(int F, int n);     // Function that returns nth Fibonacci number  int fib(int n)  {      int F = {{1, 1}, {1, 0}};      if (n == 0)          return 0;      power(F, n - 1);         return F;  }     // Optimized version of power() in method 4  void power(int F, int n)  {      if(n == 0 || n == 1)         return;      int M = {{1, 1}, {1, 0}};             power(F, n / 2);      multiply(F, F);             if (n % 2 != 0)          multiply(F, M);  }     void multiply(int F, int M)  {      int x = F * M + F * M;      int y = F * M + F * M;      int z = F * M + F * M;      int w = F * M + F * M;             F = x;      F = y;      F = z;      F = w;  }     // Driver code  int main()  {      int n = 9;             cout << fib(9);      getchar();             return 0;  }

Output

34

Time Complexity: O(Logn)
Extra Space: O(Logn) if we consider the function call stack size, otherwise O(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 the 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
from two different coefficients of the matrix product)

FmFn + Fm-1Fn-1 = Fm+n-1         ---------------------------(1)

By putting n = n+1 in equation(1),
FmFn+1 + Fm-1Fn = Fm+n             --------------------------(2)

Putting m = n in equation(1).
F2n-1 = Fn2 + Fn-12
Putting m = n in equation(2)

F2n = (Fn-1 + Fn+1)Fn = (2Fn-1 + Fn)Fn (Source: Wiki)   --------
( By putting Fn+1 = Fn + Fn-1 )
To get the formula to be proved, we simply need to do the following
If n is even, we can put k = n/2
If n is odd, we can put k = (n+1)/2

Below is the implementation of the above idea.

## C++

 // C++ Program to find n'th fibonacci Number in  // with O(Log n) arithmetic operations  #include  using namespace std;     const int MAX = 1000;     // 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];  }     // Driver program to test above function   int main()  {      int n = 9;      printf("%d ", fib(n));      return 0;  }

Output

34

Time complexity: O(Log n), as we divide the problem in half in every recursive call.

Method 7: (Another approach(Using Binet’s formula))
In this method, we directly implement the formula for the nth term in the Fibonacci series.
Fn = {[(√5 + 1)/2] ^ n} / √5

Note: Above Formula gives correct result only upto for n<71. Because as we move forward from n>=71 , rounding error becomes significantly large . Although , using floor function instead of round function will give correct result for n=71 . But after from n=72 , it also fails.

Example: For N=72 , Correct result is 498454011879264 but above formula gives 498454011879265.

## C++

 // C++ Program to find n'th fibonacci Number  #include  #include     int fib(int n) {    double phi = (1 + sqrt(5)) / 2;    return round(pow(phi, n) / sqrt(5));  }     // Driver Code  int main ()  {    int n = 9;    std::cout << fib(n) << std::endl;    return 0;  }

Output

34

Time Complexity: O(logn), this is because calculating phi^n takes logn time
Auxiliary Space: O(1)

Method 8: DP using memoization(Top down approach)

We can avoid the repeated work done in method 1 by storing the Fibonacci numbers calculated so far. We just need to store all the values in an array.

## C++

 #include  using namespace std;  int dp;  int fib(int n)  {      if (n <= 1)          return n;         // temporary variables to store      //  values of fib(n-1) & fib(n-2)      int first, second;         if (dp[n - 1] != -1)          first = dp[n - 1];      else         first = fib(n - 1);         if (dp[n - 2] != -1)          second = dp[n - 2];      else         second = fib(n - 2);         // memoization      return dp[n] = first + second;  }     // Driver Code  int main()  {      int n = 9;         memset(dp, -1, sizeof(dp));         cout << fib(n);      getchar();      return 0;     }

Output

34

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