Generalized Fibonacci Numbers

Last Updated : 06 Apr, 2021

We all know that Fibonacci numbers (Fn) are defined by the recurrence relation

Fibonacci Numbers (Fn) = F(n-1) + F(n-2)
with seed values
F0 = 0 and F1 = 1

Similarly, we can generalize these numbers. Such a number sequence is known as Generalized Fibonacci number (G).
Generalized Fibonacci number (G) is defined by the recurrence relation

Generalized Fibonacci Numbers (Gn) = (c * G(n-1)) + (d * G(n-2))
with seed values
G0 = a and G1 = b

Finding Nth term

Given the four constant values of Generalized Fibonacci Numbers as a, b, c, and d and an integer N, the task is to find the Nth term of the Generalized Fibonacci Numbers, i.e. Gn.

Examples:

Input: N = 2, a = 0, b = 1, c = 2, d = 3
Output: 2
Explanation:
As a = 0 -> G(0) = 0
b = 1 -> G(1) = 1
So, G(2) = 2 * G(1) + 3 * G(0) = 2

Input: N = 3, a = 0, b = 1, c = 2, d = 3
Output: 7

Naive Approach: Using the given values, find each term of the series till the Nth term and then print the Nth term.
Time Complexity: O(2^N)

Another Approach: The idea is to use DP tabulation to find all the terms till the Nth terms and then print the Nth term.
Time Complexity: O(N)

Efficient Approach: Using matrix multiplication we can solve the given problem in log(N) time.

\small {Given\, G(0)=a\, and\, G(1)=b, \, therefore \, G(2)=c*a+d*b } \\ \small {} \\ \begin{bmatrix} G(2) & G(1)\\ G(1) & G(0) \end{bmatrix} = \begin{bmatrix} d*b+c*a & b\\ b & a \end{bmatrix} \newline \text {Multiplying matrices on RHS will eventually give us our LHS }\\ \begin{bmatrix} G(3) & G(2)\\ G(2) & G(1) \end{bmatrix} = \begin{bmatrix} G(2) & G(1)\\ G(1) & G(0) \end{bmatrix} * \begin{bmatrix} d & 1\\ c & 0 \end{bmatrix} \newline \text {One more step } \\ \begin{bmatrix} G(4) & G(3)\\ G(3) & G(2) \end{bmatrix} = \begin{bmatrix} G(3) & G(2)\\ G(2) & G(1) \end{bmatrix} * \begin{bmatrix} d & 1\\ c & 0 \end{bmatrix} \newline \text {So by looking at the sequence above we can generalize the Nth term as... }\\ \begin{bmatrix} G(N) & G(N-1)\\ G(N-1) & G(N-2) \end{bmatrix} = \begin{bmatrix} G(2) & G(1)\\ G(1) & G(0) \end{bmatrix} * \begin{bmatrix} d & 1\\ c & 0 \end{bmatrix} ^{\!N-2}\newline \text{Now }\begin{bmatrix} d & 1\\ c & 0 \end{bmatrix} ^{\!N-2}} \text{ can be solved in log(N) time complexity }

Below is the implementation of the above approach:

C++

// C++ program to implement the  // Generalised Fibonacci numbers  #include <bits/stdc++.h>  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[2][2], int M[2][2]);   // Helper function that calculates F[][]  // raised to the power N  // and puts the result in F[][]  void power(int F[2][2], int N, int m, int n);   // Function to find the Nth term  int F(int N, int a, int b, int m, int n)  {      // m 1      // n 0      int F[2][2] = { { m, 1 }, { n, 0 } };       if (N == 0)          return a;       if (N == 1)          return b;       if (N == 2)          return m * b + n * a;       int initial[2][2]          = { { m * b + n * a, b },              { b, a } };       power(F, N - 2, m, n);       // Discussed above     multiply(initial, F);       return F[0][0];  }   // Function that multiplies  // 2 matrices F and M of size 2*2,  // and puts the multiplication  // result back to F[][]  void multiply(int F[2][2], int M[2][2])  {      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;  }   // Function that calculates F[][]  // raised to the power N  // and puts the result in F[][]  void power(int F[2][2], int N, int m, int n)  {      int i;      int M[2][2] = { { m, 1 }, { n, 0 } };       for (i = 1; i <= N; i++)          multiply(F, M);  }   // Driver code  int main()  {      int N = 2, a = 0, b = 1, m = 2, n = 3;      printf("%d\n", F(N, a, b, m, n));       N = 3;      printf("%d\n", F(N, a, b, m, n));       N = 4;      printf("%d\n", F(N, a, b, m, n));       N = 5;      printf("%d\n", F(N, a, b, m, n));       return 0;  }

Java

// Java program to implement the  // Generalised Fibonacci numbers  import java.util.*;  class GFG{  // Function to find the Nth term  static int F(int N, int a, int b,               int m, int n)  {      // m 1      // n 0      int[][] F = { { m, 1 }, { n, 0 } };       if (N == 0)          return a;      if (N == 1)          return b;      if (N == 2)          return m * b + n * a;       int[][] initial = { { m * b + n * a, b },                          { b, a } };                               power(F, N - 2, m, n);       // Discussed below      multiply(initial, F);       return F[0][0];  }   // 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;  }   // Function that calculates F[][]  // raised to the power N  // and puts the result in F[][]  static void power(int[][] F, int N,                    int m, int n)  {      int i;      int[][] M = { { m, 1 }, { n, 0 } };       for(i = 1; i <= N; i++)          multiply(F, M);  }   // Driver code public static void main(String[] args) {     int N = 2, a = 0, b = 1, m = 2, n = 3;      System.out.println(F(N, a, b, m, n));          N = 3;      System.out.println(F(N, a, b, m, n));           N = 4;      System.out.println(F(N, a, b, m, n));           N = 5;      System.out.println(F(N, a, b, m, n));  } }  // This code is contributed by offbeat

Python3

# Python3 program to implement the # Generalised Fibonacci numbers  # Function to find the Nth term def F(N, a, b, m, n):          # m 1      # n 0      F = [[ m, 1 ], [ n, 0 ]]      if(N == 0):         return a     if(N == 1):         return b     if(N == 2):         return m * b + n * a      initial = [[ m * b + n * b, b ],                            [ b, a ]]      power(F, N - 2, m, n)      multiply(initial, F)      return F[0][0]  # Function that multiplies # 2 matrices F and M of size 2*2, # and puts the multiplication # result back to F[][] def multiply(F, M):          x = (F[0][0] * M[0][0] +           F[0][1] * M[1][0])      y = (F[0][0] * M[0][1] +          F[0][1] * M[1][1])      z = (F[1][0] * M[0][0] +           F[1][1] * M[1][0])      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  # Function that calculates F[][] # raised to the power N # and puts the result in F[][] def power(F, N, m, n):      M = [[ m, 1 ], [ n, 0 ]]     for i in range(1, N + 1):         multiply(F, M)  # Driver code if __name__ == '__main__':      N, a, b, m, n = 2, 0, 1, 2, 3     print(F(N, a, b, m, n))      N = 3     print(F(N, a, b, m, n))      N = 4     print(F(N, a, b, m, n))      N = 5     print(F(N, a, b, m, n))  # This code is contributed by Shivam Singh

// C# program to implement the  // Generalised Fibonacci numbers  using System; class GFG{   // Function to find the Nth term  static int F(int N, int a, int b,               int m, int n)  {      // m 1      // n 0      int[,] F = { { m, 1 }, { n, 0 } };      if (N == 0)          return a;      if (N == 1)          return b;      if (N == 2)          return m * b + n * a;      int[,] initial = { { m * b + n * a, b },                          { b, a } };      power(F, N - 2, m, n);        // Discussed below      multiply(initial, F);      return F[0, 0];  }    // 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;  }    // Function that calculates F[,]  // raised to the power N  // and puts the result in F[,]  static void power(int[,] F, int N,                    int m, int n)  {      int i;      int[,] M = { { m, 1 }, { n, 0 } };      for(i = 1; i <= N; i++)          multiply(F, M);  }    // Driver code public static void Main(String[] args) {     int N = 2, a = 0, b = 1, m = 2, n = 3;      Console.WriteLine(F(N, a, b, m, n));     N = 3;      Console.WriteLine(F(N, a, b, m, n));      N = 4;      Console.WriteLine(F(N, a, b, m, n));      N = 5;      Console.WriteLine(F(N, a, b, m, n));  } }   // This code is contributed by shikhasingrajput

JavaScript

<script>  // Javascript program to implement the  // Generalised Fibonacci numbers   // Function to find the Nth term  function F(N, a, b, m, n) {     var F = [ [ m, 1 ], [ n, 0 ] ]          if (N == 0)          return a;      if (N == 1)          return b;      if (N == 2)          return m * b + n * a;           var initial = [ [ m * b + n * a, b ], [ b, a ] ]          power(F, N - 2, m, n);           // Discussed below      multiply(initial, F);           return F[0][0];  }  // Function that multiplies  // 2 matrices F and M of size 2*2,  // and puts the multiplication  // result back to F[][]  function multiply(F, M)  {      var x = F[0][0] * M[0][0] +              F[0][1] * M[1][0];      var y = F[0][0] * M[0][1] +              F[0][1] * M[1][1];      var z = F[1][0] * M[0][0] +              F[1][1] * M[1][0];      var 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;  }   // Function that calculates F[][]  // raised to the power N  // and puts the result in F[][]  function power(F, N, m, n)  {      var i;      var M = [ [ m, 1 ], [ n, 0 ] ];       for(i = 1; i <= N; i++)          multiply(F, M);  }   // Driver code var N = 2, a = 0, b = 1, m = 2, n = 3; document.write(F(N, a, b, m, n) + "</br>");  N = 3; document.write(F(N, a, b, m, n) + "</br>");  N = 4; document.write(F(N, a, b, m, n) + "</br>");  N = 5; document.write(F(N, a, b, m, n));      // This code is contributed by Ankita saini     </script>

Output:

2 7 20 61

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Generalized Fibonacci Numbers

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