Find Nth term (A matrix exponentiation example) in C++

C++Server Side ProgrammingProgramming

In this problem, we are given an integer N and a recursive function that given Nth term as a function of other terms. Our task is to create a program to Find Nth term (A matrix exponentiation example).

The function is

T(n) = 2*( T(n-1) ) + 3*( T(n-2) )
Initial values are
   T(0) = 1 , T(1) = 1

Let’s take an example to understand the problem,

Input

N = 4

Output

41

Explanation

T(4) = 2* (T(3)) + 3*(T(2))
T(4) = 2* ( 2*(T(2)) + 3*(T(1)) ) + 3*( 2* (T(1)) + 3*(T(0)) )
T(4) = 2*( 2*(2* (T(1)) + 3*(T(0))) + 3*(1) ) + 3*(2*(1) + 3*(1))
T(4) = 2*(2 * (2 *(1) + 3*(1) )) + 3 ) + 3 * (5)
T(4) = 2*(2 * (2 + 3) + 3) + 15
T(4) = 2*(2 * (5) + 3) + 15
T(4) = 2*(10 + 3) + 15
T(4) = 2*(13) + 15 = 26 + 15 = 41

Solution Approach

A simple approach to solve the problem is using recursion or iteration. We can find nth term as a recursive call to previous terms and using initial values the result can be found.

Program to illustrate the working of our solution,

Example

 Live Demo

#include <iostream>
using namespace std;
long calcNthTerm(long n) {
   if(n == 0 || n == 1)
      return 1;
   return ( ( 2*(calcNthTerm(n-1)) ) + ( 3*(calcNthTerm(n-2)) ) );
}
int main() {
   long n = 5;
   cout<<n<<"th term of found using matrix exponentiation is "<<calcNthTerm(n);
   return 0;
}

Output

5th term of found using matrix exponentiation is 121

Efficient approach

An efficient approach to solve the problem is using the concept of matrix exponentiation. In this method, we will use a transform matrix to find Nth term.

For this we need to find the transform matrix. The matrix is dependent on the number of dependent terms which happens to be 2 here. And the initial values which are T(0) = 1, T(1)= 1.

The transform matrix is of the size k*k. Which when multiplied with the initial matrix of size k*1, returns the next term.

Here are the values,

initial matrix =

$$\begin{bmatrix}1 \\0 \end{bmatrix}$$

Transform matrix =

$$\begin{bmatrix}2&3 \\1&0 \end{bmatrix}$$

The values of Tn is given as TM(n-1)*IM

$$\begin{bmatrix}2&3 \\1&0 \end{bmatrix}^{n-1}*\begin{bmatrix}2 \\3 \end{bmatrix}$$

Program to illustrate the working of our solution,

Example

 Live Demo

#include <iostream>
using namespace std;
#define MOD 1000000009
long calcNthTerm(long n) {
   if (n <= 1)
      return 1;
   n--;
   long resultantMat[2][2] = { 1, 0, 0, 1 };
   long transMat[2][2] = { 2, 3, 1, 0 };
   while (n) {
      long tempMat[2][2];
      if (n & 1) {
         tempMat[0][0] = (resultantMat[0][0] * transMat[0][0] +
         resultantMat[0][1] * transMat[1][0]) % MOD;
         tempMat[0][1] = (resultantMat[0][0] * transMat[0][1] +
         resultantMat[0][1] * transMat[1][1]) % MOD;
         tempMat[1][0] = (resultantMat[1][0] * transMat[0][0] +
         resultantMat[1][1] * transMat[1][0]) % MOD;
         tempMat[1][1] = (resultantMat[1][0] * transMat[0][1] +
         resultantMat[1][1] * transMat[1][1]) % MOD;
         resultantMat[0][0] = tempMat[0][0];
         resultantMat[0][1] = tempMat[0][1];
         resultantMat[1][0] = tempMat[1][0];
         resultantMat[1][1] = tempMat[1][1];
      }
      n = n / 2;
      tempMat[0][0] = (transMat[0][0] * transMat[0][0] +
      transMat[0][1] * transMat[1][0]) % MOD;
      tempMat[0][1] = (transMat[0][0] * transMat[0][1] +
      transMat[0][1] * transMat[1][1]) % MOD;
      tempMat[1][0] = (transMat[1][0] * transMat[0][0] +
      transMat[1][1] * transMat[1][0]) % MOD;
      tempMat[1][1] = (transMat[1][0] * transMat[0][1] +
      transMat[1][1] * transMat[1][1]) % MOD;
      transMat[0][0] = tempMat[0][0];
      transMat[0][1] = tempMat[0][1];
      transMat[1][0] = tempMat[1][0];
      transMat[1][1] = tempMat[1][1];
   }
   return (resultantMat[0][0] * 1 + resultantMat[0][1] * 1) % MOD;
}
int main() {
   long n = 5;
   cout<<n<<"th term of found using matrix exponentiation is "<<calcNthTerm(n);
   return 0;
}

Output

5th term of found using matrix exponentiation is 121
raja
Published on 13-Mar-2021 13:16:20

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