Program for Perrin numbers

The Perrin numbers are the numbers in the following integer sequence.
3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17, 22, 29, 39 …

In mathematical terms, the sequence p(n) of Perrin numbers is defined by the recurrence relation

 P(n) = P(n-2) + P(n-3) for n > 2, 

with initial values
    P(0) = 3, P(1) = 0, P(2) = 2.

Write a function int per(int n) that returns p(n). For example, if n = 0, then per() should return 3. If n = 1, then it should return 0 If n = 2, then it should return 2. For n > 2, it should return p(n-2) + p(n-3)

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 ( Use recursion : Exponential )
Below is simple recursive implementation of above formula.

C++

// n'th perrin number using Recursion' 
#include <bits/stdc++.h> 
using namespace std; 
  
int per(int n) 
{ 
    if (n == 0) 
        return 3; 
    if (n == 1) 
        return 0; 
    if (n == 2) 
        return 2; 
    return per(n - 2) + per(n - 3); 
} 
  
// Driver code 
int main() 
{ 
    int n = 9; 
    cout << per(n); 
    return 0; 
} 
  
// This code is contributed  
// by Akanksha Rai 

C

// n'th perrin number using Recursion' 
#include <stdio.h> 
int per(int n) 
{ 
    if (n == 0) 
        return 3; 
    if (n == 1) 
        return 0; 
    if (n == 2) 
        return 2; 
    return per(n - 2) + per(n - 3); 
} 
  
// Driver code 
int main() 
{ 
    int n = 9; 
    printf("%d", per(n)); 
    return 0; 
} 

Java

// Java code for n'th perrin number 
// using Recursion' 
import java.io.*; 
  
class GFG { 
  
    static int per(int n) 
    { 
        if (n == 0) 
            return 3; 
        if (n == 1) 
            return 0; 
        if (n == 2) 
            return 2; 
        return per(n - 2) + per(n - 3); 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
  
        int n = 9; 
  
        System.out.println(per(n)); 
    } 
} 
  
// This code is contributed by vt_m. 

Python3

# Python3 code for n'th perrin  
# number using Recursion' 
  
# function return n'th 
# perrin number 
def per(n): 
  
    if (n == 0): 
        return 3; 
    if (n == 1): 
        return 0; 
    if (n == 2): 
        return 2; 
    return per(n - 2) + per(n - 3); 
  
# Driver Code 
n = 9; 
print(per(n)); 
      
# This code is contributed mits 

C#

// C# code for n'th perrin number 
// using Recursion' 
using System; 
  
class GFG { 
  
    static int per(int n) 
    { 
        if (n == 0) 
            return 3; 
        if (n == 1) 
            return 0; 
        if (n == 2) 
            return 2; 
        return per(n - 2) + per(n - 3); 
    } 
  
    // Driver code 
    public static void Main() 
    { 
  
        int n = 9; 
  
        Console.Write(per(n)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP code for n'th perrin  
// number using Recursion' 
  
// function return n'th 
// perrin number 
function per($n) 
{ 
    if ($n == 0) 
        return 3; 
    if ($n == 1) 
        return 0; 
    if ($n == 2) 
        return 2; 
    return per($n - 2) +  
           per($n - 3); 
} 
  
    // Driver Code 
    $n = 9; 
    echo per($n); 
      
#This code is contributed ajit. 
?> 

Output:

We see that in this implementation a lot of repeated work in the following recursion tree.

                           per(8)   
                       /           \     
               per(6)             per(5)   
              /      \             /     \
        per(4)      per(3)        per(3)    per(2)
       /     \        /    \        /  \  
   per(2)   per(1)  per(1) per(0) per(1) per(0)

Method 2: ( Optimized : Linear)

C++

// Optimized C++ program for n'th perrin number 
#include <bits/stdc++.h> 
using namespace std; 
int per(int n) 
{ 
    int a = 3, b = 0, c = 2, i; 
    int m; 
    if (n == 0) 
        return a; 
    if (n == 1) 
        return b; 
    if (n == 2) 
        return c; 
    while (n > 2) { 
        m = a + b; 
        a = b; 
        b = c; 
        c = m; 
        n--; 
    } 
    return m; 
} 
  
// Driver code 
int main() 
{ 
    int n = 9; 
    cout << per(n); 
    return 0; 
} 
  
// This code is contributed  
// by Akanksha Rai 

C

// Optimized C program for n'th perrin number 
#include <stdio.h> 
int per(int n) 
{ 
    int a = 3, b = 0, c = 2, i; 
    int m; 
    if (n == 0) 
        return a; 
    if (n == 1) 
        return b; 
    if (n == 2) 
        return c; 
    while (n > 2) { 
        m = a + b; 
        a = b; 
        b = c; 
        c = m; 
        n--; 
    } 
    return m; 
} 
  
// Driver code 
int main() 
{ 
    int n = 9; 
    printf("%d", per(n)); 
    return 0; 
} 

Java

// Optimized Java program for n'th perrin number 
import java.io.*; 
  
class GFG { 
  
    static int per(int n) 
    { 
        int a = 3, b = 0, c = 2, i; 
        int m = 0; 
        if (n == 0) 
            return a; 
        if (n == 1) 
            return b; 
        if (n == 2) 
            return c; 
        while (n > 2) { 
            m = a + b; 
            a = b; 
            b = c; 
            c = m; 
            n--; 
        } 
        return m; 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        int n = 9; 
  
        System.out.println(per(n)); 
    } 
} 
  
// This code is contributed by vt_m. 

C#

// Optimized C# program for n'th perrin number 
using System; 
  
class GFG { 
  
    static int per(int n) 
    { 
        int a = 3, b = 0, c = 2; 
  
        // int i; 
        int m = 0; 
        if (n == 0) 
            return a; 
        if (n == 1) 
            return b; 
        if (n == 2) 
            return c; 
  
        while (n > 2) { 
            m = a + b; 
            a = b; 
            b = c; 
            c = m; 
            n--; 
        } 
  
        return m; 
    } 
  
    // Driver code 
    public static void Main() 
    { 
  
        int n = 9; 
  
        Console.WriteLine(per(n)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// Optimized PHP program for  
// n'th perrin number 
  
// function return the  
// n'th perrin number 
function per($n) 
{ 
    $a = 3; $b = 0;  
    $c = 2; $i; 
    $m; 
    if ($n == 0) 
        return $a; 
    if ($n == 1) 
        return $b; 
    if ($n == 2) 
        return $c; 
    while ($n > 2)  
    { 
        $m = $a + $b; 
        $a = $b; 
        $b = $c; 
        $c = $m; 
        $n--; 
    } 
    return $m; 
} 
  
    // Driver code 
    $n = 9; 
    echo per($n); 
      
// This code is contributed by ajit 
?> 

Output:

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

Method 3: (Further Optimized : Logarithmic)
We can further optimize using Matrix Exponentiation. The matrix power formula for n’th Perrin number is

${\Huge \begin{pmatrix} 0& 1 & 0\\ 0& 0&1 \\ 1 &1 & 0 \\ \end{pmatrix}^n \begin{pmatrix} 3\\ 0\\ 2 \end{pmatrix} = \begin{pmatrix} P(n)\\ P(n+1)\\ P(n+2) \end{pmatrix}}$

We can implement this method similar to implementation of method 5 of Fibonacci numbers. Since we can compute n’th power of a constant matrix in O(Log n), time complexity of this method is O(Log n)

Application :
The number of different maximal independent sets in an n-vertex cycle graph is counted by the nth Perrin number for n > 1

Related Article :
Sum of Perrin Numbers

Reference:
https://en.wikipedia.org/wiki/Perrin_number

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

C

Java

Python3

C#

PHP

C++

C

Java

C#

PHP

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