Given a positive integer N, the task is to find all the prime triplets {P, Q, R} such that P = R – Q and P, Q and R is less than N.
Examples:
Input: N = 8
Output:
2 3 5
2 5 7
Explanation:
The only 2 prime triplets satisfying the given conditions are:
- {2, 3, 5}: P = 2, Q = 3, R = 5. Therefore, P, Q and R are prime numbers and P = R – Q.
- {2, 5, 7}: P = 2, Q = 5, R = 7. Therefore, P, Q and R are prime numbers and P = R – Q.
Input: N = 5
Output: 2 3 5
Approach: The given problem can be solved based on the following observations:
- By rearranging the given equation, it can be observed that P + Q = R, and the sum of two odd numbers is even and the sum of one odd and one even is odd.
- As there is only one even prime i.e., 2. Let P is odd prime and Q is odd prime then R can never be a prime number, so it is necessary that P should be always 2 which is even prime and Q is odd prime then R should be an odd prime. So there is necessary to find prime Q such that Q > 2 and R = P + Q ≤ N (where P = 2) and R should be prime.
Follow the steps below to solve the problem:
- Precompute all the prime numbers over the range [1, N] using Sieve of Eratosthenes.
- Initialize a vector of vectors, say V, to stores all the resultant triplets.
- Iterate over the range [3, N] using a variable, say i. If i and (i + 2) are prime and 2 + i ≤ N, then store the current triplets in the vector V.
- After completing the above steps, print all the triplets stored in V[].
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Stores 1 and 0 at indices which // are prime and non-prime respectively bool prime[100000];
// Function to find all prime // numbers from the range [0, N] void SieveOfEratosthenes( int n)
{ // Consider all numbers to prime initially
memset (prime, true , sizeof (prime));
// Iterate over the range [2, sqrt(N)]
for ( int p = 2; p * p <= n; p++) {
// If p is a prime
if (prime[p] == true ) {
// Update all tultiples
// of p as false
for ( int i = p * p;
i <= n; i += p) {
prime[i] = false ;
}
}
}
} // Function to find all prime triplets // satisfying the given conditions void findTriplets( int N)
{ // Generate all primes up to N
SieveOfEratosthenes(N);
// Stores the triplets
vector<vector< int > > V;
// Iterate over the range [3, N]
for ( int i = 3; i <= N; i++) {
// Check for the condition
if (2 + i <= N && prime[i]
&& prime[2 + i]) {
// Store the triplets
V.push_back({ 2, i, i + 2 });
}
}
// Print all the stored triplets
for ( int i = 0; i < V.size(); i++) {
cout << V[i][0] << " "
<< V[i][1] << " "
<< V[i][2] << "\n" ;
}
} // Driver Code int main()
{ int N = 8;
findTriplets(N);
return 0;
} |
Java
// Java program for the above approach import java.util.*;
class GFG{
// Stores 1 and 0 at indices which // are prime and non-prime respectively static boolean [] prime = new boolean [ 100000 ];
static void initialize()
{ for ( int i = 0 ; i < 100000 ; i++)
prime[i] = true ;
} // Function to find all prime // numbers from the range [0, N] static void SieveOfEratosthenes( int n)
{ // Iterate over the range [2, sqrt(N)]
for ( int p = 2 ; p * p <= n; p++)
{
// If p is a prime
if (prime[p] == true )
{
// Update all tultiples
// of p as false
for ( int i = p * p; i <= n; i += p)
{
prime[i] = false ;
}
}
}
} // Function to find all prime triplets // satisfying the given conditions static void findTriplets( int N)
{ // Generate all primes up to N
SieveOfEratosthenes(N);
// Stores the triplets
ArrayList<ArrayList<Integer>> V = new ArrayList<ArrayList<Integer>>();
// List<List<int> > V = new List<List<int>>();
// Iterate over the range [3, N]
for ( int i = 3 ; i <= N; i++)
{
// Check for the condition
if ( 2 + i <= N && prime[i] && prime[ 2 + i])
{
// Store the triplets
ArrayList<Integer> a1 = new ArrayList<Integer>();
a1.add( 2 );
a1.add(i);
a1.add(i + 2 );
V.add(a1);
}
}
// Print all the stored triplets
for ( int i = 0 ; i < V.size(); i++)
{
System.out.println(V.get(i).get( 0 ) + " " +
V.get(i).get( 1 ) + " " +
V.get(i).get( 2 ));
}
} // Driver Code public static void main(String args[])
{ initialize();
int N = 8 ;
findTriplets(N);
} } // This code is contributed by ipg2016107 |
Python3
# Python3 program for the above approach from math import sqrt
# Stores 1 and 0 at indices which # are prime and non-prime respectively prime = [ True for i in range ( 100000 )]
# Function to find all prime # numbers from the range [0, N] def SieveOfEratosthenes(n):
# Iterate over the range [2, sqrt(N)]
for p in range ( 2 , int (sqrt(n)) + 1 , 1 ):
# If p is a prime
if (prime[p] = = True ):
# Update all tultiples
# of p as false
for i in range (p * p, n + 1 , p):
prime[i] = False
# Function to find all prime triplets # satisfying the given conditions def findTriplets(N):
# Generate all primes up to N
SieveOfEratosthenes(N)
# Stores the triplets
V = []
# Iterate over the range [3, N]
for i in range ( 3 , N + 1 , 1 ):
# Check for the condition
if ( 2 + i < = N and prime[i] and prime[ 2 + i]):
# Store the triplets
V.append([ 2 , i, i + 2 ])
# Print all the stored triplets
for i in range ( len (V)):
print (V[i][ 0 ], V[i][ 1 ], V[i][ 2 ])
# Driver Code if __name__ = = '__main__' :
N = 8
findTriplets(N)
# This code is contributed by bgangwar59.
|
C#
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Stores 1 and 0 at indices which // are prime and non-prime respectively static bool [] prime = new bool [100000];
static void initialize()
{ for ( int i = 0; i < 100000; i++)
prime[i] = true ;
} // Function to find all prime // numbers from the range [0, N] static void SieveOfEratosthenes( int n)
{ // Iterate over the range [2, sqrt(N)]
for ( int p = 2; p * p <= n; p++)
{
// If p is a prime
if (prime[p] == true )
{
// Update all tultiples
// of p as false
for ( int i = p * p; i <= n; i += p)
{
prime[i] = false ;
}
}
}
} // Function to find all prime triplets // satisfying the given conditions static void findTriplets( int N)
{ // Generate all primes up to N
SieveOfEratosthenes(N);
// Stores the triplets
List<List< int >> V = new List<List< int >>();
// Iterate over the range [3, N]
for ( int i = 3; i <= N; i++)
{
// Check for the condition
if (2 + i <= N && prime[i] ==
true && prime[2 + i])
{
// Store the triplets
List< int > a1 = new List< int >();
a1.Add(2);
a1.Add(i);
a1.Add(i + 2);
V.Add(a1);
}
}
// Print all the stored triplets
for ( int i = 0; i < V.Count; i++)
{
Console.WriteLine(V[i][0] + " " +
V[i][1] + " " +
V[i][2]);
}
} // Driver Code public static void Main()
{ initialize();
int N = 8;
findTriplets(N);
} } // This code is contributed by SURENDRA_GANGWAR |
Javascript
<script> // Javascript program for the above approach // Stores 1 and 0 at indices which // are prime and non-prime respectively var prime = new Array(100000);
// Function to find all prime // numbers from the range [0, N] function SieveOfEratosthenes(n)
{ // Consider all numbers to prime initially
prime.fill( true );
// Iterate over the range [2, sqrt(N)]
for ( var p = 2; p * p <= n; p++)
{
// If p is a prime
if (prime[p] == true )
{
// Update all tultiples
// of p as false
for ( var i = p * p;
i <= n; i += p)
{
prime[i] = false ;
}
}
}
} // Function to find all prime triplets // satisfying the given conditions function findTriplets(N)
{ // Generate all primes up to N
SieveOfEratosthenes(N);
// Stores the triplets
var V = [];
// Iterate over the range [3, N]
for ( var i = 3; i <= N; i++)
{
// Check for the condition
if (2 + i <= N && prime[i] == true &&
prime[2 + i] == true )
{
// Store the triplets
var a1 = [2, i, i + 2];
V.push(a1);
}
}
// Print all the stored triplets
for ( var i = 0; i < V.length; i++)
{
document.write(V[i][0] + " " +
V[i][1] + " " +
V[i][2] + "<br>" );
}
} // Driver code N = 8; findTriplets(N); // This code is contributed by SoumikMondal </script> |
Output:
2 3 5 2 5 7
Time Complexity: O(N*log(log(N)))
Auxiliary Space: O(1)