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Find the number of pairs such that their gcd is equals to 1
  • Difficulty Level : Expert
  • Last Updated : 14 Apr, 2021

Given an array a of size N. The task is to find the number of pairs such that gcd(a[i], a[j]) where 1 ≤ i < j ≤ N
Examples: 
 

Input : a[] = {1, 2, 4, 6} 
Output :
{1, 2}, {1, 4}, {1, 6} are such pairs
Input : a[] = {1, 2, 3, 4, 5, 6} 
Output : 11 
 

 

Approach : 
The answer is to sum of μ(X) * C(D(X), 3) overall integer X. Where, μ(X) is Mobius function, C(N, K) is the selection of K things from N and D(X) is the number of integers in the given sequence that are divisible by X.
The correctness of the solution follows from the fact that we can do an inclusion-exclusion principle solution and to show that it is, in fact, equal to our answer. That means that we will add to the answer the number of pairs that are divisible by some intermediate (in the IEP) product D if D is formed by multiplication of even number of prime numbers and subtract this number of pairs otherwise. 
So, we get: 
 

  • 1 for addition, because that is Möbius function for square-free numbers with even number of prime divisors.
  • -1 for subtraction, that is Mobius function for square-free numbers with an odd number of prime divisors.
  • 0 for square-free numbers. By the definition, they can’t occur in our IEP solution.

Below is the implementation of the above approach : 
 

C++




// CPP program to find the number of pairs
// such that gcd equals to 1
#include <bits/stdc++.h>
using namespace std;
 
#define N 100050
 
int lpf[N], mobius[N];
 
// Function to calculate least
// prime factor of each number
void least_prime_factor()
{
    for (int i = 2; i < N; i++)
 
        // If it is a prime number
        if (!lpf[i])
 
            for (int j = i; j < N; j += i)
 
                // For all multiples which are not
                // visited yet.
                if (!lpf[j])
                    lpf[j] = i;
}
 
// Function to find the value of Mobius function
// for all the numbers from 1 to n
void Mobius()
{
    for (int i = 1; i < N; i++) {
 
        // If number is one
        if (i == 1)
            mobius[i] = 1;
        else {
 
            // If number has a squared prime factor
            if (lpf[i / lpf[i]] == lpf[i])
                mobius[i] = 0;
 
            // Multiply -1 with the previous number
            else
                mobius[i] = -1 * mobius[i / lpf[i]];
        }
    }
}
 
// Function to find the number of pairs
// such that gcd equals to 1
int gcd_pairs(int a[], int n)
{
    // To store maximum number
    int maxi = 0;
 
    // To store frequency of each number
    int fre[N] = { 0 };
 
    // Find frequency and maximum number
    for (int i = 0; i < n; i++) {
        fre[a[i]]++;
        maxi = max(a[i], maxi);
    }
 
    least_prime_factor();
    Mobius();
 
    // To store number of pairs with gcd equals to 1
    int ans = 0;
 
    // Traverse through the all possible elements
    for (int i = 1; i <= maxi; i++) {
        if (!mobius[i])
            continue;
 
        int temp = 0;
        for (int j = i; j <= maxi; j += i)
            temp += fre[j];
 
        ans += temp * (temp - 1) / 2 * mobius[i];
    }
 
    // Return the number of pairs
    return ans;
}
 
// Driver code
int main()
{
    int a[] = { 1, 2, 3, 4, 5, 6 };
 
    int n = sizeof(a) / sizeof(a[0]);
 
    // Function call
    cout << gcd_pairs(a, n);
 
    return 0;
}

Java




// Java program to find the number of pairs
// such that gcd equals to 1
class GFG
{
 
static int N = 100050;
 
static int []lpf = new int[N];
static int []mobius = new int[N];
 
// Function to calculate least
// prime factor of each number
static void least_prime_factor()
{
    for (int i = 2; i < N; i++)
 
        // If it is a prime number
        if (lpf[i] == 0)
 
            for (int j = i; j < N; j += i)
 
                // For all multiples which are not
                // visited yet.
                if (lpf[j] == 0)
                    lpf[j] = i;
}
 
// Function to find the value of Mobius function
// for all the numbers from 1 to n
static void Mobius()
{
    for (int i = 1; i < N; i++)
    {
 
        // If number is one
        if (i == 1)
            mobius[i] = 1;
        else
        {
 
            // If number has a squared prime factor
            if (lpf[i / lpf[i]] == lpf[i])
                mobius[i] = 0;
 
            // Multiply -1 with the previous number
            else
                mobius[i] = -1 * mobius[i / lpf[i]];
        }
    }
}
 
// Function to find the number of pairs
// such that gcd equals to 1
static int gcd_pairs(int a[], int n)
{
    // To store maximum number
    int maxi = 0;
 
    // To store frequency of each number
    int []fre = new int[N];
 
    // Find frequency and maximum number
    for (int i = 0; i < n; i++)
    {
        fre[a[i]]++;
        maxi = Math.max(a[i], maxi);
    }
 
    least_prime_factor();
    Mobius();
 
    // To store number of pairs with gcd equals to 1
    int ans = 0;
 
    // Traverse through the all possible elements
    for (int i = 1; i <= maxi; i++)
    {
        if (mobius[i] == 0)
            continue;
 
        int temp = 0;
        for (int j = i; j <= maxi; j += i)
            temp += fre[j];
 
        ans += temp * (temp - 1) / 2 * mobius[i];
    }
 
    // Return the number of pairs
    return ans;
}
 
// Driver code
public static void main (String[] args)
{
    int a[] = { 1, 2, 3, 4, 5, 6 };
 
    int n = a.length;
 
    // Function call
    System.out.print(gcd_pairs(a, n));
}
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 program to find the number of pairs
# such that gcd equals to 1
N = 100050
 
lpf = [0 for i in range(N)]
mobius = [0 for i in range(N)]
 
# Function to calculate least
# prime factor of each number
def least_prime_factor():
 
    for i in range(2, N):
 
        # If it is a prime number
        if (lpf[i] == 0):
 
            for j in range(i, N, i):
 
                # For all multiples which are not
                # visited yet.
                if (lpf[j] == 0):
                    lpf[j] = i
 
# Function to find the value of Mobius function
# for all the numbers from 1 to n
def Mobius():
 
    for i in range(1, N):
 
        # If number is one
        if (i == 1):
            mobius[i] = 1
        else:
 
            # If number has a squared prime factor
            if (lpf[ (i // lpf[i]) ] == lpf[i]):
                mobius[i] = 0
 
            # Multiply -1 with the previous number
            else:
                mobius[i] = -1 * mobius[i // lpf[i]]
 
# Function to find the number of pairs
# such that gcd equals to 1
def gcd_pairs(a, n):
 
    # To store maximum number
    maxi = 0
 
    # To store frequency of each number
    fre = [0 for i in range(N)]
 
    # Find frequency and maximum number
    for i in range(n):
        fre[a[i]] += 1
        maxi = max(a[i], maxi)
 
    least_prime_factor()
    Mobius()
 
    # To store number of pairs with gcd equals to 1
    ans = 0
 
    # Traverse through the all possible elements
    for i in range(1, maxi + 1):
        if (mobius[i] == 0):
            continue
 
        temp = 0
        for j in range(i, maxi + 1, i):
            temp += fre[j]
 
        ans += temp * (temp - 1) // 2 * mobius[i]
 
    # Return the number of pairs
    return ans
 
# Driver code
a = [1, 2, 3, 4, 5, 6]
 
n = len(a)
 
# Function call
print(gcd_pairs(a, n))
 
# This code is contributed by Mohit Kumar

C#




// C# program to find the number of pairs
// such that gcd equals to 1
using System;
 
class GFG
{
static int N = 100050;
 
static int []lpf = new int[N];
static int []mobius = new int[N];
 
// Function to calculate least
// prime factor of each number
static void least_prime_factor()
{
    for (int i = 2; i < N; i++)
 
        // If it is a prime number
        if (lpf[i] == 0)
 
            for (int j = i; j < N; j += i)
 
                // For all multiples which are not
                // visited yet.
                if (lpf[j] == 0)
                    lpf[j] = i;
}
 
// Function to find the value of Mobius function
// for all the numbers from 1 to n
static void Mobius()
{
    for (int i = 1; i < N; i++)
    {
 
        // If number is one
        if (i == 1)
            mobius[i] = 1;
        else
        {
 
            // If number has a squared prime factor
            if (lpf[i / lpf[i]] == lpf[i])
                mobius[i] = 0;
 
            // Multiply -1 with the previous number
            else
                mobius[i] = -1 * mobius[i / lpf[i]];
        }
    }
}
 
// Function to find the number of pairs
// such that gcd equals to 1
static int gcd_pairs(int []a, int n)
{
    // To store maximum number
    int maxi = 0;
 
    // To store frequency of each number
    int []fre = new int[N];
 
    // Find frequency and maximum number
    for (int i = 0; i < n; i++)
    {
        fre[a[i]]++;
        maxi = Math.Max(a[i], maxi);
    }
 
    least_prime_factor();
    Mobius();
 
    // To store number of pairs with gcd equals to 1
    int ans = 0;
 
    // Traverse through the all possible elements
    for (int i = 1; i <= maxi; i++)
    {
        if (mobius[i] == 0)
            continue;
 
        int temp = 0;
        for (int j = i; j <= maxi; j += i)
            temp += fre[j];
 
        ans += temp * (temp - 1) / 2 * mobius[i];
    }
 
    // Return the number of pairs
    return ans;
}
 
// Driver code
public static void Main (String[] args)
{
    int []a = { 1, 2, 3, 4, 5, 6 };
 
    int n = a.Length;
 
    // Function call
    Console.Write(gcd_pairs(a, n));
}
}
     
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript program to find the number of pairs
// such that gcd equals to 1   
var N = 100050;
 
    var lpf = Array(N).fill(0);
    var mobius = Array(N).fill(0);
 
    // Function to calculate least
    // prime factor of each number
    function least_prime_factor() {
        for (i = 2; i < N; i++)
 
            // If it is a prime number
            if (lpf[i] == 0)
 
                for (j = i; j < N; j += i)
 
                    // For all multiples which are not
                    // visited yet.
                    if (lpf[j] == 0)
                        lpf[j] = i;
    }
 
    // Function to find the value of Mobius function
    // for all the numbers from 1 to n
    function Mobius() {
        for (i = 1; i < N; i++) {
 
            // If number is one
            if (i == 1)
                mobius[i] = 1;
            else {
 
                // If number has a squared prime factor
                if (lpf[i / lpf[i]] == lpf[i])
                    mobius[i] = 0;
 
                // Multiply -1 with the previous number
                else
                    mobius[i] = -1 * mobius[i / lpf[i]];
            }
        }
    }
 
    // Function to find the number of pairs
    // such that gcd equals to 1
    function gcd_pairs(a , n) {
        // To store maximum number
        var maxi = 0;
 
        // To store frequency of each number
        var fre = Array(n+1).fill(0);
 
        // Find frequency and maximum number
        for (i = 0; i < n; i++) {
            fre[a[i]]++;
            maxi = Math.max(a[i], maxi);
        }
 
        least_prime_factor();
        Mobius();
 
        // To store number of pairs with gcd equals to 1
        var ans = 0;
 
        // Traverse through the all possible elements
        for (i = 1; i <= maxi; i++) {
            if (mobius[i] == 0)
                continue;
 
            var temp = 0;
            for (j = i; j <= maxi; j += i)
                temp = parseInt(temp+fre[j]);
 
            ans += parseInt(temp * (temp - 1) / 2 * mobius[i]);
        }
 
        // Return the number of pairs
        return ans;
    }
 
    // Driver code
     
        var a = [ 1, 2, 3, 4, 5, 6 ];
 
        var n = a.length;
 
        // Function call
        document.write(gcd_pairs(a, n));
 
// This code contributed by Rajput-Ji
 
</script>
Output: 
11

 

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