Summation of GCD of all the pairs up to N

4.2

Given a number N, find sum of all GCDs that can be formed by selecting all the pairs from 1 to N.

Input  : 4
Output : 7
Explanation: 
Numbers from 1 to 4 are: 1, 2, 3, 4
Result = gcd(1,2) + gcd(1,3) + gcd(1,4) + 
         gcd(2,3) + gcd(2,4) + gcd(3,4)
       = 1 + 1 + 1 + 1 + 2 + 1
       = 7

Input  : 12
Output : 105

Input  : 1
Output : 0

Input  : 2
Output : 1

A Naive approach is to run two loops one inside the other. Select all pairs one by one, find GCD of every pair and then find sum of these GCDs. Time complexity of this approach is O(N2 * log(N))

Efficient Approach is based on following concepts:

  • Euler’s Totient function Φ(n) for an input n is count of numbers in {1, 2, 3, …, n} that are relatively prime to n, i.e., the numbers whose GCD (Greatest Common Divisor) with n is 1. For example, Φ(4) = 2, Φ(3) = 2 and Φ(5) = 4. There are 2 numbers smaller or equal to 4 that are relatively prime to 4, 2 numbers smaller or equal to 3 that are relatively prime to 3. And 4 numbers smaller than or equal to 5 that are relatively prime to 5.
  • The idea is to convert given problem into sum of Euler Totient Functions.

    Sum of all GCDs where j is a part of
    pair is and j is greater element in pair:
    Sumj = ∑(i=1 to j-1) gcd(i, j)
    Our final result is 
    Result = ∑(j=1 to N) Sumj
    
    The above equation can be written as :
    Sumj = ∑ g * count(g) 
    For every possible GCD 'g' of j. Here count(g)
    represents count of pairs having GCD equals to
    g. For every such pair(i, j), we can write :
     gcd(i/g, j/g) = 1
    
    We can re-write our previous equation as
    Sumj = ∑ d * phi(j/d) 
    For every divisor d of j and phi[] is Euler
    Totient number 
    
    Example : j = 12 and d = 3 is one of divisor 
    of j so in order to calculate the sum of count
    of all pairs having 3 as gcd we can simple write
    it as 
    => 3*phi[12/3]  
    => 3*phi[4]
    => 3*2
    => 6
    
    Therefore sum of GCDs of all pairs where 12 is 
    greater part of pair and 3 is GCD.
    GCD(3, 12) + GCD(9, 12) = 6.
    
    Complete Example : 
    N = 4
    Sum1 = 0
    Sum2 = 1 [GCD(1, 2)]
    Sum3 = 2 [GCD(1, 3) + GCD(2, 3)]
    Sum4 = 4 [GCD(1, 4) + GCD(3, 4) + GCD(2, 4)]
    
    Result = Sum1 + Sum2 + Sum3 + Sum4
           = 0 + 1 + 2 + 4
           = 7
     

    Below is C++ implementation of above idea. We precompute Euler Totient Functions and result for all numbers till a maximum value. The idea used in implementation is based this post.

    // C++ approach of finding sum of GCD of all pairs
    #include<bits/stdc++.h>
    using namespace std;
    
    #define MAX 100001
    
    // phi[i] stores euler totient function for i
    // result[j] stores result for value j
    long long phi[MAX], result[MAX];
    
    // Precomputation of phi[] numbers. Refer below link
    // for details : https://goo.gl/LUqdtY
    void computeTotient()
    {
        // Refer https://goo.gl/LUqdtY
        phi[1] = 1;
        for (int i=2; i<MAX; i++)
        {
            if (!phi[i])
            {
                phi[i] = i-1;
                for (int j = (i<<1); j<MAX; j+=i)
                {
                    if (!phi[j])
                        phi[j] = j;
    
                    phi[j] = (phi[j]/i)*(i-1);
                }
            }
        }
    }
    
    // Precomputes result for all numbers till MAX
    void sumOfGcdPairs()
    {
        // Precompute all phi value
        computeTotient();
    
        for (int i=1; i<MAX; ++i)
        {
            // Iterate throght all the divisors
            // of i.
            for (int j=2; i*j<MAX; ++j)
                result[i*j] += i*phi[j];
        }
    
        // Add summation of previous calculated sum
        for (int i=2; i<MAX; i++)
            result[i] += result[i-1];
    }
    
    // Driver code
    int main()
    {
        // Function to calculate sum of all the GCD
        // pairs
        sumOfGcdPairs();
    
        int N = 4;
        cout << "Summation of " << N << " = "
             << result[N] << endl;;
        N = 12;
        cout << "Summation of " << N << " = "
            << result[N] << endl;
        N = 5000;
        cout << "Summation of " << N << " = "
            << result[N] ;
    
        return 0;
    }
    
    Output:
    Summation of 4 = 7
    Summation of 12 = 105
    Summation of 5000 = 61567426
    

    Time complexity: O(MAX*log(log MAX))
    Auxiliary space: O(MAX)

    Reference:
    https://www.quora.com/How-can-I-solve-the-problem-GCD-Extreme-on-SPOJ-SPOJ-com-Problem-GCDEX

    This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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