Count of all possible pairs of disjoint subsets of integers from 1 to N

• Last Updated : 30 Aug, 2019

Given an integer N. Consider the set of first N natural numbers A = {1, 2, 3, …, N}. Let M and P be two non-empty subsets of A. The task is to count the number of unordered pairs of (M, P) such that M and P are disjoint sets. Note that the order of M and P doesn’t matter.

Examples:

Input: N = 3
Output: 6
The unordered pairs are ({1}, {2}), ({1}, {3}),
({2}, {3}), ({1}, {2, 3}), ({2}, {1, 3}), ({3}, {1, 2}).

Input: N = 2
Output: 1

Input: N = 10
Output: 28501

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

Approach:

1. Lets assume there are only 6 elements in the set {1, 2, 3, 4, 5, 6}.
2. When you count the number of subsets with 1 as one of the element of first subset, it comes out to be 211.
3. Counting number of subsets with 2 being one of the element of first subset, it comes out to be 65, because 1’s not included as order of sets doesn’t matter.
4. Counting number of subset with 3 being one of the element of first set it comes out to be 65, here a pattern can be observed.
5. Pattern:

5 = 3 * 1 + 2
19 = 3 * 5 + 4
65 = 3 * 19 + 8
211 = 3 * 65 + 16
S(n) = 3 * S(n-1) + 2(n – 2)

6. Expanding it until n->2 (means numbers of elements n-2+1=n-1)
2(n-2) * 3(0) + 2(n – 3) * 31 + 2(n – 4) * 32 + 2(n – 5) * 33 + … + 2(0) * 3(n – 2)
From Geometric progression, a + a * r0 + a * r1 + … + a * r(n – 1) = a * (rn – 1) / (r – 1)
7. S(n) = 3(n – 1) – 2(n – 1). Remember S(n) is number of subsets with 1 as a one of the elements of first subset but to get the required result, Denoted by T(n) = S(1) + S(2) + S(3) + … +S(n)
8. It also forms a Geometric progression, so we calculate it by formula of sum of GP
T(n) = (3n – 2n + 1 + 1)/2
9. As we require T(n) % p where p = 109 + 7
We have to use Fermats’s little theorem
a-1 = a(m – 2) (mod m) for modular division

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach  #include using namespace std;#define p 1000000007  // Modulo exponentiation functionlong long power(long long x, long long y){    // Function to calculate (x^y)%p in O(log(y))    long long res = 1;    x = x % p;      while (y > 0) {        if (y & 1)            res = (res * x) % p;        y = y >> 1;        x = (x * x) % p;    }      return res % p;}  // Driver functionint main(){    long long n = 3;      // Evaluating ((3^n-2^(n+1)+1)/2)%p    long long x = (power(3, n) % p + 1) % p;      x = (x - power(2, n + 1) + p) % p;      // From  Fermats’s little theorem    // a^-1 ? a^(m-2) (mod m)      x = (x * power(2, p - 2)) % p;    cout << x << "\n";}

Java

 // Java implementation of the approachclass GFG { static int p = 1000000007;  // Modulo exponentiation functionstatic long power(long x, long y){    // Function to calculate (x^y)%p in O(log(y))    long res = 1;    x = x % p;      while (y > 0)    {        if (y % 2 == 1)            res = (res * x) % p;        y = y >> 1;        x = (x * x) % p;    }    return res % p;}  // Driver Codepublic static void main(String[] args) {    long n = 3;      // Evaluating ((3^n-2^(n+1)+1)/2)%p    long x = (power(3, n) % p + 1) % p;      x = (x - power(2, n + 1) + p) % p;      // From Fermats's little theorem    // a^-1 ? a^(m-2) (mod m)      x = (x * power(2, p - 2)) % p;    System.out.println(x);}}  // This code is contributed by Rajput-Ji

Python3

 # Python3 implementation of the approachp = 1000000007  # Modulo exponentiation functiondef power(x, y):          # Function to calculate (x^y)%p in O(log(y))    res = 1    x = x % p      while (y > 0):        if (y & 1):            res = (res * x) % p;        y = y >> 1        x = (x * x) % p      return res % p  # Driver Coden = 3  # Evaluating ((3^n-2^(n+1)+1)/2)%px = (power(3, n) % p + 1) % p  x = (x - power(2, n + 1) + p) % p  # From Fermats’s little theorem# a^-1 ? a^(m-2) (mod m)x = (x * power(2, p - 2)) % p  print(x)  # This code is contributed by Mohit Kumar

C#

 // C# implementation of the approachusing System;  class GFG{static int p = 1000000007;  // Modulo exponentiation functionstatic long power(long x, long y){    // Function to calculate (x^y)%p in O(log(y))    long res = 1;    x = x % p;      while (y > 0)    {        if (y % 2 == 1)            res = (res * x) % p;        y = y >> 1;        x = (x * x) % p;    }    return res % p;}  // Driver Codestatic public void Main (){    long n = 3;          // Evaluating ((3^n-2^(n+1)+1)/2)%p    long x = (power(3, n) % p + 1) % p;          x = (x - power(2, n + 1) + p) % p;          // From Fermats's little theorem    // a^-1 ? a^(m-2) (mod m)          x = (x * power(2, p - 2)) % p;    Console.Write(x);}}  // This code is contributed by ajit.
Output:
6

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