Weird Number

In number theory, a weird number is a natural number that is abundant but not semiperfect. In other words, the sum of the proper divisors (divisors including 1 but not itself) of the number is greater than the number, but no subset of those divisors sums to the number itself.

Given a number N, the task is to check if the number is weird or not.

Examples:

Input: 40
Output: The number is not weird
1+4+5+10+20=40, hence it is not weird.

Input: 70
Output: The number is Weird
The smallest weird number is 70. Its proper divisors are 1, 2, 5, 7, 10, 14, and 35; these sum to 74, but no subset of these sums to 70.

The number 12, for example, is abundant but not weird, because the proper divisors of 12 are 1, 2, 3, 4, and 6, which sum to 16; but 2+4+6 = 12. The first few weird numbers are 70, 836, 4030, 5830, 7192, 7912, 9272, 10430, 10570, 10792, 10990, 11410, 11690, 12110, 12530, 12670, 13370, 13510, ..



Approach: Check if the number is abundant or not. The approach has been discussed here. Once the checking has been done check if the number is semiperfect or not. The approach for checking semiperfects numbers has been discussed here.

Below is the implementation of the above approach:

C++

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// C++ program to check if the 
// number is weird or not
#include <bits/stdc++.h>
using namespace std;
  
// code to find all the factors of
// the number excluding the number itself
vector<int> factors(int n)
{
    // vector to store the factors
    vector<int> v;
    v.push_back(1);
  
    // note that this loop runs till sqrt(n)
    for (int i = 2; i <= sqrt(n); i++) {
  
        // if the value of i is a factor
        if (n % i == 0) {
            v.push_back(i);
  
            // condition to check the
            // divisor is not the number itself
            if (n / i != i) {
                v.push_back(n / i);
            }
        }
    }
    // return the vector
    return v;
}
  
// Function to check if the number 
// is abundant or not 
bool checkAbundant(int n)
{
    vector<int> v;
  
    int sum = 0;
  
    // find the divisors using function
    v = factors(n);
  
    // sum all the factors
    for (int i = 0; i < v.size(); i++) {
        sum += v[i];
    }
  
    // check for abundant or not
    if (sum > n)
        return true;
    else
        return false;
}
  
// Function to check if the
// number is semi-perfecr or not
bool checkSemiPerfect(int n)
{
    vector<int> v;
  
    // find the divisors
    v = factors(n);
  
    // sorting the vector
    sort(v.begin(), v.end());
  
    int r = v.size();
  
    // subset to check if no is semiperfect
    bool subset[r + 1][n + 1];
  
    // initialising 1st column to true
    for (int i = 0; i <= r; i++)
        subset[i][0] = true;
  
    // initialing 1st row except zero position to 0
    for (int i = 1; i <= n; i++)
        subset[0][i] = false;
  
    // loop to find whther the number is semiperfect
    for (int i = 1; i <= r; i++) {
        for (int j = 1; j <= n; j++) {
  
            // calculation to check if the
            // number can be made by summation of diviors
            if (j < v[i - 1])
                subset[i][j] = subset[i - 1][j];
            else {
                subset[i][j] = subset[i - 1][j] || 
                               subset[i - 1][j - v[i - 1]];
            }
        }
    }
  
    // if not possible to make the
    // number by any combination of divisors
    if ((subset[r][n]) == 0)
        return false;
    else
        return true;
}
  
// Function to check for 
// weird or not 
bool checkweird(int n)
{
    if (checkAbundant(n) == true && 
        checkSemiPerfect(n) == false)
        return true;
    else
        return false;
}
  
// Driver Code
int main()
{
    int n = 70;
  
    if (checkweird(n))
        cout << "Weird Number";
    else
        cout << "Not Weird Number";
    return 0;
}

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Java

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// Java program to check if  
// the number is weird or not
import java.util.*;
class GFG
// code to find all the 
// factors of the number 
// excluding the number itself
static ArrayList<Integer> factors(int n)
{
    // ArrayList to store
    // the factors
    ArrayList<Integer> v = new ArrayList<Integer>();
    v.add(1);
  
    // note that this loop 
    // runs till sqrt(n)
    for (int i = 2;
             i <= Math.sqrt(n); i++)
    {
  
        // if the value of 
        // i is a factor
        if (n % i == 0
        {
            v.add(i);
  
            // condition to check 
            // the divisor is not 
            // the number itself
            if (n / i != i) 
            {
                v.add(n / i);
            }
        }
    }
      
    // return the ArrayList
    return v;
}
  
// Function to check if the 
// number is abundant or not 
static boolean checkAbundant(int n)
{
    ArrayList<Integer> v;
  
    int sum = 0;
  
    // find the divisors
    // using function
    v = factors(n);
  
    // sum all the factors
    for (int i = 0; i < v.size(); i++) 
    {
        sum += v.get(i);
    }
  
    // check for abundant
    // or not
    if (sum > n)
        return true;
    else
        return false;
}
  
// Function to check if the
// number is semi-perfecr or not
static boolean checkSemiPerfect(int n)
{
    ArrayList<Integer> v;
  
    // find the divisors
    v = factors(n);
  
    // sorting the ArrayList
    Collections.sort(v);
  
    int r = v.size();
  
    // subset to check if 
    // no is semiperfect
    boolean subset[][] = new boolean[r + 1][n + 1];
  
    // initialising 1st
    // column to true
    for (int i = 0; i <= r; i++)
        subset[i][0] = true;
  
    // initialing 1st row except 
    // zero position to 0
    for (int i = 1; i <= n; i++)
        subset[0][i] = false;
  
    // loop to find whther 
    // the number is semiperfect
    for (int i = 1; i <= r; i++) 
    {
        for (int j = 1; j <= n; j++) 
        {
  
            // calculation to check 
            // if the number can be 
            // made by summation of 
            // diviors
            if (j < v.get(i - 1))
                subset[i][j] = subset[i - 1][j];
            else {
                subset[i][j] = subset[i - 1][j] || 
                               subset[i - 1][j - 
                                v.get(i - 1)];
            }
        }
    }
  
    // if not possible to make 
    // the number by any 
    // combination of divisors
    if ((subset[r][n]) == false)
        return false;
    else
        return true;
}
  
// Function to check 
// for weird or not 
static boolean checkweird(int n)
{
    if (checkAbundant(n) == true && 
        checkSemiPerfect(n) == false)
        return true;
    else
        return false;
}
  
// Driver Code
public static void main(String args[])
{
    int n = 70;
  
    if (checkweird(n))
        System.out.println("Weird Number");
    else
        System.out.println("Not Weird Number");
}
}
  
// This code is contributed
// by Arnab Kundu

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Python3

# Python 3 program to check if the
# number is weird or not
from math import sqrt

# code to find all the factors of
# the number excluding the number itself
def factors(n):

# vector to store the factors
v = []
v.append(1)

# note that this loop runs till sqrt(n)
for i in range(2, int(sqrt(n)) + 1, 1):

# if the value of i is a factor
if (n % i == 0):
v.append(i);

# condition to check the
# divisor is not the number itself
if (int(n / i) != i):
v.append(int(n / i))

# return the vector
return v

# Function to check if the number
# is abundant or not
def checkAbundant(n):
sum = 0

# find the divisors using function
v = factors(n)

# sum all the factors
for i in range(len(v)):
sum += v[i]

# check for abundant or not
if (sum > n):
return True
else:
return False

# Function to check if the
# number is semi-perfecr or not
def checkSemiPerfect(n):

# find the divisors
v = factors(n)

# sorting the vector
v.sort(reverse = False)
r = len(v)

# subset to check if no is semiperfect
subset = [[0 for i in range(n + 1)]
for j in range(r + 1)]

# initialising 1st column to true
for i in range(r + 1):
subset[i][0] = True

# initialing 1st row except zero position to 0
for i in range(1, n + 1):
subset[0][i] = False

# loop to find whther the number is semiperfect
for i in range(1, r + 1):
for j in range(1, n + 1):

# calculation to check if the
# number can be made by summation of diviors
if (j < v[i - 1]): subset[i][j] = subset[i - 1][j] else: subset[i][j] = (subset[i - 1][j] or subset[i - 1][j - v[i - 1]]) # if not possible to make the # number by any combination of divisors if ((subset[r][n]) == 0): return False else: return True # Function to check for # weird or not def checkweird(n): if (checkAbundant(n) == True and checkSemiPerfect(n) == False): return True else: return False # Driver Code if __name__ == '__main__': n = 70 if (checkweird(n)): print("Weird Number") else: print("Not Weird Number") # This code is contributed by # Surendra_Gangwar [tabbyending]

Output:

Weird Number

Time Complexity: O(N * number of factors)
Auxiliary Space: O(N * number of factors)



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