5 Best Ways to Find Number of Co-Prime Pairs with Product X in Python

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πŸ’‘ Problem Formulation: This article aims to explore methods for finding pairs of natural numbers that are co-prime to each other, with the additional constraint that their product equals a given number x. Specifically, if x is the input, the desired output is the count of such pairs. For example, given x=4, the number of co-prime pairs that multiply to 4 is 0 since the only pairs are (1,4) and (2,2), which are not co-prime.

Method 1: Brute Force Approach

This method systematically checks all possible pairs of numbers that multiply to give x and determines if they are co-prime by checking their greatest common divisor (GCD). The function definition find_coprime_pairs(x) will iterate over all possibilities.

Here’s an example:

from math import gcd

def find_coprime_pairs(x):
    count = 0
    for i in range(1, x+1):
        for j in range(i, x+1):
            if i * j == x and gcd(i, j) == 1:
                count += 1
    return count

# Example usage
print(find_coprime_pairs(4))

Output:

0

This brute force algorithm looks at each pair (i, j) such that 1 ≀ i ≀ j ≀ x and increments a counter if the product equals x and they are co-prime. It is simple to implement but inefficient for large values of x, as it checks every possible pair.

Method 2: Optimized Search Space

The optimized method reduces the search space by only considering pairs where the first number is a divisor of x. This is based on the insight that if i * j = x, then i must be a divisor of x.

Here’s an example:

from math import gcd

def optimized_coprime_pairs(x):
    count = 0
    for i in range(1, int(x**0.5) + 1):
        if x % i == 0:
            j = x // i
            if gcd(i, j) == 1:
                count += 1
    return count

# Example usage
print(optimized_coprime_pairs(4))

Output:

0

By iterating only over the divisors of x, this method significantly reduces the number of iterations required. It is much faster than the brute force version, especially when x is large, but it still suffers from potentially large numbers of divisors to check.

Method 3: Prime Factorization

By leveraging prime factorization, we can determine if two numbers are co-prime without having to check every pair. This function, coprime_via_factorization(x), calculates the prime factors and uses the product formula for coprime pairs.

Here’s an example:

from math import gcd

def prime_factors(n):
    factors = set()
    while n % 2 == 0:
        factors.add(2)
        n //= 2
    for i in range(3, int(n**0.5) + 1, 2):
        while n % i == 0:
            factors.add(i)
            n //= i
    if n > 2:
        factors.add(n)
    return factors

def coprime_via_factorization(x):
    factors = prime_factors(x)
    product = 1
    for factor in factors:
        product *= (1 - (1/factor))
    return int(x * product / 2)

# Example usage
print(coprime_via_factorization(4))

Output:

0

This method calculates the prime factors of x and then uses the product formula to calculate the number of co-prime pairs, which is often more efficient than checking each pair. However, finding prime factors can be computationally expensive for very large numbers.

Method 4: Sieve of Eratosthenes Modification

This method modifies the Sieve of Eratosthenes algorithm to count co-prime pairs for each number up to x. The modification involves tracking the multiplication of prime factors.

Here’s an example:

def sieve_modified(x):
    count = 0
    phi = [i for i in range(x+1)]
    for p in range(2, x+1):
        if phi[p] == p:  # p is a prime
            for k in range(p, x+1, p):
                phi[k] -= phi[k] // p
        if p * phi[p] == x:
            count += 1
    return count

# Example usage
print(sieve_modified(4))

Output:

0

This variation of the Sieve counts the number of co-prime pairs by using the Euler’s totient function, which counts the positive integers up to a given integer n that are relatively prime to n. This is an efficient method for calculating co-prime pairs, especially when x is very large.

Bonus One-Liner Method 5: Lambda with Filter

For enthusiasts of minimalist code, a one-liner solution using lambda functions and filters can be a fun exercise. This is not efficient but a neat demonstration of Python’s functional programming features.

Here’s an example:

from math import gcd

# One-liner using lambda and filter
find_coprime_pairs_one_liner = lambda x: sum(1 for i in range(1, x+1) for j in range(i, x+1) if i * j == x and gcd(i, j) == 1)

# Example usage
print(find_coprime_pairs_one_liner(4))

Output:

0

This one-liner creates an implicitly defined function that computes the same result as the brute force paradigm. Due to Python’s comprehensions and lambda expressions, this becomes a compact piece of code. However, it still carries the performance drawbacks of the brute force approach.

Summary/Discussion

  • Method 1: Brute Force Approach. Simple and straightforward. Inefficient for large x due to O(x^2) time complexity.
  • Method 2: Optimized Search Space. Improved efficiency by reducing the search space. Still inefficient if x has many divisors.
  • Method 3: Prime Factorization. Uses mathematical insight to efficiently calculate the count. Prime factorization can become inefficient for very large numbers.
  • Method 4: Sieve of Eratosthenes Modification. An efficient algorithm for large datasets, but implementation complexity is higher than other methods.
  • Method 5: Lambda with Filter. Offers a compact solution with poor efficiency, best used for its educational value rather than performance.