Explore 5 Ways to Calculate the Sum of Number of Divisors of Divisors in Python

💡 Problem Formulation: We need a program that takes a single integer as input and computes the sum of the number of divisors for each of its divisors. For instance, if the input is 10, its divisors are 1, 2, 5, 10. The respective number of divisors for each are 1, 2, 2, 4, hence the desired sum would be 1+2+2+4=9.

Method 1: Brute Force

This method calculates the number of divisors for each divisor of a given number using nested loops, which is straightforward but not the most efficient. It is suitable for small input sizes and is easily understandable for beginners.

Here’s an example:

def sum_of_divisors(n):
    def count_divisors(x):
        return sum(1 for i in range(1, x+1) if x % i == 0)
    
    return sum(count_divisors(i) for i in range(1, n+1) if n % i == 0)

# Example usage:
print(sum_of_divisors(10))

The output of this code is:

This function sum_of_divisors() iterates through all the numbers from 1 up to the given number n, checks if they are divisors of n, and if so, calculates the number of their divisors using the inner count_divisors() function. The individual counts are then summed up to get the final result.

Method 2: Sieve of Eratosthenes Concept

Using the Sieve of Eratosthenes algorithm, we can improve on the brute force approach by determining the divisor counts for all numbers up to n in a more efficient manner, leveraging precomputed information.

Here’s an example:

def sum_of_divisors_sieve(n):
    divisor_counts = [0] * (n + 1)
    for i in range(1, n + 1):
        for j in range(i, n + 1, i):
            divisor_counts[j] += 1
    return sum(divisor_counts[i] for i in range(1, n + 1) if n % i == 0)

# Example usage:
print(sum_of_divisors_sieve(10))

The output of this code is:

In this implementation, divisor_counts stores the number of divisors for all numbers up to n. We then sum up the counts only for divisors of n, which results in a performance improvement over the brute force method, particularly for larger values of n.

Method 3: Using Prime Factorization

For each divisor d of n, the number of divisors can be determined using its prime factorization. It reduces the problem to prime factor computation, which is faster than counting divisors for each number individually.

Here’s an example:

from math import prod

def prime_factors(n):
    i = 2
    factors = {}
    while i * i  1:
        factors[n] = factors.get(n, 0) + 1
    return factors

def sum_of_divisors_prime(n):
    factors = prime_factors(n)
    # For a number n with prime factorization n = p1^e1 * ... * pk^ek,
    # where pi are prime factors and ei are their respective exponents,
    # The number of divisors is (e1 + 1) * ... * (ek + 1).
    return prod(e + 1 for e in factors.values())

# Example usage:
print(sum_of_divisors_prime(10))

The output of this code is:

The function sum_of_divisors_prime() uses a helper function prime_factors() to compute the prime factors of a number n and their exponents. It then calculates the number of divisors by multiplying the increments of the exponents. This method is significantly more efficient for large numbers because it does not require a complete enumeration of all divisors.

Method 4: Optimized Prime Factorization with Precomputation

Combining the prime factorization method with precomputation further enhances its efficiency. By precomputing prime factor counts for a range of numbers, we can retrieve the number of divisors for each divisor of n more swiftly.

Here’s an example:

def optimized_prime_factors(n):
    # Similar to Method 3 but with precomputation optimization.
    pass

# This is a placeholder function that would be implemented similarly to Method 3,
# with the addition of some form of caching or sieve-based precomputation for efficiency.

# Example usage:
# This would be called in a similar manner to the earlier examples, presuming the optimized_prime_factors implemented.

As the code is a conceptual enhancement of Method 3, no specific implementation is given here. However, the idea is to use a precomputed array or dictionary that maps each integer in a range to its number of divisors, which can be populated efficiently using a modified Sieve of Eratosthenes or dynamic programming approach.

Bonus One-Liner Method 5: List Comprehension with Function

For the pythonic aficionado, here’s a one-liner that combines list comprehension with a divisor counting function to provide a succinct though potentially less readable solution.

Here’s an example:

print(sum(len([1 for divisor in range(1, divisor+1) if n % divisor == 0]) for divisor in range(1, n+1) if n % divisor == 0))

The output for n=10 is:

This one-liner nests a list comprehension inside a sum() function call, iterating through divisors of n and counting their divisors inline. It’s concise, but the nested comprehension can be confusing and may sacrifice readability and performance for brevity.

Summary/Discussion

Method 1: Brute Force. Simple to understand. Inefficient for large numbers.
Method 2: Sieve Concept. More efficient than brute force. Still not optimal for very large numbers.
Method 3: Prime Factorization. Efficient for individual large numbers. Requires an optimized prime factor computation.
Method 4: Optimized Prime Factorization. Potentially the most efficient approach. Requires precomputation which may take additional memory and time upfront.
Method 5: List Comprehension One-Liner. Clever and concise. Can be inefficient and hard to read.