solutions.problem23

Complexity

Total Complexity

5

Size/Duplication

Total Lines	98
Duplicated Lines	0 %

Importance

Changes

0

1 Function

Rating	Name	Duplication	Size	Complexity
A	solve()	0	19	5

"""
Project Euler Problem 23: Non-Abundant Sums
===========================================

.. module:: solutions.problem23
   :synopsis: My solution to problem #23.

The source code for this problem can be
`found here <https://bitbucket.org/nekedome/project-euler/src/master/solutions/problem23.py>`_.

Problem Statement
#################

A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the

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sum of the proper divisors of :math:`28` would be :math:`1 + 2 + 4 + 7 + 14 = 28`, which means that :math:`28` is a

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perfect number.

A number :math:`n` is called deficient if the sum of its proper divisors is less than :math:`n` and it is called

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abundant if this sum exceeds :math:`n`.

As :math:`12` is the smallest abundant number, :math:`1 + 2 + 3 + 4 + 6 = 16`, the smallest number that can be written

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as the sum of two abundant numbers is :math:`24`. By mathematical analysis, it can be shown that all integers greater

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than :math:`28123` can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any

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further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant

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numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

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Solution Discussion
###################

The solution can be found with the following algorithm:

1. Identify all abundant numbers
2. Iterate over all integers in :math:`[2, 28123]` and determine if it is a sum of two abundant numbers, accumulate the

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   sum of all integers that cannot be expressed as such a sum

Both steps can be optimised with the following approaches.

First, build a divisor sum sieve which can be enumerated to determine abundant numbers (i.e. :math:`n` s.t.

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:math:`s(n) \\gt n` where :math:`s(n)` is the sum of the proper divisors of :math:`n`).

.. note:: the use of a divisor sieve will be much quicker than identifying the divisors of each individual :math:`n`.

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Then, iterate through :math:`[2, 28123]` and check whether there are two abundant numbers that can sum to the current

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candidate.

* Let :math:`n` be the current candidate
* Let :math:`a_i` be the :math:`i^{th}` abundant number in :math:`[2, 28123]`
* For all :math:`a_i`, check whether there exists an :math:`a_j` s.t. :math:`a_i + a_j = n`
* If no :math:`a_i, a_j` exist, accumulate :math:`n` into the final answer

Importantly, by ordering :math:`\\lbrace a_i \\rbrace`, we don't need to check all :math:`a_j` at each step of this

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algorithm.

.. math::

    a_i + a_j &= n \\\\
    \\Rightarrow n - a_i &= a_j

Therefore, we only need to check :math:`a_i \\lt n` (otherwise :math:`n - a_i` would be negative) and since :math:`a_i`

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is positive, we only need to check :math:`0 \\lt a_j \\lt n`. This means, that we may consider the subset of

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:math:`\\lbrace a_k \\rbrace` s.t. :math:`a_k \\le n` for each candidate :math:`n`.

Solution Implementation
#######################

.. literalinclude:: ../../solutions/problem23.py
   :language: python
   :lines: 73-
"""

from lib.numbertheory import divisors_sieve


def solve():
    """ Compute the answer to Project Euler's problem #23 """

    upper_bound = 28123

    # Step1: Sieve the sum of proper divisors in [2, upper_bound]
    sum_divisors = list(divisors_sieve(upper_bound + 1, proper=True, aggregate="sum"))

    # Step2: Identify integers not expressible as the sum of two abundant numbers, accumulate their sum

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    abundant_subspace = set()  # will contain only those a_i that need be considered for each n
    answer = 0
    for n in range(1, upper_bound + 1):

The name n does not conform to the variable naming conventions ((([a-z][a-z0-9_]{2,30})|(_[a-z0-9_]*))$).

        if sum_divisors[n - 1] > n:
            abundant_subspace.add(n)
        if not any((n - a in abundant_subspace) for a in abundant_subspace):
            # The any operator provides lazy evaluation, terminating on the first satisfied condition

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            answer += n  # accumulate sum

    return answer


expected_answer = 4179871

The name expected_answer does not conform to the constant naming conventions ((([A-Z_][A-Z0-9_]*)|(__.*__))$).