project_euler.problem_074.sol1
==============================

.. py:module:: project_euler.problem_074.sol1

.. autoapi-nested-parse::

   Project Euler Problem 74: https://projecteuler.net/problem=74

   The number 145 is well known for the property that the sum of the factorial of its
   digits is equal to 145:

   1! + 4! + 5! = 1 + 24 + 120 = 145

   Perhaps less well known is 169, in that it produces the longest chain of numbers that
   link back to 169; it turns out that there are only three such loops that exist:

   169 → 363601 → 1454 → 169
   871 → 45361 → 871
   872 → 45362 → 872

   It is not difficult to prove that EVERY starting number will eventually get stuck in
   a loop. For example,

   69 → 363600 → 1454 → 169 → 363601 (→ 1454)
   78 → 45360 → 871 → 45361 (→ 871)
   540 → 145 (→ 145)

   Starting with 69 produces a chain of five non-repeating terms, but the longest
   non-repeating chain with a starting number below one million is sixty terms.

   How many chains, with a starting number below one million, contain exactly sixty
   non-repeating terms?



Attributes
----------

.. autoapisummary::

   project_euler.problem_074.sol1.CACHE_SUM_DIGIT_FACTORIALS
   project_euler.problem_074.sol1.CHAIN_LENGTH_CACHE
   project_euler.problem_074.sol1.DIGIT_FACTORIALS


Functions
---------

.. autoapisummary::

   project_euler.problem_074.sol1.chain_length
   project_euler.problem_074.sol1.solution
   project_euler.problem_074.sol1.sum_digit_factorials


Module Contents
---------------

.. py:function:: chain_length(n: int, previous: set | None = None) -> int

   Calculate the length of the chain of non-repeating terms starting with n.
   Previous is a set containing the previous member of the chain.
   >>> chain_length(10101)
   11
   >>> chain_length(555)
   20
   >>> chain_length(178924)
   39


.. py:function:: solution(num_terms: int = 60, max_start: int = 1000000) -> int

   Return the number of chains with a starting number below one million which
   contain exactly n non-repeating terms.
   >>> solution(10,1000)
   28


.. py:function:: sum_digit_factorials(n: int) -> int

   Return the sum of the factorial of the digits of n.
   >>> sum_digit_factorials(145)
   145
   >>> sum_digit_factorials(45361)
   871
   >>> sum_digit_factorials(540)
   145


.. py:data:: CACHE_SUM_DIGIT_FACTORIALS

.. py:data:: CHAIN_LENGTH_CACHE

.. py:data:: DIGIT_FACTORIALS