project_euler.problem_145.sol1
==============================

.. py:module:: project_euler.problem_145.sol1

.. autoapi-nested-parse::

   Project Euler problem 145: https://projecteuler.net/problem=145
   Author: Vineet Rao, Maxim Smolskiy
   Problem statement:

   Some positive integers n have the property that the sum [ n + reverse(n) ]
   consists entirely of odd (decimal) digits.
   For instance, 36 + 63 = 99 and 409 + 904 = 1313.
   We will call such numbers reversible; so 36, 63, 409, and 904 are reversible.
   Leading zeroes are not allowed in either n or reverse(n).

   There are 120 reversible numbers below one-thousand.

   How many reversible numbers are there below one-billion (10^9)?



Attributes
----------

.. autoapisummary::

   project_euler.problem_145.sol1.EVEN_DIGITS
   project_euler.problem_145.sol1.ODD_DIGITS


Functions
---------

.. autoapisummary::

   project_euler.problem_145.sol1.benchmark
   project_euler.problem_145.sol1.reversible_numbers
   project_euler.problem_145.sol1.slow_reversible_numbers
   project_euler.problem_145.sol1.slow_solution
   project_euler.problem_145.sol1.solution


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

.. py:function:: benchmark() -> None

   Benchmarks


.. py:function:: reversible_numbers(remaining_length: int, remainder: int, digits: list[int], length: int) -> int

   Count the number of reversible numbers of given length.
   Iterate over possible digits considering parity of current sum remainder.
   >>> reversible_numbers(1, 0, [0], 1)
   0
   >>> reversible_numbers(2, 0, [0] * 2, 2)
   20
   >>> reversible_numbers(3, 0, [0] * 3, 3)
   100


.. py:function:: slow_reversible_numbers(remaining_length: int, remainder: int, digits: list[int], length: int) -> int

   Count the number of reversible numbers of given length.
   Iterate over possible digits considering parity of current sum remainder.
   >>> slow_reversible_numbers(1, 0, [0], 1)
   0
   >>> slow_reversible_numbers(2, 0, [0] * 2, 2)
   20
   >>> slow_reversible_numbers(3, 0, [0] * 3, 3)
   100


.. py:function:: slow_solution(max_power: int = 9) -> int

   To evaluate the solution, use solution()
   >>> slow_solution(3)
   120
   >>> slow_solution(6)
   18720
   >>> slow_solution(7)
   68720


.. py:function:: solution(max_power: int = 9) -> int

   To evaluate the solution, use solution()
   >>> solution(3)
   120
   >>> solution(6)
   18720
   >>> solution(7)
   68720


.. py:data:: EVEN_DIGITS
   :value: [0, 2, 4, 6, 8]


.. py:data:: ODD_DIGITS
   :value: [1, 3, 5, 7, 9]