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@wo0dyn
Last active Aug 9, 2018
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Math Puzzle

One possible solution: brute force

import collections
import itertools


def check(numbers, code):
    """
    Compute correct numbers and numbers in the wrong position.

    >>> check([1, 2, 3], [1, 2, 3])
    (3, 0)
    >>> check([3, 2, 1], [1, 2, 3])
    (1, 2)
    >>> check([3, 1, 2], [1, 2, 3])
    (0, 3)
    >>> check([4, 5, 6], [1, 2, 3])
    (0, 0)
    """
    
    stats = collections.Counter()
    for index, number in enumerate(numbers):
        if number == code[index]:
            stats['correct'] += 1
        elif number in code:
            stats['wrong-position'] += 1

    return stats['correct'], stats['wrong-position']


for code in itertools.product(range(10), repeat=3):
    # Testing (0, 0, 0), (0, 0, 1), …, (9, 9, 9)
    try:
        assert check(code, [2, 9, 1]) == (1, 0)
        assert check(code, [2, 4, 5]) == (0, 1)
        assert check(code, [4, 6, 3]) == (0, 2)
        assert check(code, [5, 7, 8]) == (0, 0)
        assert check(code, [5, 6, 9]) == (0, 1)
    except AssertionError:
        continue
    else:
        print(f'Code is {code}')
$ python3 puzzle.py
Code is (3, 9, 4)
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