Advent of Code - 2015

This is a solution to Day 4 of Advent of Code 2015.

Day 4 - The Ideal Stocking Stuffer

Santa needs help mining some AdventCoins (very similar to bitcoins) to use as gifts for all the economically forward-thinking little girls and boys.

To do this, he needs to find MD5 hashes which, in hexadecimal, start with at least five zeroes. The input to the MD5 hash is some secret key (your puzzle input, given below) followed by a number in decimal. To mine AdventCoins, you must find Santa the lowest positive number (no leading zeroes: 1, 2, 3, ...) that produces such a hash.

For example:

• If your secret key is abcdef, the answer is 609043, because the MD5 hash of abcdef609043 starts with five zeroes (000001dbbfa...), and it is the lowest such number to do so.
• If your secret key is pqrstuv, the lowest number it combines with to make an MD5 hash starting with five zeroes is 1048970; that is, the MD5 hash of pqrstuv1048970 looks like 000006136ef....

Read input

from utils import read_input

secret_key = read_input(4)[0]

Part 1

I'm doing a very naive solution here: start with 0 and try every number, creating the md5 hash (using built-in hashlib library's md5 function) and seeing if it starts with five zeroes.

import hashlib


def calculate_smallest_number(secret_key, zeroes=5):
    number = 0
    hexa = ''
    while not hexa.startswith('0' * zeroes):
        number += 1
        hexa = hashlib.md5(f'{secret_key}{number}'.encode()).hexdigest()
    return number

result = calculate_smallest_number(secret_key)

print(f'Solution: {result}')
assert result == 254575

Part 2

Now find one that starts with six zeroes.

When the change was this small, I was expecting the number to grow so much it would generate the naive solution too inefficient to run but it wasn't.

So I added an extra argument to the above function and called it with 6 to get the right answer.

result = calculate_smallest_number(secret_key, zeroes=6)

print(f'Solution: {result}')