Advent of Code 2015 Day 11

— Day 11: Corporate Policy —

Santa’s previous password expired, and he needs help choosing a new one.

Given Santa’s current password (your puzzle input), what should his next password be?

import re
import string

with open('data-2015-11.txt', 'r') as f:
    inp = f.read()
    
inp = list(inp)

Set up the requirements

def req1(st):
    # must contain three consecutive letters
    return any([q for q in seq if q in st])
  
def req2(st):
    # must not contain i, o, l
    # I think we can set this up as part of iteration
    pass
  
def req3(st):
    # must contain two double letters
    regex = r'([a-z])\1.*([a-z])\2'
    if re.search(regex, st, re.IGNORECASE):
        return True
    else: 
        return False

Set up our letters

# use this handy dandy string
letters = list(string.ascii_lowercase)

# create sequence of consecutive letters
seq = [letters[n] + letters[n+1] + letters[n+2] for n in range(len(letters) - 2)]
# these cannot be in our answer so remove from sequence
seq = [m for m in seq if 'i' not in m and 'l' not in m and 'o' not in m]

# remove from the list for iteration purposes
letters.remove('i')
letters.remove('l')
letters.remove('o')

# one letter goes on to the next, this seems like a good idea
vals = letters[1:] + [letters[0]]
# usage: password[i] = next[password[i]]
next = dict(zip(letters, vals))

There’s probably some nice way to do this, but let’s make a big nested loop

a = 0
# change the input to part one answer
inp = list('hxbxxzaa')


while not (req1(''.join(inp)) and req3(''.join(inp))): 
    a += 1
    # print(inp)
    if inp[-1] == 'z':
        if inp[-2] == 'z':
            if inp[-3] == 'z':
                if inp[-4] == 'z':
                    if inp[-5] == 'z':
                        if inp[-6] == 'z':
                            if inp[-7] == 'z':
                                inp[-8] = next[inp[-8]]    
                            inp[-7] = next[inp[-7]]       
                        inp[-6] = next[inp[-6]]        
                    inp[-5] = next[inp[-5]]
                inp[-4] = next[inp[-4]]
            inp[-3] = next[inp[-3]]
        inp[-2] = next[inp[-2]]
    inp[-1] = next[inp[-1]]
    
    # if a > 100000:
    #     break
      
    

a
''.join(inp)

'hxcaabcc'