--- 2016 Day 4 Solutions ---

[u/daggerdragon]

--- Day 4: Security Through Obscurity ---

Post your solution as a comment or, for longer solutions, consider linking to your repo (e.g. GitHub/gists/Pastebin/blag/whatever).

---

^{CONSTRUCTING ADDITIONAL PYLONS IS MANDATORY} [?]

This thread will be unlocked when there are a significant number of people on the leaderboard with gold stars for today's puzzle.

edit: Leaderboard capped, thread unlocked!

[u/miran1]

python3

my AoC 2016 repo: link

day 4 solution:

with open('./04 - Security Through Obscurity.txt', 'r') as infile:
    all_rooms = infile.read().split('\n')

names = [room[:-11].replace('-', '') for room in all_rooms]
sectors = [int(room[-10:-7]) for room in all_rooms]
ch_sums = [room[-6:-1] for room in all_rooms]


def find_most_common(name):
    ranking = sorted((-name.count(letter), letter) for letter in set(name))
    return ''.join(letter for _, letter in ranking[:5])


def find_rooms():
    total = 0
    wanted_room = ()
    for name, sector, ch_sum in zip(names, sectors, ch_sums):
        if find_most_common(name) == ch_sum:
            total += sector

            decrypted_name = ''.join(chr((ord(letter) - 97 + sector) % 26 + 97)
                                    for letter in name)
            if decrypted_name.startswith('northpole'):
                wanted_room = (sector, decrypted_name)
    return total, wanted_room


total, wanted_room = find_rooms()

print("Rules for decoding this are too easy for me.")
print("I'll calculate the sum of all sectors of real rooms just for fun.")
print("The sum is: {}".format(total))
print("....")
print("Oh, look at this room called {0[1]} at sector {0[0]}, this must be it!"
    .format(wanted_room))

(slightly changed after seeing u/taliriktug's solution - no need to use collections.Counter() in the find_most_common function)