wordly-guesser.py

'''
Straregy: for the first guess, find the 5 most common letters and construct
a word with those. Most common letters have most chance of revealing _some_
information about the target word.

With around 6100 five letter words taken from the Linux wordlist, guessing
'esaro' reduces the search space by 5x, on average, so to 1200. Repeating the
process 6 times (including constructing a letter frequency counter) will likely
yield a result, considering that: 6100 / (5^6) < 1

TODO: apply it recursively :)
'''

import random
from collections import Counter
from string import ascii_lowercase

legit_five_letter_words = []
with open('/usr/share/dict/words') as f:
    for word in f:
        word = word.lower().replace('\n', '')
        if len(word) != 5:
            continue
        no_funny_letters = all(letter in ascii_lowercase for letter in word)
        if no_funny_letters:
            legit_five_letter_words.append(word)
dictionary = legit_five_letter_words

def get_clues(guess, chosen_word):
    clues = []
    for i, letter in enumerate(guess):
        if letter == chosen_word[i]:
            clues.append('Y')
        elif letter in chosen_word:
            clues.append('P')
        else:
            clues.append('.')
    return ''.join(clues)

def filter_by_clue(dictionary, guess, clue):
    reduced_dictionary = []
    for dict_word in dictionary:
        possible_word = True
        for i, hint in enumerate(clue):
            clue_letter = guess[i]
            if hint == 'Y' and clue_letter != dict_word[i]:
                possible_word = False
            elif hint == 'P' and clue_letter not in dict_word:
                possible_word = False
        if possible_word:
            reduced_dictionary.append(dict_word)
    return reduced_dictionary

reductions = []
for _ in range(1000):
    chosen_word = random.choice(dictionary)
    guess = 'esaro'
    clues = get_clues(guess, chosen_word)
    filtered_dict = filter_by_clue(dictionary, guess, clues)
    reduction_in_size_percent = 100 * (len(filtered_dict) / len(dictionary))
    reductions.append(reduction_in_size_percent)


print('average reduction', sum(reductions) / len(reductions))