My daughter recently challenged me to an instant messenging Anagrams game where the goal is to generate the most words using a starting set of letters. Each word scores points and longer words are worth more points.
Somehow, I managed to lose. Inconceivable. Which brings us to some words of wisdom:
Old age and treachery will always defeat youth and skill.
English Proverb
Below is a snippet of Python code that identifies words in the Unix word dictionary file (available on Mac OS X as well) that can be created with a given set of letters. The words are returned from longest to shortest, to gain the most points as quickly as possible.
# Return words that can be constructed with a set of letters
# - uses the words list available in most Unix systems, including Mac OS X
# Define constants
wordfile = '/usr/share/dict/words' # source of wordlist
min_word_length = 4
# Create dictionary of how many of each available letter
starting = input("Letters available: ")
avail_letters = {} # dictionary with key = letter and value = # available
for letter in starting.strip().lower():
avail_letters[letter] = avail_letters.get(letter, 0) + 1 # default 0
# Build list of possible words
words = []
with open(wordfile) as f:
for line in f:
word = line.strip().lower()
success = True
if len(word) < min_word_length: # word too short, skip it
continue # go to next word in file
for letter in set(word):
if avail_letters.get(letter, 0) < word.count(letter): # not enough letters
success = False
break
if success:
words.append(word) # add to found word list
if len(words) >= 100: # don't return more than 100 words
break
# Return nested sort of unique words
# - by longest first, and alphabetically within same length
words = sorted(sorted(list(set(words))), key = lambda x: -len(x))
for word in words:
print(len(word), word)Sample Output
Letters available: hutocs
6 schout
6 scouth
5 chous
5 couth
5 hocus
5 scout
5 shout
5 socht
5 south
5 thuoc
5 touch
4 chou
4 chut
4 cosh
4 cost
4 coth
4 cush
4 host
4 huso
4 ocht
4 otus
4 ouch
4 oustThe result
