Small script to give you the longest words you can play in Letterpress for iOS. Used it for demonstration purposes. Cheating spoils the fun.

letterpress.py

from collections import Counter


class Letterpress(object):

    DICTIONARY = "Words/Words/en.txt"

    def __init__(self, available):
        """Inialize letterpress object"""
        self.words = self.build_words()
        self.available = Counter(available.lower())

    def build_words(self):
        """Build list of words"""
        with open(self.DICTIONARY) as f:
            words = f.read().splitlines()
        filtered_words = [word for word in words if self.valid_word(word)]
        return filtered_words

    def valid_word(self, word):
        """Discard short words and proper nouns"""
        return len(word) > 3 and word[0].islower()

    def can_form(self, word):
        """Define words that can be formed with the available characters"""
        word = Counter(word)
        return all(word[k] <= self.available[k] for k in word)

    def solve(self, count=200):
        """Search all possible words that can be formed and sort them by length"""
        possible_words = [word for word in self.words if self.can_form(word)]
        possible_words.sort(key=len, reverse=True)
        return possible_words[:count]

    def solve_filtered(self, letters, count=200):
        """Search for words with the given characters"""
        return [word for word in self.solve() if
                all(c in word for c in list(letters.lower()))]


if __name__ == "__main__":
    from sys import argv

    letterpress = Letterpress(argv[1])

    print "Solution for board: " + argv[1].upper()
    if len(argv) > 2:
        print letterpress.solve_filtered(argv[2])
    else:
        print letterpress.solve()


# Run it with python letterpress.py ABCDEABCDEABCDE
# Optionally provide a second parameter to filter words containing specific letters

The official Letterpress dictionary can now be found at https://github.com/atebits/Words