Generate words of varying length give n number of characters

Question

I need to generate all possible words of Ki lengths given n characters for example:

given

LNDJOBEAWRL

do bear

I cant come up with the word of len 5 but this is the idea

n = 11
k1 = 2
k2 = 4
k3 = 5 

So basically all words of length 2 4 and 5 but without reusing characters. What would be the best way to do it?

---

My dictionary structure looks likes this:

{
    3: [{u'eit': u' "eit":0'}], 
    5: [{u'doosw': u' "woods": 4601, '}, {u'acenr': u' "caner": 0, '}, {u'acens': u' "canes": 0, '}, {u'acden': u' "caned": 0, '}, {u'aceln': u' "canel": 0,'}], 
    6: [{u'abeill': u' "alible": 0, '}, {u'cdeeit': u' "deciet":0,'}, {u'demoor': u' "mooder": 0, '}], 
    7: [{u'deiprss': u' "spiders": 0, '}, {u'deiprsy': u' "spidery": 0, '}, {u'cersttu': u' "scutter": 0, '}], 
    8: [{u'chiiilst': u' "chilitis": 0, '}, {u'agilnrtw': u' "trawling": 0, '}, {u'abdeemns': u' "beadsmen": 0, '}], 
    9: [{u'abeiilnns': u' "biennials": 0, '}, {u'bclooortu': u' "oblocutor": 0, '}, {u'aabfiinst': u' "fabianist": 0, '}, {u'acdeiituz': u' "diazeutic": 0, '}, {u'aabfiimns': u' "fabianism": 0, '}, {u'ehnoooppt': u' "optophone": 0, '}], 
    10: [{u'aiilnoprtt': u' "tripolitan": 0, '}, {u'eeilprrsty': u' "sperrylite": 0, '}, {u'gghhiilttt': u' "lighttight": 0, '}, {u'aeegilrruz': u' "regularize": 0, '}, {u'ellnprtuuy': u' "purulently": 0, '}], 
    11: [{u'cdgilnoostu': u' "outscolding": 0, '}], 
    12: [{u'ceeeilnostuy': u' "leucosyenite": 0, '}, {u'aacciloprsst': u' "sarcoplastic": 0, '}], 
    13: [{u'acdeimmoprrsu': u' "cardiospermum": 0, '}, {u'celnnooostuvy': u' "noncovetously": 0, '}], 
    14: [{u'adeejmnnoprrtu': u' "preadjournment": 0, '}]
}

And my modified code looks like this:

wlen = self.table[pos]
if pos == 0:
    # See if the letters remaining in the bag are a valid word
    key = ''.join(sorted(bag.elements()))

    for d in wlen:
        if key in d.keys():
            yield solution + [key]
else:
    pos -= 1
    for dic in wlen:
        print(len(dic))
        for key in dic.keys():

Answer 1

First thing is standardize the words such that two words that are anagrams of each other will be handled exactly the same. We can do this by converting to lowercase and sorting the letters of the word. The next step is to distinguish between multiple occurrences of a give letter. To do this, we'll map each letter to a symbol containing the letter, and a number representing its occurrence in the string.

target = "LNDJOBEAWRL".lower()
symbols = sorted([c + str(target[i+1:].count(c)) for i, c in enumerate(target)])

Now that we have a standard representation for each word, we need a quick way to check if any permutations will match them. For this we use the trie datastructure. Here is some starter code for one:

class Trie:
    def __init__(self, symbol):
        self.symbol = symbol
        self.words = []
        self.children = dict()

    def add_word(self, word):
        self.words.append(word)

    def add_child(self, symbol, trie):
        self.children[symbol] = trie

Now you need to make an empty trie as the root, with anything as a symbol, dedicated to holding all top-level tries. Then iterate over each word we transformed earlier, and for the first symbol we produced, check if the root trie has a child with that symbol. If it does not, create a trie for it and add it. If it does, proceed to the next symbol, and check if a trie with that symbol is in the previous trie. Proceed in this fashion until you've exhausted all symbols, in which case the current trie node represents the standardized form for that word we transformed. Store the original word in this trie, and proceed to the next word.

Once that's done, your entire word list will be contained in this trie datastructure. Then, you can just do something like:

def print_words(symbols, node):
    for word in node.words:
        print(word)
    for sym in node.children:
        if sym in symbols:
            print_words(symbols, node.children[sym])

print_words(symbols, root_trie)

To print all words that can be composed from the symbols for the target word.

Answer 2

The code below uses a recursive generator to build solutions. To store the target letters we use a collections.Counter, this acts like a set that permits repeated items.

To simplify searching we create a dictionary for each word length that we want, storing each of those dictionaries in a dictionary called all_words, with the word length as the key. Each sub-dictionary stores lists of words containing the same letters, with the sorted letters as the key, eg 'aet': ['ate', 'eat', 'tea'].

I use the standard Unix '/usr/share/dict/words' word file. If you use a file with a different format you may need to modify the code that puts words into all_words.

The solve function starts its search with the smallest word length, and works up to the largest. That's probably the most efficient order if the set containing the longest words is the biggest, since the final search is performed by doing a simple dict lookup, which is very fast. The previous searches have to test each word in the sub-dictionary of that length, looking for keys that are still in the target bag.

#!/usr/bin/env python3

''' Create anagrams from a string of target letters and a list of word lengths '''

from collections import Counter
from itertools import product

# The Unix word list
fname = '/usr/share/dict/words'

# The target letters to use
target = 'lndjobeawrl'

# Word lengths, in descending order
wordlengths = [5, 4, 2]

# A dict to hold dicts for each word length.
# The inner dicts store lists of words containing the same letters,
# with the sorted letters as the key, eg 'aet': ['ate', 'eat', 'tea']
all_words = {i: {} for i in wordlengths}

# A method that tests if a word only contains letters in target
valid = set(target).issuperset

print('Scanning', fname, 'for valid words...')
count = 0
with open(fname) as f:
    for word in f:
        word = word.rstrip()
        wlen = len(word)
        # Only add words of the correct length, with no punctuation.
        # Using word.islower() eliminates most abbreviations.
        if (wlen in wordlengths and word.islower()
        and word.isalpha() and valid(word)):
            sorted_word = ''.join(sorted(word))
            # Add this word to the list in all_words[wlen],
            # creating the list if it doesn't exist
            all_words[wlen].setdefault(sorted_word, []).append(word)
            count += 1

print(count, 'words found')
for k, v in all_words.items():
    print(k, len(v))
print('\nSolving...')

def solve(pos, bag, solution):
    wlen = wordlengths[pos]
    if pos == 0:
        # See if the letters remaining in the bag are a valid word
        key = ''.join(sorted(bag.elements()))
        if key in all_words[wlen]:
            yield solution + [key]
    else:
        pos -= 1
        for key in all_words[wlen].keys():
            # Test that all letters in key are in the bag
            newbag = bag.copy()
            newbag.subtract(key)
            if all(v >= 0 for v in newbag.values()):
                # Add this key to the current solution and 
                # recurse to find the next key
                yield from solve(pos, newbag, solution + [key])

# Find all lists of keys that produce valid combinations
for solution in solve(len(wordlengths) - 1, Counter(target), []):
    # Convert solutions to tuples of words
    t = [all_words[len(key)][key] for key in solution]
    for s in product(*t):
        print(s)

output

Scanning /usr/share/dict/words for valid words...
300 words found
5 110
4 112
2 11

Solving...
('ad', 'jell', 'brown')
('do', 'jell', 'brawn')
('ow', 'jell', 'brand')
('re', 'jowl', 'bland')

---

FWIW, here are the results for

target = 'nobigword'
wordlengths = [4, 3, 2]

output

Scanning /usr/share/dict/words for valid words...
83 words found
4 31
3 33
2 7

Solving...
('do', 'big', 'worn')
('do', 'bin', 'grow')
('do', 'nib', 'grow')
('do', 'bow', 'grin')
('do', 'bow', 'ring')
('do', 'gin', 'brow')
('do', 'now', 'brig')
('do', 'own', 'brig')
('do', 'won', 'brig')
('do', 'orb', 'wing')
('do', 'rob', 'wing')
('do', 'rib', 'gown')
('do', 'wig', 'born')
('go', 'bid', 'worn')
('go', 'bin', 'word')
('go', 'nib', 'word')
('go', 'bow', 'rind')
('go', 'din', 'brow')
('go', 'now', 'bird')
('go', 'own', 'bird')
('go', 'won', 'bird')
('go', 'orb', 'wind')
('go', 'rob', 'wind')
('go', 'rib', 'down')
('go', 'row', 'bind')
('id', 'bog', 'worn')
('id', 'gob', 'worn')
('id', 'orb', 'gown')
('id', 'rob', 'gown')
('id', 'row', 'bong')
('in', 'bog', 'word')
('in', 'gob', 'word')
('in', 'dog', 'brow')
('in', 'god', 'brow')
('no', 'bid', 'grow')
('on', 'bid', 'grow')
('no', 'big', 'word')
('on', 'big', 'word')
('no', 'bow', 'gird')
('no', 'bow', 'grid')
('on', 'bow', 'gird')
('on', 'bow', 'grid')
('no', 'dig', 'brow')
('on', 'dig', 'brow')
('or', 'bid', 'gown')
('or', 'big', 'down')
('or', 'bog', 'wind')
('or', 'gob', 'wind')
('or', 'bow', 'ding')
('or', 'wig', 'bond')
('ow', 'bog', 'rind')
('ow', 'gob', 'rind')
('ow', 'dig', 'born')
('ow', 'don', 'brig')
('ow', 'nod', 'brig')
('ow', 'orb', 'ding')
('ow', 'rob', 'ding')
('ow', 'rid', 'bong')
('ow', 'rig', 'bond')

---

This code was written for Python 3. You can use it on Python 2.7 but you will need to change

yield from solve(pos, newbag, solution + [key])

to

for result in solve(pos, newbag, solution + [key]):
    yield result