Eg if input string is helloworld I want the output to be like:
do
he
we
low
hell
hold
roll
well
word
hello
lower
world
...
all the way up to the longest word that is an anagram of a substring of helloworld. Like in Scrabble for example. The input string can be any length, but rarely more than 16 chars.
I've done a search and come up with structures like a trie, but I am still unsure of how to actually do this.
The structure used to hold your dictionary of valid entries will have a huge impact on efficiency. Organize it as a tree, root being the singular zero letter "word", the empty string. Each child of root is a single first letter of a possible word, children of those being the second letter of a possible word, etc., with each node marked as to whether it actually forms a word or not.
Your tester function will be recursive. It starts with zero letters, finds from the tree of valid entries that "" isn't a word but it does have children, so you call your tester recursively with your start word (of no letters) appended with each available remaining letter from your input string (which is all of them at that point). Check each one-letter entry in tree, if valid make note; if children, re-call tester function appending each of remaining available letters, and so on.
So for example, if your input string is "helloworld", you're going to first call your recursive tester function with "", passing the remaining available letters "helloworld" as a 2nd parameter. Function sees that "" isn't a word, but child "h" does exist. So it calls itself with "h", and "elloworld". Function sees that "h" isn't a word, but child "e" exists. So it calls itself with "he" and "lloworld". Function sees that "e" is marked, so "he" is a word, take note. Further, child "l" exists, so next call is "hel" with "loworld". It will next find "hell", then "hello", then will have to back out and probably next find "hollow", before backing all the way out to the empty string again and then starting with "e" words next.
I couldn't resist my own implementation. It creates a dictionary by sorting all the letters alphabetically, and mapping them to the words that can be created from them. This is an O(n) start-up operation that eliminates the need to find all permutations. You could implement the dictionary as a trie in another language to attain faster speedups.
The "getAnagrams" command is also an O(n) operation which searches each word in the dictionary to see if it is a subset of the search. Doing getAnagrams("radiotelegraphically")" (a 20 letter word) took approximately 1 second on my laptop, and returned 1496 anagrams.
# Using the 38617 word dictionary at
# http://www.cs.umd.edu/class/fall2008/cmsc433/p5/Usr.Dict.Words.txt
# Usage: getAnagrams("helloworld")
def containsLetters(subword, word):
wordlen = len(word)
subwordlen = len(subword)
if subwordlen > wordlen:
return False
word = list(word)
for c in subword:
try:
index = word.index(c)
except ValueError:
return False
word.pop(index)
return True
def getAnagrams(word):
output = []
for key in mydict.iterkeys():
if containsLetters(key, word):
output.extend(mydict[key])
output.sort(key=len)
return output
f = open("dict.txt")
wordlist = f.readlines()
f.close()
mydict = {}
for word in wordlist:
word = word.rstrip()
temp = list(word)
temp.sort()
letters = ''.join(temp)
if letters in mydict:
mydict[letters].append(word)
else:
mydict[letters] = [word]
An example run:
>>> getAnagrams("helloworld")
>>> ['do', 'he', 'we', 're', 'oh', 'or', 'row', 'hew', 'her', 'hoe', 'woo', 'red', 'dew', 'led', 'doe', 'ode', 'low', 'owl', 'rod', 'old', 'how', 'who', 'rho', 'ore', 'roe', 'owe', 'woe', 'hero', 'wood', 'door', 'odor', 'hold', 'well', 'owed', 'dell', 'dole', 'lewd', 'weld', 'doer', 'redo', 'rode', 'howl', 'hole', 'hell', 'drew', 'word', 'roll', 'wore', 'wool','herd', 'held', 'lore', 'role', 'lord', 'doll', 'hood', 'whore', 'rowed', 'wooed', 'whorl', 'world', 'older', 'dowel', 'horde', 'droll', 'drool', 'dwell', 'holed', 'lower', 'hello', 'wooer', 'rodeo', 'whole', 'hollow', 'howler', 'rolled', 'howled', 'holder', 'hollowed']
The data structure you want is called a Directed Acyclic Word Graph (dawg), and it is described by Andrew Appel and Guy Jacobsen in their paper "The World's Fastest Scrabble Program" which unfortunately they have chosen not to make available free online. An ACM membership or a university library will get it for you.
I have implemented this data structure in at least two languages---it is simple, easy to implement, and very, very fast.
What you want is an implementation of a power set.
Also look at Eric Lipparts blog, he blogged about this very thing a little while back
EDIT:
Here is an implementation I wrote of getting the powerset from a given string...
private IEnumerable<string> GetPowerSet(string letters)
{
char[] letterArray = letters.ToCharArray();
for (int i = 0; i < Math.Pow(2.0, letterArray.Length); i++)
{
StringBuilder sb = new StringBuilder();
for (int j = 0; j < letterArray.Length; j++)
{
int pos = Convert.ToInt32(Math.Pow(2.0, j));
if ((pos & i) == pos)
{
sb.Append(letterArray[j]);
}
}
yield return new string(sb.ToString().ToCharArray().OrderBy(c => c).ToArray());
}
}
This function gives me the powersets of chars that make up the passed in string, I then can use these as keys into a dictionary of anagrams...
Dictionary<string,IEnumerable<string>>
I created my dictionary of anagrams like so... (there are probably more efficient ways, but this was simple and plenty quick enough with the scrabble tournament word list)
wordlist = (from s in fileText.Split(new string[] { Environment.NewLine }, StringSplitOptions.RemoveEmptyEntries)
let k = new string(s.ToCharArray().OrderBy(c => c).ToArray())
group s by k).ToDictionary(o => o.Key, sl => sl.Select(a => a));
A simple-minded approach is to generate all the "substrings" and, for each of them, check whether it's an element of the set of acceptable words. E.g., in Python 2.6:
import itertools
import urllib
def words():
f = urllib.urlopen(
'http://www.cs.umd.edu/class/fall2008/cmsc433/p5/Usr.Dict.Words.txt')
allwords = set(w[:-1] for w in f)
f.close()
return allwords
def substrings(s):
for i in range(2, len(s)+1):
for p in itertools.permutations(s, i):
yield ''.join(p)
def main():
w = words()
print '%d words' % len(w)
ss = set(substrings('weep'))
print '%d substrings' % len(ss)
good = ss & w
print '%d good ones' % len(good)
sgood = sorted(good, key=lambda w:(len(w), w))
for aword in sgood:
print aword
main()
will emit:
38617 words
31 substrings
5 good ones
we
ewe
pew
wee
weep
Of course, as other responses pointed out, organizing your data purposefully can greatly speed-up your runtime -- although the best data organization for a fast anagram finder could well be different... but that will largely depend on the nature of your dictionary of allowed words (a few tens of thousands, like here -- or millions?). Hash-maps and "signatures" (based on sorting the letters in each word) should be considered, as well as tries &c.
I've been playing a lot of Wordfeud on my phone recently and was curious if I could come up with some code to give me a list of possible words. The following code takes your availble source letters (* for a wildcards) and an array with a master list of allowable words (TWL, SOWPODS, etc) and generates a list of matches. It does this by trying to build each word in the master list from your source letters.
I found this topic after writing my code, and it's definitely not as efficient as John Pirie's method or the DAWG algorithm, but it's still pretty quick.
public IList<string> Matches(string sourceLetters, string [] wordList)
{
sourceLetters = sourceLetters.ToUpper();
IList<string> matches = new List<string>();
foreach (string word in wordList)
{
if (WordCanBeBuiltFromSourceLetters(word, sourceLetters))
matches.Add(word);
}
return matches;
}
public bool WordCanBeBuiltFromSourceLetters(string targetWord, string sourceLetters)
{
string builtWord = "";
foreach (char letter in targetWord)
{
int pos = sourceLetters.IndexOf(letter);
if (pos >= 0)
{
builtWord += letter;
sourceLetters = sourceLetters.Remove(pos, 1);
continue;
}
// check for wildcard
pos = sourceLetters.IndexOf("*");
if (pos >= 0)
{
builtWord += letter;
sourceLetters = sourceLetters.Remove(pos, 1);
}
}
return string.Equals(builtWord, targetWord);
}
Like Tim J, Eric Lippert's blog posts where the first thing to come to my mind. I wanted to add that he wrote a follow-up about ways to improve the performance of his first attempt.
