Find word sounds like

I have an algorithm that generates strings based on a list of input words. How do I separate only the strings that sounds like English words? ie. discard RDLO while keeping LORD.

EDIT: To clarify, they do not need to be actual words in the dictionary. They just need to sound like English. For example KEAL would be accepted.

You can build a markov-chain of a huge english text.

Afterwards you can feed words into the markov chain and check how high the probability is that the word is english.

See here: http://en.wikipedia.org/wiki/Markov_chain

At the bottom of the page you can see the markov text generator. What you want is exactly the reverse of it.

In a nutshell: The markov-chain stores for each character the probabilities of which next character will follow. You can extend this idea to two or three characters if you have enough memory.

answered Sep 18, 2008 at 12:23

You could approach this by tokenizing a candidate string into bigrams—pairs of adjascent letters—and checking each bigram against a table of English bigram frequencies.

• Simple: if any bigram is sufficiently low on the frequency table (or outright absent), reject the string as implausible. (String contains a «QZ» bigram? Reject!)
• Less simple: calculate the overall plausibility of the whole string in terms of, say, a product of the frequencies of each bigram divided by the mean frequency of a valid English string of that length. This would allow you to both (a) accept a string with an odd low-frequency bigram among otherwise high-frequency bigrams, and (b) reject a string with several individual low-but-not-quite-below-the-threshold bigrams.

Either of those would require some tuning of the threshold(s), the second technique more so than the first.

Doing the same thing with trigrams would likely be more robust, though it’ll also likely lead to a somewhat more strict set of «valid» strings. Whether that’s a win or not depends on your application.

Bigram and trigram tables based on existing research corpora may be available for free or purchase (I didn’t find any freely available but only did a cursory google so far), but you can calculate a bigram or trigram table from yourself from any good-sized corpus of English text. Just crank through each word as a token and tally up each bigram—you might handle this as a hash with a given bigram as the key and an incremented integer counter as the value.

English morphology and English phonetics are (famously!) less than isometric, so this technique might well generate strings that «look» English but present troublesome prounciations. This is another argument for trigrams rather than bigrams—the weirdness produced by analysis of sounds that use several letters in sequence to produce a given phoneme will be reduced if the n-gram spans the whole sound. (Think «plough» or «tsunami», for example.)

answered Sep 18, 2008 at 18:31

answered Sep 18, 2008 at 12:22

You should research «pronounceable» password generators, since they’re trying to accomplish the same task.

A Perl solution would be Crypt::PassGen, which you can train with a dictionary (so you could train it to various languages if you need to). It walks through the dictionary and collects statistics on 1, 2, and 3-letter sequences, then builds new «words» based on relative frequencies.

answered Sep 18, 2008 at 12:44

I’d be tempted to run the soundex algorithm over a dictionary of English words and cache the results, then soundex your candidate string and match against the cache.

Depending on performance requirements, you could work out a distance algorithm for soundex codes and accept strings within a certain tolerance.

Soundex is very easy to implement — see Wikipedia for a description of the algorithm.

An example implementation of what you want to do would be:

def soundex(name, len=4):
    digits = '01230120022455012623010202'
    sndx = ''
    fc = ''

    for c in name.upper():
        if c.isalpha():
            if not fc: fc = c
            d = digits[ord(c)-ord('A')]
            if not sndx or (d != sndx[-1]):
                sndx += d

    sndx = fc + sndx[1:]
    sndx = sndx.replace('0','')
    return (sndx + (len * '0'))[:len]

real_words = load_english_dictionary()
soundex_cache = [ soundex(word) for word in real_words ]

if soundex(candidate) in soundex_cache:
    print "keep"
else:
    print "discard"

Obviously you’ll need to provide an implementation of read_english_dictionary.

EDIT: Your example of «KEAL» will be fine, since it has the same soundex code (K400) as «KEEL». You may need to log rejected words and manually verify them if you want to get an idea of failure rate.

answered Sep 18, 2008 at 12:30

Metaphone and Double Metaphone are similar to SOUNDEX, except they may be tuned more toward your goal than SOUNDEX. They’re designed to «hash» words based on their phonetic «sound», and are good at doing this for the English language (but not so much other languages and proper names).

One thing to keep in mind with all three algorithms is that they’re extremely sensitive to the first letter of your word. For example, if you’re trying to figure out if KEAL is English-sounding, you won’t find a match to REAL because the initial letters are different.

answered Sep 18, 2008 at 12:53

Do they have to be real English words, or just strings that look like they could be English words?

If they just need to look like possible English words you could do some statistical analysis on some real English texts and work out which combinations of letters occur frequently. Once you’ve done that you can throw out strings that are too improbable, although some of them may be real words.

Or you could just use a dictionary and reject words that aren’t in it (with some allowances for plurals and other variations).

answered Sep 18, 2008 at 12:25

answered Sep 18, 2008 at 12:22

answered Sep 18, 2008 at 12:26

I would probably evaluate each word using a SOUNDEX algorithm against a database of english words. If you’re doing this on a SQL-server it should be pretty easy to setup a database containing a list of most english words (using a freely available dictionary), and MSSQL server has SOUNDEX implemented as an available search-algorithm.

Obviously you can implement this yourself if you want, in any language — but it might be quite a task.

This way you’d get an evaluation of how much each word sounds like an existing english word, if any, and you could setup some limits for how low you’d want to accept results. You’d probably want to consider how to combine results for multiple words, and you would probably tweak the acceptance-limits based on testing.

answered Sep 18, 2008 at 12:32

I’d suggest a few simple rules and standard pairs and triplets would be good.

For example, english sounding words tend to follow the pattern of vowel-consonant-vowel, apart from some dipthongs and standard consonant pairs (e.g. th, ie and ei, oo, tr). With a system like that you should strip out almost all words that don’t sound like they could be english. You’d find on closer inspection that you will probably strip out a lot of words that do sound like english as well, but you can then start adding rules that allow for a wider range of words and ‘train’ your algorithm manually.

You won’t remove all false negatives (e.g. I don’t think you could manage to come up with a rule to include ‘rythm’ without explicitly coding in that rythm is a word) but it will provide a method of filtering.

I’m also assuming that you want strings that could be english words (they sound reasonable when pronounced) rather than strings that are definitely words with an english meaning.

answered Sep 18, 2008 at 12:35