Ytdyklly

First, please forgive my ignorance, I'm completely new to linguistics.

Given the IPA spelling for word, is it possible to programmatically split it into its sounds? So, for example, given the word "ingredient" and it's IPA spelling "ɪn'gridiənt", is it possible to split it into ["ɪn", "gri", "di", "ənt"]? Is there perhaps a finite list of sounds, at least in the English language, by which an IPA word can be split?

Any help appreciated

This is, in fact, possible! It's not trivial, but it is straightforward.

Your goal seems to be to break an English word (written in phonemic IPA) into syllables. There's a bit of controversy about how useful the concept of a "syllable" is in English, and a few different theories about what exactly a "syllable" is if it does exist, but the following is pretty widely accepted and should be good for your purpose.

First, the theory of syllable structure: every syllable looks something like ONC, where O is the onset, N is the nucleus, and C is the coda. The nucleus is a vowel(*), and always has to be there; the onset and coda are groups of consonants, and aren't required.

Second, the maximal onset principle: we want the onset to be as long as possible. So "tube" is /tub/, but "tuba" is /tu.ba/: the /b/ goes with the second syllable, because that makes the onset bigger.

Third, the syllable structure constraints: certain patterns of consonants aren't allowed together. This one varies by language, but in English, the onset can only be three consonants long at most, and you can't have a stop followed by a fricative in the onset, among many others. So "axle" is /ak.səl/ instead of */a.ksəl/, despite the maximal onset principle. Wikipedia has a good list of these constrains.

So if you want an algorithm for doing this:

• Locate all the nuclei (vowels)


• For each nucleus, work backward, adding sounds to the onset


• If the onset stops being valid, take a step back, then put all the rest of the sounds in the previous syllable's coda

(*) Some analyses of English have syllabic resonants, while others treat them as /ə/ plus resonant. In this answer I'm assuming you're using the version with schwa.