Wolfram Alpha Blog : Yes, That’s Dumb—A Lexicographic Footnote
blog.wolframalpha.com
blog.wolframalpha.com
Screenshot: http://minimalism.linguistics.arizona.edu/~sandiway/wnconnec...
As you can see from the screenshot, it allows you to map all kinds of relationships between words (e.g. taxonomies).
I always found WordNet to be an amazing tool, albeit one that could use some UX work.
Wordnet is one of the sources for the WordData[] function. http://reference.wolfram.com/mathematica/ref/WordData.html
The entire English language must be partitioned into "K" non-overlapping synonym networks. What is the value of "K" ?
I wonder if K = 2, ... and the other synonym network contains "white".
If K > 2, then how many words are in the "white" synonym network, and what other words are not connected with "black" or "white" ?
Whatever K is, it must be relatively small, since there are only so many words in the language, and the synonym networks seem rather large.
More widely, why must there be K non-overlapping networks that cover the entire language? Naively, I'd expect an unbounded number of non-overlapping small networks. But the two ideas of "Basic English" and of specialized vocabularies lead me to posit that every "significant" network (a network with a N number of members, where N > 5,000) overlaps with at least 1 other "significant" network. As a result, I would suspect that it would be difficult or impossible to find K non-overlapping sets that partition the entire language (where K is less than... oh, let's say 100).
You're right, of course: I was using "unbounded" in a loose fashion. There are a finite number of phonemes used in English, and I suppose there must be some upper limit to the number of phonemes in a word.
The networks are "non-overlapping", because of the definition of a synonym network. You start with a seed word, and grow it until it stops growing. The English language is finite, so K must be finite. The example network containing "black" is large, so K is expected to be small, (unless there are a few large networks, and many tiny ones).
Finding K is not difficult at all. Pick a word, "black". Create the synonym network containing "black"and all words linked to it through the "synonym" relationship. Mark those words (roughly 26,000 of them) as "covered". K = 1.
Pick any other word not already covered. Grow the next synonym network. Mark those words covered. K = K + 1
Repeat until there are no words left.
K is the number of synonym networks.