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<wob_jonas> learn Aristotle said that every illness can be cured by balancing the four vitreous humors, and everyone believed him for two thousand years, even though people still died of illnesses. It wasn\'t until the 20th century that Szent-Gy\xc3\xb6rgyi Albert realized that Aristotle didn\'t find fifth kind of vitreous humor, vitamin C, because the Greek alphabet
| author | HackBot |
|---|---|
| date | Mon, 01 Jan 2018 17:57:43 +0000 |
| parents | fe852e72f4e2 |
| children |
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2011-08-26.txt:20:09:35: <fizzie> Given that what you get from an n-gram is (n-1) words of context, I think it's pretty safe bet to say that the Markov assumption (of order n-1) will hold for most things you do with them. 2011-09-26.txt:13:03:19: <fizzie> CakeProphet: Certainly there are different ways to do language models; I just can't offhand figure out how to make a (sensible) language model that would use n-grams but not have the (n-1)-order Markov assumption. 2011-09-26.txt:16:54:56: <fizzie> tehporPekaC: There's an alternative solution which will always hit the target length, and thanks to the Markov assumption really shouldn't affect the distribution of the last characters of a word: when generating a word of length K with trigrams, first generate K-2 characters so that you ignore all "xy " entries. For the penultimate character, only consider such trigrams "xyz" for which any trigram "z? " exists. For the final character, only consider such trigr 2011-12-23.txt:09:46:31: <fizzie> "säänellaan" -- broken vowel harmony 1, Markov assumption 0. 2012-05-17.txt:14:19:28: <elliott> `pastlog markov assumption 0 2012-05-17.txt:14:20:05: <elliott> `pastlog markov assumption 2012-05-17.txt:14:20:16: <HackEgo> 2011-08-26.txt:20:09:35: <fizzie> Given that what you get from an n-gram is (n-1) words of context, I think it's pretty safe bet to say that the Markov assumption (of order n-1) will hold for most things you do with them. 2012-05-17.txt:14:20:32: <elliott> `pastlog markov assumption 2012-05-17.txt:14:20:39: <HackEgo> 2011-08-26.txt:20:09:35: <fizzie> Given that what you get from an n-gram is (n-1) words of context, I think it's pretty safe bet to say that the Markov assumption (of order n-1) will hold for most things you do with them. 2012-05-17.txt:14:20:43: <elliott> How many things involving the Markov assumption can you say, you speech recognition researcher? 2012-05-17.txt:14:20:45: <elliott> `pastlog markov assumption 2012-05-17.txt:14:20:52: <HackEgo> 2011-09-26.txt:16:54:56: <fizzie> tehporPekaC: There's an alternative solution which will always hit the target length, and thanks to the Markov assumption really shouldn't affect the distribution of the last characters of a word: when generating a word of length K with trigrams, first generate K-2 characters so that you ignore all "xy " entries. For the penultimate character, only consider such trigrams "xyz" for 2012-05-17.txt:14:21:04: <elliott> `pastlog markov assumption 2012-05-17.txt:14:21:10: <HackEgo> 2011-09-26.txt:13:03:19: <fizzie> CakeProphet: Certainly there are different ways to do language models; I just can't offhand figure out how to make a (sensible) language model that would use n-grams but not have the (n-1)-order Markov assumption. 2012-05-17.txt:14:22:01: <elliott> `pastlog markov assumption 2012-05-17.txt:14:22:09: <HackEgo> 2011-09-26.txt:16:54:56: <fizzie> tehporPekaC: There's an alternative solution which will always hit the target length, and thanks to the Markov assumption really shouldn't affect the distribution of the last characters of a word: when generating a word of length K with trigrams, first generate K-2 characters so that you ignore all "xy " entries. For the penultimate character, only consider such trigrams "xyz" for 2012-05-17.txt:14:22:18: <elliott> `pastelogs markov assumption