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1 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.
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2 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.
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3 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
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4 2011-12-23.txt:09:46:31: <fizzie> "säänellaan" -- broken vowel harmony 1, Markov assumption 0.
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5 2012-05-17.txt:14:19:28: <elliott> `pastlog markov assumption 0
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6 2012-05-17.txt:14:20:05: <elliott> `pastlog markov assumption
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7 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.
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8 2012-05-17.txt:14:20:32: <elliott> `pastlog markov assumption
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9 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.
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10 2012-05-17.txt:14:20:43: <elliott> How many things involving the Markov assumption can you say, you speech recognition researcher?
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11 2012-05-17.txt:14:20:45: <elliott> `pastlog markov assumption
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12 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
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13 2012-05-17.txt:14:21:04: <elliott> `pastlog markov assumption
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14 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.
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15 2012-05-17.txt:14:22:01: <elliott> `pastlog markov assumption
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16 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
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17 2012-05-17.txt:14:22:18: <elliott> `pastelogs markov assumption
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