Mercurial > repo
changeset 397:fe852e72f4e2
<elliott> pastelogs markov assumption
author | HackBot |
---|---|
date | Thu, 17 May 2012 14:22:27 +0000 |
parents | 15dec66d8533 |
children | bdb161c7f923 |
files | paste/paste.11442 |
diffstat | 1 files changed, 17 insertions(+), 0 deletions(-) [+] |
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--- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/paste/paste.11442 Thu May 17 14:22:27 2012 +0000 @@ -0,0 +1,17 @@ +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