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Speech Recognition  Observe: Acoustic signal (A=a 1,…,a n )  Challenge: Find the likely word sequence  But we also have to consider the context Starting.

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Presentation on theme: "Speech Recognition  Observe: Acoustic signal (A=a 1,…,a n )  Challenge: Find the likely word sequence  But we also have to consider the context Starting."— Presentation transcript:

1 Speech Recognition  Observe: Acoustic signal (A=a 1,…,a n )  Challenge: Find the likely word sequence  But we also have to consider the context Starting at this point, we need to be able to model the target language

2 LML Speech Recognition 2008 2 Language Modeling

3  Description: After determining word boundaries, the speech recognition process matches an array of possible word sequences from spoken audio  Issues to consider  determine the intended word sequence  resolve grammatical and pronunciation errors  Implementation: Establish word sequence probabilities  Use existing corpora  Train program with run-time data

4  Problem: Our recognizer translates the audio to a possible string of text. How do we know the translation is correct?  Problem: How do we handle a string of text containing words that are not in the dictionary?  Problem: How do we handle strings with valid words, but which do not form sentences with semantics that makes sense?

5  Problem: Resolving words not in the dictionary  Question: How different is a recognized word from those that are in the dictionary?  Solution: Count the single step transformations necessary to convert one word into another.  Example: caat  cat with removal of one letter  Example: flpc  fireplace requires adding the letters ire after f and a before c and e at the end

6 Simple: *Every word follows every other word w/ equal probability (0-gram) – Assume |V| is the size of the vocabulary – Likelihood of sentence S of length n is = 1/|V| × 1/|V| … × 1/|V| – If English has 100,000 words, probability of each next word is 1/100000 =.00001 Simple vs. Smart Smarter: Probability of each next word is related to word frequency – Likelihood of sentence S = P(w1) × P(w2) × … × P(wn) – Assumes probability of each word is independent of probabilities of other words. Even smarter: Look at probability given previous words – Likelihood of sentence S = P(w1) × P(w2|w1) × … × P(wn|wn-1) – Assumes probability of each word is dependent on probabilities of other words. n times

7  What’s the probability of “canine”?  What’s the probability of “canine tooth” or tooth | canine?  What’s the probability of “canine companion”?  P(tooth|canine) = P(canine & tooth)/P(canine)  Sometimes we can use counts to deduce probabilities.  Example: According to google:  P(canine): occurs 1,750,000 times  P(canine tooth): 6280 times  P(tooth | canine): 6280/1750000 =.0035  P(companion | canine):.01  So companion is the more likely next word after canine Detecting likely word sequences using counts/table look up

8  Limitation: ignores context  We might need to factor in the surrounding words - Use P(need|I) instead of just P(need) - Note: P(new|I) < P(need|I) WordP(O|w)P(w)P(O|w)P(w) new.36.001.00036 neat.52.00013.000068 need.11.00056.000062 knee1.00.000024 P([ni]|new)P(new) P([ni]|neat)P(neat) P([ni]|need)P(need) P([ni]|knee)P(knee) Single word probability  Compute likelihood P([ni]|w), then multiply

9  What is the most likely word sequence? 'botik-'spen-siv'pre-z & ns boatexcessivepresidents baldexpensivepresence boldexpressivepresents boughtinactivepress

10  Conditional Probability P(A 1,A 2 ) = P(A 1 ) · P(A 2 |A 1 )  The Chain Rule generalizes to multiple events  P(A 1, …,A n ) = P(A 1 ) P(A 2 |A 1 ) P(A 3 |A 1,A 2 )…P(A n |A 1 …A n-1 )  Examples:  P(the dog) = P(the) P(dog | the)  P(the dog bites) = P(the) P(dog | the) P(bites| the dog)  Conditional probability applies more than individual relative word frequencies because they consider the context  Dog may be relatively rare word in a corpus  But if we see barking, P(dog|barking) is much more likely 1 n In general, the probability of a complete string of words w 1 …w n is: P(w ) = P(w1)P(w2|w1)P(w3|w1..w2)…P(wn|w1…wn-1) = Detecting likely word sequences using probabilities

11  0 gram: Every word’s likelihood probability is equal  Each word of a 300,000 word corpora has.000033 frequency probabilities  Uni-gram: A word’s likelihood depends on frequency counts  The word, ‘the’ occurs 69,971 in the Brown corpus of 1,000,000 words  Bi-gram: word likelihood determined by the previous word  P(w|a) = P(w) * P(w|w i-1 )  The appears with frequency.07, rabbit appears with frequency.00001  Rabbit is a more likely word that follows the word white than the is  Tri-gram: word likelihood determined by the previous two words  P(w|a) = P(w) * P(w|w i-1 & w i-2 )  N-gram  A model of word or phoneme prediction that uses the previous N-1 words or phonemes to predict the next How many previous words should we consider?

12  Generating sentences: random unigrams...  Every enter now severally so, let  Hill he late speaks; or! a more to leg less first you enter  With bigrams...  What means, sir. I confess she? then all sorts, he is trim, captain.  Why dost stand forth thy canopy, forsooth; he is this palpable hit the King Henry.  Trigrams  Sweet prince, Falstaff shall die.  This shall forbid it should be branded, if renown made it empty.

13  Quadrigrams (Output now is Shakespeare)  What! I will go seek the traitor Gloucester.  Will you not tell me who I am?  Comments  The accuracy of an n-gram model increases with increasing n because word combinations are more and more constrained  Higher n-gram models are more and more sparse. Shakespeare produced 0.04% of 844 million possible bigrams.  There is a tradeoff between accuracy and computational overhead and memory requirements

14 Unigrams (SWB): Most Common: “I”, “and”, “the”, “you”, “a” Rank-100: “she”, “an”, “going” Least Common: “Abraham”, “Alastair”, “Acura” Bigrams (SWB): Most Common: “you know”, “yeah SENT!”, “!SENT um-hum”, “I think” Rank-100: “do it”, “that we”, “don’t think” Least Common:“raw fish”, “moisture content”, “Reagan Bush” Trigrams (SWB): Most Common: “!SENT um-hum SENT!”, “a lot of”, “I don’t know” Rank-100: “it was a”, “you know that” Least Common:“you have parents”, “you seen Brooklyn”

15  Non-word detection (easiest) Example: graffe => (giraffe)  Isolated-word (context-free) error correction  A correction is not possible when the error word is in the dictionary  Context-dependent (hardest)Example: your an idiot => you’re an idiot (the mistyped word happens to be a real word)

16 Mispelled word: acress Candidates – with probabilities of use and use within context Context Context * P(c)

17  Word frequency percentage is not enough  We need p(typo|candidate) * p(candidate)  How likely is the particular error?  Deletion of a t after a c and before an r  Insertion of an a at the beginning  Transpose a c and an a  Substitute a c for an r  Substitute an o for an e  Insert an s before the last s, or after the last s  Context of the word within a sentence or paragraph Misspelled word: accress

18  They are leaving in about fifteen minuets  The study was conducted manly be John Black.  The design an construction of the system will take more than a year.  Hopefully, all with continue smoothly in my absence.  Can they lave him my messages?  I need to notified the bank of….  He is trying to fine out. Spell check without considering context will fail Difficulty: Detecting grammatical errors, or nonsensical expressions

19  Definitions  Maximum likelihood: Finding the most probable sequence of tokens based on the context of the input  N-gram sequence: A sequence of n words whose context speech algorithms consider  Training data: A group of probabilities computed from a corpora of text data  Sparse data problem: How should algorithms handle n-grams that have very low probabilities?  Data sparseness is a frequently occurring problem  Algorithms will make incorrect decisions if it is not handled  Problem 1: Low frequency n-grams  Assume n-gram x occurs twice and n-gram y occurs once  Is x really twice as likely to occur as y?  Problem 2: Zero counts  Probabilities compute to zero for n-grams not seen in the corpora  If n-gram y does not occur, should its probability is zero?

20 An algorithm that redistributes the probability mass Discounting: Reduces probabilities of n- grams with non-zero counts to accommodate the n-grams with zero counts (that are unseen in the corpora). Definition: A corpora is a collection of written or spoken material in machine-readable form

21  The Naïve smoothing technique  Add one to the count of all seen and unseen n-grams  Add the total increased count to the probability mass  Example: uni-grams  Un-smoothed probability for word w: uni-grams  Add-one revised probability for word w:  N = number of words encountered, V = vocabulary size, c(w) = number of times word, w, was encountered

22 P(w n |w n-1 ) = C(w n-1 w n )/C(w n-1 ) P +1 (w n |w n-1 ) = [C(w n-1 w n )+1]/[C(w n-1 )+V] Note: This example assumes bi-gram counts and a vocabulary V = 1616 words Note: row = times that word in column precedes word on left, or starts a sentence Note: C(I)=3437, C(want)=1215, C(to)=3256, C(eat)=938, C(Chinese)=213, C(food)=1506, C(lunch)=459

23 C(W I ) I3437 Want1215 To3256 Eat938 Chinese213 Food1506 Lunch459 c’(w i,w i-1 ) =(c(w i,w i-1 ) i +1) * c(w i,w i-1 ) Original Counts Revised Counts Note: High counts reduce by approximately a third for this example Note: Low counts get larger Note : N = c(w i-1 ), V = vocabulary size = 1616

24  Advantage:  Simple technique to implement and understand  Disadvantages:  Too much probability mass moves to the unseen n-grams  Underestimates the probabilities of the common n-grams  Overestimates probabilities of rare (or unseen) n-grams  Relative smoothing of all unseen n-grams is the same  Relative smoothing of rare n-grams still incorrect  Alternative:  Use a smaller add value  Disadvantage: Does not fully solve this problem

25  Compute the probability of a first time encounter of a new word  Note: Every one of O observed words had a first encounter  How many Unseen words: U = V – O  What is the probability of encountering a new word?  Answer: P( any newly encountered word ) = O/(V+O)  Equally add this probability across all unobserved words  P( any specific newly encountered word ) = 1/U * O/(V+O)  Adjusted counts = V * 1/U*O/(V+O))  Discount each encountered word i to preserve probability space  Probability From: count i /V To: count i /(V+O)  Discounted Counts From: count i To: count i * V/(V+O) Add probability mass to un-encountered words; discount the rest O = observed words, U = words never seen, V = corpus vocabulary words

26  Consider the bi-gram w n w n-1  O(w n-1 ) = number of uniquely observed bi-grams starting with w n-1  V(w n-1 ) = count of bi-grams starting with w n-1  U(w n-1 ) = number of un-observed bi-grams starting with w n-1  Compute probability of a new bi-gram (bi n-1 ) starting with w n-1  Answer: P( any newly encountered bi-gram ) = O(w n-1 )/(V(w n-1 ) +O(w n-1 ))  Note: We observed O(w n-1 ) bi-grams in V(w n-1 )+O(w n-1 ) events  Note: An event is either a bi-gram or a first time encounter  Divide this probability among all unseen bi-grams (new(w n-1 ))  Adjusted P(new(w n-1 )) = 1/U(w n-1 )*O(w n-1 )/(V(w n-1 )+O(w n-1 ))  Adjusted count = V(w n-1 ) * 1/U(w n-1 ) * O(w n-1 )/(V(w n-1 )+O(w n-1 ))  Discount observed bi-grams gram(w n-1 ) to preserve probability space  Probability From: c(w n-1 w n )/V(w n-1 ) To: c(w n-1 w n )/(V(w n-1 ) + O(w n-1 ))  Counts From: c(w n-1 w n ) To: c(w n-1 w n ) * V(w n-1 )/(V(w n-1 )+O(w n-1 )) Add probability mass to un-encountered bi-grams; discount the rest O = observed bi-gram, U = bi-gram never seen, V = corpus vocabulary bi-grams

27 c′(w n,w n-1 )= (c(w n,w n-1 )+1) c(w n,w n-1 ) c′(w n,w n-1 ) = O/U if c(w n, w n-1 )=0 c(w n,w n-1 ) otherwise Original Counts Adjusted Add-One Counts Adjusted Witten-Bell Counts V, O and U values are on the next slide VN V  Note: V, O, U refer to w n-1 counts VN V  VN V 

28 O(w n-1 )U(W n-1 )V(w n-1 ) I951,5213437 Want761,5401215 To1301,4863256 Eat1241,492938 Chinese201,596213 Food821,5341506 Lunch451,571459 O(w n-1 ) = number of observed bi-grams starting with w n-1 V(w n-1 ) = count of bi-grams starting with w n-1 U(w n-1 ) = number of un-observed bi-grams starting with

29  Estimates probability of already encountered grams to compute probabilities for unseen grams  Smaller impact on probabilities of already encountered grams  Generally computes reasonable probabilities

30  The general Concept  Consider the trigram (w n,w n-1, w n-2 )  If c(w n-1, w n-2 ) = 0, consider the ‘back-off’ bi-gram (w n, w n-1 )  If c(w n-1 ) = 0, consider the ‘back-off’ unigram w n  Goal is to use a hierarchy of approximations  trigram > bigram > unigram  Degrade gracefully when higher level grams don’t exist  Given a word sequence fragment: w n-2 w n-1 w n …  Utilize the following preference rule  1.p(w n |w n-2 w n-1 ) if c(w n-2 w n-1 w n )  0  2.  1 p(w n |w n-1 ) if c(w n-1 w n )  0  3.  2 p(w n ) Note:  1 and  2 are values carefully computed to preserve probability mass

31  Goal: Reduce the trainable units that the recognizer needs to process  Approach:  HMMs represent sub-phonetic units  A tree structure Combine sub- phonetic units  Phoneme recognizer searches tree to find HMMs  Nodes partition with questions about neighbors  Performance:  Triphones reduces error rate by:15%  Senones reduces error rate by 24% Definition: A cluster of similar Markov States Is left phone sonorant or nasal? Is right a back-R?Is left s, z, sh, zh? Is left a back-L? Is right voiced?

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