Resolving Word Ambiguities 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 Applications: spell-checking, allophonic variations of pronunciation, automatic speech recognition Implementation: Establish word sequence probabilities – Use existing corpora – Train program with run-time data

Entropy Questions answered if we can compute entropy? How much information there is in a particular grammar, words, parts of speech, phonemes, etc? How predictive is a language in computing the next word based on previous words? How difficult is a speech recognition task? Entropy is measured in bits What is the least number of bits required to encode a piece of information Measures the quantity of information in a signal stream  p i lgp i i  1 r  H(X) = X is a random variable that can assume r values Entropy of spoken languages could focus implementation possibilities

Entropy Example Eight Horses are in an upcoming race –We want to take bets on the winner –Naïve approach is to use 3 bits –Better approach is to use less bits for the horses bet on more frequently Entropy –What is the minimum number of bits needed? = -∑ i=1,8 p(i)log p(i) = - ½ log ½ - ¼ log ¼ - 1/8 log 1/8 – 1/16 log 1/16 – 4 * 1/64 log 1/64 = 1/2 + 2/4 +3/8 + 4/ * 6/64 = ( )/64 = 2 The table to the right shows the optimal coding scheme Question: What if the odd were all equal (1/8) HorseoddsCode 1½0 2¼10 31/ / / / / /

Entropy of words and languages What is the entropy of a sequence of words? H(w 1,w 2,…,w n ) = -∑ p(w i n ) log p (w i n ) – w i n ε L – P(w i n ) = probability that w i n is in a sequence of n words What is the entropy of a word appearing in an n word sequence? H(w 1 n ) = - 1/n ∑ p(w i n ) log p (w i n ) What is the entropy of a language? H(L) = lim n=∞ (- 1/n * ∑ p(w i n ) log p (w i n ))

Cross Entropy We want to know the entropy of a language L, but don’t know its distribution We model L by an approximation to its probability distribution We take sequences of words, phonemes, etc from the real language but use the following formula H(p,m) = lim n->∞ - 1/n log m(w 1,w 2,…,w n ) Cross entropy will always be an upper bound for the actual language Example – Trigram model of 583 million words of English – Corpus of 1,014,312 tokens – Character entropy computed based on a tri-gram grammar – Result 1.75 bits per character

Probability Chain Rule 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

Counts 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/ =.0035 – P(companion | canine):.01 – So companion is the more likely next word after canine Detecting likely word sequences using counts/table look up

Single Word Probabilities WordP(O|w)P(w)P(O|w)P(w) new neat need knee P([ni]|new)P(new) P([ni]|neat)P(neat) P([ni]|need)P(need) P([ni]|knee)P(knee) 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) Single word probability  Compute likelihood P([ni]|w), then multiply

Word Prediction Approaches 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/ = n times 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.

Common Spelling Errors 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: Detect grammatical errors, or nonsensical expressions

N-grams 0 gram: Every word’s likelihood probability is equal –Each word of a 300,000 word corpora has frequency probability Uni-gram: A word’s likelihood depends on frequency counts –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 –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 ) Question: How many previous words should we consider? –Test: Generate random sentences from Shakesphere –Results: Trigram sentences start looking like those of Shakesphere –Tradeoffs: Computational overhead and memory requirements How many previous words should we consider?

The Sparse Data Problem 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?

Smoothing 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

Add-One Smoothing 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

Add-One Smoothing Example 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

Add-One Discounting 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 C(W I ) I3437 Want1215 To3256 Eat938 Chinese213 Food1506 Lunch459

Evaluation of Add-One Smoothing 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

Unigram Witten-Bell Discounting 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

Bi-gram Witten-Bell Discounting 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

Witten-Bell Smoothing 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 

Bi-gram Counts for Example O(w n-1 )U(W n-1 )V(w n-1 ) I951, Want761, To1301, Eat1241, Chinese201, Food821, Lunch451, 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

Evaluation of Witten-Bell Estimates probability of already encountered grams to compute probabilities for unseen grams Smaller impact on probabilities of already encountered grams Generally computes reasonable probabilities

Back-off Discounting 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

N-grams for Spell Checks Non-word detection (easiest) Example: graffe => (giraffe) –Isolated-word (context-free) error correction –by definition cannot correct when error word is a valid word Context-dependent (hardest)Example: your an idiot => you’re an idiot when the mis-typed word happens to be a real word 15% Peterson (1986), 25%-40% Kukich (1992)