Processing/Speech, NLP and the Web

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1 CS460/626 : Natural Language Processing/Speech, NLP and the Web (Lecture 25 Probabilistic Parsing) Pushpak Bhattacharyya CSE Dept., IIT Bombay 14 th March, 2011

2 Bracketed Structure: Treebank Corpus [ S1[ S[ S[ VP[ VBCome][ NP[ NNPJuly]]]] [,,] [ CC and] [ S [ NP [ DT the] [ JJ IIT] [ NN campus]] [ VP [ AUX is] [ ADJP [ JJ abuzz] [ PP [ IN with] [ NP [ ADJP [ JJ new] [ CC and] [ VBG returning]] [ NNS students]]]]]] [..]]]

3 Noisy Channel Modeling Source sentence Noisy Channel Target parse T*= argmax [P(T S)] T = argmax [P(T).P(S T)] ( )] T = argmax [P(T)], since given the parse the T sentence is completely determined and P(S T)=1

4 Corpus A collection of text called corpus, is used for collecting various language data With annotation: more information, but manual labor intensive Practice: label automatically; correct manually The famous Brown Corpus contains 1 million tagged words. Switchboard: very famous corpora 2400 conversations, 543 speakers, many US dialects, annotated with orthography and phonetics

5 Discriminative vs. Generative Model W * = argmax (P(W SS)) W Discriminativei i i Model Generative Model Compute directly from P(W SS) Compute from P(W).P(SS W)

6 Language Models N-grams: sequence of n consecutive words/characters Probabilistic / Stochastic Context Free Grammars: Simple probabilistic models capable of handling recursion A CFG with probabilities attached to rules Rule probabilities how likely is it that a particular rewrite rule is used?

7 PCFGs Why PCFGs? Intuitive probabilistic models for tree-structured languages Algorithms are extensions of HMM algorithms Better than the n-gram model for language modeling.

8 Formal Definition of PCFG A PCFG consists of A set of terminals {w k }, k = 1,.,V {w k } = { child, teddy, bear, played } A set of non-terminals {N i }, i = 1,,n {N i } = { NP, VP, DT } A designated start symbol N 1 A set of rules {N i ζ j }, where ζ j is a sequence of terminals & non-terminals NP DT NN A corresponding set of rule probabilities bili i

9 Rule Probabilities Rule probabilities are such that i j i P(N ζ ) = 1 i Eg E.g., P( NP DT NN) = 0.2 P( NP NN) = 0.5 P( NP NP PP) = P( NP DT NN) = 0.2 Means 20 % of the training data parses Means 20 % of the training data parses use the rule NP DT NN

10 Probabilistic Context Free Grammars S NP VP DT the NP DT NN 0.5 NN gunman 0.5 NP NNS NN building 0.5 NP NP PP 0.2 VBD sprayed 1.0 PP PNP 1.0 NNS bullets 1.0 VP VP PP 0.6 VP VBD NP 0.4

11 Example Parse t 1` The gunman sprayed the building with bullets. S 1.0 NP 0.5 VP 0.6 P (t 1 ) = 1.0 * 0.5 * 1.0 * 0.5 * 0.6 * 0.4 * 1.0 * 0.5 * 1.0 * 0.5 * 1.0 * 1.0 * 0.3 * 1.0 = DT 1.0 NN VP 0.4 PP 1.0 The gunman VBD 1.0 NP 0.5 P 1.0 NP 0.3 sprayed DT 1.0 NN 0.5 with NNS 1.0 the building bullets

12 Another Parse t 2 The gunman sprayed the building with bullets. S 1.0 P (t 2 ) = 1.0 * 0.5 * 1.0 * 0.5 * NP 0.5 VP 0.4 DT 1.0 NN 0.5 VBD 1.0 NP * 1.0 * 0.2 * 0.5 * 1.0 * 0.5 * 1.0 * 1.0 * 0.3 * 1.0 = The gunman sprayed NP 0.5 PP 1.0 DT 1.0 NN 0.5 P 1.0 NP 0.3 th building with NNS 1.0 e bullet s

13 Probability of a sentence Notation : w ab subsequence w a.w b N j NP N j dominates w a.w b or yield(n j ) = w a.w b w a..w b the..sweet..teddy..be Probability of a sentence = P(w 1m) ) Pw ( ) = Pw (, t) 1m 1m t = PtPw ( ) ( t) t 1m Where t is a parse tree of the sentence = Pt () Q Pw ( 1m t ) = 1 t: yield ( t) = w1 m If t is a parse tree for the sentence w 1m, this will be 1!!

14 Assumptions of the PCFG model Place invariance : P(NP DT NN) is same in locations 1 and 2 Context-free : P(NP DT NN anything outside The child ) = P(NP DT NN) Ancestor free : At 2, P(NP DT NN its ancestor is VP) =P(NP DT NN) 1 NP The child S VP 2 NP The toy

15 Probability of a parse tree Domination :We say N j dominates from k to l, symbolized as, if W k,l is derived from N j P (tree sentence) = P (tree S 1l 1,l ) where S 1,l means that the start symbol S dominates the word sequence W 1,l P (t s) approximately equals joint probability of constituent non-terminals dominating the sentence fragments (next slide)

16 Probability of a parse tree (cont.) S 1,l NP 1,2 VP 3,l P ( t s ) = P (t S 1,l ) DT 1 N 2 V 3,3 PP 4,l = P ( NP 1,2, DT 1,1, w 1, w w P NP 1 2 w 3 4,4 5,l N 2,2, w 2, VP 3,l, V 3,3, w 3, PP 4,l, P 4,4, w 4, NP 5,l, w 5 l S 1,l ) w 4 w 5 w l = P ( NP 1,2, VP 3,l S 1,l ) * P ( DT 1,1, N 2,2 NP 1,2 ) * D(w 1 DT 1,1 ) * P( (w 2 N 2,2 )*P(V 3,3, PP 4,l VP 3,l )*P( P(w 3 V 3,3 )*P(P P 4,4, NP 5,l PP 4,l ) * P(w 4 P 4,4 ) * P (w 5 l NP 5,l ) (Using Chain Rule, Context Freeness and Ancestor Freeness )

17 HMM PCFG O observed sequence w 1m sentence X state sequence t parse tree μ model G grammar Three fundamental questions

18 HMM PCFG How likely is a certain observation given the model? How likely l is a sentence given the grammar? PO ( μ ) Pw ( G ) How to choose a state sequence which best explains the observations? How to choose a parse which best supports the sentence? 1m arg max PX ( Oμ, ) arg max P( t w1 m, G) X t

19 HMM PCFG How to choose the model parameters that best explain the observed data? How to choose rule probabilities which maximize the probabilities of the observed sentences? arg max PO ( ) μ μ P w1 m G arg max ( G)

20 Interesting Probabilities N 1 What is the probability of having a NP at this position such that it will derive the building? - NP (4,5) NP Inside Probabilities The gunman sprayed the building with bullets Outside Probabilities biliti What is the probability of starting from N 1 and deriving The gunman sprayed, a NP and with bullets? - α NP (4,5)

21 Interesting Probabilities Random variables to be considered The non-terminal being expanded. E.g., NP The word-span covered by the non-terminal. E.g., (4,5) refers to words the building While calculating gp probabilities, consider: The rule to be used for expansion : E.g., NP DT NN The probabilities associated with the RHS nonterminals : E.g., DT subtree s inside/outside probabilities & NN subtree s inside/outside probabilities

22 Outside Probability α j(p,q) : The probability of beginning with N 1 & generating the non-terminal N j pq and all words outside w p..w q ( p, q) P( w, N, w G) j α j = 1( p 1) pq ( q+ 1) m N 1 N j w 1 w p-1 w p w q w q+1 w m

23 Inside Probabilities j (p,q) : The probability of generating the words w p..w q starting with the non-terminal N j pq. ( pq, ) = Pw ( N j, G) j pq pq N 1 α N j w 1 w p-1 w p w q w q+1 w m

24 Outside & Inside Probabilities: example α = NP (4,5) for "the building" P (The gunman sprayed, NP,with bullets G) (4,5) for "the building" = (the building, ) NP P NP4,5 G N 1 4,5 NP The gunman sprayed the building with bullets

25 Calculating Inside probabilities j (p,q) Base case: j j ( kk, ) = Pw ( N, G ) = PN ( w G ) j k kk kk k Base case is used for rules which derive the words or terminals directly Eg E.g., Suppose N j = NN is being considered & NN building is one of the rules with probability 0.5 (5,5) = (, ) NN Pbuilding NN5,5 G = PNN ( bildi building G ) = 0.5 5,5

26 Induction Step: Assuming Grammar in Chomsky Normal Form Induction step : ( pq, ) = Pw ( N, G) N j j j pq pq q 1 j r s N r N s, = PN ( NN)* rs d= p r ( pd, )* w p w d w d+1 w q s ( d + 1, q) Consider different splits of the words - indicated by d E.g., the huge building Split here for d=2 d=3 Consider different non-terminals to be used in the rule: NP DT NN, NP DT NNS are available options Consider summation over all these.

27 The Bottom-Up Approach The idea of induction Consider the gunman NP Base cases : Apply unary rules DT the Prob = 1.0 NN gunman Prob = 0.5 DT 1.0 NN 0.5 The gunman Induction : Prob that a NP covers these 2 words = P (NP DT NN) * P (DT deriving the word the ) * P (NN deriving the word gunman ) = 0.5 * 1.0 * 0.5 = 0.25

28 Parse Triangle A parse triangle is constructed cted for calculating lating j (p,q) Probability of a sentence using j (p,q): P( w G) = P( N w G) = P( w N, G) = (1, m) 1 1 1m 1m 1m 1m 1

29 Parse Triangle The gunman sprayed the building with bullets (1) (2) (3) (4) (5) (6) (7) 1 DT = = NN VBD = 1.0 = 1.0 DT NN = 0.5 P = 1.0 NNS = Fill diagonals with j ( kk, )

30 Parse Triangle The gunman sprayed the building with bullets (1) (2) (3) (4) (5) (6) (7) = 1.0 DT NP = 0.25 NN = 0.5 = VBD 4 DT = = Calculate l using induction formula NP = P NP1,2 G (1,2) (the gunman, ) = PNP ( DTNN)* DT (1,1) 1) * NN (2, 2) = 0.5*1.0*0.5 = 0.25 NN P = 1.0 NNS = 1.0

31 Example Parse t 1 The gunman sprayed the building with bullets. S 1.0 NP 0.5 VP 0.6 DT 1.0 NN 0.5 VP 0.4 PP 1.0 Rule used here is VP VP PP The gunman VBD 1.0 NP 0.5 P 1.0 NP 0.3 DT 1.0 NN 0.5 with NNS sprayed 1.0 th e building bullet s

32 Another Parse t 2 The gunman sprayed the building with bullets. S 1.0 Rule used here is NP 0.5 VP 0.4 VP VBD NP DT 1.0 NN 0.5 VBD 1.0 NP 0.2 The gunmansprayed NP 0.5 PP 1.0 DT 1.0 NN 0.5 P 1.0 NP 0.3 th building with NNS e bullet s

33 Parse Triangle The (1) gunman (2) sprayed (3) the (4) building (5) with (6) bullets (7) 1 DT = NP = S = NN = VBD = 1.0 VP = 1.0 VP = DT = 1.0 NP = 0.25 NP = NN = P = PP = NNS = 1.0 (3,7) = (sprayed the building with bullets, ) VP P VP37 3,7 G = PVP ( VPPP)* VP (3,5)* PP (6,7) + P ( VP VBD NP)* VBD (3,3)* NP (4,7) = 0.6*1.0* *1.0*0.015 = 0.186

34 Different Parses Consider Different splitting points : E.g., 5th and 3 rd position Using different rules for VP expansion : E.g., VP VP PP, VP VBD NP Different parses for the VP sprayed the building with bullets can be constructed this way.

35 Outside Probabilities α j (p,q) Base case: α (1, m) = 1 for start symbol 1 α j (1, m) = 0 for j 1 Inductive step for calculating α j ( pq, : ) N 1 N f pe α f ( pe, ) PN ( f NN j g ) N j pq N g (q+1) e ( q + g 1, e ) w 1 w p-1 w p w q w q+1 w e w e+1 w m Summation over f, g & e

36 Probability of a Sentence Pw (, N G) = Pw ( N, G) = α ( pq, ) ( pq, ) j 1m pq 1m pq j j j j Joint probability of a sentence w 1m and that there is a constituent spanning words w p to w q is given as: P (The gunman...bullets, N G) N 1 NP 45 4,5 = P(The gunman...bullets N, G) j = α NP j 4,5 (4,5) NP (4,5) + α (4,5) (4,5) +... VP VP The gunman sprayed the building with bullets