# 684.0222/02/20161 probabilistic context free grammars chris brew ohio state university

DESCRIPTION

/02/20163 An example s -> np vps -> np vp pp np -> det nnp -> np pp vp->v np pp->p np n->girln -> boy n -> parkn -> telescope v-> saw p-> withp -> in Sample sentence: “The boy saw the girl in the park with the telescope”TRANSCRIPT

684.02 05/05/23 1

Probabilistic Context Free Grammars

Chris Brew

Ohio State University

684.02 05/05/23 2

Context Free Grammars HMMs are sophisticated tools for

language modelling based on finite state machines.

Context-free grammars go beyond FSMs They can encode longer range

dependencies than FSMs They too can be made probabilistic

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An examples -> np vp s -> np vp ppnp -> det n np -> np ppvp -> v nppp -> p np

n -> girl n -> boy n -> park n -> telescopev -> sawp -> with p -> in

Sample sentence: “The boy saw the girl in the park with the telescope”

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Multiple analyses 2 of the 5 are

S

NP

DET

the

N

boy

VP

V

saw

NP

NP

DET

the

N

girl

PP

P

in

NP

DET

the

N

park

PP

P

with

NP

DET

the

N

telescope

S

NP

DET

the

N

boy

VP

V

saw

NP

NP

DET

the

N

girl

PP

P

in

NP

NP

DET

the

N

park

PP

P

with

NP

DET

the

N

telescope

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How serious is this ambiguity? Very serious, ambiguities in different

places multiply Easy to get millions of analyses for

simple seeming sentences Maybe we can use probabilities to

disambiguate, just as we chose from exponentially many paths through FSM

Fortunately, similar techniques apply

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Probabilistic Context Free Grammars

Same as context free grammars, with one extension– Where there is a choice of productions for a

non-terminal, give each alternative a probability.

– For each choice point, sum of probabilities of available options is 1

– i.e. Production probability is p(rhs|lhs)

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An examples -> np vp:0.8 s -> np vp pp:0.2np -> det n:0.5np -> np pp:0.5vp -> v np:1.0pp -> p np:1.0

n -> girl:0.25 n -> boy :0.25 n -> park:0.25 n -> telescope:0.25v -> saw:1.0p -> with:0.5 p -> in:0.5

Sample sentence: “The boy saw the girl in the park with the telescope”

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The “low” attachment

S

NP

DET

the

N

boy

VP

V

saw

NP

NP

DET

the

N

girl

PP

P

in

NP

NP

DET

the

N

park

PP

P

with

NP

DET

the

N

telescope

p(“np vp”|s) * p(“det n”|np) * p(“the”|det) *p(“boy”|n) *p(“v np”|vp) * p(“det n”|np) * p(“the”|det) * ...

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The “high” attachmentp(“np vp pp”|s) * p(“det n”|np) * p(“the”|det) *p(“boy”|n) *p(“v np”|vp) * p(“det n”|np) * p(“the”|det) * ...

S

NP

DET

the

N

boy

VP

V

saw

NP

NP

DET

the

N

girl

PP

P

in

NP

DET

the

N

park

PP

P

with

NP

DET

the

N

telescope

Note: I’m not claiming that this matches any particular set of psycholinguistic claims, only that the formalism allows such distinctions to be made.

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Generating from Probabilistic Context Free Grammars

Start with the distinguished symbol “s” Choose a way of expanding “s”

– This introduces new non-terminals (eg. “np” “vp”)

Choose ways of expanding these Carry on until no more non-terminals

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Issues The space of possible trees is infinite.

– But the sum of probabilities for all trees is 1

There is a strong assumption built in to the model– Expansion probability is independent of

position of non-terminal within tree– This assumption is questionable.

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Training for Probabilistic Context Free Grammars

Supervised: you have a treebank Unsupervised: you have only words In between: Pereira and Schabes

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Supervised Training Look at the trees in your corpus Count the number of times each lhs ->

rhs occurs Divide these counts by number of times

each lhs occurs Maybe smooth as described in the lecture

on probability estimation from counts

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Unsupervised Training These are Rabiner’s problems, but for

PCFGs– Calculate the probability of a corpus given

a model– Guess the sequence of states passed

through– Adapt the model to the corpus

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Hidden Trees All you see is the output:

– “The boy saw the girl in the park” But you can’t tell which of several trees led to

that sentence Each tree may have a different probability.

Although trees which use the same rules the same number of times must give the same answer.

Don’t know which state you are in.

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The three problems Probability estimation

– Given a sequence of observations O and a grammar G. Find P(O|G)

Best tree estimation– Given a sequence of observations O and a

grammar G, find a Tree which maximizes P(O,Tree|G).

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The third problem Training

– Adjust the model parameters so that P(O|G) is as large as possible for given O. Hard problem because there are so many adjustable parameters which could vary. Worse than for HMMs. More local maxima.

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Probability estimationP(O | G) P(O, Tree | G)

Tree

Easy in principle. Marginalize out the trees, leaving probability of strings.

But this involves sum over exponentially many trees.

Efficient algorithm keeps track of inside and outside probabilities.

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Inside Probability The probability that non-terminal NT

expands to the words between i and j

... i SENT A LETTER j ...

RR

NP

NP

...

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Outside probability Dual of inside probability.

NP

SENT A LETTER... i SENT A LETTER j ...A MAN

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Corpus probability Inside probability of S node and entire string

is probability of all ways of making sentences over that string

Product over all strings in corpus is corpus probability

Can also get corpus probability from outside probabilities

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Training Uses inside and outside probabilities Starts from an initial guess Improves the initial guess using data Stops at a (locally) best model Specialization of the EM algorithm

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Expected rule counts Consider p(uses rule lhs -> rhs to cover i

through j) Four things need to happen

– Generate outside words leaving hole for lhs– Choose correct rhs– Generate word seen between i and k from first

item in rhs (inside probability)– Generate words seen between k and j using

other items in rhs (more inside probailities)

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Refinements In practice there are very many local maxima,

so strategies which involve generating hundreds of thousands of rules may fail badly.

Pereira and Schabes discovered that letting the system know some limited stuff about bracketting is enough to guide it to correct answers

Different grammar formalisms (TAGs, Categorial Grammars...)

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A basic parsing algorithm The simplest statistical parsing algorithm

is called CYK or CKY. It is a statistical variant of a bottom-up

tabular parsing algorithm that you should have seen in 684.01

It (somewhat surprisingly) turns out to be closely related to the problem of multiplying matrices.

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Basic CKY (review) Assume we have organized the lexicon as a function

lexicon: string -> nonterminal set Organize these nonterminals into the relevant parts of

a two dimensional array indexed by left and right end of the itemFor I = 1 to length(sentence) dochart[I,I+1] = lexicon(sentence[i])

endfor

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Basic CKY Assume we have organized the grammar as a function

grammar: nonterminal -> nonterminal -> nonterminal set

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Basic CKY Build up new entries from existing entries, working

from shorter entries to longer onesfor l = 2 to length(sentence) do

// l is length of constituentfor s = 1 to len – l + 1 do // s is start of rhs1

for t = 1 to l-1 do (left,mid,right) = (s,s+t,s+l) chart[left,right] =

combine(chart[left,mid],chart[mid,right]) endfor endforendfor

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Basic CKY Combine is fun combine(set1,set2) result = empty for item1 in set1 do for item2 in set2 do

result = union result (grammar item1 item2) endfor endforreturn result

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Going statistical The basic algorithm tracks labels for

each substring of the input The cell contents are sets of labels A statistical version keeps track of

labels and their probabilities Now the cell contents must be weighted

sets

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Going statistical Make the grammar and lexicon produce

weighted sets. gexicon: word -> real*nt setgrammar: real*nt->real*nt -> real*nt set

We now need an operation corresponding to set union for weighted sets.

{s:0.1,np:0.2} WU {s:0.2,np:0.1} = ???

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Going statistical (one way)

{s:0.1,np:0.2} WU {s:0.2,np:0.1} = {s:0.3,np:0.3}

If we implement this, we get a parser that calculates the inside probability for each label on each span.

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Going statistical (another way)

{s:0.1,np:0.2} WU {s:0.2,np:0.1} = {s:0.2,np:0.2}

If we implement this, we get a parser that calculates the best parse probability for each label on each span.

The difference is that in one case we are combining weights with +, while in the second we use max

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Building trees Make the cell contents be sets of trees Make the lexicon be a function from

words to little trees Make the grammar be a function from

pairs of trees to sets of newly created (bigger) trees

Set union is now over sets of trees Nothing else needs to change

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Building weighted trees Make the cell contents be sets of trees,

labelled with probabilities Make the lexicon be a function from words to

weighted (little trees) Make the grammar be a function from pairs of

weighted trees to sets of newly created (bigger) trees

Set union is now over sets of weighted trees Again we have a choice of min or +, to get

either parse forest or just best parse

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Where to get more information

Roark and Sproat ch 7 Charniak chapters 5 and 6 Allen Natural Language

Understanding ch 7 Lisp code associated with Natural Language

Understanding Goodman: Semiring parsing

(http://www.aclweb.org/anthology/J99-1004)