an introduction to statistical machine translation

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An Introduction to Statistical Machine Translation. Dept. of CSIE, NCKU Yao-Sheng Chang Date: 2011.04.12. Outline. Introduction Peter Brown The Mathematics of Machine Translation: Parameter Estimation, computational linguistics , vol. 19,1993, pp.263-311. Model 1. Introduction (1). - PowerPoint PPT Presentation

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1

An Introduction to Statistical Machine

Translation

Dept. of CSIE, NCKUYao-Sheng ChangDate: 2011.04.12

2

Outline

Introduction Peter Brown

The Mathematics of Machine Translation: Parameter Estimation, computational linguistics, vol. 19,1993, pp.263-311.

Model 1

3

Introduction (1)

Machine translation is available Statistical method, information theory Faster computer, large storage Machine-readable corpora

Statistical method have proven their value Automatic speech recognition Lexicography, Natural language

processing

4

Introduction (2)

Translations involve many cultural respects We only consider the translation of individual

sentence, just acceptable sentences.

Every sentence in one language is a possible translation of any sentence in the other Assign (S,T) a probability, Pr(T|S), to be the

probability that a translator will produce T in the target language when presented with S in the source language.

5

Statistical Machine Translation(SMT)

Noise channel problem

6

Fundamental of SMT

Given a string of French f, the job of our translation system is to find the string e that the speaker had in mind when he produced f. (Baye’s theorem)

Since denominator Pr(f) here is a constant, the best e is one which has the greatest probability.

7

Practical Challenges

Computation of translation model Pr(f|e) Computation of language model Pr(e) Decoding (i.e., search for e that maximize Pr(f|e) Pr(e))

8

Alignment of case 1

9

Alignment of case 2

10

Alignment of case 3

11

Formulation of Alignment(1) Let e = e1

le1e2…el and f = f1m f1f2…fm

An alignment between a pair of strings e and f use a mapping of every word ei to some word fj

In other words, an alignment a between e and f tells that the word ei, 1 i l is generated by the word faj, aj{1,…,m}

There are (l+1)m different alignments between e and f. (Including Null – no mapping )

e = e1e2…ei…el

f = f1 f2… fj… fm

aj =i

faj =ei

12

Formulation of Alignment(2)

Probability of an alignment

13

Translation Model

The alignment, a, can be represented by a series, a1

m = ala2... am, of m values, each between 0 and l such that if the word in position j of the French string is connected to the word in position i of the English string, then aj = i , and if it is not connected to any English word, then aj = 0 (null).

14

IBM Model I (1)

15

IBM Model I (2)

The alignment is determined by specifying the values of aj for j from 1 to m, each of which can take any value from 0 to l.

16

Constrained Maximization

We wish to adjust the translation probabilities so as to maximize Pr(f|e ) subject to the constraints that for each e

17

Lagrange Multipliers (1) Method of Lagrange multipliers(拉格朗乘數法) : Lagrange multipliers with one constraint

If there is a maximum or minimum subject to the constraint g(x,y) = 0, then it will occur at one of the critical numbers of the function F defined by is called the

f(x,y) is called the objective function(目標函數) . g(x,y) is called the constrained equation(條件限制方程式) .

F(x, y, ) is called the Lagrange function(拉格朗函數) . is called the Lagrange multiplier (拉格朗乘數) .

),(),(),,( yxgyxfyxF

18

Lagrange Multipliers (2) Example 1: Maximize

Subject to Let

Set

代入 (2) 與 (3) ,可得

(5) 與 (6) 代入 (4) ,可得 ,由此可得 因此,最大值為

xyzV

024346 zyx

)24346(),,,( zyxxyzzyxF

)4(024346

)3(03

)2(04

)1(06

zyxF

xyF

xzF

yzF

z

y

x

6)1(

yz

)5(2

30

64 xyyz

xz

)6(20

63 xzyz

xy

0241824)2(32

346

xxxx

3

4 x

y z 28

3,

9

64

3

8)2(

3

4

xyzV

19

Lagrange Multipliers (3)

Following standard practice for constrained maximization, we introduce Lagrange multipliers e, and seek an unconstrained extremum of the auxiliary function

20

Derivation (1)

The partial derivative of h with respect to t(f|e) is

where is the Kronecker delta function, equal to one when both of its arguments are the same and equal to zero otherwise

21

Derivation (2)

We call the expected number of times that e connects to f in the translation (f|e) the count of f given e for (f|e) and denote it by c(f|e; f, e). By definition,

22

Derivation (3)

replacing e by ePr(f|e), then Equation (11) can be written very compactly as

In practice, our training data consists of a set of translations, (f(1) le(1)), (f(2)

le(2)), ..., (f(s) le(s)), , so this equation becomes

23

Derivation (4)

For an expression that can be evaluated efficiently.

24

Derivation (5)

Thus, the number of operations necessary to calculate a count is proportional to l + m rather than to (l + 1)m as Equation (12)

25

EM Algorithm

26

EM Algorithm

27

Introduction(1)

In statistical computing, an expectation-maximization (EM) algorithm is an algorithm for finding maximum likelihood estimates of parameters in probabilistic models, where the model depends on unobserved latent variables. EM is frequently used for data clustering in machine learning and computer vision.

28

Introduction(2)

EM alternates between performing an expectation (E) step, which computes an expectation of the likelihood by including the latent variables as if they were observed, and a maximization (M) step, which computes the maximum likelihood estimates of the parameters by maximizing the expected likelihood found on the E step. The parameters found on the M step are then used to begin another E step, and the process is repeated.

(From: http://en.wikipedia.org/wiki/Expectation-maximization_algorithm )

29

EM algorithm is a soft version of K-means clustering.

The idea is that the observed data are generated by several underlying causes.

Each cause contributes independently to the generation process, bur we only see the final mixture –without information about which cause contributed what.

30

Observable data

Each

Unobservable / hidden data

Each zij can be interpreted as cluster membership probabilities.

The component zij is 1 if object i is a member of cluster j.

}{ iX x

Timii xx ),,( 1 x

}{ izZ

Ti ikii zzzz }{ ,,2,1

31

Initial Assumption

At first , suppose we have a data set , where each is the vector

that correspond to the ith data point.

Further , assume the samples are drawn from k mixture Gaussians , .

Notice that the p.d.f. of multivariate normal distribution is

}{ iX x

Timii xx ),,( 1 x

)()(

2

1exp

)2(

1),;(p 1

j jjT

j

jm

jj

xxx

jc kj 1

A normal distribution in a x variate with mean and 2variance is a statistic distribution with probability function

32

E-step

Let be a n by k matrix , where

Notice that , if we set for then by Bayes formula we have

ijhH

kcj

1)p(

)|p(

)|p(

)p()|p(

)p()|p()|p(

11li

k

l

ji

lli

k

l

jjiij

cx

cx

ccx

ccxxc

kj 1

);|(

);();();(*1);(

1

|||

li

k

l

jiiijiijiijij

cxp

c|xpxzPxzPxzEh

33

M-step

ij

n

i

iij

n

ij

h

xh

1

1'

i

ij

n

i

ijn

ix

h

h

1

1

ij

n

i

Tjijiij

n

ij

h

xxh

1

''

1'))((

G

Tjiji

ij

n

i

ijn

ixx

h

h))(( ''

1

1

n

hh

h

h

h

h ij

n

in

i

ij

n

i

ij

k

j

n

i

ij

n

i

ij

n

i

k

j

ij

n

ij

1

1

1

11

1

11

1'

1

                                                         .

34

log likelihood

The log likelihood of the data set X given theparameters is

,

where , and is the weight of cluster j .

Notice that

),;(log)P(log)|(111

jjijj

k

j

n

ii

n

ixpxXl

),;(log11

jjijj

k

j

n

ixp

Tk),,( 1 ),,( jjjj

j

11

j

k

j

35

計算示範 (1)

假設

4.01.0

3.02.0

),;(),;(

),;(),;(

22221121

22121111

xpxp

xpxpN

8.02.0

6.04.0

4.01.0

4.0

4.01.0

1.03.02.0

3.0

3.02.0

2.0

H

0

11x

1

02x

36

計算示範 (2)

33.0

67.0

1

0

2.04.0

2.0

0

1

2.04.0

4.0'1

57.0

43.0

1

0

8.06.0

8.0

0

1

8.06.0

6.0'2

8.02.0

6.04.0H

0

11x

1

02x

22.022.0

22.022.0

33.01

67.00

33.01

67.00

2.04.0

2.0

33.00

67.01

33.00

67.01

2.04.0

4.0'1

TT

25.025.0

25.025.0

57.01

43.00

57.01

43.00

8.06.0

8.0

57.00

43.01

57.00

43.01

8.06.0

6.0'2

TT

3.02

2.04.0'1

7.0

2

8.06.0'2

48.2)4.07.01.03.0log()3.07.02.03.0log()|( Xl

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