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Linguistica : Unsupervised Learning of Natural Language Morphology Using MDLPowerPoint Presentation

Linguistica : Unsupervised Learning of Natural Language Morphology Using MDL

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### Linguistica:Unsupervised Learning of Natural Language Morphology Using MDL

### End be selected. How many ways of doing this are there?

John Goldsmith

Department of Linguistics

The University of Chicago

The Goal:

- To develop a program that learns the structure of words in any human language on the basis of a raw text.
- No human supervision, except for the naïve creation of the text.

Value

- To linguistic theory: reconstruct linguistic theory in a quantitative fashion
- Practical value:
- Information retrieval on data bases of unrestricted languages
- develop stochastic morphologies rapidly: necessary for automatic speech recognition

- A step towards syntax

The product

- Currently a C++ program that functions as a Windows-based tool for corpus-based linguistics.
- Available in beta version on web site.

What do we want?

If you give the program a computer file containing Tom Sawyer, it should tell you that the language has a category of words that take the suffixes ing,s,ed, and NULL; another category that takes the suffixes 's, s, and NULL;

If you give it Jules Verne, it tells you there's a category with suffixes:

a aient ait ant (chanta, chantaient, chantait, chantant)

Immediate queries:

- Do you tell it what language to expect? No.
- Does it have access to meaning? No.
- Does that matter? No.
- How much data does it need. ...

- You get reasonable results fast, with 5,000 words, but results are much better with 50,000, and much better with 500,000 words (length of corpus).
- 100,000 word tokens ~ 12,000 distinct words.

Game plan

- Overview of MDL = Minimum Description Length, where
- Description Length = Length of Analysis + Length of Compressed Data
- Length of data as optimal compressed length of the corpus, given probabilities derived from morphology
- Length of morphology in information theoretic terms
- MDL is dead without heuristics…(then again, heuristics without MDL lack all finesse.)

Game plan (continued)

- Heuristic 1: discover basic candidate suffixes of the language using weighted mutual information
- Heuristic 2: use these to find regular signatures;
- Now use MDL to correct errors generated by heuristics

Game plan (end)

Why using MDL is closely related to measuring the (log of the) size of the space of possible vocabularies.

- For the purposes of version 1 of Linguistica 1, I will restrict myself to Indo-European languages, and in general languages in which the average number of suffixes per word is not greater than 2. (We drop this requirement in Linguistica 2.)

Minimum Description Length (Rissanen 1989)

Basic idea:

A good analysis of a set of data is one that (1) extracts the structure found in the data, and (2) which does it without overfitting the data.

If you have a set of pointers to a bunch of objects, and a probability distribution over those pointers, then

You may act as if the information-length of each pointer =

-1* log prob (that pointer).

So for our entire corpus-- probability distribution over those pointers, then

The length of the compressed size of

each piece w is -log prob(w); so...

Total compressed length of the corpus is:

Overfitting the data: probability distribution over those pointers, then

- The Gettysburg Address can be compressed to 2 bits if you choose an eccentric encoding scheme.
- But that encoding scheme (1) will be long, and (2) will do more poorly than an encoding scheme that does not waste its probability mass on the Gettysburg Address.

Even scientific theories bow to the exigencies of MDL... probability distribution over those pointers, then

- in a sense.
- A theory is penalized if it does not capture generalizations within the observational data (e.g., predicting future observations on the basis of the initial conditions);
- It is penalized if it is more complex than it needs to be (Ockham’s Razor).

Minimum Description Length: probability distribution over those pointers, then

For a given set of data D, choose the analysis Ai to minimize the function:

Length(Compression of D using Ai)

+

Length (Ai)

Compressed length of the data using A probability distribution over those pointers, then i?

The data is the corpus.

The compressed length of the corpus is just (summing over the words)

Our morphology has two necessary properties: probability distribution over those pointers, then

- It must assign a probability to every word of the language (so that we can speak of its ability to compress the corpus) -- we’ll return to this immediately;
- And it must have a well-defined length.

Morphology assigns a frequency: probability distribution over those pointers, then

- If the morphology assigns no internal structure to a word (John, the, …), it assigns the observed frequency to the word.
- If the morphology analyzes a word (dog+s), it assigns a frequency to that word on the basis of 3 things:

1. The frequency of the suffixal pattern in which the word is found (dog-s, dog-’s, dog-NULL);

2. The frequency of the stem (dog);

3. The frequency of the suffix (-s) within that pattern (-s, -’s, -NULL)

Terminology: is found (dog-

The pattern of suffixes that a stem takes is its signature:

- NULL.ed.ing.s
- NULL.er.est.ness

Frequency of analyzed word is found (dog-

W is analyzed as belonging to

Signature s,stem T and suffix F.

[x] means the

count of x’s

in the corpus

(token count)

Where [W] is the total number of words.

Actually what we care about is the log of this:

So far: is found (dog-

- The behavior we demand of our morphology is that it assign a frequency to any given word; we need this so that we can evaluate the particular morphology’s goodness as an analysis, i.e., as a compressor.

Next, let’s see how to measure is found (dog-the length of a morphology

A morphology is a set of 3 things:

- A list of stems;
- A list of suffixes;
- A list of signatures with the associated stems.

Let’s measure the list of suffixes is found (dog-

A list of suffixes consists of:

- a piece of punctuation telling us how long the list is (of length log (size) );
- A list of pointers to the suffixes (each pointer of size - log (freq (suffix));
- A concatenation of the letters of the suffixes (we could compress this, too, or just count number of letters).

punctuation is found (dog-

~ of length log(4)

of length 3,

because p(ed) = 1/8

4

pointered

pointers

pointerNULL

pointering

ed

s

NULL

ing

of length 2,

because 2 letters long

Same for stem list: is found (dog-

Indication of size of the list (of length log (size));

List of pointers to each stem, where each pointer is of length - log freq (stem);

Concatenation of stems (sum of lengths of stems in letters)

Size of the signature list is found (dog-

What is the size of an individual signature? It consists of two subparts:

- a list of pointers to stems, and a list of pointers to suffixes.
- And we already know how to measure the size of a list of pointers.

An individual signature is found (dog-

for the words dog, dogs, cat, cats, glove, gloves

Length of a signature is found (dog-

Sum of the lengths

of the pointers

to the stems

Sum of the lengths

of the pointers

to the suffixes

I’m glossing over an important is found (dog-natural language complexity:recursive structure.

word

(Significant effects

on distribution of

probability mass

over all the words.)

word

find

ing

s

So the total length of the morphology is... is found (dog-

(iv) Signature component: is found (dog-

Signature component is found (dog-

list of pointers to signatures

<X> indicates the number

of distinct elements in X

MDL needs heuristics is found (dog-

- MDL does only one thing: it tells you which of two analyses is better.
- It doesn’t tell you how to find those analysis.

Overall strategy is found (dog-

- Use initial heuristic to establish sets of signatures and sets of stems.
- Use heuristics to propose various corrections.
- Use MDL to decide on whether proposed corrections are to be accepted or refused.

Initial Heuristic is found (dog-

1. Take top 100 ngrams based on weighted mutual information as candidate morphemes of the language:

If a word ends in a candidate morpheme, split it thusly, to form a candidate stem thereby:

sanity:

- sanit + y
- sanity
- san + ity

How to choose in ambiguous cases? form a

This turns out to be a lot harder than you’d think, given what I’ve said so far.

Short answer is a heuristic: maximize the objective function

There’s no good short explanation for this,

except this:the frequency of a single letter is a very bad

first approximation of its likelihood to be a morpheme.

For each stem, find the suffixes it appears with form a

- This forms its signature:
- NULL.ed.ing.s, for example.
Now eliminate all signatures that appear only once.

This gives us an excellent first guess for the morphology.

Stems with their signatures form a

abrupt NULL ly ness.

abs ence ent.

absent -minded NULL ia ly.

absent-minded NULL ly

absentee NULL ism

(French:)

absolu NULL e ment.

absorb ait ant e er é ée

abus ait er

abîm e es ée.

Now build up signature collection... form a

Top 10, 100K words

1 .NULL.ed.ing. 65 1214

2 .NULL.ed.ing.s. 27 1464

3 .NULL.s. 290 8184

4 .'s.NULL.s. 27 2645

5 .NULL.ed.s. 26 541

6 .NULL.ly. 128 2124

7 .NULL.ed. 87 767

8 .'s.NULL. 75 3655

9 .NULL.d.s. 14 510

10 .NULL.ing. 62 983

Verbose signature... form a

.NULL.ed.ing. 58

heap check revolt

plunder look obtain

escort proclaim arrest

gain destroy stay

suspect kill consent

knock track succeed

answer frighten glitter.…\

Find strictly regular signatures: form a

- A signature is strictly regular if it contains more than one suffix, and is found on more than one stem.
- A suffix found in a strictly regular suffix is a regular suffix.
- Keep only signatures composed of regular suffixes (=regular signatures).

Examples of non-regular signatures form a

Only one stem for this signature:

- ch.e.erial.erials.rimony.rons.uring
- el.ezed.nce.reupon.ther

Prefixes form a

Just the same, in mirror-image style. Perform either on stems or on words.

English prefixes form a

.NULL.re. 8

.NULL.dis. 7

.NULL.de. 4

.NULL.un. 4

.NULL.con. 3

.NULL.en. 3

.NULL.al. 3

.NULL.t. 3

.NULL.con.ex. 2

French prefix signatures form a

NULL.d'.l'. NULL.d'.

NULL.l'. NULL.dé.

NULL.re. d'.l'.

NULL.qu'. NULL.par.

NULL.en. NULL.in.

NULL.di. NULL.com.

NULL.s'. NULL.l'en.

NULL.n'. NULL.cou.

NULL.pro. NULL.ent.

NULL.ré. NULL.d'.s'.

Now use MDL to fix problems: form a Repair heuristics

Problems that arise:

1. “ments” problem: a suffix may really be two suffixes.

2. ted.ting.ts: a letter which occurs stem finally with high frequency may get wrongly parsed

(e.g., shou-ted, shou-ting, shou-ts).

3. Spurious signatures form a

4. Misplaced word-breaks

Repair heuristics: using MDL form a

We could compute the entire MDL in one state of the morphology; make a change; compute the whole MDL in the proposed (modified) state; and compared the two lengths.

Original morphology

+ Compressed data

Revised

morphology+

compressed data

<

>

But it’s better to have a more thoughtful approach. form a

Let’s define

Then the size of the punctuation for the 3 lists is:

Then the change of the size of the punctuation in the lists:

Size of the suffix component, form a remember:

Change in its size when we

consider a modification to the morphology:

1. Global effects of change of number of suffixes;

2. Effects on change of size of suffixes in both states;

3. Suffixes present only in state 1;

4. Suffixes present only in state 2;

Suffix component change: form a

Suffixes whose

counts change

Global effect of change

on all suffixes

Contribution of suffixes

that appear only in State1

Contribution of suffixes

that appear only in State 2

Entropy, MDL, and morphology form a

Why using MDL is closely related to measuring the complexity of the space of possible vocabularies

Consider the space of all words of length L, built from an alphabet of size b.

How many ways are there to build a vocabulary of size N?Call that U(b,L,N).

Clearly,

Compare that with the operation (choosing a set of N words of length L, alphabet size b) with the operation of choosing a set of T stems (of length t) and a set of F suffixes (of length f), where t + f = L.

If we take the complexity of each task to be measured by the log of its size, then we’re asking the size of:

is easy to approximate, however. of length L, alphabet size b) with the operation of choosing a set of T stems (of length t) and a set of F suffixes (of length f), where t + f = L.

remember:

The number of bits needed of length L, alphabet size b) with the operation of choosing a set of T stems (of length t) and a set of F suffixes (of length f), where t + f = L.

to list all the words:

the analysis

The length of all the pointers

to all the words:

the compressed corpus

Thus the log of the number of vocabularies =

description length of that vocabulary,

in the terms we’ve been using

That means that the differences in the sizes of the spaces of length L, alphabet size b) with the operation of choosing a set of T stems (of length t) and a set of F suffixes (of length f), where t + f = L.

of possible vocabularies is equal to the difference in the

description length in the two cases:

hence,

Difference of complexity of “simplex word” analysis

and complexity of analyzed word analysis=

log U(b,L,N) - U(b,t,T)-U(b,f,F)

Difference in size of

morphologies

Difference in size

of compressed data

But we’ve (over)simplified in this case by ignoring the frequencies inherent in real corpora. What’s of great interest in real life is the fact that some suffixes are used often, others rarely, and similarly for stems.

We know something about the distribution of words, but nothing about distribution of stems and especially suffixes.

But suppose we wanted to think about the statistics of vocabulary choice in which words could be selected more than once….

We want to select N words of length L, and the same word can be selected. How many ways of doing this are there?

These are like bosons: you can have any number of occurrence of a word, and 2 sets of the same number of them are indistinguishable. How many such vocabularies are there, then?

where Z(i) is the number of words of frequency be selected. How many ways of doing this are there?i.

(‘Z’ stands for “Zipf”).

We don’t know much about frequencies of suffixes,

but Zipf’s law says that

hence for a morpheme

set that obeyed

the Zipf distribution:

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