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LING 408/508: Computational Techniques for Linguists

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LING 408/508: Computational Techniques for Linguists

Lecture 23

10/15/2012

- Formal language theory: strings and languages
- Coding CFGs
- Generating strings and trees
- Short assignment #15

- Note: the following is mathematical notation. It is not Python code.

- Let Σ be a finite alphabet.
- The elements of Σ may be multi-character symbols

- Examples:
- DIGITS = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
- LETTERS = { A, B, C, …, X, Y, Z }
- ENGLISH_WORDS = { a, aa, …, zyzzogeton, zyzzyva }

- A string is a sequence of zero or more symbols taken from an alphabet.
- Example:
- DIGITS = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
- s1 = 0123456789
- s2 = 321654

- The empty string, ε, consists of zero symbols
- Pronounced as “epsilon”

- A string is produced by concatenation of symbols from an alphabet. The concatenation of symbols is written by placing two symbols immediately next to each other.
- Strings can also be concatenated.
- Example:
- DIGITS = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
- 1 є DIGITS and 2 є DIGITS. Concatenate 1 and 2 to produce the string 12
- Concatenate 123 and 456 to produce the string 123456

- Use an “exponent” with an integer value >= 0 to indicate the number of times a symbol is repeated in a string
- Zero repetitions is equivalent to є
- Examples:
- a5 = aaaaa
- a1 = a
- a0 = ε

- ab2 = abb
- (ab)2 = abab
- abc(def)2 = abcdefdef
- abc(def)1 = abcdef
- abc(def)0 = abcε = abc

- An “exponent” on an alphabet indicates a set of strings, such that each string consists of any symbols from that alphabet, and is of the length specified
- Zero repetitions is equivalent to { є }
- Example:
- Σ = { 1, 2, 3 }
- Σ0 = { ε }
- Σ1 = { 1, 2, 3 }
- Σ2 = { 11, 12, 13, 21, 22, 23, 31, 32, 33 }

- An “exponent” on an alphabet indicates a set of strings, such that each string consists of any symbols from that alphabet, and is of the length specified
- Zero repetitions is equivalent to { є }
- Closure: * indicates zero or more symbols
Σ* = Σ0υΣ1υΣ2υΣ3υ …

- Example:
- Σ = { 1, 2, 3 }
- Σ0 = { ε }
- Σ1 = { 1, 2, 3 }
- Σ2 = { 11, 12, 13, 21, 22, 23, 31, 32, 33 }
- Σ* = { ε } υ { 1, 2, 3 } υ { 11, 12, 13, 21, 22, 23, 31, 32, 33 } υ …
= { є, 1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, … }

- Finite repetition such as a5 is not available within regular expression syntax
- In regular expression, instead specify as aaaaa

- Closure on symbols such a* (in a regular expression) as is not available within formal language theory
- By “a*”, one probably means { ε, a, aa, aaa, … }, but there is another way to specify this in formal language theory

- A language is a set of strings
- May be finite or infinite

- Examples:
- L1 = { apples, bananas, pears }
- L2 = { 0, 1, 2, 3, 4, … }
- L3 = { ε, a, b, aa, ab, ba, bb, aaa, aab, aba, … }

- Finite sets are easy to describe because you can list the elements.
- With infinite languages, the use of “…” is not precise. We need better mathematical notation to describe these sets.
- Example:
- L4 = { 1, 2, 3, … }
- L4 could be { 1, 2, 3, 4, 5, 6, … }, or it could be { 1, 2, 3, 11, 12, 13, … }, or many other sets

- Describe a set through a pattern and remarks on the properties of that pattern
- Read “|” as “such that”

- Examples:
- { x | x is a positive integer } = {1, 2, 3, 4, 5, … }
- { an | n is an integer >= 0 } = { ε, a, aa, aaa, … }
- { an | n is an integer and 1 <= n <= 5 } = { a, aa, aaa, aaaa, aaaaa }
- { anbn | n >= 0 } = { ε, ab, aabb, aaabbb, … }
- { anbm | n >= 0 and m > n } = { b, bb, …, abb, abbb, …, aabbb, aabbbb, aabbbbb, …, aaabbbb, aaabbbbb, …}

- A predicate definition of a language tells us about the properties of the strings in a language, but it does not tell us a procedure by which the language can be generated
- Alternatives:
- Inductive definitions
- Grammar-based definitions

- State base case(s). At least one base case.
- State inductive case(s). Zero or more.
- Example:
- L = { anbn | n >= 0 } = { є, ab, aabb, aaabbb, … }
- Base case: ε є L
- Inductive case: if x є L, then a x b є L

- Base case: ε є L
- Inductive case: if x є L, then a x b є L
- L = { }
- Begin with empty set.
- From the base case, ε є L. So we place ε in L
- ε is a string in L. From the inductive case, a ε b = a b є L
- abis a string in L. From the inductive case, a ab b = aabbє L
- etc.
- This inductive definition generates
L = { anbn | n >= 0 } = { є, ab, aabb, aaabbb, … }

ε

, ab

, aabb

- A grammar is a formalism (with rules for computation) that generates a set of strings
- Examples:
- Regular expression
- Finite-state automaton
- Context-free grammar
- Pushdown automaton
- etc.

- Some languages can(not) be generated by certain types of grammars
- Examples
L1 = { an | n >= 0 } = { є, a, aa, aaa, … }

- Can be generated by regular expression
- Can be generated by context-free grammar
L2 = { anbn | n >= 0 } = { є, ab, aabb, aaabbb, … }

- Cannot be generated by regular expression
- Can be generated by context-free grammar
L3 = { anbncn | n >= 0 } = { є, abc, aabbcc, aaabbbccc, … }

- Cannot be generated by regular expression
- Cannot be generated by context-free grammar

- Formal language theory: strings and languages
- Coding CFGs
- Generating strings and trees
- Short assignment #15

toy_cfg_str = """

S -> NP VP

NP -> DT N

VP -> V

DT -> a

N -> flight

V -> left

V -> arrived

"""

expr_cfg_str = """

E -> ( E )

E -> E + E

E -> E - E

E -> E * E

E -> E / E

E -> 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9

"""

- Example:
NONTERM1 -> a b

NONTERM2 -> c

NONTERM2 -> d

NONTERM3 -> a | a NONTERM3

- LHS (left-hand side) of rule is a nonterminal
- RHS is the sequence of symbols that the rule rewrites as
- May be multiple rules for a single nonterminal (NONTERM2)
- Multiple rules may be written as a disjunction (NONTERM3)

- Map each (LHS) nonterminal to a list of a list of strings
- List of strings: symbols (nonterminals and terminals) on RHS of a particular rule
- List of list of strings: there may be multiple rules for a single nonterminal

- Grammar:
NONTERM1 -> a b

NONTERM2 -> c

NONTERM2 -> d

NONTERM3 -> a | a NONTERM3

- Represent as dictionary:
cfg = {}

cfg['NONTERM1'] = [['a', 'b']]

cfg['NONTERM2'] = [['c'], ['d']]

cfg['NONTERM3'] = [['a'], ['a', 'NONTERM3']]

- Want to read string for CFG, return dictionary representation of the grammar
- Parsing code will also allow for:
- Empty lines in the grammar
- Comments

- Apply split to RHS of a rule to get disjunction of rules

>>> s = """

S -> NP VP # comment: S is the start symbol

V -> eat

"""

>>> s

'\nS -> NP VP\n\nV -> eat\n'

>>> s.split() # don't want this as the result

['S', '->', 'NP', 'VP', 'V', '->', 'eat']

>>> s.split('\n') # each rule is in separate string

['', 'S -> NP VP', '', 'V -> eat', '']

- When there is a disjunction of rules in a line,
need to get list of strings for each rule

>>> s = 'a | a NONTERM3'

>>> s.split('|')

['a ', ' a NONTERM3']

>>> for x in s.split('|'): # split again to get

print(x.split()) # rid of whitespace

['a']

['a', 'NONTERM3']

def read_grammar(cfg_str):

cfg = {}

for ln in cfg_str.split('\n'):

comment_idx = ln.find('#') # ignore comments

if comment_idx!=-1: ln = ln[:comment_idx]

if ln=='': continue # ignore empty lines

lhs = ln[:ln.index('->')] # split the line

rhs = ln[ln.index('->')+3:] # into lhs and rhs

rhs_rules = cfg.get(lhs, []) # assign rules to lhs

for symbols in rhs.split('|'): # disjunct. of rhs

rhs_rules.append(symbols.split())

cfg[lhs] = rhs_rules

return cfg

def print_cfg(cfg):

for (lhs, rhs_rules) in cfg.items():

for symbols in rhs_rules:

s = '{} -> {}'.format(lhs, ' '.join(symbols))

print(s)

toy_cfg = read_grammar(toy_cfg_str)

expr_cfg = read_grammar(expr_cfg_str)

print('toy cfg:')

print_cfg(toy_cfg)

print('\nexprcfg:')

print_cfg(expr_cfg)

toy cfg:

N -> flight

DT -> a

VP -> V

S -> NP VP

V -> left

V -> arrived

NP -> DT N

exprcfg:

E -> ( E )

E -> E + E

E -> E - E

E -> E * E

E -> E / E

E -> 0

E -> 1

E -> 2

E -> 3

E -> 4

E -> 5

E -> 6

E -> 7

E -> 8

E -> 9

def get_nonterminals(cfg):

return set(cfg.keys())

toy_cfg_nonterms = get_nonterminals(toy_cfg)

expr_cfg_nonterms = get_nonterminals(expr_cfg)

print('\ntoy_cfg_nonterms:', toy_cfg_nonterms)

print('expr_cfg_nonterms:', expr_cfg_nonterms)

# Output:

# toy_cfg_nonterms: set(['N', 'DT', 'VP', 'S',

# 'V', 'NP'])

# expr_cfg_nonterms: set(['E'])

- Formal language theory: strings and languages
- Coding CFGs
- Generating strings and trees
- Short assignment #15

- Procedure:
- For a given nonterminal, choose a rule at random
- For each symbol in the RHS of that rule:
- If it is a terminal symbol, add to string
- If it is a nonterminal, recursively generate a string from that nonterminal

- Generate a string from the grammar through the start symbol

- Generate a string from this CFG
A -> x Y z

Y -> c

- Begin with start symbol A
- Look up rule for nonterminal: A -> x Y z
- Generate items on right-hand side of rule
- Generate terminal: x
- Recursively generate string for nonterminalY
- Look up rule for nonterminal: Y -> c
- Generate items on right-hand side of rule
- Generate terminal: c

- Generate terminal: z

- Final string generated: x c z

import random

def generate_string(cfg, lhs, nonterms):

s = ''

nonterm_rules = cfg[lhs]

# randomly choose a rule

r_idx = random.randint(0,len(nonterm_rules)-1)

rule = nonterm_rules[r_idx]

for sym in rule: # step through symbols in rule

if sym in nonterms: # recursive case

s += generate_string(cfg, sym, nonterms) + ' '

else: # terminal symbol, concatenate to string

s += sym + ' '

return s[:-1] # remove last ' '

print '\nRandom strings generated by toy cfg:'

for i in range(10):

print(generate_string(toy_cfg,'S',toy_cfg_nonterms))

print '\nRandom strings generated by exprcfg:'

for i in range(10):

print(generate_string(expr_cfg,'E',expr_cfg_nonterms))

Random strings from toy cfg:

a flight left

a flight left

a flight arrived

a flight left

a flight left

a flight arrived

a flight arrived

a flight arrived

a flight arrived

a flight left

Random strings from exprcfg:

( 6 ) - 5 * 5

5

9

5 / 5

1

2 - ( 0 - 0 + 3 ) + 1

4

3 + 9 / 0 / 5 / 0 + 7

3

2

print('\nLong strings generated by exprcfg:')

for i in range(1000):

s = generate_string(expr_cfg,'E',expr_cfg_nonterms)

if len(s) > 50:

print(s)

Output:

random strings from exprcfg:

3 * 9 * 6 / ( 0 + 6 - ( 9 / 8 - 2 ) + 3 / 1 - 4 ) * 4

4 + 5 * 5 + 1 / 5 - ( 1 ) + ( 4 ) / ( 9 ) + 6 + 2 - 1

- Let’s modify our tree representation to allow an arbitrary number of children:
(value, list-of-children)

- Parent node: (nonterminal, list-of-child-nodes)
- Leaf node: (terminal, [])

- A is the parent of b and C and d. C is the parent of e.
('A', [('b', []),

('C', [('e', [])]),

('d', [])])

- Parent node: (nonterminal, list-of-child-nodes)
- Leaf node: (terminal, [])

import random

def generate_tree(cfg, lhs, nonterms):

# randomly choose a rule

nonterm_rules = cfg[lhs]

r_idx = random.randint(0,len(nonterm_rules)-1)

rule = nonterm_rules[r_idx]

children = []

for sym in rule:

if sym in nonterms: # recursive case

ch_node = generate_tree(cfg, sym, nonterms)

children.append(ch_node)

else: # base case: leaf node

children.append((sym, []))

parent = (lhs, children)

return parent

def pretty_print(node):

pass

# this is your homework problem

toy_tree = generate_tree(toy_cfg, 'S', toy_cfg_nonterms)

pretty_print(toy_tree)

expr_tree = generate_tree(expr_cfg, 'E',expr_cfg_nonterms)

pretty_print(expr_tree)

S

NP

DT

a

N

flight

VP

V

arrived

E

E

E

0

+

E

E

4

*

E

1

+

E

E

8

/

E

7

- Given a string (sentence) generated by an arbitrary CFG, determine its parse tree
- Or parse trees, if the string (sentence) is ambiguous

- Formal language theory: strings and languages
- Coding CFGs
- Generating strings and trees
- Short assignment #15

- Download short assignment from course web page
- #1: i, l, m, q
- #2: a, b, c, g
- #3: a, b, e, f