Lecture 13 Grammars

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Lecture 13Grammars
CSCE 355 Foundations of Computation

• Topics
• Context Free Grammars
• Parses, derivations, parse trees
• Language generated by a grammar

October 13, 2008
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• Last Time Readings section 3.1,3.2(skip DFA?
  RE), 3.3
• Ruby Matching in files (fileArgv.rb)
• Unix Shell regular expressions
• Examples of Sets of indistinguishable prefixes
• Pumping Lemma
• New
• If L is regular is the Reversal LR regular?

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Homework

• 1. Modify fileArgv.rb to substitute xxx-xxxx
  for every occurrence of phone number in every
  file in a directory
• 4.4.2 minimization problem

0 1
?A B E
B C F
C D H
E F I
F G B
G H B
H I C
I A E
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Minimization of DFA

• First eliminate states not reachable from the
  start state
• Partition the states into sets of
  indistinguishable states (equivalence classes)
• Figure 4.12 Minimum state DFA equivalent to DFA
  in figure 4.8 (slide 10)

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Proofs Languages are not Regular

• Show L is not regular
• Sets of distinguishable prefixes S S1, S2,
  Sn
• S gtn means any DFA for L would need at least n
  states
• So if we can construct a set of distinguishable
  prefixes of arbitrary size (unbounded)
• Examples
• L 0n1n n gt 1
• S 0, 02, 03, 0n
• For i ! j, 0i is distinguished for L from 0j
  by the string 1i
• Thus there can be a DFA that recognizes L
• L is not regular

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Example L 0n1n n gt 1

• L 0n1n n gt 1
• S 0, 02, 03, 0n
• For i ! j, 0i is distinguished for L from 0j
  by the string 1i
• Thus there can be a DFA (with n states) that
  recognizes L
• L is not regular

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Example L 0n n is power of 2

• S ________________________
• For i ! j,
• ____ is distinguished by L from ___ by the
  string
• Thus there can be a DFA (with n states) that
  recognizes L
• L is not regular

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Languages Regular or Not

• Reference Bumping the Pumping Lemma
• L 0n1m n gt m
• L 0n1m n gt 100, m lt 100
• L 0m m k2
• L ww w in a,b
• Trick question L uww u,w in a,b
• L w in a,b w viewed as binary is
  divisible by 7
• Trick question L uww u,w in a,b

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Pumping Lemma Theorem 4.1

• Let L be a regular language. Then there exists a
  constant n (which depends on n, i.e. number of
  states in a DFA) such that for every string w e L
  such that w gt n, we can break into three
  strings,
• w xyz, such that
• y ! e
• xy lt n
• For all k gt 0, the string xyiz is also in L.

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Proof of the Pumping Lemma
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Pumping Lemma as Adversarial Game

1. Player 1 picks the language L to be proved
   nonregular.
2. Player 2 picks n, but does not reveal it.
3. Player 1 picks w, which may depend on n which
   will w gtn
4. Player 2 divides w into x, y and z obeying 1,2,
   and 3 satisfying constraints (does not tell
   specific x, y and z)
5. Player 1 wins by picking k, which may be a
   function of n, x, y and z such that xykz is not
   in L.

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Chapter 5 Grammars

• Grammar for Palindromes
• P ? e
• P ? 0
• P ? 1
• P ? 0 P 0
• P ? 1 P 1

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Definitions G (N, T, P, S)

• G (N, T, P, S)
• N set of nonterminal symbols
• T set of terminals (or sometime tokens)
• S start symbol (a nonterminal)
• P a set of Productions, rewrite rules of the
  form
• X ? a1 a2 an-1 an
• Where each ai is from N U T

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Derivations

• a ? ß
• a ? ß derives in zero or more steps

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Parse Trees
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Grammar for Expressions
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