Chapter 6 Simplification of CFGs and Normal Forms

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Chapter 6Simplification of CFGs and Normal Forms
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6.1 Methods for Transforming Grammars (1) A
Useful Substitution Rule

• Theorem 6.1
• This intuitive theorem allows us to simplify
  grammars.
• Let G (NT, T, S, P) be a context-free grammar.
  Suppose that P contains a production rule of the
  form
• A ? xBz
• Assume that A and B are different NT and that
• B ? y1 y2 ... yn is the set of all
  productions in P which have B as the left side.
• Let G (NT, T, S, P) be the grammar in which
  P is constructed from P by replacing rule
• A ? xBz with A ? x y1z xy2z ... xynz
• Then L(G) L(G)

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6.1 Methods for Transforming Grammars (2) A
Useful Substitution Rule

• Let G be
• S ? a aaS abBc
• B ? abbS b
• Applying theorem 6.1 results in
• S ? a aaS ababbSc abbc
• B ? abbS b
• The rules B ? abbS b, which are still part of
  the grammar, no longer serve any purpose.
• Both of these useless rules may be deleted
  without effectively changing the grammar.

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6.1 Methods for Transforming Grammars (3)
Removing Useless Productions

• A non-terminal A is useful (it occurs in at least
  one derivation.) if
• it is reachable occurs in a sentential form S ?
  aAb
• it is live generates a terminal string A ? w ?
  T
• A non-terminal A is useless if
• A does not occur in any sentential form
• It cannot be reached from start symbol
• OR
• A does not generate any string of terminals.
• It cannot derive a terminal string
• A terminal is useful if it occurs in a sentence w
  ? L(G)
• Any production involving a useless symbol is a
  useless production.

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6.1 Methods for Transforming Grammars (4)
Removing Useless Productions

• To eliminate useless symbols
• First Find the set TERM that contains all
  non-terminals that derive a terminal string
• A ? w, where w ? T
• Non-terminals NOT in TERM are useless, they
  cannot contribute to generate strings in L(G)
• Second Find the set REACH that contains all
  non-terminals A ? TERM that are reachable from S
• S ? aAb

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6.1 Methods for Transforming Grammars (5)
Removing Useless Productions

• Example 1
• G S ? AC BS B
• A ? aA aF
• B ? CF b ?
• C ? cC D
• D ? aD BD C
• E ? aA BSA
• F ? bB b ?
• L(G) is b
• B, F ? TERM, since both generate terminals
• S ? TERM, since S ? B and hence S ? b
• A ? TERM, since A ? aF and hence A ? ab
• E ? TERM, since E ? aA and hence E ? aab

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6.1 Methods for Transforming Grammars (6)
Removing Useless Productions

• C and D do not belong to TERM, so all rules
  containing C and D are removed
• The new grammar is
• GT S ? BS B
• A ? aA aF
• B ? b
• E ? aA BSA
• F ? bB b
• All non-terminals in GT derive terminal strings
• Now, we must remove the non-terminals that do not
  occur in sentential forms of the grammar
• A set REACH is built that contains all
  non-terminals ? TERM derivable from S

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6.1 Methods for Transforming Grammars (7)
Removing Useless Productions

• GT S ? BS B
• A ? aA aF
• B ? b
• E ? aA BSA
• F ? bB b
• S ? REACH, since it is the start symbol
• B ? REACH, since S ? SB, and hence B is derivable
  from S
• A, E, and F can not be derived from S or B, so
  all rules containing A, E and F are removed

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6.1 Methods for Transforming Grammars (8)
Removing Useless Productions

• The new grammar is
• GU S ? BS B
• B ? b
• L(GU) b
• The set of terminals of GU is b, a is removed
  since it does not occur in any string in the
  language of GU
• The order is important
• Applying Second step (REACH) before First Step
  (TERM) may not remove all useless symbols.

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6.1 Methods for Transforming Grammars (9)
Removing Useless Productions

• Home exercise Remove all useless productions.
• S ? AB CD ADF CF EA
• A ? abA ab
• B ? bB aD BF aF
• C ? cB EC Ab
• D ? bB FFB
• E ? bC AB
• F ? abbF baF bD BB
• G ? EbE CE ba
• Let G (S, A, B, C, a, b, S, S ? aS A
  C, A ? a, B ? aa, C ? aCb) be a CFG.
• Remove all useless productions
• Final grammar is
• G (S, a, S, S ? aS a)

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6.1 Methods for Transforming Grammars (10)
Removing e-Productions

• Let G be S ? SaB aB B ? bB e
• A non-terminal symbol that can derive the null
  string (e) is called nullable.
• For example, in G above, B is nullable since B ?
  e
• A grammar without nullable non-terminals is
  called non-contracting
• G, above, is not non-contracting, since it has
  one nullable non-terminal, which is B.

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6.1 Methods for Transforming Grammars (11)
Removing e-Productions

• How to find nullable non-terminals?
• Mark all non-terminals A for which there exists a
  production of the form A ? ?
• Repeat
• Mark non-terminal X for which there exists X ? ?
  and all symbols in ? have been marked as nullable
• Until no new non-terminal is marked
• Read Theorem 6.3

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6.1 Methods for Transforming Grammars (12)
Removing e-Productions

• The set of nullable non-terminals of the grammar
• S ? ACA
• A ? aAa B C
• B ? bB b
• C ? cC e
• is S, A, C
• C is nullable
• since C ? e and hence C ? e
• A is nullable
• since A ? C, and C is nullable
• S is nullable
• since S ? ACA, and A and C are nullable

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6.1 Methods for Transforming Grammars (13)
Removing e-Productions

• Find nullable non-terminals.
• S ? aS SS bA
• A ? BB
• B ? CC ab aAbC
• C ? ?

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6.1 Methods for Transforming Grammars (14)
Removing e-Productions

• If ? ? L(G), we can eliminate all productions A ?
  ?
• For every B referring to A
• For example, if B ? e and A ? BABa
• Then after eliminating the rule B ? ?, new rules
  for A will be added
• A ? BABa
• A ? ABa
• A ? BAa
• A ? Aa

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6.1 Methods for Transforming Grammars (15)
Removing e-Productions

• Let G be
• S ? SaB aB
• B ? bB e
• After removing e-productions, the new grammar
  will be
• S ? SaB Sa aB a
• B ? bB b
• The removal of e-productions increases the
  number of rules but reduces the length of
  derivations.

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6.1 Methods for Transforming Grammars (16)
Removing e-Productions

• Let G S ? ACA
• A ? aAa B C
• B ? bB b
• C ? cC e
• The equivalent essentially non-contracting
  grammar GL is
• GL S ? ACA CA AA AC A C e
• A ? aAa aa B C
• B ? bB b
• C ? cC c
• Since S ? e in G, the rule S ? e is allowed in
  GL, but all other e-productions are replaced
• A grammar satisfying these conditions is called
  essentially non-contracting (only start symbol is
  nullable)

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6.1 Methods for Transforming Grammars (17)
Removing e-Productions

• Let G be
• S ? aS SS bA
• A ? BB
• B ? ab aAbC aAb CC
• C ? ?
• We eliminate C ? ? by replacing
• B ? CC into B ? CC, B ? C, and B ? ?
• B ? aAbC into B ? aAbC and B ? aAb
• Since C ? ? is only C production
• only B ? ? and B ? aAb retained.
• The new grammar
• S ? aS SS bA
• A ? BB
• B ? ? ab aAb

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6.1 Methods for Transforming Grammars (18)
Removing e-Productions

• The new grammar
• S ? aS SS bA
• A ? BB
• B ? ? ab aAb
• We eliminate B ? ? by replacing
• A ? BB into A ? BB, A ? B, and A ? ?
• Since there are other B productions, these are
  all retained
• The new grammar
• S ? aS SS bA
• A ? BB B ?
• B ? ab aAb

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6.1 Methods for Transforming Grammars (19)
Removing e-Productions

• The new grammar
• S ? aS SS bA
• A ? BB B ?
• B ? ab aAb
• Finally we eliminate A ? ? by replacing
• B ? aAb into B ? aAb, B ? ab
• S ? bA into S ? bA b
• The final CFG is
• S ? aS SS bA b
• A ? BB B
• B ? ab aAb

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6.1 Methods for Transforming Grammars (20)
Removing of Unit Rules

• Rules having this form A ? B are called unit
  rules
• Consider the rules
• A ? aA a B
• B ? bB b C
• The unit rule A ? B indicates that any string
  derivable from B is also derivable from A
• The removal of unit rules increases the number of
  rules but reduces the length of derivations.

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6.1 Methods for Transforming Grammars (21)
Removing of Unit Rules

• To eliminate the unit rule, add A rules that
  directly generate the same strings as B
• Add a rule A ? u for each B ? u and deleting A ?
  B from the grammar
• Read Theorem 6.4

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6.1 Methods for Transforming Grammars (22)
Removing of Unit Rules

• Consider the rules
• A ? aA a B
• B ? bB b C
• The new rules after eliminating the unit rule A ?
  B
• A ? aA a bB b C
• B ? bB b C
• We add new rules to A by replacing B in A with
  all its RHS rules

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6.1 Methods for Transforming Grammars (23)
Removing of Unit Rules

• GL S ? ACA CA AA AC A C e
• A ? aAa aa B C
• B ? bB b
• C ? cC c
• The new equivalent grammar (without unit rules)
• GC S ? ACA CA AA AC
• aAa aa bB b cC c e
• A ? aAa aa bB b cC c
• B ? bB b
• C ? cC c

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6.1 Methods for Transforming Grammars (24)
Removing of Unit Rules

• Remove unit rules
• S ? T S T
• T ? F F T
• F ? a (S)
• S ? T S T
• T ? a (S) F T
• F ? a (S)
• S ? a (S) F T S T
• T ? a (S) F T
• F ? a (S)

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6.2 Chomsky Normal Form (1)

• The Chomsky normal form places restrictions on
  the length and the composition of the right-hand
  side of a rule
• Definition 6.4
• A CFG is in Chomsky normal form if each
  production rule has one of the following forms
• A ? a
• A ? BC
• S ? e
• where B, C ? NT
• Read Theorem 6.6

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6.2 Chomsky Normal Form (2)

• Algorithm Step 1
• Make sure that the following are satisfied
• No e-productions (other than S ? e)
• No chain rules
• No useless symbols

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6.2 Chomsky Normal Form (3)

• Algorithm Step 2
• Eliminate terminals from RHS of productions
• For each production A ? X1X2Xm
• where Xi ? NT ? T
• If m ? 1, replace each terminal a ? RHS of A
• Add (if needed) Ca ? a for each a ? T, where each
  Ca is new non-terminal.
• In production A, replace terminal a with
  corresponding Ca

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6.2 Chomsky Normal Form (4)

• Algorithm Step 3
• Eliminate productions with long RHS
• For each production
• A ? B1B2Bm, m ? 2, where Bi ? NT
• replace with productions
• A ? B1D1
• D1 ? B2D2
• Dm-2 ? Bm-1Bm
• where D1Dm-2 are new non-terminals.

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6.2 Chomsky Normal Form (5)

• Original grammar (no chain rules, useless
  symbols, or e-productions)
• S ? X a Y Y b
• X ? Y X a Y a
• Y ? S S a X b
• Grammar after eliminating terminals from RHSs
• S ? X A Y Y B A ? a
• X ? Y X A Y a B ? b
• Y ? S S A X b
• Grammar after eliminating long RHSs
• S ? X T Y B T ? A Y A ? a
• X ? Y F a F ? X G B ? b
• Y ? S S A X b G ? A Y
• Note Could simplify by combining redundant
  variables T and G

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6.2 Chomsky Normal Form (6)

• Original grammar (no chain rules, useless
  symbols, or e-productions)
• S ? aXYZ a X ? aX a
• Y ? bcY bc Z ? cZ c
• Grammar after eliminating terminals from RHSs
• S ? AXYZ a A ? a
• X ? AX a B ? b
• Y ? BCY BC C ? c
• Z ? CZ c
• Grammar after eliminating long RHSs
• S ? AF a A ? a F? XG
• X ? AX a B ? b G ? YZ
• Y ? BH BC C ? c H ? CY
• Z ? CZ c
• See Example 6.8

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6.2 Greibach Normal Form (7)

• A context-free grammar is in Greibach Normal Form
  if every production is of the form A ? aX
• where A ? NT, X ? NT, and a ? S
• Examples
• G1 (S, A, a, b, S, S ? aSA a, A ? aA
  b)
• GNF
• G2 (S, A, a, b, S, S ? AS AAS, A ? SA
  aa)
• not GNF
• This grammar S ? AB A ? aA bB b B ? b
• is not in GNF
• This grammar S ? aAB bBB bB
• A ? aA bB b B ? b
• is in GNF