Simplification of CFG and Normal Forms Wen-Guey Tzeng Computer - - PowerPoint PPT Presentation

Simplification of CFG and Normal Forms

Wen-Guey Tzeng Computer Science Department National Chiao Tung University

Normal Forms

• We want a cfg with either Chomsky or

Greibach normal form

– Chomsky normal form

• Aa, ABC

– Greibach normal form

• Aax, xV*

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• CFG with normal forms are easier for parsing

– The membership problem – Given a grammar G and a string w, find the parsing tree for w if a parsing tree exists.

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w = x+y*z 

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• -free languages

– A language that does not contain 

• We consider CFG G such that L(G) is -free
• For any cfg G, there is G’ such that L(G’)=L(G)-{}

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Transformation to normal forms: steps

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CFG G=(V, T, P, S) (-free context- free language) Remove (1) -productions (2) unit-productions (3) useless productions from P to get G’ Convert G’ to normal forms

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A substitution rule

• For AB,

A x1Bx2, By1|y2|…|yn is equivalent to Ax1y1x2|x1y2x2|…|x1ynx2, By1|y2|…|yn

• Example

– Aa|aaA|abBc, BabbA|b is equivalent to Aa|aaA|ababbAc|abbc, BabbA|b

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Remove -productions

• -production: A
• Nullable variable A: A* 
• Steps
• 1. Find the nullable variable set VN
• 2. For each Ax1x2…xm, xiVT,
• For each combination xi, xj, …, xk of variables in VN

add Ax1 …xi-1 xi+1… xj-1 xj+1 ... xk-1 xk+1…xm

• Note: don’t add A, if all xi are in VN

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Example

• SABaC, ABC, Bb|, CD|, Dd
• Nullable set VN={A, B, C}
• Add productions

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Remove unit-productions

• unit-production: AB
• Steps

– Remove AA immediately – Draw dependency graph for variables A and B with: A*B – For A*B and By1|y2|…|yn

• Add Ay1|y2|…|yn

– Remove all AB, where A and B are in dependency graph

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Example

• S Aa|B, BA|bb, Aa|bc|B
• 1. Draw dependency graph
• 2. Add

Sbb|a|bc Abb Ba|bc

• 3. Remove SB, BA, AB
• 4. Finally

Sa|bc|bb|Aa Aa|bc|bb Ba|bc|bb

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Remove useless productions

• A variable AV is useful if S can generate

some terminal string through it.

– That is, S * xAy * w, wT*

• Example

– SaSb|AB|Ba, AaA, Bb|Bb, CcB|c – S  Ba  ba. Thus, B is useful. – S is useful. – But, A and C are not useful (useless)

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• Two cases for useless variables

– Case 1: variables that cannot generate strings in T*

• SaSb|AB|Ba, AaA, Bb|Bb, CcB|c
• Algorithm (finding variables that generate strings)
• 1. V1={}
• 2. For rule Ax, x(TV1)*, add A to V1
• 3. Repeat 2 until no rules can be added to V1
• V1={S, B, C}
• SaSb|Ba, Bb|Bb, CcB|c

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– Case 2: variables that cannot be reached from S

• SaSb|Ba, Bb|Bb, CcB|c
• Algorithm: dependency graph
• C is un-reachable from S.
• SaSb|Ba, Bb|Bb

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S B C

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• Algorithm (removing useless productions)

Input: G=(V, T, P, S)

• 1. Find the useless variables in Case 1 and remove

related useless productions.

• 2. Find the useless (un-reachable) variables in Case

2 and remove the related useless productions

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Chomosy normal form

• A cfg is in Chomsky normal form (CNF) if all

productions are of form ABC, or Aa

• Example

– SAS|a, ASA|b

• Every cfg G, with L(G), has an equivalent

CNF grammar.

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Converting into CNF

• 1. Apply the rules of removing -, unit-, and

useless-productions

• 2. Convert the productions into the form

AC1C2…Cn, or Aa

• 3. Convert AC1C2…Cn into

AC1D1, D1C2D2, …, Dn-2Cn-1Cn

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Example

• SABa, Aaab, BAc
• Step 2:
• Step 3:

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Greibach normal form

• A cfg is in Greibach normal form (GNF) if all

productions are of form AaB1B2…Bn, n0

• Example

– SaBC, BaBA, Aa|bBSC

• Every cfg G, with L(G), has an equivalent

GNF grammar.

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Example

• Example

– SAB, AaA|bB|b, Bb – Result

• SaAB|bBB|bB, AaA|bB|b, Bb
• Example

– SabSb|aa – Result

• SaBSB|aA, Bb, Aa

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Parsing (membership)

• Question: Given a CFG G in Chomsky normal

form and a string w, determine whether wL(G)

• Idea: the dynamic programming technique

– A large problem is decomposed into smaller problems – Combine solutions to smaller problems into a solution for the large problem

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• Assume w=a1a2…an
• Main problem: V1n={V: V * w}
• wL(G) if and only SV1n
• Use the dynamic programming technique

– Vij={ V : V* aiai+1…aj}: variables that can generate substring aiai+1…aj

• Solve smaller problems Vik, Vk+1,j, for k=i, i+1,…, j-1
• Combine them to compute Vij

– Vij = {A：ABC, BVik, CVk+1,j, ik<j}

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w = a1 a2 a3 … ai ai+1 … aj-1 aj … an Vij contains the variables that generate aiai+1…aj-1aj ai ai+1 … ak ak+1 ak+2 … aj-1 aj

Vi k Vk+1 j Vk+2 j Vi k+1

. . . . . .

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Vi i Vi+1 i

CYK Algorithm

• Input: G=(V, T, S, P) is in CNF and w=a1a2…an

– Compute Vij={ AV : A* aiai+1…aj}

• V11, V22, …, Vnn
• V12, V23, …, Vn-1n
• …
• V1n
• 1. Smallest problem: add A to Vii
• if Aai is a production in P
• 2. Bigger problem: add to A to Vij if
• For some k, ikj-1, ABC in P, B in Vik, C in Vk+1 j
• 3. wL(G) if and only if SV1n

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Example

• SAB, ABB|a, BAB|b
• w=aabbb
• Steps

– V11={A}, V22={A}, V33={B}, V44={B}, V55={B} – V12=, V23={S, B}, V34={A}, V45={A} – V13={S, B}, V24={A}, V35={S, B} – V14={A}, V25={S, B} – V15={S, B}

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• Triangular table (n=5)

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V1,5 V1,4 V2,5 V1,3 V2,4 V3,5 V1,2 V2,3 V3,4 V4,5 V1,1 V2,2 V3,3 V4,4 V5,5

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Sum up

• Context-free grammars are used in designing

programming languages, such as , C, PSACAL, etc.

• Membership problem in CFG is equivalent to

the parsing problem in programming languages

• Normal forms are needed for “automatically”

generating a “parser” for the programming language

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