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Chomsky Normal Form

Chomsky Normal Form. Chomsky Normal Form (CNF). Definition If a CFG has only productions of the form Non-Terminal  string of exactly two non-terminals OR Non-Terminal  one terminal It is said to be in Chomsky Normal Form. Chomsky Normal Form (CNF).

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Chomsky Normal Form

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  1. Chomsky Normal Form

  2. Chomsky Normal Form (CNF) Definition If a CFG has only productions of the form Non-Terminal  string of exactly two non-terminals OR Non-Terminal  one terminal It is said to be in Chomsky Normal Form.

  3. Chomsky Normal Form (CNF) • It is to be noted that any CFG can be converted to be in CNF, if the null productions and unit productions are removed. • Also if a CFG contains nullable productions as well, then the corresponding new production are also to be added.

  4. Theorem 25 • If L is a language generated by some CFG, then there exist another CFG that generate all the non-null words of L, all of whose productions are of one of the two basic forms: Non-Terminal  string of only Non-terminals Or Non-Terminal  one Terminal

  5. Proof • The proof is a constructive algorithm by converting a given CFG in a CFG of the production of designed format. • Let us suppose in given CFG the non-terminal are S, X1, X2, X3, … . Let us also assume {a,b} are two terminals. STEP-1 We now add two new non-terminals A and B and the productions A  a B  b

  6. Proof Theorem 25 contd… STEP-2 now for every previous production involving terminals, we replace each ‘a’ with the non-terminal A and each ‘b’ with B. e.g. X3 X4aX1SbbX7a Becomes X3  X4AX1SBBX7A Which is a string of solid non-terminals X6  abaab Becomes X6 ABAAB

  7. S  X1 | X2aX2 | aSb | b X1  X2X2 | b X2  aX2 | aaX2 S  X1 | X2AX2 | ASB | B X1  X2X2 | B X2  AX2 | AAX2 A  a B  b Example

  8. Example • Right sides of other production may contain terminals and non-terminals e.g. N  aM | NaMb • We need to have string of solid non-terminal on right side we introduce now products as X a Y  b And apply them to all rules getting N  XM | NXMY

  9. Proof contd. • Introduce new non-terminals to restrict the length to 2 N  XM X  a N  NR1 Y  b R1 XR2 R2  MY

  10. Theorem 26 Theorem 26: For any context-free language L, the non-Λ words of L can be generated by a grammar in which all productions are in CNF.

  11. Proof of Theorem 26 • The proof will be by constructive algorithm. • From Theorems 23 and 24, we know that there is a CFG for L (or for all L except Λ) that has no Λ-productions and no unit productions. • Suppose further that we have made the CFG fit the form specified in Theorem 25. • For productions of the form Nonterminal  one terminal we leave them alone. • For each production of the form Nonterminal  string of Nonterminals we use the following expansion that involves new nonterminals R1, R2, ...

  12. Proof of Theorem 26 contd. • For instance, the production S  X1X2X3X8 should be replaced by S  X1R1 where R1 X2R2 and where R2 X3X8. • Thus, we convert a production with long string of non-terminals into a sequence of productions with exactly two non-terminals. • If we need to convert another production, we introduce new R’s with different subscripts. • The new grammar generates the same language as the old grammar with the possible exception of the word Λ.

  13. Remarks • Theorem 26 has enabled a form of algebra of productions by introducing non-terminals R1, R2, R3 … • The derivation tree associated with a derivation in Chomsky Normal Form grammar is a binary tree.

  14. Example S  aSa | bSb | a | b | aa | bb S  AR2 R2  SA S  BR1 R1 SB S  AA S  BB S  a S  b A  a B  b S  ASA S  BSB S  AA S  BB

  15. Example Consider the following CFG S → ABAB A → a|Λ B → b|Λ Here S → ABAB is nullable production and A → Λ, B → Λ are null productions. Removing the null productions A → Λ and B → Λ, and introducing the new productions as S → BAB|AAB|ABB|ABA|AA|AB|BA|BB|A|B Now S → A|B are unit productions to be eliminated as shown below S → A gives S → a (using A → a) S → B gives S → b (using B → b)

  16. Example Thus the new resultant CFG takes the form S → BAB|AAB|ABB|ABA|AA|AB|BA|BB|a|b, A → a, B → b. Introduce the nonterminal C where C → AB, so that S → BAB|AAB|ABB|ABA|AA|AB|BA|BB|a|b S → BC|AC|CB|CA|AA|C|BA|BB|a|b A → a B → b C → AB is the CFG in CNF.

  17. Leftmost Derivation Definition: • The leftmost non-terminal in a working string is the first non-terminal that we encounter when we scan the string from left to right. Example: In the string abNbaXYa, the leftmost non terminal is N. Definition: • If a word w is generated by a CFG such that at each step in the derivation, a production is applied to the leftmost non-terminal in the working string, then this derivation is called a leftmost derivation.

  18. Example: Left derivation S  aSX | b X  Xb | a aababa S  aSX  aaSXX  aabXX  aabXbX  aababX  aababa

  19. Parsing Definition: the process of finding the derivation of word generated by particular grammar is called parsing. There are different parsing techniques, containing the following three • Top down parsing. • Bottom up parsing. • Parsing technique for particular grammar of arithmetic expression.

  20. Top-Down Parsing • The parse tree is created top to bottom. • Top-down parser • Recursive-Descent Parsing • Backtracking is needed (If a choice of a production rule does not work, we backtrack to try other alternatives.) • It is a general parsing technique, but not widely used. • Not efficient • Predictive Parsing • no backtracking • efficient • needs a special form of grammars (LL(1) grammars). • Recursive Predictive Parsing is a special form of Recursive Descent parsing without backtracking. • Non-Recursive (Table Driven) Predictive Parser is also known as LL(1) parser.

  21. Recursive-Descent Parsing (uses Backtracking) • Backtracking is needed. • It tries to find the left-most derivation. S  aBc B  bc | b S S input: abc a Bc a B c b c b fails, backtrack

  22. Left recursion • Any productions of the form A  Aα A  BA where B*ε Are called left recursive productions

  23. Left Recursion Removal • Consider the grammar A  Aa | b • Halting condition of top-down parsing depend upon the generation of terminal prefixes to discover dead ends. • Repeated application of above rule fail to generate a prefix that can terminate the parse.

  24. Removing left recursion • For each rule of the form • A -> A α1 | A α2| ………| A αn | β 1 |β 2 | …β n • A is a left-recursive nonterminal. • α is a sequence of nonterminals and terminals. • β is a sequence of nonterminals and terminals that does not start with A. • Replace the A-production by the production: • A -> β 1 A’ |….| β n A’ • And create a new nonterminal • A’ -> Λ| α1 A’ |….|αn A’

  25. Removing left recursion • To remove left recursion from A, the A rules are divided into two groups. Left recursive and others A  AU1 | AU2 | AU3 | ….| AUj A  V1 | V2 | V3 | … | Vk Solution: A  V1 | V2 | V3 | … | Vk| V1Z | V2Z | … | VkZ Z  u1Z | u2Z | u3Z | … | ujZ | u1 | u2 | u3 | … | uj

  26. A  Aa | b Solution: A  bZ | b Z  aZ | a A  Aa | Ab | b | c A  bZ | cZ | b | c Z  aZ | bZ | a | b Example

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