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CS 301 Lecture 10 Chomsky Normal Form Stephen Checkoway February - PowerPoint PPT Presentation

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CS 301 Lecture 10 Chomsky Normal Form Stephen Checkoway February 19, 2018 1 / 23 More CFLs A = { a i b j c k i j or i = k } B = { w w { a , b , c } contains the same number of a s as b s and c s combined } C

  1. CS 301 Lecture 10 – Chomsky Normal Form Stephen Checkoway February 19, 2018 1 / 23

  2. More CFLs • A = { a i b j c k ∣ i ≤ j or i = k } • B = { w ∣ w ∈ { a , b , c } ∗ contains the same number of a s as b s and c s combined } • C = { 1 m + 1 n = 1 m + n ∣ m, n ≥ 1 } ; Σ = { 1 , + , = } • D = ( abb ∗ ∣ bbaa ) ∗ • E = { w ∣ w ∈ { 0 , 1 } ∗ and w R is a binary number not divisible by 5 } 2 / 23

  3. Another proof that regular languages are context-free We can encode the computation of a DFA on a string using a CFG Give a DFA M = ( Q, Σ , δ, q 0 , F ) , we can construct an equivalent CFG G = ( V, Σ , R, S ) where 3 / 23

  4. Another proof that regular languages are context-free We can encode the computation of a DFA on a string using a CFG Give a DFA M = ( Q, Σ , δ, q 0 , F ) , we can construct an equivalent CFG G = ( V, Σ , R, S ) where • states of M are variables in G 3 / 23

  5. Another proof that regular languages are context-free We can encode the computation of a DFA on a string using a CFG Give a DFA M = ( Q, Σ , δ, q 0 , F ) , we can construct an equivalent CFG G = ( V, Σ , R, S ) where • states of M are variables in G • q 0 is the start variable, and 3 / 23

  6. Another proof that regular languages are context-free We can encode the computation of a DFA on a string using a CFG Give a DFA M = ( Q, Σ , δ, q 0 , F ) , we can construct an equivalent CFG G = ( V, Σ , R, S ) where • states of M are variables in G • q 0 is the start variable, and • transitions δ ( q, t ) = r become rules q → tr 3 / 23

  7. Another proof that regular languages are context-free We can encode the computation of a DFA on a string using a CFG Give a DFA M = ( Q, Σ , δ, q 0 , F ) , we can construct an equivalent CFG G = ( V, Σ , R, S ) where • states of M are variables in G • q 0 is the start variable, and • transitions δ ( q, t ) = r become rules q → tr If on input w = w 1 w 2 ⋯ w n , M goes through states r 0 , r 1 , . . . , r n , then r 0 ⇒ w 1 r 1 ⇒ w 1 w 2 r 2 ⇒ ⋯ ⇒ w 1 w 2 ⋯ w n r n 3 / 23

  8. Another proof that regular languages are context-free We can encode the computation of a DFA on a string using a CFG Give a DFA M = ( Q, Σ , δ, q 0 , F ) , we can construct an equivalent CFG G = ( V, Σ , R, S ) where • states of M are variables in G • q 0 is the start variable, and • transitions δ ( q, t ) = r become rules q → tr If on input w = w 1 w 2 ⋯ w n , M goes through states r 0 , r 1 , . . . , r n , then r 0 ⇒ w 1 r 1 ⇒ w 1 w 2 r 2 ⇒ ⋯ ⇒ w 1 w 2 ⋯ w n r n So G has derived the string wr n but this still has a variable What additional rules should we add to end up with a string of terminals? For each state q ∈ F , add a rule q → ε 3 / 23

  9. Formally Proof. Given a DFA M = ( Q, Σ , δ, q 0 , F ) , we can construct an equivalent CFG G = ( V, Σ , R, S ) where V = Q S = q 0 R = { q → tr ∶ δ ( q, t ) = r } ∪ { q → ε ∶ q ∈ F } 4 / 23

  10. Formally Proof. Given a DFA M = ( Q, Σ , δ, q 0 , F ) , we can construct an equivalent CFG G = ( V, Σ , R, S ) where V = Q S = q 0 R = { q → tr ∶ δ ( q, t ) = r } ∪ { q → ε ∶ q ∈ F } If r 0 , r 1 , . . . , r n is the computation of M on input w = w 1 w 2 ⋯ w n , then r 0 = q 0 and δ ( r i − 1 , w i ) = r i for 1 ≤ i ≤ n ∗ By construction r 0 ⇒ w 1 r 1 ⇒ w 1 w 2 r 2 ⇒ w 1 w 2 ⋯ w n r n ∗ Therefore, w ∈ L ( M ) iff r n ∈ F iff r n ⇒ ε iff q 0 ⇒ w iff w ∈ L ( G ) 4 / 23

  11. Returning to our language E = { w ∣ w ∈ { 0,1 } ∗ and w R is a binary number not divisible by 5 } Q 1 0 1 0 Q 0 Q 2 1 0 1 1 Q 4 Q 3 1 0 0 5 / 23

  12. Returning to our language E = { w ∣ w ∈ { 0,1 } ∗ and w R is a binary number not divisible by 5 } Q 1 0 Q 0 → 0 Q 0 ∣ 1 Q 2 1 0 Q 1 → 0 Q 3 ∣ 1 Q 0 ∣ ε Q 0 Q 2 Q 2 → 0 Q 1 ∣ 1 Q 3 ∣ ε 1 Q 3 → 0 Q 4 ∣ 1 Q 1 ∣ ε 0 1 Q 4 → 0 Q 2 ∣ 1 Q 4 ∣ ε 1 Q 4 Q 3 1 0 0 5 / 23

  13. Chomsky Normal Form (CNF) A CFG G = ( V, Σ , R, S ) is in Chomsky Normal Form if all rules have one of these forms • S → ε where S is the start variable • A → BC where A ∈ V and B, C ∈ V ∖ { S } • A → t where A ∈ V and t ∈ Σ Note • The only rule with ε on the right has the start variable on the left • The start variable doesn’t appear on the right hand side of any rule 6 / 23

  14. CNF example Let A = { w ∣ w ∈ { a , b } ∗ and w = w R } . CFG in CNF Derivation of baaab S → AU ∣ BV ∣ a ∣ b ∣ ε S T → AU ∣ BV ∣ a ∣ b U → TA V → TB A → a B → b 7 / 23

  15. CNF example Let A = { w ∣ w ∈ { a , b } ∗ and w = w R } . CFG in CNF Derivation of baaab S → AU ∣ BV ∣ a ∣ b ∣ ε S ⇒ BV T → AU ∣ BV ∣ a ∣ b U → TA V → TB A → a B → b 7 / 23

  16. CNF example Let A = { w ∣ w ∈ { a , b } ∗ and w = w R } . CFG in CNF Derivation of baaab S → AU ∣ BV ∣ a ∣ b ∣ ε S ⇒ BV T → AU ∣ BV ∣ a ∣ b ⇒ b V U → TA V → TB A → a B → b 7 / 23

  17. CNF example Let A = { w ∣ w ∈ { a , b } ∗ and w = w R } . CFG in CNF Derivation of baaab S → AU ∣ BV ∣ a ∣ b ∣ ε S ⇒ BV T → AU ∣ BV ∣ a ∣ b ⇒ b V U → TA ⇒ b TB V → TB A → a B → b 7 / 23

  18. CNF example Let A = { w ∣ w ∈ { a , b } ∗ and w = w R } . CFG in CNF Derivation of baaab S → AU ∣ BV ∣ a ∣ b ∣ ε S ⇒ BV T → AU ∣ BV ∣ a ∣ b ⇒ b V U → TA ⇒ b TB V → TB ⇒ b AUB A → a B → b 7 / 23

  19. CNF example Let A = { w ∣ w ∈ { a , b } ∗ and w = w R } . CFG in CNF Derivation of baaab S → AU ∣ BV ∣ a ∣ b ∣ ε S ⇒ BV T → AU ∣ BV ∣ a ∣ b ⇒ b V U → TA ⇒ b TB V → TB ⇒ b AUB A → a ⇒ ba UB B → b 7 / 23

  20. CNF example Let A = { w ∣ w ∈ { a , b } ∗ and w = w R } . CFG in CNF Derivation of baaab S → AU ∣ BV ∣ a ∣ b ∣ ε S ⇒ BV T → AU ∣ BV ∣ a ∣ b ⇒ b V U → TA ⇒ b TB V → TB ⇒ b AUB A → a ⇒ ba UB B → b ⇒ ba TAB 7 / 23

  21. CNF example Let A = { w ∣ w ∈ { a , b } ∗ and w = w R } . CFG in CNF Derivation of baaab S → AU ∣ BV ∣ a ∣ b ∣ ε S ⇒ BV T → AU ∣ BV ∣ a ∣ b ⇒ b V U → TA ⇒ b TB V → TB ⇒ b AUB A → a ⇒ ba UB B → b ⇒ ba TAB ⇒ baa AB 7 / 23

  22. CNF example Let A = { w ∣ w ∈ { a , b } ∗ and w = w R } . CFG in CNF Derivation of baaab S → AU ∣ BV ∣ a ∣ b ∣ ε S ⇒ BV T → AU ∣ BV ∣ a ∣ b ⇒ b V U → TA ⇒ b TB V → TB ⇒ b AUB A → a ⇒ ba UB B → b ⇒ ba TAB ⇒ baa AB ⇒ baaa B 7 / 23

  23. CNF example Let A = { w ∣ w ∈ { a , b } ∗ and w = w R } . CFG in CNF Derivation of baaab S → AU ∣ BV ∣ a ∣ b ∣ ε S ⇒ BV T → AU ∣ BV ∣ a ∣ b ⇒ b V U → TA ⇒ b TB V → TB ⇒ b AUB A → a ⇒ ba UB B → b ⇒ ba TAB ⇒ baa AB ⇒ baaa B ⇒ baaab 7 / 23

  24. Converting to CNF Theorem Every context-free language A is generated by some CFG in CNF. Proof. Given a CFG G = ( V, Σ , R, S ) generating A , we construct a new CFG G ′ = ( V ′ , Σ , R ′ , S ′ ) in CNF generating A . There are five steps. START Add a new start variable BIN Replace rules with RHS longer than two with multiple rules each of which has a RHS of length two DEL- ε Remove all ε -rules ( A → ε ) UNIT Remove all unit-rules ( A → B ) TERM Add a variable and rule for each terminal ( T → t ) and replace terminals on the RHS of rules 8 / 23

  25. Proof continued In the following x ∈ V ∪ Σ and u ∈ ( Σ ∪ V ) + START Add a new start variable S ′ and a rule S ′ → S 9 / 23

  26. Proof continued In the following x ∈ V ∪ Σ and u ∈ ( Σ ∪ V ) + START Add a new start variable S ′ and a rule S ′ → S BIN Replace each rule A → xu with the rules A → xA 1 and A 1 → u and repeat until the RHS of every rule has length at most two 9 / 23

  27. Proof continued In the following x ∈ V ∪ Σ and u ∈ ( Σ ∪ V ) + START Add a new start variable S ′ and a rule S ′ → S BIN Replace each rule A → xu with the rules A → xA 1 and A 1 → u and repeat until the RHS of every rule has length at most two DEL- ε For each rule of the form A → ε other than S ′ → ε remove A → ε and update all rules with A in the RHS 9 / 23

  28. Proof continued In the following x ∈ V ∪ Σ and u ∈ ( Σ ∪ V ) + START Add a new start variable S ′ and a rule S ′ → S BIN Replace each rule A → xu with the rules A → xA 1 and A 1 → u and repeat until the RHS of every rule has length at most two DEL- ε For each rule of the form A → ε other than S ′ → ε remove A → ε and update all rules with A in the RHS • B → A . Add rule B → ε unless B → ε has already been removed 9 / 23

  29. Proof continued In the following x ∈ V ∪ Σ and u ∈ ( Σ ∪ V ) + START Add a new start variable S ′ and a rule S ′ → S BIN Replace each rule A → xu with the rules A → xA 1 and A 1 → u and repeat until the RHS of every rule has length at most two DEL- ε For each rule of the form A → ε other than S ′ → ε remove A → ε and update all rules with A in the RHS • B → A . Add rule B → ε unless B → ε has already been removed • B → AA . Add rule B → A and if B → ε has not already been removed, add it 9 / 23

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