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Nathan Brunelle Department of Computer Science University of Virginia www.cs.virginia.edu/~njb2b/theory Theory of Computation CS3102 – Spring 2014 A tale of computers, math, problem solving, life, love and tragic death

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Closure Properties of CFLs Theorem: The context-free languages are closed under union. Hint: Derive a new grammar for the union. Let G 1 =(V 1, Σ,R 1, S 1 ), G 2 =(V 2, Σ,R 2, S 2 ) Then G 1+2 =(V 1 +V 2 +{S}, Σ, R 1 + R 2 +{S→ S 1 | S 2 }, S) S→ S 1 | S 2 S 1 →… S 2 →…

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Closure Properties of CFLs Theorem: The CFLs are closed under Kleene closure. Hint: Derive a new grammar for the Kleene closure. Let G =(V, Σ,R, S) Then G * =(V, Σ,R+ {S → SS|ε}, S) S → SS|ε|…

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Closure Properties of CFLs Theorem: The CFLs are closed under with regular langs. Hint: Simulate PDA and FA in parallel. Let FSA M fsa =(Q f, , f, q f, F f ), PDA M pda =(Q p, , , p, q p, F p ) Then M ∩ =(Q f ×Q p, , (q f,q p ), F f ×F p ) Q f ×Q p ( { }) 2 Q s.t. q f, q p, f q f, p q p, M pda a b a stack M fsa ×

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Closure Properties of CFLs Theorem: The CFLs are not closed under intersection. Hint: Find a counter example. L={a n b n c n } is not context free (Involves pumping lemma of CFLs, we’ll get to it) Let L 1 ={a i b j c j }, L 2 ={a i b i c j } L 1 ∩L 1 = L

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Closure Properties of CFLs Theorem: The CFLs are not closed under complementation. Hint: Use De Morgan’s law. Assume FSORC that CFLs are closed under complementation. Since we know that they are closed under union we can find

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Decidable PDA / CFG Problems Given an arbitrary pushdown automata M the following problems are decidable (i.e., have algorithms): Q 1 : Is L(M) = Ø ? Q 5 : Is L(G) = Ø ? Q 2 :Is L(M) finite ? Q 6 :Is L(G) finite ? Q 3 :Is L(M) infinite ? Q 7 :Is L(G) infinite ? Q 4 : Is w L(M) ? Q 8 : Is w L(G) ? ≡ (or CFG G) Extra-credit: prove each!

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Theorem: the following are undecidable (i.e., there exist no algorithms to answer these questions): Q: Is PDA M minimal ? Q: Are PDAs M 1 and M 2 equivalent ? Q: Is CFG G minimal ? Q: Is CFG G ambiguous ? Q: Is L(G 1 ) = L(G 2 ) ? Q: Is L(G 1 ) L(G 2 ) = Ø ? Q: Is CFL L inherently ambiguous ? Undecidable PDA / CFG Problems ≡ Extra-credit: prove each!

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PDA Enhancements Theorem: 2-way PDAs are more powerful than 1-way PDAs. Hint: Find an example non-CFL accepted by a 2-way PDA. Theorem: 2-stack PDAs are more powerful than 1-stack PDAs. Hint: Find an example non-CFL accepted by a 2-stack PDA. Theorem: 1-queue PDAs are more powerful than 1-stack PDAs. Hint: Find an example non-CFL accepted by a 1-queue PDA. Theorem: 2-head PDAs are more powerful than 1-head PDAs. Hint: Find an example non-CFL accepted by a 2-head PDA. Theorem: Non-determinism increases the power of PDAs. Hint: Find a CFL not accepted by any deterministic PDA. Extra-credit: prove each!

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PDA = CFG Theorem: If a Language is derivable by some context free grammar then it is accepted by some PDA Proof idea: Construct a PDA which checks if there is a way of reversing the substitution rules of the grammar to produce the start variable. Machine description: 1.Put the symbol ‘$’ onto the stack 2.Put the start symbol onto the stack 3.Repeat the following: If the top of the stack has a variable V, nondeterministically choose a substitution rule. Pop the V and push the rhs of the rule onto the stack If the top is a terminal symbol ‘a’ then check if the next input symbol is also ‘a’. If so then pop the ‘a’, if not then reject. If the top of the stack is ‘$’ then enter the accept state, accept if all input has been read. Loop startend ε, ↓$ Var S ε, ↓S ε, ↑V↓R a, ↑a ε, ↑$

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PDA = CFG Example: S→aSb|ε Loop startend ε, ↓$ Var S ε, ↓S ε, ↑S ε, ↑$ a, ↑a b, ↑b ε, ↑S↓bSa

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PDA = CFG Example: S→aSaa|B B→bB|ε Loop startend ε, ↓$ Var S ε, ↓S ε, ↑S↓aaSa ε, ↑$ ε, ↑S↓B a, ↑a b, ↑b ε, ↑B↓Bb ε, ↑B

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PDA = CFG Theorem: If a Language is accepted by some PDA then it is derivable by some CFG Proof idea: For every pair of states (p,q) in the machine construct a variable. This variable generates all strings that cause the machine to transition from p with an empty stack to q on an empty stack. The variable for (start, accept) is the start variable. Machine restrictions: There is a unique accept state The stack is empty before accepting Each transition performs at most 1 stack operation Construction: There are two cases for states (p,q): The last item popped is the same as the first item pushed, apply the rule V pq →a V rs b where ‘a’ matches the transition out of p to state r, and ‘b’ matches the transition out of state s to state q. The stack was emptied previously in in the path. For the latter we add rule V pq →V pr V rq where r is the state where the stack becomes empty.

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PDA = CFG Example: V start,start → ε V start,a → a V start,b →ab V start,end → a V a,b a a‘s a end b‘s start b a V a,a → ε V a,b → b V a,end →ba V b,b → ε V b,end →a V end,end → ε

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PDA = CFG Example: V start,start → ε V start,a → V start,b → V start,end → ε| ε V a,b ε V a,start → V a,a → ε a‘s ε, ↓$ end b‘s a, ↓a start b, ↑a ε, ↑$ b, ↑a ε V a,b → a V a,b b|a V a,a b V a,end → V b,start → V b,a → V b,b → ε V b,end → V end, start → V end,a → V end,b → V end,end → ε

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PDA = CFG Example: V start,start → ε V start,a → V start,b → V start,end → ε V a,b ε V a,start → V a,a → ε V a,b → a V a,b a| b V a,b b| ε V a,end → V b,start → V b,a → V b,b → ε V b,end → V end, start → V end,a → V end,b → V end,end → ε a‘s ε, ↓$ end b‘s a, ↓a b, ↓b start ε ε, ↑$ a, ↑ a b, ↑ b

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Parse Tree Describes derivation of a string from a CFG Examples: S→aSaa|B B→BB|b|ε S S S B BB aa aa b b S a for each a L S (S) | SS | S* | S+S S S S SS (a+b)*

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Pumping Lemma for Context Free Languages If L is a CFL then there is some number p such that for any string s in L where |s|>p we can divide s into 5 substrings s=uvxyz in a way to satisfy all of the following: 1.For each i≥0, 2.|vy|>0 3.|vxy|≤p

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Pumping Lemma for Context Free Languages 1.For each i≥0, 2.|vy|>0 3.|vxy|≤p Proof: Consider the CFG for L. The grammar has a finite number of production rules. If a string is long enough, some variable must have been reached twice. S R R uvxyz S R u x z S uz R vy R vxy R

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Pumping Lemma for Context Free Languages 1.For each i≥0, 2.|vy|>0 3.|vxy|≤p Example: Consider: If any of a,b,c are not part of vy then we definitely fail rule 1. Because of rule 3 if v contains the letter ‘a’ then y cannot contain the letter ‘b’

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Pumping Lemma for Context Free Languages 1.For each i≥0, 2.|vy|>0 3.|vxy|≤p Example:. Consider: If neither a,c or b,d are not part of vy then we definitely fail rule 1. Because of rule 3 if v contains the letter ‘a’ then y cannot contain the letter ‘c’ If v contains the letter ‘b’ then y cannot contain the letter ‘d’

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Pumping Lemma for Context Free Languages 1.For each i≥0, 2.|vy|>0 3.|vxy|≤p Example: Consider: If I modify the first set of ‘a’s then I must also modify the second set. If v contains one of the first set of ‘a’s then y cannot contain one of the second set due to rule 3. The case for the ‘b’s is similar.

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Here Be an extra slide

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Pumping Lemma for Context Free Languages 1.For each i≥0, 2.|vy|>0 3.|vxy|≤p Proof: Consider the CFG for L. The grammar has a finite number of production rules. If a string is long enough, some rule must have been applied twice. S R R uvxyz S R u x z S R R uvxyz R vy

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