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Context free languages (∑, V, S, R) (you have a stack) ∑ = Set of terminals, V = Set of variables, V ∑ = S = Start variable,PowerPoint Presentation

Context free languages (∑, V, S, R) (you have a stack) ∑ = Set of terminals, V = Set of variables, V ∑ = S = Start variable,

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Context free languages (∑, V, S, R) (you have a stack) ∑ = Set of terminals, V = Set of variables, V ∑ = S = Start variable, R V x (V ∑) * production rules. ( Ex: A a B C a b )

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- Context free languages (∑, V, S, R) (you have a stack)
- ∑ = Set of terminals,
- V = Set of variables, V ∑ =
- S = Start variable,
- R V x (V ∑)* production rules.
- (Ex: A a B C a b )
- We say x | y if x = u A v, y = u w v, A w where A V and u, w, v (V ∑*).
- We say x |* y if x | z1, z1 | z2, …, zk-1 | zk, zk | y
- for some z1, z2, …zk V ∑*, k Z+
- Given G, L(G) is the language generated by G.
- L(G) = {w ∑* | S|* w}. (G = (V, ∑, S, R))
- Note: L(G) ∑*

A B

a A b B

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If we have rules:

SA B, A aA, A , B bB, B ,

We might have:S | AB | aAB | aAbB | …

Which we represent as: (because order doesn't matter)

Notice this language is a* b* .

- Context Free Grammar (CFG)
- G = (V, ∑, R, S)
- V = Variables, ∑ = Terminals, R = Production Rules,
- Where S = Start State, S V, and R V x (V ∑)*
- L = {an bn | n N {0}} is not regular. Is it context-free? Yes. The grammar G = (V, ∑, R, S), where
- ∑ = {a, b}, V = {S}, R = { S | a S b | } generates L !
- L = {an bm | n m}: R = {S | a S b | a S | }.
- L = {an bm | n > m}: R = {S | a A ; A | a A b | | a A}.

- Thm: Every regular language is context free.
- Pf: Let L be regular. Then a finite automaton
- (Q, ∑, , q0, F). We will build a CFG for L.
- The grammar is ( V, ∑1, R, S1),
- where V = Q S1, ∑1 = ∑, S1 = Sq0,
- and if Q = {q0, q1, …}, V = {Sq0, Sq1, …}.
- Finally, for every transition (qi, ) = qj ,
- write Sqi Sqj as a rule. Finally, add the rule
- Sqk if qk F.
- Now we need to argue that this grammar accepts L.
- Why is this the case?

Thm: The union of two context free languages is also

context free.

Pf: Let G1 = (V1, ∑1, R1, S1) and G2 = (V2, ∑2, R2, S2) be CFG's.

We want to create G = (V, ∑, R, S) s.t. L (G) = L(G1) L (G2).

where the rules are: R = {S S1 | S2} R1 R2 ,

∑ = ∑1 ∑2, and V = S V1 V2 . x

S1 S2

L1L2

- Thm: Concatenation of context-free languages is context free.
- Pf: Given G1 = (V1, ∑1, R1, S1) and G2 = (V2, ∑2, R2, S2), CFGs, define a new grammar G = (V, ∑, R, S):
- R = {SS1 S2} R1 R2,
- ∑ = ∑1 ∑2 , V = V1 V2 S. x
- Thm: CFG's closed under *
- Pf: Given G=(V, S, R, S), define a new grammar
- G = (V, ∑, R1, S) where
- R1 = {S S S | } R. x

- Note:
- L = { w wR | w ∑*} is context free since: S a S a | b S b | generates it.
- L = { w w | w ∑*} is not context free.
- L = { an bn cn | n Z+} is not context free.
- (we’ll prove these next time!)
- Careful:
- L3 = {an bn cm | n, m N {0}} is context free:
- S a S1 b S2 | | S2S2 c S2 | (*or use concatenation!)S1 a S1 b |
- So is L4 = {am bn cn | n, m N {0}}.
- ButL5 = L3 L4 = { an bn cn | N {0}}, so this shows CFG's
- aren't closed under intersection.
- (And thus they aren’t closed under complement either.)

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