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

1 / 7

# Context free languages (?, V, S, R) (you have a stack) ? = Set of terminals, V = Set of variables, V ? ? = - PowerPoint PPT Presentation

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 )

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Context free languages (?, V, S, R) (you have a stack) ? = Set of terminals, V = Set of variables, V ? ? =' - Faraday

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

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)  ∑*

S

A B

a A b B

.

.

.

.

.

.

.

.

.

.

.

.

If we have rules:

SA 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

S

S1 S2

L1L2

• 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 = {SS1 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.)