1. CS 3240 – Chapter 8

2.  Is a n b n c n context-free? CS 3240 - Properties of Context-Free Languages2

3.  a n b n c n is not context-free  Neither is ww  although ww R is!  We will develop a pumping lemma for context-free languages  (oh joy! :-)  as before, can only be used to show that a language is not CF CS 3240 - Properties of Context-Free Languages3

4.  How can you tell by looking at a CFG whether its language is infinite or not? CS 3240 - Properties of Context-Free Languages4

5.  S → aaB  A → bBb | λ  B → Aa  Consider the derivation: S ⇒ aaB ⇒ aaAa ⇒ aabBba ⇒ aabAaba ⇒ aabbBbaba ⇒ aabbAababa … CS 3240 - Properties of Context-Free Languages5

6.  Grammars for infinite CFL must reuse a variable in some derivation  S ⇒ * uAz ⇒ * uvAyz  That is, A ⇒ * vAy  u,v,y,z,are derivable strings of characters and variables  We can repeat the same choices for A again:  S ⇒ * uAz ⇒ * uvAyz ⇒ * uvvAyyz  And again and again… (finally stopping with x)  S ⇒ * uv n Ay n z ⇒ * uv n xy n z  So for any sufficiently long string, s, we have  s = uv n xy n z CS 3240 - Properties of Context-Free Languages6 ⇒

7.  Hint: think of derivation trees, and the relationship between the depth and the number of leaves in a tree CS 3240 - Properties of Context-Free Languages7

8.  If a complete binary has depth d, how many leaves does it have? CS 3240 - Properties of Context-Free Languages8

9.  Consider a path from the root of a tree (S) to a leaf.  It is all variables, except for the leaf  The longer the string, the deeper the path  Eventually a variable must be repeated! CS 3240 - Properties of Context-Free Languages9

10.  Based on a repeated variable (a type of loop)  For sufficiently-long strings (≥ p = 2 v ), some variable will be a descendant of itself in the parse  Every string of sufficient length from an infinite CFL can be written as uvxyz, and pumped as uv i xy i z, i ≥ 0:  |v| + |y| > 0  |vxy| <= p

11.  Intuitively: We’ve already used up the stack to coordinate the a n b n prefix  Must consider all cases for a proof  CFLPL-8.PDF CFLPL-8.PDF

12.  Use the pumping lemma with a p b p a p b p

13.  (N)CFLs are closed under:  Union  Concatenation  Kleene Star  “Regular Intersection” (CF ∩ R = CF)  (N)CFLs are not closed under:  intersection  complement CS 3240 - Properties of Context-Free Languages13

14.  Let S 1 be the start symbol for L 1, and S 2 for L 2  Just have a new start symbol point to the OR of the old ones:  S => S 1 | S 2 S 1 => … S 2 => …

15.  S => S 1 S 2 S 1 => … S 2 => …

16.  Rename the old start variable to S 1  S => S 1 S | λ S 1 => …

17.  Let L 1 = a n b n c m  The concatenation of a n b n and c m  Let L 2 = a n b m c m  The concatenation of a n and b m c m  These are both context-free  L 1 ∩ L 2 = a n b n c n  We already showed this is not context free

18.  Let R be a regular language and L a context- free language  The R ∩ L is context-free  Why? CS 3240 - Properties of Context-Free Languages18

19. CS 3240 - Properties of Context-Free Languages19 a,λb,X ± ss/Xf/λ + fΦf/λ ab ± ABC BAD CDA DCB

20. CS 3240 - Properties of Context-Free Languages20 a,λb,X ± sAs/X; Bf/λ; C sBs/X; Af/λ; D fC; Df/λ; A fD; Cf/λ; B + fA; Bf/λ; C fB; Af/λ; D

21. CS 3240 - Properties of Context-Free Languages21 jail

22.  Proof by contradiction, derived from the formula for intersection: L 1 ∩ L 2 = (L 1 ' + L 2 ')' Since the intersection is not closed, but union is, then the complement cannot be.  (Otherwise we could compute the intersection, which in general is not CF)

23.  Non-determinism is the problem  Remember NFAs?  To find the complement, we needed to first convert to a DFA, then flip the states  Since some CFLs are inherently non- deterministic, they have no deterministic equivalent to “flip”  But… what does this say about deterministic CFLs? CS 3240 - Properties of Context-Free Languages23

24.  DCFLs are closed under:  Complement!  Concatenation  Kleene Star  “Regular Intersection” (CF ∩ R = CF)  DCFLs are not closed under:  Intersection  Union! CS 3240 - Properties of Context-Free Languages24

25.  Consider: L 1 = {a i b j c k | i = j} L 2 = {a i b j c k | j = k}  Each of these is DCF  (Easy to show – your 7.1 homework was similar)  The union is not!  It requires non-determinism  It’s still context-free, but not Deterministic CF

26.  DCFLs always have an associated CFG that is unambiguous

27.  Closed under Union, Concatenation, Kleene Star  Not closed under intersection, complement  CFL ∩ Regular = CFL  DCFLs are closed under complement  But not union!  Swap those two

28.  Unanswerable questions  Answerable questions

29.  Do 2 arbitrary CFGs generate the same language?  Is a CFG ambiguous?  Is a given NCFL’s complement also CF?  Is the intersection of 2 given CFLs CF?  Do 2 CFLs have a common word?

30.  Is a non-terminal ever used in a productive derivation?  Draw the connectivity graph ✔  Does a CFG generate any words?  Substitute each “terminating production” (RHS is all terminals) throughout and see what happens ▪ “back substitution method”  Is a CFL finite or infinite?  Procedure to detect useful, repeated variables

31. CS 3240 - Properties of Context-Free Languages31 S → aA | bB | λ A → a | aCA | bDA | bBa | aAa B → b | aAb | aCB | bDB | bBb C → aCC | bDC D → aCD | bDD First remove useless variables…

32. CS 3240 - Properties of Context-Free Languages32 S → aA | bB | λ A → a | bBa | aAa B → b | aAb | bBb Pick a non-empty, terminal rule: A → a Back-substitute that rule: S → aa | bB | λ A → a | bBa | aaa B → b | aab | bBb Keep going until we have S →, or we can’t continue. We have S → aa. STOP.

33. CS 3240 - Properties of Context-Free Languages33 S → aA | bB | λ A → a | bBa | aAa B → b | aAb | bBb All variables are useful. Let’s see if A is repeated, for instance. First, mark all A’s on the right: S → aA | bB | λ A → a | bBa | aAa B → b | aAb | bBb Now mark all variables affected on the left: S → aA | bB | λ A → a | bBa | aAa B → b | aAb | bBb Since A was marked on the left, it is repeated.

34. CS 3240 - Properties of Context-Free Languages34 S ➞ aA | SBMark on left: A ➞ baB | λ B ➞ bB | bAS ➞ aA | SB A ➞ baB | λ Mark A’s on right:B ➞ bB | bA S ➞ aA | SBA was marked on left. DONE. A ➞ baB | λ B ➞ bB | bA Now mark corresponding variables on left: S ➞ aA | SB A ➞ baB | λ B ➞ bB | bA Repeat marking on right: S ➞ aA | SB A ➞ baB | λ B ➞ bB | bA