1. 1 Context Free Languages October 2, 2001

2. 2 Announcement HW 3 due now

3. 3 Agenda Context Free Grammars Derivations Parse-trees CFGParse Ambiguity

4. 4 Motivating Example pal = { x  {0,1}* | x =x R } = { , 0,1, 00,11, 000,010, 101,111, 0000,0110,1001,1111, … } Saw last time that pal non-regular. But there is a pattern. Q: If you have one palindrome, how can you generate another?

5. 5 Motivating Example pal = { x  {0,1}* | x =x R } A: Can generate pal recursively as follows. BASE CASE: , 0 and 1 are palindromes RECURSE: if x is a palindrome, then so are 0x 0 and 1x 1.

6. 6 Motivating Example pal = { x  {0,1}* | x =x R } SUMMARY: In pal any x can be replaced by any one of { , 0, 1, 0x 0, 1x 1}. NOTATION: x   |0|1|0x 0|1x 1 Each pipe “|” is an or, just ad in UNIX regexp’s. In fact, all elements of pal can be generated from  by using these rules  Q: How would you generate 11011011 starting from the variable x ?

7. 7 Motivating Example A: Generate the string from outside-in 11011011 = 1(1(0(1()1)0)1)1 so that: x  1x1  11x11  110x 011  1101x1011  1101  1011 = 11011011

8. 8 Context Free Grammars. Definition DEF: A context free grammar consists of (V, , R, S ) with: V –a finite set of variables (or symbols, or non- terminals)   –a finite set set of terminals (or the alphabet ) R –a finite set of rules (or productions) of the form v  w with v  V, and w  (   V )* (read “v yields w ” or “v produces w ” ) S  V –the start symbol. Q: What are (V, , R, S ) for pal ?

9. 9 Context Free Grammars. Definition A: V = {x}  = {0,1}, R = {x  , x  0, x  1, x  0x 0, x  1x 1} S = x Standardize pal, reset: V = {S }, S = S R = {S  , S  0, S  1, S  0S 0, S  1S 1}

10. 10 Derivations DEF: The derivation symbol “  ”, read “1-step derives” or “1-step produces” is a relation between strings in (   V )*. We write x  y if x and y can be broken up as x = svt and y = swt with v  w a production in R. Q: What possible y satisfy (in pal) S 0000S  y ?

11. 11 Derivations A: S 0000S  y : Any one of: 0000S, 00000S, 10000S, 0S00000S, 1S10000S, S0000, S00000, S00001, S00000S0, S00001S1

12. 12 Derivations DEF: The derivation symbol “  *”, read “derives” or “produces” or “yields” is a relation between strings in (   V )*. We write x  *y if there is a sequence of 1-step productions from x to y. I.e., there are strings x i with i ranging from 0 to n such that x = x 0, y = x n and x 0  x 1, x 1  x 2, x 2  x 3, …, x n-1  x n Q: Which of LHS’s yield which of RHS’s in pal? 01S, SS  ?  01, 0, 01S, 01110111, 0100111

13. 13 Derivations A: Not all answers are unique. 01S  * 01 SS  * 0 01S  * 01S SS  * 01110111 Nothing yields 0100111

14. 14 Language Generated by a CFG DEF: Let G be a context free grammar. The language generated by G is the set of all terminal strings which are derivable from the start symbol. Symbolically: L(G ) = {w   * | S  * w}

15. 15 Example. Infix Expressions Infix expressions involving {+, , a, b, c, (, )} E stands for an expression (most general) F stands for factor (a multiplicative part) T stands for term (a product of factors) V stands for a variable, I.e. a, b, or c Grammar is given by: E  T | E+T T  F | T  F F  V | (E ) V  a | b | c CONVENTION: Start variable is the first one (E)

16. 16 Example. Infix Expressions EG: Consider the string u given by a  b + (c + (a + c ) ) This is a valid infix expression. Should be able to generated it from E. 1. A sum of two expressions, so first production must be E  E +T 2. Sub-expression a  b is a product, so a term so generated by sequence E +T  T +T  T  F +T  * a  b +T 3. Second sub-expression is a factor only because a parenthesized sum. a  b +T  a  b +F  a  b +(E )  a  b +(E +T )

17. 17 Example. Infix Expressions Continuing on in this fashion and summarizing: E  E +T  T +T  T  F +T  F  F +T  V  F+T  a  F+T  a  V+T  a  b +T  a  b +F  a  b + (E )  a  b + (E +T )  a  b+(T+T)  a  b+(F+T)  a  b+(V+T )  a  b + (c +T )  a  b + (c +F )  a  b + (c + (E ) )  a  b + (c +(E +T ))  a  b+(c+(T+T ) )  a  b +(c + (F +T ) )  a  b+(c+(a+T ) )  a  b +(c +(a + F ) )  a  b+(c+(a+V ) )  a  b + (c + (a+c ) ) This is a so-called left-most derivation.

18. 18 Right-most derivation In a right-most derivation, the variable most to the right is replaced. E  E +T  E + F  E + (E )  E + (E +T )  E + (E +F )  etc. There is a lot of ambiguity involved in how a string is derived. However, if decide that derivations are left-most, or right-most, each derivation is not unique. Another way to describe a derivation in a unique way is using derivation trees.

19. 19 Derivation Trees In a derivation tree (or parse tree) each node is symbol. Each parent is a variable whose children spell out the production from left to right. For, example v  abcdefg: The root is the start variable. The leaves spell out the derived string from left to right. v abcdefg

20. 20 Derivation Trees EG, a derivation tree for a  b + (c + (a + c ) ) Advantage. Derivation trees also help understanding semantics! You can tell how expression should be evaluated from the tree.

21. 21 CFGParse CFGParse is a tool that I wrote that helps you play with context free grammars and create derivation trees. Yoav Hirsch (TA) also added some functionality that allows you to convert CFG’s into various forms that we’ll learn about next week. USAGE: java CFGParse pdflatex Second command only works on CUNIX, unless install LaTeX plus some libraries (ecltree).

22. 22 Ambiguity  | with   | and  |  Hannibal | Clarice | rice | onions  ate | played  with | and | or  | Clarice played with Hannibal Clarice ate rice with onions Hannibal ate rice with Clarice Q: Are there any suspect sentences?

23. 23 Ambiguity A: Consider “Hannibal ate rice with Clarice”. Could either mean Hannibal and Clarice ate rice together. Hannibal ate rice and ate Clarice. And this is not absurd, given what we know about Hannibal! This ambiguity arises from the fact that the sentence has two different parse-trees, and therefore two different interpretations:

24. 24 Hannibal and Clarice Ate noun Hannibal Clarice subject sentence activity with actionsubject objectverb aterice

25. 25 noun Hannibal Claric e subject sentence activity action objectverb ate with Hannibal the Cannibal object rice prepnoun

26. 26 Ambiguity. Definition DEF: A string x is said to be ambiguous relative the grammar G if there are two essentially different ways to derive x in G. I.e. x admits two (or more) different parse-trees (equivalently, x admits different left-most [resp. right-most] derivations). A grammar G is said to be ambiguous if there is some string x in L(G ) which is ambiguous. Q: Is the grammar S  ab | ba | aSb | bSa |SS ambiguous? What language is generated?

27. 27 Ambiguity A: L(G ) = the language with equal no. of a’ s and b’ s Yes, the language is ambiguous: SSSS a a b b a a b b S S S S S

28. 28 CFG’s Proving Correctness The recursive nature of CFG’s means that they are especially amenable to correctness proofs. For example let’s consider the grammar G = ( S   | ab | ba | aSb | bSa | SS ) with purported generated language L(G ) = { x  {a,b}* | n a (x) = n b (x) }. Here n a (x) is the number of a’s in x, and n b (x) is the number of b’s.

29. 29 CFG’s Proving Correctness Proof. There are two parts. Let L be the purported language generated by G. We want to show that L = L(G ). The usual way to prove that sets are equal is to show both inclusions: I. L  L(G ). Every string with purported can be generated by G. II. L  L(G ). G only generate strings of the purported pattern.

30. 30 Proving Correctness L  L(G ) I. L  L(G ): Show that every string x with the same number of a’s as b’s is generated by G. Prove by induction on the length n = | x |. Base case: n = 0. So x is empty so derived by the production S  . Inductive hypothesis: Assume n > 0. Let u be the smallest non-empty prefix of x which is also in L. There are two case: 1) u = x 2) u is a proper prefix

31. 31 Proving Correctness L  L(G ) I.1) u = x : Notice that u can’t start and end in the same letter. E.g., if it started and ended with a then write u = ava. This means that v must have 2 more b’s than a’s. So somewhere in v the b’s of u catch up to the a’s which means that there’s a smaller prefix in L, contradicting the definition of u as the smallest prefix in L. Thus for some string v in L we have u = avb ORu = bva The length of v is shorter than u so by induction we can assume that a derivation S  * v exists. Therefore u can be derived by one of: S  aSb  * avb = xOR S  aSb  * avb = x

32. 32 Proving Correctness L  L(G ) I.2) u is a proper prefix. So can write x = uv with both u and v in L As each of u, v are shorter than x, can apply inductive hypothesis to assume that S  *u and S  *v. To derive x just use: S  SS  * uS  * uv = x

33. 33 Proving Correctness L  L(G ) II) Everything derivable from L conforms to the pattern. We prove something more general: Any string in {S,a,b}* derivable from S contains the same number of a’s as b’s. Again induction is used, but this time on the number of steps n in the derivation. Base case n = 0: The only thing derivable in 0 steps is S which has 0 a’s and 0 b’s. OK! Intuctive step: Assume that u is any string derivable from S in n steps. I.e., S  *u. Any further step would have to utilize on of S   | ab | ba | aSb | bSa | SS. But each of these productions preserves the relative number of a’s vs. b’s. Thus any string derivable from S in n+1 steps must also have the same number of a’s as b’s. //QED

34. 34 CFG Exercises On black-board