1. CSCI 2670 Introduction to Theory of Computing September 15, 2005

2. Agenda Yesterday –Introduce context-free grammars Today –No quiz! Quiz postponed until Monday –Regular expressions, pumping lemma, CFG’s –Build CFG’s

3. Context-free grammar definition A context-free grammar is a 4-tuple (V, ,R,S), where 1.V is a finite set called the variables, 2.  is a finite set, disjoint from V, called the terminals, 3.R is a finite set of rules, with each rule being a variable and a string of variables and terminals, and 4.S  V is the start variable.

4. More definitions If u, v, and w are strings of variables and terminals, and A  w is a rule of the grammar, we say uAv yields uwv –Denoted uAv  uwv If a sequence of rules leads from u to v – i.e., u  u 1  u 2  …  v, we denote this u  * v (I can’t do the actual notation in powerpoint – the * should be over the double bars) The language of the grammar is {w   * | s  * w}

5. Example A  Ab | Bb B  aBb | ab V = {A,B}  = {a,b} R is the set of rules listed above S = A The language of this grammar is {w  {a,b} * | w = a n b m, m > n > 0}

6. Designing CFG’s Requires creativity There are some guidelines to help –Union of two CFG’s –Converting a DFA to a CFG –Linked terminals –Recursive behavior

7. Designing the union of two CFG’s For the union of k CFG’s, design each CFG separately with starting variables S 1, S 2, …, S k and combine using the rule S  S 1 | S 2 | … | S k What is the CFG for the following language {a i b j c k | i, j, k ≥ 0 and i = j or j = k} {a i b j c k | i, j, k ≥ 0 and i = j}  {a i b j c k | i, j, k ≥ 0 and j = k}

8. Example First design {a i b j c k | i, j, k ≥ 0 and i=j} S 1  S 1 c | A A  aAb |  Then design {a i b j c k | i, j, k ≥ 0 and j=k} –(use different variables) S 2  aS 2 | B B  bBc |  Finally, add the “unifying” rule S  S 1 | S 2

9. Converting DFA’s into CFG’s For each state q i in the DFA, make a variable R i for your CFG. For each transition rule  (q i,a)=q k in your DFA, add the rule R i  aR k to your CFG For each accept state q a in your DFA, add the rule R a  ε If q 0 is the start state in your DFA, then R 0 is the starting variable in your CFG

10. Example q1q1 0, 1 q2q2 0 1 0 q3q3 1 V = {R 1, R 2, R 3 } R 1  0R 3 | 1R 2 R 2  0R 1 | 1R 3 R 3  1R 3 | 1R 3 R 2  ε R 1 is the start state

11. Linked terminals Terminals may be “linked” to one another in that they have the same number of occurrences (or a related number) –{0 n 1 n | n ≥ 0}, {x n y 2n | n > 0} Add terminals simultaneously –S  0S1 | ε –S  xSyy | xyy

12. Recursive behavior Some languages may be built of pieces that are within the language –Legal pairing of parenthesis For these languages, you will want a recursive rule –S  SS Not all recursive rules will be that easy!

13. Example Construct a CFG accepting all strings in {0,1} * that have equal numbers of 0’s and 1’s S  S0S1S | S1S0S | ε

14. Ambiguity Consider the CFG ({S},{0,1},R,S), where the rules of R are S  0 | 1 | S + S | S * S Derive the string 0 * 1 + 1

15. Ambiguity S  0 | 1 | S + S | S * S 0 * 1 + 1 S SS* S+S 0 11 S SS+ S*S 1 01 Different parse trees!

16. Definition of ambiguity A context-free grammar G generates a string w if there are two different parse trees that generate w –Different derivations that differ only in order do not indicate ambiguity

17. Derivation & ambiguity A derivation of a string w in a grammar G is a leftmost derivation if every step of the derivation replaced the leftmost variable A string is derived ambiguously in CFG G if it has two or more different leftmost derivations The grammar G is ambiguous if it generates some string ambiguously –Some grammars are inherently ambiguous