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Midterm 1 Breakdown >79 3 >29 7 >69 5 >19 5 >59 7 >49 9 >39 7.

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Presentation on theme: "Midterm 1 Breakdown >79 3 >29 7 >69 5 >19 5 >59 7 >49 9 >39 7."— Presentation transcript:

1 Midterm 1 Breakdown >79 3 >29 7 >69 5 >19 5 >59 7 >49 9 >39 7

2 Project 1

3 CS 3240: Languages and Computation Context-Free Languages

4 What are Grammars? A grammar G is defined by (V, T, P, S), where  V is a finite set of variables  T is a finite set of terminal symbols  P is a finite set of productions or ules  S  V is the special start variable Each grammar G defines a language L(G), which is the set of strings in T* (=Σ*) that G can generate from S. Automata are actuated by the transition table, grammars by the production rules.

5 Context-Free Grammars A context-free grammar (V, T, P, S) is a grammar where all production rules are of the form: A → x, with A  V and x  (V  T)* A string in (V  T)* is called sentential form E.g., let G = ({S}, {a,b}, P, S) with for P:  S→aSa  S→bSb  S→ . Some derivations from this grammar:  S  aSa  aaSaa  aabSbaa  aabbaa  S  bSb  baSab  baab, and so on.  In general S  ….  ww R for w  {a,b}*.

6 Beyond regular Languages

7 Formal definition of CF Grammar & Language

8 Another CFG Example Consider the CFG G=({S,Z},{0,1}, P, S) with P: S  0S1 | 0Z1 Z  0Z |  What is the language generated by this G? Answer: L(G) = {0 i 1 j | i  j } Specifically, S yields the 0 j+k 1 j according to: S  0S1  …  0 j S1 j  0 j Z1 j  0 j 0Z1 j  …  0 j+k Z1 j  0 j+k ε1 j = 0 j+k 1 j

9 Exercise Design CFGs for the following languages: a) { 0 n 1 n : n ≥ 0} b) { 0 n 1 m : n,m ≥ 0} c) { (0|1)* : # of 0s > # of 1s} Answers: a) S → 0S1 |  b) S → 0S | R and R → 1R |  c) S → T0T and T → TT | 0T1 | 1T0 | 0 | 

10 derivations

11 Leftmost righmost

12 Sentential forms


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