COSC 3340 Introduction to Theory of Computation University

COSC 3340: Introduction to Theory of Computation University of Houston Dr. Verma Lecture 14 1 Lecture 14 Uof. H - COSC 3340 - Dr. Verma

PDA’s and CFG’s l l 2 For every CFG G there is a PDA M such that L(G) = L(M) For every PDA M there is a CFG G such that L(M) = L(G) Lecture 14 Uof. H - COSC 3340 - Dr. Verma

CFG PDA l l Given CFG G = (V, , R, S) Let PDA M = (Q, , V {$}, , qstart, {qaccept}) – l contains transitions for the form 1. 2. 3. 4. 3 Q = {qstart, qloop, qaccept} , S$ ((qstart, , ), (qloop, S$)) For each rule A w R(G) there is a transition ((qloop, , A), (qloop, w)) *** For each symbol ((qloop, , ), (qloop, )) , $ ((qloop, , $), (qaccept, )) Lecture 14 qstart , , A w qloop qaccept Uof. H - COSC 3340 - Dr. Verma

CFG PDA l 1. 2. The PDA simulates a leftmost derivation of the string. Place the marker symbol $ and the start variable on the stack. Repeat the following steps forever (a) If the top of stack is a variable symbol A, nondeterministically select on of the rules for A and substitute A by the string on the right-hand side of the rule. (b) If the top of stack is terminal symbol , read the next symbol from the input and compare it to . If they match, repeat. If they do not match, reject on this branch of the nondeterminism. (c) If the top of stack is the symbol $, enter the accept state. Doing so accepts the input if it has all been read. 4 Lecture 14 Uof. H - COSC 3340 - Dr. Verma

Example: S a. Sb | Z is used instead of $ Instead of q 0 = qstart q 1 = qloop q 2 = qaccept 5 Lecture 14 Uof. H - COSC 3340 - Dr. Verma

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Idea of PDA CFG l First, we simplify our task by modifying P slightly to give it the following three features 1. 2. 3. 32 It has a single accept state, qaccept. It empties its stack before accepting. Each transition either pushes a symbol onto the stack (a push move) or pops one off the stack (a pop move), but does not do both at the same time. Lecture 14 Uof. H - COSC 3340 - Dr. Verma

PDA CFG l l l 33 Suppose P = {Q, , , Δ, q 0, {qaccept}} to construct G. The variables of G are {Apq | p, q Q}. Apq generates all the strings that can take P from p with an empty stack to q with an empty stack. Two possibilities occur during P’s computation on x. Either the symbol popped at the end is the symbol pushed at the beginning or not. First, simulated by Type 1 rules on next slide and the second by Type 2 rules. Lecture 14 Uof. H - COSC 3340 - Dr. Verma

PDA CFG l The start variable is Aq 0 qaccept. Now we describe G’s rules. – – 34 [Type 1] For each p, q, r, s Q, t , and a, b , if ((p, a, ), (r, t)) is in Δ and ((s, b, t), (q, )) is in Δ, put the rule Apq a. Arsb in G. In other words, find pairs of transitions in the PDA such that the first transition in the pair pushes a symbol t and the second transition pops the same symbol t. Each such pair of transitions gives a Type 1 rule. The states p, q, r, s, and the symbols a, b are determined by looking at the transitions in the pair. Lecture 14 Uof. H - COSC 3340 - Dr. Verma

PDA CFG – – 35 [Type 2] For each p, q, r Q put the rule Apq Apr. Arq in G. [Type 3] Finally, for each p Q put the rule App in G. Lecture 14 Uof. H - COSC 3340 - Dr. Verma

Example l Let M be the PDA for {anbn | n > 0} Note that n cannot be 0, which makes the example a little simpler. M = {{p, q}, {a, b}, {a}, , p, {q}}, where = {((p, a, ), (p, a)), ((p, b, a), (q, )), ((q, b, a), (q, ))} – p 36 q Lecture 14 Uof. H - COSC 3340 - Dr. Verma

Example: contd. l l CFG, G = (V, {a, b}, Apq, R) corresponding to M has V = {App, Apq, Aqp, Aqq}. R contains the following rules: Type I: – – l Type II: – – l App App | Apq Aqp Apq App Apq | Apq Aqp App | Aqq Aqp Aqq Aqp Apq | Aqq Type III: – – 37 Apq a. Appb Apq a. Apqb App Aqq Lecture 14 We can discard all rules containing the variables Aqq and Aqp. And we can also simplify the rules containing App and get the grammar with just two rules Apq ab and Apq a. Apqb. Uof. H - COSC 3340 - Dr. Verma

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