7.3.1. PDAs and Context Free Languages¶

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Now we want to show that given an NPDA, we can construct a CFG. But first, we will show a result to make the next proof easier.

Theorem: Given a NPDA \(M\), \(\exists\) a NPDA \(M'\) such that all transitions have the form \(\delta(q_i, a, A) = \{c_1, c_2, \ldots c_n\}\) where

Each move either increases or decreases stack contents by a single symbol.

Proof: (sketch)

Theorem: If \(L = L(M)\) for some NPDA \(M\), then \(L\) is a CFL.

Proof:

Given NPDA \(M\), first, construct an equivalent NPDA \(M'\) that will be easier to work with. Construct \(M'\) such that

1. \(M'\) accepts if stack is empty

2. Each move increases or decreases stack content by a single symbol. (Can only push 2 variables or no variables with each transition.)

\(M' = (Q, \Sigma, \Gamma, \delta, q_0, z, F)\)

Construct \(G = (V,\Sigma, S, P)\) where

\(V = \{(q_icq_j)\ |\ q_i, q_j \in Q, c \in \Gamma \}\)

(Some of these variables will be useless.)

\((q_icq_j)\) represents “starting at state \(q_i\) the stack contents are \(cw, w \in \Gamma^*\), some path is followed to state \(q_j\) and the contents of the stack are now \(w\)”.

Goal: \((q_0zq_f)\) which will be the start symbol in the grammar.

Meaning: We start in state \(q_0\) with \(z\) on the stack and process the input tape. Eventually we will reach the final state \(q_f\) and the stack will be empty. (Along the way we may push symbols on the stack, but these symbols will be popped from the stack).

(NOTE: Machine accepts by empty stack, but it is such that there is only 1 final state in which the machine accepts by final state.)

To construct the productions in P:

1) Replace

by

\[(q_iAq_j) \rightarrow a\]

where the stack changes are:

\[\begin{split}\begin{array}{lcclc} & q_i & \ \ (\mbox{some path}\ \rightarrow) \ \ & &q_j \\ \\ \mbox{stack:} & A && \mbox{stack:} & \\ & X_1 & && X_1 \\ & X_2 &&& X_2 \\ & \underline{X_n} &&& \underline{X_n} \\ \end{array}\end{split}\]

2) Replace

by

\[(q_iAq_k) \rightarrow a(q_jBq_l)(q_lCq_k)\ \mbox{for all}\ q_l, q_k \in Q\]

\[\begin{split}\begin{array}{ccccccc} q_i & \ \ (\mbox{path}\ \rightarrow) \ \ & q_j &\ \ (\mbox{path}\ \rightarrow) \ \ & q_l &\ \ (\mbox{path}\ \rightarrow) \ \ & q_k \\ \\ &&B&& \\ A && C &&C \\ X_1 & & X_1 & & X_1 & & X_1 \\ X_2 && X_2 && X_2 && X_2 \\ \underline{X_n} &&\underline{X_n} &&\underline{X_n} &&\underline{X_n}\\ \end{array}\end{split}\]

This will create some useless variables, but that’s ok.

Must show that the constructed grammar \(G\) is such that \(L(G) = L(M')\). That is, \(w \in L(G) \mbox{iff}\ w \in L(M)\). (see book) QED.

Example 7.3.1

\(L(M) = \{aa^*b\}\), \(M = (Q, \Sigma, \Gamma, \delta, q_0, z, F)\), \(Q = \{q_0, q_1, q_2, q_3\}\), \(\Sigma = \{a, b\}, \Gamma = \{A, z\}\), \(F = \{\}\). \(M\) accepts by empty stack.

Construct the grammar \(G = (V,T,S,P)\),

\(V = \{(q_0Aq_0), (q_0zq_0), (q_0Aq_1), (q_0zq_1), \ldots \}\)

NOTE: some variables may be useless.

\(T = \Sigma\)

\(S = (q_0zq_2)\)

\(P =\)

\[\begin{split}\begin{array}{crl} \mbox{From transition 1} & (q_0Aq_1) \rightarrow & b \\ \\ \mbox{From transition 2} & (q_1zq_2) \rightarrow & \lambda \\ \\ \mbox{From transition 3} & (q_0Aq_3) \rightarrow & a \\ \\ \mbox{From transition 4} & (q_0zq_0) \rightarrow & a(q_0Aq_0)(q_0zq_0)| \\ & & a(q_0Aq_1)(q_1zq_0)| \\ & & a(q_0Aq_2)(q_2zq_0)| \\ & & a(q_0Aq_3)(q_3zq_0) \\ & (q_0zq_1) \rightarrow & a(q_0Aq_0)(q_0zq_1)| \\ & & a(q_0Aq_1)(q_1zq_1)| \\ & & a(q_0Aq_2)(q_2zq_1)| \\ & & a(q_0Aq_3)(q_3zq_1) \\ & (q_0zq_2) \rightarrow & a(q_0Aq_0)(q_0zq_2)| \\ & & a(q_0Aq_1)(q_1zq_2)| \\ & & a(q_0Aq_2)(q_2zq_2)| \\ & & a(q_0Aq_3)(q_3zq_2) \\ & (q_0zq_3) \rightarrow & a(q_0Aq_0)(q_0zq_3)| \\ & & a(q_0Aq_1)(q_1zq_3)| \\ & & a(q_0Aq_2)(q_2zq_3)| \\ & & a(q_0Aq_3)(q_3zq_3) \\ \mbox{From transition 5} & (q_3zq_0) \rightarrow & (q_0Aq_0)(q_0zq_0)| \\ & & (q_0Aq_1)(q_1zq_0)| \\ & & (q_0Aq_2)(q_2zq_0)| \\ & & (q_0Aq_3)(q_3zq_0) \\ & (q_3zq_1) \rightarrow & (q_0Aq_0)(q_0zq_1)| \\ & & (q_0Aq_1)(q_1zq_1)| \\ & & (q_0Aq_2)(q_2zq_1)| \\ & & (q_0Aq_3)(q_3zq_1) \\ & (q_3zq_2) \rightarrow & (q_0Aq_0)(q_0zq_2)| \\ & & (q_0Aq_1)(q_1zq_2)| \\ & & (q_0Aq_2)(q_2zq_2)| \\ & & (q_0Aq_3)(q_3zq_2) \\ & (q_3zq_3) \rightarrow & (q_0Aq_0)(q_0zq_3)| \\ & & (q_0Aq_1)(q_1zq_3)| \\ & & (q_0Aq_2)(q_2zq_3)| \\ & & (q_0Aq_3)(q_3zq_3) \\ \end{array}\end{split}\]

Recognizing aaab in M:

\[\begin{split}\begin{eqnarray*} (q_0,aaab,z) & \vdash & (q_0,aab,Az) \\ & \vdash & (q_3,ab,z) \\ & \vdash & (q_0,ab,Az) \\ & \vdash & (q_3,b,z) \\ & \vdash & (q_0,b,Az) \\ & \vdash & (q_1, \lambda, z) \\ & \vdash & (q_2, \lambda, \lambda) \\ \end{eqnarray*}\end{split}\]

At this point stack is empty.

Derivation of string aaab in G:

\[\begin{split}\begin{eqnarray*} (q_0zq_2) & \Rightarrow & a(q_0Aq_3)(q_3zq_2) \\ & \Rightarrow & aa(q_3zq_2) \\ & \Rightarrow & aa(q_0Aq_3)(q_3zq_2) \\ & \Rightarrow & aaa(q_3zq_2) \\ & \Rightarrow & aaa(q_0Aq_1)(q_1zq_2) \\ & \Rightarrow & aaab(q_1zq_2) \\ & \Rightarrow & aaab \\ \end{eqnarray*}\end{split}\]

Meaning of first line in derivation is: \((q_0zq_2) \stackrel{*}{\Rightarrow} axy\) where \((q_0Aq_3)\stackrel{*} {\Rightarrow} x\) (which in the example above will eventually derive \(a\)) and \((q_3zq_2)\stackrel{*}{\Rightarrow} y\) (which in the example above will eventually derive \(ab\)).

Must show that the constructed grammar \(G\) is such that \(L(G) = L(M')\). That is, \(w \in L(G)\) iff \(w \in L(M)\). (see book) QED.