1. PDA to CFG Toqa Manasrah

2. Example Convert the PDA P={{p,q}, {0,1}, {X,Z}, q, Z, { }, δ} to a CFG if δ is given by: 1. δ(q,1,Z)={(q,XZ)} 2. δ(q,1,X)={(q,XX)} 3. δ(q,0,X)={(q,X)} 4. δ(p,0,Z)={(q,Z)} 5. δ(q,Λ,X)={(q, Λ)} 6. δ(p,1,X)={(p, Λ)} 2 stack symbols 1 stack symbols 0 stack symbols States Q: q,p Stack symbols: x,z inputs: 0,1 Each stack symbol  one path

3. Initial production S for each variable S  [q 0 z 0 q] S  [q 0 z 0 p] Step 1 Variables: q,p

4.  (q, a,Z) = (p,X) q p q p ? a, Z  X process X  consume a pop Z push X move to state p X not yet processed consume a pop Z push X move to state p Since we don’t know which state the PDA will be in after processing X, define a production [qZr]  a[pXr] that ends in each possible state r

5. Step 2 Λ transition 5. δ (q Λ x ) = {( q, Λ )} [ q x q ]  Λ 6. δ( p, 1, x ) = {( p, Λ )} [ p x p ]  1 Pop x, consume Λ Pop x, consume 1

6. Step 3 Regular transition 3. δ(q,0,X)={(q,X)} [ q x r ]  0[ p x r ] [ q x q ]  0[ p x q ] [ q x p ]  0[ p x p ] r = all states in Q Pop x, consume 0, push x

7. Step 3 Regular transition 4. δ(p,0,Z)={(q,Z)} [ p z r ]  0[ q z r ] [ p z q ]  0[ q z q ] [ p z p ]  0[ q z p ] r = p or r = q

8. Step 3 Regular transition 1. δ(q,1,Z)={(q,XZ)} [qzr]  1[qxk][kzr] [qzq]  1[qxq][qzq] [qzq]  1[qxp][pzq] [qzp]  1[qxq][qzp] [qzp]  1[qxp][pzp] [qzq]  1[qxk][kzq] [qzp]  1[qxk][kzp] Pop z, consume 1, move to state q, push x, mover to k, push z, move to r

9. Step 3 Regular transition 1. δ(q,1,X)={(q,XX)} [qxr]  1[qxk][kxr] [qxq]  1[qxq][qxq] [qxq]  1[qxp][pxq] [qxp]  1[qxq][qxp] [qxp]  1[qxp][pxp] [qzq]  1[qxk][kxq] [qzp]  1[qxk][kxp]

10. [qzq]  1[qxq][qzq] [qzq]  1[qxp][pzq] [qzp]  1[qxq][qzp] [qzp]  1[qxp][pzp] [qxq]  1[qxq][qxq] [qxq]  1[qxp][pxq] [qxp]  1[qxq][qxp] [qxp]  1[qxp][pxp] S  [q 0 z 0 q] S  [q 0 z 0 p] [ q x q ]  0[ p x q ] [ q x p ]  0[ p x p ] [ p z q ]  0[ q z q ] [ p z p ]  0[ q z p ] [ q x q ]  Λ [ p x p ]  1