Lecture 19 - IITKhome.iitk.ac.in/~akasha/mth401/lectures/Lecture19.pdfCFG PDM Akash Anand Ø Let L = L(G) Ø Construct PDM P such that N(P) = L Ø P has : q One state q q Input symbols

MTH401ATheory of Computation

Lecture 19

CFG and PDM

Akash Anand

Equivalence of Context Free Languages and Pushdown

Machines

Conversion of CFG to PDMConversion of PDM to CFG

CFG ➜ PDM

Akash Anand

Ø Let L = L(G)Ø Construct PDM P such that N(P) = L

CFG ➜ PDM

Akash Anand

Ø Let L = L(G)Ø Construct PDM P such that N(P) = L

Ø P has :q One state qq Input symbols = terminals of Gq Stack symbols = all symbols of Gq Start symbol = start symbol of Gq Transitions ??

CFG ➜ PDM

Akash Anand

Ø P has :q Transitions :

1. δ(q, a, a) = {(q, ε)}.2. If A → α is a production of G then

δ(q, ε, A) contains (q, α).

CFG ➜ PDM

Akash Anand

Example : Grammar, G : S → 0S0 | 1S1 | ε.

Pushdown Machine, P :States, Q = {q}Input Symbols, Σ= {0, 1}Stack Symbols, Γ= {0,1, S}Start State, q0 = qFinal State, F = {q}

CFG ➜ PDM

Akash Anand

Example : Grammar, G : S → 0S0 | 1S1 | ε.

Pushdown Machine, P :Start Symbol, Z0 = S

CFG ➜ PDM

Akash Anand

Example : Grammar, G : S → 0S0 | 1S1 | ε.

Pushdown Machine, P :Start Symbol, Z0 = STransition Function, δ :

1. δ(q, a, a) = {(q, ε)}.2. If A → α is a production of G then δ(q, ε, A) contains (q, α).

CFG ➜ PDM

Akash Anand

Example : Grammar, G : S → 0S0 | 1S1 | ε.

Pushdown Machine, P :Start Symbol, Z0 = STransition Function, δ :δ(q, 0, 0) = {(q, ε)}δ(q, 1, 1) = {(q, ε)}δ(q, ε, S) = {(q, 0S0), (q, 1S1), (q, ε)}

1. δ(q, a, a) = {(q, ε)}.2. If A → α is a production of G then δ(q, ε, A) contains (q, α).

CFG ➜ PDM

Akash Anand

S

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1 0 1 1 0 1

q

CFG ➜ PDM

Akash Anand

S

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1 0 1 1 0 1

q

δ(q, ε, S) = {(q,0S0), (q,1S1), (q, ε)}

CFG ➜ PDM

Akash Anand

1

S

1

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1 0 1 1 0 1

q

CFG ➜ PDM

Akash Anand

1

S

1

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1 0 1 1 0 1

q

δ(q, 1, 1) = {(q, ε)}

CFG ➜ PDM

Akash Anand

1

S

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

0 1 1 0 1

q

CFG ➜ PDM

Akash Anand

1

S

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

0 1 1 0 1

q

δ(q, ε, S) = {(q,0S0), (q,1S1), (q, ε)}

CFG ➜ PDM

Akash Anand

1

0

S

0

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

0 1 1 0 1

q

CFG ➜ PDM

Akash Anand

1

0

S

0

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

0 1 1 0 1

q

δ(q, 0, 0) = {(q, ε)}

CFG ➜ PDM

Akash Anand

1

0

S

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1 1 0 1

q

CFG ➜ PDM

Akash Anand

1

0

S

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1 1 0 1

q

δ(q, ε, S) = {(q,0S0), (q,1S1), (q, ε)}

CFG ➜ PDM

Akash Anand

1

0

1

S

1

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1 1 0 1

q

CFG ➜ PDM

Akash Anand

1

0

1

S

1

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1 1 0 1

q

δ(q, 1, 1) = {(q, ε)}

CFG ➜ PDM

Akash Anand

1

0

1

S

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1 0 1

q

CFG ➜ PDM

Akash Anand

1

0

1

S

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1 0 1

q

δ(q, ε, S) = {(q,0S0), (q,1S1), (q, ε)}

CFG ➜ PDM

Akash Anand

1

0

1

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1 0 1

q

CFG ➜ PDM

Akash Anand

1

0

1

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1 0 1

q

δ(q, 1, 1) = {(q, ε)}

CFG ➜ PDM

Akash Anand

1

0

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

0 1

q

CFG ➜ PDM

Akash Anand

1

0

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

0 1

q

δ(q, 0, 0) = {(q, ε)}

CFG ➜ PDM

Akash Anand

1

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1

q

CFG ➜ PDM

Akash Anand

1

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

1

q

δ(q, 1, 1) = {(q, ε)}

CFG ➜ PDM

Akash Anand

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

q

CFG ➜ PDM

Akash Anand

Example : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

q

Accept !

CFG ➜ PDM

Akash Anand

Example (continued) : 101101 : S⇒1S1⇒10S01⇒101S101⇒101101

(q, 101101, S) ⊦ (q, 101101, 1S1) ⊦(q, 01101, S1) ⊦ (q, 01101, 0S01) ⊦(q, 1101, S01) ⊦ (q, 1101, 1S101) ⊦(q, 101, S101) ⊦ (q, 101, 101) ⊦(q, 01, 01) ⊦ (q, 1, 1) ⊦ (q, ε, ε).

CFG ➜ PDM

Akash Anand

Ø Formally :

(q,wx,S) ⊦* (q,x,α) ⇔ S ⇒* wα

CFG ➜ PDM

Akash Anand

Ø Formally :

(q,wx,S) ⊦* (q,x,α) ⇔ S ⇒* wα

( ⇒ ) is an induction on the number of steps made by P.

CFG ➜ PDM

Akash Anand

Ø Formally :

(q,wx,S) ⊦* (q,x,α) ⇔ S ⇒* wα

( ⇒ ) is an induction on the number of steps made by P.

Basis : (0 steps)Then, (q,wx,S) ⊦* (q,x,α) andS ⇒* wα are true with α=S, w=ε.

CFG ➜ PDM

Akash Anand

( ⇒ ) continued …

Induction : Consider n moves of P (q,wx,S) ⊦* (q,x,α) and assume the IH for n-1 moves.

CFG ➜ PDM

Akash Anand

( ⇒ ) continued …

Induction : Consider n moves of P (q,wx,S) ⊦* (q,x,α) and assume the IH for n-1 moves.

There are two cases, depending on whether the last move uses a type 1 or type 2 rule.

CFG ➜ PDM

Akash Anand

( ⇒ ) continued …

Induction : Consider n moves of P (q,wx,S) ⊦* (q,x,α) and assume the IH for n-1 moves.

There are two cases, depending on whether the last move uses a type 1 or type 2 rule.

1. δ(q, a, a) = {(q, ε)}.2. If A → α is a production of G then δ(q, ε, A) contains (q, α).

CFG ➜ PDM

Akash Anand

( ⇒ ) continued …

1. The move sequence must be of the form (q,yax,S) ⊦* (q,ax,aα) ⊦(q,x,α), where w = ya.

1. δ(q, a, a) = {(q, ε)}.2. If A → α is a production of G then δ(q, ε, A) contains (q, α).

CFG ➜ PDM

Akash Anand

( ⇒ ) continued …

1. The move sequence must be of the form (q,yax,S) ⊦* (q,ax,aα) ⊦(q,x,α), where w = ya.By IH (applied to first n-1 steps), S ⇒* yaα but w = ya so S ⇒* wα.

1. δ(q, a, a) = {(q, ε)}.2. If A → α is a production of G then δ(q, ε, A) contains (q, α).

CFG ➜ PDM

Akash Anand

( ⇒ ) continued …

2. The move sequence must be of the form (q,wx,S) ⊦* (q,x,Aβ) ⊦(q,x,γβ), where A → γ is a production and α = γβ.

1. δ(q, a, a) = {(q, ε)}.2. If A → α is a production of G then δ(q, ε, A) contains (q, α).

CFG ➜ PDM

Akash Anand

( ⇒ ) continued …

2. The move sequence must be of the form (q,wx,S) ⊦* (q,x,Aβ) ⊦(q,x,γβ), where A → γ is a production and α = γβ.By IH (applied to first n-1 steps), S ⇒* wAβ. Thus S ⇒* wγβ = wα.

1. δ(q, a, a) = {(q, ε)}.2. If A → α is a production of G then δ(q, ε, A) contains (q, α).

CFG ➜ PDM

Akash Anand

Ø (q,wx,S) ⊦* (q,x,α) ⇔ S ⇒* wα

(⇐) is an induction on the number of steps in the leftmost derivation.

CFG ➜ PDM

Akash Anand

Ø (q,wx,S) ⊦* (q,x,α) ⇔ S ⇒* wα

(⇐) is an induction on the number of steps in the leftmost derivation.

Similar idea! Complete the proof.

CFG ➜ PDM

Akash Anand

Ø (q,wx,S) ⊦* (q,x,α) ⇔ S ⇒* wα

(⇐) is an induction on the number of steps in the leftmost derivation.

Similar idea! Complete the proof.ü If it helps, you can assume that

the grammar is in Chomsky Normal form.

PDM ➜ GFG

Akash Anand

PDM ➜ GFG

Akash Anand

Ø Let L = N(P)Ø Construct CFG G such that L(G) = L

PDM ➜ GFG

Akash Anand

Ø Let L = N(P)Ø Construct CFG G such that L(G) = L

Ø If P accepts w by (p,w,X) ⊦* (q,ε,ε) then we need to have a variable that generates w and its productions :q Let’s call this variable [pXq]q [pXq] generates all and only those

strings w such that (p,w,X) ⊦* (q,ε,ε)

PDM ➜ GFG

Akash Anand

X

Simplest Case :

p

δ(p, a, X) = {…, (q, ε), …}

a

PDM ➜ GFG

Akash Anand

Simplest Case :

q

PDM ➜ GFG

Akash Anand

Simplest Case :

Then the production is : [pXq] → a

q

PDM ➜ GFG

Akash Anand

Next Simplest Case :

δ(p, a, X) = {…, (r, Y), …}

αn

:

α2

α1

X

p

a w1 w2… wn

PDM ➜ GFG

Akash Anand

αn

:

α2

α1

Y

Next Simplest Case :

r

w1 w2 … wn

PDM ➜ GFG

Akash Anand

αn

:

α2

α1

Y

Next Simplest Case :

Then the production is : [pXq] → a [rYq]

r

w1 w2 … wn