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# CS402 Theory of Automata Assignment#06 Solution & Discussion Spring 2010

Question:                                                                                                                                     Mark:5+5+5+5

Consider the following Context Free Grammar (CFG)

S → aY | Ybb | Y

X → Λ | a

Y → aXY | bb | XXa

b)      Draw a Total Language Tree (TLT) for the given CFG.

c)      Convert the CFG into CNF.

d)      Build the PDA corresponding to the CFG (in CNF) of part (c).

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### Replies to This Discussion

if any one have GDB solution then plx share i want some idea plx..
yes an FA recognizes all the languages accepted by a PDA Actually PDA is a new form to express a language like as FA. The new machines which are to be defined are more powerful and can be constructed with the help of FAs with new format. Languages accepted by FA or NFA or TG also accepted by deterministic PDA.
no FA doesnot recognize all the languages accepted by PDA. FA has low memory problem. so it does not accept the language which require large memory.PDA has stack so it has no memory problem
NO, an Fa does not recognize all the languages accepted by PDA. A PDA can make use of a stack to remember how many times a symbol has appeared in a given word.That means the language a^nb^n can be accepted by a PDA, but a FA has no way of keeping track of how many times a symbol has occured, so it can't accept a language a^nb^n.
plz yar send assignment solution.................
Question No 1:
Write CFG for the language given by the following Regular Expression
(ab+ba)(aa+b)*(aaa+bba)
Solution;
S-àabA│ baA
AàaaA│ bA │aB│ bB
BàaC│ bC
D→∆
bhai plz upload the solution of 6th assignment of cs 402 plz
bhai plz give the solution of GDB of automata thanx today is last date
assigment solution tips required..........................share plz
first ka answer mjy lgta ha yes it is ambiguous coz one string can be formed from different formats. 2nd ques me tree ki diagram buht complicated hoti ha ne bun rai
1) this CFG Is not ambiguous because any string in not shown two derivation tree.
2) second is its total language tree may be arbitrary wide as well as infinity long.
3) CFG Into CNF removing null and unit production
s-> Ay| yBB | b
x-> a | a
y-> Axy | Ax |BB | xxa| xx | xA | a
A->a
B->b

If there is else solution kindly, discuss it.
This is wrong.I was not able to convert it into CNF, but I am very close to do it so!
S->aY|Ybb|Y
X->^|a
Y->aXY|bb|XXa
null production
Y->aY|bb|Xa|a
after removing null production
S->aY|Ybb|Y
X->a
Y->aY|bb|Xa|a
Unit production
S->Y
S->Y->aY gives S->abb (Y->bb)
S->Y->aY gives S->aa (Y->a)
S->Y->aY->aXa gives S->aaa (X->a)
New CFG will be
S->aY|Ybb|abb|aa|aaa
X->a
Y->aY|bb|Xa|a
(i am stuck here)

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