GATE_2000

THEORY OF COMPUTATION

HINTS AND SOLUTIONS TO PROBLEMS

Q1.4 ANSWER-----------(C)

(a+b*) is the same as (a+b+b*)=(a+b)*

Q1.5 ANSWER--------(B)

L is (00)+ as the 0's always occur in pairs.

(D) is not correct as the grammar is a cfg.

(A) is excluded as 00 is in the language generated.

Q2.9 ANSWER-----------(A)

For P1 simply trace the path through the state diagram of the finite automata.

For P2 take the reduced grammar in the CNF and draw the graph of the grammar. If cycles exist the cfl is infinite else it is finite.

Q.CS7(a). The finite automat state transition diagram is given below.

By examination we find all the states can be distinguished on some input string. So the minimal automata is the same as that given above.

Q3.7(b)By examination it seen that the first production will demand the same number of b's as a's.A can optionally yield extra b's. So we have j>i as the required condition.

Q.CS8.(a) q0 is a push state which pushes 0's and 1's onto the stack. So we have

del(q0,2,s)=(q0,2.s)

q1 is a pop state so we have

del(q1,2,2.s)=(q1,s)

QCS8(b) q0 and q1 will be push states where for every 0 in the input we nondeterministially push one or two 0's onto the stack.

del(q0,0,s)={(q0,0.s),(q2,0.s)}

del(q2,e,s)={(q0,0.s)}

When a 1 occurs we nondeterministically unstack one or two 0's from the stack.

del(q0,1,0.s)={(q1,s)}

del(q0,e,s)={(q1,s)}

In q1 the pop state we unstack one or two 0's for a 1 in the input.

del(q1,1,0.s)={(q1,s),(q2,s)}

del(q2,1,0.s)={(q1,s)}

ANS
1987	1988	1989
1990	1991	1992
1993	1994	1995
1997	1998	1999
2000	2001	2002
2003	2004	1987
QUES
1987	1988	1989
1990	1991	1992
1993	1994	1995
1997	1998	1999
2000	2001	2002
2003	2004	1987

Go to:

2005, GATE THEORY OF COMPUTATION