CLICK ON A PICTURE BELOW FOR PREVIOUS GATE QUESTIONS 1987 ONWARDS IN THE THEORY OF COMPUTATION


GATE_2004
THEORY OF COMPUTATION
HINTS AND SOLUTIONS TO PROBLEMS
v
Q30. ANSWER(C)
The problem 3SAT is NPcomplete wheras the problem 2_SAT has a polynomial time algorithm.
Q86. ANSWER(A)
The machine accepts set of all strings where the 1's are divisible by three and the 0's divisible by 2.
Try 111 it is accepted, this throws out choices (B),(C) and (D)
Q87. ANSWER(B)
One can easily design a dpda which pushes a's and b's onto the stack and for every c in the input it pops and a or a b.
Q88. ANSWER(C)
S>b+ by the first set of rules S>bSS, this eliminates choices (A) and (D)
S>aaa by S>*aA>*aaB>aaa and this eliminates (B)
Q89. ANSWER(A)
If L1 is recursive then we enumerate L2 and find out wj in a finite amount of time, and then enumerate strings <wj in lexicographic order in a finite amount of time.
If L2 is recursive enumerate it in lexicographic order of wj etc.
Q7(IT) ANSWER(c)
(a* +b* +c*)=(a+b+c+other strings)*=(a+b+c)*
for (B) we have the standard identity (r*s*)*=(r+s)*
for (D) we have (a*b*+c*)=(a+b+c+other strings)*=(a+b+c)*
for C we have a is not a substring
Q9(IT) ANSWER(C)
Nondeterminism makes a difference to the pda. The language {wwR} is accepted by a nondeterministic pda but not by a deterministic pda.
(A) is the standard theorem: there exist inherently ambiguous languages
(B) is the definition of ambiguity
(D) all finite sets are regular sets
Q40(IT) ANSWER(B)
The fastest way is to trace out all the strings.
Q41(IT) ANSWER(B)
The move
ð(A,a)=A demands the rule A>aA and this excludes (A) and
(C)
The move
ð(A,b)=B demands the rule A>bB and this excludes (D)
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 