GATE CS 2008

The given set theory expression can be converted into an equivalent boolean algebra expression as follows:

=p.q.r + p'.q.r + q'+r'
=qr(p+p')+q'+r'
=qr+q'+r'
=(q+q') . (r+q') +r'
=r+r'+q'
=1+q'
=1
=U

Option (D) is correct

Augment the given matrix as
1 1 2 | 1
1 2 3 | 2
1 4 a | 4
Apply R2 <- R2 - R1 and R3 <- R3 - R1
1 1 2 | 1
0 1 1 | 1
0 3 a-2 | 3
Apply R3 <- R3 - 3R2
1 1 2 | 1
0 1 1 | 1
0 0 a-5 | 0
So for the system of equations to have a unique solution,
a - 5 != 0
or
a != 5
or
a = R - {5} 

One group consists of (0000, 0010, 1000, 1010) which gives b’d’ Other group is (0000, 0001, 1000, 1001) which gives a’d’ So, solution is b’d’ + a’d’

As we can see 121 contains digit ‘2’ which can’t be represented directly in base ‘2’ (as digits should be less than base) so number system must have   “r>2”. Ans (D) part.

From logic diagram  we have f=f1.f2+f3 f=m(4,5,6,7,8).f2+m(1,6,15)----(1) from eq(1) we need to find such f2 so that we can get  f=m(1,6,8,15) eq(1)  says we can get m(1,6,15) from f3 ,so only  8 left now from option (a,b,d) we get (4,6), (4,8) ,(4,6,8) respectively for m(4,5,6,7,8)f2 which is not required as m4 is undesired. But option (C) m(4,5,6,7,8)(6,8)+(1,6,15)

• m(6,8)+m(1,6,15)
• m(1,6,8,15)an

Option (C) is correct.

Turing machine can be designed for a^p using the concept of ‘Sieve of Eratosthenes’.
 
Suppose we are given an integer ‘n’ and we want to find out all the prime numbers less than or equal to ‘n’. We repeat the following steps : We find the smallest number in the list, declare it prime and eliminate all the multiples of that number from the list. We keep doing this until each element has been declared prime or eliminated from the list.
 
Now, if p = 0 or p = 1, we reject the input. Else, we replace the first and the last ‘a’ with symbol $.
 
In the above steps, what we do is we find the first non-black symbol from the left. Let this symbol occur at position ‘x’. Suppose ‘x’ is a prime number. If this non-blank symbol is $, input string will be accepted. But, if the symbol is ‘a’, we mark it as a* and replace all the multiples of ‘x’ with the symbol ‘blank’. If at the end, symbol $ is replaced with \'blank’, we reject the input string (because p will be multiple of some ‘x’ in that case). Else, we go back and repeat the steps.
 
Thus, the input is neither regular nor context-free, but is accepted by a Turing machine.
 
Please comment below if you find anything wrong in the above post.

(A) Intersection of two regular languages is regular and checking if a regular language is infinite is decidable. (B) Deciding regularity of a context free language is undecidable. We check if L(CFG) contains any string with length between n and 2n−1 , where n is the pumping lemma constant. If so, L(CFG) is infinite otherwise it is finite. (C) Equality problem is undecidable for all languages except in case of finite automata i.e. for regular languages. (D) We have to check if the grammar obeys the rules of CFG. If, it obeys such rules then it is decidable.  Thus, option (B) is correct. Please comment below if you find anything wrong in the above post.

Let\'s first understand the terminology used here. LR Parsing - Here \'L\' stands for Left to Right screening of input string, and \'R\' stands for Right Most Derivation in Reverse ( because it is about bottom-up parsing). Sentential Form - Suppose For a given Context Free Grammar G, we have a start symbol S, then to define Language generated by Grammar G, i.e. L(G), we start the derivation starting from S using the production rules of the grammar. After one complete derivation we get a string w which consists of only terminal symbols, i.e. w belongs to L(G). Then we can say that w is a sentence of the Grammar G. Now, while the derivation process, if it gets some form q, where q may contain some Non-terminal then we say that q is a sentential form of the Grammar G. Even the start symbol S is also the sentential form of the Grammar G ( because it also contains Non-terminal S).

For Ex :

Grammar is :

S-> aAcBe

A->Ab|b

B->d

Input string : abbcde

Derivation : ( Top-Down, Right Most Derivation)

S->aAcBe
->aAcde
->aAbcde
->abbcde 

Here { abbcde } is the sentence of the Grammar( because it contains only terminal symbols), and { S, aAcBe, aAcde, aAbcde } are the sentential forms of the G ( because these forms contain non-terminals during the derivation process ) Now, let\'s look at the question. The question is related to LR parsing which is a bottom-up parsing. Let\'s take the same grammar above, as it is a bottom up parsing we need to start from the string \"abbcde\" and try to get S using production rules.

: abbcde

->aAbcde ( using A-> b )

->aAcde ( using A-> Ab )

->aAcBe ( using B -> d )

->S ( using S-> aAcBe ) 

The above process is called reduction. The RHS of the Production which is being replaced by their LHS is called Handle, So , { b, Ab, d, aAcBe} are handles, and replacing it with its LHS is called Handle-Pruning. Hence option D suits best.