# Recent posts tagged iit-interviews

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i think the language would be L= 00(0000)* if L= 00 + (0000)* then 0000 cant reaches the final state .
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Yeah why not you have found a solution it cab be one of the option but always choose the best one
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return res * x; where x is a pointer while multiplication on pinters is not valid so it is still ambigious
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If you put "Null" as introducer it will work. But by self introduction I mean "john" introducing himself- introducer will be "john" for customer "john". But I guess this needn't be considered unless specified as by the meaning of "introduction" we can assume it is a different person.
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A)The n^{th} statement in a list of 100 statements is: "Exactly n of the statements in this list are false." 1)Exactly 1 of the statements in this list are false. 2)Exactly 2 of the statements in this list are false. 3)Exactly 3 of the statements in this list are false. . . ... (100) will be false. B)Same way we can solve it. So we will get 1-50 are true and 51-100 are false. C)This cannot happen.
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Barber is one out of set of affected people.. if barber is in trouble then above statement is illogical.. if he is in trouble then der is a barber who save him but that saver barber is in trouble. i.e. no such barber exists.
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bhai integration is area only http://www.mathsisfun.com/calculus/integration-introduction.html
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If $g(x) = 1 - x$ and $h(x) = \frac{x}{x-1}$, then $\frac{g(h(x))}{h(g(x))}$ is: $\frac{h(x)}{g(x)}$ $\frac{-1}{x}$ $\frac{g(x)}{h(x)}$ $\frac{x}{(1-x)^{2}}$
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Is there any SQL command which belongs to both DDL and DML?
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Answer is D. $L_1$ is context-free and hence recursive also. Recursive set being closed under complement, $L_1$' will be recursive. $L_1$' being recursive it is also recursively enumerable and Recursively Enumerable set is closed under Union. So, $L_1' \cup L_2$ is recursively enumerable. ... $L_2$')' $= L_2$ is also recursive which is not the case here. So, $II$ is also false.
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$\lim_{x\rightarrow \infty } x^{ \tfrac{1}{x}}$ is $\infty$ 0 1 Not defined
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2 is correct answer. fig 2, b and c have {f,g} as upperbound. but for the graph to be a lattice it should have a least upper bound. Since b and c have two upper bounds they cannot have a least upper bound<which is always unique for a pair for vertices>. In ... {f,g}. therefore it is not a join semilattice(every pair of element should have a least upper bound). henceforth it is also not a lattice
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L2 is surely regular and I would choose L1 to be non-regular. Is the answer correct?
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$(C012.25)_H - (10111001110.101)_B$ $= 1100\;0000\;0001\;0010.\;0010\;0101$ $- 0000\;0101\;1100\;1110.\;1010\;0000$ $= 1011\;1010\;0100\;0011.\;1000\;0101$ $= 1\;011\;101\;001\;000\;011 .\;100\;001\;010$ $= (135103.412)_o$ Binary subtraction is like decimal subtraction:$0-0 = 0, 1-1 = 0, 1-0 = 1, 0-1 = 1$ with $1$ borrow. Correct Answer: $A$
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SLR paper is more powerful than LALR . False . LALR parser is more powerful than Canonical LR parser . False . Canonical LR parser is more powerful than LALR parser. True. The parsers SLR, Canonical CR, and LALR have the same power. False. answer - C
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How query 2 will not produce the correct result? I am getting the desired result from query 2.
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Instead of finding a function 'f' for the first MUX (in terms of z,x and y) and the second MUX (in terms of f , x and y), we can logically find out the answer.We can put in 4 values of A and B and analyse the output C. We can get the values appropriate values of x and y by drawing and comparing with a 2x1 MUX when A and B are given as inputs respectively.
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Later I got another idea to solve it A' + AB',apply distributive property, that comes to (A'+A)(A'+B') = A'+B'
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Solution 48: spills to m/m while execution of program how many times m/m will be hit, we have 2 regiters Program Execution c = a + b; R2 <- R1 + R2 d = c * a; [m/m spill] <- R2*R1(we need R & R1 further value cant be replaced) e = c + a; [m/m spill] <- R2+R1 x = c * c; R2 < ... } else { d = d * d; R3<- R3*R3 e = e * e; R4<- R4*R4 } If 4 registers are used, m/m spills will be 0, hence answer is (B)
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Time without pipeline $=6 \text{ stages}=6 \text{ cycles}$ Time with pipeline $=1+\text{stall freqency}\times \text{stall cycle}$ $=1+.25\times 2$ $=1.5$ Speed up $=\dfrac{6}{1.5}=4$
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(c) Every finite subset of a regular set is regular this is false example:a^n b^n from regular set (a+b)* is not regular
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Level order traversal of a rooted tree can be done by starting from the root and performing preorder traversal in-order traversal depth first search breadth first search
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