# Recent posts tagged gate2021

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Show that in a group of five people (where any two people are either friends or enemies), there are not necessarily three mutual friends or three mutual enemies.
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Let $f(n)$ and $g(n)$ be asymptotically non-negative functions. Which of the following is correct? $\theta (f(n)^*g(n))=min (f(n),g(n))$ $\theta (f(n)^*g(n))=max(f(n),g(n))$ $\theta (f(n)+g(n))=min(f(n),g(n))$ $\theta (f(n)+g(n))=max(f(n),g(n))$
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(&forall;x&forall;yP(x,y))&rarr;(&forall;x&forall;yP(y,x)) Here the LHS will be true only if P(x,y) is true for every possible x and y. Since both x and y are coming from the same universe, if this is true, then their order doesn't matter for P(x, y).
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You should know the precedence rule. $\to$ has higher precedence than quantification.
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For empty string as language, q0 as final states gives 16 DFAs. But when both q0 and q1 are final states we must ensure none of the transitions go to q1 from q0. This reduces the possible DFAs to $4$ as we have 2 choices for each {0, 1} transitions from q1 (either to q0 or to q1 itself). So, totally 16 + 4 = 20 DFAs.
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I doubt the correctness of this solution, the diagram below; in Red Rectangle shows what this solution is suggesting, on the other hand the image in the Green Rectangle shows what is expected by the question. This solutions does not consider the fact that after $L_1$ on miss, the request is forwarded to $L_3$. How does this solution keeps that in perspective?
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SIR ACCORDING TO UR WIKIPEDIA its answer should be inner join because in question it is mentioned that we r selecting table1.column & table2.column so only same column produced two times in inner join and in equi join only same column produced only one times sir plzzz clear my doubts
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answer of of 4.12 is c) 10 as in DVR the routers rely completely on their neighbour Routers .So N2 , N4 will send their distance vector to N3 and N3 will find out that in order to reach N1 the option I have are: 1) infinity (from N1)+ 2(N1 N2 distance ) =infinity 2)8 (from N4 )+ 2 (N4 N3 distance)= 10 N3 will choose the minimum of two which is 10.
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Thank you sir...and as the signals are getting lost their corresponding iterations will be missed out in Process Y. Am I correct?
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more details plzzz
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Yes. You are correct. But not just the "max element" cause problem. All the elements in the last level can cause this problem. So, only binary search tree is the answer here.
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