# Featured posts

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Ans is c Large memory overhead would there.To reduce it use multi level paging will be used.
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False : Consider $E: \{1,2,3,4\}$, $F:\{3,4,5,6\}$ $f(E) = \{5,6,7,8\}$ $f(F)=\{7,8,11,6\}$ $f(E \cap F) = \{7,8\}$ where as, $f(E) \cap f(F) = \{6,7,8\}$ Becomes TRUE only when $f$ ... $f(E \cap F) \subseteq f(E) \cap f(F)$ $\subseteq$ becomes $=$ only when $f$ is one-one - when the mappings are unique for each element.
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He got M=&Sigma;* as an example to show that it L$\bigcap$M can be CFL but not regular . L$\bigcap$M will always be CFL but may or may not be regular.
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All 3 test() are inside "PAR BEGIN, PAR END" so they execute concurrently..
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All College Wise Interview Experience Blog Links.

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Hello Everyone....!!

Here is the most awaited post of GATE Overflow TEST SERIES for GATE CSE - 2021.

You can check the complete details of the Test Series in the below link....!! If you have any doubts can drop a comment to this post..!!

## Schedule

EDIT: The test series will be as per the updated syllabus for GATE 2021 and will be having Multiple Select Questions from mid August...

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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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Sl.No:

Institute

Programme

Specialization

Year

1

IIIT Bengaluru

MS

2018

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As it is already written that x is a intermediate node so it cann't be the last node. Hence, complexity should be O(1) by the algo that gate_asp have given below. Tell me if i am wrong?
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C and D only out of AB,AF,AE,C,D
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In batch processing processes halt in between executions and and there is no preemptive scheduling.So option d.
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Here bound of the loop are constants, therefore compiler will do the loop unrolling(If compiler won't then prefetcher will do) to increase the instruction level parallelism. And after loop unrolling $23$ cycles are required for execution. Therefore, correct answer would be (B). PS: We ...
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yes. Exactly. But a main correction. Then L "MAY BE" regular. Otherwise L is not regular. Because, if pumping lemma is not obeyed, language is not regular. But if the language satisfies pumping lemma, that doesn't mean language is regular.
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A grammer having left recursion generates a cycle. And no left recursive grammer is LL(1) grammer.
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We wish to schedule three processes $P1$, $P2$ and $P3$ ... -preemptive scheduling respectively? $30$ sec, $30$ sec $30$ sec, $10$ sec $42$ sec, $42$ sec $30$ sec, $42$ sec
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Let $G$ be a weighted undirected graph and e be an edge with maximum weight in $G$. Suppose there is a minimum weight spanning tree in $G$ containing the edge $e$. Which of the following statements is always TRUE? There exists a cutset in $G$ having all edges of maximum ... in $G$ having all edges of maximum weight. Edge $e$ cannot be contained in a cycle. All edges in $G$ have the same weight.
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$P\left(x\right) = \left(\neg \left(x=1\right)\wedge \forall y \left(\exists z\left(x=y*z\right)\right) \Rightarrow \left(y=x\right) \vee \left(y=1\right) \right)$
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The formula used to compute an approximation for the second derivative of a function $f$ at a point $X_0$ is $\dfrac{f(x_0 +h) + f(x_0 – h)}{2}$ $\dfrac{f(x_0 +h) - f(x_0 – h)}{2h}$ $\dfrac{f(x_0 +h) + 2f(x_0) + f(x_0 – h)}{h^2}$ $\dfrac{f(x_0 +h) - 2f(x_0) + f(x_0 – h)}{h^2}$
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Implement a circuit having the following output expression using an inverter and a nand gate $Z=\overline{A} + \overline{B} +C$