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Ans (C). $\left(P \rightarrow Q\right) \leftrightarrow \left(\neg P \vee Q\right)$ (D) is wrong as shown below. Let $S = \left\{2, 3, 4, 5\right\}$ and $P(x,y)$ be $x < y$. Now, $P(2,3)$ is true but $P(3,2), P(4,2)$ etc are false and hence the implication also. This is ... If P(x,y) is true for all (x,y), then P(y,x) is true for all (x,y).
posted Jun 21, 2019 in Interview Experience akshay_b 1,801 views
C) P - iii, Q - i, R - iv, S - ii Ref:
posted Apr 25, 2019 in Interview Experience Pooja Khatri 760 views
In a ---> there are consecutive writes by T2 and T1 on the same item, i.e. B In c --->there are consecutive writes by T2 on A and T1 on B. Here though there are consecutive writes but on different data items. Now read my above comment of Thomas rule statement.
posted Feb 9, 2019 in Others Naruto Uzumaki 841 views
The sequence &hellip;&hellip;&hellip;&hellip;&hellip; is an optimal non-preemptive scheduling sequence for the following jobs which leaves the CPU idle for &hellip;&hellip;&hellip;&hellip;&hellip;&hellip;&hellip; unit(s) of time. ---------------------------- Job Arrival_Time Burst_Time ------------------ ... (2,1,3},0 (c) {3,2,1),0 (d) {1,2,3},5 Ans: option (a) please explain why answer is a)....
posted Jun 21, 2018 in Interview Experience priyendu mori 1 1,873 views
Acc. to me it should be (C) because: according to condition, out of all, one philosopher will get both the forks. So, deadlock should not be there.
posted Apr 20, 2018 in Motivation Dilip Puri 7 1,116 views
14, 26 and 44 mod 6 = 2 and hence in any input sequence they must come in sequence (otherwise their sequence changes in the end result). The other 3 numbers hash to distinct locations and hence they can be inserted in any order. So, number of possible insertion sequence $=\frac{\text{Total insertion sequences}}{\text{Invalid insertion sequences}}$ $=\frac{6!}{3!}$ $= \frac{720}{6} = 120$
posted Mar 17, 2018 in Preparation Advice Akhilesh Singla 524 views
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