# Recent posts tagged iiit-h

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Yes . You are right.
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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).
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Option A is correct because height $5$ means level $6$ so maximum node $= 2^l -1 =2^6 -1=63$ and for minimum, at each level only single node so total $6$.
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C) P - iii, Q - i, R - iv, S - ii Ref: https://en.wikipedia.org/wiki/Software_testing
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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.
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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)....
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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.
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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 $=\frac{\text{Total insertion sequences}}{\text{Invalid insertion sequences}}$ $=\frac{6!}{3!}$ $= \frac{720}{6} = 120$
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