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No. $\begin{array}{|c|c|c|} \hline \text {A} & \text {B} & \text {C} \\\hline\text {1} & \text {5} & \text {6} \\\hline\text {2} & \text {4} & \text {7}\\\hline \text {3} & \text {4} & \text {5}\\\hline \end{array}$ Suppose this is the relational ... hold for $R.$ PS: If we have a single instance where $A \to BC$ is not holding, it is enough to say $A \to BC$ does not hold for the relation $R.$
posted Aug 12, 2020 in Interview Experience sharan21 394 views
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R1,R2 has common attribute B. B is key in R2. so R1(AB) R2(BC) composed into R12(ABC) Between R12(ABC) and R3(CD) common attribute is C and C is key in R3. Hence the decomposition is lossless. Checking dependency preservation. R1(AB) R2(BC) R3(CD) A->B B->C C->D We ... if D->A is covered by G D+={DA} D+ in G ={DCBA} hence D->A is also preserved. Hence loss-less as well as dependency preserving.
posted Aug 9, 2020 in Interview Experience Raghavendra_NV 410 views
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Which of the following describes the minimum condition for ambiguity in a grammar? a) Every derived word must have atleast 2 rightmost derivation b) some word must have more than 1 leftmost derivation c)A derived word has one rightmost and another leftmost derivation d)each derived word has 2 leftmost and 2 rightmost derivation
posted Aug 7, 2020 in Interview Experience Raghavendra_NV 934 views
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lets rename the attributes a) Let Bookid=A,subject-category-of-book=B,Name-of-author=C,Nationality-of-Author=D A is the Pk hence A+={BCD} F .D is A->BCD Since there is one attributed CK,there is no prime attribute hence,no partial dependency,so Relation is 2NF No transitive ... b)Let Book-Title=E Author-Address=F CE->ABD ce is PK There is a non trivial dependency and LHS of which is SK so BCNF.
posted Aug 4, 2020 in Interview Experience aditya19 556 views
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option A C$\Rightarrow$ a or, C $\Rightarrow$b or, C $\Rightarrow$aCb$\Rightarrow$aaCbb$\Rightarrow$ aaaCbbb .. soon at last you have to put either C$\rightarrow$ a or C$\rightarrow$ b so production C is used to derive $a^{n+1}b^{n}$ or $a^{n}b^{n+1}$ $n \geq 0$ S$\rightarrow$ AC [Aanbn+1] ... $\rightarrow$ CB will generate $a^{n}b^{n}b^{+}$ i.e $a^{i}b^{j} \ i<j$ option D is right .
posted Jul 20, 2020 in Interview Experience goxul 365 views
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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.
posted Jul 18, 2020 in Interview Experience vijayp_ 680 views
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Answer: B Round Trip Time $= 80ms$ Frame size $=32\times 8\text{ bits}$ Bandwidth $=128\text{ kbps}$ Transmission Time $=\dfrac{32\times 8}{128}\ ms = 2\ ms$ Let $n$ be the window size. Utilization $=\dfrac{n}{1+2a}$ where $\large a = \dfrac{\text{Propagation Time}}{\text{Transmission Time}}$ $ =\dfrac{n}{1+\dfrac{2 \times 40}{2}}$ For maximum utilization: $n = 41$ which is close to option (B).
posted Jul 9, 2020 in Interview Experience goxul 1,249 views
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starting with 001100110011 means alternative sequence of process P and Q.. Process P should start execution so at W, P(s) where S=1.. to get alternate sequence X and Y are operation on same semaphore i.e. T. option B or C. bt process Q shouldn't start execution before process P .. means Initial value T=0 W : P(s) X : V(T) Y : P(S) Z : V(S) S = 1 T = 0
posted Apr 2, 2020 in Interview Experience rishabh100gpt 1,199 views
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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: $\frac{h(x)}{g(x)}$ $\frac{-1}{x}$ $\frac{g(x)}{h(x)}$ $\frac{x}{(1-x)^{2}}$
posted Apr 25, 2019 in Interview Experience Pooja Khatri 977 views
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C) P - iii, Q - i, R - iv, S - ii Ref: https://en.wikipedia.org/wiki/Software_testing
posted Apr 25, 2019 in Interview Experience Pooja Khatri 768 views
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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.
posted Apr 25, 2019 in Interview Experience Pooja Khatri 964 views
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