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Hello sir How can we find the middle for given string ? here they mentioned Even length palindrome according to that we find middle Is this correct ?

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GATE Overflow Book for GATE CSE 2022
[closed]

answer is A) 5.56E+004 8.12e+268 e Scientific notation (mantissa/exponent), lowercase E Scientific notation (mantissa/exponent), uppercase

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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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Delay for generating C is 1 AND + 1 OR = 1.2 + 1.2 = 2.4 ms y 1 and here ??? shd 2 and 3 * delay for carry this 3 due to 4 bit binary adder ryt ???

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Yes , context switch takes less time . threads are smaller and so the contain of the thread is also small . so the swapping during context switch take less time , its not the main reason though ( partial reason :P ) . the main reason i think when context switch ... program context switch occurs without any help taken from OS (means mode change does`nt required ) , it is done by the program . peace

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0 means lowest vertex in the Hasse diagram (f in this case). 1 means highest vertex in the Hasse diagram (a in this case).

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@arjun sir is fully correct now I am going to discuss about why frame transmission time must be >= 2 * prop. Time In csma/cd the sender doesn't holds a copy after it has sent a frame so the sender sends a frame now if a collision is detected it resends after ... If the sender is still transmitting then only it can resend it else not. That's reason why frame transmission time must be > 2 prop. Time

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Not reflexive - $(4,4)$ not present. Not irreflexive - $(1, 1)$ is present. Not symmetric - $(2, 1)$ is present but not $(1, 2)$. Not antisymmetric - $(2, 3)$ and $(3, 2)$ are present. Not Asymmetric - asymmetry requires both antisymmetry and irreflexivity. It is transitive. So, the correct option is $B$. transitive.

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A[2] is a second element of array A or third element of A in C ...

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Answer is (c) Xb must be 1 because both P(S) and V(S) operations perform Pb(Xb) first. So if Xb= 0 then all the processes performing these operations will be blocked. Yb must be 0. otherwise two processes can be in critical section at the same time, when s=1 and Yb=1 . So Yb must not be 1.

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sir normally instructions are sequential (untill unless branch) thats why m taking instructions as example. same concept for accessing sequential data i.e. array access.. Eements of an array are neighbors in memory, and are likely to be referenced one after the other.

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record size = 64B no of records = 16k total size of records = 64*16k no of block to store records = 64*16k/512 = 2k key size = 14B Address size = 2B no of block for 1st level index = no of keys *key size / block size = 2k * (14+2)/ 512 = 64 block no block for second level index = 64*16/512 = 2 no ofof block for 3rd level index = 2*16/512 = 1 block total 3 level indexing.

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When window scaling happens, a 14 bit shift count is used in TCP header. sir why 14 bit only ?? any reason or just an assumption ??

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Consider the relation R(ABCD) with FD's set F={ AB->CD,C->.A,D->B} which of the following is false ? 1.C->A is a partial Dependency 2.C->A is a transitive Dependency 3.D->B IS A Partial dependency 4. All of these

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Answer: B For correction:$\left\lfloor \dfrac{(\text{Hamming Distance} - 1)}{2}\right\rfloor$ $=\big\lfloor1.5\big\rfloor=1\text{ bit error}$. For detection:$\text{ Hamming Distance - 1 = 3 bit error}$.

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Okay sir :) yes i read that in morris manno (R compliment ) compliment is 10^n - number where n is no digit in number (Integer part ) And suppose if i want to find diminished compliement (R-1 )then it is 10^n- 10^-m- number where n is no of digit in integer part m is the number of digit in fractional part