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Pursuing M.Tech in Computer Science & Engineering at IIT Kharagpur and Editor at Techtud.

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Alma Mater:

B.tech in Compter Science and Engineering
Kamla Nehru Institute of Technology, Sultanpur
2012 to 2016

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Techtud Editor
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5 months 2 weeks ago
All the Best for GATE 2018 :)
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Techtud Editor
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5 months 3 weeks ago
Tomorrow is the last date for attempting VirtualGATE-2.
http://virtualgate.in/login/index.php
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Techtud Editor
Example text
6 months 6 days ago

c stands for Complement here.

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Techtud Editor
Example text
6 months 1 week ago

@vikash957 at x= -3, function is not defind ( devide by zero).

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Techtud Editor
Multichoice
6 months 1 week ago
@reenakandari
Question is about non-preemptive kernel design, where a running process can be preempted only when it is being forced for it.
Timer interrupts and Disk interrupts will wait untill the process is running. A blocking system call or an exit request are the high priority kernel requests, so these can not be hold on wait.
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@vaisbhat, I am unable to find any better method to get upper triangular matrix.
I must say, use other techniques to solve such problems instead of kirchoff rule.

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Taking m=2,
k=0;
for i1 = 1 to n 
 for i2 =1 to i1
  k=k+1

for i=1, k is incrementing  1 time
for i=2, k is incrementing 2 times
for i=3, k is incrementing  3 times
........................................................
for i=n, k is incrementing n times.
Total increment in k = 1+2+3+................+n = n(n+1)/2 = selecting 2 integers with repetition among n integers.
Taking m=3,
for i=1, k is incrementing 1 time. (1 times)
for i=2, k is incrementing 3 times. (1+2 times)
for i=3, k is incrementing 5 times. (1+2+3 times)
.........................................................................
for i=n, k is incrementing (1+2+3+....n) times.
Total increment in k = 1 + (1+2) + (1+2+3) + (1+2+3+4) +.............(1+2+3......+n)
 = n(n+1)(n+2) / 6  =  selecting 3 integers with repetition among n integers.
Similarly, Take m=k, K will be C(n+k-1,k) =  selecting m integers with repetition among n integers.

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@vaisbhat none of the image is visible..please change the image properties.

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According to the Kirchhoff matrix theorem, all the cofactors of the matrix are equal and equal to the number of spanning trees of the graph. So you need to find only one cofactor and it is not a big deal with 6x6 matrix.

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@shweta1920, we are working on that. Sorry for the delay.

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There are many approaches to solve these type of questions. This is mine.
Consider each case:
1. TLB Hit + Cache Hit:
T1 = TLB Hit probability* Cache Hit probability*  access time
T1 = 0.96 * 0.9 * (1+1)  = 1.728 ns
2. TLB Hit + Cache Miss:
T2 = TLB Hit probability* Cache Miss probability*  access time (Main memoery will be accessed)
T2 = 0.96 * 0.1 * (1+1+10) = 1.152 ns
3. TLB Miss + Cache Hit:
T3 = TLB Miss probability* Cache Hit probability*  access time (2 page table will be accessed)
T3 = 0.04 * 0.9 * (1+1+10+10) = 0.792 ns
4. TLB Miss + Cache Miss:
T4 = TLB Miss probability* Cache Miss probability*  access time 
T4 = 0.04 * 0.1 * (1+10+1+10+10) = 0.128 ns

Total avg. time = 3.8 ns ≅ 4ns.

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Apply divide and conquer approach to find minimum no. of comparisons.
T(n) = 0, if n=0 (only one element is present)
        = 1, if n=2 (if only two elements are present )
        = floor[T(n/2)] + ceil[T(n/2)] + 2, otherwise .
If you will solve the recursion, you will get 3n/2 - 2 as the answer.

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I think you are not considering negative numbers.
Consider an array with elements like, 3, 5, -1, -7 .
Here you can see that sum of sub array 3, 5 is greater than sum of all the elements.
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10 May 2017 - 5:52pm

no. of indegee is zero?
Can you explain the question a little bit ?
Topological Sorting requires DFS [O(V+E)] to find the order of the vertices.
Have a look here:
https://www.youtube.com/watch?time_continue=401&v=Q9PIxaNGnig

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10 May 2017 - 3:25pm

We can have any number of combinations. Keep one thing in mind that a candidate key should not be the subset of any other candidate key. 
Example: Consider the relation R(A,B,C,D,E,F) and the following functional dependencies,
F = { A→ BCDEF
         BC →ADEF
         B→F
         D→E }
Candidate Keys = {A, BC}

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