About

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

Role

Alma Mater:

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

Experience:

Not updated.
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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i is getting incremented in the loop.
So it will check First Option:1<3, 2<3 and 3<3 (Total 3 Checking)

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19 Nov 2017 - 12:44am

Digital Logic:
1 day, Gates and Minimization.
1 day, Number System
1 day, Combinational Circuits
1 day, Sequential Circuits
So you can cover Digital in 4+1days. (It's very easy so do this 1st)

Discrete Maths: 
2 days proposition logic + First order logic
(You can skip other parts and cover them later)

CN:

1st cover IP addresing part. (Most important)
Cover reamining parts as time permits.

 

P.S - It's my personal suggestions. You can make your own schedule better.

 

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29 Oct 2017 - 2:00pm

We are working for ECE but till now only CSE contents are published.

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26 Oct 2017 - 9:31pm
I will highly suggest you to do not waste time specially in pre increment and post increment type questions. We all know these things are system dependent .
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25 Oct 2017 - 7:30pm

@setgate
After 1 you can not accpet 0. (You will go to C, which is a dead state)
 

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25 Oct 2017 - 6:48pm

0*1*

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Suppose you have attributes 10 attributes in a relation R as below

R(A, B, C, D, E, F, G, H, I, J )

Now you start finding the candidate key. 
Candidate key:- The mininal set of attribute which can uniquely determine all the other attributes.(Note - NOT Minimum set).

You started by taking one attribute.

Suppose B can determine all the other (9) attributes then B is our candidate fine. (It's minimal)

Now, you know that any superset of B will also determine all the other attributes[ Consider AB, It will determine all the attributes because of B.
Now it can not be the candidate key because it's minimal version that it B, is already a candidate key ]. So now we can check taking 2 attributes at a time [example, CD , AD, IJ,....](but not BD, BA...).

You found that { GH, IJ } both the pair can determine all the other attributes. So both GH and IJ will also be our candidate keys. {Note- these are minimal version of itself because you can not remove I or H}.

And so on. 

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How will it be negation of the given statement?

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"None of my friends are perfect."
negation of above statement will be "atleast one of my friend  is perfect." 

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