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
Multichoice
6 months 1 week ago

@ujjwal10  @rohittulasyan
The number of odd degree vertices in any graph is even. Since we already have four vertices of odd degree, the degree of the last vertex cannot be 3 or 5. It cannot be 0 either, since there is one vertex with degree 7, which means that it is a neighbour to all the other vertices, which implies that there is no isolated vertex in the graph. Thus the only possible degree of the last vertex is 4.

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Techtud Editor
Multichoice
6 months 1 week ago

Lattice with gcd as meet and lcm as join is distributive 

gcd(a, lcm(bc)) = lcm(gcd(ab), gcd(ac))

lcm(a, gcd(bc)) = gcd(lcm(ab), lcm(ac)).

it is complete semi-meet lattice as there is no upper bound it is not complete semi-join lattice.

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Techtud Editor
Example text
6 months 1 week ago

Evaluate the integral\(\int \limits_0^1 (1 + y^2)^{-1.5} dy\)

(A) 1
(B) -1
(C) 1/2
(D) 1/√2

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Techtud Editor
Example text
6 months 1 week ago

There few grammars where you can not remove ambiguity by applying anything.
An example of an inherently ambiguous language is the union of \{a^{n}b^{m}c^{m}d^{n}|n,m>0\} with \{a^{n}b^{n}c^{m}d^{m}|n,m>0\}. This set is context-free, since the union of two context-free languages is always context-free. But it is  proved that there is no way to unambiguously parse strings in the (non-context-free) common subset \{a^{n}b^{n}c^{n}d^{n}|n>0\}.

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Techtud Editor
Example text
6 months 1 week ago

How many 32x16 ROMs with enable will be required to construct a 128x16 ROM and What type of decoder will you need for this design? 

(A) 4, 2 to 4 decoder
(B) 4, 4 to 16 decoder
(C) 8, 2 to 4 decoder
(D) 8, 4 to 16 decoder

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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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