About

Pritam is the co-founder of OpenSense Labs and an alumnus of IIT Kharagpur. He qualified GATE-13 with percentile 99.93 . He also possess more than 3 years of industry experience of working for services based companies like TCS, AZRI Solutions & Enova Technologies.

Role

Alma Mater:

M. Tech.
IIT Kharagpur
2013 to 2015
B. E.
University Institute of Technology, RGPV, Bhopal
2007 to 2011

Experience:

Director
Perfecist Technologies
2015
Application Developer
Azri Solutions
2012 to 2013
Business Analyst
Tata Consultancy Services (TCS)
2011 to 2012
Web Developer
E-nova Technologies, Okhla, New Delhi
2009 to 2010
Tech
Post
1 month 2 weeks ago
Most often freshers mention the difficulty they face when they join the industry after completing the graduation. For some of them, it becomes the real challenge to adapt to the new environment where your capability is not judged just by looking at a single piece of paper, but by many factors like your problem-solving skills, team management, on-time delivery, quick learning skills and many other significant factors.

I believe, Internship is one of the ways where you get an opportunity to have a close understanding of how differently companies operate and what is the most suitable career path for you.

Do not miss an internship this summer!!

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7 Jun 2016 - 11:02pm

Image is broken, can you check it ?

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6 Jun 2016 - 2:11pm

This is quite obvious, isn't it ?

Functional Dependencies says "Using left side attributes you can uniquely identify right hand side attributes". So, in a set of FDs if some attribute is missing in RHS, it simply means it can not be determined using any set of attributes which are present in LHS. In contrary, candidate key must determine all the attributes uniquely.

Hence, while forming candidate key we keep these attributes which can not be determined by any FD.

Understand it like this. If US can not be defeated by all countries together and you want to form a group which can defeat all countries, then it's mandatory to include US in the group.

This answers why D should be present in candidate key.

Now if you have doubt that "is it mandatory to pair 'D' up with some attribute ?", then the answer is NO.
If D alone can determine all the attribute then D will be the CK, you need not to pair it up with anything else.

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6 Jun 2016 - 7:55am

I am not aware of this book, I just varified with with Korth (chapter 15, page- 672) which specifies that the "Data items may be unlocked at any time".

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5 Jun 2016 - 11:03am

The question mean to say A, B & C are infinite sets, it doesn't mean that there exist infinite number of sets A, B, C which satisfy such condition.

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5 Jun 2016 - 8:18am

I would try to find one possible condition which satisfies. If I can find one, as it is 'there exist' so I can say it to be TRUE.

S1

A: Set of all even numbers
B: Set of all +ve odd numbers
C: Set of all -ve odd numbers

Now,  A ∩ (B ∪ C) = Ø which is a finite set.

S2

x = 3 - √5   and  y = 3 + √5
x + y = 6 (Rational Number)

Hence both statements S1 and S2 are CORRECT.

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abhinandansh, following your logic written in the question If I assume a straight line, I can color it using only two color but, that will not make answer as 2.

When you have to generalize the statement you have to consider all possible cases.

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3 Jun 2016 - 7:54am

Given:

ξ(G) = Σd id x d, where id is the number of vertices of degree d in G. It simply means the sum of the degree of vertices.

For a tree, Sum of total degree = 2(Number of edges) = 2(|V| - 1)

Here, ξ(S) = ξ(T)
⇒ 2(|S| - 1) = 2(|T| - 1)
⇒ |S| = |T|

Following is a simple example for ξ(S) = ξ(T) = 4
 

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2 Jun 2016 - 12:56pm

Hradesh, writing 'my try' is not required (as it is obvious). If you can give some description of finding it, your answer can be marked as best answer

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31 May 2016 - 9:10pm

The answer above is correct. However below is some improvement in the solution:

I simplify the eqn and take it as: T(n) = 2(n-1) + T(n-1)

T(n) = 2(n-1) + T(n-1)

        = 2(n-1) + 2(n-2) + T(n-2)

        = 2(n-1) + 2(n-2) + 2(n-3) + T(n-3)

        ........

        = 2(n-1) + 2(n-2) + 2(n-3) + 2(n-4) + 2(n-5) + 2(n-6) + ........ + 2(n-(n-1)) + T(1)

        = 2[ (n-1) + (n-2) + (n-3) + (n-4) + .....+ 3 + 2 + 1] ( ignoring T(1) as it constant )

        = 2[ n(n-1)/2]

        = n(n-1)

        = O(n2)

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31 May 2016 - 6:39pm

That's how I thought it but as answer is different, we need to rethink.

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