About

Pritam is the director of Perfecist Technologies 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
Answer
2 months 3 weeks ago

You need to mark a topic as completed in the status section after completion.

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Tech
Answer
3 months 1 day ago

Let us apply the recursion-tree method for solving this recurrence problem:

The cost of the root is n because it is the combined cost to divide and merge back the results from the conquered subprocesses. Now here you can see that the height of the subtrees will be largest for T(n-1). It is equivalent to the fact mentioned by @arpitdhuriya that T(n-1) is asymptotically larger.

Hence the complexity would be O(n2)
More precisely, Ω(n2)

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Tech
Answer
3 months 1 week ago

Yes, None of the options is correct.

We can apply a bottom up approach (as you said) for better understanding:

1. SELECT P.pid FROM Parts P WHERE P.color<> 'blue' : Non-Blue parts id
2. SELECT C.sid FROM Catalog C WHERE C.pid NOT IN (1):  Supplier ID who supply at least one blue part
3. SELECT S.sname FROM Suppliers S WHERE S.sid NOT IN (2): Supplier name who supplies only non-blue parts. 

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29 Aug 2017 - 8:35pm

You need to mark a topic as completed in the status section after completion.

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Let us apply the recursion-tree method for solving this recurrence problem:

The cost of the root is n because it is the combined cost to divide and merge back the results from the conquered subprocesses. Now here you can see that the height of the subtrees will be largest for T(n-1). It is equivalent to the fact mentioned by @arpitdhuriya that T(n-1) is asymptotically larger.

Hence the complexity would be O(n2)
More precisely, Ω(n2)

more less

Why don't you give it a try and we correct.

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Yes, None of the options is correct.

We can apply a bottom up approach (as you said) for better understanding:

1. SELECT P.pid FROM Parts P WHERE P.color<> 'blue' : Non-Blue parts id
2. SELECT C.sid FROM Catalog C WHERE C.pid NOT IN (1):  Supplier ID who supply at least one blue part
3. SELECT S.sname FROM Suppliers S WHERE S.sid NOT IN (2): Supplier name who supplies only non-blue parts. 

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This is a simple one.

when you draw an  n x n matrix like below:

So, you have (1+ 2 + 3 + 4 + .... n-1 + n = ) n(n+1)/2 places where you have elements present. How the size depends on the data type being stored in the array.

Hence size = n(n+1)/2 * sizeof(data_type)

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Well Done!

That's how one should always approach to the problem. Remember, getting the solution is not our goal. Our aim should be to develop the problem-solving ability.

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Why don't you give it a try, I will help you to reach to the answer.

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Answer would be d) b & c

I want you to give it a try with the following hint:

  1. For broadcast, host ids will be all 1
  2. Assuming that the host bit N would be the host bit try to find the possible IPs

I suggest you to solve it by your own.

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Techtud test series are found at http://virtualgate.in

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Thanks a mistake, thanks for informing.

In transaction T1 all the subscripts should be "1". So, it is:
T1: R1(A) W1(A) R1(B) W1(B)

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