About

This is Vivek Vikram Singh. Currently pursuing Master Of Engineering in Computer Science stream from BITS, Pilani. I strongly believe in contribution-based education and helping out fellow/upcoming graduates with the experience I have. I have shared my written and interview experience in various known institutes/ PSUs for my M.Tech. admission in 2014 Find me on my blog at: https://vivekvsingh14.wordpress.com/

Role

Alma Mater:

Master of Engineering
Birla Institute of Technology and Science, Pilani
2014 to 2016
Bachelor of Engineering
Bansal Institute of Science and Technology, Bhopal
2009 to 2013

Experience:

Intern
EMC Corporation
2016
ASE
EMC Corporation
2016
TudLead
Example text
1 year 10 months ago

A = O(nlogn), B= O(n)

It is very trivial if you know the complexity concepts.

n=O(nlogn)

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TudLead
Answer
1 year 10 months ago

#include #include int main() { int i; for (i=0; i<3; i++) { pid_t pid = fork(); if (pid == 0) printf("parent:[%d] current:[%d] i=%d\n", getppid(), getpid(), i); else printf("FORK Failed for current: %d\n", getpid()); printf("*\n"); } printf("HI \n"); return 0; }

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TudLead
Answer
1 year 10 months ago

Tricky problem.

Please search for Fork() bomb.

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10 May 2015 - 7:42pm

I would suggest to go with Kurose- Ross and Peterson-Davie. Every topic is really good explained in Kurose- Ross book. Peterson Davie is good with questions. Whichever book you choose, just stick with that at a time, finish it and then only jump to another one,if you do.

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DBMS is general term which refers to a family while RDBMS i.e. Relational DBMS is a member of the family, with other types of DBMS such as Network DBMS, Hierarchical DBMS etc.

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22 Feb 2015 - 11:47pm

Arul,small correction. Speed up is a factor so that is not measured in percentages.

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Try to search material on Youtube,coursera and edx. NPTEL will have lectures on Microprocessor and Computer Graphics . They will provide the reference books. Get those books and start practicing .

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for string of even number of 0's , all 1's is a valid string as it contains zero number of 0's which is an even number .right?? eg, 1,11,111,1111 and so on.

In option A, if you try to generate string, there will be at least two 0's but you can not generate any string with zero 0's,which should be there.

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Sorry Shimpy,error.Corrected my answer.

L(R3) will contain at least four 1's.

and Yes, D is not correct too as explained by Arul above. 

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L(R1) = set of strings which start from 11.

L(R2)= Set of strings which end with 11.

Biggest mistake to consider L(R3) is set of strings which starts and ends with 11.

L(R3) is NOT set of strings which starts and end with 11, as it does not contain 11,which starts and ends with 11.

B option ) L(R1) U L(R2) will contain strings like {11,110,011} which is not in L(R3). So B is wrong.

C ) L(R1) will contain string eg 110 which is not in L(R3) AS L(R3) will contain at least fours 1's ,2 start and 2 end. So C is wrong.  

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15 Feb 2015 - 2:23pm

Most common example but easy to understand too.

Suppose L1 is defined over ∑ = {a,b} and L1= {a+b} which is all the strings of a and b. L2 = {anbn, n>=0}.

L2 is subset of L1 but L2 is not Regular. So option A is wrong.

L1∩L2 = L2 which is same as A.

L2 is CFL and concatenation of it with itself "n" times will result in CFL. So C is wrong too.

 

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You need to carefully observe that Every string in L1 will have only one C,whereas in L2 ,only one A. 

Intersection suggests that A should be equal to B and B should be equal to C.

So ABC is the correct answer.D option.

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