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)

moreless
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; }

moreless
TudLead
Answer
1 year 10 months ago

Tricky problem.

Please search for Fork() bomb.

moreless
narayanapot's picture
Lakshminarayana Potukuchi
virtualgate's picture
Virtual GATE
priyesh's picture
Priyesh Priyatam
ranita's picture
Ranita Biswas
pritam's picture
Pritam Prasun
pritam's picture
Pritam Prasun
905
Rajeev
914
priyesh's picture
Priyesh Priyatam
923
Himanshi
976
mnlcht's picture
jhilik
979
pshall's picture
shailendra joshi
998
maheshkumars's picture
Mahesh Kumar
1017
chandanchawda's picture
Chandan Chawda
1019
kaushalmaurya's picture
kaushal
1022
shabinmuhammed's picture
Shabin Muhammed
1065
tar_gate's picture
TarGate
1165
vineetkumar's picture
vineet
1238
rahulkumar's picture
Rahul Kumar
1246
kalpishsinghal's picture
Kalpish Singhal
1474
targetgate's picture
Target Gate
1518

Pages

C is correct.

A) it will not contain only 1's which true for even number of 0s.It will surely have two 0's.

B) same story as A.

C) Correct.

D) Kind of same as A and B.

more less

If some one say FA that is implicitly NFA, while DFA is DFA.

more less
11 Feb 2015 - 2:18pm

Here is th link to previous papers of ISRO but do not take them as they are. See and observe them carefully and prepare for the worst case. All the best

http://www.techtud.com/computer-science#overlay=resource-share/isro-previous-year-paper-cs

more less
5 Feb 2015 - 6:10pm

Ans 2.

ID stage will detect hazard which will be related to ID stage functions that is Instruction Decoding and Register Read. If You carefully think then you will get to know that RAW  and WAR data hazards are something which are related to ID stage.In that too,WAR hazard will be related to Writing(4th stage) the Data ,by current instruction,into registers which is Already been read(2nd stage) by previous instruction. So no chance of hazard detection in this case.

You can easily see that there are 2 RAW hazard between ADD-SUB(r1) and SUB-MUL(r2).That is ID stage of SUB will detecting that result in R1 is not yet updated. Same for SUB-MUL.

So answer is  2.

more less
5 Feb 2015 - 6:02pm

You can check doubt sections and all the questions have been put there. People have already started discussing and mentors will be answering the official answers too.

 

more less
3 Feb 2015 - 10:53pm

Yeah, ca not see the picture.

more less
3 Feb 2015 - 5:47pm

 
In CSMA/CD , to transmit a frame, transmitter must seize the channel. In order to seize the channel , it must propagate Jamming signal upto RTT. This applies to receiver as well for acknowledgement.
So total time = Seize time of sender + TT of sender + PT +  Seize time of receiver + TT of receiver i.e. Ack transmission time  + PT.
Seize time  = 2* PT .
PT =  distance / Speed of signal
Signal speed =  200m/microsec = 2* 10^8 m/sec
PT = 10 microsec
Seize time for both sender and receiver = 2* PT =  20 microsec
TT of sender = 51.2 microsec
TT of receiver = 3.2 microsec 
Total time = 20 + 51.2 + 10 + 20 + 3.2 + 10 = 114.4 microsec
useful time = TT of sender , exclduing header as it is also an overhead = (512- 32)/10Mbps = 48.0 microsec
efficiency = 48.0 / 114.4 = .4210
Effective data rate = 4.210 Mbps

more less

thats true. Is it Ace Academy test problem?

more less

Please take a look at this example and try to solve accordingly.

http://www.techtud.com/doubts#overlay=doubt/virtual-gate-2015-question-38

more less
30 Jan 2015 - 3:58pm

If nothing mentioned about Flag, Addressing Mode field etc, then the solution goes like this above. If anything given , then you need to consider those bits too for solution.

more less

Pages