About

Pursuing M.Tech in Computer Science & Engineering at IIT Kharagpur and Editor at Techtud.

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Alma Mater:

B.tech in Compter Science and Engineering
Kamla Nehru Institute of Technology, Sultanpur
2012 to 2016

Experience:

Not updated.
Techtud Editor
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5 months 2 weeks ago
All the Best for GATE 2018 :)
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Techtud Editor
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5 months 3 weeks ago
Tomorrow is the last date for attempting VirtualGATE-2.
http://virtualgate.in/login/index.php
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Techtud Editor
Example text
6 months 6 days ago

c stands for Complement here.

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Techtud Editor
Example text
6 months 1 week ago

@vikash957 at x= -3, function is not defind ( devide by zero).

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Techtud Editor
Multichoice
6 months 1 week ago
@reenakandari
Question is about non-preemptive kernel design, where a running process can be preempted only when it is being forced for it.
Timer interrupts and Disk interrupts will wait untill the process is running. A blocking system call or an exit request are the high priority kernel requests, so these can not be hold on wait.
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x should be same in your discussion.
∃x [ P(x) ∨ Q(x) ] ⇔ ∃xP(x) ∨ ∃xQ(x)
To prove this we have to check two cases:
1. ∃x [ P(x) ∨ Q(x) ] → ∃xP(x) ∨ ∃xQ(x)  
2. ∃xP(x) ∨ ∃xQ(x) → ∃x [ P(x) ∨ Q(x) ]
If any one of the case is false, then equivalency can not be applied.
Take, 
x be the domain of students in a school.
P(x) = x is in Mathematics department.
Q(x) = x is in Biology department.
Check case 1: 
∃x [ P(x) ∨ Q(x) ] → ∃xP(x) ∨ ∃xQ(x)
To make it false, we have to make LHS as True and RHS as False.
Here, LHS says, there is a student who is either in Mathematics department or in Biology department. If we will make this as True, we can not make RHS as False. So this case 1  is True.
Check case 2:
∃xP(x) ∨ ∃xQ(x) → ∃x [ P(x) ∨ Q(x) ]
Here, if we will make LHS as True, then there is at least one student who is either in Mathematics or Biology department. Now, we can not make RHS as False. So case2 is True.
Hence, Formula is correct.

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26 Jun 2017 - 2:23pm

Very nice explanation is here
http://www.techtud.com/doubt/frequency

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25 Jun 2017 - 12:21am

An ambiguous grammar is a context-free grammar for which there exists a string that can have more than one leftmost derivation (or rightmost derivation).
Having both leftmost and rightmost derivation can not imply ambiguity. 

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(B) 6561

it can be simply seen by below figure :

Also ,  the catch here is , x is passed by reference , so each loop will change the value of x.

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@nandini1508 In this code, there will be 4 recursive calls with parameters (6,4), (7,3), (8,2) and (9,1). The last call returns 1. But due to pass by reference, x in all the previous functions is now 9. Hence, the value returned by f(p,p) will be 9 * 9 * 9 * 9 * 1 = 6561.

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Yeah, answer will be  option (a).

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@shweta1920,
Shortest job first is the way to minimize average response time.
if x <3, order will be: x, 3, 5, 6
if 3 < x < 5 , order will be: 3, x, 5, 6
if 5 < x < 6, order will be: 3, 5, x, 6
if x >6, order will be: 3, 5, 6, x.

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