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

### Alma Mater:

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

### Experience:

Not updated.
Techtud Editor
Multichoice
6 months 1 week ago

@ujjwal10  @rohittulasyan
The number of odd degree vertices in any graph is even. Since we already have four vertices of odd degree, the degree of the last vertex cannot be 3 or 5. It cannot be 0 either, since there is one vertex with degree 7, which means that it is a neighbour to all the other vertices, which implies that there is no isolated vertex in the graph. Thus the only possible degree of the last vertex is 4.

moreless
Techtud Editor
Multichoice
6 months 1 week ago

Lattice with gcd as meet and lcm as join is distributive

gcd(a, lcm(bc)) = lcm(gcd(ab), gcd(ac))

lcm(a, gcd(bc)) = gcd(lcm(ab), lcm(ac)).

it is complete semi-meet lattice as there is no upper bound it is not complete semi-join lattice.

moreless
Techtud Editor
Example text
6 months 1 week ago

Evaluate the integral$\int \limits_0^1 (1 + y^2)^{-1.5} dy$

(A) 1
(B) -1
(C) 1/2
(D) 1/√2

moreless
Techtud Editor
Example text
6 months 1 week ago

There few grammars where you can not remove ambiguity by applying anything.
An example of an inherently ambiguous language is the union of  with . This set is context-free, since the union of two context-free languages is always context-free. But it is  proved that there is no way to unambiguously parse strings in the (non-context-free) common subset .

moreless
Techtud Editor
Example text
6 months 1 week ago

How many 32x16 ROMs with enable will be required to construct a 128x16 ROM and What type of decoder will you need for this design?

(A) 4, 2 to 4 decoder
(B) 4, 4 to 16 decoder
(C) 8, 2 to 4 decoder
(D) 8, 4 to 16 decoder

moreless
S_raj
piyushi
Vaibhav Gupta
Niket Gangwar
Anshu Agarwal
Nirdesh Kumar Joshiya
Prashant Sharma
Prof R.K. Dhawan
Sheshang Ajwalia
shivani
KIRAN KUMAR
KIRAN KUMAR
Narasimha Karumanchi

## Pages

Pritam Prasun
905
Shabin Muhammed
1065
Rajesh Kumar Pandey
1522
Vinita Sangwan
2169
partha
2584
Sayan Bose
4083
Rohit Kumar
4589
techtud
6230
Biduro Maharana
6492
Virtual GATE
6911
shashi kant verma
7507
Rakhi Mahanta
7572
Anjan
9527
10290
phanindra
10613

## Pages

16 Dec 2017 - 10:17pm

i is getting incremented in the loop.
So it will check First Option:1<3, 2<3 and 3<3 (Total 3 Checking)

more less
16 Dec 2017 - 2:03pm

3 times.

more less
19 Nov 2017 - 12:44am

Digital Logic:
1 day, Gates and Minimization.
1 day, Number System
1 day, Combinational Circuits
1 day, Sequential Circuits
So you can cover Digital in 4+1days. (It's very easy so do this 1st)

Discrete Maths:
2 days proposition logic + First order logic
(You can skip other parts and cover them later)

CN:

1st cover IP addresing part. (Most important)
Cover reamining parts as time permits.

P.S - It's my personal suggestions. You can make your own schedule better.

more less
29 Oct 2017 - 2:00pm

We are working for ECE but till now only CSE contents are published.

more less
26 Oct 2017 - 9:31pm
I will highly suggest you to do not waste time specially in pre increment and post increment type questions. We all know these things are system dependent .
more less
25 Oct 2017 - 7:30pm

@setgate
After 1 you can not accpet 0. (You will go to C, which is a dead state)

more less
25 Oct 2017 - 6:48pm

0*1*

more less
4 Oct 2017 - 11:50pm

Suppose you have attributes 10 attributes in a relation R as below

R(A, B, C, D, E, F, G, H, I, J )

Now you start finding the candidate key.
Candidate key:- The mininal set of attribute which can uniquely determine all the other attributes.(Note - NOT Minimum set).

You started by taking one attribute.

Suppose B can determine all the other (9) attributes then B is our candidate fine. (It's minimal)

Now, you know that any superset of B will also determine all the other attributes[ Consider AB, It will determine all the attributes because of B.
Now it can not be the candidate key because it's minimal version that it B, is already a candidate key ]. So now we can check taking 2 attributes at a time [example, CD , AD, IJ,....](but not BD, BA...).

You found that { GH, IJ } both the pair can determine all the other attributes. So both GH and IJ will also be our candidate keys. {Note- these are minimal version of itself because you can not remove I or H}.

And so on.

more less
29 Aug 2017 - 7:17pm

How will it be negation of the given statement?

more less
29 Aug 2017 - 5:17pm

"None of my friends are perfect."
negation of above statement will be "atleast one of my friend  is perfect."

more less