The actions of the virus can be represented using the following context-free grammar: where V is the virus, a and b are acid and base molecules respectively.
V→ aVb
V→ bVa
V→ VV
V→ ε
The language of the above grammar is the set of strings consisting of an equal number of a’s and b’s. A sequence of acid and base molecules is just a string over {a, b}. To check if a single virus could have produced the sequence, you can use CYK algorithm to check membership in the above context-free grammar.

5 Jun 2017 - 12:10pm

it is better and safer to use (!list) than (list==NULL) as while implementation there might be some memory - segmentation fault errors.

2 Jun 2017 - 2:49pm

Can you share the link of the video?

9 May 2017 - 11:44am

@shweta1920, you can try running this codes on your machine too, by downloading the standard c-compiler

see http://www.mingw.org/ : for windows 

and any linux distribution has c-compiler (GCC) by default .

How to run programs using GCC ?

see: http://pages.cs.wisc.edu/~beechung/ref/gcc-intro.html

and http://www.akira.ruc.dk/~keld/teaching/CAN_e14/Readings/How%20to%20Compi...

 

17 Apr 2017 - 1:47pm

Please ask you questions in the doubt section of techtud , rather than commenting on a previously asked question. This will increase the visibility of your question .

 

Now, 

Let 60 % of mangoes weigh X grams and,

      40 % of mangoes weigh Y grams 

Let total number of mangoes be "n".

 

A/Q

\(350 = {X \over 0.6n}\)  ⇒  \(350*0.6 = {X \over n}\)   ⇒  \(210 = {X \over n}\)     ------------------------- (1)

\(400 = {Y \over 0.4n}\)   ⇒ \(400*0.4 = {Y \over n}\)   ⇒   \(160 = {Y \over n}\)     --------------------------(2)

(1) + (2)

\(210 + 160 = { X + Y \over n}\)

⇒ \(370 = {Total Wt. \over n}\)

Therefore, Average Weight of mangoes = 370 grams

6 Mar 2017 - 1:12pm

Correct answer is (C) 1, 0

we call with x = 1 and y = 1, then ptr points to the variable x and we change its value to 0. Now, we put that value using *ptr to the variable y, hence y gets 0. Now, we change back the value of ptr i.e. x to 1. So, finally x = 1 and y = 0.

14 Feb 2017 - 3:52pm

Correct answer (B) III only

Only address space is shared by all the threads in a process, all the others can be thread specific.

14 Feb 2017 - 3:10pm

Yes, D is correct. Edited now.

2 Feb 2017 - 1:46pm

Answer updated

2 Sep 2016 - 7:59pm

Answer updated.

2 Sep 2016 - 7:04pm

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