How was it ? :)

I think aptitude was easy, lengthy paper but medium.

If you remember any answers and sure about it.. Please post :p

I think aptitude was easy, lengthy paper but medium.

If you remember any answers and sure about it.. Please post :p

No need to discuss answers now. Better discuss them after the response sheet is released.

^Exactly. Technical was not even medium. 5 matrices were given for matrix multiplication that also with complex options.

huge lengthy questions,one paragraph two paragraph size..........I know one question which I gave wrong answer...the "hi" "bye" one...silly mistake,realized I marked wrong answer just at the last second.No of heaps....which I knew before but could not derive the count in short time...Gate is removing all types of pattern match easy questions like in set 2 last year.Of all the recent old papers...I think 2016 set 2,2017 set 2 was the most easy ones.Little depressed with my performance...looking forward to other exams which are there...

Yes .. either lengthy or difficult :/

If neither then they would be twisted -_-

The only question I found easy was number of states in DFA for corresponding NFA :p

If neither then they would be twisted -_-

The only question I found easy was number of states in DFA for corresponding NFA :p

compare this with set 2 paper last year....for Gate you need to be prepared really well that is true....bt it is easy to score more on easy paper with direct questions.....that requires luck sometimes...

I don't remember.

Btw what was the answer for that instruction length question? 4 type of instructions.. length 2 byte.. Integers floating points ...What was that question :/

Btw what was the answer for that instruction length question? 4 type of instructions.. length 2 byte.. Integers floating points ...What was that question :/

^{th} March. it is better not to discuss ... Everyone did their best ... time to rest ... :P

I did 38....1 question wrong already.....good to know the ESI answer is correct...most of 2 marks questions were looking like 5 marks questions as you get like in TIFR exam...great way to eliminate average people...

forget about questions now....take some time off ....and get back to work if you want to crack bits,pgee,isi etc...

Thanks Hira for sharing that link. I answered it as 3 and was tensed when everyone here was saying that it was 2.

in deque operation tail pointer is given ... so we have to traverse till that then delete the node pointed by tail...i think:)so deq is O(n)

I think :-

Linkedlist with queue:- head and tail pointer both were given. So to follow the property of queue enque should be from start (where head pointer was a) andque from end (where tail pointer was) .

So I think both can be in Our(1).

Minm spanning tree when x=4 , bcoz it will give more choice. And due to this there will be 2 minm spanning tree.

Linkedlist with queue:- head and tail pointer both were given. So to follow the property of queue enque should be from start (where head pointer was a) andque from end (where tail pointer was) .

So I think both can be in Our(1).

Minm spanning tree when x=4 , bcoz it will give more choice. And due to this there will be 2 minm spanning tree.

I think x=5, because if you take x = 4, then the edge adjacent to it which had weight 5 will never get chosen in an MST

I think x=4 bcoz when when we have same weight then choice will increased as per kruskal . There was three edge with 4 and of x=4 then we have 4 edges of same weight. There was an edge of 5 and If x=5 then there will be two edges with edge with 5. So comparatively less choice for 5 and maxm no of mst can be obtained with weight 4.

It's my opinion dear.. it may be possible that u r ryt..

It's my opinion dear.. it may be possible that u r ryt..

Linked list Insertion deletion O(1) and insertion O(1)

MST _ 4

Guys are you all sure chromatic no will be 2 and not 3 ? pls reply !!!

MST _ 4

Guys are you all sure chromatic no will be 2 and not 3 ? pls reply !!!

Yar chromatic no mera 3 aaya h.. 2 to hona hi nai chahiye.. bcoz there were three adjacent edges Nd to color them atleast 3 color will be required..

#chromatic no :-

Bhai I am quite sure that it will not be 2. It will be more than 2.. and I got 3. If I am not wrong Aisa hi tha kuch..

i dont remember the exact diagram but people are saying that from the middle if you start then may be the chromatic number will be 2 !!! :-( bad news for guys who marked 3... not sure

If anyone have idea about figure of Minm spanning tree then plz provide .. that will help to make more clear ..

acha what will be hi hi bye bye option ? pls anyone !!!! anyone does remember the correct option pls reply!! and anymoreC questions... counter value print question... pls guys post more answers!!

"hi bye hi bye" Both the times it was only the formal addresses that were getting swapped .. not the data inside.

Correct?

Correct?

but it was a double pointer the second time .... in order to change the data we shoulld access it using double pointer, right? but they used single pointer.

Thanks bro for the counter answer but i asked for C programming print count ,,, can anybody answer !!!

my answer for hi bye is wrong i answered option C :-(

my answer for hi bye is wrong i answered option C :-(

hi bye bye hi, is correct, second time using double pointer actual(one that were inside main()) single pointer got swapped.

**D flip-flop**

2 states should be the answer

as state 00 on in=0 was staying on 00

and state 11 on in=1 was staying on 11

C program ans:-

(What I did )

1) 1023*10=10230

2) hi bye bye hi (bcoz second tym double pointer was there )

3) for loop wala

J=40

Hence count value will be =5

Bcoz for

40 1

20 -> 2

10-> 3

5-> 4

2-> 5

1-> Condn false

Hence 5

(What I did )

1) 1023*10=10230

2) hi bye bye hi (bcoz second tym double pointer was there )

3) for loop wala

J=40

Hence count value will be =5

Bcoz for

40 1

20 -> 2

10-> 3

5-> 4

2-> 5

1-> Condn false

Hence 5

#include<stdio.h>

/* Swaps strings by swapping pointers */

void swap1(char **str1_ptr, char **str2_ptr)

{

char *temp = *str1_ptr;

*str1_ptr = *str2_ptr;

*str2_ptr = temp;

}

int main()

{

char str1 = "hi";

char str2 = "bye";

swap1(str1,str2);

swap1(&str1, &str2);

printf("%s %s", str1, str2);

getchar();

return 0;

}

is this the hi,bye string questions?

/* Swaps strings by swapping pointers */

void swap1(char **str1_ptr, char **str2_ptr)

{

char *temp = *str1_ptr;

*str1_ptr = *str2_ptr;

*str2_ptr = temp;

}

int main()

{

char str1 = "hi";

char str2 = "bye";

swap1(str1,str2);

swap1(&str1, &str2);

printf("%s %s", str1, str2);

getchar();

return 0;

}

is this the hi,bye string questions?

I think we're mixing two questions above..... the one with 2^40 is 5

There was another question with an increment of a variable counter .... right? Or am I imagining things?

There was another question with an increment of a variable counter .... right? Or am I imagining things?

I can't remember the exact values but I considered cylinder movements.

They have given 20mw for 100 movwmwnts so one will take 0.2 and then I calculated everything and added 15 whenever turned in opp direction

They have given 20mw for 100 movwmwnts so one will take 0.2 and then I calculated everything and added 15 whenever turned in opp direction

Bro .. if I am not wrong ..

It was 2^40 and in each loop of i it was decreased by I/2 and due to this every time j was increased .. so finally j=40

And there was a for loop with variable j where count value was increased by one every tym and loop was decreased by j/2.

Hence for j =40 it was giving count =5 ...

It was 2^40 and in each loop of i it was decreased by I/2 and due to this every time j was increased .. so finally j=40

And there was a for loop with variable j where count value was increased by one every tym and loop was decreased by j/2.

Hence for j =40 it was giving count =5 ...

Ans I got :- (F1(F2(F3F4)F5)

According to que explicit multiplication that means direct multiplaction..

So (F3F4) will be answer...

According to que explicit multiplication that means direct multiplaction..

So (F3F4) will be answer...

any one remember #include<stdio.h>

/* Swaps strings by swapping pointers */

void swap1(char **str1_ptr, char **str2_ptr)

{

char *temp = *str1_ptr;

*str1_ptr = *str2_ptr;

*str2_ptr = temp;

}

int main()

{

char str1 = "hi";

char str2 = "bye";

swap1(str1,str2);

swap1(&str1, &str2);

printf("%s %s", str1, str2);

getchar();

return 0;

}

in this question i marked A option

am i right ??

/* Swaps strings by swapping pointers */

void swap1(char **str1_ptr, char **str2_ptr)

{

char *temp = *str1_ptr;

*str1_ptr = *str2_ptr;

*str2_ptr = temp;

}

int main()

{

char str1 = "hi";

char str2 = "bye";

swap1(str1,str2);

swap1(&str1, &str2);

printf("%s %s", str1, str2);

getchar();

return 0;

}

in this question i marked A option

am i right ??

the question of co: type 1 ,2 3 ,4 instructions .. type 4 had 6 bit register and it was 16 bits ....??

@ Pranav Madhani are you talking about the co question ?

i known this is the ans but i confirm the option i think i marked wrong option so please anyone one remember the option than tell me

Yes, I am kinda very sure that answer for that question was

"Hi Bye Hi bye"

And

For the verbal ability question's answers eas "**Meandering**"

And another answer for C Programming was, IMO

'0 ', 'c'

Because, it was clearly written printf("%c %c", ...) Which means whatever you give it, if will convert it to corresponding char.

Thus, not 0 c but '0' 'c' should be the answers.

did anyone notice one thing there was no as such good question from Algorithm rather i will say i didnot find any question from algorithm except that knapsack prob!!! same ignorance for algorithm like 2010 !!!

ADITYA CHAURASIYA 5 I Think producer consumer is D .. PLEASE TELL!!

Pranav Madhani the computer organization prob... 4 types of instruction,...??

Que was :- guest are invited out of which 60% were male and 40% were female.

Total 80% of the guest invited were present in party and 100% female were present ?

Ratio(male/female ) ??

Let total =100

Attendance =80

Total female present=40

Total male present =40

Hence ratio 1:1

Total 80% of the guest invited were present in party and 100% female were present ?

Ratio(male/female ) ??

Let total =100

Attendance =80

Total female present=40

Total male present =40

Hence ratio 1:1

Producer consumer problem :-

Empty =0 , full =N , muted =1

Producer :-

Wait(P)

Signal(Q)

Consumer :-

Wait(R)

Signal(S)

Then what will be P,Q,R,S ?

Now what will be the correct solutions ?

And how ??

Empty =0 , full =N , muted =1

Producer :-

Wait(P)

Signal(Q)

Consumer :-

Wait(R)

Signal(S)

Then what will be P,Q,R,S ?

Now what will be the correct solutions ?

And how ??

oh thanks.. Warlock lord ..:)

WHAT WAS Answer to that question dead lock safe or not and if not what has to be added in instance. 3 process given of which 3 instange of E and F were there and nothing was mentioned about G.

Pranav Madhani it's in safe state.

Anyone remember the answer of IPV4 next header, UDP portno. ,

I remember it was QTP something!!!

anyone?

I remember it was QTP something!!!

anyone?

Does anyone know these

1) Computer organization Type 4 , the no of instructions? is it 1024?

2) φ(7 variables are given ) tuple relation calculus ... is this answer that it cannot have <=3 variables (B)..MIN 4 variables .

1) Computer organization Type 4 , the no of instructions? is it 1024?

2) φ(7 variables are given ) tuple relation calculus ... is this answer that it cannot have <=3 variables (B)..MIN 4 variables .

@ Vasu Srivastava please dont get me wrong ... im no1 to control you but may be the administrators of this site like Arjun Sir or others does not like the "F" word here cause if one use others will also start.... hope you get my point !! if you please edit your comment and remove that word !!!

Diameter = x√2 (diagonal of square of side 'd')

i.e d√2

Radius =(d√2)/2 = d/√2

Area =πr^2

= (πd^2)/2

I got it..this will be correct answer.. I think I have done it wrong..

i.e d√2

Radius =(d√2)/2 = d/√2

Area =πr^2

= (πd^2)/2

I got it..this will be correct answer.. I think I have done it wrong..

Diag of square is under root(2d), not d under root 2. You did a mistake in using Pythagoras theorem. Correct answer is pi.d/2

Okk so I think most of the que has been discussed.. so what will be the good attempt ?

No of que and marks ???

No of que and marks ???

I think AIR 1 can't have more than 65 actual marks. So 50-60 marks should be the range for IITs(or maybe even IISc) this year. But that's just my opinion.

This year gate paper was easy but bit length que were asked... But length que were so easy that you can see it and answer it... For example dbms query que were extremely easy topic was joints left right ..

Programming que were moderately tought and one que on pointer was so confusing ( that "hi bye hi bye "one)

Maths que were length to evaluate and very conceptual base on graph theory and eigen values

TOC was cake walk..

CN bits of Ip mac and UDp header were asked and also wrap around que was asked with a bit of twist..I felt like who have done CN all the topics where well they were able to do it in a fraction of second ...

DL was very easy

COA moderatly tought I was surprised to see a que kway set associative and direct mapping concept in one que which I asked on gateoverflow as my test series doubt.. the que was very similar so thanx to GO

Thanx to GO ,Arjun suresh sir and team for making awesomely cool website it helped me aaaaalot...

Programming que were moderately tought and one que on pointer was so confusing ( that "hi bye hi bye "one)

Maths que were length to evaluate and very conceptual base on graph theory and eigen values

TOC was cake walk..

CN bits of Ip mac and UDp header were asked and also wrap around que was asked with a bit of twist..I felt like who have done CN all the topics where well they were able to do it in a fraction of second ...

DL was very easy

COA moderatly tought I was surprised to see a que kway set associative and direct mapping concept in one que which I asked on gateoverflow as my test series doubt.. the que was very similar so thanx to GO

Thanx to GO ,Arjun suresh sir and team for making awesomely cool website it helped me aaaaalot...

heap was like this

1

2 3

4 5 6 7

the first level node has to be the same

second level nodes can be arranged in 2! ways

3rd level nodes can be arranged in 4! ways = 24

hence total no of binary heaps should be 2*24 i.e. 48

many people are saying the answer is 70, what is wrong with this method?

1

2 3

4 5 6 7

the first level node has to be the same

second level nodes can be arranged in 2! ways

3rd level nodes can be arranged in 4! ways = 24

hence total no of binary heaps should be 2*24 i.e. 48

many people are saying the answer is 70, what is wrong with this method?

woah ... thank god it had no negative marking ... thank you for clarification @severustux

p = q^{x}

r = p^{z}

q = r^{y }

log(p) = x.log(q)

log(r) = z.log(p)

log(q) = y.log(r)

x.y.z = log(p)/log(q) * log(q)/log(r) * log(r)/log(p) = 1

if they asked for the value of x.y.z

Subset sum problem :- maxm capacity =11

Ans we need to choose one element that too with maxm value.

So one item with weight 10 was being considered and then we need to calculate value -no of item

Ans we need to choose one element that too with maxm value.

So one item with weight 10 was being considered and then we need to calculate value -no of item

type checking is done before parsing was the incorrect one. as type checking is done in semantic analysis

solution for knapsack problem

V_{greedy} = 44

V_{optimal} = 60

answer = 60-44 = 16

anyone else got the same ans ?

what was the solution to first technical question?

TCP waala qn in which BW and SEQ NO was given ...

and for this qn too

fragment offset value was 144 right ?

TCP waala qn in which BW and SEQ NO was given ...

and for this qn too

fragment offset value was 144 right ?

the answer to that was 2nd schema ... it was in 3NF as 2nd dependency had a prime attribute in RHS but left one was just an attribute so it was definitely not in BCNF.

Drawing out of memory

Apti

meandering

they’re their there

2 12 60 …

smalles no div by 7

pi d/2

green red faces

200000

angle

1:!

x y z

1 marks

1)chromatic no

2)integration

3)sub groups

4)RISC control unit

5)TCP seg no

6)IP next header

7)type checking

8)deadlock process

9)NOR gate

10)digital no of states

11)NFA DFA

12)r.e complement

13)eigen value

14)count(81,4)

15# ofpage fault

16)post order

17)interrupt

18)probability

19(queue LL

20)generating function

21)full outer join

22)string pointer

23)slow start cwnd

24)DRAM refresh

25)ER model

2 marks

1)regular expression

2)CFG

3)countability

4)unrestricted grammars

5)IP fragmentation

6)CSMA

7)1023*10

8)long int

9)bye hi hi bye

10)# of min heaps

11)logic

12)graph 100!

13)MST

14)DFS BFS

15)matrix mult

16)greedy/optimal

17)eigen vector

18)cond. probability

19)synchronisation

20)safe state

21)cache

22)# of inst.

23)3NF BCNF

24)LOJ

25)lexical analysis

26)precedence associativity

27)SSTF

28)Prime implicants

29)Pipelining

30)Fixed point representation

Apti

meandering

they’re their there

2 12 60 …

smalles no div by 7

pi d/2

green red faces

200000

angle

1:!

x y z

1 marks

1)chromatic no

2)integration

3)sub groups

4)RISC control unit

5)TCP seg no

6)IP next header

7)type checking

8)deadlock process

9)NOR gate

10)digital no of states

11)NFA DFA

12)r.e complement

13)eigen value

14)count(81,4)

15# ofpage fault

16)post order

17)interrupt

18)probability

19(queue LL

20)generating function

21)full outer join

22)string pointer

23)slow start cwnd

24)DRAM refresh

25)ER model

2 marks

1)regular expression

2)CFG

3)countability

4)unrestricted grammars

5)IP fragmentation

6)CSMA

7)1023*10

8)long int

9)bye hi hi bye

10)# of min heaps

11)logic

12)graph 100!

13)MST

14)DFS BFS

15)matrix mult

16)greedy/optimal

17)eigen vector

18)cond. probability

19)synchronisation

20)safe state

21)cache

22)# of inst.

23)3NF BCNF

24)LOJ

25)lexical analysis

26)precedence associativity

27)SSTF

28)Prime implicants

29)Pipelining

30)Fixed point representation

I want to bang my head on a wall so badly! I did some really really stupid mistakes, like where y'll are getting 10230.. I ended up getting 10240(didn't subtract -1 from function... *Face palm*) and whoever is getting height of tree as 4... I got 3.....

Not to mention 2-3 stupid mistakes in Apti now...

Plus, I forgot that Knapsack approach(lack of revision I suppose)

Better be careful next time....

Not to mention 2-3 stupid mistakes in Apti now...

Plus, I forgot that Knapsack approach(lack of revision I suppose)

Better be careful next time....

Both will be O(1) https://www.geeksforgeeks.org/queue-set-2-linked-list-implementation/

You didn't read question carefully ... In question it was given that enqueue was at front i.e. start of ll while dequeue was at the rear I.e. end of linked list ... While that geeksforgeeks link did the exact opposite of this .... I.e. enqueue at end of ll ... And dequeue at start ... That's why as per gate question ... Complexities are O(1) and O(N)

Does anyone know the answer to the parsing question ??? They said match longest prefix ... So I marked option D ... T3T3 ...

Oh no buddy, you are forgetting the question, we had both Head and tail pointers.....

So we aren't supposed to traverse all the way to tail, as there's a pointer to the tail, hence both O(1)

So we aren't supposed to traverse all the way to tail, as there's a pointer to the tail, hence both O(1)

1->2->3->4->5->6

Head ->1

Tail -> 6

After deleting 6 please tell me how will you make it point to 5 in O(1)

Head ->1

Tail -> 6

After deleting 6 please tell me how will you make it point to 5 in O(1)

In this Question , I suppose ; Vopt were no of item taken by to get Optimal Weight (not Actual Optimal Weight) ; similarly VGreedy .

Solution : Vopt=1 Vgreedy = 2 Ans : -1

I have read question twice to see ; and previously I was also reducing the weights. Hope I am correct.

Solution : Vopt=1 Vgreedy = 2 Ans : -1

I have read question twice to see ; and previously I was also reducing the weights. Hope I am correct.

not fair they should give marks to both -1 and 16

or else give options :/ Question dhyan se word to word padhte rahe ya solve bhi re :p

or else give options :/ Question dhyan se word to word padhte rahe ya solve bhi re :p

Hey @goravkc or @ anyone what is the answer of superset in DBMS query? do u remeber? is it anyhow option d Q4 where it was full outer join?

I don't remember what the question was ... But if it is what he said and ans is -1 then they will not give any marks for 16 as not reading question properly is our mistake not theirs ... So you can't expect marks for that ...

3rd one is definitely not context free ... You can't represent a^nb^nc^n using pda ... As far as I remember answer to that was 1 and 2 ... Problem with ashish's solution is a's and b's are not being compared ... Their count should be same too ...

Vopt and Vgreedy were total values/profit that each algo could get. It is also indicated by the use of alphabet "V". I also read that question twice and I am quite sure its answer is 16. No offense to anyone!

yes we cant represent but check for full condition. we have to select such strings in in which **m=n=p && p !=q. In such case we can push for m pop for n and do nothing for p specifically we have to make p more or less but not equal to q so in stack either p should be there or q should be there.**

option m=o and n = p what ever alternate comparision can be solved by linear bounded automata so thats not possible for sure.

i answered -1 for V optimal - V greedy

i dont remember exactly the ques but i read the ques twice . i may be wrong or may not

i dont remember exactly the ques but i read the ques twice . i may be wrong or may not

I think knapsack que was like ?

We have capacity given =11.

And

Item weight value

1. 10. 60

2. 8. 30

3. 4. (Don't remember)

4. 2. 12

So we can include only first item to get more value that too within capacity.

And that value was V=60

Then they were asking no of item included =only one item

Finally they were asking diff between value and no of item included =60-1 =59 will be the answer.

Note :- data might be some change but concept was same what I have done here..

We have capacity given =11.

And

Item weight value

1. 10. 60

2. 8. 30

3. 4. (Don't remember)

4. 2. 12

So we can include only first item to get more value that too within capacity.

And that value was V=60

Then they were asking no of item included =only one item

Finally they were asking diff between value and no of item included =60-1 =59 will be the answer.

Note :- data might be some change but concept was same what I have done here..

Yea I think so. But the concept was same I think . That they were asking the diff between value obtained and no of items included to obtained that..

@Manis I think the question was Vopt - Vgreedy.

Vopt - The total value taken if we follow optimal approach.

Vgreedy - The total value taken if we follow greedy approach. And i was getting 21 or 19 as answer.

Vopt - The total value taken if we follow optimal approach.

Vgreedy - The total value taken if we follow greedy approach. And i was getting 21 or 19 as answer.

@Harish Kumar.... Actually I don't remember exact que that what it was ?. But I think there was something tricky.. don't go with name. Let them declare the que ?..

Argh... Chill guys, what's done is done, let's rest and chill with our buddies and enjoy the time, discussing stuff will bring unnecessary anxiety followed by depression, and whatever we have marked can't be changed, so it's better if we cut our selves some slack and enjoy the time until official answer keys have been declared... (Sic)

Lol it got posted now... I commented it on the very same day when reponse sheet got released... But when I clicked on Add comment button it was showing it will reviewed and then be added... And after so long final my comment was added

This commet was posted by me on that day itself... But now its showing 3 hr ago... Feeling like I m on gov website

This commet was posted by me on that day itself... But now its showing 3 hr ago... Feeling like I m on gov website