I had appeared in GATE 2008 and 2009. After that I had not been interested much in GATE. However this time (GATE 2012) a close friend of mine appeared for the same, which got me interested again. Here is a solution set to the GATE 2012 Computer Science paper. I have solved question from Algorithms, Automata, C programming, Probability ... in short, the topics that I enjoyed enough to remember them after 2 years.

As I mentioned earlier, I have not appeared in for the examination. I got the question GATE 2012 CS and IT question set C from a fellow netizen. Please download the paper first as I will be using that as the reference for objective type answers.

2 years earlier I had posted solutions to GATE 2009 CS question paper. Comments and suggestions from that post made me realize the importance of explaining every answer. Feel free to ask questions if you need further clarifications.

GATE 2012 Computer Science and IT solutions to question set C

Please leave a comment if you were helped by my effort. Feel free to ask questions if you need further clarifications or have found some error in my solution.

As I mentioned earlier, I have not appeared in for the examination. I got the question GATE 2012 CS and IT question set C from a fellow netizen. Please download the paper first as I will be using that as the reference for objective type answers.

2 years earlier I had posted solutions to GATE 2009 CS question paper. Comments and suggestions from that post made me realize the importance of explaining every answer. Feel free to ask questions if you need further clarifications.

GATE 2012 Computer Science and IT solutions to question set C

- b
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- b
- Draw the sine graph for the interval. Only one minima at 3*pi/2 where sin(x) = -1
- c
- fork() will spawn off a new child process every call. Three calls to fork() results in 8 processes, out of which one is original, other 7 are children.
- b
- this is the hidden one concept
- a
- Obviously. Look at f(X,Y) and compare it with X
- c
- Balanced binary search trees require log (number of elements) time to search
- log (n*2^n) = log (n) + log(2^n) = O(n)
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- c
- Standard switch case behaviour called "fall through", when the break statement is not there
- c
- Look into any DB book, I studied Navathe.
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- c
- Do the checking manually
- c or d
- 1 is false, 3 and 4 are true
- not sue about 2
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- c
- follows from definition of CDF
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- d
- 4 bit multiplier requires 2x4 = 8 bit input, i.e. 2^8 set of inputs
- multiplication of two 4-bit numbers will result in at most 8-bit result, so 8 bits for every multiplication
- 2^8 * 8 = 2^11 bits = 2 Kbits
- c
- Average case is at lease as good as worst case
- A(n) better than W(n) :: Quicksort
- A(n) same as W(n) :: Heapsort
- c
- Draw it and find out. They key thing is "any embedding of G on the plane"
- d
- T(n-1) for moving n-1 disk to aux from source
- 1 for moving the n-th disk to destination
- T(n-1) for moving n-1 disk back to source from aux
- b
- S is correct
- regarding P or Q, there is some confusion. It seems that the SQL syntax allows it. You can test it out in an SQL validator.
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- c
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- b
- examine the cases:
- P(6 in first throw) = 1/6
- P(1 in first throw followed by 5 or 6 in second throw) = (1/6)*(2/6)
- P(2 in first throw followed by 4, 5 or 6 in second throw) = (1/6)*(3/6)
- P(3 in first throw followed by 3,4, 5 or 6 in second throw) = (1/6)*(4/6)
- summing up = (1/6)*(1+2/6+3/6+4/6) = 5/12
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- b
- b'd' -- group the 4 corners
- b'c' -- group the a'b' columns to 2 cells with the top 2 cells in ab' column
- a
- MST algorithm does not depend on the value of the weights, it depends only on the relative ordering of the edge weights.
- Ranking remains unchanged as all the weights are >= 1
- b
- I did it by hand, don't know if there is a shortcut
- b
- Observe the graphs
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- d
- observe that q is a dead state, so anything beginning with 000 should go to q -- A and B gets eliminated by this logic
- after 10 if there is a 0 the state should move to 00 state, so D is correct
- a
- look into any Data Structure book
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- d
- Dijkstra's algorithm probes the best solution so far
- Thanks to Shivam for pointing this out
- b
- merge-sort takes O(n log n) comparisons
- here to compare two strings you need O(n) time, so O(n^2 log n)
- c
- represent a cycle by a permutation of 4 labeled vertices
- cycles are identical if one is a rotation of another
- same cycle of length 4 can have 4 different representations each starting with a unique letter
- (6 P 4) / 4 = 90
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- c
- line 2 has a static variable, which retains the old value when he function is called again
- d
- "auto int" is same as normal "int"
- "register int" is same as normal "int" as far as value retention is considered
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- a
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- a
- maxima-minima problem
- maximize the profit where, profit = 50q-5q^2
- b
- b
- Given:
- P(X) = 0.6
- P(Y) = 0.4
- P(R | X) = 0.96
- P(R | Y) = 0.72
- P(R) = P(X) * P(R | X) + P(Y) * P(R | Y) = 0.864

- Applying Bayes Rule:
- P(Y | R) = [ P(R | Y) * P(Y) ] / P(R) = 0.333
- c
- follows from fundamental definitions of mean and standard deviation
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- b
- another maxima-minima problem and the function is already given, just maximize y

Please leave a comment if you were helped by my effort. Feel free to ask questions if you need further clarifications or have found some error in my solution.