GATE CS 2013
Question 1 |



Commutative but not associative | |
Both commutative and associative | |
Associative but not commutative | |
Neither commutative nor associative |
Discuss it








Question 2 |
Which one of the following does NOT equal to


C | |
D | |
B | |
A |
Discuss it
First of all, you should know the basic properties of determinants before approaching
For these kind of problems.
1) Applying any row or column transformation does not change the determinant
2) If you interchange any two rows, sign of the determinant will change
A = | 1 x x^2 |
| 1 y y^2 |
| 1 z z^2 |
To prove option (b)
=> Apply column transformation C2 -> C2+C1
C3 -> C3+C1
=> det(A) = | 1 x+1 x^2+1 |
| 1 y+1 y^2+1 |
| 1 z+1 z^2+1 |
To prove option (c),
=> Apply row transformations R1 -> R1-R2
R2 -> R2-R3
=> det(A) = | 0 x-y x^2-y^2 |
| 0 y-z y^2-z^2 |
| 1 z z^2 |
To prove option (d),
=> Apply row transformations R1 -> R1+R2
R2 -> R2+R3
=> det(A) = | 2 x+y x^2+y^2 |
| 2 y+z y^2+z^2 |
| 1 z z^2 |
Question 3 |
-256 | |
-128 | |
-127 | |
0 |
Discuss it
Question 4 |

Priority encoder | |
Decoder | |
Multiplexer | |
Demultiplexer |
Discuss it
Question 5 |
O(log n) | |
O(n) | |
O(nLogn) | |
O(n^2) |
Discuss it
Question 6 |
O(1) | |
O(Logn) | |
O(n) | |
O(nLogn) |
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To insert 50, we will have to traverse all nodes. 10 \ 20 \ 30 \ 40
Question 7 |
A | |
B | |
C | |
D |
Discuss it









Question 8 |
n/2 | |
n-1 | |
2n-1 | |
2n |
Discuss it
S->aB B->bC C->cdThe Right Most Derivation for the above is:
S -> aB ( Reduction 3 ) -> abC ( Reduction 2 ) -> abcd ( Reduction 1 )We can see here that no production is for unit or epsilon. Hence 3 reductions here. We can get less number of reductions with some other grammar which also does't produce unit or epsilon productions,
S->abA A-> cdThe Right Most Derivation for the above as:
S -> abA ( Reduction 2 ) -> abcd ( Reduction 1 )Hence 2 reductions. But we are interested in knowing the maximum number of reductions which comes from the 1st grammar. Hence total 3 reductions as maximum, which is ( n - 1) as n = 4 here. Thus, Option B.
Question 9 |
This algorithm is equivalent to the first-come-first-serve algorithm | |
This algorithm is equivalent to the round-robin algorithm. | |
This algorithm is equivalent to the shortest-job-first algorithm.. | |
This algorithm is equivalent to the shortest-remaining-time-first algorithm |
Discuss it
Question 10 |
GROUP I GROUP II (P) Service oriented computing (1) Interoperability (Q) Heterogeneous communicating systems (2) BPMN (R) Information representation (3) Publish-find-bind (S) Process description (4) XML
P-1, Q-2, R-3, S-4 | |
P-3, Q-4, R-2, S-1 | |
P-3, Q-1, R-4, S-2 | |
P-4, Q-3, R-2, S-1 |
Discuss it