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Answers by rishu_darkshadow

1 votes
41
GATE CSE 1995 | Question: 6
What is the number of binary trees with $3$ nodes which when traversed in post-order give the sequence $A, B, C ?$ Draw all these binary trees.
What is the number of binary trees with $3$ nodes which when traversed in post-order give the sequence $A, B, C ?$ Draw all these binary trees.
5 votes
42
GATE CSE 1995 | Question: 1.25
The minimum number of edges in a connected cyclic graph on $n$ vertices is: $n-1$ $n$ $n+1$ None of the above
The minimum number of edges in a connected cyclic graph on $n$ vertices is:$n-1$$n$$n+1$None of the above
0 votes
44
GATE CSE 1994 | Question: 3.5
Match the following items ...
Match the following items$$\begin{array}{ll|ll}\hline \text{(i)} & \text{Backus-Naur form} & \text{(a)} & \text{Regular expressions} \\\hline \text{(ii)} & \text{Lexical...
2 votes
46
GATE CSE 1994 | Question: 2.5
The number of edges in a regular graph of degree $d$ and $n$ vertices is ____________
The number of edges in a regular graph of degree $d$ and $n$ vertices is ____________
3 votes
47
GATE CSE 1994 | Question: 17a
State whether the following statements are True or False with reasons for your answer: Coroutine is just another name for a subroutine.
State whether the following statements are True or False with reasons for your answer:Coroutine is just another name for a subroutine.
–3 votes
48
GATE CSE 1994 | Question: 1.15
The number of substrings (of all lengths inclusive) that can be formed from a character string of length $n$ is $n$ $n^2$ $\frac{n(n-1)}{2}$ $\frac{n(n+1)}{2}$
The number of substrings (of all lengths inclusive) that can be formed from a character string of length $n$ is$n$$n^2$$\frac{n(n-1)}{2}$$\frac{n(n+1)}{2}$
6 votes
49
GATE CSE 1994 | Question: 3.6
State True or False with reason There is always a decomposition into Boyce-Codd normal form (BCNF) that is lossless and dependency preserving.
State True or False with reasonThere is always a decomposition into Boyce-Codd normal form (BCNF) that is lossless and dependency preserving.
16 votes
50
GATE CSE 1992 | Question: 02,viii
A non-planar graph with minimum number of vertices has $9$ edges, $6$ vertices $6$ edges, $4$ vertices $10$ edges, $5$ vertices $9$ edges, $5$ vertices
A non-planar graph with minimum number of vertices has$9$ edges, $6$ vertices$6$ edges, $4$ vertices$10$ edges, $5$ vertices$9$ edges, $5$ vertices
1 votes
51
GATE CSE 1992 | Question: 02,vii
A $2-3$ tree is such that All internal nodes have either $2$ or $3$ children All paths from root to the leaves have the same length The number of internal nodes of a $2-3$ tree having $9$ leaves could be $4$ $5$ $6$ $7$
A $2-3$ tree is such thatAll internal nodes have either $2$ or $3$ childrenAll paths from root to the leaves have the same lengthThe number of internal nodes of a $2-3$ t...
2 votes
55
GATE CSE 1988 | Question: 1iii
Quicksort is ________ efficient than heapsort in the worst case.
Quicksort is ________ efficient than heapsort in the worst case.
2 votes
56
GATE CSE 1987 | Question: 2f
State whether the following statements are TRUE or FALSE: Every infinite cyclic group is isomorphic to the infinite cyclic group of integers under addition.
State whether the following statements are TRUE or FALSE:Every infinite cyclic group is isomorphic to the infinite cyclic group of integers under addition.
3 votes
58
UGC NET CSE | December 2008 | Part 2 | Question: 3
The total number of spanning trees that can be drawn using five labeled vertices is: 125 64 36 16
The total number of spanning trees that can be drawn using five labeled vertices is:125 64 36 16
0 votes
64
UGC NET CSE | December 2008 | Part 2 | Question: 21
N processes are waiting for $I/O$. A process spends a fraction $p$ of its time in $I/O$ wait state. The $CPU$ utilization is given by : 1-p-N 1-pN pN p-N
N processes are waiting for $I/O$. A process spends a fraction $p$ of its time in $I/O$ wait state. The $CPU$ utilization is given by :1-p-N 1-pN p...
0 votes
65
UGC NET CSE | December 2008 | Part 2 | Question: 15
If $4$ input multiplexers drive a $4$ input multiplexer, we get a: $16$ input $MUX$ $8$ input $MUX$ $4$ input $MUX $ $2$ input $MUX$
If $4$ input multiplexers drive a $4$ input multiplexer, we get a:$16$ input $MUX$ $8$ input $MUX$$4$ input $MUX $ $2$ input $MUX$
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