Ans .
B
Dataflow diagram comes under Logical View of the system. A data flow diagram (DFD) is a graphical representation of the "flow" of data through an information system, modelling its process aspects. A DFD is often used as a preliminary step to create an overview of the system without going into great detail, which can later be elaborated.
Ans .
A
Maintenance and Evaluation is the phase where all upgrade and bug fixed reported is implemented on the system.
Ans .
B
In Review phase the quality and performance of a system is measured and commented on.
Ans .
C
Requirements determination is the beginning sub phase of analysis. In this sub phase, analysts should gather information on what the system should do from as many sources as possible.
Ans .
D
Basically all the stake holders who are involved in the developement
Ans .
C
Self Exploratory
Ans .
A
Self Exploratory
Ans .
A
DSS - Analysis, decision Support MIS - Information processing EIS - Status Access
Ans .
C
Common mistake: using ∧ as the main connective with ∀
Ans .
D
Ans .
A
This is minimal finite automata. Since its endings with therefore n+1 states.
Ans .
D
Deterministic pushdown automata can recognize all deterministic context-free languages while nondeterministic ones can recognize all context-free languages. Mainly the former are used in parser design. Deterministic context-free languages (DCFL) are a proper subset of context-free languages. Non-deterministic finite automata and Deterministic finite automata, both accept same set of languages as NFAs can be translated to equivalent DFAs using the subset construction algorithm.
Ans .
B
We need to find the number of different configurations possible for memory and each of these will be a state in FSM. (At any time memory will be in one configuration and in next instance it either remains same or goes to a different configuration) A word is of n bits. And we have m such words. So, total number of bits = m*n. We need a separate state for each bit combination. So, no. of states = 2^(mn).
Ans .
A
X is undecidable but partially decidable. We have the TM M. Just make the state q the final state and make all other final states non-final and get a new TM M'. Give input w to M'. If w would have taken M to state q (yes case of the problem), our new TM M' would accept it. So, the given problem is partially decidable.
Ans .
C
L1 - CSL L2 - Palindrome so CFL L3 - Linear Power and regular expression can be stated as (00)∗ L4 - non linear power So CSL
Ans .
B
In DFA any subset of the N states (for N element set 2^N subsets possible) can become a new state and they can remain even when the DFA is minimized. So, maximum we can get 2^N states for the minimized DFA. (at least in question must be a typo for at most).
Ans .
A
Babel Fish was a free Web-based multilingual translation application.
Ans .
A
The minimum criterion takes the least value in the set given. In an AND operation all the values have to be true in order for the whole expression to be true. Hence taking the least of the values ensures that the range of all values are met.
Ans .
A
A knowledge base (KB) is a technology used to store complex structured and unstructured information used by a computer system. So it contains Rules, facts and relationships.
Ans .
B
Self Explanatory
Ans .
C
Let the amount of B1 be x (in kg) Let the amount of B2 be y (in kg) Minimize 5x + 8y Subject To: x + y = 5 (since the weight of the brick must be 5 kg) x ≤ 4 (since there must be no more than 4 kg of B1) y ≥ 2 (since there must be a min. of 2 kg of B2) x ≥ 0
Ans .
C
Self Explanatory
Information for question 23 and 24 : A modern IT park has 7 buildings viz. B1,.....,B7 and 9 roads containing them. B1 is centrally located and is connected to all the 6 buildings B2,.....,B7 surrounding it. The pair (B2,B3),(B4,B5) and (B6,B7) are connected. cleaning of the roads in the IT park is done by cleaning truck and the depot of the vehicle is located on the road joining B4 and B5. The truck operator needs some help in planning the cleaning tour.Help him by identifying the correct statements about the tour
Ans .
C
Solve the problem using Hamaltonion Path.
Ans .
B
Since it does not contain any hamaltonion cycle and B1 is least costly path therefore its obvious it would be visited more.
Ans .
A
n(AᴜBᴜC) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(C∩A) + n(A∩B∩C)
Information for question 26 and 27 : A multi-layer feed-forward artificial neural network has 16 neurons in the input layer. The number of neurons in every following layer is half of that in previous layer . There is only one 1 neuron in the output layer.
Ans .
B
Ans .
A
Ans .
A
Ans .
D
Ans .
B
P(B|A)/P(B)
Ans .
B
There is no such file as user32.exe user16.exe and user.dll in 32 but windows
Ans .
C
$$ contains current shell process id.
Ans .
B
.o is a simple program in passive state, .h is c++ header file format while .a is unix static library extension
Ans .
B
Loader's job is to bring a program from pasive state to active by loading it and running it from memory
Ans .
A
Shift reduce conflict and since its shift reduce conflict its also reduce conflict
Ans .
B/C/D
Ans .
A
Ans .
C
Ans .
B
Ans .
D
Ans .
B
Ans .
C
Ans .
B
Ans .
B
Ans .
B
Ans .
C
To insert an element, we need to search for its place first. The search operation may take O(n)
Ans .
A
HTTP is an application layer protocol. Since firewal is at layer 4, it cannot block HTTP data.
Ans .
A
Lexical analysis is the first step in compilation. In lexical analysis, program is divided into tokens. Lexical analyzers are typically based on finite state automata. Tokens can typically be expressed as different regular expressions: An identifier is given by [a-zA-Z][a-zA-Z0-9]* The keyword if is given by if. Integers are given by [+-]?[0-9]+.
Ans .
B
Minimum spanning tree for 2 nodes would be (v1) _ (v2) Total weight 3 Minimum spanning tree for 3 nodes would be (v1) _ (v2) | (v3) Total weight= 3 + 4 = 7 Minimum spanning tree for 4 nodes would be (v1) _ (v2) _ (v4) | (v3) Total weight= 3 + 4 + 6 = 13 Minimum spanning tree for 5 nodes would be (v1) _ (v2) _ (v4) | (v3) | (v5) Total weight= 3 + 4 + 6 + 8 = 21 Minimum spanning tree for 6 nodes would be (v1) _ (v2) _ (v4) _ (v6) | (v3) | (v5) Total weight= 3 + 4 + 6 + 8 + 10 = 31 We can observe from above examples that when we add kth node, the weight of spanning tree increases by 2k-2. Let T(n) be the weight of minimum spanning tree. T(n) can be written as T(n) = T(n-1) + (2n-2) for n > 2 T(1) = 0, T(2) = 0 and T(2) = 3 The recurrence can be written as sum of series (2n – 2) + (2n-4) + (2n-6) + (2n-8) + …. 3 and solution of this recurrence is n^2 – n + 1.
Ans .
D
If the graph is planar, then it must follow below Euler’s Formula for planar graphs v - e + f = 2 v is number of vertices e is number of edges f is number of faces including bounded and unbounded 10 - 15 + f = 2 f = 7 There is always one unbounded face, so the number of bounded faces = 6
Ans .
B
There can be total 6C4 ways to pick 4 vertices from 6. The value of 6C4 is 15. Note that the given graph is complete so any 4 vertices can form a cycle. There can be 6 different cycle with 4 vertices. For example, consider 4 vertices as a, b, c and d. The three distinct cycles are cycles should be like this (a, b, c, d,a) (a, b, d, c,a) (a, c, b, d,a) (a, c, d, b,a) (a, d, b, c,a) (a, d, c, b,a) and (a, b, c, d,a) and (a, d, c, b,a) (a, b, d, c,a) and (a, c, d, b,a) (a, c, b, d,a) and (a, d, b, c,a) are same cycles. So total number of distinct cycles is (15*3) = 45.
Ans .
C
Lossy compression
Ans .
B
The main idea for using Spanning Trees is to avoid loops.
Ans .
B
2^(n-1)