U.G.C. NET Exam. June 2023 Paper-II (COMPUTER SCIENCE)

To find a key for relation scheme R, we need to identify a minimal set of attributes that uniquely determine all other attributes in the reason. Given the dependencies
C → F
E → A
EC → D
A → B
We can start with attribute sets, and check if they satisfy the closure property to include all attributes.
Now, Let's try {E, C}:
{E, C}+ = {C, E, A, F}
(from C → F and E → A)
It includes all attributes, So {E, C} is a candidate key. We can verify that no subset of {E, C} uniquely determines all attributes, Hence, {E, C} is the key for relation scheme R.

In given code, each element in the array 'a' is initialized with a pointer to the next element's 'z' member. So,
'a [0], z' refers to the string " Nagpur", which is the 'z' value of the first structure.
'ptr → z' also gives " Nagpur" because 'ptr' was intialized to point to the first structure in the array.
'a[2]. p → z' also gives "Nagpur" because 'p' of the third structure was initialized to point to the first structure in the array.
Therefore, the output of this code would be Nagpur Nagpur Nagpur.

A transaction is a single logical unit of work that accesses and possibly modifies the contents of a database. Transactions access data using read and write operations. In order to maintain consistency in a database, before and after the transaction, certain properties are followed. These are called ACID properties.
A = Atomicity
C = Consistency
I = Isolation
D = Durability

Analyze each statements-
(A) The maximum transfer unit (MTU) value of the link between P and R is 1500 bytes and between R and Q is 820 bytes, and the IP header size is 20 bytes, the maximum payload size for the second fragment would be 800 bytes (820–20) R should have received a full datagram of size 1420 bytes (1400 + 20). The other with the remaining data 600 bytes (1400 – 800). the second fragment would indeed be 620 bytes (600 + 20). So, the statement (A) is true.
(B) R does not have the capability to resend lost fragments. It's the responsibility of the end points (P and Q) to handle packet loss and retransmission. so, the statement (B) is false.
(C) TCP relies or acknowledgments to ensure reliable delivery. If a fragment is lost and not acknowledged by the receiver, TCP will trigger a retransmission of the entire segment. So, the statement (c) is true. (
D) The destination port is a TCP header field which is located in the first fragment. The second fragment does not contain this information So, the statement (D) is false.

Dendral is indeed an expert system but the reason provided doesn't accurately describe the characteristics of an expert system. The rationality of an agent is often related to its reaction to the environment, as expert systems like dendral typically exhibit rational behavior in responding to input data or queries within their domain of expertise. So, the Assertion A is true but Reason R is false.

In genetic algorithms, solution representations typically include, Binary - valued representation uses strings of binary digits to encode solutions. Real-valued representation encodes solutions as vectors of real numbers. Permutation representation represents solutions as permutations of set of elements, However, 'combinations' are not typically used as a direct representation in genetic algorithms.

Syntax analysis is the second phase of a compiler, also known as parsing. This phase takes the stream to tokens generated by the lexical analysis phase and checks whether they conform to the grammar of the programming language. The output of this phase is usually an Abstract syntax tree (AST).