Let LOC of L1=X, so LOC of L2=3X
→ (X/1,000)*70,000 + 10*1,00,000 = (3X/1,000)*90,000 + 10*40,000
→70X+1000000=270X+400000
→200X=600000
→X=3000

Explanation: Hadoop Distributed File System(HDFS) has a master/slave architecture. An HDFS cluster consists:
1. NameNode, a master server that manages the file system namespace and regulates access to files by clients.
2. DataNodes, usually one per node in the cluster, which manage storage attached to the nodes that they run on. HDFS exposes a file system namespace and allows user data to be stored in files.
Internally, a file is split into one or more blocks and these blocks are stored in a set of DataNodes. The NameNode executes file system namespace operations like opening, closing, and renaming files and directories. It also determines the mapping of blocks to DataNodes. The DataNodes are responsible for serving read and write requests from the file system’s clients. The DataNodes also perform block creation, deletion, and replication upon instruction from the NameNode.

Option 4 is incorrect because CFL includes both DCFL and NDCFL and NDCFL are accepted by only non-deterministic pushdown automata, they can’t be accepted by Deterministic PDA.

A language is Regular language only if it creates an AP series.
Here only statement 2 makes an AP series. Example: L be a lanusage where i=1 and Z>= 0 then L ={ epsilon, a, aa, aaa, aaaa,..................} which is an AP. And Finite automata for this will be

➡ Cohesion Cohesion is a measure of the relative functional strength of a module.
➡ Types of cohesion :
➡ Coincidentally cohesive(LOW) a module that performs a set of tasks that relate to each other loosely, if at all. Such modules are termed coincidentally cohesive.
➡ Logically cohesive A module that performs tasks that are related logically (e.g., a module that produces all output regardless of type) is logically cohesive.
➡ Temporal cohesion When a module contains tasks that are related by the fact that all must be executed with the same span of time, the module exhibits temporal cohesion
➡ Procedural cohesion When processing elements of a module are related and must be executed in a specific order, procedural cohesion exists.
➡ Communicational cohesion When all processing elements concentrate on one area of a data structure, communicational cohesion is present.
➡ Sequential Cohesion If the function of the module forms part of sequence, where the output of one function is the input to another function.
➡ Functional Cohesion(HIGH): In this, different elements of the module cooperate with each other to achieve a single function.

1. (R ∧E) ↔C
This is not equivalent. It says that all (and only) conservatives are radical and electable.
2. R →(E ↔C)
This one is equivalent. if a person is a radical then they are electable if and only if they are conservative.
3. R →((C →E) ∨¬E)
This one is vacuous. It’s equivalent to ¬R ∨ (¬C ∨ E ∨ ¬E), which is true in all interpretations.
4.R ⇒ (E ⇐⇒ C) ≡ R ⇒ ((E ⇒ C) ∧ (C ⇒ E))
≡ ¬R ∨ ((¬E ∨ C) ∧ (¬C ∨ E))
≡ (¬R ∨ ¬E ∨ C) ∧ (¬R ∨ ¬C ∨ E))

In cryptography, a Caesar cipher, also known as Caesar's cipher, the shift cipher, Caesar's code or Caesar shift, is one of the simplest and most widely known encryption techniques. It is a type of substitution cipher in which each letter in the plaintext is replaced by a letter with some fixed number of positions down the alphabet. For example, with a left shift of 3, D would be replaced by A, E would become B, and so on. The method is named after Julius Caesar, who used it in his private correspondence.
Diffie–Hellman key exchange is a method of securely exchanging cryptographic keys over a public channel and was one of the first public-key protocols as conceived by Ralph Merkle and named after Whitfield Diffie and Martin Hellman

Effective memory access time =0.8(15+150)+0.2(15+150+150)
=132+63
=195 ns

Time required to execute 500 tasks without pipelining =500*50 ns=25000 ns
Time required to execute 500 tasks with 6-stage pipelining =(1*6*10)+(499*1*10)
=60+4990
=5050

Statement 1 is true. For example: Let n= 2 then n⁵-n will be 30 which is divisible by 5. Hence This statement is correct.
Statement 2 is also correct. For example: Let n= 2 then n³-n will be 6 which is divisible by 6. Hence This statement is correct.

The availability of high-capacity networks, low-cost computers and storage devices as well as the widespread adoption of hardware virtualization, service-oriented architecture and autonomic and utility computing has led to growth in cloud computing.
Cloud computing is the on-demand availability of computer system resources, especially data storage (cloud storage) and computing power, without direct active management by the user. The term is generally used to describe data centers available to many users over the Internet

n= P*q
n= 3*11
n=33
ф(n) = (p-1)(q-1)
ф(n) = 2*10
ф(n) = 20
E * d=1 mod ф(n)
e*7=1 mod 20
Now find the value of “e” in such a way that if we multiply the value of “e” with 7 then by dividing it with 20 mod value should come as “1”.
So if e=3 then the above condition can be satisfied hence option C is the right answer. Encrypted value = (Message)^e mod n = (19)³ mod 33 = 28

r={a,aa,ab,aaa,aab,aba,aaab,......}
s={aa,ab,aaa,aab,aba,aaab,......}
t={ab,aab,aaab,....}
If you notice then r⊇s⊇t

In computing, a firewall is a network security system that monitors and controls incoming and outgoing network traffic based on predetermined security rules. A firewall typically establishes a barrier between a trusted network and an untrusted network, such as the Internet.
Packets may be filtered by source and destination IP addresses, protocol, source and destination ports. The bulk of Internet communication in 20th and early 21st century used either Transmission Control Protocol (TCP) or User Datagram Protocol (UDP) in conjunction with well-known ports, enabling firewalls of that era to distinguish between specific types of traffic such as web browsing, remote printing, email transmission, file transfer.

As we know that “n” fork calls will lead to 2n-1child processes. Hence Total no. of process will be 2ⁿ-1+1(We added 1 for the parent process).
Now since in P1 all 3 fork calls are done before the printf statement so all 8( 2³-1+1) processes will execute the Printf statement. Hence in P1 Printf statement will be executed for 8 times.

I. TRUE: Undirected graphs do not have cross edges in DFS. But can have cross edges in directed graphs.
II. FALSE: Just draw a triangle ABC. Source is A. Vertex B and C are at the same level at distance 1. There is an edge between B and C too. So here |i - j| = |1 - 1| = 0.

Abstraction: the process of removing physical, spatial, or temporal details or attributes in the study of objects or systems to focus attention on details of greater importance; it is similar in nature to the process of generalization.
Data extraction is the act or process of retrieving data out of (usually unstructured or poorly structured) data sources for further data processing or data storage (data migration).
Data mining is a process of discovering patterns in large data sets involving methods at the intersection of machine learning, statistics, and database systems.

In Boolean logic, a formula is in conjunctive normal form (CNF) or clausal normal form if it is a conjunction of one or more clauses, where a clause is a disjunction of literals; otherwise put, it is a product of sums or an AND of ORs.

To find the lexicographic order of given strings will be:
0001<001<010<0101<011
Note: Lexicographical order is nothing but dictionary based order.

Hardwired Unit:
It can be optimized to produce fast modes of operations.
It uses a large number of registers and because of that it is costly.
Instruction size in the Hardwired unit is fixed.
It has a small number of addressing modes.
Indirect Addressing mode:
The address part is the binary equivalent of 300. The control goes to address 300 to find the address of the operand. The address of the operand in this case is 1350. The operand found in address 1350 is then added to the content of AC. The indirect address instruction needs two references to memory to fetch an operand. The first reference is needed to read the address of the operand; the second is for the operand itself.