→ Named pipe is a special instance of a file that has no contents on the filesystem.
→ A FIFO special file (a named pipe) is similar to a pipe, except that it is accessed as part of the filesystem. It can be opened by multiple processes for reading or writing. When processes are exchanging data via the FIFO, the kernel passes all data internally without writing it to the filesystem. Thus, the FIFO special file has no contents on the filesystem; the filesystem entry merely serves as a reference point so that processes can access the pipe using a name in the filesystem.
→ The kernel maintains exactly one pipe object for each FIFO special file that is opened by at least one process. The FIFO must be opened on both ends (reading and writing) before data can be passed. Normally, opening the FIFO blocks until the other end is opened also.

Step-1: Given, 5 rows with 8 seats in each row.
Total number of seats = 5*40
= 40 seats
Total number of students= 30
Step-2: Given constraint that, 6th seat in the fifth row is empty
When we are deleting 6th seat, 30 students have 39 choices of seats
Step-3: Total number of choices = ³⁹C₃₀
Total ways to choose = ⁴⁰C₃₀
Step-4: Final probability = ³⁹C₃₀ / ⁴⁰C₃₀
= 1/4

→ The range of sinx value lies between -1 and 1 and whereas log(sinx) value will be the positive value.
→ So the domain of log(log sin(x)) is undefined which is empty Set.

The above problem clearly specifies that sum of subset problem. The sum of the subset problem using recursion will give exponential time. Using dynamic programming we can get pseudo-polynomial time.
Sum of subset problem:
Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with the sum equal to the given sum.

Example:
Array set[ ]= {1,2,3,4,5} and sum=6
Here there are possible subsets (2,4),(1,2,3) and (5,1).

An abelian group is a commutative group. As per the above options, option C is the correct answer because it follows commutative property.
(x * y )² = x² * y², for any x, y belonging to G

As per code segment,
Step-1: The ith loop will compute from value 1 to 9. If the condition is true, then it computes j=j*i statement.
Step-2: The statement j=j*i is nothing but calculating 9!. If the condition fails automatically while loop will be terminated. It means maximum we can execute 9 values.
Step-3: when x becomes 10 then maximum I value will be 9. So, we can calculate up to 9! And I become 10 then it terminates the condition.

→ Binary numbers are maximum 2n-1.
→ But in question they are given ternary numbers, it means 3x-1.
→ Both will take different no. of bits to represent the same number.
3^x -1 = 2ⁿ -1
3^x = 2ⁿ
Apply log on both side
x= log₃( 2ⁿ)
x=n*log₃2 .

Breadth-first search can be used to solve many problems in graph theory

Examples
1. Copying garbage collection, Cheney's algorithm
2. Find the diameter of the graph
3. Testing bipartiteness of a graph.
4. Finding the shortest path between two nodes u and v, with path length measured by the number of edges (an advantage over depth-first search)
5. (Reverse) Cuthill–McKee mesh numbering
6. Ford–Fulkerson method for computing the maximum flow in a flow network
7. Serialization/Deserialization of a binary tree vs serialization in sorted order, allows the tree to be reconstructed in an efficient manner.
8. Construction of the failure function of the Aho-Corasick pattern matcher.

→ The Microprogram is computing a sequence of microinstructions that controls the operation of an arithmetic and logic unit so that machine code instructions are executed.
→ The Microprogram is set of microinstructions that define the individual operations in response to a machine-language instruction

Here the maximum number of comparisons is m+n-1.
For example: Take 2 sub-arrays of size 3 such as 1,3,5 and another subarray 2,4,6. Then in this case number of comparisons is 5 which is m+n-1 can be taken as O(m+n).

Given data
Step-1: 90% of code were tested,
Testing with the probability=0.9
Step-2: We can write tested data into 0.9 because it is given in percentages.
Step-3: Reliability of the module = Tested data * probability
= 0.9 * 0.9
= 0.81(at most)

Explanation:
→ If there are K search-key values in the file, the path is no longer than ⌈log⌈n/2⌉(K)⌉.
→ A node is generally the same size as a disk block, typically 4 kilobytes, and n is typically around 100 (40 bytes per index entry).
→ With 1 million search key values and n = 100, at most log50(1000000)=4 nodes are accessed in a lookup.
Note: Contrast this with a balanced binary free with 1 million search key values around 20 nodes are accessed in a lookup. Above difference is significant since every node access may need a disk I/O, costing around 20 milliseconds!

→ The primary index is created for the primary key of a table. So, records are usually clustered according to the primary key. It can be sparse.
→ In the sparse index, index records are not created for every search key. An index record here contains a search key and an actual pointer to the data on the disk. To search a record, we first proceed by index record and reach the actual location of the data. If the data we are looking for is not where we directly reach by following the index, then the system starts a sequential search until the desired data is found.
→ The secondary index usually dense.
→ In the dense index, there is an index record for every search key value in the database. This makes searching faster but requires more space to store index records itself. Index records contain search key value and a pointer to the actual record on the disk.

→ It is the best example of referential integrity. In referential integrity, we are using a foreign key.
→ If we want to delete any entity in the dominant entity(Like primary key of a table) then subordinate entity(derived attribute) also deleted.
Note: If we want to delete any subordinate entity in entity then the dominant entity is not going to delete.

→ Memory-mapped I/O uses the same address space to address both memory and I/O devices.
→ The memory and registers of the I/O devices are mapped to (associated with) address values.
→ So when an address is accessed by the CPU, it may refer to a portion of physical RAM, or it can instead refer to the memory of the I/O device. Thus, the CPU instructions used to access the memory can also be used for accessing devices.
→ Each I/O device monitors the CPU's address bus and responds to any CPU access of an address assigned to that device, connecting the data bus to the desired device's hardware register.
→ To accommodate the I/O devices, areas of the addresses used by the CPU must be reserved for I/O and must not be available for normal physical memory.
→ The reservation may be permanent, or temporary (as achieved via bank switching).

It only satisfied statement I. because increasing the memory size increases the rate at which requests are satisfied but can not alter the possibility of deadlock and neither does it play any role in implementation.

Let the element just before x be y and that just after be z. In order to link y to z,
we need x→ Bwd→ Fwd = x.
Fwd and in order to link z to y, we need x→ Fwd→ Bwd=x→ Bwd.

→ Pre order traverses Root, Left, Right.
→ Post order traverses Left, Right, Root.
→ In-order traverses Left, Root, Right.
Tree-1: Post order: GJIHEFBDCA
Tree-2: In order: GJIHEFBDCA