Let us assume a set with 4 elements
S={1,2,3,4}
→ If a set is said to be equivalence, then the set must be
i) Reflexive
ii) Symmetric
iii) Transitive
i) Reflexive Relation: A relation ‘R’ on a set ‘A’ is said to be reflexive if (xRx)∀x∈A.
A = {1,2,3}

R = {(1,1), (2,2), (3,3)}
R = {(1,1), (2,2)} It is false.
R = {(1,1), (2,2), (3,3), (1,2)}

ii) Symmetric Relation: A relation on a set A is said to be symmetric if (xRy). Then (yRx)∀x,y∈A i.e., if ordered pair (x,y)∈R. Then (y,x)∈R ∀x,y∈A.
A={1,2,3}
R1={(1,2), (2,1)}
R2={(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)}

Transitive Relation:
A relation ‘R’ on a set ‘A’ is said to be transitive if (xRy) and (yRz), then (xRz)∀(x,y,z∈A).
A={1,2,3}
R1={ }
R2={(1,1)}
R3={(1,2), (3,1)}
R4={(1,2), (2,1), (1,1)}
⇾ A={1,2,3,4}

Largest ordered set is
S×S={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)}
⇒ Total = 16 = 42 = n2
Smallest ordered set = {(1,1),(2,2),(3,3),(4,4)}
⇒ Total=4=n
Note: In question, they are clearly mentioned that Rank of an Equivalence relation is equal to the number of induced Equivalence classes. Since we have maximum number of ordered pairs(which are reflexive, symmetric and transitive ) in largest Equivalence relation, its rank is always 1.

Reflexive Relation:
A relation ‘R’ on a set ‘A’ is said to be reflexive if (xRx)∀x∈A.
Example: 4 D 8, 4 D 12, 4 D16, 4 D20…….(Here, D means divide)
8 D 16, 8 D 24……….
In this example, we didn’t get 4D4. So, it is not reflexive.
AntiSymmetric Relation:
For all x ∈ I, R(x,y) and R(y,x) then x=y is antisymmetric. We can easily make a violation as R(-2,2) and R(2,-2) are not antisymmetric.
It is violating. So, not antisymmetric relation.
Transitive relation:
A relation ‘R’ on a set ‘A’ is said to be transitive if (xRy) and (yRz), then (xRz)∀(x,y,z∈A).
Example: 4D8, 4D12, 4D16, 4D20…….(Here, D means divide)
8D16, 8D24……….
{ 4D8, 8D16, 1D16}. So, it is satisfied.

Given data,
Step-1: Bag contains 19 Red(R) and 19 Blue(B) balls.
BB (or) RR happen we are discarded.
If we get BR (or) RB then B is discarded and R is returned.
Step-2: There are some conditions that,
→ If black balls will either come with black then both black balls are discarder.
→ If it will come with red then only black balls will be discarded.
→ Suppose 2 red balls will come together means we are discarding both red balls.
Step-3: As per the above constraints, total 19 Red balls it means odd number.
→ Among 19 only 18 will be discarded.
Step-3: Final content of bag at second last trail will be either R,B (or) R,R,R and finally in last
trail bag will left with one red ball in both the cases.

Given data,
Step-1: x= -1 and x=2
f(x) = α log |x| + β x2 + x
f'(x)= α/x + 2βx + 1 = 0
Step-2: for extreme points f'(x)=0
α/x + 2βx + 1=0
Step-3: For x= -1 then we will get α+2β= 1 → (i)
For x= 2: then we will get α+8β= 2 → (ii)
from (i) and (ii) we can get the value of α=2 and β= -1/2

Given data,
Step-1: f(x) = log|x| and given range is [-∞ to +∞]
g(x) = sin(x) and given range is [-1,1]
Step-2: Given 2 variables are A and B
A= f(g(x))
= log|g(x)|
= log|sin(x)|
So, we will get A range is [-∞ ,0]
Step-3: B= g(f(x))
= sin(f(x))
= sin(log|x|)
So, we will get B range is [-1, 1]
Step-4: Common in both A and B is A∩B
A∩B = [-1, 0]
Key point: Ranges [ -1 ≤ sin(x) ≤ 1 and -∞ ≤ log|x| ≤ ∞ ]

Statement-A: When 'a' will comes as input. We will push it on the top of stack and when 'b' will comes as input after 'a' we will pop one 'a' for each 'b'.
This way language 'A' can be accepted by the push down automat. Hence A is CFL.
Statement-B: When last 'b' i.e., bⁿ after a^m comes as input then top of the stack will contain a^m
So we can't compare aⁿ with aⁿbⁿ. Hence it is not CFL
Statement-C: It is a CFL

NP-Hard- if it can be solved in polynomial time then all NP-Complete can be solved in polynomial time. If NP Hard problem is reducible to another problem in Polynomial Time, then that problem is also NP Hard which means every NP problem can be reduced to this problem in Polynomial Time.

Finding number of binary tree, we are using catalan numbers formula

Here, n=4.
Total number of binary trees are 14.

Encryption: Source has to encrypt with its private key for forming Digital signature for Authentication. Source has to encrypt the (M, σ) with Y’s public key to send it confidentially. Decryption: Destination Y has to decrypt first with its private key, then decrypt using source public key.

To generate a message digest by the network security protocols we need SHA and MD5.
→ SHA-256 and SHA-512 are novel hash functions computed with 32-bit and 64-bit words, respectively. They use different shift amounts and additive constants, but their structures are otherwise virtually identical, differing only in the number of rounds.
→ The MD5 message-digest algorithm is a widely used hash function producing a 128-bit hash value. Although MD5 was initially designed to be used as a cryptographic hash function, it has been found to suffer from extensive vulnerabilities. It can still be used as a checksum to verify data integrity, but only against unintentional corruption. It remains suitable for other non-cryptographic purposes, for example for determining the partition for a particular key in a partitioned database.

→ IP address belonging to class C are assigned to small-sized networks.
The network ID is 24 bits long.
The host ID is 8 bits long.
→ The higher order bits of the first octet of IP addresses of class C are always set to 110. The remaining 21 bits are used to determine network ID.
→ The 8 bits of host ID is used to determine the host in any network. The default subnet mask for class C is 255.255.255.x. Class C has a total of:
2²¹ = 2097152 network address
2⁸ – 2 = 254 host address
IP addresses belonging to class C ranges from 192.0.0.x – 223.255.255.x.

Source and destination are hosts. The hosts have all layers but routers have only 3 layers i.e physical,data link and network layer. The purpose of routers is to just pass on the packet between hosts or networks.
Therefore ,

→ UDP is cheap.
→ UDP itself is not reliable, but higher level protocols - as DNS - may maintain reliability, e.g. by repeating the UDP datagram in the case of no response.
→ But the last is not the case for DNS. DNS itself uses sometimes besides UDP (as its primary protocol) the reliable Transmission Control Protocol (TCP).
→ The last is used when the response data size exceeds 512 bytes, and for tasks which require the reliable delivery (e.g. zone transfers).
→ UDP is much faster. TCP is slow as it requires 3 way handshake. The load on DNS servers is also an important factor. DNS servers (since they use UDP) don’t have keep connections.
→ DNS requests are generally very small and fit well within UDP segments.
→ UDP is not reliable, but reliability can added on application layer. An application can use UDP and can be reliable by using timeout and resend at application layer.

→ SMTP is used for connecting to outbound servers to send email while POP3 and IMAP are used to connect to incoming servers to retrieve messages.
→ POP download email headers from mailbox and retrieve mails from servers to a cache.
→ HTTPS checking email through a web browser.

Given data,
-- Round trip delay between A and B = 40ms
-- Station A using frame size = 32 bytes. 32 bytes=32*8 bits
-- Bottleneck bandwidth on the path A and B = 64kbps
-- Window size=?
Step-1: First we have to find transmission time
Transmission time= Frame size/bandwidth
= 32*8/(64) ms
= 4 ms
Step-2: We have to find window size.
Standard Utilization formula is = n/(1+2a)
where ‘a’ is Propagation time / transmission time
= n/(1+2a)
= n/(1+40/4)
= n/11
Maximum utilization is ‘n’ = 11

2-way set associative it means 2 cache lines in one set
Physical Address =4GB
Cache Size =16KB
Block Size = 8 words = 8* 32 bits = 256 bits = 32 Bytes -------- because 1 word =32 bits
No.of cache lines /blocks = 16KB/ 32 B = 2^14/ 2^5= 2^9 i.e 512 lines
No of sets = No of cache lines / 2 ---------- because it is 2-way set-associative cache
= 2^9 /2 = 2^8 sets => 8 set bits
Now, because it is set-associative cache memory physical address generated by CPU is divided into 3 parts - first part indicate number of tag bits.
Second part indicates -no of bits required to address each set.
Third part indicates - no. of bits for block offset.
Let suppose tag bits = X
Therefore , we get
X+8+5 = 32 => X= 19 are the tag bits

→ Access A in row major order.
→ No. of cache blocks=Cache size/Block size = 32KB / 128B = 32×2¹⁰B / 128B = 2¹⁵ / 2⁷ = 256
→ No. of array elements in each block = Block size / Element size = 128B / 8B =16
→ Total misses for (P1)=Array size * No. of array elements / No. of cache blocks = (512×512) * 16 / 256=16384

Range of ‘n’ bit 2’s complement binary number is always –2^(n-1) to 2^(n-1)-1.