→ Initially calculate the possible cases of a relation i.e., R={(A,B), (C,B), (A,C)}
→ Now check whether the relation is transitive or commutative or etc..,
If a relation is commutative then if (A,B) present in the relation then (B,A) also need to exists. but in the relation (B,A) is not present. so given relation is not commutative.

In graph theory, a circle graph is the intersection graph of a set of chords of a circle.
→ A circle graph is an undirected graph whose vertices can be associated with chords of a circle such that two vertices are adjacent if and only if the corresponding chords cross each other.
→ Circle does not have vertices.

Message M: “You are good students” is stored as M'
The message length is "n".
The following string is "Youare areyou aregood goodare goodstudents studentsgood" divided as follows
You are good students --- n
are you students good ---- n
are good ----- n
good are ----- n
Then total messages length is 4n.

→ Set A is a subset of another set B if all elements of the set A are elements of the set B Program is a collection of statements and that collection may be subset so the option(a) is true.
→ In the program, the execution of instructions will be performed sequentially.So option (b) also true.
→ A partially ordered set (also poset) formalizes and generalizes the intuitive concept of an ordering, sequencing, or arrangement of the elements of a set. According to the above definition, the option (c) also true.

(i) string-> MIUIUIUIUIU
It has M (5 times ) IU
So start with MIU
Production 2 says: M.x-> M.x.x
So
MU-> MUU again using it
MUU-> MUUU -> MUUUU ->MUUUUU
Now using prod 1: Last U-> IU
MUUUUU -> MIUUUUU
MIUUUUU -> MIUIUUUU
MIUIUUU ----------------> MIUIUIUIUIU

Given boolean expression is (X+Y+XY) (X+Z)
Step-1: (X (1+Y)+Y) (X+Z) /* To take common as X */
Step-2: (X+Y) (X+Z) /* To take common as X */
Step-3: (X+YZ)

We can solve this problem by two methods
1. Using Boolean expression
2. Using truth table

Emitter Coupled Logic (ECL)
The storage time is eliminated as the transistors are used in difference amplifier mode and are never driven into saturation.
1. Fastest among all logic families
2. Lowest propagation delay.
Complementary metal oxide semiconductor(CMOS)
The power dissipation is usually 10nW per gate depending upon the power supply voltage, output load etc.
1. Lowest power dissipation
2. Excellent noise immunity
3. High packing density
4. Wide range of supply voltage
5. Highest fan out among all logic families

Given expression is 3 × 512 + 7 × 64 + 5 × 8 + 3
We can write above statement based on precedence is (3 * 512) + (7 * 64) + (5 * 8) + 3
Step-1: 1536+448+40+3
Step-2: (2027)₁₀
Step-3: Equivalent binary number is (‭011111101011‬)₂
Note: Total number of 1’s is 9.

Recursive functions are executed in a last in first out-order. It is using stack data structure to evaluate recursive functions.

The input given in string type. But we need to print the value in integer type. For this we have to use type casting (or) atoi function. So, it will give output is “error”
Note: argv[1], argv[2] and argv[3] is string type.