Step-1: Given compound proposition is
(~(p ∧ q))↔(~ p ∨ ~ q)
It is clearly specifying that
↔ is a root
(~(p ∧ q)) is left subtree
(~p ∨ ~q) is right subtree.
Step-2: Finally the tree is looks like

Step-3: Prefix operation traverse through Root,left and Right ↔~∧pq∨ ~ p~q Step-4: Postfix operation traverse through Left,Right and Root.
pq∧~p~q~∨↔

Step-1: Let A and B be sets in a finite universal set U.
Step-2: |A–B| we can also write into |A|-|A ∩ B| and equivalent Venn diagram is

Step-3: |A⊕B| We can also write into |A|+|B|- 2|A∩B| and equivalent Venn diagram is

Step-4: |A| + |B| We can represented into |A| + |B| and equivalent Venn diagram is

Step-5: |A∪B| We can also write into |A|+|B|- |A∩B| and equivalent Venn diagram is

Step-6: |A – B| ≤ |A ⊕ B| ≤ |A ∪ B| ≤ |A| + |B| is correct order.

Palindrome is a number that remains the same when its digits are reversed.
Ex: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121...,
Step-1:​ Here, string length is 10. If we consider first 5 numbers as random choices like 0 or 1.
Remaining 5 numbers are fixed. Total number of possibilities are 2​ ¹⁰​ . But we are only considering first 5 choices. The probability is 2​ ⁵​ .
Step-2:​ The probability 2​ ⁵​ /2^{​ 10}
= 1/2​ ⁵
= 1/32

Step-1: G1 have odd number of vertices. So, it is not euler circuit.
Step-2: G2 also have odd number of vertices. So, it not euler circuit.

(326.4) ₈ = ( ?)
Step1: First convert given octal no. to binary number because it will be easier to solve this way.

Step 2: Now convert above binary no. into decimal .
(011010110.100) ₂
= ( 1 * 2 ⁷ ) + ( 1 × 2 ⁶ ) + ( 1 × 2 ⁴ ) + ( 1 × 2 ² ) + ( 1 × 2 ¹ ) · [ 1 × ( 1/2) ]
= (214.5) ₁₀
(326.4) ₈ = (?) ₁₆ Step 1: Convert given octal no. into binary no.

Step 2: Now convert above binary no. into decimal .
(011010110.100) ₂
= ( 1 * 2 ⁷ ) + ( 1 × 2 ⁶ ) + ( 1 × 2 ⁴ ) + ( 1 × 2 ² ) + ( 1 × 2 ¹ ) · [ 1 × ( 1/2) ]
= (214.5) ₁₀
(326.4) ₈ = (?) ₁₆

2’s complement has single representation for zero , but Sign-magnitude, 1’s complement and 9’s complement have two representations for 0 (i.e., both positive zero and negative zero). While doing arithmetic operations like addition or subtraction using 1's complement(or 9's complement), we have to add an extra carry bit, i.e 1 to the result to get the correct answer. 2's complement doesn't require such extra calculation.

a. Controlled Inverter : A circuit that transmits a binary word or its1’s complement.
b. Full adder : A circuit that can add 3 bits. It adds two data bits(ai,bi) and also carry bit(ci).
c. Half adder : A logic circuit that adds 2 bits(ai,bi).
d. Binary adder : A circuit that can add two binary numbers(A and B). Collection of Full adders form a binary adder.

Step-1: Evaluate x ​ * 3 because multiplication has more priority than remaining operators x * 3→ 1.5
Step-2: && is logical AND. Both the statements are TRUE, then it returns 1 otherwise 0.
1.5 && 3 is TRUE. So, it returns 1.
Step-3: j | k is bitwise OR operator. It returns -1.
Step-4: ((x * 3) && 3) || (j | k) islogical OR operator. It returns 1
Note: The precedence is ((x * 3) && 3) || (j | k)

int *f [ ] ( ); It declares array of functions returning pointers to integers.