√3 + √7=1.7320508075688772935274463415059 + 2.6457513110645905905016157536393
=4.3778021186334678840290620951451
√10=3.1622776601683793319988935444327
So, √3 + √7 > √10 is true.

A given connected graph is a Euler Graph if and only if all vertices of are of even degree.
Proof:
→ Let G(V, E) be an Euler graph. Thus G contains an Euler line Z, which is a closed walk.
→ Let this walk start and end at the vertex u ∈ V. Since each visit of Z to an intermediate vertex v of Z contributes two to the degree of v and since Z traverses each edge exactly once, d(v) is even for every such vertex.
→ Each intermediate visit to u contributes two to the degree of u, and also the initial and final edges of Z contribute one each to the degree of u. So the degree d(u) of u is also even.

The set of vertices has size n, the number of such subsets is given by the binomial coefficient C(n,2)⋅C(n,2)=n(n-1)/2.

A + AB’+ AB’C = A(1+B’+B’C) = A
No GATE is required to implement the function A.

If the switches are in parallel then use “+”, and if they are serial then use “ ”.
A(B+C) + AB + (A+B)C = AB + AC + AB + AC + BC
= (AB+AB) + (AC+AC) + BC
= AB + AC + BC

The function table for the Half Subtractor is as follows
A-B= D= A’B + AB’
X= A’B

W’ + W’Z + Z’XY = W’(1+Z) + Z’XY
= W’ + Z’XY
The term W’Z is redundant which is represented by GATE 2.

→ A dynamic hazard is the possibility of an output changing more than once as a result of a single input change.
→ Dynamic hazards often occur in larger logic circuits where there are different routes to the output (from the input).
→ If each route has a different delay, then it quickly becomes clear that there is the potential for changing output values that differ from the required / expected output. e.g.
→ A logic circuit is meant to change output state from 1 to 0, but instead changes from 1 to 0 then 1 and finally rests at the correct value 0. This is a dynamic hazard.
→ As a rule, dynamic hazards are more complex to resolve, but note that if all static hazards have been eliminated from a circuit, then dynamic hazards cannot occur.

X1= Y1
X2= Y1⊕ Y2
X3= Y2 ⊕ Y3

The square wave has alternating amplitudes(0 and 1) with duty cycle 1.
An odd number of cascaded NOT gates produce a square wave.

Note: Duty cycle= Ratio of durations in which the circuit is ON and OFF in a cycle.

Given, (12A7C)₁₆ = (0001 0010 1010 0111 1100)₂
MAke blocks of 3 bits each from LSB to MSB.
(Note: In the last block append zeros (as MSBs) if number bits is not three)
(000 010 010 101 001 111 100)
Each of the above blocks represents a digit in base 8 and they can be converted to base 8 as shown below.
= (0 2 2 5 1 7 4)₈

Excess-3 code is also called Self-Complementing Code. Because 1’s complement of an excess-3 number is equivalent to 9’s complement of the corresponding decimal digit.
→ In excess-3 code, each of the 4-bit numbers represents decimal digit which is 3 less than the actual decimal digit. So the bits have no fixed weight.
Excess-3 code is neither CRC nor Algebraic Code which is used for error detection and/or correction.

Hint: Number of bits=4
(Decimal value of maximum 4-bit number +1 )/2= (15+1)/2=8

-> JK flip flop accepts input J=K=1. When J=K=1, the state of the flip-flop gets complimented. But it's not a valid input in SR flip-flop.
-> JK flip flop doesn’t have a feedback path.

→ Given that the processor clock rate = 2.5 GHz, the processor takes 2.5 G cycles in one second.
→ Time taken to complete one cycle = (1 / 2.5 G) seconds
→ Since it is given that average number of cycles per instruction = 4, the time taken for completing one instruction=(4/2.5 G) = 1.6 ns
→ In the pipelined case we know in the ideal case CPI = 1, and the clock speed = 2 GHz.
→ Time taken for one instruction in the pipelined case = (1 / 2 G) = 0.5 ns
→ Speedup = 1.6/0.5 = 3.2

If we are given p instead of *p then it prints the address of k but we are printing the value of k.
The same rule applies in **m.
Output is 5 5 5

→ Given that the disk pack has 16 surfaces, 128 tracks per surface, 256 sectors per track and each sector size is 512 bytes.
→ So the disk pack capacity = 16*128*256*512 bytes = 256 MB
→ To specify a sector we need the information about surface number, track number and sector number within a track.
→ Surface number needs 4 bits as there are 16 surfaces(2⁴), track number needs 7 bits as there are 128 tracks(2⁷) within a surface, within a track the sector number needs 8 bits as there are 256 sectors (2⁸).
→ Total number bits needed to specify a particular sector = 4+7+8 = 19 bits.