→ Sometimes, BCNF may not be functional dependency preserving.
→ lossless join should be compulsory in any normal form.

The extended notation would be (0+1)⁴ but however, we may allow some or all the factors to be ε. Thus ε needs to be included in the given regular expression.

Given data.
root= X⁴-X-1.
Root lies Between 1 and 2,
After second iteration=?
Using bisection method.
f(1)=X⁴-X-1
=1-1-1
= -1
f(2)=X⁴-X-1
= 2⁴ -2 -1
=13
Given constraint that “root lies between 1 and 2”
Iteration-1: x1=(a+b)/2
=(1+2)/2
= 1.5
f(1.5) = 2.5625
Iteration-2: x2=(a+b)/2
=(1+1.5)/2
=1.25
f(1.25)=0.19140625 >0
Root may lie in between (1, 1.25)
Algorithm - Bisection Scheme
Given a function f (x) continuous on an interval [a,b] and f (a) * f (b) < 0
Do
c=(a+b)/2
if f(a)*f(c)< 0 then b=c
else a=c
while (none of the convergence criteria C1, C2 or C3 is satisfied)
More info:
Bisection method is the simplest among all the numerical schemes to solve the transcendental equations. This scheme is based on the intermediate value theorem for continuous functions .
Consider a transcendental equation f(x)=0 which has a zero in the interval [a,b] and f(a)*f(b)<0. Bisection scheme computes the zero, say c, by repeatedly halving the interval [a,b]. That is, starting with
c = (a+b) / 2
The interval [a,b] is replaced either with [c,b] or with [a,c] depending on the sign of f (a) * f (c) . This process is continued until the zero is obtained. Since the zero is obtained numerically the value of c may not exactly match with all the decimal places of the analytical solution of f (x) = 0 in the interval [a,b]. Hence any one of the following mechanisms can be used to stop the bisection iterations :
C1. Fixing a priori the total number of bisection iterations N i.e., the length of the interval or the maximum error after N iterations in this case is less than | b-a | / 2N.
C2. By testing the condition | c_i - c _i-1| (where i are the iteration number) less than some tolerance limit, say epsilon, fixed a priori.
C3. By testing the condition | f (c_i ) | less than some tolerance limit alpha again fixed a priori.
http://mat.iitm.ac.in/home/sryedida/public_html/caimna/transcendental/bracketing%20methods/bisection/bisection.html

If S is the number of sets in our cache, then the set index has s=log₂ S bits.
→ Note that in a fully-associative cache, there is only 1 set so the set index will not exist. The remaining bits are used for the tag.
→ If ℓ is the length of the address (in bits), then the number of tag bits is t = ℓ − b − s.

Zero-Address Instructions:
A stack-organized computer does not use an address field for the instructions ADD and MUL. The PUSH and POP instructions, however, need an address field to specify the operand that communicates with the stack. The following program shows how X=(A+B)*(C+D) will be written for a stack-organized computer. (TOS stands for top of stack).

To evaluate arithmetic expressions in a stack computer, it is necessary to convert the expression into reverse Polish notation. The name “zero-address” is given to this type of computer because of the absence of an address field in the computational instructions.

Given data,
W₁ and W₂ are 2 strings,
|W₁|=|W₂| means same length
Example CFG grammar:
S→ Cbb | cc
C→ a
W₁=cc W₂=Cbb
As per the grammar, W₁ require only one derivation S→ cc
W₂ requires two derivations S→ Cbb→ abb
It means W₁ is smaller than W₂.

→ If we apply binary search to find the first occurrence of 1 in the list, it will give us the smallest index i where 1 is stored.
→ As in this array sequence of 0’s is followed by sequence of 1’s, the array is sorted. We can apply binary search directly without sorting it.
→ So number of probes = ⌈(log₂ 31)⌉
= ⌈4.954196310386876⌉
⇒5

Prime factorize: 162
⇒162 =2×3×3×3×3 = 3³×(2×3)
For (2×3) to be a perfect cube, it should be multiplied by (2²×3²)
∴ Required number = y = 2²×3² = 36

Given regular expression is (a+b*c). It means either a or zero or more occurence of b followed by c. But according to option A, they given “one or more occurences of b”. So, it is false.

Decidable Problems
A problem is decidable if we can construct a Turing machine which will halt in finite amount of time for every input and give answer as ‘yes’ or ‘no’. A decidable problem has an algorithm to determine the answer for a given input.
Examples
Equivalence of two regular languages:Given two regular languages, there is an algorithm and Turing machine to decide whether two regular languages are equal or not.
Finiteness of regular language: Given a regular language, there is an algorithm and Turing machine to decide whether regular language is finite or not.
Emptiness of context free language: Given a context free language, there is an algorithm whether CFL is empty or not.

Undecidable Problems
A problem is undecidable if there is no Turing machine which will always halt in finite amount of time to give answer as ‘yes’ or ‘no’. An undecidable problem has no algorithm to determine the answer for a given input.
Examples
Ambiguity of context-free languages: Given a context-free language, there is no Turing machine which will always halt in finite amount of time and give answer whether language is ambiguous or not.
Equivalence of two context-free languages: Given two context-free languages, there is no Turing machine which will always halt in finite amount of time and give answer whether two context free languages are equal or not.
Everything or completeness of CFG: Given a CFG and input alphabet, whether CFG will generate all possible strings of input alphabet (∑*)is undecidable.
Regularity of CFL, CSL, REC and REC: Given a CFL, CSL, REC or REC, determining whether this language is regular is undecidable.

Given, FCFS it means first come first serve. And also it is pure preemptive scheduling.

Average Turnaround Time =(8+13.4+14.2+15.6)/4=(51.2)/4=12.8

Given data,
Non pipelined machine with 6 stages,
Length of each stage=20,10,30,25,40,15 ns.
Implementing the pipelining the machine adds each stage=5ns overhead.
Speed up factor of the pipelining system=?

Step-1: Non pipeline for 1 instruction to all stages=20+10+30+25+40+15 ns
=140 ns
Step-2: Per cycle adds 5ns overhead to each stage =25,15,35,30,45,20 ns
= 45 ns
Step-3: Speedup factor= Time non pipelining / Time with pipeline
= 140/45
= 3.11 ns

→ First fix one boy and place other 7 in alternative seats so total ways is 7! Because they are seated in circular table. (n-1)!.
→ Now place each girl between a pair of boys. So, total ways of seating arrangement of girls is 8!
→ Finally, 7!8! Possible ways are possible.

A dirty bit (or) modified bit is a bit that is associated with a block of computer memory and indicates whether or not the corresponding block of memory has been modified. The dirty bit is set when the processor writes to (modifies) this memory.
→ The bit indicates that its associated block of memory has been modified and has not been saved to storage yet. When a block of memory is to be replaced, its corresponding dirty bit is checked to see if the block needs to be written back to secondary memory before being replaced or if it can simply be removed.
→ Dirty bits are used by the CPU cache and in the page replacement algorithms of an operating system.
→ Dirty bits can also be used in Incremental computing by marking segments of data that need to be processed or have yet to be processed.