→ Consecutive words in consecutive memory modules in low order interleaving as the lower order bits determine the module.
→ Consecutive words in same memory module in high order interleaving as the higher order bits determine the module.

Hexadecimal representation of a given no. is,
(9753)₁₆
It's binary representation is,
1001011101010011
∴ The no. of 1's is 9.

⇒ c₁ = b₁a₀ ⊕ a₁b₀

Minimum no. of comparisons
= ⌈log(n!)⌉
= ⌈log(5!)⌉
= ⌈log(120)⌉
= 7

Inorder traversal is
(Left, Root, Right)
So, the order will be
4, 1, 6, 7, 3, 2, 5, 8

The recurrence relation for the no. of calls is
T(n) = T(n-1) + T(n-2) + 2
T(0) = T(1) = 0 (for fib(0) and fib(1), there are no extra recursive calls)
T(2) = 2
T(3) = 4
T(4) = 8
T(5) = 14
T(6) = 24
T(7) = 40
Counting the initial call, we get
40+1 = 41

** is used for exponentiation.
So, in total 3 registers are required and 6 memory operations in total to fetch all operands.

A given set of processes canf th be implemented by using only parbegin/parends statement, if the presedence graph of these processes is properly nested.

Suppose the length of the longest chain in a partial order is n. Then the elements in the poset can be partitioned into a disjoint antichains.

N vertices and K components.
To get maximum, take one vertex each for each component, except last component.
Now, k-1 components have 1 vertex each and so on edges.
The last component has (n-(k-1)) vertices. So make the last component complete, i.e., it has ^n-(n-1)C₂
= (n-k)(n-k+1)/2
= maximum no. of edges

A-R
B-P
C-S
D-Q
Buddy system: The buddy system is a technique to allocate the memory by dividing the memory into partitions to try to satisfy a memory request as suitably as possible.
Interpretation: Runtime.
Pointer type: Garbage collector dereference the dangling pointer.
Virtual memory: Segmentation.

Note: Out of syllabus.

In synchronization, there is a less chance of hazards but it can increase the delay. Then the advantage is ease of avoiding problems due to hazards.

Virtual memory is independent of size of main memory and depends on available secondary storage.
MAR can hold the address that is generated from CPU and limits the total virtual memory address space.

Note: Out of syllabus.

Note: Out of syllabus.

Though the edges are sorted still due to finding union operation complexity is O(m log n).
→ Where m is no. of edges and n is number of vertices then n = O(m²)
→ O(m logn) < O(mn)

PUSH(10), PUSH(20), POP = 20 (i)
→ PUSH(10), PUSH(10), PUSH(20), POP = 20 (ii)
→ PUSH(10), PUSH(10), POP = 10 (iii)
→ PUSH(10), POP = 10 (iv)
→ PUSH(20)
→ PUSH(20), POP = 20 (v)
⇒ 20, 20, 10, 10, 20

Note: Out of syllabus.

Linked editor can be able to perform
1) external symbol resolution
2) relocation

(A) It is true. Left recursive grammar if used directly in recursive descent parsing causes an infinite loop. So, left recursion must be removed before giving to a recursive descent parser.
(B) False.
(C) It is false. The language can have dynamic data types which required dynamically growing memory when data type size increases.
(D) Is true and using macro we can do this.
(E) Out of syllabus now.

(A) Is true.
(B) Is false, because one of the solution is Peterson's solution which is purely software based solution without use of hardware.
(C) Is true.
(D) False, memory can get fragmented with best fit technique.

For the propositional formula A→B to be tautology the T→F condition should never arise.
So, in option (B) it is saying that F< sub>1 ∧ F₂ is not satisfiable means F₁ ∧ F₂ is always false.
And False → anything is always true.

L(s) ⊆ L(r), because 'r' generates all strings which 's' does but 'r' also generates '101' which 's' does not generate.
L(s) ⊆ L(t), because 't' generates all the strings which 's' generates but 't' also generates '0' which 's' do not generates.

(B), (C) and (D) are true. But the strongest answer would be (D), a regular language. Because every finite language is a regular language.
And, regular language ⊂ context-free ⊂ context-sensitive ⊂ Turing recognizable, would imply that regular language is the strongest answer.