Algorithms
Question 1 
A  
B  
C  
D 
Let assume n=512
Method1:
Using standard recursive algorithm:
MaxMin is a recursive algorithm that finds the maximum and minimum of the set of elements {a(i), a(i+1), ..., a(j)}. The situation of set sizes one (i=j) and two (i=j1) are handled separately. For sets containing more than two elements, the midpoint is determined ( just as in binary search) and two new subproblems are generated. When the maxima and minima of these subproblems are determined, the two maxima are compared and the two minima are compared to achieve the solution for the entire set.
To find the number of element comparisons needed for Maxmin, T(n) represents this number, then the resulting recurrence relation is
When n is a power of two, n = 2^{k} for some positive integer k, then
T(n)=2T(n/2)+2
=2(2T(n/4)+2)+2
=4T(n/4)+4+2
፧
=2k1T(2)+1ik12i
=2k1+2k2
=3n/22
→ The given example n=512
Apply into 3n/2 2
= (3*512)/2 2
= 7682
= 766
Method2:
Find the minimum and maximum independently, using n1 comparisons for each, for a total of 2n2 comparisons. In fact, at most 3⌊n/2⌋ comparisons are sufficient to find both the minimum and the maximum. The strategy is to maintain the minimum and maximum elements seen thus far. Rather than processing each element of the input by comparing it against the current minimum and maximum, at a cost of 2 comparisons per element, we process elements in pairs. We compare pairs of elements from the input first with each other, and then we compare the smaller to the current minimum and the larger to the current maximum, at a cost of 3 comparisons for every 2 elements.
Setting up initial values for the current minimum and maximum depends on whether n is odd or even. If n is odd, we set both the minimum and maximum to the value of the first element,and then we process the rest of the elements in pairs. If n is even, we perform 1 comparison on the first 2 elements to determine the initial values of the minimum and maximum, and then process the rest of the elements in pairs as in the case for odd n.
Let us analyze the total number of comparisons. If n is odd, then we perform 3⌊n/2⌋ comparisons. If n is even, we perform 1 initial comparison followed by 3(n2)/2 comparisons, for a total of (3n/2)2. Thus, in either case, the total number of comparisons is at most 3⌊n/2⌋.
Given an even number of elements. So, 3n/2 2 comparisons.
= (3*512)/2 2
= 7682
= 766
Method3:
By using Tournament Method:
Step1: To find the minimum element in the array will take n1 comparisons.
We are given 512 elements. So, to find the minimum element in the array will take 5121= 511
Step2: To find the largest element in the array using the tournament method.
 After the first round of Tournament , there will be exactly n/2 numbers =256 that will lose the round.
 The biggest loser (the largest number) should be among these 256 loosers.To find the largest number will take (n/2)−1 comparisons =2561 = 255
Total 511+255= 766
Question 2 
Which algorithm out of the following options uses the least number of comparisons (among the array elements) to sort the above array in ascending order?
A  Quicksort using the last element as pivot 
B  Insertion sort 
C  Selection sort 
D  Mergesort 
Quick sort(with last element as pivot) → will give the worst case time complexity as O(n^2).
Merge Sort → The merge sort will not depend upon the input order and will give O(nlogn) time complexity.
Insertion Sort → Insertion sort will perform best case time complexity when the input array is in sorted order. If the array is already sorted then the inversion count is 0 and If the array is sorted in reverse order that inversion count is the maximum.
Note: Inversion formal definition is two elements A[i] and A[j] form an inversion if A[i] > A[j] and i < j.
The number of comparisons will not take more than “n” for the given input array.
Selection Sort → Selection sort will not depend upon the input order and will give O(n^2) time complexity.
Question 3 
Which one of the following options is correct?
A  
B  
C  
D 
Question 4 
The number of minimumweight spanning trees of the graph is _______
A  3 
To find the number of spanning trees using 2 methods:
 If graph is complete, use n^n2 formula
 If graph is not complete, then use kirchhoff theorem.
Steps in Kirchoff’s Approach:
(i) Make an Adjacency matrix.
(ii) Replace all nondiagonal is by – 1.
(iii) Replace all diagonal zero’s by the degree of the corresponding vertex.
(iv) Cofactors of any element will give the number of spanning trees.
Using the Kirchhoff theorem will take lot of time because the number of vertices are 9.
So, we use brute force technique to solve the problem with the help of Kruskal's algorithm.
Question 5 
p[1] = 1, p[2] = 5, p[3] = 8, p[4] = 9, p[5] = 10, p[6] = 17, p[7] = 18
Which of the following statements is/are correct about R_{7}?
A  R_{7is achieved by three different solutions.
} 
B  R_{7}=18 
C  R_{7}=19 
D  R_{7}cannot be achieved by a solution consisting of three pieces. 
There are 3 different possible ways to get the maximum amount.
P[6] + P[1] → 17+1 = 18
P[2] + P[2] + P[3] → 5+5+8 = 18
P[7] → 18 = 18
Question 6 
Which one of the following options arranges the functions in the increasing order of asymptotic growth rate?
A  f_{2}, f_{3}, f_{1} 
B  f_{3}, f_{2}, f_{1} 
C  f_{2}, f_{1}, f_{3} 
D  f_{1}, f_{2}, f_{3} 
The asymptotic notation order should be
Constant < logarithmic < linear < polynomial < exponential < factorial
F2 and F3 → Polynomial
F1 → Exponential
By the order of asymptotic notations F1 is definitely larger.
Method1:
Consider n=100
F1 : 100 ^100 ⇒ 1.e+200
F2 : N^log(100) base 2 ⇒ 100 ^ log(100) base 2 ⇒ 100^6.6438561897747 = 1,93,96,00,91,15,564.181300016469223466
F3 : N^Sqrt(n) ====> 100^Sqrt(100) ⇒ 100^10 ⇒ 10,00,00,00,00,00,00,00,00,000
Method2:
We can apply "log" on both sides.
log(F1)=nlog10 (base 2)
log(F2)=(logn)^2 = logn * logn (base 2)
log(F3)=sqrt(n)logn (base 2)
Answer: F2< F3< F1
Question 7 
An independent set in a graph is a subset of vertices such that no two vertices in the subset are connected by an edge. An incomplete scheme for a greedy algorithm to find a maximum independent set in a tree is given below:
V: Set of all vertices in the tree; I:=φ; While V ≠ φdo begin select a vertex u; ∈ V such that V:= V – {u}; if u is such that then 1:= I ∪ {u} end; output(I);
(a) Complete the algorithm by specifying the property of vertex u in each case
(b) What is the time complexity of the algorithm.
A  Theory Explanation. 
Question 8 
Consider the following recursive function:
function fib (1:integer);integer; begin if (n=0) or (n=1) then fib:=1 else fib:=fib(n1) + fib(n2) end;
The above function is run on a computer with a stack of 64 bytes. Assuming that only return address and parameter and passed on the stack, and that an integer value and an address takes 2 bytes each, estimate the maximum value of n for which the stack will not overflow. Give reasons for your answer.
A  Theory Explanation. 
Question 9 
(a) Use the patterns given to prove that
(You are not permitted to employ induction)
(b) Use the result obtained in (a) to prove that
A  Theory Explanation. 
Question 10 
An array A contains n integers in locations A[0],A[1], …………… A[n1]. It is required to shift the elements of the array cyclically to the left by K places, where 1≤K≤n1. An incomplete algorithm for doing this in linear time, without using another is given below. Complete the algorithm by filling in the blanks. Assume all variables are suitably declared.
min:=n; i=0; while _____do begin temp:=A[i]; j:=i; while _____do begin A[j]:=_____; j:=(j+K) mod n; if j
A  Theory Explanation. 
Question 11 
Which of the following algorithm design techniques is used in the quicksort algorithm?
A  Dynamic programming 
B  Backtracking 
C  Divide and conquer 
D  Greedy method 
Question 12 
In which one of the following cases is it possible to obtain different results for callby reference and callbyname parameter passing methods?
A  Passing a constant value as a parameter 
B  Passing the address of an array as a parameter 
C  Passing an array element as a parameter 
D  Passing an array following statements is true 
{ ........
a[ ] = {1, 2, 3, 4}
i = 0
fun(a[i]);
print a[0];
}
fun(int x)
{
int i = 1;
x = 8;
}
O/p:
Callbyreference = 8
Callbyvalue = 1
Question 13 
Which one of the following statements is false?
A  Optimal binary search tree construction can be performed efficiently using dynamic programming. 
B  Breadthfirst search cannot be used to find connected components of a graph. 
C  Given the prefix and postfix walks over a binary tree, the binary tree cannot be uniquely constructed. 
D  Depthfirst search can be used to find connected components of a graph. 
Question 14 
Consider the following two functions:
Which of the following is true?
A  g_{1}(n) is O(g_{2}(n)) 
B  g_{1} (n) is O(^{3}) 
C  g_{2} (n) is O(g_{1} (n)) 
D  g_{2} (n) is O(n) 
E  Both A and B 
Growth rate of g_{1} is less than that of g_{2} i.e., g_{1}(n) = O(g_{2}(n)) = O(n).
Question 15 
The recurrence relation that arises in relation with the complexity of binary search is:
A  T(n) = T(n/2) + k, k a constant 
B  T(n) = 2T(n/2) + k, k a constant 
C  T(n) = T(n/2) + log n 
D  T(n) = T(n/2) + n 
∴ T(n) = 2T(n/2) + k, k a constant
Question 16 
FORTRAN implementation do not permit recursion because
A  they use static allocation for variables 
B  they use dynamic allocation for variables 
C  stacks are not available on all machines 
D  it is not possible to implement recursion on all machines 
→ Recursion requires dynamic allocation of data.
Question 17 
How many minimum spanning trees does the following graph have? Draw them. (Weights are assigned to the edge).
A  Theory Explanation. 
Question 18 
Consider the following sequence of numbers
92, 37, 52, 12, 11, 25
Use bubble sort to arrange the sequence in ascending order. Give the sequence at the end of each of the first five passes.
A  Theory Explanation. 
Question 19 
The postfix expression for the infix expression
A + B*(C + D)/F + D*E is:
A  AB + CD + *F/D + E* 
B  ABCD + *F/DE*++ 
C  A *B + CD/F *DE++ 
D  A + *BCD/F* DE++ 
E  None of the above 
A B C D + * F / + D E * +
Question 20 
Which of the following statements is true?
 I. As the number of entries in a hash table increases, the number of collisions increases.
II. Recursive programs are efficient
III. The worst case complexity for Quicksort is O(n^{2})
A  I and II 
B  II and III 
C  I and IV 
D  I and III 
Recursive program requires stack for storing the function state. Any recursive program can be rewritten as an iterative one. Advantage of recursion is "less programming effort", while it always lags behind iterative one in terms of performance.
Question 21 
Merge sort uses
A  Divide and conquer strategy 
B  Backtracking approach 
C  Heuristic search 
D  Greedy approach 
Question 22 
Consider a graph G = (V, E), where V = {v_{1}, v_{2}, …, v_{100}}, E = {(v_{i}, v_{j})  1 ≤ i < j ≤ 100}, and weight of the edge (v_{i}, v_{j}) is i  j. The weight of the minimum spanning tree of G is ______.
A  99 
• N =100
• Edge weight is ij for Edge (vi,vj) {1<=i<=100}
• The weight of edge(v1,v2) is 1 , edge(v5,v6) is 1.
• So, 99 edges of weight is 99.
Question 23 
Let G = (V,E) be a weighted undirected graph and let T be a Minimum Spanning Tree (MST) of G maintained using adjacency lists. Suppose a new weighted edge (u,v) ∈ V×V is added to G. The worst case time complexity of determining if T is still an MST of the resultant graph is
A  θ(E+V) 
B  θ(E logV) 
C  θ(EV) 
D  θ(V) 
• As T is a minimum spanning tree and we need to add a new edge to existing spanning tree.
• Later we need to check still T is a minimum spanning tree or not, So we need to check all vertices whether there is any cycle present after adding a new edge.
• All vertices need to traverse to confirm minimum spanning tree after adding new edge then time complexity is O(V).
Method2:
Time Complexity:
Total vertices: V, Total Edges : E
• O(logV) – to extract each vertex from the queue. So for V vertices – O(VlogV)
• O(logV) – each time a new pair object with a new key value of a vertex and will be done at most once for each edge. So for total E edge – O(ElogV)
• So overall complexity: O(VlogV) + O(ElogV) = O((E+V)logV) = O(ElogV)
Note: Method1 is the most appropriate answer for giving a question.
Question 24 
For parameters a and b, both of which are ω(1), T(n) = T(n^{1/a})+1, and T(b)=1.
Then T(n) is
A  θ(log_{a} log_{b} n) 
B  θ(log_{b} log_{a} n)

C  θ(log_{2} log_{2} n)

D  θ(log_{ab} n)

T(n) = [T(n^{1/a2})+1] + 1
= [T(n^{1/a3})+1] + 2
= [T(n^{1/a3})] + 3
= [T(n^{1/ak})] + b
= log_{b} n = a^{k}
= log log_{b} n = k log a
= k= log_{a} log_{b} n
T(n)=1+log_{a} log_{b} n
T(n)=O(log_{a} log_{b} n)
Question 25 
Consider the directed graph shown in the figure below. There are multiple shortest paths between vertices S and T. Which one will be reported by Dijkstra's shortest path algorithm? Assume that, in any iteration, the shortest path to a vertex v is updated only when a strictly shorter path to v is discovered.
A  SDT 
B  SBDT 
C  SACDT 
D  SACET 
The shortest path between S to T is SBDT also but if you follow Dijkstra shortest path algorithm then the shortest path you will getting from S to T is only SACET. We suggest you to apply Dijkstra algorithm on S and find the shortest path between S to all vertices. Then the path you will get from S to T is SACET.
Here we will draw edge from E to T not D to S because we updated the T value to 10 after selecting vertex E.
So, path is S, A, C, E, T.
Here D will get 7 only through S. So, SBDT is not possible and SDT is not possible because T will get 10 only after selecting E. So, path is SACET.
Question 26 
A list of n strings, each of length n, is sorted into lexicographic order using the mergesort algorithm. The worst case running time of this computation is
A  O (n log n) 
B  O (n^{2} log n) 
C  O (n^{2} + log n) 
D  O (n^{2}) 
2.The length of the string is n, the time taken is to be O(n) for each comparison.
3. For level 1(bottomup order): Every element has to be compared with other element the number of comparisons is O(n/2).
4. For level 1(bottomup order): Total time taken by one level is O(n^{2}).
5. For copying level to level the time taken by O(n^{2}).
So, For level 1= O(n^{2})+ O(n^{2})
6. Second level O(n^{2})+ O(n^{2}).
;
;
;
Final n level (logn)*O(n^{2}) = O(n^{2} logn)
Question 27 
Let G be a weighted graph with edge weights greater than one and G' be the graph constructed by squaring the weights of edges in G. Let T and T' be the minimum spanning trees of G and G', respectively, with total weights t and t'. Which of the following statements is TRUE?
A  T' = T with total weight t' = t^{2} 
B  T' = T with total weight t' 
C  T' ≠ T but total weight t' = t^{2} 
D  None of the above 
Then MST for G is,
Now let's square the weights,
Then MST for G' is,
So, from above we can see that T is not necessarily equal to T' and moreover (t^{1}) < (t^{2}).
So option (D) is correct answer.
Question 28 
Let W(n) and A(n) denote respectively, the worst case and average case running time of an algorithm executed on an input of size n. Which of the following is ALWAYS TRUE?
A  A(n) = Ω (W(n)) 
B  A(n) = Θ (W(n)) 
C  A(n) = O (W(n)) 
D  A(n) = o (W(n)) 
So, A(n) would be upper bounded by W(n) and it will not be strict upper bound as it can even be same (e.g. Bubble Sort and merge sort).
A(n) = O(W(n))
Note: Option A is wrong because A(n) is not equal to Ω(w(n)) .
Question 29 
Assuming P ≠ NP, which of the following is TRUE?
A  NPcomplete = NP 
B  NPcomplete ∩ P = ∅ 
C  NPhard = NP 
D  P = NPcomplete 
The definition of NPcomplete is,
A decision problem p is NPcomplete if:
1. p is in NP, and
2. Every problem in NP is reducible to p in polynomial time.
It is given that assume P ≠ NP , hence NPcomplete ∩ P = ∅ .
This is due to the fact that, if NPcomplete ∩ P ≠ ∅ i.e. there are some problem (lets say problem P1) which is in P and in NPcomplete also, then it means that P1 (NPcomplete problem) can be solved in polynomial time also (as it is also in P class) and this implies that every NP problem can be solve in polynomial time, as we can convert every NP problem into NPcomplete problem in polynomial time.
Which means that we can convert every NP problem to P1 and solve in polynomial time and hence P = NP, which is contradiction to the given assumption that P ≠ NP.
Question 30 
Consider the following program that attempts to locate an element x in a sorted array a[] using binary search. Assume N>1. The program is erroneous. Under what conditions does the program fail?
var i,j,k: integer; x: integer; a:= array; [1...N] of integer; begin i:= 1; j:= N; repeat k:(i+j) div 2; if a[k] < x then i:= k else j:= k until (a[k] = x) or (i >= j); if (a[k] = x) then writeln ('x is in the array') else writeln ('x is not in the array') end;
A  Theory Explanation. 
Question 31 
Insert the characters of the string K R P C S N Y T J M into a hash table of size 10.
Use the hash function
h(x) = (ord(x) – ord("a") + 1) mod10
and linear probing to resolve collisions.
(a) Which insertions cause collisions?
(b) Display the final hash table.
A  Theory Explanation. 
Question 32 
A complete, undirected, weighted graph G is given on the vertex {0, 1,...., n−1} for any fixed ‘n’. Draw the minimum spanning tree of G if
(a) the weight of the edge (u,v) is ∣u − v∣
(b) the weight of the edge (u,v) is u + v
A  Theory Explanation. 
Question 33 
Let G be the directed, weighted graph shown in below figure.
We are interested in the shortest paths from A.
(a) Output the sequence of vertices identified by the Dijkstra’s algorithm for single source shortest path when the algorithm is started at node A.
(b) Write down sequence of vertices in the shortest path from A to E.
(c) What is the cost of the shortest path from A to E?
A  Theory Explanation. 
Question 34 
A two dimensional array A[1...n][1...n] of integers is partially sorted if
∀i, j ∈ [1...n−1], A[i][j] < A[i][j+1] and A[i][j] < A[i+1][j]
Fill in the blanks:
(a) The smallest item in the array is at A[i][j] where i=............and j=..............
(b) The smallest item is deleted. Complete the following O(n) procedure to insert item x (which is guaranteed to be smaller than any item in the last row or column) still keeping A partially sorted.
procedure insert (x: integer); var i,j: integer; begin (1) i:=1; j:=1, A[i][j]:=x; (2) while (x > ...... or x > ......) do (3) if A[i+1][j] < A[i][j] ......... then begin (4) A[i][j]:=A[i+1][j]; i:=i+1; (5) end (6) else begin (7) ............ (8) end (9) A[i][j]:= ............. end
A  Theory Explanation. 
Question 35 
Quicksort is run on two inputs shown below to sort in ascending order taking first element as pivot,
(i) 1,2,3,...,n (ii) n,n1,n2,...,2,1
Let C_{1} and C_{2} be the number of comparisons made for the inputs (i) and (ii) respectively. Then,
A  C_{1} < C_{2} 
B  C_{1} > C_{2} 
C  C_{1} = C_{2} 
D  we cannot say anything for arbitrary n. 
So, option is (C) is correct.
Question 36 
The recurrence relation
T(1) = 2 T(n) = 3T(n/4)+n
has the solution, T(n) equals to
A  O(n) 
B  O(log n) 
C  O(n^{3/4}) 
D  None of the above 
Question 37 
The average number of key comparisons done on a successful sequential search in list of length n is
A  log n 
B  n1/2 
C  n/2 
D  n+1/2 
= No. of comparisons if element present in 1^{st} position + No. of comparisons if element present in 2^{nd} position + ............. + No. of comparisons if element present in n^{th} position
= 1 + 2 + 3 + ... + n
= n(n+1)/2
Since there are n elements in the list, so average no. of comparisons
= Total comparisons/Total no. of elements
= (n(n+1)/2)/n
= n+1/2
Question 38 
Which of the following is false?
A  
B  
C  
D 
Question 39 
Let T(n) be the function defined by T(1)= 1, T(n)= 2T(⌊n/2⌋) + √n for n≥2. Which of the following statement(s) is true?
A  T(n) = O(√n) 
B  T(n) = O(n) 
C  T(n) = O(log n) 
D  None of the above 
Question 40 
The correct matching for the following pairs is
(A) All pairs shortest path (1) Greedy (B) Quick Sort (2) DepthFirst search (C) Minimum weight spanning tree (3) Dynamic Programming (D) Connected Components (4) Divide and and Conquer
A  A – 2 B – 4 C – 1 D – 3 
B  A – 3 B – 4 C – 1 D – 2 
C  A – 3 B – 4 C – 2 D – 1 
D  A – 4 B – 1 C – 2 D – 3 
Quick sort  Divide and Conquer
Minimum weight Spanning tree  Greedy
Connected components  DepthFirst search
Question 41 
(a) Solve the following recurrence relation:
x_{n} = 2x_{n1}  1, n>1 x_{1} = 2
(b) Consider the grammar
S → Aa  b A → Ac  Sd  ε
Construct an equivalent grammar with no left recursion and with minimum number of production rules.
A  Theory Explanation. 
Question 42 
Which one of the following algorithm design techniques is used in finding all pairs of shortest distances in a graph?
A  Dynamic programming 
B  Backtracking 
C  Greedy 
D  Divide and Conquer 
Question 43 
Give the correct matching for the following pairs:
A. O(log n) 1. Selection sort B. O(n) 2. Insertion sort C. O(nlog n) 3. Binary search D. O(n^{2}) 4. Merge sort
A  A – R B – P C – Q D – S 
B  A – R B – P C – S D – Q 
C  A – P B – R C – S D – Q 
D  A – P B – S C – R D – Q 
Selection = O(n)
Merge sort = O(n log n)
Insertion sort = O(n^{2})
Question 44 
Let A be an n×n matrix such that the elements in each row and each column are arranged in ascending order. Draw a decision tree which finds 1^{st}, 2^{nd} and 3^{rd} smallest elements in minimum number of comparisons.
A  Theory Explanation. 
Question 45 
(a) Consider the following algorithm. Assume procedure A and procedure B take O(1) and O(1/n) unit of time respectively. Derive the time complexity of the algorithm in Onotation.
algorithm what (n) begin if n = 1 then call A else begin what (n1); call B(n) end end.
(b) Write a constant time algorithm to insert a node with data D just before the node with address p of a singly linked list.
A  Theory Explanation. 
Question 46 
The minimum number of record movements required to merge five files A (with 10 records), B (with 20 records), C (with 15 records), D (with 5 records) and E (with 25 records) is:
A  165 
B  90 
C  75 
D  65 
10, 20, 15, 5, 25
Merge 5 & 10:
5+10 = 15 movements
Now the list is
15, 20, 15, 25
Merge 15 & 15:
15+15 = 30 movements
Now the list is
30, 20, 25
Merge 20 & 25:
20+25 = 45 movements
Now the list is
30, 45
Merge 30 & 45:
30+45 = 75 movements
∴ Total no. of movements
= 15+30+45+75
= 165
Question 47 
If T_{1} = O(1), give the correct matching for the following pairs:
(M) T_{n} = T_{n−1} + n (U) T_{n} = O(n) (N) T_{n} = T_{n/2} + n (V) T_{n} = O(nlogn) (O) T_{n} = T_{n/2} + nlogn (W) T = O(n^{2}) (P) T_{n} = T_{n−1} + logn (X) T_{n} = O(log^{2}n)
A  M – W N – V O – U P  X 
B  M – W N – U O – X P  V 
C  M – V N – W O – X P  U 
D  None of the above 
(N) Apply Master's theorem
T(n) = θ(n) = O(n)
(O) Apply Master's theorem
T(n) = θ(n logn) = O(n logn)
(P) Here we are adding the log of firstn natural numbers.
So,
T_{n} = log1 + log2 + log3 + ... + logn
= log (1×2×...n)
= log(n!)
= θ(n logn)
Question 48 
If one uses straight twoway merge sort algorithm to sort the following elements in ascending order:
20, 47, 15, 8, 9, 4, 40, 30, 12, 17
then the order of these elements after second pass of the algorithm is:
A  8, 9, 15, 20, 47, 4, 12, 17, 30, 40 
B  8, 15, 20, 47, 4, 9, 30, 40, 12, 17 
C  15, 20, 47, 4, 8, 9, 12, 30, 40, 17 
D  4, 8, 9, 15, 20, 47, 12, 17, 30, 40 
Question 49 
If n is a power of 2, then the minimum number of multiplications needed to compute a* is
A  log_{2} n 
B  √n 
C  n1 
D  n 
We require 4 multiplications to calculate a^{16} .....(I)
→ Like that 3 multiplications requires to calculate a^{8} .....(II)
I, II are satisfied with the option A.
Question 50 
A sorting technique is called stable if
A  it takes O (nlog n) time 
B  it maintains the relative order of occurrence of nondistinct elements 
C  it uses divide and conquer paradigm 
D  it takes O(n) space 