Algorithms

Question 1

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
Question 1 Explanation: 
FORTRAN implementation do not permit recursion because they use the static allocation for variables.
→ Recursion requires dynamic allocation of data.
Question 2

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
Question 2 Explanation: 
In binary search, search for the half of the list and constant time for comparing. So,
∴ T(n) = 2T(n/2) + k, k a constant
Question 3

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 3 Explanation: 
In quick sort, we use divide and conquer technique.
Question 4

In which one of the following cases is it possible to obtain different results for call-by reference and call-by-name 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
Question 4 Explanation: 
Passing an array element as a parameter then it gives different output values for the call-by-reference and call-by-name parameters.
{ ........
a[ ] = {1, 2, 3, 4}
i = 0
fun(a[i]);
print a[0];
}
fun(int x)
{
int i = 1;
x = 8;
}
O/p:
Call-by-reference = 8
Call-by-value = 1
Question 5

Which one of the following statements is false?

A
Optimal binary search tree construction can be performed efficiently using dynamic programming.
B
Breadth-first 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
Depth-first search can be used to find connected components of a graph.
Question 5 Explanation: 
In BFS algorithm, we can randomly select a source vertex and then run, after that whether we need to check distance to each and every vertex from source is still infinite (or) not. If we find any vertex having infinite distance then the graph is not connected.
Question 6

Consider the following two functions:

Which of the following is true?

A
g1(n) is O(g2(n))
B
g1 (n) is O(3)
C
g2 (n) is O(g1 (n))
D
g2 (n) is O(n)
E
Both A and B
Question 6 Explanation: 
In asymptotic complexity, we assume sufficiently large n. So, g1(n) = n2 and g2(n) = n3.
Growth rate of g1 is less than that of g2 i.e., g1(n) = O(g2(n)) = O(n).
Question 7

An array A contains n integers in locations A[0],A[1], …………… A[n-1]. It is required to shift the elements of the array cyclically to the left by K places, where 1≤K≤n-1. 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 8

(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 9

Consider the following recursive function:

 function fib (1:integer);integer;
 begin
 if (n=0) or (n=1) then fib:=1
 else fib:=fib(n-1) + fib(n-2)
 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 10

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 11

Merge sort uses

A
Divide and conquer strategy
B
Backtracking approach
C
Heuristic search
D
Greedy approach
Question 11 Explanation: 
Merge sort uses the divide and conquer strategy.
Question 12

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
Question 12 Explanation: 
The postfix expression will be,
A B C D + * F / + D E * +
Question 13

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(n2)
A
I and II
B
II and III
C
I and IV
D
I and III
Question 13 Explanation: 
Binary search using linked list is not efficient as it will not give O(log n), because we will not be able to find the mid in constant time. Finding mid in linked list takes O(n) time.
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 14

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 15

How many minimum spanning trees does the following graph have? Draw them. (Weights are assigned to the edge).

A
Theory Explanation.
Question 16

For parameters a and b, both of which are ω(1), T(n) = T(n1/a)+1, and T(b)=1.

Then T(n) is

A
θ(loga logb n)
B
θ(logb loga n)
C
θ(log2 log2 n)
D
θ(logab n)
Question 16 Explanation: 
T(n) = T(n1/a+1, T(b) = 1
T(n) = [T(n1/a2)+1] + 1
= [T(n1/a3)+1] + 2
= [T(n1/a3)] + 3
= [T(n1/ak)] + b
= logb n = ak
= log logb n = k log a
= k= loga logb n
T(n)=1+loga logb n
T(n)=O(loga logb n)
Question 17

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| log|V|)
C
θ(|E||V|)
D
θ(|V|)
Question 17 Explanation: 
Method-1:
• 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).
Method-2:
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: Method-1 is the most appropriate answer for giving a question.
Question 18

Consider a graph G = (V, E), where V = {v1, v2, …, v100}, E = {(vi, vj) | 1 ≤ i < j ≤ 100}, and weight of the edge (vi, vj) is |i - j|. The weight of the minimum spanning tree of G is ______.

A
99
Question 18 Explanation: 
• If there are n vertices in the graph, then each spanning tree has n − 1 edges.
• N =100
• Edge weight is |i-j| 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 19

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 20

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 21

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 22

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 23

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 24

(a) Solve the following recurrence relation:

  xn = 2xn-1 - 1, n>1
  x1 = 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 25

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 1st, 2nd and 3rd smallest elements in minimum number of comparisons.

A
Theory Explanation.
Question 26

(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 O-notation.

         algorithm what (n)      
             begin 
                  if n = 1 then call A 
             else begin
                   what (n-1);
                   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 27

The root directory of a disk should be placed

A
at a fixed address in main memory
B
at a fixed location on the disk
C
anywhere on the disk
D
at a fixed location on the system disk
E
anywhere on the system disk
Question 27 Explanation: 
Root directory can points to the various user directories. Then they will be stored in a way that user can't be easily modify them. Then they should be at fixed location on the disk.
Question 28

Consider a simple connected graph G with n vertices and n-edges (n>2). Then, which of the following statements are true?

A
G has no cycles.
B
The graph obtained by removing any edge from G is not connected.
C
G has at least one cycle.
D
The graph obtained by removing any two edges from G is not connected.
E
Both C and D.
Question 28 Explanation: 
If a graph have n vertices and n edges (n>2) then it is to be cyclic graph. Then it have atleast one cycle and if we remove two edges then it is not connected.
For example let us consider, n=3.
Question 29

where O(n) stands for order n is:

A
O(n)
B
O(n2)
C
O(n3)
D
O(3n2)
E
O(1.5n2)
F
B, C, D and E
Question 29 Explanation: 

⇒ In this 'n' is constant. So, n is added to n times itself which is O(n2).
Hence, (a) is wrong. And rest (B), (C), (D), (E) are correct.
Question 30

Consider the recursive algorithm given below:

 procedure bubblersort (n);
 var i,j: index; temp : item;
 begin
    for i:=1 to n-1 do
    if A[i] > A [i+1] then
 begin
    temp : A[i];
    A[i]:=A[i+1]; 
    A[i+1]:=temp
    end;
   bubblesort (n-1)
 end 

Let an be the number of times the ‘if…then….’ Statement gets executed when the algorithm is run with value n. Set up the recurrence relation by defining an in terms of an-1. Solve for an.

A
Theory Explanation.
Question 31

Complexity of Kruskal’s algorithm for finding the minimum spanning tree of an undirected graph containing n vertices and m edges if the edges are sorted is __________

A
O(m log n)
Question 31 Explanation: 
Though the edges are to be sorted still due to union find operation complexity is O(m log n).
Question 32

Following algorithm(s) can be used to sort n integers in the range [1...n3] in O(n) time

A
Heapsort
B
Quicksort
C
Mergesort
D
Radixsort
Question 32 Explanation: 
As no comparison based sort can ever do any better than nlogn. So option (A), (B), (C) are eliminated. O(nlogn) is lower bound for comparison based sorting.
As Radix sort is not comparison based sort (it is counting sort), so can be done in linear time.
Question 33

Assume that the last element of the set is used as partition element in Quicksort. If n distinct elements from the set [1…..n] are to be sorted, give an input for which Quicksort takes maximum time.

A
n
Question 33 Explanation: 
For n distinct elements the algorithm will take maximum time when:
→ The array is already sorted in ascending order.
→ The array is already sorted in descending order.
Question 34

Consider the function F(n) for which the pseudo code is given below:

     Function F(n)
     begin
     F1 ← 1
     if(n=1) then F ← 3
     else For i = 1 to n do
                       begin
                       C ← 0
     For               j = 1 to F(n – 1) do
                       begin C ← C + 1 end
                       F1 = F1 * C
     end
                       F = F1
     end
     [n is a positive integer greater than zero] 

(a) Derive a recurrence relation for F(n)
(b) Solve the recurrence relation for a closed form solutions of F(n).

A
Theory Explanation.
Question 35

The minimum number of comparisons required to sort 5 elements is _____

A
7
Question 35 Explanation: 
Minimum no. of comparisons
= ⌈log(n!)⌉
= ⌈log(5!)⌉
= ⌈log(120)⌉
= 7
Question 36
 

The weighted external path length of the binary tree in figure is _____

A
144
Question 36 Explanation: 
Question 37

If the binary tree in figure is traversed in inorder, then the order in which the nodes will be visited is ______

A
4, 1, 6, 7, 3, 2, 5, 8
Question 37 Explanation: 
Inorder traversal is
(Left, Root, Right)
So, the order will be
4, 1, 6, 7, 3, 2, 5, 8
Question 38

Match the pairs in the following questions by writing the corresponding letters only.

(A) The number distinct binary trees with n nodes       (P) n!/2        
(B) The number of binary strings of length of 2n        (Q) (3n/n)
    with an equal number of 0’s and 1’s                                     
(C) The number of even permutations of n objects        (R) (2n/n) 
(D) The number of binary strings of length 6m which     (S) 1/n+1(2n/n) 
    are palindromes with 2n 0’s.
A
A-S, B-R, C-P, D-Q
Question 39

Choose the correct alternatives (more than one may be correct) and write the corresponding letters only: Kruskal’s algorithm for finding a minimum spanning tree of a weighted graph G with vertices and m edges has the time complexity of:

A
O(n2)
B
O(mn)
C
O(m+n)
D
O(m log n)
E
O(m2
F
B, D and E
Question 39 Explanation: 
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)
Question 40

Choose the correct alternatives (more than one may be correct) and write the corresponding letters only: The following sequence of operations is performed on a stack:

 PUSH (10), PUSH (20), POP, PUSH (10), PUSH (20), POP, POP, POP, PUSH (20), POP
The sequence of values popped out is:
A
20, 10, 20, 10, 20
B
20, 20, 10, 10, 20
C
10, 20, 20, 10, 20
D
20, 20, 10, 20, 10
Question 40 Explanation: 
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
Question 41

Which one of the following options arranges the functions in the increasing order of asymptotic growth rate?
A

f2, f3, f1

B
f3, f2, f1
C
f2, f1, f3
D
f1, f2, f3
Question 41 Explanation: 

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. 

Method-1: 

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

Method-2:

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 42
 Define Rn< >to be the maximum amount earned by cutting a rod of length n metres into one or more pieces of integer length and selling them. For i > 0, let p[i] denote the selling price of a rod whose length is i meters. Consider the array of prices:
      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 R7?
A
R7is achieved by three different solutions.
B
R7=18
C
R7=19
D
R7cannot be achieved by a solution consisting of three pieces.
Question 42 Explanation: 
The rod length of R7 is 18.

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 43
Consider the following undirected graph with edge weights as shown:

The number of minimum-weight spanning trees of the graph is _______
A
3
Question 43 Explanation: 

To find the number of spanning trees using 2 methods:

  1. If graph is complete, use n^n-2 formula
  2. If graph is not complete, then use kirchhoff theorem.

Steps in Kirchoff’s Approach:

(i) Make an Adjacency matrix.

(ii) Replace all non-diagonal is by – 1.

(iii) Replace all diagonal zero’s by the degree of the corresponding vertex.

(iv) Co-factors 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 44
Consider the following recurrence relation.

Which one of the following options is correct?
A
B
C
D
Question 44 Explanation: 
Question 45
Consider the following array.

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
Question 45 Explanation: 

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 46
Let P be an array containing n integers. Let t be the lowest upper bound on the number of comparisons of the array elements, required to find the minimum and maximum values in an arbitrary array of n elements. Which one of the following choices is correct?
A
B
C
D
Question 46 Explanation: 

Let assume n=512

Method-1: 

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=j-1) 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 = 2k for some positive integer k, then

T(n)=2T(n/2)+2

        =2(2T(n/4)+2)+2

        =4T(n/4)+4+2

           ፧

        =2k-1T(2)+1ik-12i

        =2k-1+2k-2

        =3n/2-2 

→ The given example n=512

Apply into 3n/2 -2

= (3*512)/2 -2

= 768-2

= 766

Method-2: 

Find the minimum and maximum independently, using n-1 comparisons for each, for a total of 2n-2 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(n-2)/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

= 768-2

= 766

 

Method-3:

By using Tournament Method: 

Step-1: To find the minimum element in the array will take n-1 comparisons. 

We are given 512 elements. So, to find the minimum element in the array will take 512-1= 511 

Step-2: To find the largest element in the array using the tournament method. 

  1. After the first round of Tournament , there will be exactly n/2 numbers =256 that will lose the round.
  2. The biggest loser (the largest number) should be among these 256 loosers.To find the largest number will take (n/2)−1 comparisons =256-1 = 255

Total 511+255= 766

Question 47
What is the worst-case number of arithmetic operations performed by recursive binary search on a sorted array of size n?
A
B
C
D
Question 47 Explanation: 

In this question they given three main constraints

  1. The input array is in sorted order
  2. Use recursive binary search
  3. Worst case number of operations

If the array is in sorted order then the worst case time complexity is O(logn)

Ex: 10, 20, 30

The binary search approach is using either recursive or iterative method is

Step-1: element = middle, the element is found and return the index.

Step-2: element > middle, search for the element in the sub-array starting from middle+1 index to n.

Step-3: element < middle, search for element in the sub-array starting from 0 index to middle -1.

Note: The worst case happens when we want to find the smallest/largest farthest node. So, it will not take more than O(logn) time. 

Question 48
For constants a ≥ 1 and b > 1, consider the following recurrence defined on the non-negative integers:

Which one of the following options is correct about the recurrence T(n)?
A
B
C
D
Question 48 Explanation: 
Question 49
Let G be a connected undirected weighted graph. Consider the following two statements.
S1: There exists a minimum weighted edge in G which is present in every minimum spanning tree of G.
S2:If every edge in G has distinct weight, then G has a unique minimum spanning tree.
Which of the following options is correct?
A
S1is false and S2is true.
B
S1is true and S2is false.
C
Both S1 and S2are false.
D
Both S1 and S2are true.
Question 49 Explanation: 

Statement-1: FALSE: The given statement is not valid for all the cases because they are not mentioned, edge weights are in distinct or duplicate. So, we can take any random edge weights for the given statement. 

Example:


Statement-2: TRUE: Using the kruskal’s (or) prim's algorithm we get a unique MST when there is a unique edge weight. 

Example:


Based on the above graph, we get the MST is
Question 50
 In a directed acyclic graph with a source vertex s, the quality-score of s directed path is defined to be the product of the weights of the edges on the path. Further, for a vertex v other than s, the quality-score of v is defined to be the maximum among the quality-scores of all the paths from s to v. The quality-score of s is assumed to be 1.

The sum of the quality-scores of all the vertices in the graph shown above is _________.
A
929
There are 50 questions to complete.

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