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Test: Algorithm Analysis & Asymptotic Notation- 1 - Computer Science Engineering (CSE) MCQ


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10 Questions MCQ Test - Test: Algorithm Analysis & Asymptotic Notation- 1

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Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 1

Sorting is useful for

Detailed Solution for Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 1

Sorting is useful for making search easier and efficient and in report generation but not minimize the space needed.

Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 2

The order of an algorithm that finds whether a given Boolean function of ‘n’ variables, produces a 1 is

Detailed Solution for Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 2

In the worst case it has to check all the 2n possible input combinations, which is exponential.

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Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 3

The order of the binary search algorithm is

Detailed Solution for Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 3

Let there be ‘n' items to be searched, After the first search the list is divided into two, each of length n/2 . After the next search, 2 lists, each of length n/4  and so on. This successive division has to stop when the length of list becomes 1. Let it happen after k steps.
After the k steps,

Hence the order is log(n).

Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 4

For merging two sorted lists of sizes m and n into a sorted list of size m + n. Find out the time complexity of this merging process.

Detailed Solution for Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 4

Each comparison will append one item to the existing merge list. In the worst case one needs m + n -1 comparisons which is of order m+ n.

Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 5

A machine needs a minimum of 100 sec to sort 1000 names by quick sort. The minimum time needed to sort 100 names will be approximately

Detailed Solution for Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 5

In the best case quick sort algorithm makes O(nlog (n)) comparisons.
So 1000 x log (1000) = 9000 comparisons, which takes 100 sec.
To sort 100 names a minimum of 100 (log 100) = 600 comparisons are-needed.

Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 6

A machine took 200 sec to sort 200 names, using bubble sort. In 800 sec, it can approximately sort

Detailed Solution for Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 6

For sorting 200 names bubble sort makes   comparisons. The time needed  for 1 comparison is 200 sec (approximately). In 800 sec it can make 80,000 comparisons. We have to find n, such that 

Solving, n is approximately 400.

Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 7

Unrestricted use of goto is harmful, because it

Detailed Solution for Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 7

Unrestricted use of goto statement is harmful because it makes more difficult to verify programs i.e. use of goto can results in unstructured code and there can be blocks with multiple entry and exit which can cause difficulty which debugging of program.

Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 8

The recurrence relation that arises in relation with the complexity of binary search is

Detailed Solution for Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 8

Since, with every iteration of Binary Search, data set to be searched is divided into half of the original/earlier dataset. Also, same constant time is lapsed during each iteration. So, recurrence relation for Binary Search would be:

Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 9

Consider the following two functions: 

Which of the following is/are true?
1. f(n) is O(n3)
2. g(n) is O (n3)
3. 0(f(n)) is same as O(g(n))
4. g(n) is O{n2)

Detailed Solution for Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 9

In this question total possibility is four but only two possibility will give solution and other two possibilities will not give solutions. 
g(n) = O (f(n))
In case of 0 ≤ n < 100 & & 0 ≤n< 10,000 
(it is also not possible)
in case of otherwise From the above two equations the conclusion is O(f (n)) is same as O(g(m)) and g(n) and f(n) will be O(n2).

Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 10

Suppose f, g, h, k : N→ N.
If f  =  then

Detailed Solution for Test: Algorithm Analysis & Asymptotic Notation- 1 - Question 10

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