CS502-Midterm
1 / 50
_________ is one of the few problems, where provable lower bounds exist on how fast we can sort.
2 / 50
A RAM is an idealized algorithm with takes an infinitely large random-access memory.
3 / 50
In analysis of f (n) =n (n/5) +n-10 log n, f (n) is asymptotically equivalent to ________.
4 / 50
A point p in 2-dimensional space is usually given by its integer coordinate(s)____________
5 / 50
For the heap sort, access to nodes involves simple _______________ operations.
6 / 50
The number of nodes in a complete binary tree of height h is
7 / 50
The reason for introducing Sieve Technique algorithm is that it illustrates a very important special case of,
8 / 50
The O-notation is used to state only the asymptotic ________bounds.
9 / 50
In Sieve Technique we do not know which item is of interest
10 / 50
For Chain Matrix Multiplication we can not use divide and conquer approach because,
11 / 50
A RAM is an idealized machine with ______________ random-access memory.
12 / 50
The analysis of Selection algorithm shows the total running time is indeed ________in n,
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14 / 50
The definition of Theta-notation relies on proving ___________asymptotic bound.
15 / 50
The sieve technique is a special case, where the number of sub problems is just
16 / 50
Quick sort is
17 / 50
Quick sort is best from the perspective of Locality of reference.
18 / 50
Sieve Technique applies to problems where we are interested in finding a single item
from a larger set of _____________
19 / 50
In analysis, the Upper Bound means the function grows asymptotically no faster than its largest term.
20 / 50
Which may be a stable sort?
21 / 50
Due to left complete nature of binary tree, the heap can be stored in
22 / 50
Is it possible to sort without making comparisons?
23 / 50
We cannot make any significant improvement in the running time which is better than that of brute-force algorithm.
24 / 50
F (n) and g (n) are asymptotically equivalent. This means that they have essentially the same __________ for large n.
25 / 50
For the sieve technique we solve the problem,
26 / 50
Which sorting algorithm is faster
27 / 50
The array to be sorted is not passed as argument to the merge sort algorithm.
28 / 50
For the worst-case running time analysis, the nested loop structure containing one “for” and one “while” loop, might be expressed as a pair of _________nested summations.
29 / 50
Consider the following code:
For(j=1; j
For(k=1; k<15;k++)
For(l=5; l
{
Do_something_constant();
}
What is the order of execution for this code.
30 / 50
How many elements do we eliminate in each time for the Analysis of Selection algorithm?
31 / 50
Brute-force algorithm uses no intelligence in pruning out decisions.
32 / 50
Algorithm is a mathematical entity, which is independent of a specific machine and operating system.
33 / 50
In analysis, the Lower Bound means the function grows asymptotically at least as fast as its largest term.
34 / 50
Floor and ceiling are ____________ to calculate while analyzing algorithms.
35 / 50
Counting sort has time complexity:
36 / 50
Analysis of Selection algorithm ends up with,
37 / 50
What is the solution to the recurrence T(n) = T(n/2)+n .
38 / 50
In Heap Sort algorithm, the maximum levels an element can move upward is _________
39 / 50
In Heap Sort algorithm, we build _______ for ascending sort.
40 / 50
In RAM model instructions are executed
41 / 50
The running time of quick sort depends heavily on the selection of
42 / 50
One example of in place but not stable algorithm is
43 / 50
One Example of in place but not stable sort is
44 / 50
In pseudo code, the level of details depends on intended audience of the algorithm.w
45 / 50
46 / 50
In Quick Sort Constants hidden in T(n log n) are
47 / 50
Sieve Technique can be applied to selection problem?
48 / 50
In the analysis of Selection algorithm, we eliminate a constant fraction of the array with each phase; we get the convergent _______________ series in the analysis,
49 / 50
In 2d-space a point is said to be ________if it is not dominated by any other point in that space.
50 / 50
The function f(n)= n(logn+1)/2 is asymptotically equivalent to n log n. Here Upper Bound means the function f(n) grows asymptotically ____________ faster than n log n.
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