Question

. Big O Notation.Thanks to Reges, Building Java Programs, 2nd edition.

Estimate the big-O complexity for each of these algorithms, and justify your answer.

To confirm your calculations, answers are provided at the end of the rubric. Your justification can be mathematical or written, formal or informal.

Rubric:
Correct Big-O classification of four problems
Justification of four problems
Big-O categories: 3.1. O(log n). 3.2. O(n). 3.3. O(n²). 3.4. O(1)

Problem	Code fragment
3.1	int sum = 0; int j = 1; while (j <= n) { sum++; j *= 2; }
	Big-O Category
	Justification (why did you pick the way you did?)
3.2	int sum = 0; for (int j = 1; j < n; j++) { sum++; if (j % 2 == 0) {     sum++; } }
	Big-O Category
	Justification (why did you pick the way you did?)
3.3	int sum = 0; for (int i = 1; i <= n * 2; i++) { for (int j = 1; j <= n; j++) {     sum++; } }
	Big-O Category
	Justification (why did you pick the way you did?)
3.4	for (int j = 1; j < 100; j++) { sum++; sum++; }
	Big-O Category
	Justification (why did you pick the way you did?)

Answer 1

3.1) The complexity of the algorithm is O(log(n))

All the single step execution path are considered as execution time of O(1), so

int sum = 0;------->O(1)

int j = 1;--------------> O(1)

Then while loop is executed, the value of j is the series, 1, 2,2^2,2^3,.....,2^k

If we put k equals to Log₂n, we get 2^Log₂n which is n.

ie the time complexity of a code isO( log(n)) when loop variable is multiplied or divided by a constant value

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3.2) Here the for loop executes n times and so the code inside the for loop.

So the final time complexity will be O(n) + O(1) = O(n)

Note here O(1) is the time complexity of code outside for loop

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3.3)Time complexity of code outside nested for loops= O(1)

Here we have nested for loop, so to find the time complexity , find the number of time inner for loop is executed.

Inner for loop is executed 2*n times

and complexity of inner for loop =O(n) as it executes codes inside it n times.

So total time complexity= O(1)+O(2n*n)=O(2n^2)= O(n^2), we can eliminate constant 2

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3.4)

In this case the loop variable i executes 100 times and so the code inside the for loop

So the time complexity of algo =O(100)= O(1), because complexity of any algorithm executing a constant number of times i can be considered as O(1)