I have gone through Google and Stack Overflow search, but nowhere I was able to find a clear and straightforward explanation for how to calculate time complexity.
What do I know already?
Say for code as simple as the one below:
char h = 'y'; // This will be executed 1 time
int abc = 0; // This will be executed 1 time
Say for a loop like the one below:
for (int i = 0; i < N; i++) {
Console.Write('Hello, World!!');
}
int i=0; This will be executed only once.
The time is actually calculated to i=0 and not the declaration.
i < N; This will be executed N+1 times
i++ This will be executed N times
So the number of operations required by this loop are {1+(N+1)+N} = 2N+2. (But this still may be wrong, as I am not confident about my understanding.)
OK, so these small basic calculations I think I know, but in most cases I have seen the time complexity as O(N), O(n^2), O(log n), O(n!), and many others.
How to find time complexity of an algorithm
You add up how many machine instructions it will execute as a function of the size of its input, and then simplify the expression to the largest (when N is very large) term and can include any simplifying constant factor.
For example, lets see how we simplify 2N + 2 machine instructions to describe this as just O(N).
Why do we remove the two 2s ?
We are interested in the performance of the algorithm as N becomes large.
Consider the two terms 2N and 2.
What is the relative influence of these two terms as N becomes large? Suppose N is a million.
Then the first term is 2 million and the second term is only 2.
For this reason, we drop all but the largest terms for large N.
So, now we have gone from 2N + 2 to 2N.
Traditionally, we are only interested in performance up to constant factors.
This means that we don't really care if there is some constant multiple of difference in performance when N is large. The unit of 2N is not well-defined in the first place anyway. So we can multiply or divide by a constant factor to get to the simplest expression.
So 2N becomes just N.
This is an excellent article: Time complexity of algorithm
The below answer is copied from above (in case the excellent link goes bust)
The most common metric for calculating time complexity is Big O notation. This removes all constant factors so that the running time can be estimated in relation to N as N approaches infinity. In general you can think of it like this:
statement;
Is constant. The running time of the statement will not change in relation to N.
for ( i = 0; i < N; i++ )
statement;
Is linear. The running time of the loop is directly proportional to N. When N doubles, so does the running time.
for ( i = 0; i < N; i++ ) {
for ( j = 0; j < N; j++ )
statement;
}
Is quadratic. The running time of the two loops is proportional to the square of N. When N doubles, the running time increases by N * N.
while ( low <= high ) {
mid = ( low + high ) / 2;
if ( target < list[mid] )
high = mid - 1;
else if ( target > list[mid] )
low = mid + 1;
else break;
}
Is logarithmic. The running time of the algorithm is proportional to the number of times N can be divided by 2. This is because the algorithm divides the working area in half with each iteration.
void quicksort (int list[], int left, int right)
{
int pivot = partition (list, left, right);
quicksort(list, left, pivot - 1);
quicksort(list, pivot + 1, right);
}
Is N * log (N). The running time consists of N loops (iterative or recursive) that are logarithmic, thus the algorithm is a combination of linear and logarithmic.
In general, doing something with every item in one dimension is linear, doing something with every item in two dimensions is quadratic, and dividing the working area in half is logarithmic. There are other Big O measures such as cubic, exponential, and square root, but they're not nearly as common. Big O notation is described as O ( <type> ) where <type> is the measure. The quicksort algorithm would be described as O (N * log(N )).
Note that none of this has taken into account best, average, and worst case measures. Each would have its own Big O notation. Also note that this is a VERY simplistic explanation. Big O is the most common, but it's also more complex that I've shown. There are also other notations such as big omega, little o, and big theta. You probably won't encounter them outside of an algorithm analysis course. ;)
Taken from here - Introduction to Time Complexity of an Algorithm
1. Introduction
In computer science, the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function of the length of the string representing the input.
2. Big O notation
The time complexity of an algorithm is commonly expressed using big O notation, which excludes coefficients and lower order terms. When expressed this way, the time complexity is said to be described asymptotically, i.e., as the input size goes to infinity.
For example, if the time required by an algorithm on all inputs of size n is at most 5n3 + 3n, the asymptotic time complexity is O(n3). More on that later.
A few more examples:
1 = O(n)
n = O(n2)
log(n) = O(n)
2 n + 1 = O(n)
3. O(1) constant time:
An algorithm is said to run in constant time if it requires the same amount of time regardless of the input size.
Examples:
array: accessing any element
fixed-size stack: push and pop methods
fixed-size queue: enqueue and dequeue methods
4. O(n) linear time
An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i.e. time grows linearly as input size increases.
Consider the following examples. Below I am linearly searching for an element, and this has a time complexity of O(n).
int find = 66;
var numbers = new int[] { 33, 435, 36, 37, 43, 45, 66, 656, 2232 };
for (int i = 0; i < numbers.Length - 1; i++)
{
if(find == numbers[i])
{
return;
}
}
More Examples:
Array: Linear Search, Traversing, Find minimum etc
ArrayList: contains method
Queue: contains method
5. O(log n) logarithmic time:
An algorithm is said to run in logarithmic time if its time execution is proportional to the logarithm of the input size.
Example: Binary Search
Recall the "twenty questions" game - the task is to guess the value of a hidden number in an interval. Each time you make a guess, you are told whether your guess is too high or too low. Twenty questions game implies a strategy that uses your guess number to halve the interval size. This is an example of the general problem-solving method known as binary search.
6. O(n2) quadratic time
An algorithm is said to run in quadratic time if its time execution is proportional to the square of the input size.
Examples:
Bubble Sort
Selection Sort
Insertion Sort
7. Some useful links
Big-O Misconceptions
Determining The Complexity Of Algorithm
Big O Cheat Sheet
Several examples of loop.
O(n) time complexity of a loop is considered as O(n) if the loop variables is incremented / decremented by a constant amount. For example following functions have O(n) time complexity.
// Here c is a positive integer constant
for (int i = 1; i <= n; i += c) {
// some O(1) expressions
}
for (int i = n; i > 0; i -= c) {
// some O(1) expressions
}
O(nc) time complexity of nested loops is equal to the number of times the innermost statement is executed. For example, the following sample loops have O(n2) time complexity
for (int i = 1; i <=n; i += c) {
for (int j = 1; j <=n; j += c) {
// some O(1) expressions
}
}
for (int i = n; i > 0; i += c) {
for (int j = i+1; j <=n; j += c) {
// some O(1) expressions
}
For example, selection sort and insertion sort have O(n2) time complexity.
O(log n) time complexity of a loop is considered as O(log n) if the loop variables is divided / multiplied by a constant amount.
for (int i = 1; i <=n; i *= c) {
// some O(1) expressions
}
for (int i = n; i > 0; i /= c) {
// some O(1) expressions
}
For example, [binary search][3] has _O(log n)_ time complexity.
O(log log n) time complexity of a loop is considered as O(log log n) if the loop variables is reduced / increased exponentially by a constant amount.
// Here c is a constant greater than 1
for (int i = 2; i <=n; i = pow(i, c)) {
// some O(1) expressions
}
//Here fun is sqrt or cuberoot or any other constant root
for (int i = n; i > 0; i = fun(i)) {
// some O(1) expressions
}
One example of time complexity analysis
int fun(int n)
{
for (int i = 1; i <= n; i++)
{
for (int j = 1; j < n; j += i)
{
// Some O(1) task
}
}
}
Analysis:
For i = 1, the inner loop is executed n times.
For i = 2, the inner loop is executed approximately n/2 times.
For i = 3, the inner loop is executed approximately n/3 times.
For i = 4, the inner loop is executed approximately n/4 times.
…………………………………………………….
For i = n, the inner loop is executed approximately n/n times.
So the total time complexity of the above algorithm is (n + n/2 + n/3 + … + n/n), which becomes n * (1/1 + 1/2 + 1/3 + … + 1/n)
The important thing about series (1/1 + 1/2 + 1/3 + … + 1/n) is around to O(log n). So the time complexity of the above code is O(n·log n).
References:
1
2
3
Time complexity with examples
1 - Basic operations (arithmetic, comparisons, accessing array’s elements, assignment): The running time is always constant O(1)
Example:
read(x) // O(1)
a = 10; // O(1)
a = 1,000,000,000,000,000,000 // O(1)
2 - If then else statement: Only taking the maximum running time from two or more possible statements.
Example:
age = read(x) // (1+1) = 2
if age < 17 then begin // 1
status = "Not allowed!"; // 1
end else begin
status = "Welcome! Please come in"; // 1
visitors = visitors + 1; // 1+1 = 2
end;
So, the complexity of the above pseudo code is T(n) = 2 + 1 + max(1, 1+2) = 6. Thus, its big oh is still constant T(n) = O(1).
3 - Looping (for, while, repeat): Running time for this statement is the number of loops multiplied by the number of operations inside that looping.
Example:
total = 0; // 1
for i = 1 to n do begin // (1+1)*n = 2n
total = total + i; // (1+1)*n = 2n
end;
writeln(total); // 1
So, its complexity is T(n) = 1+4n+1 = 4n + 2. Thus, T(n) = O(n).
4 - Nested loop (looping inside looping): Since there is at least one looping inside the main looping, running time of this statement used O(n^2) or O(n^3).
Example:
for i = 1 to n do begin // (1+1)*n = 2n
for j = 1 to n do begin // (1+1)n*n = 2n^2
x = x + 1; // (1+1)n*n = 2n^2
print(x); // (n*n) = n^2
end;
end;
Common running time
There are some common running times when analyzing an algorithm:
O(1) – Constant time
Constant time means the running time is constant, it’s not affected by the input size.
O(n) – Linear time
When an algorithm accepts n input size, it would perform n operations as well.
O(log n) – Logarithmic time
Algorithm that has running time O(log n) is slight faster than O(n). Commonly, algorithm divides the problem into sub problems with the same size. Example: binary search algorithm, binary conversion algorithm.
O(n log n) – Linearithmic time
This running time is often found in "divide & conquer algorithms" which divide the problem into sub problems recursively and then merge them in n time. Example: Merge Sort algorithm.
O(n2) – Quadratic time
Look Bubble Sort algorithm!
O(n3) – Cubic time
It has the same principle with O(n2).
O(2n) – Exponential time
It is very slow as input get larger, if n = 1,000,000, T(n) would be 21,000,000. Brute Force algorithm has this running time.
O(n!) – Factorial time
The slowest!!! Example: Travelling salesman problem (TSP)
It is taken from this article. It is very well explained and you should give it a read.
When you're analyzing code, you have to analyse it line by line, counting every operation/recognizing time complexity. In the end, you have to sum it to get whole picture.
For example, you can have one simple loop with linear complexity, but later in that same program you can have a triple loop that has cubic complexity, so your program will have cubic complexity. Function order of growth comes into play right here.
Let's look at what are possibilities for time complexity of an algorithm, you can see order of growth I mentioned above:
Constant time has an order of growth 1, for example: a = b + c.
Logarithmic time has an order of growth log N. It usually occurs when you're dividing something in half (binary search, trees, and even loops), or multiplying something in same way.
Linear. The order of growth is N, for example
int p = 0;
for (int i = 1; i < N; i++)
p = p + 2;
Linearithmic. The order of growth is n·log N. It usually occurs in divide-and-conquer algorithms.
Cubic. The order of growth is N3. A classic example is a triple loop where you check all triplets:
int x = 0;
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
for (int k = 0; k < N; k++)
x = x + 2
Exponential. The order of growth is 2N. It usually occurs when you do exhaustive search, for example, check subsets of some set.
Loosely speaking, time complexity is a way of summarising how the number of operations or run-time of an algorithm grows as the input size increases.
Like most things in life, a cocktail party can help us understand.
O(N)
When you arrive at the party, you have to shake everyone's hand (do an operation on every item). As the number of attendees N increases, the time/work it will take you to shake everyone's hand increases as O(N).
Why O(N) and not cN?
There's variation in the amount of time it takes to shake hands with people. You could average this out and capture it in a constant c. But the fundamental operation here --- shaking hands with everyone --- would always be proportional to O(N), no matter what c was. When debating whether we should go to a cocktail party, we're often more interested in the fact that we'll have to meet everyone than in the minute details of what those meetings look like.
O(N^2)
The host of the cocktail party wants you to play a silly game where everyone meets everyone else. Therefore, you must meet N-1 other people and, because the next person has already met you, they must meet N-2 people, and so on. The sum of this series is x^2/2+x/2. As the number of attendees grows, the x^2 term gets big fast, so we just drop everything else.
O(N^3)
You have to meet everyone else and, during each meeting, you must talk about everyone else in the room.
O(1)
The host wants to announce something. They ding a wineglass and speak loudly. Everyone hears them. It turns out it doesn't matter how many attendees there are, this operation always takes the same amount of time.
O(log N)
The host has laid everyone out at the table in alphabetical order. Where is Dan? You reason that he must be somewhere between Adam and Mandy (certainly not between Mandy and Zach!). Given that, is he between George and Mandy? No. He must be between Adam and Fred, and between Cindy and Fred. And so on... we can efficiently locate Dan by looking at half the set and then half of that set. Ultimately, we look at O(log_2 N) individuals.
O(N log N)
You could find where to sit down at the table using the algorithm above. If a large number of people came to the table, one at a time, and all did this, that would take O(N log N) time. This turns out to be how long it takes to sort any collection of items when they must be compared.
Best/Worst Case
You arrive at the party and need to find Inigo - how long will it take? It depends on when you arrive. If everyone is milling around you've hit the worst-case: it will take O(N) time. However, if everyone is sitting down at the table, it will take only O(log N) time. Or maybe you can leverage the host's wineglass-shouting power and it will take only O(1) time.
Assuming the host is unavailable, we can say that the Inigo-finding algorithm has a lower-bound of O(log N) and an upper-bound of O(N), depending on the state of the party when you arrive.
Space & Communication
The same ideas can be applied to understanding how algorithms use space or communication.
Knuth has written a nice paper about the former entitled "The Complexity of Songs".
Theorem 2: There exist arbitrarily long songs of complexity O(1).
PROOF: (due to Casey and the Sunshine Band). Consider the songs Sk defined by (15), but with
V_k = 'That's the way,' U 'I like it, ' U
U = 'uh huh,' 'uh huh'
for all k.
For the mathematically-minded people: The master theorem is another useful thing to know when studying complexity.
O(n) is big O notation used for writing time complexity of an algorithm. When you add up the number of executions in an algorithm, you'll get an expression in result like 2N+2. In this expression, N is the dominating term (the term having largest effect on expression if its value increases or decreases). Now O(N) is the time complexity while N is dominating term.
Example
For i = 1 to n;
j = 0;
while(j <= n);
j = j + 1;
Here the total number of executions for the inner loop are n+1 and the total number of executions for the outer loop are n(n+1)/2, so the total number of executions for the whole algorithm are n + 1 + n(n+1/2) = (n2 + 3n)/2.
Here n^2 is the dominating term so the time complexity for this algorithm is O(n2).
Other answers concentrate on the big-O-notation and practical examples. I want to answer the question by emphasizing the theoretical view. The explanation below is necessarily lacking in details; an excellent source to learn computational complexity theory is Introduction to the Theory of Computation by Michael Sipser.
Turing Machines
The most widespread model to investigate any question about computation is a Turing machine. A Turing machine has a one dimensional tape consisting of symbols which is used as a memory device. It has a tapehead which is used to write and read from the tape. It has a transition table determining the machine's behaviour, which is a fixed hardware component that is decided when the machine is created. A Turing machine works at discrete time steps doing the following:
It reads the symbol under the tapehead.
Depending on the symbol and its internal state, which can only take finitely many values, it reads three values s, σ, and X from its transition table, where s is an internal state, σ is a symbol, and X is either Right or Left.
It changes its internal state to s.
It changes the symbol it has read to σ.
It moves the tapehead one step according to the direction in X.
Turing machines are powerful models of computation. They can do everything that your digital computer can do. They were introduced before the advent of digital modern computers by the father of theoretical computer science and mathematician: Alan Turing.
Time Complexity
It is hard to define the time complexity of a single problem like "Does white have a winning strategy in chess?" because there is a machine which runs for a single step giving the correct answer: Either the machine which says directly 'No' or directly 'Yes'. To make it work we instead define the time complexity of a family of problems L each of which has a size, usually the length of the problem description. Then we take a Turing machine M which correctly solves every problem in that family. When M is given a problem of this family of size n, it solves it in finitely many steps. Let us call f(n) the longest possible time it takes M to solve problems of size n. Then we say that the time complexity of L is O(f(n)), which means that there is a Turing machine which will solve an instance of it of size n in at most C.f(n) time where C is a constant independent of n.
Isn't it dependent on the machines? Can digital computers do it faster?
Yes! Some problems can be solved faster by other models of computation, for example two tape Turing machines solve some problems faster than those with a single tape. This is why theoreticians prefer to use robust complexity classes such as NL, P, NP, PSPACE, EXPTIME, etc. For example, P is the class of decision problems whose time complexity is O(p(n)) where p is a polynomial. The class P do not change even if you add ten thousand tapes to your Turing machine, or use other types of theoretical models such as random access machines.
A Difference in Theory and Practice
It is usually assumed that the time complexity of integer addition is O(1). This assumption makes sense in practice because computers use a fixed number of bits to store numbers for many applications. There is no reason to assume such a thing in theory, so time complexity of addition is O(k) where k is the number of bits needed to express the integer.
Finding The Time Complexity of a Class of Problems
The straightforward way to show the time complexity of a problem is O(f(n)) is to construct a Turing machine which solves it in O(f(n)) time. Creating Turing machines for complex problems is not trivial; one needs some familiarity with them. A transition table for a Turing machine is rarely given, and it is described in high level. It becomes easier to see how long it will take a machine to halt as one gets themselves familiar with them.
Showing that a problem is not O(f(n)) time complexity is another story... Even though there are some results like the time hierarchy theorem, there are many open problems here. For example whether problems in NP are in P, i.e. solvable in polynomial time, is one of the seven millennium prize problems in mathematics, whose solver will be awarded 1 million dollars.
So the algorithm for which the time complexity is to estimated is this
int a = 0;
for (i = 0; i < n; i++) //runs from i-n;
for (j = i; j < i * i; j++) // runs from i - i*i;
for (k = 0; k < j; k++) //runs from 0 - j;
a++;
I have commented the basic details that i have understood about the algorithm.
The outer loop clearly runs O(n) times and first inner loop runs from 'i' to 'i*i'
The second inner loop runs '0' to 'j'
I also know i have to apply the rule of products.
But I am unable to calculate the time complexity of the two inner loops relative to 'n'.
Please correct me if I'm wrong with my explanation.
Whenever there is a doubt always use mathematics as the mathematical proofs are more palatable like strong intuitions.
Let n = size of input.
The innermost loop or lets call it as the 3rd loop runs for j+1 number of times.
The inner loop or 2nd loop runs for i*i - i number of times.
Finally, the outer loop for n times.
So, the number of iterations for the entire program can be expressed in mathematical terms as shown in the following figure:
When you have multiple levels of for loops, analyze the complexity of each loop in isolation and then multiply them together.
In this example, the complexity of the first loop is O(n), like you said.
The complexity of the second loop is O(n^2) because in the worst case, the number of operations you have to perform is on the order of i*i which could be as big as n^2. It doesn't matter that it doesn't start at zero either because in an expression like O(n^2 - n), everything but the highest order term gets ignored.
The third loop also takes O(n^2) because in the worst case, you could have as many as j operations would again could be as big as n^2.
Lastly a++ happens in constant time. Multiply everything together and you have a complexity of O(n^5).
I've been trying to determine the time complexity of nested for loops
here is the problem:
I've done some work , but yet can't reach a decisive answer. I've reached a sequence of(log(100)+ log(200)+....log(n/4)), yet can't determine the sum of it.
Update
I have an idea, I'm uploading it it as a picture, please tell me if it's correct:
Basically, what you have here is a series. To be able to calculate the sum, you should write the series in a uniform notation and either calculate the sum yourself or use some tool (such as Wolfram Alpha) to compute the series for you.
In this case, the format is: sum from i = 1 to (n/400) of log(100 * i).
https://www.wolframalpha.com/input/?i=sum+log_2(100+*+i),+i+%3D+1+to+n%2F400
I think you might be over thinking it. You have two nested loops. Clearly the outer one is O(n):
for (int i = 0; i < n / 4; i += 100)
It has a low constant factor, a constant factor that is making it faster, but that does not mean that it is not O(n)
In the inner loop
for (int j = 1; j < i; j *= 2)
j keeps doubling until it is larger than i so it runs k times if k is the first power of two greater than i, so 2^k ≥ i, or k is the base-2 log of i. That means this loop is O(log n) since i is linearly related to n. Thus the whole thing is O(n log n)
I'm having some trouble finding the big O for the if statement in the code below:
public static boolean areUnique (int[] ar)
{
for (int i = 0; i < ar.length-1; i++) // O(n)
{
for (int j = i+1; j < ar.length-1; j++) // O(n)
{
if (ar[i] == ar[j]) // O(???)
return false; // O(1)
}
}
return true; //O(1)
}
I'm trying to do a time complexity analysis for the best, worst, and average case
Thank you everyone for answering so quickly! I'm not sure if my best worst and average cases are correct... There should be a case difference should there not because of the if statement? But when I do my analysis I have them all ending up as O(n2)
Best: O(n) * O(n) * [O(1) + O(1)] = O(n2)
Worst: O(n) * O(n) * [O(1) + O(1) + O(1)] = n2
Average: O(n) * O(n) * [O(1) + O(1) + O(1)] = O(n2)
Am I doing this right? My textbook is not very helpful
For starters, this line
if (ar[i] == ar[j])
always takes time Θ(1) to execute. It does only a constant amount of work (a comparison plus a branch), so the work done here won't asymptotically contribute to the overall runtime.
Given this, we can analyze the worst-case behavior by considering what happens if this statement is always false. That means that the loop runs as long as possible. As you noticed, since each loop runs O(n) times, the total work done is Θ(n2) in the worst-case.
In the best case, however, the runtime is much lower. Imagine any array where the first two elements are the same. In that case, the function will terminate almost instantly when the conditional is encountered for the first time. In this case, the runtime is Θ(1), because a constant number of statements will be executed.
The average-case, however, is not well-defined here. Average-case is typically defined relative to some distribution - the average over what? - and it's not clear what that is here. If you assume that the array consists of truly random int values and that ints can take on any integer value (not a reasonable assumption, but it's fine for now), then the probability that a randomly-chosen array has a duplicate is 0 and we're back in the worst-case (runtime Θ(n2)). However, if the values are more constrained, the runtime changes. Let's suppose that there are n numbers in the array and the integers range from 0 to k - 1, inclusive. Given a random array, the runtime depends on
Whether there's any duplicates or not, and
If there is a duplicate, where the first duplicated value appears in the array.
I am fairly confident that this math is going to be very hard to work out and if I have the time later today I'll come back and try to get an exact value (or at least something asymptotically appropriate). I seriously doubt this is what was expected since this seems to be an introductory big-O assignment, but it's an interesting question and I'd like to look into it more.
Hope this helps!
the if itself is O(1);
this is because it does not take into account the process within the ALU or the CPU, so if(ar[i] == ar[j]) would be in reality O(6), that translates into O(1)
You can regard it as O(1).
No matter what you consider as 'one' step,
the instructions for carrying out a[i] == a[j] doesn't depend on the
value n in this case.
I am taking up algorithm course on coursera,and I am stuck on this particular problem. I am supposed to find the time complexity of this code.
int sum = 0
for (int i = 1; i <= N*N; i = i*2)
{
for (int j = 0; j < i; j++)
sum++;
}
I checked it in eclipse itself, for any value of N the number of times sum statement is executed is less than N
final value of sum:
for N=8 sum=3
for N=16 sum=7
for N=100000 sum=511
so the time complexity should be less than N
but the answer that is given is N raised to the power 2, How is it possible?
What I have done so far:
the first loop will run log(N^ 2) times, and consequently second loop will be execute 1,2,3.. 2 logN
The first inner loop will be 1 + 2 + 4 + 8 .. 2^M where 2^M is <= N * N.
The sum of powers of 2 up to N * N is approximately 2 * N * N or O(N ^ 2)
Note: When N=100000 , N*N will overflow so its result is misleading. If you consider overflow to be part of the problem, the sum is pretty random for large numbers so you could argue its O(1), i.e. if N=2^15, N^2 = 2^30 and the sum will be Integer.MAX_VALUE. No higher value of N will give a higher sum.
There is a lot of confusion here, but the important thing is that Big-O notation is all about growth rate, or limiting behavior as mathematicians will say. That a function will execute in O(n*n) means that the time to execute will increase faster than, for example n, but slower than, for example 2^n.
When reasoning with big-O notation, remember that constants "do not count". There is a few querks in this particular question.
The N*N expression it-self would lead to a O(log n*n) complexity if the loop was a regular for-loop...
...however, the for-loop increment is i = i*2 causing the outer loop to be executed approximately log n and the function would be in O(log n) if the contents of the inner loop run in a time independent of n.
But again, the inner-loop run-time depends on n, but it doesn't do n*n runs, rather it does roughly log (n*n)/2 loops. Remembering that "constants don't count" and that we end up in O(n*n).
Hope this cleared things up.
So sum ++ will be executed 1 + 2 + 4 + 8 + ... + N*N, total log2(N*N) times. Sum of geometrical progression 1 * (1 - 2 ^ log2(N*N)/(1 - 2) = O(N*N).
Your outer loop is log(N^2)->2*log(N)->log(N), your inner loop is N^2/2->N^2. So, the time complexity is N^2*log(N).
About the benchmark, values with N=8 or N=16 are ridiculous, the time in the loop is marginal in relation with setting JVM, cache fails, and so on. You must:
Begin with biggest N, and check how it evaluate.
Make multiple runs with each value of N.
Think that the time complexity is a measure of how the algorithm works when N becomes really big.