Complexity ramblings


A few totally-not authoritative notes on evaluating the complexity of an algorithm.

Mohammad Sajid Anwar expressed a couple of times doubts about how to calculate the complexity of an algorithm. I hope these few notes might be of help, without being wrong.

In a nutshell, calculating the complexity of an algorithm is about evaluating how much resources it takes to perform its workings. As algorithms usually operate on data, this evaluation is usually related to how much data there is.

Resources: time and space

The two main resources that are usually considered are time and space (usually in terms of memory).

Sometimes algorithms can trade one for another. As an example, if we have a lot of memory we might adopt an algorithm that consumes a lot of space but gives us better times. On the other hand, if memory is a scarce resource we might compensate by doing more and taking more time.

As an example, consider the calculation of the $n$-th Fibonacci number. We might keep a very long list of them, and just take the item from the list when we need it, or we might calculate it on the spot. The first approach is faster, the second approach uses a low amount of memory.

Normally we talk about time complexity, but considering space complexity is always advisable too.

Let’s start simple, and talk granularity

Let’s take a simple program to write out stuff from the input:

while (<>) { print }

This reads a line, writes a line, reads a line, …

Reading a line means reading all characters up to the end of the line. So we can say that we will do this read operation for each character. Writing has a similar consideration, so we do a write operator for each character too.

In memory terms, how much do we need? If our file is a single, long line… then our space complexity scales like the file size. You have a 1 MB file with one line only? You will need about 1 MB RAM. Your file is 100 MB? You might need 100 MB RAM in the worst case where there’s only a single line.

Of course not all inputs have one, single line. So the estimation above is the worst case, but it also makes sense to consider the average case, or even the best case. This is why we read about them. In a normal text file where the author wraps lines at about 80 characters, we will need about 80 characters of buffer to do the read/write cycles, so much less than the worst case.

It might also happen that we don’t really mind about individual characters, especially if our lines do not deviate too much (or too often) from the average. At this point we might simplify the problem and just reason in terms of “lines” instead of “characters”. Hence, instead of saying:

  • read characters up to a newline or the end of file
  • write all characters read in the previous line

we might “simplify” like this:

  • read one line
  • write one line.

Hence, we have one read and one write operation for each line. The more the lines, the more stuff we will have to do.

Of course, the read and write operations are not simpler than before. It’s just that we’re switching gears, and we are considering lines as our granularity level, instead of individual characters. Whether this makes sense or not depends on the problem and… on us.

We touched upon the space complexity, and we know that the average space needed is that to hold an average line, with the worst case being the full file size. If we are dealing with “normal” text files, we can just say that the space complexity is constant, because - well - a memory allocation that is about X character long should be fine.

What about time complexity? Well, this is probably simpler, because we have to do one read and one write operation for each character, or for each line (depending on the chosen granularity level). Hence, whatever we put it, the complexity increases linearly with the input. Do we have 1 MB of file? It will take this long. Is it a 100 MB file? It will take about 100 times more.

Except when it does not, because maybe reading X bytes all together takes exactly the same time as reading only one, thanks to optimizations in the memory transfer routines. But let’s keep it simple!

Let’s go detailed

If we have an estimation of how much time an operation takes, then we can make the calculation quite exact. Let’s say that a read operation on a line takes 1 unit of time, while a write takes 2 units of time; in this case, for a file that is $N$ long, we have:

  • $N$ units of time for reading
  • $2 \cdot N$ units of time for writing
  • $0.01 \cdot N$ units of time for checks, etc

The one about loop checks is an estimation, of course; it might be different. Hence, we have that the total amount of time resources would be:

\[T = N + 2 \cdot N + 0.01 \cdot N \\ T = 3.01 \cdot N\]

Another example

Let’s now consider a simple sorting algorithm, bubble sort (pseudo-algorithm taken from Wikipedia):

procedure bubbleSort(A : list of sortable items)
    n := length(A)
        swapped := false
        for i := 1 to n-1 inclusive do
            /* if this pair is out of order */
            if A[i-1] > A[i] then
                /* swap them and remember something changed */
                swap(A[i-1], A[i])
                swapped := true
            end if
        end for
    until not swapped
end procedure

In the best case, we have a single sweep but no swapping happens, which means that after the iteration from $1$ to $n - 1$ the algorithm will exit. Hence, the best case has a time complexity equal to:

\[T = (n - 1)\]

if a single comparison takes one unit of time. On the other hand, the worst case would require us to go through the whole list multiple times, making a lot of swap operations. The amount of operations would be this:

\[T = n \cdot (n - 1) + 4 \cdot \frac{(n - 1)(n - 2)}{2}\\ T = n^2 - n + 2n^2 - 6n + 4 \\ T = 3n^2 - 7n + 4\]

assuming, of course, that a swap operation takes $4$ units of time. Am I throwing numbers? Yes I am.

But wait! there’s a length operation at the beginning… if we have to count all items, the time will be proportional to the amount of items $n$, let’s assume it takes $0.5$ units of time for each count operation:

\[T = 0.5 n + 3n^2 - 7n + 4 \\ T = 3n^2 -6.5n + 4\]

The interesting thing is that there is a quadratic term and a linear term, in addition to a constant.

Making things simpler

Let’s suppose that we have two algorithms, with the following estimations:

\[T_1 = 3n^2 + n + 10 \\ T_2 = 800n\]

Which of the two is better? Well… of course it depends. For little values of $n$, the first one clearly wins, On the other hand, as $n$ grows, the first algorithm grows faster than the second one, up to the point where its resource requirements overcomes those of the second one.

This is where the concept of asymptoticity kicks in. We figure that our inputs will be larger and larger, hence we want algorithms that can address larger and larger inputs without requiring too many resources.

For this reason, we usually keep only the fastest growing term in the equation. What makes the difference and makes $T_1$ worse than $T_2$ for large inputs is the square term. Hence, we neglect all other terms and simplify the two estimations like this:

\[T'_1 = 3n^2 \\ T'_2 = 800 n\]

Another thing that is usually neglected are multiplicative constants. I always failed to fully get this point, but it helps us get a gist of how an algorithm scales without too many details. This leads us to the so-called Big-O notation:

\[C_1 = O(n^2) \\ C_2 = O(n)\]

So there it is - this is a representation of how fast the resources requirements grows. At this point it’s easy to say that, for bigger and bigger inputs, the second algorithm is definitely the way to go.

This procedure applies both to the time and to the space estimations. Saying that an algorithm goes like $O(n^3)$ space-wise means that we can expect that for an input that is $n$ long, we will need an amount of memory that grows like $n^3$, so it’s better than $n^4$ but definitely worse than a linear growth.

Doing quick estimations

To quickly estimate the complexity of an algorithm, we usually have to look for loops.

Do we act upon each item a few times? Then it’s $O(n)$. Do we have two nested loops to compare each item in an array with a significant portion of the array itself? Then it’s $O(n^2)$. Do we want to evaluate a tri-dimensional function over $n$ points on each dimension? Then we will have to calculate $n^3$ values, for a complexity of $O(n^3)$. You get the idea.

Sometimes we don’t even have to act upon each item. As an example, when we have an array of $n$ items, and we already know that it is sorted, then we can use a binary search and complete the search in no more than about $log(n) + 1$ operations. So… we have $O(log(n))$, because the $log(n)$ part grows with $n$ and the constant $1$ does not, so we ignore it and only keep the logarithm.


Here we are, practical considerations and a whole load of stuff to read about what Big-O means, what Little-o means, what Theta and Omega are for… Intrigued?!?

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