Why is the constant always dropped from big O analysis?

Question

I'm trying to understand a particular aspect of Big O analysis in the context of running programs on a PC.

Suppose I have an algorithm that has a performance of O(n + 2). Here if n gets really large the 2 becomes insignificant. In this case it's perfectly clear the real performance is O(n).

However, say another algorithm has an average performance of O(n² / 2). The book where I saw this example says the real performance is O(n²). I'm not sure I get why, I mean the 2 in this case seems not completely insignificant. So I was looking for a nice clear explanation from the book. The book explains it this way:

"Consider though what the 1/2 means. The actual time to check each value is highly dependent on the machine instruction that the code translates to and then on the speed at which the CPU can execute the instructions. Therefore the 1/2 doesn't mean very much."

And my reaction is... huh? I literally have no clue what that says or more precisely what that statement has to do with their conclusion. Can somebody spell it out for me please.

Thanks for any help.

Answer 1

There's a distinction between "are these constants meaningful or relevant?" and "does big-O notation care about them?" The answer to that second question is "no," while the answer to that first question is "absolutely!"

Big-O notation doesn't care about constants because big-O notation only describes the long-term growth rate of functions, rather than their absolute magnitudes. Multiplying a function by a constant only influences its growth rate by a constant amount, so linear functions still grow linearly, logarithmic functions still grow logarithmically, exponential functions still grow exponentially, etc. Since these categories aren't affected by constants, it doesn't matter that we drop the constants.

That said, those constants are absolutely significant! A function whose runtime is 10¹⁰⁰n will be way slower than a function whose runtime is just n. A function whose runtime is n² / 2 will be faster than a function whose runtime is just n². The fact that the first two functions are both O(n) and the second two are O(n²) doesn't change the fact that they don't run in the same amount of time, since that's not what big-O notation is designed for. O notation is good for determining whether in the long term one function will be bigger than another. Even though 10¹⁰⁰n is a colossally huge value for any n > 0, that function is O(n) and so for large enough n eventually it will beat the function whose runtime is n² / 2 because that function is O(n²).

In summary - since big-O only talks about relative classes of growth rates, it ignores the constant factor. However, those constants are absolutely significant; they just aren't relevant to an asymptotic analysis.

Answer 2

Big O notation is most commonly used to describe an algorithm's running time. In this context, I would argue that specific constant values are essentially meaningless. Imagine the following conversation:

Alice: What is the running time of your algorithm?

Bob: 7n²

Alice: What do you mean by 7n²?

• What are the units? Microseconds? Milliseconds? Nanoseconds?
• What CPU are you running it on? Intel i9-9900K? Qualcomm Snapdragon 845? (Or are you using a GPU, an FPGA, or other hardware?)
• What type of RAM are you using?
• What programming language did you implement the algorithm in? What is the source code?
• What compiler / VM are you using? What flags are you passing to the compiler / VM?
• What is the operating system?
• etc.

So as you can see, any attempt to indicate a specific constant value is inherently problematic. But once we set aside constant factors, we are able to clearly describe an algorithm's running time. Big O notation gives us a robust and useful description of how long an algorithm takes, while abstracting away from the technical features of its implementation and execution.

Now it is possible to specify the constant factor when describing the number of operations (suitably defined) or CPU instructions an algorithm executes, the number of comparisons a sorting algorithm performs, and so forth. But typically, what we're really interested in is the running time.

None of this is meant to suggest that the real-world performance characteristics of an algorithm are unimportant. For example, if you need an algorithm for matrix multiplication, the Coppersmith-Winograd algorithm is inadvisable. It's true that this algorithm takes O(n^2.376) time, whereas the Strassen algorithm, its strongest competitor, takes O(n^2.808) time. However, according to Wikipedia, Coppersmith-Winograd is slow in practice, and "it only provides an advantage for matrices so large that they cannot be processed by modern hardware." This is usually explained by saying that the constant factor for Coppersmith-Winograd is very large. But to reiterate, if we're talking about the running time of Coppersmith-Winograd, it doesn't make sense to give a specific number for the constant factor.

Despite its limitations, big O notation is a pretty good measure of running time. And in many cases, it tells us which algorithms are fastest for sufficiently large input sizes, before we even write a single line of code.

Answer 3

Big-O notation only describes the growth rate of algorithms in terms of mathematical function, rather than the actual running time of algorithms on some machine.

Mathematically, Let f(x) and g(x) be positive for x sufficiently large. We say that f(x) and g(x) grow at the same rate as x tends to infinity, if

now let f(x)=x^2 and g(x)=x^2/2, then lim(x->infinity)f(x)/g(x)=2. so x^2 and x^2/2 both have same growth rate.so we can say O(x^2/2)=O(x^2).

As templatetypedef said, hidden constants in asymptotic notations are absolutely significant.As an example :marge sort runs in O(nlogn) worst-case time and insertion sort runs in O(n^2) worst case time.But as the hidden constant factors in insertion sort is smaller than that of marge sort, in practice insertion sort can be faster than marge sort for small problem sizes on many machines.

Answer 4

You are completely right that constants matter. In comparing many different algorithms for the same problem, the O numbers without constants give you an overview of how they compare to each other. If you then have two algorithms in the same O class, you would compare them using the constants involved.

But even for different O classes the constants are important. For instance, for multidigit or big integer multiplication, the naive algorithm is O(n^2), Karatsuba is O(n^log_2(3)), Toom-Cook O(n^log_3(5)) and Schönhage-Strassen O(n*log(n)*log(log(n))). However, each of the faster algorithms has an increasingly large overhead reflected in large constants. So to get approximate cross-over points, one needs valid estimates of those constants. Thus one gets, as SWAG, that up to n=16 the naive multiplication is fastest, up to n=50 Karatsuba and the cross-over from Toom-Cook to Schönhage-Strassen happens for n=200.

In reality, the cross-over points not only depend on the constants, but also on processor-caching and other hardware-related issues.

Answer 5

Big O without constant is enough for algorithm analysis.

First, the actual time does not only depend how many instructions but also the time for each instruction, which is closely connected to the platform where the code runs. It is more than theory analysis. So the constant is not necessary for most case.

Second, Big O is mainly used to measure how the run time will increase as the problem becomes larger or how the run time decrease as the performance of hardware improved.

Third, for situations of high performance optimizing, constant will also be taken into consideration.

Answer 6

The time required to do a particular task in computers now a days does not required a large amount of time unless the value entered is very large.

Suppose we wants to multiply 2 matrices of size 10*10 we will not have problem unless we wants to do this operation multiple times and then the role of asymptotic notations becomes prevalent and when the value of n becomes very big then the constants don't really makes any difference to the answer and are almost negligible so we tend to leave them while calculating the complexity.

Answer 7

Time complexity for O(n+n) reduces to O(2n). Now 2 is a constant. So the time complexity will essentially depend on n.
Hence the time complexity of O(2n) equates to O(n).
Also if there is something like this O(2n + 3) it will still be O(n) as essentially the time will depend on the size of n.
Now suppose there is a code which is O(n^2 + n), it will be O(n^2) as when the value of n increases the effect of n will become less significant compared to effect of n^2.
Eg:

n = 2  => 4 + 2 = 6
n = 100 => 10000 + 100 => 10100
n = 10000 => 100000000 + 10000 => 100010000 

As you can see the effect of the second expression as lesser effect as the value of n keeps increasing. Hence the time complexity evaluates to O(n^2).