比较O（2 / n）和O（1）的时间复杂度

Question

If there are 2 functions with time complexity of O(2/n) and O(100). which function has lesser execution time ?. Is there any real function with time complexity of 2/n ?. (found this in some algorithm question paper)

Answer 1

Firstly, in the O notation (as well as Theta and Omega), you can dismiss any constants, because the definition already includes the part "for some constant k ".

So, basically, O(100) is equivalent to O(1), while O(2/n) is equivalent to O(1/n). Which has faster execution time -- depends on the n . If I presume that 100 and 2/n are directly used to calculate execution time, than the execution time is:

• 100 (units of time) for O(100) in all cases
• more than 100 (units of time) for O(2/n) for n < 0.02
• less than 100 (units of time) for O(2/n) for n >= 0.02

Now, I hope this question is purely theoretical, because in reality there is no algorithm with O(1/n) complexity -- it would mean that it takes less time (and that's note time per amount of data, that's just time) the more data it needs to process. I hope this is clear, that there is no algorithm that could take 0 time for infinite amount of data.

The other complexity, O(100) is an algorithm that takes the same steps no matter what the input data is, and thus always has constant time of execution (actually, it only has to be bounded by a constant, it can run faster that that sometimes). An example would be a program that reads an input from a file of integers, and then returns the sum of the first 100 numbers in that file or of all the numbers if there isn't a 100 numbers present. Since it always reads at most a 100 numbers (the rest can be ignored) and sums them, it is bounded by a constant number of steps.

Answer 2

Surely it depends on the value of N. The O(100) is basically fixed time, and may run faster or slower than the O(2/N) depending on what n is.

I can't think of an O(2/N) algorithm off hand, something that gets faster with more data... sounds a bit weird.

Answer 3

An O(2/n) algorithm that does anything is practically impossible.

An algorithm is a finite sequence of steps that produces a result. Since a "step" on a computer takes a certain amount of time (for example at least one CPU cycle), the only way an algorithm can have O(2/n) time is if it takes zero time for sufficiently large n . Hence it does nothing.

Leaving aside algorithms and time complexity: an O(2/n) function is "less than" constant, in the sense that a O(2/n) function necessarily tends to 0 as n tends to infinity, whereas an O(1) function doesn't necessarily do that.

A remark on the text of the question from this paper: any function that is O(100) is also O(1) , and any function that is O(2/n) is also O(1/n) . It doesn't really make much sense to write big-O with unnecessary constants in there, but since it's an examination perhaps it's there to potentially confuse you.

Answer 4

I am not familiar with any algorithm that has O(2/n) running time and I doubt one can exist, but let's look at the mathematical qustion.

The mathematical question should be: (1) is O(2/n) a subset of O(1) ¹ (2) Is O(1) a subset of O(2/n) ?

1. yes. Let f(n) be a function in O(2/n) - that means that there are constants c,N such that for each n > N : c*f(n) < 2/n . There are also constants c2,N2 such that c*2/n < 1 for each n > N2 , and thus min{c1,c2} * f(n) < 1 for each n > max{N1,N2} . so O(2/n) is subset of O(1)

2. No. Since lim(2/n) = 0 at infinity, for each c,N , there is n>N such that 2/n < 1*c and thus f(n) = 2/n is not in O(1) , while it is in O(2/n)

Conclusion: a function that is O(2/n) is also O(1) - but not the other way around.
It means - each function in O(2/n) scales "smaller" then 1.

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(1) It is identical to O(100) , since O(1) = O(100)