## Big-oh is not a total order

Wed 31 Dec 1969 by mskala Tags used: compsciAnyone who's studied computer science should be familiar with the Landau asymptotic notation, or "big-oh" notation. But how well do you really know it?

Given two functions it seems intuitive that one at least one ought to be
big-oh of the other; in some cases maybe they are both big-oh of each other,
in which case they are theta of each other and in a certain sense can be
said to be equivalent. It sure feels like a total order, with a place for
everything and everything in its place, and it is a total order on all the
kinds of functions we normally care about. I've certainly marked down
plenty of students for incorrectly claiming that
*n* log *n* can't be compared either way to things like
*n* and *n*^{2}.

But today I had occasion to wonder if it's really true that *O*() induces a total order on all functions for which it is applicable. I looked on the Net and couldn't find anyone discussing this issue; I did find this archived final exam in which someone asked it of students as a true/false question and apparently was looking for the answer "true." Unfortunately, the correct answer is "false."

Counterexample below the cut. If you've got a computer science background you might want to think about it a bit yourself before going further. It shouldn't really be hard if you look carefully at the definition.

*f*(*n*) = 1+*n*+*n*(-1)^{n}

*g*(*n*) = 1+*n*-*n*(-1)^{n}

*f*(*n*) ≠ *O*(*g*(*n*))

*g*(*n*) ≠ *O*(*f*(*n*))

These two functions alternate between being equal to 1 and being equal to 2*n*+1. The function *f* is 1 for odd *n* and 2*n*+1 for even *n*, and the function *g* is the reverse. The gap between them grows arbitrarily large as *n* grows, and its direction alternates. No matter what nonnegative constants you multiply the functions by, each will exceed the other half the time when *n* gets large enough.

Do note that each of my functions is defined for all positive integer inputs, and for positive integer inputs the function values are always positive integers. I'm not cheating with weird functions here. If you wanted them to have positive real values for all positive real inputs then you might have to play with the definitions a bit, because my definitions above will end up producing imaginary numbers for some non-integer values of *n*, but it's easy to see how you could use a scaled sine wave to interpolate the definitions above with something that would be everywhere continuous, positive, and real. And many people's definitions of *O*() only care about positive integers anyway.

This issue isn't directly discussed by Cormen, Leiserson, Rivest, and Stein (*Introduction to Algorithms*, Third Edition; it's a standard text for this kind of thing and the one I chose for a course I recently taught.) Their Problem 3-5 seems to be aimed at the question of total ordering, however, because it talks about an alternative definition for big-omega that helps force functions into a total order. I also checked Knuth (*The Art of Computer Programming*, volume 1) and was a little surprised to find that he doesn't seem to discuss the question of whether big-oh is a total order at all.

So there you have it. I'm not sure it's really all that big a deal - anyone with a reason to care strongly about whether big-oh is a total order should be able to figure it out for themselves easily enough once they think about it - but I, at least, had never had a reason to think about it seriously until today. My students are probably lucky I didn't, because I might have put it on the COMP 3170 final.

The question I was actually interested in before I got onto this one was the slightly more complicated question of *how many* distinct theta-classes of functions exist. It's easy to see that there are as many as there are real numbers, but when we start throwing in things like logs, can we make them more numerous than that? The answer seems to be that we can construct a distinct class for every function we allow as input, but don't cite me on that because I haven't finished formalizing the proof. It might make a good final exam question for a course a bit more advanced than 3170.

15 comments

It's true that we ordinarily apply O() to functions that are asymptotically monotonic - either always increasing or always decreasing once n gets large enough - and I suspect that that may force it to be a total order. So far I haven't found a counterexample using only monotonic functions. But the most common official definition does not require monotonic functions.

**Matt**- 2011-07-21 16:20

**Matt**- 2011-07-21 16:24

F(n) = sqrt(n)*product_{1 to n} f(i)

G(n) = product_{1 to n} g(i)

I haven't checked the exact formula there, but the point is that the ratio F(n)/G(n) should alternate between O(sqrt(n)) and O(1/sqrt(n)).

If I am reading the final exam you linked correctly, it doesn't assert that O() induces a total ordering, it asserts that it induces an

*equivalence relation*. Your f and g don't constitute a counterexample because they are incomparable.

**T. Rev**- 2011-07-21 16:55

As for the exam, I'm looking at question 5.(f) and I think you're looking at question 6.(a).

**Matt**- 2011-07-21 17:05

**T. Rev**- 2011-07-21 19:57

Then although I'm not certain of the details, I would think it should be possible to make a continuous version of your "product" trick above: something like exp integral_0^x log f(t) dt. The log might not be necessary or even desirable, and multiplying or adding constants just inside the exp might also be necessary to sync f() with g(). The point would be that like a plain old exponential, this mess would have all its derivatives monotonically increasing, but it still alternates relatively fast and slow parts and we could arrange two of them with those alternations out of phase.

**Matt**- 2011-07-21 21:30

**kiwano**- 2011-07-22 09:20

**Matt**- 2011-07-22 11:18

ln(F(n)) = sum_{1 to n} ln(f(i))

ln(G(n)) = sum_{1 to n} ln(g(i))

We can pass to the continuous case by using the continous (and wiggly) sin-based versions of f and g:

ln(F(x)) = integral_1^x ln(f(i)) di

ln(G(x)) = integral_1^x ln(g(i)) di

Taking derivatives we get F'(x)/F(x) = f(x) and G'(x)/G(x) = g(x), from which it is easy to show that not all derivatives of F and G are monotonic.

On the other hand, we know that O() does give a total ordering on the usual suspect functions--polynomials, e^x, ln(x), x^c--and they all share the property that all their derivatives are monotonic. So it seems like 'all derivatives monotonic' might be a good starting criterion for further investigation.

**T. Rev**- 2011-07-23 01:44

*increasing*isn't the right criterion, since e.g. the second derivative of ln(x) is (monotonic) decreasing!

**T. Rev**- 2011-07-23 01:46

If f(x) != O(g), that means for all constants M and real x_0, there exists x > x_0 such that f(x)/g(x) > M, and likewise if g(x) != O(f) we have a similar condition on g(x)/f(x). So both ratios f/g and g/f get arbitarily large.

Let's define L(x) = ln(f(x)) - ln(g(x)). From the above we can say that if f and g are incomparable, then for any constant M and real x_0, there exist x and x' > x_0 such that L(x) > M and L(x') < M. Less formally, it means as x -> infinity, L(x) is always oscillating between positive and negative values of arbitrarily large magnitude. From that, we can conclude that if f and g are incomparable, at least one of them is 'wiggly'.

**T. Rev**- 2011-07-23 03:03

I'm not too interested in requiring continuity; the functions we put inside O() in computer science are generally functions of integers anyway. I'd be more interested in finding as large a function family as possible for which it's a total order. Pretty often we use polynomials, which clearly work (cannot be incomparable). Pretty often we use exponentials c^n where c is a *positive real* constant. Those seem to work as well; but allowing negative real bases like c=-1, or complex numbers (which would be how you'd get to sin and cos), clearly doesn't work. Also, we often use logs, including iterated logs. Something like (sqrt(n) log n)/log log n might be typically seen.

If we take the closure of add, subtract, exp, log, and function composition, but only allowing real values (the function just becomes undefined at that point if you try to take the log of a negative number, but such functions would still be allowed if they're defined for all sufficiently large n) is that family of functions totally ordered by O()? I think it might be, and it's big enough to cover pretty much all the things I normally expect to see inside O() in computer science.

**Matt**- 2011-07-23 09:22

As far as restating the definitions, try looking at it this way:

A pair of functions f and g are incomparable, if and only if f/g is incomparable to a constant, if and only if f/g has the 'wiggly' property of attaining arbitrarily large and small values as x -> infinity. This follows trivially from the definition, but it suggests that looking at pairs of functions is unnecessarily complicated, and continuity is also irrelevant: it's enough to consider the functions that are incomparable to a constant, and those functions are easy to characterize.

**T. Rev**- 2011-07-23 15:07

I think the intuition built into big-O notation assumes that computational complexity isn't simply positive, but increasing.

T. Rev- 2011-07-21 15:09