Test your intuition: logarithmic time

Here is a question to test your intuition for logarithms. NO CALCULATORS allowed, and don’t take a long time to work out an answer in your head! Just go with your gut. After you have committed to a choice (say, by writing it down), then go get a calculator and a pencil and see if you can work out whether you were right!

On a certain input, an O(n) algorithm runs in one-tenth of a second. On the same input, an O(n^2) algorithm takes one and a half weeks to run. Approximately how long would an O(n \log n) algorithm take on the same input?

  1. a few seconds?
  2. a few minutes?
  3. a few hours?
  4. a few days?

I’m pretty sure it wasn’t until quite a few years out of undergrad that I would have gotten this right.


About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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16 Responses to Test your intuition: logarithmic time

  1. mp says:

    Well, the correct answer is “What are the constant factors of these various algorithms?”, isn’t it?

    • Brent says:

      No, it’s not supposed to be a trick question. Let’s say they all have a constant factor of 1 (or near enough that it makes no difference).

      • mp says:

        Yeah, this was more of “the pedantically correct answer”. The factor ratios would have to be immense to really matter. The correct correct answer is far more interesting. I got it right, but before going for the calculator I was only slightly convinced that I got it right.

      • Ganesh Sittampalam says:

        Taking a logarithm also introduces a new constant factor (as you have to choose the base), but you’d have to make a pretty strange choice for it to make a difference to the answer.

  2. mdibaiee says:

    Is this the correct solution?

    n = 1.5week / 0.1s = 9072000
    time = n * log(n) = 0.1 * log (9072000) = 2.3112989209604318

    • Brent says:

      Yep, that’s right! Did you have the right intuition?

      • mdibaiee says:

        Yep! I find your blog really interesting and insightful, thanks a lot!

      • phyl says:

        Can you explain the different values of n here? I realize this is an intuition, just trying to follow. Thank you in advance.

        • Brent says:

          Well, if Cn^2 = 1.5 weeks and Cn = 0.1s (assuming they have the same constant factor), we can divide Cn^2/Cn = 1.5w/0.1s to find n. Does that help?

          • Caleb says:

            I think the confusion was about “time = n * log(n) = 0.1 * log (9072000)”, where one n turns into 0.1 and the other turns into 9072000. It would be better expressed as “time = C * n * log(n) “.

  3. Ultra Bongo says:

    Hmm I got this right but it felt plausible to me that the answer *could* have been minutes for a big enough n, but after doing the numbers it’s would have to be a hell of a large n, so that intuition was not quite right.

    (still giving myself some points for the right answer, though!)

  4. sha says:

    I recently had to work with an algorithm that runs in O(n log log n). The professor explained, “log log n grows very slowly. I just regard it as 5.”

  5. martin says:

    It’s a trick question.

    The big O notation puts an upper bound on the rate of growth of a function: f = O(g) if, for all values of input greater than an arbitrary point in the domain and a constant scaling factor C, f(x) < Cg(x).

    If all you know about an algorithm is that it is O(n log(n)), you would need a time of one input to know the maximum time of another input. The running times of unrelated algorithms tells you nothing.

    Consider the following functions:

    f( 2 ) takes 2 seconds
    f( 4 ) takes 8 seconds
    f( 8 ) takes 24 seconds

    f'( 2 ) takes 2 minutes
    f'( 4 ) takes 8 minutes
    f'( 8 ) takes 24 minutes

    f''( 2 ) takes 2 hours
    f''( 4 ) takes 8 hours
    f''( 8 ) takes 24 hours

    f'''( 2 ) takes 2 days
    f'''( 4 ) takes 8 days
    f'''( 8 ) takes 24 days

    f,f',f'',f''' are all O( n log(n) ), and can take seconds, minutes, hours, or days to complete for the same input.

    • Brent says:

      Yes, you’re absolutely right in theory, but it was not intended to be a trick question. Practically speaking, for “real-world” algorithms, the constant factors are similar enough that it does actually make sense to compare different algorithms like this.

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