Empirically estimating big-oh time efficiency
In order to get started, you have to make a couple of assumptions.
nis large compared to any constant terms.- You can effectively randomize your input data
- You can sample with sufficient density to get a good handle on the distribution of runtimes
In particular, (3) is difficult to achieve in concert with (1). So you may get something with an exponential worst case, but never run into that worst case, and thus think your algorithm is much better than it is on average.
With that said, all you need is any standard curve fitting library. Apache Commons Math has a fully adequate one. You then either create a function with all the common terms that you want to test (e.g. constant, log n, n, n log n, nn, nn*n, e^n), or you take the log of your data and fit the exponent, and then if you get an exponent not close to an integer, see if throwing in a log n gives a better fit.
(In more detail, if you fit C*x^a for C and a, or more easily log C + a log x, you can get the exponent a; in the all-common-terms-at-once scheme, you'll get weights for each term, so if you have n*n + C*n*log(n) where C is large, you'll pick up that term also.)
You'll want to vary the size by enough so that you can tell the different cases apart (might be hard with log terms, if you care about those), and safely more different sizes than you have parameters (probably 3x excess would start being okay, as long as you do at least a dozen or so runs total).
Edit: Here is Scala code that does all this for you. Rather than explain each little piece, I'll leave it to you to investigate; it implements the scheme above using the C*x^a fit, and returns ((a,C),(lower bound for a, upper bound for a)). The bounds are quite conservative, as you can see from running the thing a few times. The units of C are seconds (a is unitless), but don't trust that too much as there is some looping overhead (and also some noise).
class TimeLord[A: ClassManifest,B: ClassManifest](setup: Int => A, static: Boolean = true)(run: A => B) {
@annotation.tailrec final def exceed(time: Double, size: Int, step: Int => Int = _*2, first: Int = 1): (Int,Double) = {
var i = 0
val elapsed = 1e-9 * {
if (static) {
val a = setup(size)
var b: B = null.asInstanceOf[B]
val t0 = System.nanoTime
var i = 0
while (i < first) {
b = run(a)
i += 1
}
System.nanoTime - t0
}
else {
val starts = if (static) { val a = setup(size); Array.fill(first)(a) } else Array.fill(first)(setup(size))
val answers = new Array[B](first)
val t0 = System.nanoTime
var i = 0
while (i < first) {
answers(i) = run(starts(i))
i += 1
}
System.nanoTime - t0
}
}
if (time > elapsed) {
val second = step(first)
if (second <= first) throw new IllegalArgumentException("Iteration size increase failed: %d to %d".format(first,second))
else exceed(time, size, step, second)
}
else (first, elapsed)
}
def multibench(smallest: Int, largest: Int, time: Double, n: Int, m: Int = 1) = {
if (m < 1 || n < 1 || largest < smallest || (n>1 && largest==smallest)) throw new IllegalArgumentException("Poor choice of sizes")
val frac = (largest.toDouble)/smallest
(0 until n).map(x => (smallest*math.pow(frac,x/((n-1).toDouble))).toInt).map{ i =>
val (k,dt) = exceed(time,i)
if (m==1) i -> Array(dt/k) else {
i -> ( (dt/k) +: (1 until m).map(_ => exceed(time,i,first=k)).map{ case (j,dt2) => dt2/j }.toArray )
}
}.foldLeft(Vector[(Int,Array[Double])]()){ (acc,x) =>
if (acc.length==0 || acc.last._1 != x._1) acc :+ x
else acc.dropRight(1) :+ (x._1, acc.last._2 ++ x._2)
}
}
def alpha(data: Seq[(Int,Array[Double])]) = {
// Use Theil-Sen estimator for calculation of straight-line fit for exponent
// Assume timing relationship is t(n) = A*n^alpha
val dat = data.map{ case (i,ad) => math.log(i) -> ad.map(x => math.log(i) -> math.log(x)) }
val slopes = (for {
i <- dat.indices
j <- ((i+1) until dat.length)
(pi,px) <- dat(i)._2
(qi,qx) <- dat(j)._2
} yield (qx - px)/(qi - pi)).sorted
val mbest = slopes(slopes.length/2)
val mp05 = slopes(slopes.length/20)
val mp95 = slopes(slopes.length-(1+slopes.length/20))
val intercepts = dat.flatMap{ case (i,a) => a.map{ case (li,lx) => lx - li*mbest } }.sorted
val bbest = intercepts(intercepts.length/2)
((mbest,math.exp(bbest)),(mp05,mp95))
}
}
Note that the multibench method is expected to take about sqrt(2)nm*time to run, assuming that static initialization data is used and is relatively cheap compared to whatever you're running. Here are some examples with parameters chosen to take ~15s to run:
val tl1 = new TimeLord(x => List.range(0,x))(_.sum) // Should be linear
// Try list sizes 100 to 10000, with each run taking at least 0.1s;
// use 10 different sizes and 10 repeats of each size
scala> tl1.alpha( tl1.multibench(100,10000,0.1,10,10) )
res0: ((Double, Double), (Double, Double)) = ((1.0075537890632216,7.061397125245351E-9),(0.8763463348353099,1.102663784225697))
val longList = List.range(0,100000)
val tl2 = new TimeLord(x=>x)(longList.apply) // Again, should be linear
scala> tl2.alpha( tl2.multibench(100,10000,0.1,10,10) )
res1: ((Double, Double), (Double, Double)) = ((1.4534378213477026,1.1325696181862922E-10),(0.969955396265306,1.8294175293676322))
// 1.45?! That's not linear. Maybe the short ones are cached?
scala> tl2.alpha( tl2.multibench(9000,90000,0.1,100,1) )
res2: ((Double, Double), (Double, Double)) = ((0.9973235607566956,1.9214696731124573E-9),(0.9486294398193154,1.0365312207345019))
// Let's try some sorting
val tl3 = new TimeLord(x=>Vector.fill(x)(util.Random.nextInt))(_.sorted)
scala> tl3.alpha( tl3.multibench(100,10000,0.1,10,10) )
res3: ((Double, Double), (Double, Double)) = ((1.1713142886974603,3.882658025586512E-8),(1.0521099621639414,1.3392622111121666))
// Note the log(n) term comes out as a fractional power
// (which will decrease as the sizes increase)
// Maybe sort some arrays?
// This may take longer to run because we have to recreate the (mutable) array each time
val tl4 = new TimeLord(x=>Array.fill(x)(util.Random.nextInt), false)(java.util.Arrays.sort)
scala> tl4.alpha( tl4.multibench(100,10000,0.1,10,10) )
res4: ((Double, Double), (Double, Double)) = ((1.1216172965292541,2.2206198821180513E-8),(1.0929414090177318,1.1543697719880128))
// Let's time something slow
def kube(n: Int) = (for (i <- 1 to n; j <- 1 to n; k <- 1 to n) yield 1).sum
val tl5 = new TimeLord(x=>x)(kube)
scala> tl5.alpha( tl5.multibench(10,100,0.1,10,10) )
res5: ((Double, Double), (Double, Double)) = ((2.8456382116915484,1.0433534274508799E-7),(2.6416659356198617,2.999094292838751))
// Okay, we're a little short of 3; there's constant overhead on the small sizes
Anyway, for the stated use case--where you are checking to make sure the order doesn't change--this is probably adequate, since you can play with the values a bit when setting up the test to make sure they give something sensible. One could also create heuristics that search for stability, but that's probably overkill.
(Incidentally, there is no explicit warmup step here; the robust fitting of the Theil-Sen estimator should make it unnecessary for sensibly large benchmarks. This also is why I don't use any other benching framework; any statistics that it does just loses power from this test.)
Edit again: if you replace the alpha method with the following:
// We'll need this math
@inline private[this] def sq(x: Double) = x*x
final private[this] val inv_log_of_2 = 1/math.log(2)
@inline private[this] def log2(x: Double) = math.log(x)*inv_log_of_2
import math.{log,exp,pow}
// All the info you need to calculate a y value, e.g. y = x*m+b
case class Yp(x: Double, m: Double, b: Double) {}
// Estimators for data order
// fx = transformation to apply to x-data before linear fitting
// fy = transformation to apply to y-data before linear fitting
// model = given x, slope, and intercept, calculate predicted y
case class Estimator(fx: Double => Double, invfx: Double=> Double, fy: (Double,Double) => Double, model: Yp => Double) {}
// C*n^alpha
val alpha = Estimator(log, exp, (x,y) => log(y), p => p.b*pow(p.x,p.m))
// C*log(n)*n^alpha
val logalpha = Estimator(log, exp, (x,y) =>log(y/log2(x)), p => p.b*log2(p.x)*pow(p.x,p.m))
// Use Theil-Sen estimator for calculation of straight-line fit
case class Fit(slope: Double, const: Double, bounds: (Double,Double), fracrms: Double) {}
def theilsen(data: Seq[(Int,Array[Double])], est: Estimator = alpha) = {
// Use Theil-Sen estimator for calculation of straight-line fit for exponent
// Assume timing relationship is t(n) = A*n^alpha
val dat = data.map{ case (i,ad) => ad.map(x => est.fx(i) -> est.fy(i,x)) }
val slopes = (for {
i <- dat.indices
j <- ((i+1) until dat.length)
(pi,px) <- dat(i)
(qi,qx) <- dat(j)
} yield (qx - px)/(qi - pi)).sorted
val mbest = slopes(slopes.length/2)
val mp05 = slopes(slopes.length/20)
val mp95 = slopes(slopes.length-(1+slopes.length/20))
val intercepts = dat.flatMap{ _.map{ case (li,lx) => lx - li*mbest } }.sorted
val bbest = est.invfx(intercepts(intercepts.length/2))
val fracrms = math.sqrt(data.map{ case (x,ys) => ys.map(y => sq(1 - y/est.model(Yp(x,mbest,bbest)))).sum }.sum / data.map(_._2.length).sum)
Fit(mbest, bbest, (mp05,mp95), fracrms)
}
then you can get an estimate of the exponent when there's a log term also--error estimates exist to pick whether the log term or not is the correct way to go, but it's up to you to make the call (i.e. I'm assuming you'll be supervising this initially and reading the numbers that come off):
val tl3 = new TimeLord(x=>Vector.fill(x)(util.Random.nextInt))(_.sorted)
val timings = tl3.multibench(100,10000,0.1,10,10)
// Regular n^alpha fit
scala> tl3.theilsen( timings )
res20: tl3.Fit = Fit(1.1811648421030059,3.353753446942075E-8,(1.1100382697696545,1.3204652930525234),0.05927994882343982)
// log(n)*n^alpha fit--note first value is closer to an integer
// and last value (error) is smaller
scala> tl3.theilsen( timings, tl3.logalpha )
res21: tl3.Fit = Fit(1.0369167329732445,9.211366397621766E-9,(0.9722967182484441,1.129869067913768),0.04026308919615681)
(Edit: fixed the RMS computation so it's actually the mean, plus demonstrated that you only need to do timings once and can then try both fits.)
I don't think your approach will work in general.
The problem is that "big O" complexity is based on a limit as some scaling variable tends to infinity. For smaller values of that variable, the performance behavior can appear to fit a different curve entirely.
The problem is that with an empirical approach you can never know if the scaling variable is large enough for the limit to be apparent in the results.
Another problem is that if you implement this in Java / Scala, you have to go to considerable lengths to eliminate distortions and "noise" in your timings due to things like JVM warmup (e.g. class loading, JIT compilation, heap resizing) and garbage collection.
Finally, nobody is going to place much trust in empirical estimates of complexity. Or at least, they wouldn't if they understood the mathematics of complexity analysis.
FOLLOWUP
In response to this comment:
Your estimate's significance will improve drastically the more and larger samples you use.
This is true, though my point is that you (Daniel) haven't factored this in.
Also, runtime functions typically have special characteristics which can be exploited; for example, algorithms tend to not change their behaviour at some huge n.
For simple cases, yes.
For complicated cases and real world cases, that is a dubious assumption. For example:
Suppose some algorithm uses a hash table with a large but fixed-sized primary hash array, and uses external lists to deal with collisions. For N (== number of entries) less than the size of the primary hash array, the behaviour of most operations will appear to be
O(1). The trueO(N)behaviour can only be detected by curve fitting when N gets much larger than that.Suppose that the algorithm uses a lot of memory or network bandwidth. Typically, it will work well until you hit the resource limit, and then performance will tail off badly. How do you account for this? If it is part of the "empirical complexity", how do you make sure that you get to the transition point? If you want to exclude it, how do you do that?