Big Oh Analysis simplification - big-o

I am trying to get a better understanding of big oh algorithm analysis.
Is there a way to simplify
5n⋅(log(n))³.
I am thinking it simplifies to:
n⋅(log(n))³

That's correct. The basic idea here is to remove constant terms where they have no effect of the rate of growth.
In this case, you have:
(5) x (n) x (log(n)^3)
and, since the 5 would simply change 1,2,3,4,5 into 5,10,15,20,25, it has no effect on the growth rate.
So it can be removed.

Related

Algorithm for finding Time Complexity of Algorithm [duplicate]

I wonder whether there is any automatic way of determining (at least roughly) the Big-O time complexity of a given function?
If I graphed an O(n) function vs. an O(n lg n) function I think I would be able to visually ascertain which is which; I'm thinking there must be some heuristic solution which enables this to be done automatically.
Any ideas?
Edit: I am happy to find a semi-automated solution, just wondering whether there is some way of avoiding doing a fully manual analysis.
It sounds like what you are asking for is an extention of the Halting Problem. I do not believe that such a thing is possible, even in theory.
Just answering the question "Will this line of code ever run?" would be very difficult if not impossible to do in the general case.
Edited to add:
Although the general case is intractable, see here for a partial solution: http://research.microsoft.com/apps/pubs/default.aspx?id=104919
Also, some have stated that doing the analysis by hand is the only option, but I don't believe that is really the correct way of looking at it. An intractable problem is still intractable even when a human being is added to the system/machine. Upon further reflection, I suppose that a 99% solution may be doable, and might even work as well as or better than a human.
You can run the algorithm over various size data sets, and you could then use curve fitting to come up with an approximation. (Just looking at the curve you create probably will be enough in most cases, but any statistical package has curve fitting).
Note that some algorithms exhibit one shape with small data sets, but another with large... and the definition of large remains a bit nebulous. This means that an algorithm with a good performance curve could have so much real world overhead that (for small data sets) it doesn't work as well as the theoretically better algorithm.
As far as code inspection techniques, none exist. But instrumenting your code to run at various lengths and outputting a simple file (RunSize RunLength would be enough) should be easy. Generating proper test data could be more complex (some algorithms work better/worse with partially ordered data, so you would want to generate data that represented your normal use-case).
Because of the problems with the definition of "what is large" and the fact that performance is data dependent, I find that static analysis often is misleading. When optimizing performance and selecting between two algorithms, the real world "rubber hits the road" test is the only final arbitrator I trust.
A short answer is that it's impossible because constants matter.
For instance, I might write a function that runs in O((n^3/k) + n^2). This simplifies to O(n^3) because as n approaches infinity, the n^3 term will dominate the function, irrespective of the constant k.
However, if k is very large in the above example function, the function will appear to run in almost exactly n^2 until some crossover point, at which the n^3 term will begin to dominate. Because the constant k will be unknown to any profiling tool, it will be impossible to know just how large a dataset to test the target function with. If k can be arbitrarily large, you cannot craft test data to determine the big-oh running time.
I am surprised to see so many attempts to claim that one can "measure" complexity by a stopwatch. Several people have given the right answer, but I think that there is still room to drive the essential point home.
Algorithm complexity is not a "programming" question; it is a "computer science" question. Answering the question requires analyzing the code from the perspective of a mathematician, such that computing the Big-O complexity is practically a form of mathematical proof. It requires a very strong understanding of the fundamental computer operations, algebra, perhaps calculus (limits), and logic. No amount of "testing" can be substituted for that process.
The Halting Problem applies, so the complexity of an algorithm is fundamentally undecidable by a machine.
The limits of automated tools applies, so it might be possible to write a program to help, but it would only be able to help about as much as a calculator helps with one's physics homework, or as much as a refactoring browser helps with reorganizing a code base.
For anyone seriously considering writing such a tool, I suggest the following exercise. Pick a reasonably simple algorithm, such as your favorite sort, as your subject algorithm. Get a solid reference (book, web-based tutorial) to lead you through the process of calculating the algorithm complexity and ultimately the "Big-O". Document your steps and results as you go through the process with your subject algorithm. Perform the steps and document your progress for several scenarios, such as best-case, worst-case, and average-case. Once you are done, review your documentation and ask yourself what it would take to write a program (tool) to do it for you. Can it be done? How much would actually be automated, and how much would still be manual?
Best wishes.
I am curious as to why it is that you want to be able to do this. In my experience when someone says: "I want to ascertain the runtime complexity of this algorithm" they are not asking what they think they are asking. What you are most likely asking is what is the realistic performance of such an algorithm for likely data. Calculating the Big-O of a function is of reasonable utility, but there are so many aspects that can change the "real runtime performance" of an algorithm in real use that nothing beats instrumentation and testing.
For example, the following algorithms have the same exact Big-O (wacky pseudocode):
example a:
huge_two_dimensional_array foo
for i = 0, i < foo[i].length, i++
for j = 0; j < foo[j].length, j++
do_something_with foo[i][j]
example b:
huge_two_dimensional_array foo
for j = 0, j < foo[j].length, j++
for i = 0; i < foo[i].length, i++
do_something_with foo[i][j]
Again, exactly the same big-O... but one of them uses row ordinality and one of them uses column ordinality. It turns out that due to locality of reference and cache coherency you might have two completely different actual runtimes, especially depending on the actual size of the array foo. This doesn't even begin to touch the actual performance characteristics of how the algorithm behaves if it's part of a piece of software that has some concurrency built in.
Not to be a negative nelly but big-O is a tool with a narrow scope. It is of great use if you are deep inside algorithmic analysis or if you are trying to prove something about an algorithm, but if you are doing commercial software development the proof is in the pudding, and you are going to want to have actual performance numbers to make intelligent decisions.
Cheers!
This could work for simple algorithms, but what about O(n^2 lg n), or O(n lg^2 n)?
You could get fooled visually very easily.
And if its a really bad algorithm, maybe it wouldn't return even on n=10.
Proof that this is undecidable:
Suppose that we had some algorithm HALTS_IN_FN(Program, function) which determined whether a program halted in O(f(n)) for all n, for some function f.
Let P be the following program:
if(HALTS_IN_FN(P,f(n)))
{
while(1);
}
halt;
Since the function and the program are fixed, HALTS_IN_FN on this input is constant time. If HALTS_IN_FN returns true, the program runs forever and of course does not halt in O(f(n)) for any f(n). If HALTS_IN_FN returns false, the program halts in O(1) time.
Thus, we have a paradox, a contradiction, and so the program is undecidable.
A lot of people have commented that this is an inherently unsolvable problem in theory. Fair enough, but beyond that, even solving it for any but the most trivial cases would seem to be incredibly difficult.
Say you have a program that has a set of nested loops, each based on the number of items in an array. O(n^2). But what if the inner loop is only run in a very specific set of circumstances? Say, on average, it's run in aprox log(n) cases. Suddenly our "obviously" O(n^2) algorithm is really O(n log n). Writing a program that could determine if the inner loop would be run, and how often, is potentially more difficult than the original problem.
Remember O(N) isn't god; high constants can and will change the playing field. Quicksort algorithms are O(n log n) of course, but when the recursion gets small enough, say down to 20 items or so, many implementations of quicksort will change tactics to a separate algorithm as it's actually quicker to do a different type of sort, say insertion sort with worse O(N), but much smaller constant.
So, understand your data, make educated guesses, and test.
I think it's pretty much impossible to do this automatically. Remember that O(g(n)) is the worst-case upper bound and many functions perform better than that for a lot of data sets. You'd have to find the worst-case data set for each one in order to compare them. That's a difficult task on its own for many algorithms.
You must also take care when running such benchmarks. Some algorithms will have a behavior heavily dependent on the input type.
Take Quicksort for example. It is a worst-case O(n²), but usually O(nlogn). For two inputs of the same size.
The traveling salesman is (I think, not sure) O(n²) (EDIT: the correct value is 0(n!) for the brute force algotithm) , but most algorithms get rather good approximated solutions much faster.
This means that the the benchmarking structure has to most of the time be adapted on an ad hoc basis. Imagine writing something generic for the two examples mentioned. It would be very complex, probably unusable, and likely will be giving incorrect results anyway.
Jeffrey L Whitledge is correct. A simple reduction from the halting problem proves that this is undecidable...
ALSO, if I could write this program, I'd use it to solve P vs NP, and have $1million... B-)
I'm using a big_O library (link here) that fits the change in execution time against independent variable n to infer the order of growth class O().
The package automatically suggests the best fitting class by measuring the residual from collected data against each class growth behavior.
Check the code in this answer.
Example of output,
Measuring .columns[::-1] complexity against rapid increase in # rows
--------------------------------------------------------------------------------
Big O() fits: Cubic: time = -0.017 + 0.00067*n^3
--------------------------------------------------------------------------------
Constant: time = 0.032 (res: 0.021)
Linear: time = -0.051 + 0.024*n (res: 0.011)
Quadratic: time = -0.026 + 0.0038*n^2 (res: 0.0077)
Cubic: time = -0.017 + 0.00067*n^3 (res: 0.0052)
Polynomial: time = -6.3 * x^1.5 (res: 6)
Logarithmic: time = -0.026 + 0.053*log(n) (res: 0.015)
Linearithmic: time = -0.024 + 0.012*n*log(n) (res: 0.0094)
Exponential: time = -7 * 0.66^n (res: 3.6)
--------------------------------------------------------------------------------
I guess this isn't possible in a fully automatic way since the type and structure of the input differs a lot between functions.
Well, since you can't prove whether or not a function even halts, I think you're asking a little much.
Otherwise #Godeke has it.
I don't know what's your objective in doing this, but we had a similar problem in a course I was teaching. The students were required to implement something that works at a certain complexity.
In order not to go over their solution manually, and read their code, we used the method #Godeke suggested. The objective was to find students who used linked list instead of a balansed search tree, or students who implemented bubble sort instead of heap sort (i.e. implementations that do not work in the required complexity - but without actually reading their code).
Surprisingly, the results did not reveal students who cheated. That might be because our students are honest and want to learn (or just knew that we'll check this ;-) ). It is possible to miss cheating students if the inputs are small, or if the input itself is ordered or such. It is also possible to be wrong about students who did not cheat, but have large constant values.
But in spite of the possible errors, it is well worth it, since it saves a lot of checking time.
As others have said, this is theoretically impossible. But in practice, you can make an educated guess as to whether a function is O(n) or O(n^2), as long as you don't mind being wrong sometimes.
First time the algorithm, running it on input of various n. Plot the points on a log-log graph. Draw the best-fit line through the points. If the line fits all the points well, then the data suggests that the algorithm is O(n^k), where k is the slope of the line.
I am not a statistician. You should take all this with a grain of salt. But I have actually done this in the context of automated testing for performance regressions. The patch here contains some JS code for it.
If you have lots of homogenious computational resources, I'd time them against several samples and do linear regression, then simply take the highest term.
It's easy to get an indication (e.g. "is the function linear? sub-linear? polynomial? exponential")
It's hard to find the exact complexity.
For example, here's a Python solution: you supply the function, and a function that creates parameters of size N for it. You get back a list of (n,time) values to plot, or to perform regression analysis. It times it once for speed, to get a really good indication it would have to time it many times to minimize interference from environmental factors (e.g. with the timeit module).
import time
def measure_run_time(func, args):
start = time.time()
func(*args)
return time.time() - start
def plot_times(func, generate_args, plot_sequence):
return [
(n, measure_run_time(func, generate_args(n+1)))
for n in plot_sequence
]
And to use it to time bubble sort:
def bubble_sort(l):
for i in xrange(len(l)-1):
for j in xrange(len(l)-1-i):
if l[i+1] < l[i]:
l[i],l[i+1] = l[i+1],l[i]
import random
def gen_args_for_sort(list_length):
result = range(list_length) # list of 0..N-1
random.shuffle(result) # randomize order
# should return a tuple of arguments
return (result,)
# timing for N = 1000, 2000, ..., 5000
times = plot_times(bubble_sort, gen_args_for_sort, xrange(1000,6000,1000))
import pprint
pprint.pprint(times)
This printed on my machine:
[(1000, 0.078000068664550781),
(2000, 0.34400010108947754),
(3000, 0.7649998664855957),
(4000, 1.3440001010894775),
(5000, 2.1410000324249268)]

Issue with time-analysis

I'm teaching myself data structures, and am on a section giving a brief outline of time analysis. The following problem was given:
"Each of the following are formulas for the number of operations in some
algorithm. Express each formula in big-O notation."
The problem then goes on to give multiple scenarios. One was:
g.) The number of times that n can be divided by 10 before dropping below
1.0.
(Note: It doesn't state what n is exactly, so I'm assuming it's just some input size. But I don't think it matters in terms of how the problem is stated)
I reasoned that as this would relate to its order of magnitude, it should just be log n. However, the text says that it should be quadratic. Is there something I am missing?
Any help to help my thinking would be greatly appreciated.

Show whether f is Big Oh of g with logarithmic equations

Hello, how would you do whether f is Big Oh of g. I know how to do this for simple exponential functions but when it comes to logs, I have no idea what to do. Can anyone help?
You should use the properties of the logarithm. For example, for your second case, you can use the fact that:
So in big O notation:
Regarding the first case, my advice is to read this document, specially the section 4, labeled "Growth Rate of Standard Functions", where you will find a bullet titled "Any polylog grows slower than any polynomial".

Understanding Big-O Notation [duplicate]

This question already has answers here:
What is a plain English explanation of "Big O" notation?
(43 answers)
Closed 8 years ago.
I am reading a book "Beginning Algorithms" by Simon Harris and James Ross.
In the early pages, there is a section on Understanding Big-O Notation. I read this section, and re-read this section maybe about a dozen times. I still am not able to wrap my head around a couple of things. I appreciate any help to get rid of my confusion.
The author / authors state "The precise number of operations is not actually that important. The complexity of an algorithm is usually defined in terms of the order of magnitude of the number of operations required to perform a function, denoted by a capital O for order of followed by an expression representing some growth relative to the size of the problem denoted by the letter N."
This really made my head hurt, and unfortunately, everything else following this paragraph made no sense to me because this paragraph is supposed to lay the foundation for the next reading.
The book does not define "order of magnitude." I googled this and the results just told me that order of magnitude are defined in powers of 10. But what does that even mean? Do you take the number of operations and define that number in a power of 10, and that equals the complexity? Also, what is considered the "size of the problem?" Is the size of the problem the number of operations? Or is the size of the problem the "order of magnitude of the number of operations required to perform a function."
Any practical examples and a proper explanation of this would really help.
Thanks!
Keep it simple!
Just think of the Big-O as a way to express the performance of the algorythm. That performance will depend on the number of elements the algorythm is handeling = n.
An example, when you have to make a sum. You need one statement for the first addition, one statement for the second addition, and so on... So the performance will be linear with the number of elements = O(n).
Imagine a sort algorythm, which is very smart, and for each element of handles it automatically shortens the sort for the next element. This will be logarithmic with the number of elements = O(log(n)).
Or a complex formula with parameters, and with each extra parameter, the execution time multiplies. This will be exponential = O(10^n).

Programmatically obtaining Big-O efficiency of code

I wonder whether there is any automatic way of determining (at least roughly) the Big-O time complexity of a given function?
If I graphed an O(n) function vs. an O(n lg n) function I think I would be able to visually ascertain which is which; I'm thinking there must be some heuristic solution which enables this to be done automatically.
Any ideas?
Edit: I am happy to find a semi-automated solution, just wondering whether there is some way of avoiding doing a fully manual analysis.
It sounds like what you are asking for is an extention of the Halting Problem. I do not believe that such a thing is possible, even in theory.
Just answering the question "Will this line of code ever run?" would be very difficult if not impossible to do in the general case.
Edited to add:
Although the general case is intractable, see here for a partial solution: http://research.microsoft.com/apps/pubs/default.aspx?id=104919
Also, some have stated that doing the analysis by hand is the only option, but I don't believe that is really the correct way of looking at it. An intractable problem is still intractable even when a human being is added to the system/machine. Upon further reflection, I suppose that a 99% solution may be doable, and might even work as well as or better than a human.
You can run the algorithm over various size data sets, and you could then use curve fitting to come up with an approximation. (Just looking at the curve you create probably will be enough in most cases, but any statistical package has curve fitting).
Note that some algorithms exhibit one shape with small data sets, but another with large... and the definition of large remains a bit nebulous. This means that an algorithm with a good performance curve could have so much real world overhead that (for small data sets) it doesn't work as well as the theoretically better algorithm.
As far as code inspection techniques, none exist. But instrumenting your code to run at various lengths and outputting a simple file (RunSize RunLength would be enough) should be easy. Generating proper test data could be more complex (some algorithms work better/worse with partially ordered data, so you would want to generate data that represented your normal use-case).
Because of the problems with the definition of "what is large" and the fact that performance is data dependent, I find that static analysis often is misleading. When optimizing performance and selecting between two algorithms, the real world "rubber hits the road" test is the only final arbitrator I trust.
A short answer is that it's impossible because constants matter.
For instance, I might write a function that runs in O((n^3/k) + n^2). This simplifies to O(n^3) because as n approaches infinity, the n^3 term will dominate the function, irrespective of the constant k.
However, if k is very large in the above example function, the function will appear to run in almost exactly n^2 until some crossover point, at which the n^3 term will begin to dominate. Because the constant k will be unknown to any profiling tool, it will be impossible to know just how large a dataset to test the target function with. If k can be arbitrarily large, you cannot craft test data to determine the big-oh running time.
I am surprised to see so many attempts to claim that one can "measure" complexity by a stopwatch. Several people have given the right answer, but I think that there is still room to drive the essential point home.
Algorithm complexity is not a "programming" question; it is a "computer science" question. Answering the question requires analyzing the code from the perspective of a mathematician, such that computing the Big-O complexity is practically a form of mathematical proof. It requires a very strong understanding of the fundamental computer operations, algebra, perhaps calculus (limits), and logic. No amount of "testing" can be substituted for that process.
The Halting Problem applies, so the complexity of an algorithm is fundamentally undecidable by a machine.
The limits of automated tools applies, so it might be possible to write a program to help, but it would only be able to help about as much as a calculator helps with one's physics homework, or as much as a refactoring browser helps with reorganizing a code base.
For anyone seriously considering writing such a tool, I suggest the following exercise. Pick a reasonably simple algorithm, such as your favorite sort, as your subject algorithm. Get a solid reference (book, web-based tutorial) to lead you through the process of calculating the algorithm complexity and ultimately the "Big-O". Document your steps and results as you go through the process with your subject algorithm. Perform the steps and document your progress for several scenarios, such as best-case, worst-case, and average-case. Once you are done, review your documentation and ask yourself what it would take to write a program (tool) to do it for you. Can it be done? How much would actually be automated, and how much would still be manual?
Best wishes.
I am curious as to why it is that you want to be able to do this. In my experience when someone says: "I want to ascertain the runtime complexity of this algorithm" they are not asking what they think they are asking. What you are most likely asking is what is the realistic performance of such an algorithm for likely data. Calculating the Big-O of a function is of reasonable utility, but there are so many aspects that can change the "real runtime performance" of an algorithm in real use that nothing beats instrumentation and testing.
For example, the following algorithms have the same exact Big-O (wacky pseudocode):
example a:
huge_two_dimensional_array foo
for i = 0, i < foo[i].length, i++
for j = 0; j < foo[j].length, j++
do_something_with foo[i][j]
example b:
huge_two_dimensional_array foo
for j = 0, j < foo[j].length, j++
for i = 0; i < foo[i].length, i++
do_something_with foo[i][j]
Again, exactly the same big-O... but one of them uses row ordinality and one of them uses column ordinality. It turns out that due to locality of reference and cache coherency you might have two completely different actual runtimes, especially depending on the actual size of the array foo. This doesn't even begin to touch the actual performance characteristics of how the algorithm behaves if it's part of a piece of software that has some concurrency built in.
Not to be a negative nelly but big-O is a tool with a narrow scope. It is of great use if you are deep inside algorithmic analysis or if you are trying to prove something about an algorithm, but if you are doing commercial software development the proof is in the pudding, and you are going to want to have actual performance numbers to make intelligent decisions.
Cheers!
This could work for simple algorithms, but what about O(n^2 lg n), or O(n lg^2 n)?
You could get fooled visually very easily.
And if its a really bad algorithm, maybe it wouldn't return even on n=10.
Proof that this is undecidable:
Suppose that we had some algorithm HALTS_IN_FN(Program, function) which determined whether a program halted in O(f(n)) for all n, for some function f.
Let P be the following program:
if(HALTS_IN_FN(P,f(n)))
{
while(1);
}
halt;
Since the function and the program are fixed, HALTS_IN_FN on this input is constant time. If HALTS_IN_FN returns true, the program runs forever and of course does not halt in O(f(n)) for any f(n). If HALTS_IN_FN returns false, the program halts in O(1) time.
Thus, we have a paradox, a contradiction, and so the program is undecidable.
A lot of people have commented that this is an inherently unsolvable problem in theory. Fair enough, but beyond that, even solving it for any but the most trivial cases would seem to be incredibly difficult.
Say you have a program that has a set of nested loops, each based on the number of items in an array. O(n^2). But what if the inner loop is only run in a very specific set of circumstances? Say, on average, it's run in aprox log(n) cases. Suddenly our "obviously" O(n^2) algorithm is really O(n log n). Writing a program that could determine if the inner loop would be run, and how often, is potentially more difficult than the original problem.
Remember O(N) isn't god; high constants can and will change the playing field. Quicksort algorithms are O(n log n) of course, but when the recursion gets small enough, say down to 20 items or so, many implementations of quicksort will change tactics to a separate algorithm as it's actually quicker to do a different type of sort, say insertion sort with worse O(N), but much smaller constant.
So, understand your data, make educated guesses, and test.
I think it's pretty much impossible to do this automatically. Remember that O(g(n)) is the worst-case upper bound and many functions perform better than that for a lot of data sets. You'd have to find the worst-case data set for each one in order to compare them. That's a difficult task on its own for many algorithms.
You must also take care when running such benchmarks. Some algorithms will have a behavior heavily dependent on the input type.
Take Quicksort for example. It is a worst-case O(n²), but usually O(nlogn). For two inputs of the same size.
The traveling salesman is (I think, not sure) O(n²) (EDIT: the correct value is 0(n!) for the brute force algotithm) , but most algorithms get rather good approximated solutions much faster.
This means that the the benchmarking structure has to most of the time be adapted on an ad hoc basis. Imagine writing something generic for the two examples mentioned. It would be very complex, probably unusable, and likely will be giving incorrect results anyway.
Jeffrey L Whitledge is correct. A simple reduction from the halting problem proves that this is undecidable...
ALSO, if I could write this program, I'd use it to solve P vs NP, and have $1million... B-)
I'm using a big_O library (link here) that fits the change in execution time against independent variable n to infer the order of growth class O().
The package automatically suggests the best fitting class by measuring the residual from collected data against each class growth behavior.
Check the code in this answer.
Example of output,
Measuring .columns[::-1] complexity against rapid increase in # rows
--------------------------------------------------------------------------------
Big O() fits: Cubic: time = -0.017 + 0.00067*n^3
--------------------------------------------------------------------------------
Constant: time = 0.032 (res: 0.021)
Linear: time = -0.051 + 0.024*n (res: 0.011)
Quadratic: time = -0.026 + 0.0038*n^2 (res: 0.0077)
Cubic: time = -0.017 + 0.00067*n^3 (res: 0.0052)
Polynomial: time = -6.3 * x^1.5 (res: 6)
Logarithmic: time = -0.026 + 0.053*log(n) (res: 0.015)
Linearithmic: time = -0.024 + 0.012*n*log(n) (res: 0.0094)
Exponential: time = -7 * 0.66^n (res: 3.6)
--------------------------------------------------------------------------------
I guess this isn't possible in a fully automatic way since the type and structure of the input differs a lot between functions.
Well, since you can't prove whether or not a function even halts, I think you're asking a little much.
Otherwise #Godeke has it.
I don't know what's your objective in doing this, but we had a similar problem in a course I was teaching. The students were required to implement something that works at a certain complexity.
In order not to go over their solution manually, and read their code, we used the method #Godeke suggested. The objective was to find students who used linked list instead of a balansed search tree, or students who implemented bubble sort instead of heap sort (i.e. implementations that do not work in the required complexity - but without actually reading their code).
Surprisingly, the results did not reveal students who cheated. That might be because our students are honest and want to learn (or just knew that we'll check this ;-) ). It is possible to miss cheating students if the inputs are small, or if the input itself is ordered or such. It is also possible to be wrong about students who did not cheat, but have large constant values.
But in spite of the possible errors, it is well worth it, since it saves a lot of checking time.
As others have said, this is theoretically impossible. But in practice, you can make an educated guess as to whether a function is O(n) or O(n^2), as long as you don't mind being wrong sometimes.
First time the algorithm, running it on input of various n. Plot the points on a log-log graph. Draw the best-fit line through the points. If the line fits all the points well, then the data suggests that the algorithm is O(n^k), where k is the slope of the line.
I am not a statistician. You should take all this with a grain of salt. But I have actually done this in the context of automated testing for performance regressions. The patch here contains some JS code for it.
If you have lots of homogenious computational resources, I'd time them against several samples and do linear regression, then simply take the highest term.
It's easy to get an indication (e.g. "is the function linear? sub-linear? polynomial? exponential")
It's hard to find the exact complexity.
For example, here's a Python solution: you supply the function, and a function that creates parameters of size N for it. You get back a list of (n,time) values to plot, or to perform regression analysis. It times it once for speed, to get a really good indication it would have to time it many times to minimize interference from environmental factors (e.g. with the timeit module).
import time
def measure_run_time(func, args):
start = time.time()
func(*args)
return time.time() - start
def plot_times(func, generate_args, plot_sequence):
return [
(n, measure_run_time(func, generate_args(n+1)))
for n in plot_sequence
]
And to use it to time bubble sort:
def bubble_sort(l):
for i in xrange(len(l)-1):
for j in xrange(len(l)-1-i):
if l[i+1] < l[i]:
l[i],l[i+1] = l[i+1],l[i]
import random
def gen_args_for_sort(list_length):
result = range(list_length) # list of 0..N-1
random.shuffle(result) # randomize order
# should return a tuple of arguments
return (result,)
# timing for N = 1000, 2000, ..., 5000
times = plot_times(bubble_sort, gen_args_for_sort, xrange(1000,6000,1000))
import pprint
pprint.pprint(times)
This printed on my machine:
[(1000, 0.078000068664550781),
(2000, 0.34400010108947754),
(3000, 0.7649998664855957),
(4000, 1.3440001010894775),
(5000, 2.1410000324249268)]

Resources