Related
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)]
So, I have tried to solve some big o questions and I had some troubles with some of them. I don't quite understand them.
like for eg the dominant term of 10MlogM + (N/2) log (N/2) + N/4
and M log (N) + M log (M ). I have trouble understanding big o expressions with 'log' in them. Any help would be appreciated and thanks in advance.
I have trouble understanding big o expressions with 'log' in them
It's quite easy to understand Big-O of log (or any function) if we draw its graph:
Now since you are facing an equation having multiple terms (and one of them is log), so maybe it would confuse you out. If we plot first term (I assume its NlogN and discard the constant (10)):
Similarly for other term (N/2 * log(N/2)):
I am sure you will already know, but this graph easily demonstrates that only major term (N logN) defines the Big O (as its just to show the upper bound of function) as I plot the graph of whole function:
What does N log N mean
This link provides very good explanation with this example:
O(n log n): There was a mix-up at the printer's office, and our phone book had all its pages inserted in a random order. Fix the ordering so that it's correct by looking at the first name on each page and then putting that page in the appropriate spot in a new, empty phone book.
For the below examples, we're now at the printer's office. Phone books
are waiting to be mailed to each resident or business, and there's a
sticker on each phone book identifying where it should be mailed to.
Every person or business gets one phone book.
O(n log n): We want to personalize the phone book, so we're going to
find each person or business's name in their designated copy, then
circle their name in the book and write a short thank-you note for
their patronage.
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.
I do not know a whole lot about math, so I don't know how to begin to google what I am looking for, so I rely on the intelligence of experts to help me understand what I am after...
I am trying to find the smallest string of equations for a particular large number. For example given the number
"39402006196394479212279040100143613805079739270465446667948293404245721771497210611414266254884915640806627990306816"
The smallest equation is 64^64 (that I know of) . It contains only 5 bytes.
Basically the program would reverse the math, instead of taking an expression and finding an answer, it takes an answer and finds the most simplistic expression. Simplistic is this case means smallest string, not really simple math.
Has this already been created? If so where can I find it? I am looking to take extremely HUGE numbers (10^10000000) and break them down to hopefully expressions that will be like 100 characters in length. Is this even possible? are modern CPUs/GPUs not capable of doing such big calculations?
Edit:
Ok. So finding the smallest equation takes WAY too much time, judging on answers. Is there anyway to bruteforce this and get the smallest found thus far?
For example given a number super super large. Sometimes taking the sqaureroot of number will result in an expression smaller than the number itself.
As far as what expressions it would start off it, well it would naturally try expressions that would the expression the smallest. I am sure there is tons of math things I dont know, but one of the ways to make a number a lot smaller is powers.
Just to throw another keyword in your Google hopper, see Kolmogorov Complexity. The Kolmogorov complexity of a string is the size of the smallest Turing machine that outputs the string, given an empty input. This is one way to formalize what you seem to be after. However, calculating the Kolmogorov complexity of a given string is known to be an undecidable problem :)
Hope this helps,
TJ
There's a good program to do that here:
http://mrob.com/pub/ries/index.html
I asked the question "what's the point of doing this", as I don't know if you're looking at this question from a mathemetics point of view, or a large number factoring point of view.
As other answers have considered the factoring point of view, I'll look at the maths angle. In particular, the problem you are describing is a compressibility problem. This is where you have a number, and want to describe it in the smallest algorithm. Highly random numbers have very poor compressibility, as to describe them you either have to write out all of the digits, or describe a deterministic algorithm which is only slightly smaller than the number itself.
There is currently no general mathemetical theorem which can determine if a representation of a number is the smallest possible for that number (although a lower bound can be discovered by understanding shannon's information theory). (I said general theorem, as special cases do exist).
As you said you don't know a whole lot of math, this is perhaps not a useful answer for you...
You're doing a form of lossless compression, and lossless compression doesn't work on random data. Suppose, to the contrary, that you had a way of compressing N-bit numbers into N-1-bit numbers. In that case, you'd have 2^N values to compress into 2^N-1 designations, which is an average of 2 values per designation, so your average designation couldn't be uncompressed. Lossless compression works well on relatively structured data, where data we're likely to get is compressed small, and data we aren't going to get actually grows some.
It's a little more complicated than that, since you're compressing partly by allowing more information per character. (There are a greater number of N-character sequences involving digits and operators than digits alone.) Still, you're not going to get lossless compression that, on the average, is better than just writing the whole numbers in binary.
It looks like you're basically wanting to do factoring on an arbitrarily large number. That is such a difficult problem that it actually serves as the cornerstone of modern-day cryptography.
This really appears to be a mathematics problem, and not programming or computer science problem. You should ask this on https://math.stackexchange.com/
While your question remains unclear, perhaps integer relation finding is what you are after.
EDIT:
There is some speculation that finding a "short" form is somehow related to the factoring problem. I don't believe that is true unless your definition requires a product as the answer. Consider the following pseudo-algorithm which is just sketch and for which no optimization is attempted.
If "shortest" is a well-defined concept, then in general you get "short" expressions by using small integers to large powers. If N is my integer, then I can find an integer nearby that is 0 mod 4. How close? Within +/- 2. I can find an integer within +/- 4 that is 0 mod 8. And so on. Now that's just the powers of 2. I can perform the same exercise with 3, 5, 7, etc. We can, for example, easily find the nearest integer that is simultaneously the product of powers of 2, 3, 5, 7, 11, 13, and 17, call it N_1. Now compute N-N_1, call it d_1. Maybe d_1 is "short". If so, then N_1 (expressed as power of the prime) + d_1 is the answer. If not, recurse to find a "short" expression for d_1.
We can also pick integers that are maybe farther away than our first choice; even though the difference d_1 is larger, it might have a shorter form.
The existence of an infinite number of primes means that there will always be numbers that cannot be simplified by factoring. What you're asking for is not possible, sorry.
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)]