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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)]
This part of code is from dotproduct method of a vector class of mine. The method does inner product computing for a target array of vectors(1000 vectors).
When vector length is an odd number(262145), compute time is 4.37 seconds. When vector length(N) is 262144(multiple of 8), compute time is 1.93 seconds.
time1=System.nanotime();
int count=0;
for(int j=0;j<1000;i++)
{
b=vektors[i]; // selects next vector(b) to multiply as inner product.
// each vector has an array of float elements.
if(((N/2)*2)!=N)
{
for(int i=0;i<N;i++)
{
t1+=elements[i]*b.elements[i];
}
}
else if(((N/8)*8)==N)
{
float []vek=new float[8];
for(int i=0;i<(N/8);i++)
{
vek[0]=elements[i]*b.elements[i];
vek[1]=elements[i+1]*b.elements[i+1];
vek[2]=elements[i+2]*b.elements[i+2];
vek[3]=elements[i+3]*b.elements[i+3];
vek[4]=elements[i+4]*b.elements[i+4];
vek[5]=elements[i+5]*b.elements[i+5];
vek[6]=elements[i+6]*b.elements[i+6];
vek[7]=elements[i+7]*b.elements[i+7];
t1+=vek[0]+vek[1]+vek[2]+vek[3]+vek[4]+vek[5]+vek[6]+vek[7];
//t1 is total sum of all dot products.
}
}
}
time2=System.nanotime();
time3=(time2-time1)/1000000000.0; //seconds
Question: Could the reduction of time from 4.37s to 1.93s (2x as fast) be JIT's wise decision of using SIMD instructions or just my loop-unrolling's positive effect?
If JIT cannot do SIMD optimizaton automatically, then in this example there is also no unrolling optimization done automatically by JIT, is this true?.
For 1M iterations(vectors) and for vector size of 64, speedup multiplier goes to 3.5X(cache advantage?).
Thanks.
Your code has a bunch of problems. Are you sure you're measuring what you think you're measuring?
Your first loop does this, indented more conventionally:
for(int j=0;j<1000;i++) {
b=vektors[i]; // selects next vector(b) to multiply as inner product.
// each vector has an array of float elements.
}
Your rolled loop involves a really long chain of dependent loads and stores. Your unrolled loop involves 8 separate chains of dependent loads and stores. The JVM can't turn one into the other if you're using floating-point arithmetic because they're fundamentally different computations. Breaking dependent load-store chains can lead to major speedups on modern processors.
Your rolled loop iterates over the whole vector. Your unrolled loop only iterates over the first (roughly) eighth. Thus, the unrolled loop again computes something fundamentally different.
I haven't seen a JVM generate vectorised code for something like your second loop, but I'm maybe a few years out of date on what JVMs do. Try using -XX:+PrintAssembly when you run your code and inspect the code opto generates.
I have done a little research on this (and am drawing from knowledge from a similar project I did in C with matrix multiplication), but take my answer with a grain of salt as I am by no means an expert on this topic.
As for your first question, I think the speedup is coming from your loop unrolling; you're making roughly 87% fewer condition checks in terms of the for loop. As far as I know, JVM supports SSE since 1.4, but to actually control whether your code is using vectorization (and to know for sure), you'll need to use JNI.
See an example of JNI here: Do any JVM's JIT compilers generate code that uses vectorized floating point instructions?
When you decrease the size of your vector to 64 from 262144, cache is definitely a factor. When I did this project in C, we had to implement cache blocking for larger matrices in order to take advantage of the cache. One thing you might want to do is check your cache size.
Just as a side note: It might be a better idea to measure performance in flops rather than seconds, just because the runtime (in seconds) of your program can vary based on many different factors, such as CPU usage at the time.
I'd like to have a solid understanding of when (ignoring available memory space) it makes sense to store the result of a comparison instead of recalculating it. What is the tipping point for justifying the time cost incurred by storage? Is it 2, 3, or 4 comparisons? More?
For example, in this particular case, which option (in general) will perform better in terms of speed?
Option 1:
int result = id.compareTo(node.id);
return result > 0 ? 1 : result < 0 ? -1 : 0;
Option 2:
return id.compareTo(node.id) > 0 ? 1 : id.compareTo(node.id) < 0 ? -1 : 0;
I tried to profile the two options myself in order to answer my own question, but I don't have much experience with this sort of performance testing and, as such, would rather get a more definitive answer from someone with either more experience or else a better grasp of the theoretical elements involved.
I know it's not a big deal and that most of the time the difference will be negligible. However, I'm a perfectionist, and I'd really just like to resolve this particular issue so that I can get on with my life, haha.
Additionally, I think the answer is likely to prove enlightening in regards to similar situations I may encounter in the future wherein the difference might very well be significant (such as when the cost of a comparison or memory allocation is either unable to be incurred or else complex enough to cause a real issue concerning performance).
Answers should be relevant to programming with Java and not other languages, please.
I know I've mentioned it a few times already, but PLEASE focus answers ONLY on the SPEED DIFFERENCE! I am well aware that many other factors can and should be taken into account when writing code, but here I want just a straight-forward argument for which is FASTER.
Experience tells me that option 1 should be faster, because you're making just one call to the compare method and storing the result for reuse. Facts that support this belief are that local variables live on the stack and making a method call involves a lot more work from the stack than just pushing a value onto it. However profiling is the best and safest way to compare two implementations.
The first thing to realise is that the java compiler and JVM together may optimise your code how it wishes to get the job done most efficiently (as long as certain rules are followed). Chances are there is no difference in performance, and chances are also that what is actually executed is not what you think it is.
One really important difference however is in debugging: if you put a break point on the return statement for the store-in-variable version, you can see what was returned from the call, otherwise you can't see that in a debugger. Even more handy is when you seemingly uselessly store the value to be returned from the method in a variable, then return it, so you may see what's going to be returned from a method while debugging, otherwise there's no way to see it.
Option 1 cannot be slower than 2, if the compiler optimizes then both could be equal, but then still 1) is more readable, compacter, and better testable.
So there is no argument for Option 2).
If you like you could change to final int result = ....
Although i expect that the compiler is so clever that the final keyword makes no difference in this case, and the final makes the code a bit less readable.
option1 one always preferred one ,because here the real world scenarion
----->ok lets
1) thread exceution at id.compareTo(node.id) > 0 ? 1 , in this process some other thread
changes the value of node.id right after id.compareTo(node.id) > 0 ? 1 before going to
id.compareTo(node.id) < 0 ? -1 : 0 this check , the result not identical?
performance wise option1 has more performance when there is bit of functionality exisist in checking.
When does it make sense to store the result of a comparison versus recalculating a comparison in terms of speed?
Most of the time, micro-optimizations like option #1 versus option #2 don't make any significant difference. Indeed, it ONLY makes a significant performance difference if:
the comparison is expensive,
the comparison is performed a large number of times, AND
performance matters.
Indeed, the chances are that you have alrady spent more time and money thinking about this than will be saved over the entire useful lifetime of the application.
Instead of focussing on performance, you should be focussing on making your code readable. Think about the next person who has to read and modify the code, and make it so that he/she is less likely to misread it.
In this case, the first option is more readable than the second one. THAT is why you should use it, not performance reasons. (Though, if anything, the first version is probably faster.)
My question is if there are some smart ways of debugging complicated recursive algorithms.
Assume that we have a complicated one (not a simple case when recursion counter is decreased in each 'nested iteration').
I mean something like recursive traversing of a graph when loops are possible.
I need to check if I am not getting endless loop somewhere. And doing this just using a debugger gives not certain answer (because I am not sure if an algorithm is in endless loop or just process as it should).
It's hard to explain it without concrete example. But what I need is...
'to check if the endless loops don't occur in let's say complicated recursive algorithm'.
You need to form a theory for why you think the algorithm does terminate. Ideally, prove the theory as a mathematical theorem.
You can look for a function of the problem state that does reduce on each recursive call. For example, see the following discussion of Ackermann's function, from Wikipedia
It may not be immediately obvious that the evaluation of A(m, n) always terminates. However, the recursion is bounded because in each recursive application either m decreases, or m remains the same and n decreases. Each time that n reaches zero, m decreases, so m eventually reaches zero as well. (Expressed more technically, in each case the pair (m, n) decreases in the lexicographic order on pairs, which is a well-ordering, just like the ordering of single non-negative integers; this means one cannot go down in the ordering infinitely many times in succession.) However, when m decreases there is no upper bound on how much n can increase — and it will often increase greatly.
That is the type of reasoning you should be thinking of applying to your algorithm.
If you cannot find any way to prove your algorithm terminates, consider looking for a variation whose termination you can prove. It is not always possible to decide whether an arbitrary program terminates or not. The trick is to write algorithms you can prove terminate.
Best is proving finiteness by pre and post conditions, variants and invariants. If you can specify a (virtual) formula which value increases on every call you have a guarantee.
This is the same as proving loops to be finite. Furthermore it might make complex algorithms more tackable.
You need to count the depth of recursive calls ... and then throw an exception if the depth of recursive calls reaches a certain threshold.
For example:
void TheMethod(object[] otherParameters, int recursiveCallDepth)
{
if (recursiveCallDepth > 100) {
throw new Exception("...."); }
TheMethod(otherParameters, ++recursiveCallDepth);
}
if you want to check for endless loops,
write a System.out.println("no its not endless"); at the next line of calling the recursive function.
if the loop would be endless, this statement wont get print, if otherwise you will see the output
One suggestion is the following:
If you have endless loop then in the graph case you will obtain a path with number of vertices greater than the total number of vertices in the graph. Assuming that the number of vertices in the graph is a global variable (which, I think, is the most common case) you can do a conditional breakpoint in the beginning of the recursion if the depth is already above the total number of vertices.
Here is a link how you do conditional breakpoints for java in Eclipse.
I had a job interview today, we were given a programming question, and were asked to solve it using c/c++/Java, I solved it in java and its runtime was 3 sec (the test was more 16000 lines, and the person accompanying us said the running time was reasonable), another person there solved it in c and the runtime was 0.25 sec, so I was wondering, is a factor of 12 normal?
Edit:
As I said, I don't think there was really much room for algorithm variation except maybe in one little thing, anyway, there was this protocol that we had to implement:
A (client) and B (server) communicate according to some protocol p, before the messages are delivered their validity is checked, the protocol is defined by its state and the text messages that can be sent when it is in a certain state, in all states there was only one valid message that could be sent, except in one state where there was like 10 messages that can be sent, there are 5 states and the states transition is defined by the protocol too.
so what I did with the state from which 10 different messages can be sent was storing their string value in an ArrayList container, then when I needed to check the message validity in the corresponding state i checked if arrayList.contains(sentMessageStr); I would think that this operation's complexity is O(n) although I think java has some built-in optimization for this operation, although now that I am thinking about it, maybe I should've used a HashSet container.I suppose the c implementation would have been storing those predefined legal strings lexicographically in an array and implementing a binary search function.
thanks
I would guess that it's likely the jvm took a significant portion of that 3 seconds just to load. Try running your java version on the same machine 5 times in a row. Or try running both on a dataset 500 times as large. I suspect you'll see a significant constant latency for the Java version that will become insignificant when runtimes go into the minutes.
Sounds more like a case of insufficient samples and unequal implementations (and possibly unequal test beds).
One of the first rules in measurement is to establish enough samples and obtain the mean of the samples for comparison. Even a couple of runs of the same program is not sufficient. You need to tax the machine enough to obtain samples whose values can be compared. That's why test-beds need to be warmed up, so that there are little or no variables at play, except for the system under observation.
And of course, you also have different people implementing the same requirement/algorithm in different manners. It counts. Period. Unless the algorithm implementations have been "normalized", obtaining samples and comparing them are the same as comparing apples and watermelons.
I don't think I need to expand on the fact that the testbeds could have been of varying configurations, or under varying loads.
It's almost impossible to say without seeing the code - you may have a better algorithm for example that scales up much better for larger input but has a greater overhead for small input sizes.
Having said that, this kind of 12x difference is roughly what I would expect if you coded the solution using "higher level" constructs such as ArrayLists / boxed objects and the C solution was basically using optimised, low level pointer arithmetic into a pre-allocated memory region.
I'd rather maintain the higher level solution, but there are times when only hand-optimised low level code will do.....
Another potential explanation is that the JIT had not yet warmed up on your code. In general, you need to reach "steady state" (typically a few thousand iterations of every code path) before you will see top performance in JIT-compiled code.
Performance depends on implementation. Without knowing exactly what you code and what your competitor did, it's very difficult to tell exactly what happened.
But let's say for isntance, that you used objects like vectors or whatever to solve the problem and the C guy used arrays[], his implementation is going to be faster than yours for sure.
C code can be translated very efficiently into assembly instructions, while Java on the other hand, relies on a bunch of stuff (like the JVM) that might make the bytecode of your program fatter and probably a little bit slower.
You will be hard pressed to find something that can execute faster in Java than in C. Its true that an order of magnitude is a big difference but in general C is more performant.
On the other hand you can produce a solution to any given problem much quicker in Java (especially taking into account the richness of the libraries).
So at the end of the day, if there is a choice at all, it comes down as a dilemma between performance and productivity.
That depends on the algorithm. Java is of course generally slower than C/C++ as it's a virtual machine but for most common applications its speed is sufficient. I would not call a factor of 12 normal for common applications.
Would be nice if you posted the C and Java codes for comparison.
A factor of 12 can be normal. So could a factor of 1 or 1/2. As some commentators mentioned, a lot has to do with how you coded your solution.
Dont forget that java programs have to run in a jvm (unless you compile to native machine code), so any benchmarks should take that into account.
You can google for 'java and c speed comparisons' for some analysis
Back in the days I'd say that there's nothing wrong with your Java code being 12 times slower. But nowadays I'd rather say that the C guy implemented it more efficiently. Obviously I might be wrong, but are you sure you used proper data structures and well simply coded it well?
Also did you measure the memory usage? This might sound silly, but last year at the uni we had a programming challenge, don't really remember what it was but we had to solve a graph problem in whatever language we wanted - I did two implementations of my algorithm one in C and one in Java, the Java one was ~1,5-2x slower, BUT for instance I knew I didn't have to worry about memory management (I knew exactly how big the input will be and how many test samples will be run from the teacher) so I simply didn't free any memory (which took way too much time in a programme that run for ~1-2seconds on a graph with ~15k nodes, or was it 150k?) so my Java code was better memory wise but it was slower. I also parsed the input myself in C (didn't do that in Java) which saved me really A LOT of time (~20-25% boost, I was amazed myself). I'd say 1,5-2x is more realistic than 12x.
Most likely the algorithm used in the implementation was different.
For instance ( an over simplification ) if you want to add a number N , M number of times one implementation could be:
long addTimes( long n, long m ) {
long r = 0;
long i;
for( i = 0; i < m ; i++ ) {
r += n;
}
return r;
}
And another implementation could simply be:
long addTimes( long n, long m ) {
return n * m;
}
Both, will run mostly the same in Java and C (you don't even have to change the code ) and still, one implementation will run way lot faster than the other.