I'm measuring the execution time of a program that should run in O(n^2). To get the expected running time I would calculate n^2 from the input size I assume. But I taken the execution time using another program I get the time in milliseconds. So my question is how to compare that to n^2. For n^2 I get a larger number. How would I convert this to miliseconds? I know this question may not be worded as good as you might like. Hopefully, you know what I mean.
It doesn't measure time in any unit. It describes how the time will change if n changes.
For example, O (n^2) is defined to mean: "There is a constant c such that the time is at most c * n^2 for all but the first few n". When you run the algorithm on different computers, it could be 5 n^2 nanoseconds on one, and 17 n^2 milliseconds on another computer. You just have c = 5 nanoseconds in one case, c = 17 milliseconds in the other case.
It bounds (to within a constant factor) the asymptotic runtime in numbers of steps of an abstract model of a computer, often a Register Machine.
Big O doesn't measure a unit of time. It expresses the runtime of the code in terms of input of size n. You can't accurately use a big O notation to compute runtime in milliseconds as this may vary per machine that you run the code on.
You can estimate how long it takes if you know the runtime of each of the operations in the algorithm but that's not really what the big O notation is meant for.
Related
If I have a program that runs over some data in O(n) time, can I semi-accurately guestimate the O(n^3) runtime from my O(n) run?
**O(n) = 5 million iterations # 2 minutes total runtime**
**O(n^2) = ??**
(5 million)^2 = 2.5+13
2.5+13 / 5 million = 5 million minutes
5 million / 60 = 83,333 hours = 3,472 days = 9.5 years
**O(n^3) = ??**
(5 million)^3 = 1.25e+20
1.25e+20 / 5 million = 2.5e+13 minutes
2.5e+13 / 60 = 416666666667 hours = 17361111111.1 days = 47,564,688 years
Technically knowing O(...) doesn't tell you anything about any execution time for specific finite inputs.
Practically, you can make an estimation for example in the way you did, but the caveat is that it will only give you the order-of-magnitude under the assumptions that 1. the constant scaling factor omitted in the O(...) notations is roughly 1 in whatever units you chose (number of iterations here) in both programs/algorithms and 2. that the input value is large enough so that higher-order terms omitted by the O(...) notation are not relevant anymore.
Whether these assumptions are good assumptions will depend on the particular programs/algorithms you are looking at. It is trivial to come up with examples where this is a really bad approximation, but there are also many cases where such an estimate may be reasonable.
If you just want to estimate whether the alternate program will execute in a non-absurd time frame (e.g. hours vs centuries) I think it will often be a good enough for that, assuming you did not choose a weird unit and assuming there is nothing in the program that would explicitly increase the asymptotic scaling, like e.g. an inner loop with exactly 10000000 iterations.
If I have a program that runs over some data in O(n) time, can I semi-accurately guestimate the O(n^3) runtime from my O(n) run?
No.
There is no the O(n3) runtime, nor either any the O(n) time. Asymptotic complexity speaks to how the behavior of a particular program or subprogram scales with input size. You can use that to estimate the performance of the same program for one input size from appropriate measurements of the performance of that program for other input sizes, but that does not give you any information about any other program's specific performance for a given input size.
In particular, your idea seems to be that the usually-ignored coefficient of the bounding function is a property of the machine, but this is not at all the case. The coefficient is mostly a property of the details of the program. If you estimate it for one program then you know it only for that program. Forget programs with different asymptotic complexity: two programs with the same asymptotic complexity can be constructed that have arbitrarily different absolute performance for any given input size.
I want to know, given a computer and the big O of the running time of an algorithm, to know the actual aproximated time that the algorithm will take in that computer.
For example, let's say that I have an algorithm of complexity O(n) and a computer with one processor of 3.00 GHz, one core, 32-bits and 4 GB RAM. How can I estimate the actual seconds that will take this algorithm.
Thanks
There is no good answer to this question, simply because big O notation was never meant to answer that sort of question.
What big O notation does tell us is the following:
Big O notation characterizes functions according to their growth rates: different functions with the same growth rate may be represented using the same O notation.
Note that this means that functions that could hold very different values can be assigned the same big O value.
In other words, big O notation doesn't tell you much as to how fast an algorithm runs on a particular input, but rather compares the run times on inputs as their sizes approach infinity.
I know that O(log n) refers to an iterative reduction by a fixed ratio of the problem set N (in big O notation), but how do i actually calculate it to see how many iterations an algorithm with a log N complexity would have to preform on the problem set N before it is done (has one element left)?
You can't. You don't calculate the exact number of iterations with BigO.
You can "derive" BigO when you have exact formula for number of iterations.
BigO just gives information how the number iterations grows with growing N, and only for "big" N.
Nothing more, nothing less. With this you can draw conclusions how much more operations/time will the algorithm take if you have some sample runs.
Expressed in the words of Tim Roughgarden at his courses on algorithms:
The big-Oh notation tries to provide a sweet spot for high level algorithm reasoning
That means it is intended to describe the relation between the algorithm time execution and the size of its input avoiding dependencies on the system architecture, programming language or chosen compiler.
Imagine that big-Oh notation could provide the exact execution time, that would mean that for any algorithm, for which you know its big-Oh time complexity function, you could predict how would it behave on any machine whatsoever.
On the other hand, it is centered on asymptotic behaviour. That is, its description is more accurate for big n values (that is why lower order terms of your algorithm time function are ignored in big-Oh notation). It can reasoned that low n values do not demand you to push foward trying to improve your algorithm performance.
Big O notation only shows an order of magnitude - not the actual number of operations that algorithm would perform. If you need to calculate exact number of loop iterations or elementary operations, you have to do it by hand. However in most practical purposes exact number is irrelevant - O(log n) tells you that num. of operations will raise logarythmically with a raise of n
From big O notation you can't tell precisely how many iteration will the algorithm do, it's just estimation. That means with small numbers the different between the log(n) and actual number of iterations could be differentiate significantly but the closer you get to infinity the different less significant.
If you make some assumptions, you can estimate the time up to a constant factor. The big assumption is that the limiting behavior as the size tends to infinity is the same as the actual behavior for the problem sizes you care about.
Under that assumption, the upper bound on the time for a size N problem is C*log(N) for some constant C. The constant will change depending on the base you use for calculating the logarithm. The base does not matter as long as you are consistent about it. If you have the measured time for one size, you can estimate C and use that to guesstimate the time for a different size.
For example, suppose a size 100 problem takes 20 seconds. Using common logarithms, C is 10. (The common log of 100 is 2). That suggests a size 1000 problem might take about 30 seconds, because the common log of 1000 is 3.
However, this is very rough. The approach is most useful for estimating whether an algorithm might be usable for a large problem. In that sort of situation, you also have to pay attention to memory size. Generally, setting up a problem will be at least linear in size, so its cost will grow faster than an O(log N) operation.
I'm trying to understand a particular aspect of Big O analysis in the context of running programs on a PC.
Suppose I have an algorithm that has a performance of O(n + 2). Here if n gets really large the 2 becomes insignificant. In this case it's perfectly clear the real performance is O(n).
However, say another algorithm has an average performance of O(n2 / 2). The book where I saw this example says the real performance is O(n2). I'm not sure I get why, I mean the 2 in this case seems not completely insignificant. So I was looking for a nice clear explanation from the book. The book explains it this way:
"Consider though what the 1/2 means. The actual time to check each value
is highly dependent on the machine instruction that the code
translates to and then on the speed at which the CPU can execute the instructions. Therefore the 1/2 doesn't mean very much."
And my reaction is... huh? I literally have no clue what that says or more precisely what that statement has to do with their conclusion. Can somebody spell it out for me please.
Thanks for any help.
There's a distinction between "are these constants meaningful or relevant?" and "does big-O notation care about them?" The answer to that second question is "no," while the answer to that first question is "absolutely!"
Big-O notation doesn't care about constants because big-O notation only describes the long-term growth rate of functions, rather than their absolute magnitudes. Multiplying a function by a constant only influences its growth rate by a constant amount, so linear functions still grow linearly, logarithmic functions still grow logarithmically, exponential functions still grow exponentially, etc. Since these categories aren't affected by constants, it doesn't matter that we drop the constants.
That said, those constants are absolutely significant! A function whose runtime is 10100n will be way slower than a function whose runtime is just n. A function whose runtime is n2 / 2 will be faster than a function whose runtime is just n2. The fact that the first two functions are both O(n) and the second two are O(n2) doesn't change the fact that they don't run in the same amount of time, since that's not what big-O notation is designed for. O notation is good for determining whether in the long term one function will be bigger than another. Even though 10100n is a colossally huge value for any n > 0, that function is O(n) and so for large enough n eventually it will beat the function whose runtime is n2 / 2 because that function is O(n2).
In summary - since big-O only talks about relative classes of growth rates, it ignores the constant factor. However, those constants are absolutely significant; they just aren't relevant to an asymptotic analysis.
Big O notation is most commonly used to describe an algorithm's running time. In this context, I would argue that specific constant values are essentially meaningless. Imagine the following conversation:
Alice: What is the running time of your algorithm?
Bob: 7n2
Alice: What do you mean by 7n2?
What are the units? Microseconds? Milliseconds? Nanoseconds?
What CPU are you running it on? Intel i9-9900K? Qualcomm Snapdragon 845? (Or are you using a GPU, an FPGA, or other hardware?)
What type of RAM are you using?
What programming language did you implement the algorithm in? What is the source code?
What compiler / VM are you using? What flags are you passing to the compiler / VM?
What is the operating system?
etc.
So as you can see, any attempt to indicate a specific constant value is inherently problematic. But once we set aside constant factors, we are able to clearly describe an algorithm's running time. Big O notation gives us a robust and useful description of how long an algorithm takes, while abstracting away from the technical features of its implementation and execution.
Now it is possible to specify the constant factor when describing the number of operations (suitably defined) or CPU instructions an algorithm executes, the number of comparisons a sorting algorithm performs, and so forth. But typically, what we're really interested in is the running time.
None of this is meant to suggest that the real-world performance characteristics of an algorithm are unimportant. For example, if you need an algorithm for matrix multiplication, the Coppersmith-Winograd algorithm is inadvisable. It's true that this algorithm takes O(n2.376) time, whereas the Strassen algorithm, its strongest competitor, takes O(n2.808) time. However, according to Wikipedia, Coppersmith-Winograd is slow in practice, and "it only provides an advantage for matrices so large that they cannot be processed by modern hardware." This is usually explained by saying that the constant factor for Coppersmith-Winograd is very large. But to reiterate, if we're talking about the running time of Coppersmith-Winograd, it doesn't make sense to give a specific number for the constant factor.
Despite its limitations, big O notation is a pretty good measure of running time. And in many cases, it tells us which algorithms are fastest for sufficiently large input sizes, before we even write a single line of code.
Big-O notation only describes the growth rate of algorithms in terms of mathematical function, rather than the actual running time of algorithms on some machine.
Mathematically, Let f(x) and g(x) be positive for x sufficiently large.
We say that f(x) and g(x) grow at the same rate as x tends to infinity, if
now let f(x)=x^2 and g(x)=x^2/2, then lim(x->infinity)f(x)/g(x)=2. so x^2 and x^2/2 both have same growth rate.so we can say O(x^2/2)=O(x^2).
As templatetypedef said, hidden constants in asymptotic notations are absolutely significant.As an example :marge sort runs in O(nlogn) worst-case time and insertion sort runs in O(n^2) worst case time.But as the hidden constant factors in insertion sort is smaller than that of marge sort, in practice insertion sort can be faster than marge sort for small problem sizes on many machines.
You are completely right that constants matter. In comparing many different algorithms for the same problem, the O numbers without constants give you an overview of how they compare to each other. If you then have two algorithms in the same O class, you would compare them using the constants involved.
But even for different O classes the constants are important. For instance, for multidigit or big integer multiplication, the naive algorithm is O(n^2), Karatsuba is O(n^log_2(3)), Toom-Cook O(n^log_3(5)) and Schönhage-Strassen O(n*log(n)*log(log(n))). However, each of the faster algorithms has an increasingly large overhead reflected in large constants. So to get approximate cross-over points, one needs valid estimates of those constants. Thus one gets, as SWAG, that up to n=16 the naive multiplication is fastest, up to n=50 Karatsuba and the cross-over from Toom-Cook to Schönhage-Strassen happens for n=200.
In reality, the cross-over points not only depend on the constants, but also on processor-caching and other hardware-related issues.
Big O without constant is enough for algorithm analysis.
First, the actual time does not only depend how many instructions but also the time for each instruction, which is closely connected to the platform where the code runs. It is more than theory analysis. So the constant is not necessary for most case.
Second, Big O is mainly used to measure how the run time will increase as the problem becomes larger or how the run time decrease as the performance of hardware improved.
Third, for situations of high performance optimizing, constant will also be taken into consideration.
The time required to do a particular task in computers now a days does not required a large amount of time unless the value entered is very large.
Suppose we wants to multiply 2 matrices of size 10*10 we will not have problem unless we wants to do this operation multiple times and then the role of asymptotic notations becomes prevalent and when the value of n becomes very big then the constants don't really makes any difference to the answer and are almost negligible so we tend to leave them while calculating the complexity.
Time complexity for O(n+n) reduces to O(2n). Now 2 is a constant. So the time complexity will essentially depend on n.
Hence the time complexity of O(2n) equates to O(n).
Also if there is something like this O(2n + 3) it will still be O(n) as essentially the time will depend on the size of n.
Now suppose there is a code which is O(n^2 + n), it will be O(n^2) as when the value of n increases the effect of n will become less significant compared to effect of n^2.
Eg:
n = 2 => 4 + 2 = 6
n = 100 => 10000 + 100 => 10100
n = 10000 => 100000000 + 10000 => 100010000
As you can see the effect of the second expression as lesser effect as the value of n keeps increasing. Hence the time complexity evaluates to O(n^2).
I want to know, how i can estimate the time that my program will take to execute on my machine (for example a 2.5 Ghz machine), if i have an estimation of its worst case time complexity?
For Example : - If I have a program which is O(n^2), in worst case, and n<100000, how can i know /estimate before writing the actual program/procedure, the time that it will take to execute in seconds?
Wouldn't it be good to know how a program actually performs, and it will also save writing code which eventually turns out to be inefficient!
Help greatly appreciated.
Since big O complexity ignores linear coefficients and smaller terms, it is impossible to estimate the performance of an algorithm given only its big o complexity.
In fact, for any specific N, you cannot predict which of two given algorithms will execute faster.
For example, O(N) is not always faster than O(N*N) since an algorithm that takes 100000000*n steps is O(N) is slower than an algorithm than takes N*N steps for many small values of N.
These linear coefficients and asymptotically smaller terms vary from platform to platform and even amongst algorithms of the same equivalence class (in terms of big O measure). 3
The problem you are trying to use big O notation for is not the one it is designed to solve.
Instead of dealing with complecity, you might want to have a look at Worst Case Execution Time (WCET). This area of research most likely corresponds to what you are looking for.
http://en.wikipedia.org/wiki/Worst-case_execution_time
Multiply N^2 by the time You spend in an iteration of the innermost loop, and You have a ballpark estimate.