Related
I am trying to learn analysis of algorithms and I am stuck with relation between asymptotic notation(big O...) and cases(best, worst and average).
I learn that the Big O notation defines an upper bound of an algorithm, i.e. it defines function can not grow more than its upper bound.
At first it sound to me as it calculates the worst case.
I google about(why worst case is not big O?) and got ample of answers which were not so simple to understand for beginner.
I concluded it as follows:
Big O is not always used to represent worst case analysis of algorithm because, suppose a algorithm which takes O(n) execution steps for best, average and worst input then it's best, average and worst case can be expressed as O(n).
Please tell me if I am correct or I am missing something as I don't have anyone to validate my understanding.
Please suggest a better example to understand why Big O is not always worst case.
Big-O?
First let us see what Big O formally means:
In computer science, big O notation is used to classify algorithms
according to how their running time or space requirements grow as the
input size grows.
This means that, 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. Here, O means order of the function, and it only provides an upper bound on the growth rate of the function.
Now let us look at the rules of Big O:
If f(x) is a sum of several terms, if there is one with largest
growth rate, it can be kept, and all others omitted
If f(x) is a product of several factors, any constants (terms in the
product that do not depend on x) can be omitted.
Example:
f(x) = 6x^4 − 2x^3 + 5
Using the 1st rule we can write it as, f(x) = 6x^4
Using the 2nd rule it will give us, O(x^4)
What is Worst Case?
Worst case analysis gives the maximum number of basic operations that
have to be performed during execution of the algorithm. It assumes
that the input is in the worst possible state and maximum work has to
be done to put things right.
For example, for a sorting algorithm which aims to sort an array in ascending order, the worst case occurs when the input array is in descending order. In this case maximum number of basic operations (comparisons and assignments) have to be done to set the array in ascending order.
It depends on a lot of things like:
CPU (time) usage
memory usage
disk usage
network usage
What's the difference?
Big-O is often used to make statements about functions that measure the worst case behavior of an algorithm, but big-O notation doesn’t imply anything of the sort.
The important point here is we're talking in terms of growth, not number of operations. However, with algorithms we do talk about the number of operations relative to the input size.
Big-O is used for making statements about functions. The functions can measure time or space or cache misses or rabbits on an island or anything or nothing. Big-O notation doesn’t care.
In fact, when used for algorithms, big-O is almost never about time. It is about primitive operations.
When someone says that the time complexity of MergeSort is O(nlogn), they usually mean that the number of comparisons that MergeSort makes is O(nlogn). That in itself doesn’t tell us what the time complexity of any particular MergeSort might be because that would depend how much time it takes to make a comparison. In other words, the O(nlogn) refers to comparisons as the primitive operation.
The important point here is that when big-O is applied to algorithms, there is always an underlying model of computation. The claim that the time complexity of MergeSort is O(nlogn), is implicitly referencing an model of computation where a comparison takes constant time and everything else is free.
Example -
If we are sorting strings that are kk bytes long, we might take “read a byte” as a primitive operation that takes constant time with everything else being free.
In this model, MergeSort makes O(nlogn) string comparisons each of which makes O(k) byte comparisons, so the time complexity is O(k⋅nlogn). One common implementation of RadixSort will make k passes over the n strings with each pass reading one byte, and so has time complexity O(nk).
The two are not the same thing. Worst-case analysis as other have said is identifying instances for which the algorithm takes the longest to complete (i.e., takes the most number of steps), then formulating a growth function using this. One can analyze the worst-case time complexity using Big-Oh, or even other variants such as Big-Omega and Big-Theta (in fact, Big-Theta is usually what you want, though often Big-Oh is used for ease of comprehension by those not as much into theory). One important detail and why worst-case analysis is useful is that the algorithm will run no slower than it does in the worst case. Worst-case analysis is a method of analysis we use in analyzing algorithms.
Big-Oh itself is an asymptotic measure of a growth function; this can be totally independent as people can use Big-Oh to not even measure an algorithm's time complexity; its origins stem from Number Theory. You are correct to say it is the asymptotic upper bound of a growth function; but the manner you prescribe and construct the growth function comes from your analysis. The Big-Oh of a growth function itself means little to nothing without context as it only says something about the function you are analyzing. Keep in mind there can be infinitely many algorithms that could be constructed that share the same time complexity (by the definition of Big-Oh, Big-Oh is a set of growth functions).
In short, worst-case analysis is how you build your growth function, Big-Oh notation is one method of analyzing said growth function. Then, we can compare that result against other worst-case time complexities of competing algorithms for a given problem. Worst-case analysis if done correctly yields the worst-case running time if done exactly (you can cut a lot of corners and still get the correct asymptotics if you use a barometer), and using this growth function yields the worst-case time complexity of the algorithm. Big-Oh alone doesn't guarantee the worst-case time complexity as you had to make the growth function itself. For instance, I could utilize Big-Oh notation for any other kind of analysis (e.g., best case, average case). It really depends on what you're trying to capture. For instance, Big-Omega is great for lower bounds.
Imagine a hypothetical algorithm that in best case only needs to do 1 step, in the worst case needs to do n2 steps, but in average (expected) case, only needs to do n steps. With n being the input size.
For each of these 3 cases you could calculate a function that describes the time complexity of this algorithm.
1 Best case has O(1) because the function f(x)=1 is really the highest we can go, but also the lowest we can go in this case, omega(1). Since Omega is equal to O (the upper bound and lower bound), we state that this function, in the best case, behaves like theta(1).
2 We could do the same analysis for the worst case and figure out that O(n2 ) = omega(n2 ) =theta(n2 ).
3 Same counts for the average case but with theta( n ).
So in theory you could determine 3 cases of an algorithm and for those 3 cases calculate the lower/upper/thight bounds. I hope this clears things up a bit.
https://www.google.co.in/amp/s/amp.reddit.com/r/learnprogramming/comments/3qtgsh/how_is_big_o_not_the_same_as_worst_case_or_big/
Big O notation shows how an algorithm grows with respect to input size. It says nothing of which algorithm is faster because it doesn't account for constant set up time (which can dominate if you have small input sizes). So when you say
which takes O(n) execution steps
this almost doesn't mean anything. Big O doesn't say how many execution steps there are. There are C + O(n) steps (where C is a constant) and this algorithm grows at rate n depending on input size.
Big O can be used for best, worst, or average cases. Let's take sorting as an example. Bubble sort is a naive O(n^2) sorting algorithm, but when the list is sorted it takes O(n). Quicksort is often used for sorting (the GNU standard C library uses it with some modifications). It preforms at O(n log n), however this is only true if the pivot chosen splits the array in to two equal sized pieces (on average). In the worst case we get an empty array one side of the pivot and Quicksort performs at O(n^2).
As Big O shows how an algorithm grows with respect to size, you can look at any aspect of an algorithm. Its best case, average case, worst case in both time and/or memory usage. And it tells you how these grow when the input size grows - but it doesn't say which is faster.
If you deal with small sizes then Big O won't matter - but an analysis can tell you how things will go when your input sizes increase.
One example of where the worst case might not be the asymptotic limit: suppose you have an algorithm that works on the set difference between some set and the input. It might run in O(N) time, but get faster as the input gets larger and knocks more values out of the working set.
Or, to get more abstract, f(x) = 1/x for x > 0 is a decreasing O(1) function.
I'll focus on time as a fairly common item of interest, but Big-O can also be used to evaluate resource requirements such as memory. It's essential for you to realize that Big-O tells how the runtime or resource requirements of a problem scale (asymptotically) as the problem size increases. It does not give you a prediction of the actual time required. Predicting the actual runtimes would require us to know the constants and lower order terms in the prediction formula, which are dependent on the hardware, operating system, language, compiler, etc. Using Big-O allows us to discuss algorithm behaviors while sidestepping all of those dependencies.
Let's talk about how to interpret Big-O scalability using a few examples. If a problem is O(1), it takes the same amount of time regardless of the problem size. That may be a nanosecond or a thousand seconds, but in the limit doubling or tripling the size of the problem does not change the time. If a problem is O(n), then doubling or tripling the problem size will (asymptotically) double or triple the amounts of time required, respectively. If a problem is O(n^2), then doubling or tripling the problem size will (asymptotically) take 4 or 9 times as long, respectively. And so on...
Lots of algorithms have different performance for their best, average, or worst cases. Sorting provides some fairly straightforward examples of how best, average, and worst case analyses may differ.
I'll assume that you know how insertion sort works. In the worst case, the list could be reverse ordered, in which case each pass has to move the value currently being considered as far to the left as possible, for all items. That yields O(n^2) behavior. Doubling the list size will take four times as long. More likely, the list of inputs is in randomized order. In that case, on average each item has to move half the distance towards the front of the list. That's less than in the worst case, but only by a constant. It's still O(n^2), so sorting a randomized list that's twice as large as our first randomized list will quadruple the amount of time required, on average. It will be faster than the worst case (due to the constants involved), but it scales in the same way. The best case, however, is when the list is already sorted. In that case, you check each item to see if it needs to be slid towards the front, and immediately find the answer is "no," so after checking each of the n values you're done in O(n) time. Consequently, using insertion sort for an already ordered list that is twice the size only takes twice as long rather than four times as long.
You are right, in that you can say certainly say that an algorithm runs in O(f(n)) time in the best or average case. We do that all the time for, say, quicksort, which is O(N log N) on average, but only O(N^2) worst case.
Unless otherwise specified, however, when you say that an algorithm runs in O(f(n)) time, you are saying the algorithm runs in O(f(n)) time in the worst case. At least that's the way it should be. Sometimes people get sloppy, and you will often hear that a hash table is O(1) when in the worst case it is actually worse.
The other way in which a big O definition can fail to characterize the worst case is that it's an upper bound only. Any function in O(N) is also in O(N^2) and O(2^N), so we would be entirely correct to say that quicksort takes O(2^N) time. We just don't say that because it isn't useful to do so.
Big Theta and Big Omega are there to specify lower bounds and tight bounds respectively.
There are two "different" and most important tools:
the best, worst, and average-case complexity are for generating numerical function over the size of possible problem instances (e.g. f(x) = 2x^2 + 8x - 4) but it is very difficult to work precisely with these functions
big O notation extract the main point; "how efficient the algorithm is", it ignore a lot of non important things like constants and ... and give you a big picture
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).
When discussing computational complexity, it seems everyone generally goes straight to Big O.
Lets say for example I have a hybrid algorithm such as merge sort which uses insertion sort for smaller subarrays (I believe this is called tiled merge sort). It's still ultimately merge sort with O(n log n), but I want to discuss the behaviour/characteristics of the algorithm for small n, in cases where no merging actually takes place.
For all intents and purposes the tiled merge sort is insertion sort, executing exactly the same instructions for the domain of my small n. However, Big O deals with the large and asymptotic cases and discussing Big O for small n is pretty much an oxymoron. People have yelled at me for even thinking the words "behaves like an O(n^2) algorithm in such cases". What is the correct way to describe the algorithm's behaviour in cases of small n within the context of formal theoretical computational analysis? To clarify, not just in the case where n is small, but in the case where n is never big.
One might say that for such small n it doesn't matter but I'm interested in the cases where it does, for example with a large constant such as being executed many times, and where in practice it would show a clear trend and be the dominant factor. For example the initial quadratic growth seen in the graph below. I'm not dismissing Big O, more asking for a way to properly tell both sides of the story.
[EDIT]
If for "small n", constants can easily remove all trace of a growth rate then either
only the asymptotic case is discussed, in which case there is less relevance to any practical application, or
there must be a threshold at which we agree n is no longer "small".
What about the cases where n is not "small" (n is sufficiently big that the growth rate will not to affected significantly by any practical constant), but not yet big enough to show the final asymptotic growth rate so only sub growth rates are visible (for example the shape in the image above)?
Are there no practical algorithms that exhibit this behaviour? Even if there aren't, theoretical discussion should still be possible. Do we measure instead of discussing the theory purely because that's "what one should do"? If some behaviour is observed in all practical cases, why can't there be theory that's meaningful?
Let me turn the question around the other way. I have a graph that shows segmented super-linear steps. It sounds like many people would say "this is a pure coincidence, it could be any shape imaginable" (at the extreme of course) and wouldn't bat an eyelid if it were a sine wave instead. I know in many cases the shape could be hidden by constants, but here it's quite obvious. How can I give a formal explanation of why the graph produces this shape?
I particularly like #Sneftel's words "imprecise but useful guidance".
I know Big O and asymptotic analysis isn't applicable. What is? How far can I take it?
Discuss in chat
For small n, computation complexity - how things change as n increases towards infinity - isn't meaningful as other effects dominate.
Papers I've seen which discuss behaviour for small values of n do so by measuring the algorithms on real systems, and discuss how the algorithms perform in practice rather than from a theoretical viewpoint. For example, for the graph you've added to your post I would say 'this graph demonstrates an O(N) asymptotic behaviour overall, but the growth within each tile is bounded quadratic'.
I don't know of a situation where a discussion of such behaviour from a theoretical viewpoint would be meaningful - it is well known that for small n the practical effects outweigh the effects of scaling.
It's important to remember that asymptotic analysis is an analytic simplification, not a mandate for analyzing algorithms. Take selection sort, for instance. Yes, it executes in O(n^2) time. But it is also true that it performs precisely n*(n-1)/2 comparisons, and n-1-k swaps, where k is the number of elements (other than the maximum) which start in the correct position. Asymptotic analysis is a tool for simplifying the (otherwise generally impractical) task of performance analysis, and one we can put aside if we're not interested in the "really big n" segment.
And you can choose how you express your bounds. Say a function requires precisely n + floor(0.01*2^n) operations. That's exponential time, of course. But one can also say "for data sizes up to 10 elements, this algorithm requires between n and 2*n operations." The latter describes not the shape of the curve, but an envelope around that curve, giving imprecise but useful guidance about the practicalities of the algorithm within a particular range of use cases.
You are right.
For small n, i.e. when only insertion sort is performed, the asymptotic behavior is quadratic O(n^2).
And for larger n, when tiled merge sort enters into play, the behavior switches to O(n.Log(n)).
There is no contradiction if you remember that every behavior has its domain of validity, before the switching threshold, let N, and after it.
In practice there will be a smooth blend between the curves around N. But in practice too, that value of N is so small that the quadratic behavior does not have enough "room" to manifest itself.
Another way to deal with this analysis is to say that N being a constant, the insertion sorts take constant time. But I would disagree to say that this is a must.
Let's unpack things a bit. Big-O is a tool for describing the growth rate of a function. One of the functions to which it is commonly applied is the worst-case running time of an algorithm on inputs of length n, executing on a particular abstract machine. We often forget about the last part because there is a large class of machines with random-access memory that can emulate one another with only constant-factor slowdown, and the class of problems solvable within a particular big-O running-time bound is equivalent across these machines.
If you want to talk about complexity on small inputs, then you need to measure constant factors. One way is to measure running times of actual implementations. People usually do this on physical machines, but if you're hardcore like Knuth, you invent your own assembly language complete with instruction timings. Another way is to measure something that's readily identifiable but also a useful proxy for the other work performed. For comparison sorts, this could be comparisons. For numerical algorithms, this is often floating-point operations.
Complexity is not about execution time for one n on one machine, so there is no need to consider it even if constant is large. Complexity tells you how the size of the input affects execution time. For small n, you can treat execution time as constant. This is the one side.
From the second side you are saying that:
You have a hybrid algorithm working in O(n log n) for n larger than some k and O(n^2) for n smaller than k.
The constant k is so large that algorithm works slowly.
There is no sense in such algorithm, because you could easily improve it.
Lets take Red-black tree. Operations on this tree are performed in O(n log n) time complexity, but there is a large constant. So, on normal machines, it could work slowly (i.e. slower than simpler solutions) in some cases. There is no need to consider it in analyzing complexity. You need to consider it when you are implementing it in your system: you need to check if it's the best choice considering the actual machine(s) on which it will be working and what problems it will be solving.
Read Knuth's "The Art of Computer Programming series", starting with "Volume 1. Fundamental Algorithms", section "1.2.10: Analysis of an Algorithm". There he shows (and in all the rest of his seminal work) how exact analysis can be conducted for arbitrary problem sizes, using a suitable reference machine, by taking a detailed census of every processor instruction.
Such analyses have to take into account not only the problem size, but also any relevant aspect of the input data distribution which will influence the running time. For simplification, the analysis are often limited to the study of the worst case, the expected case or the output-sensitive case, rather than a general statistical characterization. And for further simplification, asymptotic analysis is used.
Not counting the fact that except for trivial algorithms the detailed approach is mathematically highly challenging, it has become unrealistic on modern machines. Indeed, it relies on a processor behavior similar to the so-called RAM model, which assumes constant time per instruction and per memory access (http://en.wikipedia.org/wiki/Random-access_machine). Except maybe for special hardware combined to a hard real-time OS, these assumptions are nowadays completely wrong.
When you have an algorithm with a time complexity say O(n^2).And you also have an another algorithm with a time complexity, say O(n).Then from these two time complexity you can't conclude that the latter algorithm is faster than the former one for all input values.You can only say latter algorithm is asymptotically(means for sufficiently large input values)faster than the former one.Here you have to keep in mind the fact that in case of asymptotic notations constant factors are generally ignored to increase the understand-ability of the time complexity of the algorithm.As 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.
Asymptotic notation does not describe the actual running-time of an algorithm.Actual running time is dependent on machine as different machine has different architecture and different Instruction Cycle Execution time.Asymptotic notation just describes asymptotically how fast an algorithm is with respect to other algorithms.But it does not describe the behavior of the algorithm in case of small input values(n<=no).The value of no (threshold) is dependent on the hidden constant factors and lower order terms.And hidden constant factors are dependent on the machine on which it will be executed.
Whenever I consider algorithms/data structures I tend to replace the log(N) parts by constants. Oh, I know log(N) diverges - but does it matter in real world applications?
log(infinity) < 100 for all practical purposes.
I am really curious for real world examples where this doesn't hold.
To clarify:
I understand O(f(N))
I am curious about real world examples where the asymptotic behaviour matters more than the constants of the actual performance.
If log(N) can be replaced by a constant it still can be replaced by a constant in O( N log N).
This question is for the sake of (a) entertainment and (b) to gather arguments to use if I run (again) into a controversy about the performance of a design.
Big O notation tells you about how your algorithm changes with growing input. O(1) tells you it doesn't matter how much your input grows, the algorithm will always be just as fast. O(logn) says that the algorithm will be fast, but as your input grows it will take a little longer.
O(1) and O(logn) makes a big diference when you start to combine algorithms.
Take doing joins with indexes for example. If you could do a join in O(1) instead of O(logn) you would have huge performance gains. For example with O(1) you can join any amount of times and you still have O(1). But with O(logn) you need to multiply the operation count by logn each time.
For large inputs, if you had an algorithm that was O(n^2) already, you would much rather do an operation that was O(1) inside, and not O(logn) inside.
Also remember that Big-O of anything can have a constant overhead. Let's say that constant overhead is 1 million. With O(1) that constant overhead does not amplify the number of operations as much as O(logn) does.
Another point is that everyone thinks of O(logn) representing n elements of a tree data structure for example. But it could be anything including bytes in a file.
I think this is a pragmatic approach; O(logN) will never be more than 64. In practice, whenever terms get as 'small' as O(logN), you have to measure to see if the constant factors win out. See also
Uses of Ackermann function?
To quote myself from comments on another answer:
[Big-Oh] 'Analysis' only matters for factors
that are at least O(N). For any
smaller factor, big-oh analysis is
useless and you must measure.
and
"With O(logN) your input size does
matter." This is the whole point of
the question. Of course it matters...
in theory. The question the OP asks
is, does it matter in practice? I
contend that the answer is no, there
is not, and never will be, a data set
for which logN will grow so fast as to
always be beaten a constant-time
algorithm. Even for the largest
practical dataset imaginable in the
lifetimes of our grandchildren, a logN
algorithm has a fair chance of beating
a constant time algorithm - you must
always measure.
EDIT
A good talk:
http://www.infoq.com/presentations/Value-Identity-State-Rich-Hickey
about halfway through, Rich discusses Clojure's hash tries, which are clearly O(logN), but the base of the logarithm is large and so the depth of the trie is at most 6 even if it contains 4 billion values. Here "6" is still an O(logN) value, but it is an incredibly small value, and so choosing to discard this awesome data structure because "I really need O(1)" is a foolish thing to do. This emphasizes how most of the other answers to this question are simply wrong from the perspective of the pragmatist who wants their algorithm to "run fast" and "scale well", regardless of what the "theory" says.
EDIT
See also
http://queue.acm.org/detail.cfm?id=1814327
which says
What good is an O(log2(n)) algorithm
if those operations cause page faults
and slow disk operations? For most
relevant datasets an O(n) or even an
O(n^2) algorithm, which avoids page
faults, will run circles around it.
(but go read the article for context).
This is a common mistake - remember Big O notation is NOT telling you about the absolute performance of an algorithm at a given value, it's simply telling you the behavior of an algorithm as you increase the size of the input.
When you take it in that context it becomes clear why an algorithm A ~ O(logN) and an algorithm B ~ O(1) algorithm are different:
if I run A on an input of size a, then on an input of size 1000000*a, I can expect the second input to take log(1,000,000) times as long as the first input
if I run B on an input of size a, then on an input of size 1000000*a, I can expect the second input to take about the same amount of time as the first input
EDIT: Thinking over your question some more, I do think there's some wisdom to be had in it. While I would never say it's correct to say O(lgN) == O(1), It IS possible that an O(lgN) algorithm might be used over an O(1) algorithm. This draws back to the point about absolute performance above: Just knowing one algorithm is O(1) and another algorithm is O(lgN) is NOT enough to declare you should use the O(1) over the O(lgN), it's certainly possible given your range of possible inputs an O(lgN) might serve you best.
You asked for a real-world example. I'll give you one. Computational biology. One strand of DNA encoded in ASCII is somewhere on the level of gigabytes in space. A typical database will obviously have many thousands of such strands.
Now, in the case of an indexing/searching algorithm, that log(n) multiple makes a large difference when coupled with constants. The reason why? This is one of the applications where the size of your input is astronomical. Additionally, the input size will always continue to grow.
Admittedly, these type of problems are rare. There are only so many applications this large. In those circumstances, though... it makes a world of difference.
Equality, the way you're describing it, is a common abuse of notation.
To clarify: we usually write f(x) = O(logN) to imply "f(x) is O(logN)".
At any rate, O(1) means a constant number of steps/time (as an upper bound) to perform an action regardless of how large the input set is. But for O(logN), number of steps/time still grows as a function of the input size (the logarithm of it), it just grows very slowly. For most real world applications you may be safe in assuming that this number of steps will not exceed 100, however I'd bet there are multiple examples of datasets large enough to mark your statement both dangerous and void (packet traces, environmental measurements, and many more).
For small enough N, O(N^N) can in practice be replaced with 1. Not O(1) (by definition), but for N=2 you can see it as one operation with 4 parts, or a constant-time operation.
What if all operations take 1hour? The difference between O(log N) and O(1) is then large, even with small N.
Or if you need to run the algorithm ten million times? Ok, that took 30minutes, so when I run it on a dataset a hundred times as large it should still take 30minutes because O(logN) is "the same" as O(1).... eh...what?
Your statement that "I understand O(f(N))" is clearly false.
Real world applications, oh... I don't know.... EVERY USE OF O()-notation EVER?
Binary search in sorted list of 10 million items for example. It's the very REASON we use hash tables when the data gets big enough. If you think O(logN) is the same as O(1), then why would you EVER use a hash instead of a binary tree?
As many have already said, for the real world, you need to look at the constant factors first, before even worrying about factors of O(log N).
Then, consider what you will expect N to be. If you have good reason to think that N<10, you can use a linear search instead of a binary one. That's O(N) instead of O(log N), which according to your lights would be significant -- but a linear search that moves found elements to the front may well outperform a more complicated balanced tree, depending on the application.
On the other hand, note that, even if log N is not likely to exceed 50, a performance factor of 10 is really huge -- if you're compute-bound, a factor like that can easily make or break your application. If that's not enough for you, you'll frequently see factors of (log N)^2 or (logN)^3 in algorithms, so even if you think you can ignore one factor of (log N), that doesn't mean you can ignore more of them.
Finally, note that the simplex algorithm for linear programming has a worst case performance of O(2^n). However, for practical problems, the worst case never comes up; in practice, the simplex algorithm is fast, relatively simple, and consequently very popular.
About 30 years ago, someone developed a polynomial-time algorithm for linear programming, but it was not initially practical because the result was too slow.
Nowadays, there are practical alternative algorithms for linear programming (with polynomial-time wost-case, for what that's worth), which can outperform the simplex method in practice. But, depending on the problem, the simplex method is still competitive.
The observation that O(log n) is oftentimes indistinguishable from O(1) is a good one.
As a familiar example, suppose we wanted to find a single element in a sorted array of one 1,000,000,000,000 elements:
with linear search, the search takes on average 500,000,000,000 steps
with binary search, the search takes on average 40 steps
Suppose we added a single element to the array we are searching, and now we must search for another element:
with linear search, the search takes on average 500,000,000,001 steps (indistinguishable change)
with binary search, the search takes on average 40 steps (indistinguishable change)
Suppose we doubled the number of elements in the array we are searching, and now we must search for another element:
with linear search, the search takes on average 1,000,000,000,000 steps (extraordinarily noticeable change)
with binary search, the search takes on average 41 steps (indistinguishable change)
As we can see from this example, for all intents and purposes, an O(log n) algorithm like binary search is oftentimes indistinguishable from an O(1) algorithm like omniscience.
The takeaway point is this: *we use O(log n) algorithms because they are often indistinguishable from constant time, and because they often perform phenomenally better than linear time algorithms.
Obviously, these examples assume reasonable constants. Obviously, these are generic observations and do not apply to all cases. Obviously, these points apply at the asymptotic end of the curve, not the n=3 end.
But this observation explains why, for example, we use such techniques as tuning a query to do an index seek rather than a table scan - because an index seek operates in nearly constant time no matter the size of the dataset, while a table scan is crushingly slow on sufficiently large datasets. Index seek is O(log n).
You might be interested in Soft-O, which ignores logarithmic cost. Check this paragraph in Wikipedia.
What do you mean by whether or not it "matters"?
If you're faced with the choice of an O(1) algorithm and a O(lg n) one, then you should not assume they're equal. You should choose the constant-time one. Why wouldn't you?
And if no constant-time algorithm exists, then the logarithmic-time one is usually the best you can get. Again, does it then matter? You just have to take the fastest you can find.
Can you give me a situation where you'd gain anything by defining the two as equal? At best, it'd make no difference, and at worst, you'd hide some real scalability characteristics. Because usually, a constant-time algorithm will be faster than a logarithmic one.
Even if, as you say, lg(n) < 100 for all practical purposes, that's still a factor 100 on top of your other overhead. If I call your function, N times, then it starts to matter whether your function runs logarithmic time or constant, because the total complexity is then O(n lg n) or O(n).
So rather than asking if "it matters" that you assume logarithmic complexity to be constant in "the real world", I'd ask if there's any point in doing that.
Often you can assume that logarithmic algorithms are fast enough, but what do you gain by considering them constant?
O(logN)*O(logN)*O(logN) is very different. O(1) * O(1) * O(1) is still constant.
Also a simple quicksort-style O(nlogn) is different than O(n O(1))=O(n). Try sorting 1000 and 1000000 elements. The latter isn't 1000 times slower, it's 2000 times, because log(n^2)=2log(n)
The title of the question is misleading (well chosen to drum up debate, mind you).
O(log N) == O(1) is obviously wrong (and the poster is aware of this). Big O notation, by definition, regards asymptotic analysis. When you see O(N), N is taken to approach infinity. If N is assigned a constant, it's not Big O.
Note, this isn't just a nitpicky detail that only theoretical computer scientists need to care about. All of the arithmetic used to determine the O function for an algorithm relies on it. When you publish the O function for your algorithm, you might be omitting a lot of information about it's performance.
Big O analysis is cool, because it lets you compare algorithms without getting bogged down in platform specific issues (word sizes, instructions per operation, memory speed versus disk speed). When N goes to infinity, those issues disappear. But when N is 10000, 1000, 100, those issues, along with all of the other constants that we left out of the O function, start to matter.
To answer the question of the poster: O(log N) != O(1), and you're right, algorithms with O(1) are sometimes not much better than algorithms with O(log N), depending on the size of the input, and all of those internal constants that got omitted during Big O analysis.
If you know you're going to be cranking up N, then use Big O analysis. If you're not, then you'll need some empirical tests.
In theory
Yes, in practical situations log(n) is bounded by a constant, we'll say 100. However, replacing log(n) by 100 in situations where it's correct is still throwing away information, making the upper bound on operations that you have calculated looser and less useful. Replacing an O(log(n)) by an O(1) in your analysis could result in your large n case performing 100 times worse than you expected based on your small n case. Your theoretical analysis could have been more accurate and could have predicted an issue before you'd built the system.
I would argue that the practical purpose of big-O analysis is to try and predict the execution time of your algorithm as early as possible. You can make your analysis easier by crossing out the log(n) terms, but then you've reduced the predictive power of the estimate.
In practice
If you read the original papers by Larry Page and Sergey Brin on the Google architecture, they talk about using hash tables for everything to ensure that e.g. the lookup of a cached web page only takes one hard-disk seek. If you used B-tree indices to lookup you might need four or five hard-disk seeks to do an uncached lookup [*]. Quadrupling your disk requirements on your cached web page storage is worth caring about from a business perspective, and predictable if you don't cast out all the O(log(n)) terms.
P.S. Sorry for using Google as an example, they're like Hitler in the computer science version of Godwin's law.
[*] Assuming 4KB reads from disk, 100bn web pages in the index, ~ 16 bytes per key in a B-tree node.
As others have pointed out, Big-O tells you about how the performance of your problem scales. Trust me - it matters. I have encountered several times algorithms that were just terrible and failed to meet the customers demands because they were too slow. Understanding the difference and finding an O(1) solution is a lot of times a huge improvement.
However, of course, that is not the whole story - for instance, you may notice that quicksort algorithms will always switch to insertion sort for small elements (Wikipedia says 8 - 20) because of the behaviour of both algorithms on small datasets.
So it's a matter of understanding what tradeoffs you will be doing which involves a thorough understanding of the problem, the architecture, & experience to understand which to use, and how to adjust the constants involved.
No one is saying that O(1) is always better than O(log N). However, I can guarantee you that an O(1) algorithm will also scale way better, so even if you make incorrect assumptions about how many users will be on the system, or the size of the data to process, it won't matter to the algorithm.
Yes, log(N) < 100 for most practical purposes, and No, you can not always replace it by constant.
For example, this may lead to serious errors in estimating performance of your program. If O(N) program processed array of 1000 elements in 1 ms, then you are sure it will process 106 elements in 1 second (or so). If, though, the program is O(N*logN), then it will take it ~2 secs to process 106 elements. This difference may be crucial - for example, you may think you've got enough server power because you get 3000 requests per hour and you think your server can handle up to 3600.
Another example. Imagine you have function f() working in O(logN), and on each iteration calling function g(), which works in O(logN) as well. Then, if you replace both logs by constants, you think that your program works in constant time. Reality will be cruel though - two logs may give you up to 100*100 multiplicator.
The rules of determining the Big-O notation are simpler when you don't decide that O(log n) = O(1).
As krzysio said, you may accumulate O(log n)s and then they would make a very noticeable difference. Imagine you do a binary search: O(log n) comparisons, and then imagine that each comparison's complexity O(log n). If you neglect both you get O(1) instead of O(log2n). Similarly you may somehow arrive at O(log10n) and then you'll notice a big difference for not too large "n"s.
Assume that in your entire application, one algorithm accounts for 90% of the time the user waits for the most common operation.
Suppose in real time the O(1) operation takes a second on your architecture, and the O(logN) operation is basically .5 seconds * log(N). Well, at this point I'd really like to draw you a graph with an arrow at the intersection of the curve and the line, saying, "It matters right here." You want to use the log(N) op for small datasets and the O(1) op for large datasets, in such a scenario.
Big-O notation and performance optimization is an academic exercise rather than delivering real value to the user for operations that are already cheap, but if it's an expensive operation on a critical path, then you bet it matters!
For any algorithm that can take inputs of different sizes N, the number of operations it takes is upper-bounded by some function f(N).
All big-O tells you is the shape of that function.
O(1) means there is some number A such that f(N) < A for large N.
O(N) means there is some A such that f(N) < AN for large N.
O(N^2) means there is some A such that f(N) < AN^2 for large N.
O(log(N)) means there is some A such that f(N) < AlogN for large N.
Big-O says nothing about how big A is (i.e. how fast the algorithm is), or where these functions cross each other. It only says that when you are comparing two algorithms, if their big-Os differ, then there is a value of N (which may be small or it may be very large) where one algorithm will start to outperform the other.
you are right, in many cases it does not matter for pracitcal purposes. but the key question is "how fast GROWS N". most algorithms we know of take the size of the input, so it grows linearily.
but some algorithms have the value of N derived in a complex way. if N is "the number of possible lottery combinations for a lottery with X distinct numbers" it suddenly matters if your algorithm is O(1) or O(logN)
Big-OH tells you that one algorithm is faster than another given some constant factor. If your input implies a sufficiently small constant factor, you could see great performance gains by going with a linear search rather than a log(n) search of some base.
O(log N) can be misleading. Take for example the operations on Red-Black trees.
The operations are O(logN) but rather complex, which means many low level operations.
Whenever N is the amount of objects that is stored in some kind of memory, you're correct. After all, a binary search through EVERY byte representable by a 64-bit pointer can be achieved in just 64 steps. Actually, it's possible to do a binary search of all Planck volumes in the observable universe in just 618 steps.
So in almost all cases, it's safe to approximate O(log N) with O(N) as long as N is (or could be) a physical quantity, and we know for certain that as long as N is (or could be) a physical quantity, then log N < 618
But that is assuming N is that. It may represent something else. Note that it's not always clear what it is. Just as an example, take matrix multiplication, and assume square matrices for simplicity. The time complexity for matrix multiplication is O(N^3) for a trivial algorithm. But what is N here? It is the side length. It is a reasonable way of measuring the input size, but it would also be quite reasonable to use the number of elements in the matrix, which is N^2. Let M=N^2, and now we can say that the time complexity for trivial matrix multiplication is O(M^(3/2)) where M is the number of elements in a matrix.
Unfortunately, I don't have any real world problem per se, which was what you asked. But at least I can make up something that makes some sort of sense:
Let f(S) be a function that returns the sum of the hashes of all the elements in the power set of S. Here is some pesudo:
f(S):
ret = 0
for s = powerset(S))
ret += hash(s)
Here, hash is simply the hash function, and powerset is a generator function. Each time it's called, it will generate the next (according to some order) subset of S. A generator is necessary, because we would not be able to store the lists for huge data otherwise. Btw, here is a python example of such a power set generator:
def powerset(seq):
"""
Returns all the subsets of this set. This is a generator.
"""
if len(seq) <= 1:
yield seq
yield []
else:
for item in powerset(seq[1:]):
yield [seq[0]]+item
yield item
https://www.technomancy.org/python/powerset-generator-python/
So what is the time complexity for f? As with the matrix multiplication, we can choose N to represent many things, but at least two makes a lot of sense. One is number of elements in S, in which case the time complexity is O(2^N), but another sensible way of measuring it is that N is the number of element in the power set of S. In this case the time complexity is O(N)
So what will log N be for sensible sizes of S? Well, list with a million elements are not unusual. If n is the size of S and N is the size of P(S), then N=2^n. So O(log N) = O(log 2^n) = O(n * log 2) = O(n)
In this case it would matter, because it's rare that O(n) == O(log n) in the real world.
I do not believe algorithms where you can freely choose between O(1) with a large constant and O(logN) really exists. If there is N elements to work with at the beginning, it is just plain impossible to make it O(1), the only thing that is possible is move your N to some other part of your code.
What I try to say is that in all real cases I know off you have some space/time tradeoff, or some pre-treatment such as compiling data to a more efficient form.
That is, you do not really go O(1), you just move the N part elsewhere. Either you exchange performance of some part of your code with some memory amount either you exchange performance of one part of your algorithm with another one. To stay sane you should always look at the larger picture.
My point is that if you have N items they can't disappear. In other words you can choose between inefficient O(n^2) algorithms or worse and O(n.logN) : it's a real choice. But you never really go O(1).
What I try to point out is that for every problem and initial data state there is a 'best' algorithm. You can do worse but never better. With some experience you can have a good guessing of what is this intrisic complexity. Then if your overall treatment match that complexity you know you have something. You won't be able to reduce that complexity, but only to move it around.
If problem is O(n) it won't become O(logN) or O(1), you'll merely add some pre-treatment such that the overall complexity is unchanged or worse, and potentially a later step will be improved. Say you want the smaller element of an array, you can search in O(N) or sort the array using any common O(NLogN) sort treatment then have the first using O(1).
Is it a good idea to do that casually ? Only if your problem asked also for second, third, etc. elements. Then your initial problem was truly O(NLogN), not O(N).
And it's not the same if you wait ten times or twenty times longer for your result because you simplified saying O(1) = O(LogN).
I'm waiting for a counter-example ;-) that is any real case where you have choice between O(1) and O(LogN) and where every O(LogN) step won't compare to the O(1). All you can do is take a worse algorithm instead of the natural one or move some heavy treatment to some other part of the larger pictures (pre-computing results, using storage space, etc.)
Let's say you use an image-processing algorithm that runs in O(log N), where N is the number of images. Now... stating that it runs in constant time would make one believe that no matter how many images there are, it would still complete its task it about the same amount of time. If running the algorithm on a single image would hypothetically take a whole day, and assuming that O(logN) will never be more than 100... imagine the surprise of that person that would try to run the algorithm on a very large image database - he would expect it to be done in a day or so... yet it'll take months for it to finish.
What are some examples where Big-O notation[1] fails in practice?
That is to say: when will the Big-O running time of algorithms predict algorithm A to be faster than algorithm B, yet in practice algorithm B is faster when you run it?
Slightly broader: when do theoretical predictions about algorithm performance mismatch observed running times? A non-Big-O prediction might be based on the average/expected number of rotations in a search tree, or the number of comparisons in a sorting algorithm, expressed as a factor times the number of elements.
Clarification:
Despite what some of the answers say, the Big-O notation is meant to predict algorithm performance. That said, it's a flawed tool: it only speaks about asymptotic performance, and it blurs out the constant factors. It does this for a reason: it's meant to predict algorithmic performance independent of which computer you execute the algorithm on.
What I want to know is this: when do the flaws of this tool show themselves? I've found Big-O notation to be reasonably useful, but far from perfect. What are the pitfalls, the edge cases, the gotchas?
An example of what I'm looking for: running Dijkstra's shortest path algorithm with a Fibonacci heap instead of a binary heap, you get O(m + n log n) time versus O((m+n) log n), for n vertices and m edges. You'd expect a speed increase from the Fibonacci heap sooner or later, yet said speed increase never materialized in my experiments.
(Experimental evidence, without proof, suggests that binary heaps operating on uniformly random edge weights spend O(1) time rather than O(log n) time; that's one big gotcha for the experiments. Another one that's a bitch to count is the expected number of calls to DecreaseKey).
[1] Really it isn't the notation that fails, but the concepts the notation stands for, and the theoretical approach to predicting algorithm performance. </anti-pedantry>
On the accepted answer:
I've accepted an answer to highlight the kind of answers I was hoping for. Many different answers which are just as good exist :) What I like about the answer is that it suggests a general rule for when Big-O notation "fails" (when cache misses dominate execution time) which might also increase understanding (in some sense I'm not sure how to best express ATM).
It fails in exactly one case: When people try to use it for something it's not meant for.
It tells you how an algorithm scales. It does not tell you how fast it is.
Big-O notation doesn't tell you which algorithm will be faster in any specific case. It only tells you that for sufficiently large input, one will be faster than the other.
When N is small, the constant factor dominates. Looking up an item in an array of five items is probably faster than looking it up in a hash table.
Short answer: When n is small. The Traveling Salesman Problem is quickly solved when you only have three destinations (however, finding the smallest number in a list of a trillion elements can last a while, although this is O(n). )
the canonical example is Quicksort, which has a worst time of O(n^2), while Heapsort's is O(n logn). in practice however, Quicksort is usually faster then Heapsort. why? two reasons:
each iteration in Quicksort is a lot simpler than Heapsort. Even more, it's easily optimized by simple cache strategies.
the worst case is very hard to hit.
But IMHO, this doesn't mean 'big O fails' in any way. the first factor (iteration time) is easy to incorporate into your estimates. after all, big O numbers should be multiplied by this almost-constant facto.
the second factor melts away if you get the amortized figures instead of average. They can be harder to estimate, but tell a more complete story
One area where Big O fails is memory access patterns. Big O only counts operations that need to be performed - it can't keep track if an algorithm results in more cache misses or data that needs to be paged in from disk. For small N, these effects will typically dominate. For instance, a linear search through an array of 100 integers will probably beat out a search through a binary tree of 100 integers due to memory accesses, even though the binary tree will most likely require fewer operations. Each tree node would result in a cache miss, whereas the linear search would mostly hit the cache for each lookup.
Big-O describes the efficiency/complexity of the algorithm and not necessarily the running time of the implementation of a given block of code. This doesn't mean Big-O fails. It just means that it's not meant to predict running time.
Check out the answer to this question for a great definition of Big-O.
For most algorithms there is an "average case" and a "worst case". If your data routinely falls into the "worst case" scenario, it is possible that another algorithm, while theoretically less efficient in the average case, might prove more efficient for your data.
Some algorithms also have best cases that your data can take advantage of. For example, some sorting algorithms have a terrible theoretical efficiency, but are actually very fast if the data is already sorted (or nearly so). Another algorithm, while theoretically faster in the general case, may not take advantage of the fact that the data is already sorted and in practice perform worse.
For very small data sets sometimes an algorithm that has a better theoretical efficiency may actually be less efficient because of a large "k" value.
One example (that I'm not an expert on) is that simplex algorithms for linear programming have exponential worst-case complexity on arbitrary inputs, even though they perform well in practice. An interesting solution to this is considering "smoothed complexity", which blends worst-case and average-case performance by looking at small random perturbations of arbitrary inputs.
Spielman and Teng (2004) were able to show that the shadow-vertex simplex algorithm has polynomial smoothed complexity.
Big O does not say e.g. that algorithm A runs faster than algorithm B. It can say that the time or space used by algorithm A grows at a different rate than algorithm B, when the input grows. However, for any specific input size, big O notation does not say anything about the performance of one algorithm relative to another.
For example, A may be slower per operation, but have a better big-O than B. B is more performant for smaller input, but if the data size increases, there will be some cut-off point where A becomes faster. Big-O in itself does not say anything about where that cut-off point is.
The general answer is that Big-O allows you to be really sloppy by hiding the constant factors. As mentioned in the question, the use of Fibonacci Heaps is one example. Fibonacci Heaps do have great asymptotic runtimes, but in practice the constants factors are way too large to be useful for the sizes of data sets encountered in real life.
Fibonacci Heaps are often used in proving a good lower bound for asymptotic complexity of graph-related algorithms.
Another similar example is the Coppersmith-Winograd algorithm for matrix multiplication. It is currently the algorithm with the fastest known asymptotic running time for matrix multiplication, O(n2.376). However, its constant factor is far too large to be useful in practice. Like Fibonacci Heaps, it's frequently used as a building block in other algorithms to prove theoretical time bounds.
This somewhat depends on what the Big-O is measuring - when it's worst case scenarios, it will usually "fail" in that the runtime performance will be much better than the Big-O suggests. If it's average case, then it may be much worse.
Big-O notation typically "fails" if the input data to the algorithm has some prior information. Often, the Big-O notation refers to the worst case complexity - which will often happen if the data is either completely random or completely non-random.
As an example, if you feed data to an algorithm that's profiled and the big-o is based on randomized data, but your data has a very well-defined structure, your result times may be much faster than expected. On the same token, if you're measuring average complexity, and you feed data that is horribly randomized, the algorithm may perform much worse than expected.
Small N - And for todays computers, 100 is likely too small to worry.
Hidden Multipliers - IE merge vs quick sort.
Pathological Cases - Again, merge vs quick
One broad area where Big-Oh notation fails is when the amount of data exceeds the available amount of RAM.
Using sorting as an example, the amount of time it takes to sort is not dominated by the number of comparisons or swaps (of which there are O(n log n) and O(n), respectively, in the optimal case). The amount of time is dominated by the number of disk operations: block writes and block reads.
To better analyze algorithms which handle data in excess of available RAM, the I/O-model was born, where you count the number of disk reads. In that, you consider three parameters:
The number of elements, N;
The amount of memory (RAM), M (the number of elements that can be in memory); and
The size of a disk block, B (the number of elements per block).
Notably absent is the amount of disk space; this is treated as if it were infinite. A typical extra assumption is that M > B2.
Continuing the sorting example, you typically favor merge sort in the I/O case: divide the elements into chunks of size θ(M) and sort them in memory (with, say, quicksort). Then, merge θ(M/B) of them by reading the first block from each chunk into memory, stuff all the elements into a heap, and repeatedly pick the smallest element until you have picked B of them. Write this new merge block out and continue. If you ever deplete one of the blocks you read into memory, read a new block from the same chunk and put it into the heap.
(All expressions should be read as being big θ). You form N/M sorted chunks which you then merge. You merge log (base M/B) of N/M times; each time you read and write all the N/B blocks, so it takes you N/B * (log base M/B of N/M) time.
You can analyze in-memory sorting algorithms (suitably modified to include block reads and block writes) and see that they're much less efficient than the merge sort I've presented.
This knowledge is courtesy of my I/O-algorithms course, by Arge and Brodal (http://daimi.au.dk/~large/ioS08/); I also conducted experiments which validate the theory: heap sort takes "almost infinite" time once you exceed memory. Quick sort becomes unbearably slow, merge sort barely bearably slow, I/O-efficient merge sort performs well (the best of the bunch).
I've seen a few cases where, as the data set grew, the algorithmic complexity became less important than the memory access pattern. Navigating a large data structure with a smart algorithm can, in some cases, cause far more page faults or cache misses, than an algorithm with a worse big-O.
For small n, two algorithms may be comparable. As n increases, the smarter algorithm outperforms. But, at some point, n grows big enough that the system succumbs to memory pressure, in which case the "worse" algorithm may actually perform better because the constants are essentially reset.
This isn't particularly interesting, though. By the time you reach this inversion point, the performance of both algorithms is usually unacceptable, and you have to find a new algorithm that has a friendlier memory access pattern AND a better big-O complexity.
This question is like asking, "When does a person's IQ fail in practice?" It's clear that having a high IQ does not mean you'll be successful in life and having a low IQ does not mean you'll perish. Yet, we measure IQ as a means of assessing potential, even if its not an absolute.
In algorithms, the Big-Oh notation gives you the algorithm's IQ. It doesn't necessarily mean that the algorithm will perform best for your particular situation, but there's some mathematical basis that says this algorithm has some good potential. If Big-Oh notation were enough to measure performance you would see a lot more of it and less runtime testing.
Think of Big-Oh as a range instead of a specific measure of better-or-worse. There's best case scenarios and worst case scenarios and a huge set of scenarios in between. Choose your algorithms by how well they fit within the Big-Oh range, but don't rely on the notation as an absolute for measuring performance.
When your data doesn't fit the model, big-o notation will still work, but you're going to see an overlap from best and worst case scenarios.
Also, some operations are tuned for linear data access vs. random data access, so one algorithm while superior in terms of cycles, might be doggedly slow if the method of calling it changes from design. Similarly, if an algorithm causes page/cache misses due to the way it access memory, Big-O isn't going to going to give an accurate estimate of the cost of running a process.
Apparently, as I've forgotten, also when N is small :)
The short answer: always on modern hardware when you start using a lot of memory. The textbooks assume memory access is uniform, and it is no longer. You can of course do Big O analysis for a non-uniform access model, but that is somewhat more complex.
The small n cases are obvious but not interesting: fast enough is fast enough.
In practice I've had problems using the standard collections in Delphi, Java, C# and Smalltalk with a few million objects. And with smaller ones where the dominant factor proved to be the hash function or the compare
Robert Sedgewick talks about shortcomings of the big-O notation in his Coursera course on Analysis of Algorithms. He calls particularly egregious examples galactic algorithms because while they may have a better complexity class than their predecessors, it would take inputs of astronomical sizes for it to show in practice.
https://www.cs.princeton.edu/~rs/talks/AlgsMasses.pdf
Big O and its brothers are used to compare asymptotic mathematical function growth. I would like to emphasize on the mathematical part. Its entirely about being able reduce your problem to a function where the input grows a.k.a scales. It gives you a nice plot where your input (x axis) related to the number of operations performed(y-axis). This is purely based on the mathematical function and as such requires us to accurately model the algorithm used into a polynomial of sorts. Then the assumption of scaling.
Big O immediately loses its relevance when the data is finite, fixed and constant size. Which is why nearly all embedded programmers don't even bother with big O. Mathematically this will always come out to O(1) but we know that we need to optimize our code for space and Mhz timing budget at a level that big O simply doesn't work. This is optimization is on the same order where the individual components matter due to their direct performance dependence on the system.
Big O's other failure is in its assumption that hardware differences do not matter. A CPU that has a MAC, MMU and/or a bit shift low latency math operations will outperform some tasks which may be falsely identified as higher order in the asymptotic notation. This is simply because of the limitation of the model itself.
Another common case where big O becomes absolutely irrelevant is where we falsely identify the nature of the problem to be solved and end up with a binary tree when in reality the solution is actually a state machine. The entire algorithm regimen often overlooks finite state machine problems. This is because a state machine complexity grows based on the number of states and not the number of inputs or data which in most cases are constant.
The other aspect here is the memory access itself which is an extension of the problem of being disconnected from hardware and execution environment. Many times the memory optimization gives performance optimization and vice-versa. They are not necessarily mutually exclusive. These relations cannot be easily modeled into simple polynomials. A theoretically bad algorithm running on heap (region of memory not algorithm heap) data will usually outperform a theoretically good algorithm running on data in stack. This is because there is a time and space complexity to memory access and storage efficiency that is not part of the mathematical model in most cases and even if attempted to model often get ignored as lower order terms that can have high impact. This is because these will show up as a long series of lower order terms which can have a much larger impact when there are sufficiently large number of lower order terms which are ignored by the model.
Imagine n3+86n2+5*106n2+109n
It's clear that the lower order terms that have high multiples will likely together have larger significance than the highest order term which the big O model tends to ignore. It would have us ignore everything other than n3. The term "sufficiently large n' is completely abused to imagine unrealistic scenarios to justify the algorithm. For this case, n has to be so large that you will run out of physical memory long before you have to worry about the algorithm itself. The algorithm doesn't matter if you can't even store the data. When memory access is modeled in; the lower order terms may end up looking like the above polynomial with over a 100 highly scaled lower order terms. However for all practical purposes these terms are never even part of the equation that the algorithm is trying to define.
Most scientific notations are generally the description of mathematical functions and used to model something. They are tools. As such the utility of the tool is constrained and only as good as the model itself. If the model cannot describe or is an ill fit to the problem at hand, then the model simply doesn't serve the purpose. This is when a different model needs to be used and when that doesn't work, a direct approach may serve your purpose well.
In addition many of the original algorithms were models of Turing machine that has a completely different working mechanism and all computing today are RASP models. Before you go into big O or any other model, ask yourself this question first "Am I choosing the right model for the task at hand and do I have the most practically accurate mathematical function ?". If the answer is 'No', then go with your experience, intuition and ignore the fancy stuff.