Suppose that I have 2 nested for loops, and 1 array of size N as shown in my code below:
int result = 0;
for( int i = 0; i < N ; i++)
{
for( int j = i; j < N ; j++)
{
result = array[i] + array[j]; // just some funny operation
}
}
Here are 2 cases:
(1) if the constraint is that N >= 1,000,000 strictly, then we can definitely say that the time complexity is O(N^2). This is true for sure as we all know.
(2) Now, if the constraint is that N < 25 strictly, then people could probably say that because we know that definitely, N is always too small, the time complexity is estimated to be O(1) since it takes very little time to run and complete these 2 for loops WITH MODERN COMPUTERS ? Does that sound right ?
Please tell me if the value of N plays a role in deciding the outcome of the time complexity O(N) ? If yes, then how big the value N needs to be in order to play that role (1,000 ? 5,000 ? 20,000 ? 500,000 ?) In other words, what is the general rule of thumb here ?
INTERESTING THEORETICAL QUESTION: If 15 years from now, the computer is so fast that even if N = 25,000,000, these 2 for loops can be completed in 1 second. At that time, can we say that the time complexity would be O(1) even for N = 25,000,000 ? I suppose the answer would be YES at that time. Do you agree ?
tl:dr No. The value of N has no effect on time complexity. O(1) versus O(N) is a statement about "all N" or how the amount of computation increases when N increases.
Great question! It reminds me of when I was first trying to understand time complexity. I think many people have to go through a similar journey before it ever starts to make sense so I hope this discussion can help others.
First of all, your "funny operation" is actually funnier than you think since your entire nested for-loops can be replaced with:
result = array[N - 1] + array[N - 1]; // just some hilarious operation hahaha ha ha
Since result is overwritten each time, only the last iteration effects the outcome. We'll come back to this.
As far as what you're really asking here, the purpose of Big-O is to provide a meaningful way to compare algorithms in a way that is indenependent of input size and independent of the computer's processing speed. In other words, O(1) versus O(N) has nothing to with the size of N and nothing to do with how "modern" your computer is. That all effects execution time of the algorithm on a particular machine with a particular input, but does not effect time complexity, i.e. O(1) versus O(N).
It is actually a statement about the algorithm itself, so a math discussion is unavoidable, as dxiv has so graciously alluded to in his comment. Disclaimer: I'm going to omit certain nuances in the math since the critical stuff is already a lot to explain and I'll defer to the mountains of complete explanations elsewhere on the web and textbooks.
Your code is a great example to understand what Big-O does tell us. The way you wrote it, its complexity is O(N^2). That means that no matter what machine or what era you run your code in, if you were to count the number of operations the computer has to do, for each N, and graph it as a function, say f(N), there exists some quadratic function, say g(N)=9999N^2+99999N+999 that is greater than f(N) for all N.
But wait, if we just need to find big enough coefficients in order for g(N) to be an upper bound, can't we just claim that the algorithm is O(N) and find some g(N)=aN+b with gigantic enough coefficients that its an upper bound of f(N)??? THE ANSWER TO THIS IS THE MOST IMPORTANT MATH OBSERVATION YOU NEED TO UNDERSTAND TO REALLY UNDERSTAND BIG-O NOTATION. Spoiler alert. The answer is no.
For visuals, try this graph on Desmos where you can adjust the coefficients:[https://www.desmos.com/calculator/3ppk6shwem][1]
No matter what coefficients you choose, a function of the form aN^2+bN+c will ALWAYS eventually outgrow a function of the form aN+b (both having positive a). You can push a line as high as you want like g(N)=99999N+99999, but even the function f(N)=0.01N^2+0.01N+0.01 crosses that line and grows past it after N=9999900. There is no linear function that is an upper bound to a quadratic. Similarly, there is no constant function that is an upper bound to a linear function or quadratic function. Yet, we can find a quadratic upper bound to this f(N) such as h(N)=0.01N^2+0.01N+0.02, so f(N) is in O(N^2). This observation is what allows us to just say O(1) and O(N^2) without having to distinguish between O(1), O(3), O(999), O(4N+3), O(23N+2), O(34N^2+4+e^N), etc. By using phrases like "there exists a function such that" we can brush all the constant coefficients under the rug.
So having a quadratic upper bound, aka being in O(N^2), means that the function f(N) is no bigger than quadratic and in this case happens to be exactly quadratic. It sounds like this just comes down to comparing the degree of polynomials, why not just say that the algorithm is a degree-2 algorithm? Why do we need this super abstract "there exists an upper bound function such that bla bla bla..."? This is the generalization necessary for Big-O to account for non-polynomial functions, some common ones being logN, NlogN, and e^N.
For example if the number of operations required by your algorithm is given by f(N)=floor(50+50*sin(N)), we would say that it's O(1) because there is a constant function, e.g. g(N)=101 that is an upper bound to f(N). In this example, you have some bizarre algorithm with oscillating execution times, but you can convey to someone else how much it doesn't slow down for large inputs by simply saying that it's O(1). Neat. Plus we have a way to meaningfully say that this algorithm with trigonometric execution time is more efficient than one with linear complexity O(N). Neat. Notice how it doesn't matter how fast the computer is because we're not measuring in seconds, we're measuring in operations. So you can evaluate the algorithm by hand on paper and it's still O(1) even if it takes you all day.
As for the example in your question, we know it's O(N^2) because there are aN^2+bN+c operations involved for some a, b, c. It can't be O(1) because no matter what aN+b you pick, I can find a large enough input size N such that your algorithm requires more than aN+b operations. On any computer, in any time zone, with any chance of rain outside. Nothing physical effects O(1) versus O(N) versus (N^2). What changes it to O(1) is changing the algorithm itself to the one-liner that I provided above where you just add two numbers and spit out the result no matter what N is. Let's say for N=10 it takes 4 operations to do both array lookups, the addition, and the variable assignment. If you run it again on the same machine with N=10000000 it's still doing the same 4 operations. The amount of operations required by the algorithm doesn't grow with N. That's why the algorithm is O(1).
It's why problems like finding a O(NlogN) algorithm to sort an array are math problems and not nano-technology problems. Big-O doesn't even assume you have a computer with electronics.
Hopefully this rant gives you a hint as to what you don't understand so you can do more effective studying for a complete understanding. There's no way to cover everything needed in one post here. It was some good soul-searching for me, so thanks.
I have been dealing with very small data stores where it really did not matter how wasteful I was with parsing thru the data. I recently started work on a data store with records in the 100,000s, and am taking a look into optimizing my algorithms. I just reduced my time by a multiple of several hundred, and was trying to benchmark a few other solutions. I have a question about terminology:
Is there a well-defined way, like Big O notation, to say "This algorithm takes half the time as that algorithm" ?
Big O notation is a well-understood, cross-platform way to define the time complexity of an algorithm by saying things like, a binary search on an ordered table takes O(log n) time, whereas a search on an unordered table takes O(n) time. Adrian Mejia: Big O cheatsheet and examples
Big O notation (and the definition of time complexity) is about growth rates. Some algorithm that takes 2n, n, and n/2 all grow at a linear rate and are expressed by O(n). Thus, we drop the constant preceding the 'n' when we use Big O notation and accept that it it really only useful for comparing algorithms that take O(1), O(log n), O(n), O(n^2) and other exponents. StackOverflow: Why is constant always dropped from big O analysis?
The best reason I have found for this is because these constants are implementation dependent. If my WindowsXP computer from 2002 and your Windows10 computer from 2019 do that same task, it may take WindowsXP 2n time to do what your computer does in n/2 time.
Part of the optimizations that I have recently done are to the effect of the following:
I have an algorithm in my software that goes thru a list of say 100,000 data points to get max and min values. I used to iterate thru the entire list to find the max, and then iterate thru the entire list to find the min, in two different functions that were miles apart. I now iterate thru it once to find the max and the min values, and then pass around the two values until I need them. If we assume iterating thru a list is done in n time, then I used to use 2n time to iterate thru the list twice, as opposed to now doing this in n time to iterate thru the list once. It will not matter what hardware you use, 18 year old computer or brand new one. The new algorithm is done in half the time.
int minValue = int.MaxValue;
int maxValue = int.MinValue;
foreach(int entry in myList)
{
if (entry < minValue) minValue = entry;
if (entry > maxValue) maxValue = entry;
}
(If you notice it is C# / .NET and say, use LINQ instead to speed up the algorithm, you have clearly missed the point of the question)
I have not been able to find a concise, well-understood way to say this like Big O notation. Big O, Big Omega, Big Theta, little o notation all have to do with time complexity. Thus, all are only concerned with growth rates and drop any constant in from of n.
One way I thought of was to benchmark my two implementations of the algorithm to say, for 10,000 points on a Windows10 production machine, algorithm A took 15 seconds and algorithm B took 7.5 seconds. But I do not care about the timestamps, just that algorithm B runs in half the time.
I could also do away Big O notation and just say, Algorithm B requires one iteration thru the data to do the same job as algorithm A, which requires two iterations. This works, but does not use well-understood terminology. I would think well-understood terminology would be useful in white papers, where you are trying to state that your algorithm runs in 1/100 of the time of another algorithm. This need for terminology is why, I assume, people came up with Big O notation in the first place!
So, is there well-defined terminology? Or is this question silly, and anyone who wonders it should just get on with their lives?
It is possible to do this without inventing a new notation. For example, here's how Wikipedia compares the number of comparisons done by bottom-up heapsort vs. ordinary heapsort (emphasis mine):
While ordinary heapsort requires 2n log2 n + O(n) comparisons worst-case and on average, the bottom-up variant requires n log2 n + O(1) comparisons on average, and 1.5n log2 n + O(n) in the worst case.
That is, for large n, ordinary heapsort does twice as many comparisons as bottom-up heapsort in the average case. This is a slight abuse of notation because it's adding a function like n log2 n to an asymptotic term like O(1) which really represents a set of functions, but it's understood as "n log2 n plus some function in O(1)".
In the general case, we don't necessarily know what the next asymptotically smaller term should be, so instead of writing 1.5n log2 n + O(n), the weaker bound 1.5n log2 n + o(n log n) can be written using little o notation.
Note that this makes sense when we talk about the number of operations (e.g. comparisons or swaps) done by an algorithm, but static analysis cannot be used to give a non-asymptotic formula for the actual running time, because the actual running time still depends on the time it takes to perform basic operations (e.g. reading/writing memory, adding numbers), so the running time differs from the number of operations by an unknown constant factor. So, one reason for ignoring constant factors is so we can talk about running time rather than just the number of operations.
Big O, Big Omega or Big Theta notations help us reason about classes of problems and solutions. Once you have found 2 solutions to the problem within the same class, then looking at the constants is definitely desired in analysis and comparison.
Big O notations also differ for best and worst case scenarios, so further judgement and details are definitely not being looked over and bringing constants and other caveats back to the picture is not unheard of.
Therefore there is definitely value in talking about O(n) class solutions and then comparing 2 * n vs n algorithms.
Why don't you write T2 / T1 = 2 ?
This question already has answers here:
What is a plain English explanation of "Big O" notation?
(43 answers)
Closed 5 years ago.
I'm asking more about what this means to my code. I understand the concepts mathematically, I just have a hard time wrapping my head around what they mean conceptually. For example, if one were to perform an O(1) operation on a data structure, I understand that the number of operations it has to perform won't grow because there are more items. And an O(n) operation would mean that you would perform a set of operations on each element. Could somebody fill in the blanks here?
Like what exactly would an O(n^2) operation do?
And what the heck does it mean if an operation is O(n log(n))?
And does somebody have to smoke crack to write an O(x!)?
One way of thinking about it is this:
O(N^2) means for every element, you're doing something with every other element, such as comparing them. Bubble sort is an example of this.
O(N log N) means for every element, you're doing something that only needs to look at log N of the elements. This is usually because you know something about the elements that let you make an efficient choice. Most efficient sorts are an example of this, such as merge sort.
O(N!) means to do something for all possible permutations of the N elements. Traveling salesman is an example of this, where there are N! ways to visit the nodes, and the brute force solution is to look at the total cost of every possible permutation to find the optimal one.
The big thing that Big-O notation means to your code is how it will scale when you double the amount of "things" it operates on. Here's a concrete example:
Big-O | computations for 10 things | computations for 100 things
----------------------------------------------------------------------
O(1) | 1 | 1
O(log(n)) | 3 | 7
O(n) | 10 | 100
O(n log(n)) | 30 | 700
O(n^2) | 100 | 10000
So take quicksort which is O(n log(n)) vs bubble sort which is O(n^2). When sorting 10 things, quicksort is 3 times faster than bubble sort. But when sorting 100 things, it's 14 times faster! Clearly picking the fastest algorithm is important then. When you get to databases with million rows, it can mean the difference between your query executing in 0.2 seconds, versus taking hours.
Another thing to consider is that a bad algorithm is one thing that Moore's law cannot help. For example, if you've got some scientific calculation that's O(n^3) and it can compute 100 things a day, doubling the processor speed only gets you 125 things in a day. However, knock that calculation to O(n^2) and you're doing 1000 things a day.
clarification:
Actually, Big-O says nothing about comparative performance of different algorithms at the same specific size point, but rather about comparative performance of the same algorithm at different size points:
computations computations computations
Big-O | for 10 things | for 100 things | for 1000 things
----------------------------------------------------------------------
O(1) | 1 | 1 | 1
O(log(n)) | 1 | 3 | 7
O(n) | 1 | 10 | 100
O(n log(n)) | 1 | 33 | 664
O(n^2) | 1 | 100 | 10000
You might find it useful to visualize it:
Also, on LogY/LogX scale the functions n1/2, n, n2 all look like straight lines, while on LogY/X scale 2n, en, 10n are straight lines and n! is linearithmic (looks like n log n).
This might be too mathematical, but here's my try. (I am a mathematician.)
If something is O(f(n)), then it's running time on n elements will be equal to A f(n) + B (measured in, say, clock cycles or CPU operations). It's key to understanding that you also have these constants A and B, which arise from the specific implementation. B represents essentially the "constant overhead" of your operation, for example some preprocessing that you do that doesn't depend on the size of the collection. A represents the speed of your actual item-processing algorithm.
The key, though, is that you use big O notation to figure out how well something will scale. So those constants won't really matter: if you're trying to figure out how to scale from 10 to 10000 items, who cares about the constant overhead B? Similarly, other concerns (see below) will certainly outweigh the weight of the multiplicative constant A.
So the real deal is f(n). If f grows not at all with n, e.g. f(n) = 1, then you'll scale fantastically---your running time will always just be A + B. If f grows linearly with n, i.e. f(n) = n, your running time will scale pretty much as best as can be expected---if your users are waiting 10 ns for 10 elements, they'll wait 10000 ns for 10000 elements (ignoring the additive constant). But if it grows faster, like n2, then you're in trouble; things will start slowing down way too much when you get larger collections. f(n) = n log(n) is a good compromise, usually: your operation can't be so simple as to give linear scaling, but you've managed to cut things down such that it'll scale much better than f(n) = n2.
Practically, here are some good examples:
O(1): retrieving an element from an array. We know exactly where it is in memory, so we just go get it. It doesn't matter if the collection has 10 items or 10000; it's still at index (say) 3, so we just jump to location 3 in memory.
O(n): retrieving an element from a linked list. Here, A = 0.5, because on average you''ll have to go through 1/2 of the linked list before you find the element you're looking for.
O(n2): various "dumb" sorting algorithms. Because generally their strategy involves, for each element (n), you look at all the other elements (so times another n, giving n2), then position yourself in the right place.
O(n log(n)): various "smart" sorting algorithms. It turns out that you only need to look at, say, 10 elements in a 1010-element collection to intelligently sort yourself relative to everyone else in the collection. Because everyone else is also going to look at 10 elements, and the emergent behavior is orchestrated just right so that this is enough to produce a sorted list.
O(n!): an algorithm that "tries everything," since there are (proportional to) n! possible combinations of n elements that might solve a given problem. So it just loops through all such combinations, tries them, then stops whenever it succeeds.
don.neufeld's answer is very good, but I'd probably explain it in two parts: first, there's a rough hierarchy of O()'s that most algorithms fall into. Then, you can look at each of those to come up with sketches of what typical algorithms of that time complexity do.
For practical purposes, the only O()'s that ever seem to matter are:
O(1) "constant time" - the time required is independent of the size of the input. As a rough category, I would include algorithms such as hash lookups and Union-Find here, even though neither of those are actually O(1).
O(log(n)) "logarithmic" - it gets slower as you get larger inputs, but once your input gets fairly large, it won't change enough to worry about. If your runtime is ok with reasonably-sized data, you can swamp it with as much additional data as you want and it'll still be ok.
O(n) "linear" - the more input, the longer it takes, in an even tradeoff. Three times the input size will take roughly three times as long.
O(n log(n)) "better than quadratic" - increasing the input size hurts, but it's still manageable. The algorithm is probably decent, it's just that the underlying problem is more difficult (decisions are less localized with respect to the input data) than those problems that can be solved in linear time. If your input sizes are getting up there, don't assume that you could necessarily handle twice the size without changing your architecture around (eg by moving things to overnight batch computations, or not doing things per-frame). It's ok if the input size increases a little bit, though; just watch out for multiples.
O(n^2) "quadratic" - it's really only going to work up to a certain size of your input, so pay attention to how big it could get. Also, your algorithm may suck -- think hard to see if there's an O(n log(n)) algorithm that would give you what you need. Once you're here, feel very grateful for the amazing hardware we've been gifted with. Not long ago, what you are trying to do would have been impossible for all practical purposes.
O(n^3) "cubic" - not qualitatively all that different from O(n^2). The same comments apply, only more so. There's a decent chance that a more clever algorithm could shave this time down to something smaller, eg O(n^2 log(n)) or O(n^2.8...), but then again, there's a good chance that it won't be worth the trouble. (You're already limited in your practical input size, so the constant factors that may be required for the more clever algorithms will probably swamp their advantages for practical cases. Also, thinking is slow; letting the computer chew on it may save you time overall.)
O(2^n) "exponential" - the problem is either fundamentally computationally hard or you're being an idiot. These problems have a recognizable flavor to them. Your input sizes are capped at a fairly specific hard limit. You'll know quickly whether you fit into that limit.
And that's it. There are many other possibilities that fit between these (or are greater than O(2^n)), but they don't often happen in practice and they're not qualitatively much different from one of these. Cubic algorithms are already a bit of a stretch; I only included them because I've run into them often enough to be worth mentioning (eg matrix multiplication).
What's actually happening for these classes of algorithms? Well, I think you had a good start, although there are many examples that wouldn't fit these characterizations. But for the above, I'd say it usually goes something like:
O(1) - you're only looking at most at a fixed-size chunk of your input data, and possibly none of it. Example: the maximum of a sorted list.
Or your input size is bounded. Example: addition of two numbers. (Note that addition of N numbers is linear time.)
O(log n) - each element of your input tells you enough to ignore a large fraction of the rest of the input. Example: when you look at an array element in binary search, its value tells you that you can ignore "half" of your array without looking at any of it. Or similarly, the element you look at gives you enough of a summary of a fraction of the remaining input that you won't need to look at it.
There's nothing special about halves, though -- if you can only ignore 10% of your input at each step, it's still logarithmic.
O(n) - you do some fixed amount of work per input element. (But see below.)
O(n log(n)) - there are a few variants.
You can divide the input into two piles (in no more than linear time), solve the problem independently on each pile, and then combine the two piles to form the final solution. The independence of the two piles is key. Example: classic recursive mergesort.
Each linear-time pass over the data gets you halfway to your solution. Example: quicksort if you think in terms of the maximum distance of each element to its final sorted position at each partitioning step (and yes, I know that it's actually O(n^2) because of degenerate pivot choices. But practically speaking, it falls into my O(n log(n)) category.)
O(n^2) - you have to look at every pair of input elements.
Or you don't, but you think you do, and you're using the wrong algorithm.
O(n^3) - um... I don't have a snappy characterization of these. It's probably one of:
You're multiplying matrices
You're looking at every pair of inputs but the operation you do requires looking at all of the inputs again
the entire graph structure of your input is relevant
O(2^n) - you need to consider every possible subset of your inputs.
None of these are rigorous. Especially not linear time algorithms (O(n)): I could come up with a number of examples where you have to look at all of the inputs, then half of them, then half of those, etc. Or the other way around -- you fold together pairs of inputs, then recurse on the output. These don't fit the description above, since you're not looking at each input once, but it still comes out in linear time. Still, 99.2% of the time, linear time means looking at each input once.
A lot of these are easy to demonstrate with something non-programming, like shuffling cards.
Sorting a deck of cards by going through the whole deck to find the ace of spades, then going through the whole deck to find the 2 of spades, and so on would be worst case n^2, if the deck was already sorted backwards. You looked at all 52 cards 52 times.
In general the really bad algorithms aren't necessarily intentional, they're commonly a misuse of something else, like calling a method that is linear inside some other method that repeats over the same set linearly.
I try to explain by giving simple code examples in C# and JavaScript.
C#
For List<int> numbers = new List<int> {1,2,3,4,5,6,7,12,543,7};
O(1) looks like
return numbers.First();
O(n) looks like
int result = 0;
foreach (int num in numbers)
{
result += num;
}
return result;
O(n log(n)) looks like
int result = 0;
foreach (int num in numbers)
{
int index = numbers.Count - 1;
while (index > 1)
{
// yeah, stupid, but couldn't come up with something more useful :-(
result += numbers[index];
index /= 2;
}
}
return result;
O(n2) looks like
int result = 0;
foreach (int outerNum in numbers)
{
foreach (int innerNum in numbers)
{
result += outerNum * innerNum;
}
}
return result;
O(n!) looks like, uhm, to tired to come up with anything simple.
But I hope you get the general point?
JavaScript
For const numbers = [ 1, 2, 3, 4, 5, 6, 7, 12, 543, 7 ];
O(1) looks like
numbers[0];
O(n) looks like
let result = 0;
for (num of numbers){
result += num;
}
O(n log(n)) looks like
let result = 0;
for (num of numbers){
let index = numbers.length - 1;
while (index > 1){
// yeah, stupid, but couldn't come up with something more useful :-(
result += numbers[index];
index = Math.floor(index/2)
}
}
O(n2) looks like
let result = 0;
for (outerNum of numbers){
for (innerNum of numbers){
result += outerNum * innerNum;
}
}
Ok - there are some very good answers here but almost all of them seem to make the same mistake and it's one that is pervading common usage.
Informally, we write that f(n) = O( g(n) ) if, up to a scaling factor and for all n larger than some n0, g(n) is larger than f(n). That is, f(n) grows no quicker than, or is bounded from above by, g(n). This tells us nothing about how fast f(n) grows, save for the fact that it is guaranteed not to be any worse than g(n).
A concrete example: n = O( 2^n ). We all know that n grows much less quickly than 2^n, so that entitles us to say that it is bounded by above by the exponential function. There is a lot of room between n and 2^n, so it's not a very tight bound, but it's still a legitimate bound.
Why do we (computer scientists) use bounds rather than being exact? Because a) bounds are often easier to prove and b) it gives us a short-hand to express properties of algorithms. If I say that my new algorithm is O(n.log n) that means that in the worst case its run-time will be bounded from above by n.log n on n inputs, for large enough n (although see my comments below on when I might not mean worst-case).
If instead, we want to say that a function grows exactly as quickly as some other function, we use theta to make that point (I'll write T( f(n) ) to mean \Theta of f(n) in markdown). T( g(n) ) is short hand for being bounded from above and below by g(n), again, up to a scaling factor and asymptotically.
That is f(n) = T( g(n) ) <=> f(n) = O(g(n)) and g(n) = O(f(n)). In our example, we can see that n != T( 2^n ) because 2^n != O(n).
Why get concerned about this? Because in your question you write 'would someone have to smoke crack to write an O(x!)?' The answer is no - because basically everything you write will be bounded from above by the factorial function. The run time of quicksort is O(n!) - it's just not a tight bound.
There's also another dimension of subtlety here. Typically we are talking about the worst case input when we use O( g(n) ) notation, so that we are making a compound statement: in the worst case running time it will not be any worse than an algorithm that takes g(n) steps, again modulo scaling and for large enough n. But sometimes we want to talk about the running time of the average and even best cases.
Vanilla quicksort is, as ever, a good example. It's T( n^2 ) in the worst case (it will actually take at least n^2 steps, but not significantly more), but T(n.log n) in the average case, which is to say the expected number of steps is proportional to n.log n. In the best case it is also T(n.log n) - but you could improve that for, by example, checking if the array was already sorted in which case the best case running time would be T( n ).
How does this relate to your question about the practical realisations of these bounds? Well, unfortunately, O( ) notation hides constants which real-world implementations have to deal with. So although we can say that, for example, for a T(n^2) operation we have to visit every possible pair of elements, we don't know how many times we have to visit them (except that it's not a function of n). So we could have to visit every pair 10 times, or 10^10 times, and the T(n^2) statement makes no distinction. Lower order functions are also hidden - we could have to visit every pair of elements once, and every individual element 100 times, because n^2 + 100n = T(n^2). The idea behind O( ) notation is that for large enough n, this doesn't matter at all because n^2 gets so much larger than 100n that we don't even notice the impact of 100n on the running time. However, we often deal with 'sufficiently small' n such that constant factors and so on make a real, significant difference.
For example, quicksort (average cost T(n.log n)) and heapsort (average cost T(n.log n)) are both sorting algorithms with the same average cost - yet quicksort is typically much faster than heapsort. This is because heapsort does a few more comparisons per element than quicksort.
This is not to say that O( ) notation is useless, just imprecise. It's quite a blunt tool to wield for small n.
(As a final note to this treatise, remember that O( ) notation just describes the growth of any function - it doesn't necessarily have to be time, it could be memory, messages exchanged in a distributed system or number of CPUs required for a parallel algorithm.)
The way I describe it to my nontechnical friends is like this:
Consider multi-digit addition. Good old-fashioned, pencil-and-paper addition. The kind you learned when you were 7-8 years old. Given two three-or-four-digit numbers, you can find out what they add up to fairly easily.
If I gave you two 100-digit numbers, and asked you what they add up to, figuring it out would be pretty straightforward, even if you had to use pencil-and-paper. A bright kid could do such an addition in just a few minutes. This would only require about 100 operations.
Now, consider multi-digit multiplication. You probably learned that at around 8 or 9 years old. You (hopefully) did lots of repetitive drills to learn the mechanics behind it.
Now, imagine I gave you those same two 100-digit numbers and told you to multiply them together. This would be a much, much harder task, something that would take you hours to do - and that you'd be unlikely to do without mistakes. The reason for this is that (this version of) multiplication is O(n^2); each digit in the bottom number has to be multiplied by each digit in the top number, leaving a total of about n^2 operations. In the case of the 100-digit numbers, that's 10,000 multiplications.
No, an O(n) algorithm does not mean it will perform an operation on each element. Big-O notation gives you a way to talk about the "speed" of you algorithm independent of your actual machine.
O(n) means that the time your algorithm will take grows linearly as your input increase. O(n^2) means that the time your algorithm takes grows as the square of your input. And so forth.
The way I think about it, is you have the task of cleaning up a problem caused by some evil villain V who picks N, and you have to estimate out how much longer it's going to take to finish your problem when he increases N.
O(1) -> increasing N really doesn't make any difference at all
O(log(N)) -> every time V doubles N, you have to spend an extra amount of time T to complete the task. V doubles N again, and you spend the same amount.
O(N) -> every time V doubles N, you spend twice as much time.
O(N^2) -> every time V doubles N, you spend 4x as much time. (it's not fair!!!)
O(N log(N)) -> every time V doubles N, you spend twice as much time plus a little more.
These are bounds of an algorithm; computer scientists want to describe how long it is going to take for large values of N. (which gets important when you are factoring numbers that are used in cryptography -- if the computers speed up by a factor of 10, how many more bits do you have to use to ensure it will still take them 100 years to break your encryption and not just 1 year?)
Some of the bounds can have weird expressions if it makes a difference to the people involved. I've seen stuff like O(N log(N) log(log(N))) somewhere in Knuth's Art of Computer Programming for some algorithms. (can't remember which one off the top of my head)
One thing that hasn't been touched on yet for some reason:
When you see algorithms with things like O(2^n) or O(n^3) or other nasty values it often means you're going to have to accept an imperfect answer to your problem in order to get acceptable performance.
Correct solutions that blow up like this are common when dealing with optimization problems. A nearly-correct answer delivered in a reasonable timeframe is better than a correct answer delivered long after the machine has decayed to dust.
Consider chess: I don't know exactly what the correct solution is considered to be but it's probably something like O(n^50) or even worse. It is theoretically impossible for any computer to actually calculate the correct answer--even if you use every particle in the universe as a computing element performing an operation in the minimum possible time for the life of the universe you still have a lot of zeros left. (Whether a quantum computer can solve it is another matter.)
The "Intuitition" behind Big-O
Imagine a "competition" between two functions over x, as x approaches infinity: f(x) and g(x).
Now, if from some point on (some x) one function always has a higher value then the other, then let's call this function "faster" than the other.
So, for example, if for every x > 100 you see that f(x) > g(x), then f(x) is "faster" than g(x).
In this case we would say g(x) = O(f(x)). f(x) poses a sort of "speed limit" of sorts for g(x), since eventually it passes it and leaves it behind for good.
This isn't exactly the definition of big-O notation, which also states that f(x) only has to be larger than C*g(x) for some constant C (which is just another way of saying that you can't help g(x) win the competition by multiplying it by a constant factor - f(x) will always win in the end). The formal definition also uses absolute values. But I hope I managed to make it intuitive.
And does somebody have to smoke crack to write an O(x!)?
No, just use Prolog. If you write a sorting algorithm in Prolog by just describing that each element should be bigger than the previous, and let backtracking do the sorting for you, that will be O(x!). Also known as "permutation sort".
I like don neufeld's answer, but I think I can add something about O(n log n).
An algorithm which uses a simple divide and conquer strategy is probably going to be O(log n). The simplest example of this is finding a something in an sorted list. You don't start at the beginning and scan for it. You go to the middle, you decide if you should then go backwards or forwards, jump halfway to the last place you looked, and repeat this until you find the item you're looking for.
If you look at the quicksort or mergesort algorithms, you will see that they both take the approach of dividing the list to be sorted in half, sorting each half (using the same algorithm, recursively), and then recombining the two halves. This sort of recursive divide and conquer strategy will be O(n log n).
If you think about it carefully, you'll see that quicksort does an O(n) partitioning algorithm on the whole n items, then an O(n) partitioning twice on n/2 items, then 4 times on n/4 items, etc... until you get to an n partitions on 1 item (which is degenerate). The number of times you divide n in half to get to 1 is approximately log n, and each step is O(n), so recursive divide and conquer is O(n log n). Mergesort builds the other way, starting with n recombinations of 1 item, and finishing with 1 recombination of n items, where the recombination of two sorted lists is O(n).
As for smoking crack to write an O(n!) algorithm, you are unless you have no choice. The traveling salesman problem given above is believed to be one such problem.
Think of it as stacking lego blocks (n) vertically and jumping over them.
O(1) means at each step, you do nothing. The height stays the same.
O(n) means at each step, you stack c blocks, where c1 is a constant.
O(n^2) means at each step, you stack c2 x n blocks, where c2 is a constant, and n is the number of stacked blocks.
O(nlogn) means at each step, you stack c3 x n x log n blocks, where c3 is a constant, and n is the number of stacked blocks.
Most Jon Bentley books (e.g. Programming Pearls) cover such stuff in a really pragmatic manner. This talk given by him includes one such analysis of a quicksort.
While not entirely relevant to the question, Knuth came up with an interesting idea: teaching Big-O notation in high school calculus classes, though I find this idea quite eccentric.
To understand O(n log n), remember that log n means log-base-2 of n. Then look at each part:
O(n) is, more or less, when you operate on each item in the set.
O(log n) is when the number of operations is the same as the exponent to which you raise 2, to get the number of items. A binary search, for instance, has to cut the set in half log n times.
O(n log n) is a combination – you're doing something along the lines of a binary search for each item in the set. Efficient sorts often operate by doing one loop per item, and in each loop doing a good search to find the right place to put the item or group in question. Hence n * log n.
Just to respond to the couple of comments on my above post:
Domenic - I'm on this site, and I care. Not for pedantry's sake, but because we - as programmers - typically care about precision. Using O( ) notation incorrectly in the style that some have done here renders it kind of meaningless; we may just as well say something takes n^2 units of time as O( n^2 ) under the conventions used here. Using the O( ) adds nothing. It's not just a small discrepancy between common usage and mathematical precision that I'm talking about, it's the difference between it being meaningful and it not.
I know many, many excellent programmers who use these terms precisely. Saying 'oh, we're programmers therefore we don't care' cheapens the whole enterprise.
onebyone - Well, not really although I take your point. It's not O(1) for arbitrarily large n, which is kind of the definition of O( ). It just goes to show that O( ) has limited applicability for bounded n, where we would rather actually talk about the number of steps taken rather than a bound on that number.
Tell your eight year old log(n) means the number of times you have to chop a length n log in two for it to get down to size n=1 :p
O(n log n) is usually sorting
O(n^2) is usually comparing all pairs of elements
Suppose you had a computer that could solve a problem of a certain size. Now imagine that we can double the performance a few times. How much bigger a problem can we solve with each doubling?
If we can solve a problem of double the size, that's O(n).
If we have some multiplier that isn't one, that's some sort of polynomial complexity. For example, if each doubling allows us to increase the problem size by about 40%, it's O(n^2), and about 30% would be O(n^3).
If we just add to the problem size, it's exponential or worse. For example, if each doubling means we can solve a problem 1 bigger, it's O(2^n). (This is why brute-forcing a cipher key becomes effectively impossible with reasonably sized keys: a 128-bit key requires about 16 quintillion times as much processing as a 64-bit.)
Remember the fable of the tortoise and the hare (turtle and rabbit)?
Over the long run, the tortoise wins, but over the short run the hare wins.
That's like O(logN) (tortoise) vs. O(N) (hare).
If two methods differ in their big-O, then there is a level of N at which one of them will win, but big-O says nothing about how big that N is.
To remain sincere to the question asked I would answer the question in the manner I would answer an 8 year old kid
Suppose an ice-cream seller prepares a number of ice creams ( say N ) of different shapes arranged in an orderly fashion.
You want to eat the ice cream lying in the middle
Case 1 : - You can eat an ice cream only if you have eaten all the ice creams smaller than it
You will have to eat half of all the ice creams prepared (input).Answer directly depends on the size of the input
Solution will be of order o(N)
Case 2 :- You can directly eat the ice cream in the middle
Solution will be O(1)
Case 3 : You can eat an ice cream only if you have eaten all the ice creams smaller than it and each time you eat an ice cream you allow another kid (new kid everytime ) to eat all his ice creams
Total time taken would be N + N + N.......(N/2) times
Solution will be O(N2)
log(n) means logarithmic growth. An example would be divide and conquer algorithms. If you have 1000 sorted numbers in an array ( ex. 3, 10, 34, 244, 1203 ... ) and want to search for a number in the list (find its position), you could start with checking the value of the number at index 500. If it is lower than what you seek, jump to 750. If it is higher than what you seek, jump to 250. Then you repeat the process until you find your value (and key). Every time we jump half the search space, we can cull away testing many other values since we know the number 3004 can't be above number 5000 (remember, it is a sorted list).
n log(n) then means n * log(n).
I'll try to actually write an explanation for a real eight year old boy, aside from technical terms and mathematical notions.
Like what exactly would an O(n^2) operation do?
If you are in a party, and there are n people in the party including you. How many handshakes it take so that everyone has handshaked everyone else, given that people would probably forget who they handshaked at some point.
Note: this approximate to a simplex yielding n(n-1) which is close enough to n^2.
And what the heck does it mean if an operation is O(n log(n))?
Your favorite team has won, they are standing in line, and there are n players in the team. How many hanshakes it would take you to handshake every player, given that you will hanshake each one multiple times, how many times, how many digits are in the number of the players n.
Note: this will yield n * log n to the base 10.
And does somebody have to smoke crack to write an O(x!)?
You are a rich kid and in your wardrobe there are alot of cloths, there are x drawers for each type of clothing, the drawers are next to each others, the first drawer has 1 item, each drawer has as many cloths as in the drawer to its left and one more, so you have something like 1 hat, 2 wigs, .. (x-1) pants, then x shirts. Now in how many ways can you dress up using a single item from each drawer.
Note: this example represent how many leaves in a decision-tree where number of children = depth, which is done through 1 * 2 * 3 * .. * x