Big-oh vs big-theta [duplicate] - algorithm

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What is the difference between Θ(n) and O(n)?
It seems to me like when people talk about algorithm complexity informally, they talk about big-oh. But in formal situations, I often see big-theta with the occasional big-oh thrown in.
I know mathematically what the difference is between the two, but in English, in what situation would using big-oh when you mean big-theta be incorrect, or vice versa (an example algorithm would be appreciated)?
Bonus: why do people seemingly always use big-oh when talking informally?

Big-O is an upper bound.
Big-Theta is a tight bound, i.e. upper and lower bound.
When people only worry about what's the worst that can happen, big-O is sufficient; i.e. it says that "it can't get much worse than this". The tighter the bound the better, of course, but a tight bound isn't always easy to compute.
See also
Wikipedia/Big O Notation
Related questions
What is the difference between Θ(n) and O(n)?
The following quote from Wikipedia also sheds some light:
Informally, especially in computer science, the Big O notation often is
permitted to be somewhat abused to describe an asymptotic tight bound
where using Big Theta notation might be more factually appropriate in a
given context.
For example, when considering a function T(n) = 73n3+ 22n2+ 58, all of the following are generally acceptable, but tightness of bound (i.e., bullets 2 and 3 below) are usually strongly preferred over laxness of bound (i.e., bullet 1
below).
T(n) = O(n100), which is identical to T(n) ∈ O(n100)
T(n) = O(n3), which is identical to T(n) ∈ O(n3)
T(n) = Θ(n3), which is identical to T(n) ∈ Θ(n3)
The equivalent English statements are respectively:
T(n) grows asymptotically no faster than n100
T(n) grows asymptotically no faster than n3
T(n) grows asymptotically as fast as n3.
So while all three statements are true, progressively more information is contained in
each. In some fields, however, the Big O notation (bullets number 2 in the lists above)
would be used more commonly than the Big Theta notation (bullets number 3 in the
lists above) because functions that grow more slowly are more desirable.

I'm a mathematician and I have seen and needed big-O O(n), big-Theta Θ(n), and big-Omega Ω(n) notation time and again, and not just for complexity of algorithms. As people said, big-Theta is a two-sided bound. Strictly speaking, you should use it when you want to explain that that is how well an algorithm can do, and that either that algorithm can't do better or that no algorithm can do better. For instance, if you say "Sorting requires Θ(n(log n)) comparisons for worst-case input", then you're explaining that there is a sorting algorithm that uses O(n(log n)) comparisons for any input; and that for every sorting algorithm, there is an input that forces it to make Ω(n(log n)) comparisons.
Now, one narrow reason that people use O instead of Ω is to drop disclaimers about worst or average cases. If you say "sorting requires O(n(log n)) comparisons", then the statement still holds true for favorable input. Another narrow reason is that even if one algorithm to do X takes time Θ(f(n)), another algorithm might do better, so you can only say that the complexity of X itself is O(f(n)).
However, there is a broader reason that people informally use O. At a human level, it's a pain to always make two-sided statements when the converse side is "obvious" from context. Since I'm a mathematician, I would ideally always be careful to say "I will take an umbrella if and only if it rains" or "I can juggle 4 balls but not 5", instead of "I will take an umbrella if it rains" or "I can juggle 4 balls". But the other halves of such statements are often obviously intended or obviously not intended. It's just human nature to be sloppy about the obvious. It's confusing to split hairs.
Unfortunately, in a rigorous area such as math or theory of algorithms, it's also confusing not to split hairs. People will inevitably say O when they should have said Ω or Θ. Skipping details because they're "obvious" always leads to misunderstandings. There is no solution for that.

Because my keyboard has an O key.
It does not have a Θ or an Ω key.
I suspect most people are similarly lazy and use O when they mean Θ because it's easier to type.

One reason why big O gets used so much is kind of because it gets used so much. A lot of people see the notation and think they know what it means, then use it (wrongly) themselves. This happens a lot with programmers whose formal education only went so far - I was once guilty myself.
Another is because it's easier to type a big O on most non-Greek keyboards than a big theta.
But I think a lot is because of a kind of paranoia. I worked in defence-related programming for a bit (and knew very little about algorithm analysis at the time). In that scenario, the worst case performance is always what people are interested in, because that worst case might just happen at the wrong time. It doesn't matter if the actually probability of that happening is e.g. far less than the probability of all members of a ships crew suffering a sudden fluke heart attack at the same moment - it could still happen.
Though of course a lot of algorithms have their worst case in very common circumstances - the classic example being inserting in-order into a binary tree to get what's effectively a singly-linked list. A "real" assessment of average performance needs to take into account the relative frequency of different kinds of input.

Bonus: why do people seemingly always use big-oh when talking informally?
Because in big-oh, this loop:
for i = 1 to n do
something in O(1) that doesn't change n and i and isn't a jump
is O(n), O(n^2), O(n^3), O(n^1423424). big-oh is just an upper bound, which makes it easier to calculate because you don't have to find a tight bound.
The above loop is only big-theta(n) however.
What's the complexity of the sieve of eratosthenes? If you said O(n log n) you wouldn't be wrong, but it wouldn't be the best answer either. If you said big-theta(n log n), you would be wrong.

Because there are algorithms whose best-case is quick, and thus it's technically a big O, not a big Theta.
Big O is an upper bound, big Theta is an equivalence relation.

There are a lot of good answers here but I noticed something was missing. Most answers seem to be implying that the reason why people use Big O over Big Theta is a difficulty issue, and in some cases this may be true. Often a proof that leads to a Big Theta result is far more involved than one that results in Big O. This usually holds true, but I do not believe this has a large relation to using one analysis over the other.
When talking about complexity we can say many things. Big O time complexity is just telling us what an algorithm is guarantied to run within, an upper bound. Big Omega is far less often discussed and tells us the minimum time an algorithm is guarantied to run, a lower bound. Now Big Theta tells us that both of these numbers are in fact the same for a given analysis. This tells us that the application has a very strict run time, that can only deviate by a value asymptoticly less than our complexity. Many algorithms simply do not have upper and lower bounds that happen to be asymptoticly equivalent.
So as to your question using Big O in place of Big Theta would technically always be valid, while using Big Theta in place of Big O would only be valid when Big O and Big Omega happened to be equal. For instance insertion sort has a time complexity of Big О at n^2, but its best case scenario puts its Big Omega at n. In this case it would not be correct to say that its time complexity is Big Theta of n or n^2 as they are two different bounds and should be treated as such.

I have seen Big Theta, and I'm pretty sure I was taught the difference in school. I had to look it up though. This is what Wikipedia says:
Big O is the most commonly used asymptotic notation for comparing functions, although in many cases Big O may be replaced with Big Theta Θ for asymptotically tighter bounds.
Source: Big O Notation#Related asymptotic notation
I don't know why people use Big-O when talking formally. Maybe it's because most people are more familiar with Big-O than Big-Theta? I had forgotten that Big-Theta even existed until you reminded me. Although now that my memory is refreshed, I may end up using it in conversation. :)

Related

When to use big O notation and when to use big Theta notation

I understand that Big O is an upper bound and Big Theta is a tight bound when for example we consider the functions f(n)=O(g(n)) or similarly for Big Theta.
But how do we know that a particular algorithm will be better represented using the Big theta notation instead of Big O?
For example, the time complexity of selection sort is given as Big Theta of N^2 rather than Big O of N^2, why?
It's not a question of which is better but rather what do you want to studies.
If you want to study the worst case scenario then you can use the upper bound notation. Keep in mind that the tighter the bound the better, but in some cases it's difficult to calculate the tight bound.
Generally, when people speak about big-O or big-teta they mean the same think so in Selection sort you can also use big-O notation
I agree with the answer above that it is more about what you are trying to study about the algorithm. If you just want the worst case running time, meaning in the worst case the algorithm will run in at least a certain time, then it is best to use Big-O. For example, for selection sort the worst case would be O(n^2). If you want to study the worst and the best case running times, then you would also want to find the Big-Omega. If these are the same (meaning in the best and worst case the algorithm will run in the same amount of time), then you would want to use Big theta to state that this time is a tight bound for the algorithm being studied. This is usually more descriptive because it gives a better outlook on how the algorithm will run with a large dataset. For selection sort, because the best and worst case running time is n^2 (because of the nested loops), the average will be theta(n^2).

What is difference between different asymptotic notations?

I am really very confused in asymptotic notations. As far as I know, Big-O notation is for worst cast, omega is for best case and theta is for average case. However, I have always seen Big O being used everywhere, even for best case. For e.g. in the following link, see the table where time complexities of different sorting algorithms are mentioned-
https://en.wikipedia.org/wiki/Best,_worst_and_average_case
Everywhere in the table, big O notation is used irrespective of whether it is best case, worst case or average case. Then what is the use of other two notations and where do we use it?
As far as I know, Big-O notation is for worst cast, omega is for best case and theta is for average case.
They aren't. Omicron is for (asymptotic) upper bound, omega is for lower bound and theta is for tight bound, which is both an upper and a lower bound. If the lower and upper bound of an algorithm are different, then the complexity cannot be expressed with theta notation.
The concept of upper,lower,tight bound are orthogonal to the concept of best,average,worst case. You can analyze the upper bound of each case, and you can analyze different bounds of the worst case (and also any other combination of the above).
Asymptotic bounds are always in relation to the set of variables in the expression. For example, O(n) is in relation to n. The best, average and worst cases emerge from everything else but n. For example, if n is the number of elements, then the different cases might emerge from the order of the elements, or the number of unique elements, or the distribution of values.
However, I have always seen Big O being used everywhere, even for best case.
That's because the upper bound is almost always the one that is the most important and interesting when describing an algorithm. We rarely care about the lower bound. Just like we rarely care about the best case.
The lower bound is sometimes useful in describing a problem that has been proven to have a particular complexity. For example, it is proven that worst case complexity of all general comparison sorting algorithms is Ω(n log n). If the sorting algorithm is also O(n log n), then by definition, it is also Θ(n log n).
Big O is for upper bound, not for worst case! There is no notation specifically for worst case/best case. The examples you are talking about all have Big O because they are all upper bounded by the given value. I suggest you take another look at the book from which you learned the basics because this is immensely important to understand :)
EDIT: Answering your doubt- because usually, we are bothered with our at-most performance i.e. when we say, our algorithm performs in O(logn) in the best case-scenario, we know that its performance will not be worse than logarithmic time in the given scenario. It is the upper bound that we seek to reduce usually and hence we usually mention big O to compare algorithms. (not to say that we never mention the other two)
O(...) basically means "not (much) slower than ...".
It can be used for all three cases ("the worst case is not slower than", "the best case is not slower than", and so on).
Omega is the oppsite: You can say, something can't be much faster than ... . Again, it can be used with all three cases. Compared to O(...), it's not that important, because telling someone "I'm certain my program is not faster than yours" is nothing to be proud of.
Theta is a combination: It's "(more or less) as fast as" ..., not just slower/faster.
The Big-O notation is somethin like this >= in terms of asymptotic equality.
For example if you see this :
x = O(x^2) it does say x <= x^2 (in asymptotic terms).
And it does mean "x is at most as complex as x^2", which is something that you are usually interesting it.
Even when you compare Best/Average case, you can say "At best possible input, I will have AT MOST this complexity".
There are two things mixed up: Big O, Omega, Theta, are purely mathematical constructions. For example, O (f (N)) is the set of functions which are less than c * f (n), for some c > 0, and for all n >= some minimum value N0. With that definition, n = O (f (n^4)), because n ≤ n^4. 100 = O (f (n)), because 100 <= n for n ≥ 100, or 100 <= 100 * n for n ≥ 1.
For an algorithm, you want to give worst case speed, average case speed, rarely the best case speed, sometimes amortised average speed (that's when running an algorithm once does work that can be used when it's run again. Like calculating n! for n = 1, 2, 3, ... where each calculation can take advantage of the previous one). And whatever speed you measure, you can give a result in one of the notations.
For example, you might have an algorithm where you can prove that the worst case is O (n^2), but you cannot prove whether there are faster special cases or not, and you also cannot prove that the algorithm isn't actually faster, like O (n^1.9). So O (n^2) is the only thing that you can prove.

Why is theta notation never used?

I'm studying a degree in computer science and at class we're using big-theta notation much more often than big-O notation. Although while reading articles about algorithms and its running times, I hardly ever find the big-theta notation anywhere. Why isn't theta notation used to indicate worst case of running time for algorithms in a more fitted way in most books and articles?
Big-O is an upper bound.
Big-Theta is a tight bound, i.e. upper and lower bound.
When people only worry about what's the worst that can happen, big-O is sufficient; i.e. it says that "it can't get much worse than this". The tighter the bound the better, of course, but a tight bound isn't always easy to compute [1].
The following opinion [2] will give you a better understanding:
As people said, big-Theta is a two-sided bound. Strictly speaking, you should use it when you want to explain that that is how well an algorithm can do, and that either that algorithm can't do better or that no algorithm can do better. For instance, if you say "Sorting requires Θ(n(log n)) comparisons for worst-case input", then you're explaining that there is a sorting algorithm that uses O(n(log n)) comparisons for any input; and that for every sorting algorithm, there is an input that forces it to make Ω(n(log n)) comparisons.
Now, one narrow reason that people use O instead of Ω is to drop disclaimers about worst or average cases. If you say "sorting requires O(n(log n)) comparisons", then the statement still holds true for favorable input. Another narrow reason is that even if one algorithm to do X takes time Θ(f(n)), another algorithm might do better, so you can only say that the complexity of X itself is O(f(n)).
However, there is a broader reason that people informally use O. At a human level, it's a pain to always make two-sided statements when the converse side is "obvious" from context. Since I'm a mathematician, I would ideally always be careful to say "I will take an umbrella if and only if it rains" or "I can juggle 4 balls but not 5", instead of "I will take an umbrella if it rains" or "I can juggle 4 balls". But the other halves of such statements are often obviously intended or obviously not intended. It's just human nature to be sloppy about the obvious. It's confusing to split hairs.
Unfortunately, in a rigorous area such as math or theory of algorithms, it's also confusing not to split hairs. People will inevitably say O when they should have said Ω or Θ. Skipping details because they're "obvious" always leads to misunderstandings. There is no solution for that.

Still sort of confused about Big O notation

So I've been trying to understand Big O notation as well as I can, but there are still some things I'm confused about. So I keep reading that if something is O(n), it usually is referring to the worst-case of an algorithm, but that it doesn't necessarily have to refer to the worst case scenario, which is why we can say the best-case of insertion sort for example is O(n). However, I can't really make sense of what that means. I know that if the worst-case is O(n^2), it means that the function that represents the algorithm in its worst case grows no faster than n^2 (there is an upper bound). But if you have O(n) as the best case, how should I read that as? In the best case, the algorithm grows no faster than n? What I picture is a graph with n as the upper bound, like
If the best case scenario of an algorithm is O(n), then n is the upper bound of how fast the operations of the algorithm grow in the best case, so they cannot grow faster than n...but wouldn't that mean that they can grow as fast as O(log n) or O(1), since they are below the upper bound? That wouldn't make sense though, because O(log n) or O(1) is a better scenario than O(n), so O(n) WOULDN'T be the best case? I'm so lost lol
Big-O, Big-Θ, Big-Ω are independent from worst-case, average-case, and best-case.
The notation f(n) = O(g(n)) means f(n) grows no more quickly than some constant multiple of g(n).
The notation f(n) = Ω(g(n)) means f(n) grows no more slowly than some constant multiple of g(n).
The notation f(n) = Θ(g(n)) means both of the above are true.
Note that f(n) here may represent the best-case, worst-case, or "average"-case running time of a program with input size n.
Furthermore, "average" can have many meanings: it can mean the average input or the average input size ("expected" time), or it can mean in the long run (amortized time), or both, or something else.
Often, people are interested in the worst-case running time of a program, amortized over the running time of the entire program (so if something costs n initially but only costs 1 time for the next n elements, it averages out to a cost of 2 per element). The most useful thing to measure here is the least upper bound on the worst-case time; so, typically, when you see someone asking for the Big-O of a program, this is what they're looking for.
Similarly, to prove a problem is inherently difficult, people might try to show that the worst-case (or perhaps average-case) running time is at least a certain amount (for example, exponential).
You'd use Big-Ω notation for these, because you're looking for lower bounds on these.
However, there is no special relationship between worst-case and Big-O, or best-case and Big-Ω.
Both can be used for either, it's just that one of them is more typical than the other.
So, upper-bounding the best case isn't terribly useful. Yes, if the algorithm always takes O(n) time, then you can say it's O(n) in the best case, as well as on average, as well as the worst case. That's a perfectly fine statement, except the best case is usually very trivial and hence not interesting in itself.
Furthermore, note that f(n) = n = O(n2) -- this is technically correct, because f grows more slowly than n2, but it is not useful because it is not the least upper bound -- there's a very obvious upper bound that's more useful than this one, namely O(n). So yes, you're perfectly welcome to say the best/worst/average-case running time of a program is O(n!). That's mathematically perfectly correct. It's just useless, because when people ask for Big-O they're interested in the least upper bound, not just a random upper bound.
It's also worth noting that it may simply be insufficient to describe the running-time of a program as f(n). The running time often depends on the input itself, not just its size. For example, it may be that even queries are trivially easy to answer, whereas odd queries take a long time to answer.
In that case, you can't just give f as a function of n -- it would depend on other variables as well. In the end, remember that this is just a set of mathematical tools; it's your job to figure out how to apply it to your program and to figure out what's an interesting thing to measure. Using tools in a useful manner needs some creativity, and math is no exception.
Informally speaking, best case has O(n) complexity means that when the input meets
certain conditions (i.e. is best for the algorithm performed), then the count of
operations performed in that best case, is linear with respect to n (e.g. is 1n or 1.5n or 5n).
So if the best case is O(n), usually this means that in the best case it is exactly linear
with respect to n (i.e. asymptotically no smaller and no bigger than that) - see (1). Of course,
if in the best case that same algorithm can be proven to perform at most c * log N operations
(where c is some constant), then this algorithm's best case complexity would be informally
denoted as O(log N) and not as O(N) and people would say it is O(log N) in its best case.
Formally speaking, "the algorithm's best case complexity is O(f(n))"
is an informal and wrong way of saying that "the algorithm's complexity
is Ω(f(n))" (in the sense of the Knuth definition - see (2)).
See also:
(1) Wikipedia "Family of Bachmann-Landau notations"
(2) Knuth's paper "Big Omicron and Big Omega and Big Theta"
(3)
Big Omega notation - what is f = Ω(g)?
(4)
What is the difference between Θ(n) and O(n)?
(5)
What is a plain English explanation of "Big O" notation?
I find it easier to think of O() as about ratios than about bounds. It is defined as bounds, and so that is a valid way to think of it, but it seems a bit more useful to think about "if I double the number/size of inputs to my algorithm, does my processing time double (O(n)), quadruple (O(n^2)), etc...". Thinking about it that way makes it a little bit less abstract - at least to me...

Could anyone explain Big O versus Big Omega vs Big Theta? [duplicate]

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Big Theta Notation - what exactly does big Theta represent?
I understand it in theory, I guess, but what I'm having trouble grasping is the application of the three.
In school, we always used Big O to denote the complexity of an algorithm. Bubble sort was O(n^2) for example.
Now after reading some more theory I get that Big Oh is not the only measure, there's at least two other interesting ones.
But here's my question:
Big O is the upper-bound, Big Omega is the lower bound, and Big Theta is a mix of the two. But what does that mean conceptually? I understand what it means on a graph; I've seen a million examples of that. But what does it mean for algorithm complexity? How does an "upper bound" or a "lower bound" mix with that?
I guess I just don't get its application. I understand that if multiplied by some constant c that if after some value n_0 f(x) is greater than g(x), f(x) is considered O(g(x)). But what does that mean practically? Why would we be multiplying f(x) by some value c? Hell, I thought with Big O notation multiples didn't matter.
The big O notation, and its relatives, the big Theta, the big Omega, the small o and the small omega are ways of saying something about how a function behaves at a limit point (for example, when approaching infinity, but also when approaching 0, etc.) without saying much else about the function. They are commonly used to describe running space and time of algorithms, but can also be seen in other areas of mathematics regarding asymptotic behavior.
The semi-intuitive definition is as follows:
A function g(x) is said to be O(f(x)) if "from some point on", g(x) is lower than c*f(x), where c is some constant.
The other definitions are similar, Theta demanding that g(x) be between two constant multiples of f(x), Omega demanding g(x)>c*f(x), and the small versions demand that this is true for all such constants.
But why is it interesting to say, for example, that an algorithm has run time of O(n^2)?
It's interesting mainly because, in theoretical computer science, we are most interested in how algorithms behave for large inputs. This is true because on small inputs algorithm run times can vary greatly depending on implementation, compilation, hardware, and other such things that are not really interesting when analyzing an algorithm theoretically.
The rate of growth, however, usually depends on the nature of the algorithm, and to improve it you need deeper insights on the problem you're trying to solve. This is the case, for example, with sorting algorithms, where you can get a simple algorithm (Bubble Sort) to run in O(n^2), but to improve this to O(n log n) you need a truly new idea, such as that introduced in Merge Sort or Heap Sort.
On the other hand, if you have an algorithm that runs in exactly 5n seconds, and another that runs in 1000n seconds (which is the difference between a long yawn and a launch break for n=3, for example), when you get to n=1000000000000, the difference in scale seems less important. If you have an algorithm that takes O(log n), though, you'd have to wait log(1000000000000)=12 seconds, perhaps multiplied by some constant, instead of the almost 317,098 years, which, no matter how big the constant is, is a completely different scale.
I hope this makes things a little clearer. Good luck with your studies!

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