The obvious one is a constant on a linear term for example 2n, 4n and 8n are all just n or O(n).
But what about the exponential constant 1.6^n and 2^n. In this case the constant seems to have a greater effect on the time complexity.
Also there is not really a convenient way to write a catch all for exponential time complexity.
O(K^n) perhaps.
In this cheat sheet, they seem to use O(2^n) does that mean that all exponential complexities should be written that way?
Probably not.
You're right that 2n, 4n and 8n are all just O(n), and you're also right that O(1.6n) is not the same as O(2n). To understand why, we need to refer to the definition of big O notation.
The notation O(...) means a set of functions. A function f(n) is in the set O(g(n)) if and only if, for some constants c and n0, we have f(n) ≤ c * g(n) whenever n ≥ n0. Given this definition:
The function f(n) = 8n is in the set O(n) because if we choose c = 8 and n0 = 1, we have 8n ≤ 8 * n for all n ≥ 1.
The function f(n) = 2n is not in the set O(1.6n), because whichever c and n0 we choose, 2n > c * 1.6n for some sufficiently large n. We can choose n > log2 c + n0 log2 1.6 for a concrete counterexample.
Note however that f(n) = 1.6n is in the set O(2n), because 1.6n ≤ 1 * 2n for all n ≥ 1.
For a "catch-all" way of writing exponential complexity, you can write 2O(n). This includes exponential functions with arbitrary bases, e.g. the function f(n) = 16n since this equals 24n, and 4n is in the set O(n). It's an abuse of notation, since raising the number 2 to the power of a set of functions doesn't really make sense in this context, but it is common enough to be understood.
That is correct. The cheat sheet you linked to here can not show all the different complexities so it picks the most common ones.
Simply put, if you have a function growing at 3 ^ n. It can not be classified as 2 ^ n because it will break the definition of Big O.
The some what complex looking math that describes Big O is simply saying that it can't ever be bigger. And also ignore linear growth constants.
f(n) ≤ c * g(n) whenever n ≥ n0
Let's consider classic big O notation definition (proof link):
O(f(n)) is the set of all functions such that there exist positive constants C and n0 with |g(n)| ≤ C * f(n), for all n ≥ n_0.
According to this definition it is legal to do the following (g1 and g2 are the functions that describe two algorithms complexity):
g1(n) = 9999 * n^2 + n ∈ O(9999 * n^2)
g2(n) = 5 * n^2 + N ∈ O(5 * n^2)
And it is also legal to note functions as:
g1(n) = 9999 * N^2 + N ∈ O(n^2)
g2(n) = 5 * N^2 + N ∈ O(n^2)
As you can see the first variant O(9999*N^2) vs (5*N^2) is much more precise and gives us clear vision which algorithm is faster. The second one does not show us anything.
The question is: why nobody use the first variant?
The use of the O() notation is, from the get go, the opposite of noting something "precisely". The very idea is to mask "precise" differences between algorithms, as well as being able to ignore the effect of computing hardware specifics and the choice of compiler or programming language. Indeed, g_1(n) and g_2(n) are both in the same class (or set) of functions of n - the class O(n^2). They differ in specifics, but they are similar enough.
The fact that it's a class is why I edited your question and corrected the notation from = O(9999 * N^2) to ∈ O(9999 * N^2).
By the way - I believe your question would have been a better fit on cs.stackexchange.com.
I had a job interview today. And was asked about complexity of std:set_intersection. When I was answering I mentioned that
O(n+m)
is equal to:
O(max(n,m))
I was told that this is incorrect. I was unsuccessfully trying to show equivalence with:
O(0.5*(n+m)) ≤ O(max(n,m)) ≤ O(n+m)
My question is: am I really incorrect?
For all m, n ≥ 0 it is valid that max(m, n) ≤ m + n → max(m, n) in O(m + n), and m + n ≤ 2max(m, n) → m + n in O(max(m, n)).
Thus O(max(m, n)) = O(m + n).
ADDENDUM: If f belongs O(m + n) then a constant D > 0 exists, that f(n, m) < D * (m + n) for m and n large enough. Thus f(n, m) < 2 D * max(m, n), and O(m + n) must be a subset of O(max(m, n)). The proof of O(max(m, n)) is a subset of O(m + n) is made analogously.
Well you have totally right about O(n+m) is equal to O(max(n,m)),even more precise we can prove Θ(n+m)=Θ(max(n,m) which is more tight and proves your sentence. The mathematical proof is (for both big-O and Θ) very simple and easy to understand with common sense. So since we have a mathematical proof which is a way to say something but in a more well defined and strict way which doesn't leaves any ambiguity.
Though you was (wrongly) told that this is incorrect because if we want to be very precise this is not the appropriate - mathematical way to express that order of max(m,n) is same as m+n. You used the words "is equal" referring to big-O notation but what is the definition of big-O notation?
It is referred to Sets. Saying max(n+m) belongs to O(m+n) is the
most correct way and vice versa m+n belongs to O(max(m,n)). In big O
notation is commonly used and accepted to say m+n = O(max(n,m)).
The problem caused is that you didn't try to refer to the order of a function like f is O(g) but you tried to compare Sets O(f) and O(g).But proving two infinite sets are equal is not easy (and that may confused the interviewer).
We can say Sets A and B are identical(or equal) when contain same elements (we do not try to compare but instead refer to elements they contain so they must be finite). And even identification can't be easily applied when talking about Big O Sets.
Big O of F is used to notate that we are talking about the Set that
contains all functions with order greater or equal than F. How many
functions are there??
Infinite since F+c is contained and c can take infinite values.
How could you say two different Sets are identical(or equal) when they are
infinite ,well it is not that simple.
So I understand what you are thinking that n+m and max(n,m) have same
order but **the right way to express that** is by saying n+m is
O(max(n,m)) and max(n,m)is O(m+n) ( O(m+n) is equal to O(max(m,n))
may better requires a proof).
One more thing, we said that these functions have same order and this is absolutely mathematically correct but when trying to do optimization of an algorithm and you may need to take into account some lower order factors then maybe they give you slightly different results but the asymptotic behavior is proved to be the same.
CONCLUSION
As you can read in wikipedia (and in all cs courses in every university or in every algorithm book) Big O/θ/Ω/ω/ο notations helps us compare functions and find their order of growth and not for Sets of Functions and this is why you were told you were wrong. Though is easy to say O(n^2) is subset of O(n) it is very difficult to compare infinite to say if two sets are identical. Cantor have worked on categorizing infinite sets, for example we know that natural numbers are countable infinite and real numbers are uncountable infinite so real numbers are more than natural numbers even though both are infinite. It is getting very complicating when trying t order and categorize infinite sets and this would be more of a research in maths than a way of comparing functions.
UPDATE
It turns out you could simply prove O(n+m) equals to O(max(n,m)):
for every function F which belongs to O(n+m) this means that there are constant c and k such:
F <= c(n+m) for every n>=k and m>=k
then also stands:
F <= c(n+m)<= 2c*max(n,m)
so F belongs to O(max(n,m)) and as a result O(m+n) is subset of O(max(n,m)).
Now consider F belongs to O(max(n,m)) then there are constants c and k such:
F <= c*max(n+m) for every n>=k and m>=k
and we also have:
F <= c*max(n+m)<=2c(m+n) for every n>=k and m>=k
so there is c'=2c and with same k by definition: F is O(m+n) and as a result O(max(n,m)) is subset of O(n+m). Because we proved O(m+n) is subset of O(max(n,m)) we proved that O(max(m,n)) and O(m+n) are equal and this mathematical proof proves you had totally right without any doubt.
Finally note that proving that m+n is O(max(n,m)) and max(n,m) is O(m+n) doesn't proves immediately that sets are equal (we need a proof for that) as your saying it just proves that functions have same order but we didn't examine the sets. Though it is easy to see (in general case) that if f is O(g) and g is O(F) then you can easily prove in that case the big O sets equality like we did in the previous paragraph.
We'll show by rigorous Big-O analysis that you are indeed correct, given one possible choice of parameter of growth in your analysis. However, this does not necessarily mean that the viewpoint of the interviewer is incorrect, rather that his/her choice of parameter of growth differs. His/her prompt that your answer was outright incorrect, however, is questionable: you've possibly simply used two slightly different approaches to analyzing the asymptotic complexity of std::set_intersection, both leading to the general consensus that the algorithm runs in linear time.
Preparations
Lets start by looking at the reference of std::set_intersection at cppreference (emphasis mine)
http://en.cppreference.com/w/cpp/algorithm/set_intersection
Parameters
first1, last1 - the first range of elements to examine
first2, last2 - the second range of elements to examine
Complexity
At most 2·(N1+N2-1) comparisons, where
N1 = std::distance(first1, last1)
N2 = std::distance(first2, last2)
std::distance itself is naturally linear (worst case: no random access)
std::distance
...
Returns the number of elements between first and last.
We'll proceed to briefly recall the basic of Big-O notation.
Big-O notation
We loosely state the definition of a function or algorithm f being in O(g(n)) (to be picky, O(g(n)) being a set of functions, hence f ∈ O(...), rather than the commonly misused f(n) ∈ O(...)).
If a function f is in O(g(n)), then c · g(n) is an upper
bound on f(n), for some non-negative constant c such that f(n) ≤ c · g(n)
holds, for sufficiently large n (i.e. , n ≥ n0 for some constant
n0).
Hence, to show that f ∈ O(g(n)), we need to find a set of (non-negative) constants (c, n0) that fulfils
f(n) ≤ c · g(n), for all n ≥ n0, (+)
We note, however, that this set is not unique; the problem of finding the constants (c, n0) such that (+) holds is degenerate. In fact, if any such pair of constants exists, there will exist an infinite amount of different such pairs.
We proceed with the Big-O analysis of std::set_intersection, based on the already known worst case number of comparisons of the algorithm (we'll consider one such comparison a basic operation).
Applying Big-O asymptotic analysis to the set_intersection example
Now consider two ranges of elements, say range1 and range2, and assume that the number of elements contained in these two ranges are m and n, respectively.
Note! Already at this initial stage of the analys do we make a choice: we choose to study the problem in terms of two different parameters of growth (or rather, focusing on the largest one of these two). As we shall see ahead, this will lead to the same asymptotic complexity as the one stated by the OP. However, we could just as well choose to let k = m+n be the parameter of choice: we would still conclude that std::set_intersection is of linear-time complexity, but rather in terms of k (which is m+n which is not max(m, n)) than the largest of m and n. These are simply the preconditions we freely choose to set prior to proceeding with our Big-O notation/asymptotic analysis, and it's quite possibly that the interviewer had a preference of choosing to analyze the complexity using k as parameter of growth rather than the largest of its two components.
Now, from above we know that as worst case, std::set_intersection will run 2 · (m + n - 1) comparisons/basic operations. Let
h(n, m) = 2 · (m + n - 1)
Since the goal is to find an expression of the asymptotic complexity in terms of Big-O (upper bound), we may, without loss of generality, assume that n > m, and define
f(n) = 2 · (n + n - 1) = 4n - 2 > h(n,m) (*)
We proceed to analyze the asymptotic complexity of f(n), in terms of Big-O notation. Let
g(n) = n
and note that
f(n) = 4n - 2 < 4n = 4 · g(n)
Now (choose to) let c = 4 and n0 = 1, and we can state the fact that:
f(n) < 4 · g(n) = c · g(n), for all n ≥ n0, (**)
Given (**), we know from (+) that we've now shown that
f ∈ O(g(n)) = O(n)
Furthermore, since `(*) holds, naturally
h ∈ O(g(n)) = O(n), assuming n > m (i)
holds.
If we switch our initial assumption and assume that m > n, re-tracing the analysis above will, conversely, yield the similar result
h ∈ O(g(m)) = O(m), assuming m > n (ii)
Conclusion
Hence, given two ranges range1 and range2 holding m and n elements, respectively, we've shown that the asymptotic complexity of std::set_intersection applied two these two ranges is indeed
O(max(m, n))
where we're chosen the largest of m and n as the parameter of growth of our analysis.
This is, however, not really valid annotation (at least not common) when speaking about Big-O notation. When we use Big-O notation to describe the asymptotic complexity of some algorithm or function, we do so with regard to some single parameter of growth (not two of them).
Rather than answering that the complexity is O(max(m, n)) we may, without loss of generality, assume that n describes the number of elements in the range with the most elements, and given that assumption, simply state than an upper bound for the asymptotic complexity of std::set_intersection is O(n) (linear time).
A speculation as to the interview feedback: as mentioned above, it's possible that the interviewer simply had a firm view that the Big-O notation/asymptotic analysis should've been based on k = m+n as parameter of growth rather than the largest of its two components. Another possibility could, naturally, be that the interviewer simply confusingly queried about the worst case of actual number of comparisons of std::set_intersection, while mixing this with the separate matter of Big-O notation and asymptotic complexity.
Final remarks
Finally note that the analysis of worst case complexity of std::set_intersect is not at all representative for the commonly studied non-ordered set intersection problem: the former is applied to ranges that are already sorted (see quote from Boost's set_intersection below: the origin of std::set_intersection), whereas in the latter, we study the computation of the intersection of non-ordered collections.
Boost: set_intersection
Description
set_intersection constructs a sorted range that is the intersection
of the sorted ranges rng1 and rng2. The return value is the
end of the output range.
As an example of the latter, the Intersection set method of Python applies to non-ordered collections, and is applied to say sets s and t, it has an average case and a worst-case complexity of O(min(len(s), len(t)) and O(len(s) * len(t)), respectively. The huge difference between average and worst case in this implementation stems from the fact that hash based solutions generally works very well in practice, but can, for some applications, theoretically have a very poor worst-case performance.
For additional details of the latter problem, see e.g.
Intersection of two unsorted sets or lists # SE-CSTheory
Isn't O(n) an improvement over O(1 + n)?
This is my conception of the difference:
O(n):
for i=0 to n do ; print i ;
O(1 + n):
a = 1;
for i=0 to n do ; print i+a ;
... which would just reduce to O(n) right?
If the target time complexity is O(1 + n), but I have a solution in O(n),
does this mean I'm doing something wrong?
Thanks.
O(1+n) and O(n) are mathematically identical, as you can straightforwardly prove from the formal definition or using the standard rule that O( a(n) + b(n) ) is equal to the bigger of O(a(n)) and O(b(n)).
In practice, of course, if you do n+1 things it'll (usually, dependent on compiler optimizations/etc) take longer than if you only do n things. But big-O notation is the wrong tool to talk about those differences, because it explicitly throws away differences like that.
It's not an improvement because BigO doesn't describe the exact running time of your algorithm but rather its growth rate. BigO therefore describes a class of functions, not a single function. O(n^2) doesn't mean that your algorithms for input of size 2 will run in 4 operations, it means that if you were to plot the running time of your application as a function of n it would be asymptotically upper bound by c*n^2 starting at some n0. This is nice because we know how much slower our algorithm will be for each input size, but we don't really know exactly how fast it will be. Why use the c? Because as I said we don't care about exact numbers but more about the shape of the function - when we multiply by a constant factor the shape stays the same.
Isn't O(n) an improvement over O(1 + n)?
No, it is not. Asymptotically these two are identical. In fact, O(n) is identical to O(n+k) where k is any constant value.
Consider the function F: 2^(3*n) + n^2
Can the function A: 2^(3*n) be used as a Big Theta, Omega or O as a characterisation of F? Why?
I'm revising the concepts of Big Omega, Big Theta and Big O and I came across this example but don't know where to start.
No.
2^(3*n) is the leading term, but unless you're doing something very wrong it's not going to take you that long to compute. Wikipedia has a good list of time complexities of various functions. The most complicated operation you're doing is raising to a power, the complexity of which is discussed in other posts.
Since you're looking at a function of the form g(f(x)) where g(x) = 2^x and f(x) = 3x, the time to compute is going to be O(h) + O(k), where h is the time complexity of g, k is the time complexity of f. Think about it: it should never be more than this, if it were you could just break the operation in two and save time by doing the parts separately. Because h is going to dominate this sum you'll typically leave the k term off.
Yes, all three. In general, you only need to pay attention to the fastest growing term, as slower growing terms will be "swamped" by faster growing terms.
In detail:
Obviously F grows faster than A, so F \in \Omega(A) is a no-brainer. There is a positive multiple of A (namely A itself) that is smaller than F, for all sufficiently large n.
Try plotting F against 2*A. You will find that 2*A quickly gets bigger than F and stays bigger. Thus there is a positive multiple of A (namely 2*A) that is bigger than F for sufficiently large arguments. So by the definition of O, F \in O(A).
Finally, since F \in \Omega(A) and F \in O(A), F \in \Theta(A).