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).
Related
What is the difference between Big-O notation O(n) and Little-O notation o(n)?
f ∈ O(g) says, essentially
For at least one choice of a constant k > 0, you can find a constant a such that the inequality 0 <= f(x) <= k g(x) holds for all x > a.
Note that O(g) is the set of all functions for which this condition holds.
f ∈ o(g) says, essentially
For every choice of a constant k > 0, you can find a constant a such that the inequality 0 <= f(x) < k g(x) holds for all x > a.
Once again, note that o(g) is a set.
In Big-O, it is only necessary that you find a particular multiplier k for which the inequality holds beyond some minimum x.
In Little-o, it must be that there is a minimum x after which the inequality holds no matter how small you make k, as long as it is not negative or zero.
These both describe upper bounds, although somewhat counter-intuitively, Little-o is the stronger statement. There is a much larger gap between the growth rates of f and g if f ∈ o(g) than if f ∈ O(g).
One illustration of the disparity is this: f ∈ O(f) is true, but f ∈ o(f) is false. Therefore, Big-O can be read as "f ∈ O(g) means that f's asymptotic growth is no faster than g's", whereas "f ∈ o(g) means that f's asymptotic growth is strictly slower than g's". It's like <= versus <.
More specifically, if the value of g(x) is a constant multiple of the value of f(x), then f ∈ O(g) is true. This is why you can drop constants when working with big-O notation.
However, for f ∈ o(g) to be true, then g must include a higher power of x in its formula, and so the relative separation between f(x) and g(x) must actually get larger as x gets larger.
To use purely math examples (rather than referring to algorithms):
The following are true for Big-O, but would not be true if you used little-o:
x² ∈ O(x²)
x² ∈ O(x² + x)
x² ∈ O(200 * x²)
The following are true for little-o:
x² ∈ o(x³)
x² ∈ o(x!)
ln(x) ∈ o(x)
Note that if f ∈ o(g), this implies f ∈ O(g). e.g. x² ∈ o(x³) so it is also true that x² ∈ O(x³), (again, think of O as <= and o as <)
Big-O is to little-o as ≤ is to <. Big-O is an inclusive upper bound, while little-o is a strict upper bound.
For example, the function f(n) = 3n is:
in O(n²), o(n²), and O(n)
not in O(lg n), o(lg n), or o(n)
Analogously, the number 1 is:
≤ 2, < 2, and ≤ 1
not ≤ 0, < 0, or < 1
Here's a table, showing the general idea:
(Note: the table is a good guide but its limit definition should be in terms of the superior limit instead of the normal limit. For example, 3 + (n mod 2) oscillates between 3 and 4 forever. It's in O(1) despite not having a normal limit, because it still has a lim sup: 4.)
I recommend memorizing how the Big-O notation converts to asymptotic comparisons. The comparisons are easier to remember, but less flexible because you can't say things like nO(1) = P.
I find that when I can't conceptually grasp something, thinking about why one would use X is helpful to understand X. (Not to say you haven't tried that, I'm just setting the stage.)
Stuff you know: A common way to classify algorithms is by runtime, and by citing the big-Oh complexity of an algorithm, you can get a pretty good estimation of which one is "better" -- whichever has the "smallest" function in the O! Even in the real world, O(N) is "better" than O(N²), barring silly things like super-massive constants and the like.
Let's say there's some algorithm that runs in O(N). Pretty good, huh? But let's say you (you brilliant person, you) come up with an algorithm that runs in O(N⁄loglogloglogN). YAY! Its faster! But you'd feel silly writing that over and over again when you're writing your thesis. So you write it once, and you can say "In this paper, I have proven that algorithm X, previously computable in time O(N), is in fact computable in o(n)."
Thus, everyone knows that your algorithm is faster --- by how much is unclear, but they know its faster. Theoretically. :)
In general
Asymptotic notation is something you can understand as: how do functions compare when zooming out? (A good way to test this is simply to use a tool like Desmos and play with your mouse wheel). In particular:
f(n) ∈ o(n) means: at some point, the more you zoom out, the more f(n) will be dominated by n (it will progressively diverge from it).
g(n) ∈ Θ(n) means: at some point, zooming out will not change how g(n) compare to n (if we remove ticks from the axis you couldn't tell the zoom level).
Finally h(n) ∈ O(n) means that function h can be in either of these two categories. It can either look a lot like n or it could be smaller and smaller than n when n increases. Basically, both f(n) and g(n) are also in O(n).
I think this Venn diagram (adapted from this course) could help:
It's the exact same has what we use for comparing numbers:
In computer science
In computer science, people will usually prove that a given algorithm admits both an upper O and a lower bound 𝛺. When both bounds meet that means that we found an asymptotically optimal algorithm to solve that particular problem Θ.
For example, if we prove that the complexity of an algorithm is both in O(n) and 𝛺(n) it implies that its complexity is in Θ(n). (That's the definition of Θ and it more or less translates to "asymptotically equal".) Which also means that no algorithm can solve the given problem in o(n). Again, roughly saying "this problem can't be solved in (strictly) less than n steps".
Usually the o is used within lower bound proof to show a contradiction. For example:
Suppose algorithm A can find the min value in an array of size n in o(n) steps. Since A ∈ o(n) it can't see all items from the input. In other words, there is at least one item x which A never saw. Algorithm A can't tell the difference between two similar inputs instances where only x's value changes. If x is the minimum in one of these instances and not in the other, then A will fail to find the minimum on (at least) one of these instances. In other words, finding the minimum in an array is in 𝛺(n) (no algorithm in o(n) can solve the problem).
Details about lower/upper bound meanings
An upper bound of O(n) simply means that even in the worse case, the algorithm will terminate in at most n steps (ignoring all constant factors, both multiplicative and additive). A lower bound of 𝛺(n) is a statement about the problem itself, it says that we built some example(s) where the given problem couldn't be solved by any algorithm in less than n steps (ignoring multiplicative and additive constants). The number of steps is at most n and at least n so this problem complexity is "exactly n". Instead of saying "ignoring constant multiplicative/additive factor" every time we just write Θ(n) for short.
The big-O notation has a companion called small-o notation. The big-O notation says the one function is asymptotical no more than another. To say that one function is asymptotically less than another, we use small-o notation. The difference between the big-O and small-o notations is analogous to the difference between <= (less than equal) and < (less than).
I am a college student majoring in Artificial Intelligence. I feel stupid because I just can't understand Time Complexity. And my teacher won't help me out. Im failing and clueless. Can anyone help me to understand Time Complexity? For example what is the Big O of this code? or what is the Time Complexity for this binary tree. I got 4/50 on my recent exam so Im really desperate now
Big-O is basically the amount of time/space (time complexity / space complexity) that a specific algorithm runs in.
It depends on the operations of the algorithm. I will exemplify the time complexity.
E.g. the function:
## l is a list of size=n
def linear(l):
for i in l:
print(i)
iterates over a list of size=n, so the algorithm will will need to get all the n values until accomplished its purpose.
So, it's big-o is O(n) as it runs linearly with the amount of n.
The function:
## l is a list of k lists, each of them has size=n
def quadratic(matrix):
for l in matrix:
for i in l:
print(i)
Here, you have to iterate over k lists, and over n elements in each of them.
So, the algorithm runs in O(k*n).
My Lua function:
for y=userPosY+radius,userPosY-radius,-1 do
for x=userPosX-radius,userPosX+radius,1 do
local oneNeighborFound = redis.call('lrange', userPosZone .. x .. y, '0', '0')
if next(oneNeighborFound) ~= nil then
table.insert(neighborsFoundInPosition, userPosZone .. x .. y)
neighborsFoundInPositionCount = neighborsFoundInPositionCount + 1
end
end
end
Which leads to this formula: (2n+1)^2
As I understand it correctly, that would be a time complexity of O(n^2).
How can I compare this to the time complexity of the GEORADIUS (Redis) with O(N+log(M))? https://redis.io/commands/GEORADIUS
Time complexity: O(N+log(M)) where N is the number of elements inside the bounding box of the circular area delimited by center and radius and M is the number of items inside the index.
My time complexity does not have a M. I do not know how many items are in the index (M) because I do not need to know that. My index changes often, almost with every request and can be large.
Which time complexity is when better?
Assuming N and M were independent variables, I would treat O(N + log M) the same way you treat O(N3 - 7N2 - 12N + 42): the latter becomes O(N3) simply because that's the term that has most effect on the outcome.
This is especially true as time complexity analysis is not really a case of considering runtime. Runtime has to take into account the lesser terms for specific limitations of N. For example, if your algorithm runtime can be expressed as runtime = N2 + 9999999999N, and N is always in the range [1, 4], it's the second term that's more important, not the first.
It's better to think of complexity analysis as what happens as N approaches infinity. With the O(N + log M) one, think about what happens when you:
double N?
double M?
The first has a much greater impact so I would simply convert the complexity to O(N).
However, you'll hopefully have noticed the use of the word "independent" in my first paragraph. The only sticking point to my suggestion would be if M was actually some function of N, in which case it may become the more important term.
Any function that reversed the impact of the log M would do this, such as the equality M = 101010N.
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
for (i=0;i<n;i++)
{
enumerate all subsets of size i = 2^n
each subset of size i takes o(nlogn) to search a solution
from all these solution I want to search the minimum subset of size S.
}
I want to know the complexity of this algorithm it'is 2^n O(nlogn*n)=o(2^n n²) ??
If I understand you right:
You iterate all subsets of a sorted set of n numbers.
For each subset you test in O(n log n) if its is a solution. (how ever you do this)
After you have all this solutions you looking for the one with exact S elements with the smalest sum.
The way you write it, the complexity would be O(2^n * n log n) * O(log (2^n)) = O(2^n * n^2 log n). O(log (2^n)) = O(n) is for searching the minimum solution, and you do this every round of the for loop with worst case i=n/2 and every subset is a solution.
Now Im not sure if you mixing O() and o() up.
2^n O(nlogn*n)=o(2^n n²) is only right if you mean 2^n O(nlog(n*n)).
f=O(g) means, the complexity of f is not bigger than the complexity of g.
f=o(g) means the complexity of f is smaller than the complexity of g.
So 2^n O(nlogn*n) = O(2^n n logn^2) = O(2^n n * 2 logn) = O(2^n n logn) < O(2^n n^2)
Notice: O(g) = o(h) is never a good notation. You will (most likly every time) find a function f with f=o(h) but f != O(g), if g=o(h).
Improvements:
If I understand your algorithm right, you can speed it a little up. You know the size of the subset you looking for, so only look at all the subsets that have the size S. The worst case is S=n/2, so C(n,n/2) ~ 2^(n-1) will not reduce the complexity but saves you a factor 2.
You can also just save a solution and check if the next solution is smaller. this way you get the smallest solution without serching for it again. So the complexity would be O(2^n * n log n).