I have come across some of the difficulties during doing this question. The question is: sort the following functions by order of growth from slowest to fastest:
7n^3 − 10n, 4n^2, n, n^8621909, 3n, 2^(log log n), n log n, 6n log n, n!, 1.1^n
And my answer to the question is
n,3n
nlogn, 6nlogn
4n^2(which equals to n^2)
7n^3 – 10n(which equals to n^3)
n^ 8621909
2^loglogn
1.1^n(exponent 2^0.1376n)
n!
Just wondering: can I assume that 2^(loglogn) has the same growth as 2^n? Should I take 1.1^n as a constant?
Just wondering can i assume that 2^(loglogn) has the same growth as
2^n?
No. Assuming that the logs are in base 2 then 2^log(n) mathematically equals to n, so 2^(log(log(n)) = log(n). And of course it does not have the same growth as 2^n.
Should I take 1.1^n as a constant?
No. 1.1^n is still an exponent by n that cannot be ignored - of course it is not a constant.
The correct order is:
2^loglogn = log(n)
n,3n
nlogn, 6nlogn
4n^2
7n^3 – 10n
n^8621909
1.1^n
n!
I suggest to take a look at the Formal definition of the Big-O notation. But, for simplicity, the top of the list goes slower to infinity than the lower functions.
If for example, you will put two of those function on a graph you will see that one function will eventually pass the other one and will go faster to infinity.
Take a look at this comparing n^2 with 2^n. You will notice that 2^n is passing n^2 and going faster to infinity.
You might also want to check the graph in this page.
2log(x) = x, so 2log(log(n)) = log(n), which grows far more slower than 2n.
1.1n is definitely not a constant. 1.1n tends to infinity, and a constant obviously does not.
Related
For example i have f(N) = 5N +3 from a program. I want to know what is the big (oh) of this function. we say higher order term O(N).
Is this correct method to find big(oh) of any program by dropping lower orders terms and constants?
If we got O(N) by simply looking on that complexity function 5N+3. then, what is the purpose of this formula F(N) <= C* G(N)?
i got to know that, this formula is just for comparing two functions. my question is,
In this formula, F(N) <= C* G(N), i have F(N) = 5N+3, but what is this upper bound G(N)? Where it comes from? where from will we take it?
i have studied many books, and many posts, but i am still facing confusions.
Q: Is this correct method to find big(oh) of any program by dropping
lower orders terms and constants?
Yes, most people who have at least some experience with examining time complexities use this method.
Q: If we got O(N) by simply looking on that complexity function 5N+3.
then, what is the purpose of this formula F(N) <= C* G(N)?
To formally prove that you correctly estimated big-oh for certain algorithm. Imagine that you have F(N) = 5N^2 + 10 and (incorrectly) conclude that the big-oh complexity for this example is O(N). By using this formula you can quickly see that this is not true because there does not exist constant C such that for large values of N holds 5N^2 + 10 <= C * N. This would imply C >= 5N + 10/N, but no matter how large constant C you choose, there is always N larger than this constant, so this inequality does not hold.
Q: In this formula, F(N) <= C* G(N), i have F(N) = 5N+3, but what is this
upper bound G(N)? Where it comes from? where from will we take it?
It comes from examining F(N), specifically by finding its highest order term. You should have some math knowledge to estimate which function grows faster than the other, for start check this useful link. There are several classes of complexities - constant, logarithmic, polynomial, exponential.. However, in most cases it is easy to find the highest order term for any function. If you are not sure, you can always plot a graph of a function or formally prove that one function grows faster than the other. For example, if F(N) = log(N^3) + sqrt(N) maybe it is not clear at first glance what's the highest order term, but if you calculate or plot log(N^3) for N = 1, 10 and 1000 and sqrt(N) for same values, it is immediately clear that sqrt(N) grows faster, so big-oh for this function is O(sqrt(N)).
The following question was on a recent assignment in University. I would have thought the answer would be n^2+T(n-1) as I thought the n^2 would make it's asymptotic time complexity O(n^2). Where as with T(n/2)+1 its asymptotic time complexity would be O(log2(n)).
The answers were returned and it turns out the correct answer is T(n/2)+1 however I can't get my head around why this is the case.
Could someone possibly explain to me why that's the worst case time complexity of this algorithm? It's possible my understanding of time complexity is just wrong.
The asymptotic time complexity is taking n large. In the case of your example, since the question specifies that k is fixed, the only complexity relevant is the last one. See the Wikipedia formal definition, specifically:
As n grows to infinity, the recursion that dominates T(n) = T(n / 2) + 1. You can prove this as well using the formal definition, basically picking x_0 = 10 * k and showing that a finite M can be found using the first two cases. It should be clear that both log(n) and n^2 satisfy the definition, so the tighter bound is the asymptotic complexity.
What does O (f (n)) mean? It means the time is at most c * f (n), for some unknown and possibly large c.
kevmo claimed a complexity of O (log2 n). Well, you can check all the values n ≤ 10k, and let the largest value of T (n) be X. X might be quite large (about 167 k^3 in this case, I think, but it doesn't actually matter). For larger n, the time needed is at most X + log2 (n). Choose c = X, and this is always less than c * log2 (n).
Of course people usually assume that a O (log n) algorithm would be quick, and this one most certainly isn't if say k = 10,000. So you learned as well that O notation must be handled with care.
Talking about Big O notations, if one algorithm time complexity is O(N) and other's is O(2N), which one is faster?
The definition of big O is:
O(f(n)) = { g | there exist N and c > 0 such that g(n) < c * f(n) for all n > N }
In English, O(f(n)) is the set of all functions that have an eventual growth rate less than or equal to that of f.
So O(n) = O(2n). Neither is "faster" than the other in terms of asymptotic complexity. They represent the same growth rates - namely, the "linear" growth rate.
Proof:
O(n) is a subset of O(2n): Let g be a function in O(n). Then there are N and c > 0 such that g(n) < c * n for all n > N. So g(n) < (c / 2) * 2n for all n > N. Thus g is in O(2n).
O(2n) is a subset of O(n): Let g be a function in O(2n). Then there are N and c > 0 such that g(n) < c * 2n for all n > N. So g(n) < 2c * n for all n > N. Thus g is in O(n).
Typically, when people refer to an asymptotic complexity ("big O"), they refer to the canonical forms. For example:
logarithmic: O(log n)
linear: O(n)
linearithmic: O(n log n)
quadratic: O(n2)
exponential: O(cn) for some fixed c > 1
(Here's a fuller list: Table of common time complexities)
So usually you would write O(n), not O(2n); O(n log n), not O(3 n log n + 15 n + 5 log n).
Timothy Shield's answer is absolutely correct, that O(n) and O(2n) refer to the same set of functions, and so one is not "faster" than the other. It's important to note, though, that faster isn't a great term to apply here.
Wikipedia's article on "Big O notation" uses the term "slower-growing" where you might have used "faster", which is better practice. These algorithms are defined by how they grow as n increases.
One could easily imagine a O(n^2) function that is faster than O(n) in practice, particularly when n is small or if the O(n) function requires a complex transformation. The notation indicates that for twice as much input, one can expect the O(n^2) function to take roughly 4 times as long as it had before, where the O(n) function would take roughly twice as long as it had before.
It depends on the constants hidden by the asymptotic notation. For example, an algorithm that takes 3n + 5 steps is in the class O(n). So is an algorithm that takes 2 + n/1000 steps. But 2n is less than 3n + 5 and more than 2 + n/1000...
It's a bit like asking if 5 is less than some unspecified number between 1 and 10. It depends on the unspecified number. Just knowing that an algorithm runs in O(n) steps is not enough information to decide if an algorithm that takes 2n steps will complete faster or not.
Actually, it's even worse than that: you're asking if some unspecified number between 1 and 10 is larger than some other unspecified number between 1 and 10. The sets you pick from being the same doesn't mean the numbers you happen to pick will be equal! O(n) and O(2n) are sets of algorithms, and because the definition of Big-O cancels out multiplicative factors they are the same set. Individual members of the sets may be faster or slower than other members, but the sets are the same.
Theoretically O(N) and O(2N) are the same.
But practically, O(N) will definitely have a shorter running time, but not significant. When N is large enough, the running time of both will be identical.
O(N) and O(2N) will show significant difference in growth for small numbers of N, But as N value increases O(N) will dominate the growth and coefficient 2 becomes insignificant. So we can say algorithm complexity as O(N).
Example:
Let's take this function
T(n) = 3n^2 + 8n + 2089
For n= 1 or 2, the constant 2089 seems to be the dominant part of function but for larger values of n, we can ignore the constants and 8n and can just concentrate on 3n^2 as it will contribute more to the growth, If the n value still increases the coefficient 3 also seems insignificant and we can say complexity is O(n^2).
For detailed explanation refer here
O(n) is faster however you need to understand that when we talk about Big O, we are measuring the complexity of a function/algorithm, not its speed. And we measure this complexity asymptotically. In lay man terms, when we talk about asymptotic analysis, we take immensely huge values for n. So if you plot the graph for O(n) and O(2n), the values will stay in some particular range from each other for any value of n. They are much closer compared to the other canonical forms like O(nlogn) or O(1), so by convention we approximate the complexity to the canonical form O(n).
To compare the asymptotic order of the two functions, I calculated the limit of first function over second function, when n goes to infinity.
The answer was 2 (I had to use l'hopital's rule), which means that for really high values of n, log(n^2) is larger than log(5n)
My question is: is it incorrect to say that log(n^2) is asymptotically larger than log(5n)?
My friend told me that when the limit of first function over the second function is a constant, that means that their asymptotic order is equal. Can someone confirm?
Actually log (5n) = log 5 + log n, and log (n^2) = 2 log n, so log(n^2) is larger than log 5n. In addition, we can say that log(n^2) is asymptotically larger than log 5n. One definition of Asymptotic is as follows.
The term asymptotic means approaching a value or curve arbitrarily
closely (i.e., as some sort of limit is taken). A line or curve A that
is asymptotic to given curve C is called the asymptote of C.
Depending on the context we may ignore constant factors, and write they are in the same order. We may express this by existing notations such as O, Θ and Ω. According to the widely accepted definition from the algorithmic standpoint, these two particular functions are asymptotically equivalent:
We say A(n) is asymptotically larger than B(n) if
lim n→∞A(n)/B(n) = ∞
In this case, the above limit converges to 2 (or reverse version 1/2) so they are asymptotically equal.
log(n^2) = 2 * log(n) and log(5n) = log(5) + log(n). So both are asymptotically equal when speaking about algorithms.
log(n^2)=2*log(n)
Assuming log base 2
If the limit was 2, then it means that log(5n) belongs to O(log(n^2)), doesn't it...
Lets say I have a routine that scans an entire list of n items 3 times, does a sort based on the size, and then bsearches that sorted list n times. The scans are O(n) time, the sort I will call O(n log(n)), and the n times bsearch is O(n log(n)). If we add all 3 together, does it just give us the worst case of the 3 - the n log(n) value(s) or does the semantics allow added times?
Pretty sure, now that I type this out that the answer is n log(n), but I might as well confirm now that I have it typed out :)
The sum is indeed the worst of the three for Big-O.
The reason is that your function's time complexity is
(n) => 3n + nlogn + nlogn
which is
(n) => 3n + 2nlogn
This function is bounded above by 3nlogn, so it is in O(n log n).
You can choose any constant. I just happened to choose 3 because it was a good asymptotic upper bound.
You are correct. When n gets really big, the n log(n) dominates 3n.
Yes it will just be the worst case since O-notation is just about asymptotic performance.
This should of course not be taken to mean that adding these extra steps will have no effect on your programs performance. One of the O(n) steps could easily consume a huge portion of your execution time for the given range of n where your program operates.
As Ray said, the answer is indeed O(n log(n)). The interesting part of this question is that it doesn't matter which way you look at it: does adding mean "actual addition" or does it mean "the worst case". Let's prove that these two ways of looking at it produce the same result.
Let f(n) and g(n) be functions, and without loss of generality suppose f is O(g). (Informally, that g is "worse" than f.) Then by definition, there exists constants M and k such that f(n) < M*g(n) whenever n > k. If we look at in the "worst case" way, we expect that f(n)+g(n) is O(g(n)). Now looking at it in the "actual addition" way, and specializing to the case where n > k, we have f(n) + g(n) < M*g(n) + g(n) = (M+1)*g(n), and so by definition f(n)+g(n) is O(g(n)) as desired.