I'm trying to prove that this is correct for any function f and g with domain and co-domain N. I have seen it proven using limits, but apparently you can also prove it without them.
Essentially what I'm trying to prove is "If f(n) doesn't have a big-O of g(n) then g(n) must have a big-O of f(n). What I'm having trouble is trying to understand what "f doesn't have a big-O of g" means.
According to the formal definition of big-O, if f(n) = O(g(n)) then n>=N -> f(n) <= cg(n) for some N and a constant c. If f(n) != O(g(n)) I think that means there is no c that fulfills this inequality for all values of n. Yet I don't see what I can do to use that fact to prove g(n) = O(f(n)). That doesn't prove that a c' exists for g(n) <= c'f(n), which would successfully prove the question.
Not true. Let f(n) = 1 if n is odd and zero otherwise, and g(n) = 1 if n is even and zero otherwise.
To say that f is O(g) would say there is a constant C > 0 and N > 0 such that n > N implies f(n) <= C g(n). Let n = 2 * N + 1, so that n is odd. Then f(n) = 1 but g(n) = 0 so that f(n) <= C * g(n) is impossible. Thus, f is O(g) is not true.
Similarly, we can show that g is O(f) is not true.
First of all, your definition of big-O is a little bitt off. You say:
I think that means there is no c that fulfills this inequality for all values of n.
In actuality, you need to pick a value c that fulfills the inequality for any value of n.
Anyway, to answer the question:
I don't believe the statement in the question is true... Let's see if we can think of a counter-example, where f(n) ≠ O(g(n)) and g(n) ≠ O(f(n)).
note: I'm going to use n and x interchangeably, since it's easier for me to think that way.
We'd have to come up with two functions that continually cross each other as they go towards infinity. Not only that, but they'd have to continue to cross each other regardless of the constant c that we multibly them by.
So that leaves me thinking that the functions will have to alternate between two different time complexities.
Let's look at a function that alternates between y = x and y = x^2:
f(x) = .2 (x * sin(x) + x^2 * (1 - sin(x)) )
Now, if we create a similar function with a slightly offset oscillation:
g(x) = .2 (x * cos(x) + x^2 * (1 - cos(x)) )
Then these two functions will continue to cross each others' paths out to infinity.
For any number N that you select, no matter how high, there will be an x1 greater than N such that f(x) = x^2 and g(x) = x. Similarly, there will be an x2 such that g(x) = x^2 and f(x) = x.
At these points, you won't be able to choose any c1 or c2 that will ensure that f(x) < c1 * g(x) or that g(x) < c2 * f(x).
In conclusion, f(n) ≠ O(g(n)) does not imply g(n) = O(f(n)).
Related
In other words, is o(f(n)) = O(f(n)) - Θ(f(n))?
f ∈ O(g) [big O] says, essentially
For at least one choice of a constant k > 0, you can find a constant y such that the inequality 0 <= f(x) <= k g(x) holds for all x > y.
f ∈ Θ(g) [theta] says, essentially
For at least one choice of constants k1, k2 > 0, you can find a constant y such that the inequality 0 <= k1 g(x) <= f(x) <= k2 g(x) holds for all x > y.
f ∈ o(g) [little o] 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 > y.
By the definition, it is easy to realize that o(g) ⊆ O(g), and Θ(g) ⊆ O(g). And it makes sense to one complement each other. I couldn't find any counter example of function that is in O(f(n)) and not in Θ(f(n)) that is not in o(f(n)).
Surprisingly, no, this isn’t the case. Intuitively, big-O notation gives an upper bound without making any claims about lower bounds. If you subtract out the big-Θ class, you’ve removed functions that are bounded from above and below by the function. That leaves you with some extra functions that are upper-bounded by the function but not lower bounded by it.
As an example, let f(n) be n if n is even and 0 otherwise. Then f(n) = O(n) but f(n) ≠ Θ(n). However, it’s not true that f(n) = o(n).
Suppose that f(n)=4^n and g(n)=n^n, will it be right to conclude that f(n)=Θ(g(n)).
In my opinion it's a correct claim but I'm not 100% sure.
It is incorrect. f(n) = Theta(g(n)) if and only if both f(n) = O(g(n)) and g(n) = O(f(n)). It is true that f(n) = O(g(n)). We will show that it is not the case that g(n) = O(f(n)).
Assume g(n) = O(f(n)). Then there exists a positive real constant c and a positive natural number n0 such that for all n > n0, g(n) <= c * f(n). For our functions, this implies n^n <= c * 4^n. If we take the nth root of both sides of this inequality we get n <= 4c^(1/n). We are free to assume c >= 1 and n0 >= since if we found a smaller value that worked a larger value would work too. For all c > 1 and n > 1, 4c^(1/n) is strictly less than 4c. But then if we choose n > 4c, the inequality is false. So, there cannot be an n0 such that for all n at least n0 the condition holds. This is a contradiction; our initial assumption is disproven.
We are asked to indicate whether f = O(g), or f = Ω(g), or both (in which case f = Θ(g)).
To solve the big O, I found it easy by simply providing constant C = 1 in which case 2^n <= 1(2^n+1).
I was under the impression that to solve the Ω would be impossible since there is no C in which 2^n >= C(2^n+1).
Upon looking into the solutions to check my work, I found that f = Θ(g). How could this be with this problem? What constant C could satisfy this?
What is the problem if C = 0.1 as an example? Also, you can show the Theta notation by the limit of these two functions such that lim(f(n)/g(n)) when n goes to \infty is equal 1. It means we have f(n) = \Theta(g(n)).
I can't tell whether g(n) is 2^(n+1) or (2^n)+1. In either case, f(n) is Theta(g(n)).
Assume g(n) = 2^(n+1). We can rewrite this using laws of exponents as (2^n)(2^1) which is the same thing as 2*(2^n). Now we may simply choose c = 1/2 and then f(n) = c * g(n). Because there exists a c for which the functions are simply equal, immediately f(n) = Theta(g(n)).
Assume g(n) = (2^n) + 1. Since f(n) < g(n) we immediately have f(n) = O(g(n)). All we need to show is that f(n) >= c * g(n). If we choose c = 1/2 we need to find n0 such that 2^n0 >= 1/2 2^n0 + 1/2. We can guess n0 = 1 and we find that this works; 2^1 >= 1/2 2^1 + 1/2. Because f(n) also grows faster than (1/2)2^n + 1/2, we are done.
I know the definitions of both of them, but what is the reason sometimes I see O(1) and other times Θ(1) written in textbooks?
Thanks.
O(1) and Θ(1) aren't necessarily the same if you are talking about functions over real numbers. For example, consider the function f(n) = 1/n. This function is O(1) because for any n ≥ 1, f(n) ≤ 1. However, it is not Θ(1) for the following reason: one definition of f(n) = Θ(g(n)) is that the limit of |f(n) / g(n)| as n goes to infinity is some finite value L satisfying 0 < L. Plugging in f(n) = 1/n and g(n) = 1, we take the limit of |1/n| as n goes to infinity and get that it's 0. Therefore, f(n) ≠ Θ(1).
Hope this helps!
Big-O notation expresses an asymptotic upper bound, whereas Big-Theta notation additionally expresses an asymptotic lower bound. Often, the upper bound is what people are interested in, so they write O(something), even when Theta(something) would also be true. For example, if you wanted to count the number of things that are equal to x in an unsorted list, you might say that it can be done in linear time and is O(n), because what matters to you is that it won't take any longer than that. However, it would also be true that it's Omega(n) and therefore Theta(n), since you have to examine all of the elements in the list - it can't be done in sub-linear time.
UPDATE:
Formally:
f in O(g) iff there exists a c and an n0 such that for all n > n0, f(n) <= c * g(n).
f in Omega(g) iff there exists a c and an n0 such that for all n > n0, f(n) >= c * g(n).
f in Theta(g) iff f in O(g) and f in Omega(g), i.e. iff there exist a c1, a c2 and an n0 such that for all n > n0, c1 * g(n) <= f(n) <= c2 * g(n).
I came across two asymptotic function proofs.
f(n) = O(g(n)) implies 2^f(n) = O(2^g(n))
Given: f(n) ≤ C1 g(n)
So, 2^f(n) ≤ 2^C1 g(n) --(i)
Now, 2^f(n) = O(2^g(n)) → 2^f(n) ≤ C2 2^g(n) --(ii)
From,(i) we find that (ii) will be true.
Hence 2^f(n) = O(2^g(n)) is TRUE.
Can you tell me if this proof is right? Is there any other way to solve this?
2.f(n) = O((f(n))^2)
How to prove the second example? Here I consider two cases one is if f(n)<1 and other is f(n)>1.
Note: None of them are homework questions.
The attempted-proof for example 1 looks well-intentioned but is flawed. First, “2^f(n) ≤ 2^C1 g(n)” means 2^f(n) ≤ (2^C1)*g(n), which in general is false. It should have been written 2^f(n) ≤ 2^(C1*g(n)). In the line beginning with “Now”, you should explicitly say C2 = 2^C1. The claim “(ii) will be true” is vacuous (there is no (ii)).
A function like f(n) = 1/n disproves the claim in example 2 because there are no constants N and C such that for all n > N, f(n) < C*(f(n))². Proof: Let some N and C be given. Choose n>N, n>C. f(n) = 1/n = n*(1/n²) > C*(1/n²) = C*(f(n))². Because N and C were arbitrarily chosen, this shows that there are no fixed values of N and C such that for all n > N, f(n) < C*(f(n))², QED.
Saying that “f(n) ≥ 1” is not enough to allow proving the second claim; but if you write “f(n) ≥ 1 for all n” or “f() ≥ 1” it is provable. For example, if f(n) = 1/n for odd n and 1+n for even n, we have f(n) > 1 for even n > 0, and less than 1 for odd n. To prove that f(n) = O((f(n))²) is false, use the same proof as in the previous paragraph but with the additional provision that n is even.
Actually, “f(n) ≥ 1 for all n” is stronger than necessary to ensure f(n) = O((f(n))²). Let ε be any fixed positive value. No matter how small ε is, “f(n) ≥ ε for all n > N'” ensures f(n) = O((f(n))²). To prove this, take C = max(1, 1/ε) and N=N'.