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.
Related
Let's say you were given two logarithmic functions like
and you were asked to find if f(n) is O(g(n)) Ω(g(n)) or Θ(g(n)), how would you go about it? I found questions like these easier when you were comparing two exponential equations, because for example with x(n) = n^2 and p(n) = n^2 you could find a c > 0 (ex 3) where x(n) <= cp(n) for all n greater than some n>0 and that would prove that x(n) = O(p(n)). However, comparing two logarithmic functions seems much more difficult for some reason. Any help is appreciated, thanks!
f(n) is O(g(n)) iff there is a constant c and n_0 such that f(n) <= c * g(n) for each n >= n_0.
f(n) is Ω(g(n)) iff there is a constant c and n_0 such that f(n) >= c * g(n) for each n >= n_0.
Now, f(n) is Θ(g(n)) iff f(n) is O(g(n)) and f(n) is Ω(g(n)).
So, in your cases, we have:
f(n) = log (n^2) = 2logn
which means, g(n) is logn and c = 2, which means f(n) <= 2 * logn and f(n) >= 2 * logn, which makes it Ω(logn).
Btw. its also f(n) <= n and f(n) >= 1, so f(n) can be O(n), but we don't use it, since we can find a better O(g(n)). In this case we don't have the same function in both notations, to for those values we don't have Ω. However, we just need one option for g(n) to declare Ω. In cases we can't find it, we say its not Ω. Note the word "we say".
In second case, we care only for "highest growing value", logn part. Now, c = 1, and g = log(n), so in this case, its also Ω(logn).
I have been given the problem:
f(n) are asymptotically positive functions. Prove f(n) = Θ(g(n)) iff g(n) = Θ(f(n)).
Everything I have found points to this statement being invalid. For example an answer I've come across states:
f(n) = O(g(n)) implies g(n) = O(f(n))
f(n) = O(g(n)) means g(n) grows faster than f(n). It cannot imply that f(n) grows
faster than g(n). Hence not true.
Another states:
If f(n) = O(g(n)) then O(f(n)). This is false. If f(n) = 1 and g(n) = n
for all natural numbers n, then f(n) <= g(n) for all natural numbers n, so
f(n) = O(g(n)). However, suppose g(n) = O(f(n)). Then there are natural
numbers n0 and a constant c > 0 such that n=g(n) <= cf(n) = c for all n >=
n0 which is impossible.
I understand that there are slight differences between my exact question and the examples I have found, but I've only been able to come up with solutions that do not prove it. I am correct in thinking that it is not able to be proved or am I looking over some detail?
You can start from here:
Formal Definition: f(n) = Θ (g(n)) means there are positive constants c1, c2, and k, such that 0 ≤ c1g(n) ≤ f(n) ≤ c2g(n) for all n ≥ k.
Because you have that iff, you need to start from the left side and to prove the right side, and then start from the right side and prove the left side.
Left -> right
We consider that:
f(n) = Θ(g(n))
and we want to prove that
g(n) = Θ(f(n))
So, we have some positive constants c1, c2 and k such that:
0 ≤ c1*g(n) ≤ f(n) ≤ c2*g(n), for all n ≥ k
The first relation between f and g is:
c1*g(n) ≤ f(n) => g(n) ≤ 1/c1*f(n) (1)
The second relation between f and g is:
f(n) ≤ c2*g(n) => 1/c2*f(n) ≤ g(n) (2)
If we combine (1) and (2), we obtain:
1/c2*f(n) ≤ g(n) ≤ 1/c1*f(n)
If you consider c3 = 1/c2 and c4 = 1/c1, they exist and are positive (because the denominators are positive). And this is true for all n ≥ k (where k can be the same).
So, we have some positive constants c3, c4, k such that:
c3*f(n) ≤ g(n) ≤ c4*f(n), for all n ≥ k
which means that g(n) = Θ(f(n)).
Analogous for right -> left.
I was given the function 5n^3+2n+8 to prove for big-O and big-Omega. I finished big-O, but for big-Omega I end up with a single-term function. I canceled out 2n and 8 because they're positive and make my function larger, so I just end up with 5n^3. How do I choose C and n_0? or is it simply trivial in this case?
From Big-Ω (Big-Omega) notation (slightly modified):
If a running time of some function f(n) is Ω(g(n)), then for large
enough n, say n > n_0 > 0, the running time of f(n) is at least
C⋅g(n), for some constant C > 0.
Hence, if f(n) is in Ω(g(n)), then there exists some positive constants C and n_0 such at the following holds
f(n) ≥ C⋅g(n), for all n > n_0 (+)
Now, the choice of C and n_0 is not unique, it suffices that you can show one such set of constants (such that (+) holds) to be able to describe the running time using the Big-Omega notation, as posted above.
Hence, you are indeed almost there
f(n) = 5n^3+2n+8 > 5n^3 holds for all n larger than say, 1
=> f(n) ≥ 5⋅n^3 for all n > n_0 = 1 (++)
Finally, (++) is just (+) for g(n) = n^3 and C=5, and hence, by (+), f(n) is in Ω(n^3).
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'.
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)).