In proving and disproving Big O questions that explicitly say use the definition to prove and disprove, my question is, is what I am doing correct?
For example you have a question that is g(n) = O(f(n)) ... In order to prove it I was doing the following
g(n) <= C(F(n))
g(n)/F(n) <= C .. then give n=1 and solve for C , which proves it.
The contradiction that I run to when doing this is when i approach a question of disproving this stuff
for example
g(n) =O(F(n)) to disprove it I would do
g(n) >= C(F(n)) and solve for C again . However this leads me to believe that big O can be proved and disproved at once ? which is 100% wrong.
Using real world numbers (Proving)
n^2 + 3 = O(n^2)
(n^2 + 3)/n^2 <= C assume n = 1 then C >= 3
Disproving
n^2 + 3 = O(n^2)
(n^2 + 3)/n^2 >= C assume n = 1 then C <= 3
n^2 + 3 = O(n^2)
both of these say that @ n =1 and c = 3 the algorithm is O(n^2) and is NOT O(n^2).
Can anyone help me clarify my confusion and give me a help me learn a good algorithmic way of solving big O questions?
To prove big-Oh, choose values for C and k and prove n>k implies f(n) ≤ C g(n). Choose k = 1. whenever n > 1. Proving Big-Oh: Example 2 Show that f(n)=3n + 7 is O(n).
2n+1 =22n ≤ c2n where c ≥ 2. Is 22n = O(2n) ? No. 22n = 2n · 2n.
Big O Notation is a way to measure an algorithm's efficiency. It measures the time it takes to run your function as the input grows. Or in other words, how well does the function scale. There are two parts to measuring efficiency — time complexity and space complexity.
Neither of your techniques work. Let's start with the definition of big-O:
f is O(g) iff there exist C, N such that |f(x)| ≤ C |g(x)| for all x ≥ N
To prove "there exist" type statements, you need to show that, well, the things exist. In the case of big-O proofs, you usually find the things, though proofs of existence don't generally need to be constructive. To build a proof for a "for all" statement, pretend someone just handed you specific values. Be careful you make no implicit assumptions about their properties (you can explicitly state properties, such as N > 0).
In the case of proving big-O, you need to find the C and N. Showing |g(n)| ≤ C|F(n)| for a single n isn't sufficent.
For the example "n2+3 is O(n2)":
For n ≥ 2, we have: n2 ≥ 4 > 3 ⇒ n2-1 > 2 ⇒ 2(n2-1) > (n2-1)+2 ⇒ 2n2 > (n2-1)+4 = n2+3 Thus n2+3 is O(n2) for C=2, N=2.
To disprove, you take the negation of the statement: show there is no C or N. In other words, show that for all C and N, there exists an n > N such that |f(n)| > C |g(n)|. In this case, the C and N are qualified "for all", so pretend they've been given to you. Since n is qualified "there exists", you have to find it. This is where you start with the equation you wish to prove and work backwards until you find a suitable n.
Suppose we want to prove that n is not O(ln n). Pretend we're given N and C, and we need to find an n ≥ N such that n > C ln n.
For all whole numbers C, N, let M=1+max(N, C) and n = eM. Note n > N > 0 and M > 0. Thus n = eM > M2 = M ln eM = M ln n > C ln n. QED.
Proofs of x > 0 ⇒ ex > x2 and "n is not O(ln n)" ⇒ "n is not O(logb n)" left as exercises.
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