On job interview I got a question:
is it true that 40^n is O(2^n) I said yes because only the exponent counts and constant doesn't matter. Then I got question if (40n)^2 is O(n^2), here I feel like no, it's not because the differences for next n will be huge but can't formally prove it. What is the answer for both of those which won't leave any doubts?
is it true that 40^n is O(2^n) I said yes because only the exponent counts and constant doesn't matter.
That's a big shortcut, it doesn't work here. For 40^n to be in O(2^n), there would have to be a pair of constants c and n0 such that 40^n <= c * 2^n if n >= n0. But there isn't. If you try to solve that for c, it turns out c has to be 20^n, which is not a constant. The base of an exponential cannot be ignored like that.
Then I got question if (40n)^2 is O(n^2)
If you work out the square, you get 1600 n^2. Now there is a solution such that c and n0 are constants, for example c = 1600, n0 = 1. So yes, (40n)^2 is an element of O(n^2).
Use the definition of Big O:
f(x) ∈ g(x) if and only if |f(x)| <= cg(x), ∀x:x>=x0, for some c,x0
1) 40^n ∉ O(2^n): There is no constant c and choice of x0 such that 40^x <= c2^x, for all x >= x0
2) (40n)^2 ∈ O(n^2): Choose c = 1600, x0 to be arbitrary: 1600x^2 <= 1600x^2 for all x >= x0
Related
The question is:
Show or disprove that for two functions f,g, if f is not in O(g) then g is in O(f).
My counterexample:
Let f(n) be f(n) = n^2 : if n is even
or n^4 : if n is odd
Let g(n) be g(n) = n^3
This is an example for f not in O(g) and g not in O(f).
Is my example wrong? If so, why?
Do you have any other examples?
Your counterexample works. A proof might look like this:
Suppose f were O(g). Then there is a positive constant c and an n0 such that for n >= n0, f(n) <= c * g(n). Let n' be an odd integer greater than or equal to n0. Then we have n'^4 <= c * n'^3. Dividing both sides by n'^3 gives n' <= c. However, this cannot be true for all n' > n0; so there are even n greater than n0 for which the condition does not hold, a contradiction.
The proof the other way around is similar, except you divide both sides by n'^2.
I think the kind of counterexample you identified is a good one for this; a function that oscillates by an asymptotically increasing amount and a function that goes somewhere in the middle.
I'm trying to find time complexity (big O) of a recursive formula.
I tried to find a solution, you may see the formula and my solution below:
Like Brenner said, your last assumption is false. Here is why: Let's take the definition of O(n) from the Wikipedia page (using n instead of x):
f(n) = O(n) if and only if there exist constants c, n0 s.t. |f(n)| <= c |g(n)|, for alln >= n0.
We want to check if O(2^n^2) = O(2^n). Clearly, 2^n^2 is in O(2^n^2), so let's pick f(n) = 2^n^2 and check if this is in O(2^n). Put this into the above formula:
exists c, n0: 2^n^2 <= c * 2^n for all n >= n0
Let's see if we can find suitable constant values n0 and c for which the above is true, or if we can derive a contradiction to proof that it is not true:
Take the log on both sides:
log(2^n^2) <= log(c * 2 ^ n)
Simplify:
2 ^n log(2) <= log(c) + n * log(2)
Divide by log(2):
n^2 <= log(c)/log(2) * n
It's easy to see know that there is no c, n0 for which the above is true for all n >= n0, thus O(2^n^2) = O(n^2) is not a valid assumption.
The last assumption you've specified with the question mark is false! Do not make such assumptions.
The rest of the manipulations you've supplied seem to be correct. But they actually bring you nowhere.
You should have finished this exercise in the middle of your draft:
T(n) = O(T(1)^(3^log2(n)))
And that's it. That's the solution!
You could actually claim that
3^log2(n) == n^log2(3) ==~ n^1.585
and then you get:
T(n) = O(T(1)^(n^1.585))
which is somewhat similar to the manipulations you've made in the second part of the draft.
So you can also leave it like this. But you cannot mess with the exponent. Changing the value of the exponent changes the big-O classification.
I am trying to understand Big-O notation through a book I have and it is covering Big-O by using functions although I am a bit confused. The book says that O(g(n)) where g(n) is the upper bound of f(n). So I understand that means that g(n) gives the max rate of growth for f(n) at larger values of n.
and that there exists an n_0 where the rate of growth of cg(n) (where c is some constant) and f(n) have the same rate of growth.
But what I am confused is on these examples on finding Big O in mathmatical functions.
This book says findthe upper bound for f(n) = n^4 +100n^2 + 50
they then state that n^4 +100n^2 + 50 <= 2n^4 (unsure why the 2n^4)
then they some how find n_0 =11 and c = 2, I understand why the big O is O(n^4) but I am just confused about the rest.
This is all discouraging as I don't understand but I feel like this is an important topic that I must understand.
If any one is curious the book is Data Structures and Algorithms Made Easy by Narasimha Karumanchi
Not sure if this post belongs here or in the math board.
Preparations
First, lets state, loosely, the definition of f being in O(g(n)) (note: O(g(n)) is a set of functions, so to be picky, we say that f is in O(...), rather than f(n) being in O(...)).
If a function f(n) is in O(g(n)), then c · g(n) is an upper bound on
f(n), for some constant c such that f(n) is always ≤ c · g(n),
for large enough n (i.e. , n ≥ n0 for some constant n0).
Hence, to show that f(n) is in O(g(n)), we need to find a set of constants (c, n0) that fulfils
f(n) < c · g(n), for all n ≥ n0, (+)
but this set is not unique. I.e., 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.
Showing that f ∈ O(n^4)
Now, lets proceed and look at the example that confused you
Find an upper asymptotic bound for the function
f(n) = n^4 + 100n^2 + 50 (*)
One straight-forward approach is to express the lower-order terms in (*) in terms of the higher order terms, specifically, w.r.t. bounds (... < ...).
Hence, we see if we can find a lower bound on n such that the following holds
100n^2 + 50 ≤ n^4, for all n ≥ ???, (i)
We can easily find when equality holds in (i) by solving the equation
m = n^2, m > 0
m^2 - 100m - 50 = 0
(m - 50)^2 - 50^2 - 50 = 0
(m - 50)^2 = 2550
m = 50 ± sqrt(2550) = { m > 0, single root } ≈ 100.5
=> n ≈ { n > 0 } ≈ 10.025
Hence, (i) holds for n ≳ 10.025, bu we'd much rather present this bound on n with a neat integer value, hence rounding up to 11:
100n^2 + 50 ≤ n^4, for all n ≥ 11, (ii)
From (ii) it's apparent that the following holds
f(n) = n^4 + 100n^2 + 50 ≤ n^4 + n^4 = 2 · n^4, for all n ≥ 11, (iii)
And this relation is exactly (+) with c = 2, n0 = 11 and g(n) = n^4, and hence we've shown that f ∈ O(n^4). Note again, however, that the choice of constants c and n0 is one of convenience, that is not unique. Since we've shown that (+) holds for on set of constants (c,n0), we can show that it holds for an infinite amount of different such choices of constants (e.g., it naturally holds for c=10 and n0=20, ..., and so on).
Can you prove using reflexivity that f(n) equals big Theta(f(n))? It seems straight forward when thinking about it because f(n) is bounded above and below by itself. But how will I write this down? And does this apply to big Omega and big O
I believe what you are intending to ask is (w.r.t. #emory:s answer) is something along the lines:
"For some function f(n), is it true that f ∈ ϴ(f(n))?"
If you emanate from the formal definition of Big-ϴ notation, it is quite apparent that this holds.
f ∈ ϴ(g(n))
⇨ For some positive constants c1, c2, and n0, the following holds:
c1 · |g(n)| ≤ |f(n)| ≤ c2 · |g(n)|, for all n ≥ n0 (+)
Let f(n) be some arbitrary real-valued function. Set g(n) = f(n) and choose, e.g., c1=0.5, c2=2, and n0 = 1. Then, naturally, (+) holds:
0.5 · |f(n)| ≤ |f(n)| ≤ 2 · |f(n)|, for all n ≥ 1
Hence, f ∈ ϴ(f(n)) holds.
No we can not because it is not true. ϴ(f(n)) is a set. f(n) is a member of that set. f(n)+1 is also a member of that set.
I need help understanding a Big-O problem. I get the concept and have done a few practice problems already, but this one has me stumped.
Using the definition of big O, show that f(n)=anlogn+bn is O(nlogn). (a, b > 0)
I don't know how to find C or N, because if constants A or B change, then C and N have to change as well? Or am I looking at this the wrong way?
I have a test coming up, and I'd really like to understand this beforehand.
Thanks!
When you're given a statement like this one:
Prove that an log n + bn = O(n log n)
You can think of it as the following:
For any choice of a and b, prove that an log n + bn = O(n log n)
Which in turn means
For any choice of a and b, there is some choice of c and n0 such that an log n + bn ≤ cn log n for any n ≥ n0.
In other words, you first pick a and b, then show that an log n + bn = O(n log n). You're not trying to show that there are a fixed c and n0 that work in the definition of big-O notation regardless of a and b, but rather should show that no matter how someone picks a and b, you'll always be able to find a c and n0 - which probably depend on a and b - such that an log n + bn = O(n log n) using those choices of c and n0.
To see how you'd do this in this example, one observation that might be useful is that (assuming our logs are base two), 1 ≤ log n as long as n ≥ 2. Therefore, as long as we restrict n such that n ≥ 2, we get that
an log n + bn ≤ an log n + bn log n = (a + b) n log n
Given this, do you see how you might pick c and n0? We're restricting n such that n ≥ 2, so it makes sense to pick n0 = 2. Similarly, since we've just proven that an log n + bn ≤ (a + b) n log n, we can pick c = a + b.
You can think of this argument as a dialog between two people:
Someone else: I'm going to pick an a and b, but I won't tell you what they are.
You: Um, okay.
Someone else: So prove to me that there's an n0 and c such that an log n + bn ≤ cn log n whenever n ≥ n0.
You: Sure! Try picking c = a + b and n0 = 2. Does that work?
Someone else: Hey, you're right! That does work!
Notice that the dialog starts with the other party choosing a and b. That way, you can tailor your choice of c and n0 to make sure the claim holds. If you tried picking c and n0 first, they could always find an a and b that would break it.
Hope this helps!
Since A and B are constants, it's OK to express C and N in terms of A and B. For example, you might show that C=A+B and N > 2A are sufficient to prove that f(n) = O(n lg n).
Maybe I'm missing something but this doesn't seem like typical use of Big-O. Your function is a straightforward mathematical function, not an algorithm. If:
f(n)=anlog(n)+bn
Then the complexity is the higher of that of (anlog(n)) and (bn), assuming that the plus operation has negligible complexity. Since b is constant, (bn) is O(n) and likewise as a is constant, (anlog(n)) is O(nlog(n)).