f(n) = 6*2^n + n^2
big(O) = 2^n
big(Omega) = 2^n
In above equation both big(O) and big(Omega) has same value. If big (O) is upper bound and big(omega) is lower bound shouldn't big(omega) = n^2. Why the both have same value?
It's true that O and Ω are upper and lower bounds, respectively, but they are more similar to ≤ and ≥ than to < and >. Just like it's possible that, simultaneously a ≥ b and a ≤ b (without contradiction), so can a function be both O and Ω of a different function (in fact, that's one of the ways to define Θ).
Here, for large enough n,
6 2n + n2 ≤ 12 2n so 6 2n + n2 grows at most (up to a multiplicative constant)like 2n does (it is O of it).
Conversely, 6 2n + n2 ≥ 0.1 2n so 6 2n + n2 grows at least (up to a multiplicative constant) like 2n does (it is Ω of it).
Note that you don't have to use the same multiplicative constants. The conclusion is that 6 2n + n2 = Θ( 2n)
Related
I have a few asymptotic notation problems I do not entirely grasp.
So when proving asymptotic complexity, I understand the operations of finding a constant and the n0 term of which the notation will be true for. So, for example:
Prove 7n+4 = Ω(n)
In such a case we would pick a constant c, such that it is lower than 7 since this regarding Big Omega. Picking 6 would result in
7n+4 >= 6n
n+4 >= 0
n = -4
But since n0 cannot be a negative term, we pick a positive integer, so n0 = 1.
But what about a problem like this:
Prove that n^3 − 91n^2 − 7n − 14 = Ω(n^3).
I picked 1/2 as the constant, reaching
1/2n^3 - 91n^2 - 7n -14 >= 0.
But I am unsure how to continue. Also, a problem like this, I think regarding theta:
Let g(n) = 27n^2 + 18n and let f(n) = 0.5n^2 − 100. Find positive constants n0, c1 and c2 such
that c1f(n) ≤ g(n) ≤ c2f(n) for all n ≥ n0.
In such a case am I performing two separate operations here, one big O comparison and one Big Omega comparison, so that there is a theta relationship, or tight bound? If so, how would I go about that?
To show n3 − 91n2 − 7n − 14 is in Ω(n3), we need to exhibit some numbers n0 and c such that, for all n ≥ n0:
n3 − 91n2 − 7n − 14 ≥ cn3
You've chosen c = 0.5, so let's go with that. Rearranging gives:
n3 − 0.5n3 ≥ 91n2 + 7n + 14
Multiplying both sides by 2 and simplifying:
182n2 + 14n + 28 ≤ n3
For all n ≥ 1, we have:
182n2 + 14n + 28 ≤ 182n2 + 14n2 + 28n2 = 224n2
And when n ≥ 224, we have 224n2 ≤ n3. Therefore, the choice of n0 = 224 and c = 0.5 demonstrates that the original function is in Ω(n3).
I am going through the Asymptotic notations from here. I am reading this f(n) ≤ c g(n)
For example, if f(n) = 2n + 2, We can satisfy it in any way as f(n) is O (c.g(n)) by adjusting the value of n and c. Or is there any specific rule or formula for selecting the value of c and n. Will no always be 1?
There is no formula per se. You can find the formal definition here:
f(n) = O(g(n)) means there are positive constants c and k, such that 0 ≤ f(n) ≤ cg(n) for all n ≥ k. The values of c and k must be fixed for the function f and must not depend on n. (big-O notation).
What I understood from your question is, you are not getting the essence of big-O notation. If your complexity is, for example, O(n^2), then you can guarantee that there is some value of n (greater than k) after which f(n) in no case will exceed c g(n).
Let's try to prove f(n) = 2n + 2 is O(n):
As it seems from the function itself, you cannot set the value of c equal to 2 as you want to find f(n) ≤ c g(n). If you plug in c = 2, you have to find k such that f(n) ≤ c g(n) for n ≥ k. Clearly, there is no n for which 2n ≥ 2n + 2. So, we move on to c = 3.
Now, let's find the value of k. So, we solve the equation 3n ≥ 2n + 2. Solving it:
3n ≥ 2n + 2
=> 3n - 2n ≥ 2
=> n ≥ 2
Therefore, for c = 3, we found value of k = 2 (n ≥ k).
You must also understand, your function isn't just O(n). It is also O(n^2), O(n^3), O(n^4) and so on. All because corresponding values of c and k exist for g(n) = n^2, g(n) = n^3 and g(n) = n^4.
Hope it helps.
My textbook describes the relationship as follows:
There is a very nice mathematical intuition which describes these classes too. Suppose we have an algorithm which has running time N0 when given an input of size n, and a running time of N1 on an input of size 2n. We can characterize the rates of growth in terms of the relationship between N0 and N1:
Big-Oh Relationship
O(log n) N1 ≈ N0 + c
O(n) N1 ≈ 2N0
O(n²) N1 ≈ 4N0
O(2ⁿ) N1 ≈ (N0)²
Why is this?
That is because if f(n) is in O(g(n)) then it can be thought of as acting like k * g(n) for some k.
So for example if f(n) = O(log(n)) then it acts like k log(n), and now f(2n) ≈ k log(2n) = k (log(2) + log(n)) = k log(2) + k log(n) ≈ k log(2) + f(n) and that is your desired equation with c = k log(2).
Note that this is a rough intuition only. An example of where it breaks down is that f(n) = (2 + sin(n)) log(n) = O(log(n)). The oscillating 2 + sin(n) bit means that f(2n)-f(n) can be basically anything.
I personally find this kind of rough intuition to be misleading and therefore worse than useless. Others find it very helpful. Decide for yourself how much weight you give it.
Basically what they are trying to show is just basic algebra after substituting 2n for n in the functions.
O(log n)
log(2n) = log(2) + log(n)
N1 ≈ c + N0
O(n)
2n = 2(n)
N1 ≈ 2N0
O(n²)
(2n)^2 = 4n^2 = 4(n^2)
N1 ≈ 4N0
O(2ⁿ)
2^(2n) = 2^(n*2) = (2^n)^2
N1 ≈ (N0)²
Since O(f(n)) ~ k * f(n) (almost by definition), you want to look at what happens when you put 2n in for n. In each case:
N1 ≈ k*log 2n = k*(log 2 + log n) = k*log n + k*log 2 ≈ N0 + c where c = k*log 2
N1 ≈ k*(2n) = 2*k*n ≈ 2N0
N1 ≈ k*(2n)^2 = 4*k*n^2 ≈ 4N0
N1 ≈ k*2^(2n) = k*(2^n)^2 ≈ N0*2^n ≈ N0^2/k
So the last one is not quite right, anyway. Keep in mind that these relationships are only true asymptotically, so the approximations will be more accurate as n gets larger. Also, f(n) = O(g(n)) only means g(n) is an upper bound for f(n) for large enough n. So f(n) = O(g(n)) does not necessarily mean f(n) ~ k*g(n). Ideally, you want that to be true, since your big-O bound will be tight when that is the case.
According to this book, big O means:
f(n) = O(g(n)) means c · g(n) is an upper bound on f(n). Thus there exists some constant c such that f(n) is always ≤ c · g(n), for large enough n (i.e. , n ≥ n0 for some constant n0).
I have trubble understanding the following big O equation
3n2 − 100n + 6 = O(n2), because I choose c = 3 and 3n2 > 3n2 − 100n + 6;
How can 3 be a factor? In 3n2 − 100n + 6, if we drop the low order terms -100n and 6, aren't 3n2 and 3.n2 the same? How to solve this equation?
I'll take the liberty to slightly paraphrase the question to:
Why do and have the same asymptotic complexity.
For that to be true, the definition should be in effect both directions.
First:
let
Then for the inequality is always satisfied.
The other way around:
let
We have a parabola opened upwards, therefore there is again some after which the inequality is always satisfied.
Let's look at the definition you posted for f(n) in O(g(n)):
f(n) = O(g(n)) means c · g(n) is an upper bound on f(n). Thus there
exists some constant c such that f(n) is always ≤ c · g(n), for
large enough n (i.e. , n ≥ n0 for some constant n0).
So, we only need to find one 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.
Note that here I've switched to strict inequalities, which is really only a matter of taste, but I prefer this latter convention. Now, we can re-state the Big-O definition in possibly more easy-to-understand terms:
... we can say that f(n) is O(g(n)) if we can find a constant c such
that f(n) is less than c·g(n) or all n larger than n0, i.e., for all
n>n0.
Now, let's look at your function f(n)
f(n) = 3n^2 - 100n + 6 (*)
Let's describe your functions as a sum of it's highest term and another functions
f(n) = 3n^2 + h(n) (**)
h(n) = 6 - 100n (***)
We now study the behaviour of h(n) and f(n), respectively:
h(n) = 6 - 100n
what can we say about this expression?
=> if n > 6/100, then h(n) < 0, since 6 - 100*(6/100) = 0
=> h(n) < 0, given n > 6/100 (i)
f(n) = 3n^2 + h(n)
what can we say about this expression, given (i)?
=> if n > 6/100, the f(n) = 3n^2 + h(n) < 3n^2
=> f(n) < c*n^2, with c=3, given n > 6/100 (ii)
Ok!
From (ii) we can choose constant c=3, given that we choose the other constant n0 as larger than 6/100. Lets choose the first integer that fulfils this: n0=1.
Hence, we've shown that (+) golds for constant set **(c,n0) = (3,1), and subsequently, f(n) is in O(n^2).
For a reference on asymptotic behaviour, see e.g.
https://www.khanacademy.org/computing/computer-science/algorithms/asymptotic-notation/a/big-o-notation
y=3n^2 (top graph) vs y=3n^2 - 100n + 6
Consider the sketch above. By your definition, 3n^2 only needs to be bigger than 3n^2 - 100n + 6 for large enough n (i.e. , n ≥ n0 for some constant n0). Let that n0 = 5 in this case (it could be something a little smaller, but it's clear which graph is bigger by n=5 so we'll just go with that).
Clearly from the graph, 3n^2 >= 3n^2 - 100n + 6 in the range we've plotted. The only way for 3n^2 - 100n + 6 to get bigger than 3n^2 then is for it to grow more steeply.
But the gradients of 3n^2 and 3n^2 - 100n + 6 are 6n and 6n - 100 respectively, so 3n^2 - 100n + 6 can't grow more steeply, therefore must always be underneath.
So your definition holds - 3n^2 - 100n + 6 <= 3n^2 for all n>=5
I am not an expert, but this looks a lot similar to what we just had in our real analysis course.
Basically if you have something like f(n) = 3n^2 − 100n + 6, the "fastest growing" term "wins" the other terms, when you have really really big n.
So in this case 3n^2 surpasses what ever 100n is, when the n is really big.
Another example would be something like f(n) = n/n^2 or f(n) = n! * n^2.
The first one gets smaller, as n simply cannot "keep up" with n^2. In the second example n! clearly grows faster than n^2, so I guess the answer for that should be f(n) = n! then, because the n^2 technically stops mattering with big n.
And terms like +6, which have no n affecting them are constants and matter even less as they cannot grow even if n grows.
It is all about what happends when n is really big. If your n is 34934854385754385463543856, then n^2 is hell of a bigger than 100n, because n^2 = n * n = 34934854385754385463543856 * 34934854385754385463543856.
I have this:
a) f(n) = n
b) f(n) = 1045n
c) f(n) = n2 + 70
d) f(n) = 7n + 3
e) f(n) = Cn + D (where C and D are both constants)
f) f(n) = 8
g) f(n) = n3 + n + 1
h) f(n) = 4n + 2log n + 5
I want to check if the Big O notation of them is O(n).
How can I determinate it?
And how to find the Big-O notation for the functions below:
a) f(n) = 3n3 + n
b) f(n) = 3 log n + 5n
c) f(n) = 3n2 + 5n + 4
d) f(n) = 3n3 + n2 + 5n + 99
f(x) is O(g(x)) if there exists a constant c such that f(x) < c*g(x)
You should look at the "biggest" asymptoticall factor in your function (highest exponent or something like that) and that would be your big O
Example: f(n) = 2^n + n^5 + n^2 + n*log(n) + n
This function has 5 different factors that influence big O, sorted from biggest to smallest, so this one would be O(2^n). Drop the 2^n and now f(n) is O(n^5).
Constants are O(1).
Hope I explained it well
Generally the Big O notation of a function is measured by the largest power of n that appears. In your case, this would be n², since the only other factor is the 70 which is constant.
Edit: Original post only contained the function f(n) = n² + 70.
See this answer.
In short, there's no set way to determine Big O results. Strictly speaking any function which eventually (for some n) will always be bigger than your function, is Big O of it. In practice you're looking for as tight a bound as you can get. If the only components of the function are polynomial in n, then the Big O will just be the largest power of n that appears (or rather, n to that power). Another useful thing to know is that log n is of a lower order than n (but higher than constant).