How can I solve the following questions in complexity:
|O(f(n)) – O(f(n))| = ? (Complexity of absolute value of O(f(n)) – O(f(n))).
O(f(n))+ O(f(n)) = ?
|O(f(n)) – O(f(n))| = O(f(n)). That is basically the only thing you can say about it. This is because you're subtracting, from a function growing at most like f(n), another function growing at most like f(n). The result (pre the absolute value) can be even negative (e.g., when subtracting 2 f(n) = O(f(n)), but it is clearly not larger than a function growing like f(n).
O(f(n)) + O(f(n)) = O(f(n)). This is easy to prove throught the basic definition of O.
Related
Why Big-O notation can not compare algorithms in the same complexity class. Please explain, I can not find any detailed explanation.
So, O(n^2) says that this algorithm requires less or equal number of operations to perform. So, when you have algorithm A which requires f(n) = 1000n^2 + 2000n + 3000 operations and algorithm B which requires g(n) = n^2 + 10^20 operations. They're both O(n^2)
For small n the first algorithm will perform better than the second one. And for big ns second algorithm looks better since it has 1 * n^2, but first has 1000 * n^2.
Also, h(n) = n is also O(n^2) and k(n) = 5 is O(n^2). So, I can say that k(n) is better than h(n) because I know how these functions look like.
Consider the case when I don't know how functions k(n) and h(n) look like. The only thing I'm given is k(n) ~ O(n^2), h(n) ~ O(n^2). Can I say which function is better? No.
Summary
You can't say which function is better because Big O notation stays for less or equal. And following is true
O(1) is O(n^2)
O(n) is O(n^2)
How to compare functions?
There is Big Omega notation which stays for greater or equal, for example f(n) = n^2 + n + 1, this function is Omega(n^2) and Omega(n) and Omega(1). When function has complexity equal to some asymptotic, Big Theta is used, so for f(n) described above we can say that:
f(n) is O(n^3)
f(n) is O(n^2)
f(n) is Omega(n^2)
f(n) is Omega(n)
f(n) is Theta(n^2) // this is only one way we can describe f(n) using theta notation
So, to compare asymptotics of functions you need to use Theta instead of Big O or Omega.
When we talk about Θ(g) are we referring to the highest order term of g(n) or are we referring to g(n) exactly as it is?
For example if f(n) = n3. And g(n)=1000n3+n does Θ(g) mean Θ(1000n3+n) or Θ(n3)?
In this scenario can we say that f(n) is Θ(g)?
Θ(g) yields sets of functions that are all of the same complexity class. Θ(1000n3+n) is equal to Θ(n3) because both of these result in the same set.
For simplicity's sake one will usually drop the non-significant terms and multiplicative constants. The lower order additive terms don't change the complexity, nor do any multipliers, so there's no reason to write them out.
Since Θ(g) is a set, you would say that f(n) ∈ Θ(g).
NB: Many CS teachers, textbooks, and other resources muddy the waters by using imprecise notation. Lots of people say that f(n)=n3 is O(n3), rather than f(n)=n3 is in O(n3). They use = when they mean ∈.
theta(g(n)) lies between O(g(n)) and omega(g(n))
if g(n) = 1000n^3 + n
first lets find O(g(n)) upper bound
It could be n^3, n^4, n^5 but we choose the closest one which is O(n^3).
O(n^3) is valid because we can find a constant c such that for some value of n
1000n^3 + n < c.n^3
second lets see omega(g(n)) which is lower bound
omega says f(n) > c.g(n)
we can find a constant c such that
1000.n^3 + n > c.n^3
Now we have upper bound which is O(n^3) and lower bound which is omega(n^3).
therefore we have theta which bounds both upper and lower using same function.
By rule : if f(n) = O(g(n)) and f(n) = omega(g(n)) therefore f(n) = theta(g(n))
1000.n^3 + n = theta(n^3)
I know that we can apply the Master Theorem to find the running time of a divide and conquer algorithm, when the recurrence relation has the form of:
T(n) = a*T(n/b) + f(n)
We know the following :
a is the number of subproblems that the algorithm divides the original problem
b is the size of the sun-problem i.e n/b
and finally.. f(n) encompasses the cost of dividing the problem and combining the results of the subproblems.
Now we then find something (I will come back to the term "something")
and we have 3 cases to check.
The case that f(n) = O(n^log(b)a-ε) for some ε>0; Then T(n) is O(n*log(b)a)
The case that f(n) = O(n^log(b)a) ; Then T(n) is O(n^log(b)a * log n).
If n^log(b)a+ε = O(f(n)) for some constant ε > 0, and if a*f(n/b) =< cf(n) for some constant
c < 1 and almost all n, then T(n) = O(f(n)).
All fine, I am recalling the term something. How we can use general examples (i.e uses variables and not actual numbers) to decide which case the algorithm is in?
In instance. Consider the following:
T(n) = 8T(n/2) + n
So a = 8, b = 2 and f(n) = n
How I will proceed then? How can I decide which case is? While the function f(n) = some big-Oh notation how these two things are comparable?
The above is just an example to show you where I don't get it, so the question is in general.
Thanks
As CLRS suggests, the basic idea is comparing f(n) with n^log(b)a i.e. n to the power (log a to the base b). In your hypothetical example, we have:
f(n) = n
n^log(b)a = n^3, i.e. n-cubed as your recurrence yields 8 problems of half the size at every step.
Thus, in this case, n^log(b)a is larger because n^3 is always O(n) and the solution is: T(n) = θ(n^3).
Clearly, the number of subproblems vastly outpaces the work (linear, f(n) = n) you are doing for each subproblem. Thus, the intuition tells and master theorem verifies that it is the n^log(b)a that dominates the recurrence.
There is a subtle technicality where the master theorem says that f(n) should be not only smaller than n^log(b)a O-wise, it should be smaller polynomially.
I was reading Intro to Algorithms, by Thomas H. Corman when I encountered this statement (in Asymptotic Notations)
when a>0, any linear function an+b is in O(n^2) which is essentially verified by taking c = a + |b| and no = max(1, -b/a)
I can't understand why O(n^2) and not O(n). When will O(n) upper bound fail.
For example, for 3n+2, according to the book
3n+2 <= (5)n^2 n>=1
but this also holds good
3n+2 <= 5n n>=1
So why is the upper bound in terms of n^2?
Well I found the relevant part of the book. Indeed the excerpt comes from the chapter introducing big-O notation and relatives.
The formal definition of the big-O is that the function in question does not grow asymptotically faster than the comparison function. It does not say anything about whether the function grows asymptotically slower, so:
f(n) = n is in O(n), O(n^2) and also O(e^n) because n does not grow asymptotically faster than any of these. But n is not in O(1).
Any function in O(n) is also in O(n^2) and O(e^n).
If you want to describe the tight asymptotic bound, you would use the big-Θ notation, which is introduced just before the big-O notation in the book. f(n) ∊ Θ(g(n)) means that f(n) does not grow asymptotically faster than g(n) and the other way around. So f(n) ∊ Θ(g(n)) is equivalent to f(n) ∊ O(g(n)) and g(n) ∊ O(f(n)).
So f(n) = n is in Θ(n) but not in Θ(n^2) or Θ(e^n) or Θ(1).
Another example: f(n) = n^2 + 2 is in O(n^3) but not in Θ(n^3), it is in Θ(n^2).
You need to think of O(...) as a set (which is why the set theoretic "element-of"-symbol is used). O(g(n)) is the set of all functions that do not grow asymptotically faster than g(n), while Θ(g(n)) is the set of functions that neither grow asymptotically faster nor slower than g(n). So a logical consequence is that Θ(g(n)) is a subset of O(g(n)).
Often = is used instead of the ∊ symbol, which really is misleading. It is pure notation and does not share any properties with the actual =. For example 1 = O(1) and 2 = O(1), but not 1 = O(1) = 2. It would be better to avoid using = for the big-O notation. Nonetheless you will later see that the = notation is useful, for example if you want to express the complexity of rest terms, for example: f(n) = 2*n^3 + 1/2*n - sqrt(n) + 3 = 2*n^3 + O(n), meaning that asymptotically the function behaves like 2*n^3 and the neglected part does asymptotically not grow faster than n.
All of this is kind of against the typically usage of big-O notation. You often find the time/memory complexity of an algorithm defined by it, when really it should be defined by big-Θ notation. For example if you have an algorithm in O(n^2) and one in O(n), then the first one could actually still be asymptotically faster, because it might also be in Θ(1). The reason for this may sometimes be that a tight Θ-bound does not exist or is not known for given algorithm, so at least the big-O gives you a guarantee that things won't take longer than the given bound. By convention you always try to give the lowest known big-O bound, while this is not formally necessary.
The formal definition (from Wikipedia) of the big O notation says that:
f(x) = O(g(x)) as x → ∞
if and only if there is a positive constant M such that for all
sufficiently large values of x, f(x) is at most M multiplied by g(x)
in absolute value. That is, f(x) = O(g(x)) if and only if there exists
a positive real number M and a real number x0 such that
|f(x)|≤ M|g(x)| for all x > x₀ (mean for x big enough)
In our case, we can easily show that
|an + b| < |an + n| (for n sufficiently big, ie when n > b)
Then |an + b| < (a+1)|n|
Since a+1 is constant (corresponds to M in the formal definition), definitely
an + b = O(n)
Your were right to doubt.
My question refers to the big-Oh notation in algorithm analysis. While Big-Oh seems to be a math question, it's much useful in algorithm analysis.
Suppose two functions are defined below:
f(n) = 2( to the power n) when n is even
f(n) = n when n is odd
g(n) = n when n is even
g(n) = 2( to the power n) when n is odd.
For the above two functions which one is big-Oh of other? Or whether any function is not a Big-Oh of another function.
Thanks!
In this case,
f ∉ O(g), and
g ∉ O(f).
This is because no matter what constants N and k you pick,
there exists i ≥ N such that f(i) > k g(i), and
there exists j ≥ N such that g(j) > k f(j).
The Big-Oh relationship is quite specific in that one function is, after a finite n, always larger than the other.
Is this true here? If so, give such a n. If not, you should prove it.
Usually Big-O and Big-Theta notations get confused.
A layman attempt at definition could be that Big-O means that one function is growing as fast or faster than another one, i.e. that given a large enough n, f(n)<=k*g(n) where k is some constant. That means that if f(x) = 2x^3, then it is in O(x^3), O(x^4), O(2^x), O(x!) etc..
Big-Theta means that one function is growing as fast as another one, with neither one being able to "outgrow" the other, or, k1*g(n)<=f(n)<=k2*g(n) for some k1 and k2. In programming terms that means that these two functions have the same level of complexity. If f(x) = 2x^3, then it is in Θ(x^3), as for example, if k1=1, and k2=3, 1*x^3 < 2*x^3 < 3*x^3
In my experience whenever programmers are talking about Big-O, the discussion is actually about Big-Θ, as we are concerned more with the as fast as part more, than in the no faster than part.
That said, if two functions with different Θ's are combined, as in your example, the larger one - (Θ(2^n) - swallows the smaller - Θ(n), so both f and g have the exact same Big-O and Big-Θ complexities. In this case, it's both correct that
f(n) = O(g(n)), also f(n) = Θ(g(n))
g(n) = O(f(n)), also g(n) = Θ(f(n))
so, as they have the same complexity, they are O and Θ bound by each other.