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According to this page:
The statement: f(n) + o(f(n)) = theta(f(n)) appears to be true.
Where: o = little-O, theta = big theta
This does not make intuitive sense to me. We know that o(f(n)) grows asymptotically faster than f(n). How, then could it be upper bounded by f(n) as is implied by big theta?
Here is a counter-example:
let f(n) = n, o(f(n)) = n^2.
n + n^2 is NOT in theta(n)
It seems to me that the answer in the previously linked stackexchange answer is wrong. Specifically, the statement below seems as if the poster is confusing little-o with little-omega.
Since g(n) is o(f(n)), we know that for each ϵ>0 there is an nϵ such that |g(n)|<ϵ|f(n)| whenever n≥nϵ
Update: I've realized the answer to my question
I was confused as to what o(f(n)) was. I thought that o(f(n)) for f(n)=n was, for instance, f(n) = n^2.
This is not correct. o(f(n)) is a function which is upper bounded by f and not asymptotically tight with f.
For instance, if f(n)=n, then f(n)=1 might be a member of o(f(n)), but f(n)=n^2 is NOT a member of o(f(n)).
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I got a question
What would be the time complexity of this function?
Function (int n) {
for (i = 1 to n):
print("hello")
}
apparently it's exponential because of binary numbers or something??
it should be O(n) right?
This is clearly O(n). The function prints "hello" n times. So the time-complexity is O(n) and it is not exponential. It is linear.
Since for loop is running from 1 to n, therefore complexity will be O(n). It is linear.
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What is the example of polynomial time algorithm
Is polynomial time algorithm fastest?
Suppose 100 elements in array , then how can I decide algorithm is polynomial time?
Q: What is the example of polynomial time algorithm?
for (i = 0; i < n; ++i)
printf("%d", i);
This is a linear algorithm, and linear belongs to polynomial class.
Q: Is polynomial time algorithm fastest?
No, logarithmic and constant-time algorithms are asymptotically faster than polynomial algorithms.
Q: Suppose 100 elements in array , then how can I decide algorithm is
polynomial time?
You haven't specified any algorithm here, just the data structure (array with 100 elements). However, to determine whether algorithm is polynomial time or not, you should find big-o for that algorithm. If it is O(n^k), then it is polynomial time. Read more here and here.
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I am new to Rcpp so I apologize in advance if this question is simple to answer. I searched on the web but couldn't find much help and I am hoping the savviness in this forum can help me!
I have an existing code in R using Rcpp, and I need to add to this code the following. I have a quadratic function in two variables, f(x, y), and I need to find the zeros of it:
f(x, y) = (x + by + c)' W (x + by + c)
where the unknowns are x and y. That is, I am interested in finding the set of pairs (x, y) that satisfy f(x , y)=0.
Note: This is a simulation exercise where I need to find the zeros of this function for different values of a, b, c and W. Therefore, I need to code this in a mechanical way (cannot just find the solution, for instance, by graphical inspection). Both variables are continue, and I don't want to work with a grid for (x,y) to see when f(x,y)=0. I need a more general/optimization solution. I don't really know what values (x,y) can take.
Before diving into the numerical part I think you should define this question in a better way. Here I assume that x, y, and c are vectors, and b is a scalar.
A quick observation is that, if W is positive definite, then f(x, y) = 0 implies that x + by + c = 0. If both x and y are free variables, then the solution is not unique. For example, if (x, y) is a solution, then (x - b, y + 1) (element-wise operations) is also a solution.
If W is indefinite, then the equation also has multiple solutions. I just give a very simple example here. Imagine that W is a 2x2 diagonal matrix with 1 and -1 on the diagonal. Then as long as x + by + c = (t, t)' for any t, the function value is exactly zero.
In short, under my assumption on the notations, the equation has infinite number of solutions. I believe you need additional restrictions to make it unique.
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I've tried to sort these functions in asymptotic growth order and would like to know if I'm on the right track.
5000log2(n)
sqrt(n) +7
8n
n/log2(n)
4nlog2(n)
n^1/100
1/4 n^2 - 10000n
.
You can test if f(n) is asymptotically larger than g(n) by checking if
lim f(n) / g(n) = ∞
n->∞
If the limit is a non-zero constant, f(n) and g(n) are asymptotically equal. If it is zero, f(n) is asymptotically smaller than g(n).
So. The major part of your list looks correct. There are a few mistakes, though.
n/log2(n) should be between sqrt(n) + 7 and 8n.
n^(1/100) is the 100-th root of n and should be before the square-root.
The above list would be -
1) 5000log2(n)
2) n^(1/100)
3)sqrt(n)+7
4)n/log2(n)
5)8n
6)4nlog2(n)
7)1/4n^2-10000n
as per my knowledge.
For more information on the topic you can see the definition of O(n),Big-theta n and Omega - n
Correction to the above list are most welcomed
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explain all updates in the basic algorithm of differential evolution. i am not able to find all versions of this algorithm. explain all versions of this algorithm as a survey and i am not clearly understand the theory behind this algorithm as given in the Wikipedia. Wikipedia also define only basic algorithm of differential evolution but i want to all updates of this algorithm
For complete survey in Differential Evolution, I suggest you the paper entitled Differential Evolution: A Survey of the State-of-the-Art but the brief explanation is :
DE has 2 basic crossover and 5 basic mutation operators, so we have 2*5=10 basic DE variants.
Two crossover operators are Exponential and Binomial.
Exponential Crossover:
D is problem space dimensionality, n is randomly chosen from [1,D], Cr is crossover rate and L is drawn from [1,D] according to above pseudocode.
Binomial Crossover:
j is refer to j-th dimension, i is vector number and G is generation number and jrand is randomly chosen index from [1,D].
Five mutation operators are DE/rand/1 , DE/best/1 , DE/target-to-best/1 , DE/best/2 and DE/rand/2.
DE/rand/1: V(i)=X(r1)+F*(X(r2)-X(r3))
DE/best/1: V(i)=X(best)+F*(X(r1)-X(r2))
DE/target-to-best/1: V(i)=X(i)+F*(X(best)-X(i))+F*(X(r1)-X(r2))
DE/best/2: V(i)=X(best)+F*(X(r1)-X(r2))+F*(X(r3)-X(r4))
DE/rand/2: V(i)=X(r1)+F*(X(r2)-X(r3))+F*(x(r4)-X(r5))
V(i) is donor(mutant) vector for target vector X(i), F is difference vector's scale factor, r1,r2,r3,r4,r5 are mutually exclusive, randomly chosen from [1,NP] and differ from i, best is the fittest vector's index in the current population, finally NP is population size.
These are all of things you can know about basic variants of DE.
DE also has many variants for many purposes which has explained in the mentioned paper.