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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 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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There is a elliptic curve with parameters:
a = 0xb3b04200486514cb8fdcf3037397558a8717c85acf19bac71ce72698a23f635
b = 0x12f55f6e7419e26d728c429a2b206a2645a7a56a31dbd5bfb66864425c8a2320
Also the prime number is:
q = 0x247ce416cf31bae96a1c548ef57b012a645b8bff68d3979e26aa54fc49a2c297
How can I solve the equation P * 65537 = H and obtain the value of P?
P and H are points and H equals to (72782057986002698850567456295979356220866771008308693184283729159903205979695, 7766776325114464021923523189912759786515131109431296018171065280757067869793).
Note that in the equation we have Elliptic curve point multiplication!
You need to know the number of points on the curve to solve this. Let's call that number n. Then you will have to compute the inverse of 65537 modulo n and do a scalar multiply of your point H by that number.
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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.
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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)).