Sigmoid function

Googled it out lately:

class Calc {

public static double sigmoid(double x) {

return (1/( 1 + Math.pow(Math.E,(-1*x))))

}

}

For example, let’s say you have 18 and 8. You subtract 8 from 18 and get ten and pass it to the Sigmoid function. You’re returned with 0.9999546021312976. If you had a difference of 0 though, you would get .50 and if you had a negative difference, like -18, get a really tiny number that’s bigger than 0. So in short, the Sigmoid function is easy and quite interesting.

I just passed my E = X1W2 + X2W2 + … + X3W3 + W0 there.

Here is the original source, Thanks to the author.

Disclaimer: I really dont know, will it work or not. I made a class, which – yes, produces something. The question is – will it work in a neural network.

Genetic Algorithm overview (bullet-point style)

Genetic Algorithms is a part of the evolutionary computing.

• A set of given problem solutions (chromosomes, containers of certain data) form a population;

• According to their ‘fitness’, certain best chromosomes are taken from the population to produce an offspring, which would form a second generation of population (involving Crossover, Mutation, and Elitism);

• Elitism – copying a few ‘best’ chromosomes into the next generation of population. Crucial for transforming information about best option found so far to the next generation, and eventually – find the best possible solution to a given problem.

Генетический алгоритм (тривиальный)

# include <iostream>
# include <algorithm>
# include <numeric>

int main()
{

using namespace std;
srand((unsigned)time(NULL));
const int N = 1000;
int a[N];

//заполняем нулями
fill(a, a+N, 0);
for (;;)
{

//мутация в случайную сторону каждого элемента:
for (int i = 0; i < N; ++i)

if (rand()%2 == 1)

a[i] += 1;

else

a[i] -= 1;

//теперь выбираем лучших, отсортировав по возрастанию…
sort(a, a+N);
//… и тогда лучшие окажутся во второй половине массива.
//скопируем лучших в первую половину, куда они оставили потомство, а первые умерли:
copy(a+N/2, a+N, a /*куда*/);
//теперь посмотрим на среднее состояние популяции. Как видим, оно всё лучше и лучше.

cout << accumulate(a, a+N, 0) / N << endl;

}

}