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.

И тут многое стало намного понятней

Scary way:

Easy way:

double activation = 0;

for (int i=0;  i<n;  i++)

{ activation += x[i] * w[i]; }

Break with label, 2-Dim array search.

int[][] arrayOfInts = { { 32, 87, 3, 589 },

{ 12, 1076, 2000, 8 },
{ 622, 127, 77, 955 }

};

Recursive solution (confusiong one)

#include <stdio.h>

/*printd: print n in decimal */

void printd(int n) {      //receives an argument, lets say, an int 123;

if (n<0) {                //IF it is less than zero, then

putchar(‘-’);        // print a minus,

n = -n; }             // and get it over with, ok;

if (n / 10) {                        // WHY IS IT HERE?

printd (n / 10); }           // here, its ok – call the function again, but with an ‘n’ divided by ten. I have 123 / 10, it is 12.3 – and ‘.3′ is being dropped.

putchar (n % 10 + ’0′);           // Takes whatever is left from (n / 10), and prints it out. In this case, 123 / 10 = 12.3,  so it takes 3. And so on.

}