代码如下:
import java.util.Scanner;
public class Perceptron {
private static int N = 3;
private static int n = 2;
private static double[][] X = null;
private static double[] Y = null;
private static double[][] G = null;
private static double[] A = null;
private static double[] W = null;
private static double B = 0;
private static double fi = 0.5;
private static boolean check(int id) {
double ans = B;
for(int i=0;i<N;i++)
ans += A[i] * Y[i] * G[i][id];
if(ans * Y[id] > 0) return true;
return false;
}
public static void solve() {
Scanner in = new Scanner(System.in);
System.out.print("input N:"); N = in.nextInt();
System.out.print("input n:"); n = in.nextInt();
X = new double[N][n];
Y = new double[N];
G = new double[N][N];
System.out.println("input N * n datas X[i][j]:");
for(int i=0;i<N;i++)
for(int j=0;j<n;j++)
X[i][j] = in.nextDouble();
System.out.println("input N datas Y[i]");
for(int i=0;i<N;i++)
Y[i] = in.nextDouble();
for(int i=0;i<N;i++)
for(int j=0;j<N;j++) {
G[i][j] = 0;
for(int k=0;k<n;k++)
G[i][j] += X[i][k] * X[j][k];
}
A = new double[N];
W = new double[n];
for(int i=0;i<n;i++) A[i] = 0;
B = 0;
boolean ok = true;
while(ok == true) {
ok = false;
//这里在原来算法的基础上不断地将fi缩小,以避免跳来跳去一直达不到要求的点的效果。
for(int i=0;i<N;i++) {
//System.out.println("here " + i);
while(check(i) == false) {
ok = true;
A[i] += fi;
B += fi * Y[i];
//debug();
}
}
fi *= 0.5;
}
for(int i=0;i<n;i++)
W[i] = 0;
for(int i=0;i<N;i++)
for(int j=0;j<n;j++)
W[j] += A[i] * Y[i] * X[i][j];
}
public static void main(String[] args) {
solve();
System.out.print("W = [");
for(int i=0;i<n-1;i++) System.out.print(W[i] + ", ");
System.out.println(W[n-1] + "]");
System.out.println("B = " + B);
}
}
posted on 2015-03-20 11:34
marchalex 阅读(856)
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