bp1.cpp

人工智能BP问题的C语言实现

//BP算法
//学习标准sin曲线（y=sin(x)），BP网络见报告
//2004.11.23
//interma

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>


#define PI 3.14159
#define E 2.71828

#define N 20 // 隐层的节点数目

#define M 100 // 标准样本的数目（5 * M）


//共5 * M个学习样本
double in[M][5][2]; //标准样本的输入
double out[M][5][2]; //标准样本的期待输出

//权值
double wl[2][N]; 
double wr[N][2];



double F(double x); //f(x)=1/(1+e^(-x))
double Fd(double x); //f`(x)= e^(-x)/(1+e^(-x))^2
double Random(); //返回0~1之间的数
void Init(void); //初始化标准样本
void Study(int n); //学习标准样本, 输入为学习次数
bool IsIn(double x, double y); //判定是否在sin曲线上



int main(void)
{
	srand((unsigned)time( NULL));
	Init();

	int n;
	printf("请输入学习次数（推荐不要超过100次）：\n");
	scanf("%d", &n);

	Study(n);
	printf("学习完毕！\n");

	do
	{
		double x;
		printf("请输入点的X坐标：\n");
		scanf("%lg", &x);

		double y;
		printf("请输入点的Y坐标：\n");
		scanf("%lg", &y);
	

		if (IsIn(x, y))
			printf("此点在sin曲线上！\n");
		else
			printf("此点不在sin曲线上！\n");

	}
	while (true);

}

double F(double x)
{
	double d = pow(E, -x);

	return 1 / (d + 1); 
}

double Fd(double x)
{
	double d = pow(E, -x);

	return d / pow((1 + d), 2);
}

double Random()
{
	double d = (rand() * 1.0) / RAND_MAX; 
	return d;
}

void Init(void)
{
	double d = (2 * PI) / (M);  

	for (int i = 0; i < M; i++)
	{
		in[i][0][0] = d * (i);
		in[i][0][1] = sin(d * (i)) + 0.4;
		out[i][0][0] = 0;
		out[i][0][1] = 1;

		in[i][1][0] = d * (i);
		in[i][1][1] = sin(d * (i)) + 0.1;
		out[i][1][0] = 0.5;
		out[i][1][1] = 0.5;

		in[i][2][0] = d * (i);
		in[i][2][1] = sin(d * (i));
		out[i][2][0] = 1;
		out[i][2][1] = 0;

		in[i][3][0] = d * (i);
		in[i][3][1] = sin(d * (i)) - 0.1;
		out[i][3][0] = 0.5;
		out[i][3][1] = 0.5;

		in[i][4][0] = d * (i);
		in[i][4][1] = sin(d * (i)) - 0.4;
		out[i][4][0] = 0;
		out[i][4][1] = 1;
	}

	for (int i = 0; i < 2; i++)
	{
		for (int j=0; j < N; j++)
		{
			wl[i][j] = Random();
			wr[j][i] = Random();
		}
	}
}

void Study(int n)
{
	while (n-- > 0)
	{
		for (int i = 0; i < M; i++)
		{
			double e5 = 0;

			for (int j = 0; j < 5; j++)
			{

				double x1 = in[i][j][0];
				double x2 = in[i][j][1];
				
				double h[N];

				for (int k = 0; k < N; k++)
				{
					h[k] = F(x1) * wl[0][k] + F(x2) * wl[1][k];
				}

				double u = 0;
				double v = 0;

				for (int k = 0; k < N; k++)
				{
					u += F(h[k]) * wr[k][0];
					v += F(h[k]) * wr[k][1];
				}

				double y1 = F(u);
				double y2 = F(v);

				double e = (pow(out[i][j][0] - y1, 2) + pow(out[i][j][1] - y2, 2)) / 2;
				
				e5 += e;

				double du = (out[i][j][0] - y1) * Fd(u);
				double dv = (out[i][j][1] - y2) * Fd(v);

				double dh[N];
				for (int k = 0; k < N; k ++)
				{
					dh[k] = (du * wr[k][0] + dv * wr[k][1]) * Fd(h[k]);
				}
				
				for (int k = 0; k< N; k++)
				{
					wr[k][0] -= du * F(h[k]);
					wr[k][1] -= dv * F(h[k]);
					wl[0][k] -= dh[k] * F(x1);
					wl[1][k] -= dh[k] * F(x2);
				}
			}

			if (e5 < 0.0001)
				goto End;
		}

	}

End:	;
}

bool IsIn(double x, double y)
{
	double x1 = x;
	double x2 = y;
				
	double h[N];

	for (int k = 0; k < N; k++)
	{
		h[k] = F(x1) * wl[0][k] + F(x2) * wl[1][k];
	}

	double u = 0;
	double v = 0;

	for (int k = 0; k < N; k++)
	{
		u += F(h[k]) * wr[k][0];
		v += F(h[k]) * wr[k][1];
	}

	double y1 = F(u);
	double y2 = F(v);


	if (y1 >= y2)
		return true;
	else
		return false;

}