xiaqi.c

黑白棋程序： 3*3的棋盘上面下棋

//习题1.5
//
// 前提己方先下子,呵呵

#include "stdio.h"
#include "string.h"

int b[9][9]={0};			//t[number of state][number of zuobiao]
double Vtrain[9]={0.0};
//初始化线性函数
//x0自己有威胁的独立棋子数目,x1对方,
//x2自己有威胁的连子数目,x3对方
//w[5]系数
double w[5]={6,6,6,6,6};	//初始化系数,这里的系数初始化为6
int x[4]={0,0,0,0};


main()
{
	void Performance();
	int Critic();
	void Generalizer();
	int Xiaqi();

	int Count=0;			//表征训练次数
	double rate=0.0;
	
	int i,j;
	int p;

	for (Count=5;Count<=500;Count=Count+5)
	{
		rate=0;
		for (j=0;j<5;j++)
			w[j]=6;
		for (i=0;i<Count;i++)
		{
			Performance();
			p=Critic();
			Generalizer(p);
			//printf("%d-->",i);
			//for (j=0;j<5;j++)
			//	printf("%lf   ",w[j]);
			rate=rate+Xiaqi();
			//printf("\n");
		}
		rate=rate/Count;
		printf("%lf\n",rate);
	}
}



#include "math.h"
void Performance()
{
	int getX();			//获得x值
	int i,j,m;
	int qizi=1;			//1 or -1
	int best=0;
	double v,Vmax;

	//初始化棋盘
	for (i=0;i<9;i++)
		for (j=0;j<9;j++)
			b[i][j]=0;

	for (i=0;i<9;i++)
	{	
		Vmax=-99999;	//这里设置了一个"足够小"的Value

		if (i!=0)		//新的一步需要先copy前一步的棋局
			for (m=0;m<9;m++)
				b[i][m]=b[i-1][m];

		for (j=0;j<9;j++)
		{
			if (b[i][j]==0)
			{
				b[i][j]=qizi;
				if (getX(b[i],qizi)==1)
				{
					v=w[0]*x[0]+w[1]*x[1]+w[2]*x[2]+w[3]*x[3]+w[4];
					if (Vmax<v)
					{
						Vmax=v;
						best=j;
					}
				}
				else
				{
					b[i][j]=qizi;
					goto end;
				}
				b[i][j]=0;
			}
		}

		b[i][best]=qizi;

		if (qizi==1)
			qizi=-1;
		else
			qizi=1;
	}

end:	qizi=0;
}



//int型的函数,返回可供给训练的棋局数量-1,即训练样例个数-1
int Critic()
{
	int i,p;
	int getX();

	for (i=0;i<9;i++)
		Vtrain[i]=0.0;

	for (i=0;i<4;i++)
	{
		p=2*i;
		if (getX(b[p+1],1)==1)
		{
			if (getX(b[p+2],1)==1)
				Vtrain[p]=w[0]*x[0]+w[1]*x[1]+w[2]*x[2]+w[3]*x[3]+w[4];
			else
			{
				Vtrain[p]=getX(b[p+2],1);
				return (i);
			}
		}
		else
		{
			Vtrain[p]=getX(b[p+2],1);
			return(i);
		}
	}

	return (i-1);
}




//countB获得训练样例的(个数-1)
void Generalizer(int countB)
{
	int getX();	//获得x值

	int i,j;
	int xx;
	double V1,V2=0,Vx;
	for (i=0;i<=countB;i++)
	{
		V1=Vtrain[2*i];
		xx=getX(b[2*i],1);
		V2=w[0]*x[0]+w[1]*x[1]+w[2]*x[2]+w[3]*x[3]+w[4];
		Vx=V1-V2;

		for (j=0;j<4;j++)
		{
			//printf("%d:%lf->",j,w[j]);
			w[j]=w[j]+0.05*Vx*x[j];		//这里序数取作0.05
			//printf("%lf\n",w[j]);			
		}
	}

}
		

//获得棋盘当前状态,即更新x[]
int getX(int bb[9],int n)
{

	int sum;
	int i;
	int result=1;

	for (i=0;i<4;i++)
		x[i]=0;

	for (i=0;i<=6;i=i+3)
	{
		sum=bb[i]+bb[i+1]+bb[i+2];
		if (sum==3)
			result=100;
		else if (sum==-3)
			result=-100;
		else if (sum==2)
			x[2]++;
		else if (sum==-2)
			x[3]++;
		else if (sum==1 && (bb[i]==0 || bb[i+1]==0))
			x[0]++;
		else if (sum==-1 && (bb[i]==0 || bb[i+1]==0))
			x[1]++;
	}

	for (i=0;i<=2;i++)
	{
		sum=bb[i]+bb[i+3]+bb[i+6];
		if (sum==3)
			result=100;
		else if (sum==-3)
			result=-100;
		else if (sum==2)
			x[2]++;
		else if (sum==-2)
			x[3]++;
		else if (sum==1 && (bb[i]==0 || bb[i+3]==0))
			x[0]++;
		else if (sum==-1 && (bb[i]==0 || bb[i+3]==0))
			x[1]++;
	}


		sum=bb[0]+bb[4]+bb[8];
		if (sum==3)
			result=100;
		else if (sum==-3)
			result=-100;
		else if (sum==2)
			x[2]++;
		else if (sum==-2)
			x[3]++;
		else if (sum==1 && (bb[0]==0 || bb[4]==0))
			x[0]++;
		else if (sum==-1 && (bb[0]==0 || bb[4]==0))
			x[1]++;

		sum=bb[2]+bb[4]+bb[6];
		if (sum==3)
			result=100;
		else if (sum==-3)
			result=-100;
		else if (sum==2)
			x[2]++;
		else if (sum==-2)
			x[3]++;
		else if (sum==1 && (bb[2]==0 || bb[4]==0))
			x[0]++;
		else if (sum==-1 && (bb[2]==0 || bb[4]==0))
			x[1]++;

	if (n==-1)
	{
		sum=x[0];  x[0]=x[1];  x[1]=sum;
		sum=x[2];  x[2]=x[3];  x[3]=sum;
	}

	return (result);
}


//和固定函数wp[]对弈的过程
#include "math.h"
int Xiaqi()
{
	int getX();	//获得x值
	int i,j,m;
	int qizi=1;		//1 or -1
	int best=0;
	double v,Vmax;
	//double wp[5]={5.538570,-15.946661,19.276676,-19.876809,6.000000};
	double wp[5]={6,-5,9,2,66};		//定义的固定函数,对弈的对手

	//初始化棋盘
	for (i=0;i<9;i++)
		for (j=0;j<9;j++)
			b[i][j]=0;

	for (i=0;i<9;i++)		//最多可能进行九步
	{	
		Vmax=-99999;

		if (i!=0)
			for (m=0;m<9;m++)
				b[i][m]=b[i-1][m];

		for (j=0;j<9;j++)	//一步最多要搜索9个位置
		{
			if (b[i][j]==0)
			{
				b[i][j]=qizi;
				if (getX(b[i],qizi)==1)
				{
					if (qizi==1)
						v=w[0]*x[0]+w[1]*x[1]+w[2]*x[2]+w[3]*x[3]+w[4];
					else
						v=wp[0]*x[0]+wp[1]*x[1]+wp[2]*x[2]+wp[3]*x[3]+wp[4];
					if (Vmax<v)
					{
						Vmax=v;
						best=j;
					}
				}
				else
				{
					b[i][j]=qizi;
					if (qizi==1)
					{	
						//printf("You win!\n");
						return (1);
					}
					else
					{	
						//printf("You lose!\n");	
						return (0);
					}
				}
				b[i][j]=0;
			}
		}

		b[i][best]=qizi;

		if (qizi==1)
			qizi=-1;
		else
			qizi=1;
	}

	//printf(":-)!\n");	
	return (0);

}