evaluation.c

牛角棋的源码

//evaluatation.c

#include <assert.h>
#include "data.h"

int	GameOver(int wtm, int *si, int ply)
{
	if((si[RED] == 8) || (si[RED] == 9))
		//红胜
	{	
		return ((wtm==RED) ? (WIN - ply) : -(WIN - ply));
	}

	else if ((si[RED] == 0) &&		//红子被逼死在牛角尖上
		((si[BLUESTONE1] + si[BLUESTONE2]) == 3))	
		//红负
	{
		return ((wtm==RED) ? (LOSE + ply) : -(LOSE + ply));
	}
	return ((wtm==RED) ? UNKNOWN : -UNKNOWN);
}

/*
************************************************************************
Evaluation
******************************
功能描述：为局面打分。
参数：    wtm：轮到谁走棋；si：待估局面；ply：层数。
返回值：  局面的估值。
*************************************************************************
*/
int Evaluation(int wtm, int *si, int ply)
{
	int i = 0;
	int value = ((wtm==RED) ? UNKNOWN : -UNKNOWN);

#ifdef _DEBUG						//局面合法性检查
	int good = 1;
	if(stoneIntersection[REDSTONE] == 
		stoneIntersection[BLUESTONE1])
		good = 0;
	if(stoneIntersection[REDSTONE] == 
		stoneIntersection[BLUESTONE2])
		good = 0;
	if(stoneIntersection[BLUESTONE1] ==
		stoneIntersection[BLUESTONE2])
		good = 0;
	assert(good);
#endif

	if ((si[RED] > si[BLUESTONE1]) ||		//红子成功突围			
		(si[RED] > si[BLUESTONE2]))
		//红胜
	{
		value = ((wtm==RED) ? 
			(WIN - ply) : -(WIN - ply));
	}

	else if (
		(wtm == BLUE) &&  			//该蓝子走棋,且红子和两个蓝子构成三角
		((stoneIntersection[BLUESTONE1] == si[RED] + 1 ) && 
		(stoneIntersection[BLUESTONE2] == si[RED] + 2 ) || 
		(stoneIntersection[BLUESTONE1] == si[RED] + 2 ) && 
		(stoneIntersection[BLUESTONE2] == si[RED] + 1 ))
		)
		//红胜
	{
		value = WIN - ply;
	}

	else if (
		(wtm == RED) &&				//该红子走棋，且红子和蓝子构成三角
		((1 == si[REDSTONE] || 2 == si[REDSTONE] || 3 == si[REDSTONE]) &&	
		((stoneIntersection[BLUESTONE1] == si[RED] + 1 ) && 
		(stoneIntersection[BLUESTONE2] == si[RED] + 2 ) || 
		(stoneIntersection[BLUESTONE1] == si[RED] + 2 ) && 
		(stoneIntersection[BLUESTONE2] == si[RED] + 1 )))
		)
	{
		value = -WIN + ply;
	}

	return value;
}


/*
************************************************************************
HasRepetition
******************************
功能描述：检测循环局面。
参数：    posStk：保存局面的栈结构；ply：层数。
返回值：  0，无循环；1，有循环。
*************************************************************************
*/
int HasRepetition(int* posStk, int ply)
{
	//blue stone1 与 blue stone2棋子交换，还是同一个局面
	//若采用hash table来识别不同的局面，则可以避免这样的处理
	if(stoneIntersection[BLUESTONE1] > 
		stoneIntersection[BLUESTONE2])
	{
		posStk[ply] = 
			(stoneIntersection[RED] << 8) +
			(stoneIntersection[BLUESTONE1] << 4) +
			(stoneIntersection[BLUESTONE2]);
	}
	else
	{
		posStk[ply] = 
			(stoneIntersection[RED] << 8) +
			(stoneIntersection[BLUESTONE2] << 4) +
			(stoneIntersection[BLUESTONE1]);
	}

	//5步之内，不会循环
	if (ply < 5)
		return 0;

	//探测6步一循环的着法序列:a1 a2 b1 b2 a1 a2
	if ((posStk[ply] == posStk[ply - 4]) &&
		(posStk[ply - 1] == posStk[ply - 5])
		)
	{
		return 1;
	}

	return 0;
}