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Merge branch 'master' of https://github.com/lihongxun945/gobang
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lihongxun945 committed Jul 19, 2018
2 parents 69ef578 + 61d7b55 commit 8ae6087
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Showing 4 changed files with 180 additions and 104 deletions.
160 changes: 81 additions & 79 deletions src/ai/board.js
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Original file line number Diff line number Diff line change
Expand Up @@ -104,18 +104,18 @@ class Board {
for(var j=0;j<board[i].length;j++) {
// 空位，对双方都打分
if(board[i][j] == R.empty) {
if(this.hasNeighbor([i, j], 2, 2)) { //必须是有邻居的才行
var cs = scorePoint(this, [i, j], R.com)
var hs = scorePoint(this, [i, j], R.hum)
if(this.hasNeighbor(i, j, 2, 2)) { //必须是有邻居的才行
var cs = scorePoint(this, i, j, R.com)
var hs = scorePoint(this, i, j, R.hum)
this.comScore[i][j] = cs
this.humScore[i][j] = hs
}

} else if (board[i][j] == R.com) { // 对电脑打分，玩家此位置分数为0
this.comScore[i][j] = scorePoint(this, [i, j], R.com)
this.comScore[i][j] = scorePoint(this, i, j, R.com)
this.humScore[i][j] = 0
} else if (board[i][j] == R.hum) { // 对玩家打分，电脑位置分数为0
this.humScore[i][j] = scorePoint(this, [i, j], R.hum)
this.humScore[i][j] = scorePoint(this, i, j, R.hum)
this.comScore[i][j] = 0
}
}
Expand All @@ -125,52 +125,52 @@ class Board {
//只更新一个点附近的分数
// 参见 evaluate point 中的代码，为了优化性能，在更新分数的时候可以指定只更新某一个方向的分数
updateScore (p) {
var radius = 6,
var radius = 4,
board = this.board,
self = this,
len = this.board.length

function update(x, y, dir) {
var role = self.board[x][y]
if (role !== R.reverse(R.com)) {
var cs = scorePoint(self, [x, y], R.com, dir)
var cs = scorePoint(self, x, y, R.com, dir)
self.comScore[x][y] = cs
statistic.table[x][y] += cs
} else self.comScore[x][y] = 0
if (role !== R.reverse(R.hum)) {
var hs = scorePoint(self, [x, y], R.hum, dir)
var hs = scorePoint(self, x, y, R.hum, dir)
self.humScore[x][y] = hs
statistic.table[x][y] += hs
} else self.humScore[x][y] = 0

}
// 无论是不是空位 都需要更新
// -
for(var i=-radius;i<radius;i++) {
for(var i=-radius;i<=radius;i++) {
var x = p[0], y = p[1]+i
if(y<0) continue
if(y>=len) break
update(x, y, 0)
}

// |
for(var i=-radius;i<radius;i++) {
for(var i=-radius;i<=radius;i++) {
var x = p[0]+i, y = p[1]
if(x<0) continue
if(x>=len) break
update(x, y, 1)
}

// \
for(var i=-radius;i<radius;i++) {
for(var i=-radius;i<=radius;i++) {
var x = p[0]+i, y = p[1]+i
if(x<0 || y<0) continue
if(x>=len || y>=len) break
update(x, y, 2)
}

// /
for(var i=-radius;i<radius;i++) {
for(var i=-radius;i<=radius;i++) {
var x = p[0]+i, y = p[1]-i
if(x<0 || y<0) continue
if(x>=len || y>=len) continue
Expand Down Expand Up @@ -340,72 +340,74 @@ class Board {

for(var i=0;i<board.length;i++) {
for(var j=0;j<board.length;j++) {
var p = [i, j]
if(board[i][j] == R.empty) {
var neighbor = [2,2]
if(this.allSteps.length < 6) neighbor = [1, 1]
if(this.hasNeighbor([i, j], neighbor[0], neighbor[1])) { //必须是有邻居的才行

var scoreHum = p.scoreHum = this.humScore[i][j]
var scoreCom = p.scoreCom = this.comScore[i][j]
var maxScore = Math.max(scoreCom, scoreHum)
p.score = maxScore
p.role = role


total ++
/* 双星延伸，以提升性能
* 思路：每次下的子，只可能是自己进攻，或者防守对面（也就是对面进攻点）
* 我们假定任何时候，绝大多数情况下进攻的路线都可以按次序连城一条折线，那么每次每一个子，一定都是在上一个己方棋子的八个方向之一。
* 因为既可能自己进攻，也可能防守对面，所以是最后两个子的米子方向上
* 那么极少数情况，进攻路线无法连成一条折线呢?很简单，我们对前双方两步不作star限制就好，这样可以 兼容一条折线中间伸出一段的情况
*/
if (starSpread && config.star) {
var roleScore = role === R.com ? p.scoreCom : p.scoreHum
var deRoleScore = role === R.com ? p.scoreHum: p.scoreCom

if (maxScore >= S.FOUR) {
} else if (maxScore >= S.BLOCKED_FOUR && starTo(this.currentSteps[this.currentSteps.length-1]) ) {
//star 路径不是很准，所以考虑冲四防守对手最后一步的棋
} else if (
starTo(p, attackPoints) || starTo(p, defendPoints)
) {
} else {
count ++
continue
}
}

if(scoreCom >= S.FIVE) {//先看电脑能不能连成5
fives.push(p)
} else if(scoreHum >= S.FIVE) {//再看玩家能不能连成5
//别急着返回，因为遍历还没完成，说不定电脑自己能成五。
fives.push(p)
} else if(scoreCom >= S.FOUR) {
comfours.push(p)
} else if(scoreHum >= S.FOUR) {
humfours.push(p)
} else if(scoreCom >= S.BLOCKED_FOUR) {
comblockedfours.push(p)
} else if(scoreHum >= S.BLOCKED_FOUR) {
humblockedfours.push(p)
} else if(scoreCom >= 2*S.THREE) {
//能成双三也行
comtwothrees.push(p)
} else if(scoreHum >= 2*S.THREE) {
humtwothrees.push(p)
} else if(scoreCom >= S.THREE) {
comthrees.push(p)
} else if(scoreHum >= S.THREE) {
humthrees.push(p)
} else if(scoreCom >= S.TWO) {
comtwos.unshift(p)
} else if(scoreHum >= S.TWO) {
humtwos.unshift(p)
if(this.allSteps.length < 6) {
if( !this.hasNeighbor(i, j, 1, 1)) continue
} else if (!this.hasNeighbor(i, j, 2, 2)) continue

var scoreHum = this.humScore[i][j]
var scoreCom = this.comScore[i][j]
var maxScore = Math.max(scoreCom, scoreHum)

if (onlyThrees && maxScore < S.THREE) continue

var p = [i, j]
p.scoreHum = scoreHum
p.scoreCom = scoreCom
p.score = maxScore
p.role = role

total ++
/* 双星延伸，以提升性能
* 思路：每次下的子，只可能是自己进攻，或者防守对面（也就是对面进攻点）
* 我们假定任何时候，绝大多数情况下进攻的路线都可以按次序连城一条折线，那么每次每一个子，一定都是在上一个己方棋子的八个方向之一。
* 因为既可能自己进攻，也可能防守对面，所以是最后两个子的米子方向上
* 那么极少数情况，进攻路线无法连成一条折线呢?很简单，我们对前双方两步不作star限制就好，这样可以 兼容一条折线中间伸出一段的情况
*/
if (starSpread && config.star) {
var roleScore = role === R.com ? p.scoreCom : p.scoreHum
var deRoleScore = role === R.com ? p.scoreHum: p.scoreCom

if (maxScore >= S.FOUR) {
} else if (maxScore >= S.BLOCKED_FOUR && starTo(this.currentSteps[this.currentSteps.length-1]) ) {
//star 路径不是很准，所以考虑冲四防守对手最后一步的棋
} else if (
starTo(p, attackPoints) || starTo(p, defendPoints)
) {
} else {
neighbors.push(p)
count ++
continue
}
}

if(scoreCom >= S.FIVE) {//先看电脑能不能连成5
fives.push(p)
} else if(scoreHum >= S.FIVE) {//再看玩家能不能连成5
//别急着返回，因为遍历还没完成，说不定电脑自己能成五。
fives.push(p)
} else if(scoreCom >= S.FOUR) {
comfours.push(p)
} else if(scoreHum >= S.FOUR) {
humfours.push(p)
} else if(scoreCom >= S.BLOCKED_FOUR) {
comblockedfours.push(p)
} else if(scoreHum >= S.BLOCKED_FOUR) {
humblockedfours.push(p)
} else if(scoreCom >= 2*S.THREE) {
//能成双三也行
comtwothrees.push(p)
} else if(scoreHum >= 2*S.THREE) {
humtwothrees.push(p)
} else if(scoreCom >= S.THREE) {
comthrees.push(p)
} else if(scoreHum >= S.THREE) {
humthrees.push(p)
} else if(scoreCom >= S.TWO) {
comtwos.unshift(p)
} else if(scoreHum >= S.TWO) {
humtwos.unshift(p)
} else neighbors.push(p)
}
}
}
Expand Down Expand Up @@ -476,18 +478,18 @@ class Board {
return result
}

hasNeighbor (point, distance, count) {
hasNeighbor (x, y, distance, count) {
var board = this.board
var len = board.length
var startX = point[0]-distance
var endX = point[0]+distance
var startY = point[1]-distance
var endY = point[1]+distance
var startX = x-distance
var endX = x+distance
var startY = y-distance
var endY = y+distance
for(var i=startX;i<=endX;i++) {
if(i<0||i>=len) continue
for(var j=startY;j<=endY;j++) {
if(j<0||j>=len) continue
if(i==point[0] && j==point[1]) continue
if(i==x && j==y) continue
if(board[i][j] != R.empty) {
count --
if(count <= 0) return true
Expand Down
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