Question

我正在用JavaScript编写一些麻将相关的函数。

以下是我的测试用例代码。

请注意，麻将牌由数组表示，其中包含：

元素0是手中的瓷砖总数
元素1到34是手中每种类型的瓦片数
- 先是裂纹，然后是点，然后是凸起，然后是风，最后是龙

查找等待的功能运行速度非常慢。我怎样才能加快速度？

// tiles are indexed as follows:
// 1..9 = 1 crak .. 9 crak
// 10..18 = 1 dot .. 9 dot
// 19..27 = 1 bam .. 9 bam
// 28..34 = east, south, west, north, white, green, red

var wall = new Array();
set_up_wall();

function set_up_wall() {
  for (var i=1; i<=34; i++) wall[i] = 4;
  wall[0]=136;
}

// draw tile from wall
function draw() {
  var fudge = 1-(1e-14);
  var n = Math.floor(Math.random()*wall[0]*fudge);
  var i = 1;
  while (n>=wall[i]) n-=wall[i++];
  wall[i]--;
  wall[0]--;
  return i;
}

// get number of a tile (or 0 if honor)
// e.g. 8 bams = 8
function tilenum(i) {
  if (i>27) return 0;
  if (i%9==0) return 9;
  return i%9;
}

// get suit of a tile (or 0 if honor)
function tilesuit(i) {
  var eps = 1e-10;
  return Math.ceil(i/9-eps)%4;
}

// is this a well-formed hand?
function well_formed(h) {
  // this function is recursive
  if (h[0]==2) return only_pairs(h);
  if (h[0]==14) {
    if (only_pairs(h)) return true;
    if (thirteen_orphans(h)) return true;
  }
  if (h[0]%3 != 2) return false; // wrong no. of tiles in hand
  // now we start splitting up the hand
  // look for three of a kind
  for (var i=1; i<=34; i++) {
    if (h[i]>=3) {
      // create new hand minus the three of a kind
      hh = new Array();
      for (var j=0; j<=34; j++) hh[j]=h[j];
      hh[0]-=3;
      hh[i]-=3;
      if (well_formed(hh)) return true;
    }
  }
  // look for a run of three
  for (var i=1; i<=25; i++) {
    if (tilenum(i)<=7) {
      if (h[i]>=1 && h[i+1]>=1 && h[i+2]>=1) {
      // create new hand minus the run
      hh = new Array();
      for (var j=0; j<=34; j++) hh[j]=h[j];
      hh[0]-=3;
      hh[i]--; hh[i+1]--; hh[i+2]--;
      if (well_formed(hh)) return true;
      }
    }
  }
  // if we reach here, we have exhausted all possibilities
  return false;
}

// is this hand all pairs?
function only_pairs(h) {
  for (var i=1; i<=34; i++) if (h[i]==1 || h[i]>=3) return false;
  return true;
}

// thirteen orphans?
function thirteen_orphans(h) {
  var d=0;
  var c=new Array(14, 1,0,0,0,0,0,0,0,1, 1,0,0,0,0,0,0,0,1, 1,0,0,0,0,0,0,0,1, 1,1,1,1,1,1,1);
  for (var i=0; i<=34; i++) {
    if (c[i]==0 && h[i]>0) return false;
    if (h[i]!=c[i]) d++;
  }
  return d==1;
}

// this is inefficient
function waits(h) {
  var w=new Array();
  for (var j=0; j<=34; j++) w[j]=false;  
  if (h[0]%3!=1) return w; // wrong no. of tiles in hand
  // so we don't destroy h
  var hh = new Array();
  for (var j=0; j<=34; j++) hh[j]=h[j];
  for (var i=1; i<=34; i++) {
    // add the tile we are trying to test
    hh[0]++; hh[i]++;
    if (hh[i]<5) { // exclude hands waiting for a nonexistent fifth tile
      if (well_formed(hh)) {
        w[0] = true;
        w[i] = true;
      }
    }
    hh[0]--; hh[i]--;
  }
  return w;
}

function tiles_to_string(t) { // strictly for testing purposes
  var n;
  var ss="";
  var s = "x 1c 2c 3c 4c 5c 6c 7c 8c 9c 1d 2d 3d 4d 5d 6d 7d 8d 9d ";
  s += "1b 2b 3b 4b 5b 6b 7b 8b 9b Ew Sw Ww Nw Wd Gd Rd";
  s=s.split(" ");
  for (var i=1; i<=34; i++) {
    n=t[i]*1; // kludge
    while (n--) ss+=(" "+s[i]);
  }
  return ss;
}

// tests

var x;
x = new Array(13, 0,0,0,0,0,1,2,2,2, 2,2,2,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0);
document.write("Hand: "+tiles_to_string(x)+"<br />");
document.write("Waits: "+tiles_to_string(waits(x))+"<br /><br />");
x = new Array(13, 1,0,0,0,0,0,0,0,1, 1,0,0,0,0,0,0,0,1, 1,0,0,0,0,0,0,0,1, 1,1,1,1,1,1,1);
document.write("Hand: "+tiles_to_string(x)+"<br />");
document.write("Waits: "+tiles_to_string(waits(x))+"<br /><br />");
x = new Array(13, 0,0,0,0,0,0,0,0,0, 3,1,1,1,1,1,1,1,3, 0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0);
document.write("Hand: "+tiles_to_string(x)+"<br />");
document.write("Waits: "+tiles_to_string(waits(x))+"<br /><br />");
x = new Array(13, 4,0,0,3,3,3,0,0,0, 0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0);
document.write("Hand: "+tiles_to_string(x)+"<br />");
document.write("Waits: "+tiles_to_string(waits(x))+"<br /><br />");
x = new Array(13, 0,0,4,3,3,3,0,0,0, 0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0);
document.write("Hand: "+tiles_to_string(x)+"<br />");
document.write("Waits: "+tiles_to_string(waits(x))+"<br /><br />");

Answer 1

你有一个数组来保存手的内容，然后你创建一个新的数组来保存内容每次减去一组特定的瓦片 - 在递归函数中。不是创建所有这些数组，而是创建两个数组 - 一个用于保持正在考虑的手，另一个用于保持已经考虑过的手牌 - 并且只是将它们传递给它们。所以这个：

hh = new Array();
for (var j=0; j<=34; j++) hh[j]=h[j];
hh[0]-=3;
hh[i]-=3;
if (well_formed(hh)) return true;

成为这个：

h[0]-=3;
h[i]-=3;
hc[0]+=3;
hc[i]+=3;
if (well_formed(h,hc)) return true;

你传递了h和hc，并记住要重建整个手，你需要添加两个数组。但是，考虑到是否完成，这可能会在结束时出现。

编辑：我的意思更详细：编辑：现在正在工作，我希望......首次尝试错字。

// is this a well-formed hand?
function well_formed(h) {
  hc = new Array();
  for (var i=0; i<=34; i++) hc[i]=0;
  result = well_formed_recursive(h, hc);
  for (var i=0; i<=34; i++) h[i]+=hc[i];
  return result;
}

function well_formed_recursive(h, hc) {
  // this function is recursive
  if (h[0]==2) return only_pairs(h);
  if (h[0]==14) {
    if (only_pairs(h)) return true;
    if (thirteen_orphans(h)) return true;
  }
  if (h[0]%3 != 2) return false; // wrong no. of tiles in hand
  // now we start splitting up the hand
  // look for three of a kind
  for (var i=1; i<=34; i++) {
    if (h[i]>=3) {
      h[0]-=3;
      h[i]-=3;
      hc[0]+=3;
      hc[i]+=3;
      if (well_formed_recursive(h,hc)) return true;
    }
  }
  // look for a run of three
  for (var i=1; i<=25; i++) {
    if (tilenum(i)<=7) {
      if (h[i]>=1 && h[i+1]>=1 && h[i+2]>=1) {
        h[0]-=3;
        h[i]--; h[i+1]--; h[i+2]--;
        hc[0]+=3;
        hc[i]++; hc[i+1]++; hc[i+2]++;
        if (well_formed_recursive(h,hc)) return true;
      }
    }
  }
  // if we reach here, we have exhausted all possibilities
  return false;
}

Answer 2

要复制数组，请使用concat函数。

var a=[1,2,3,4];
var b=a.concat();

Answer 3

有两件事在性能上是错误的，我可以看到。

首先是David M已经注意到的：每次你在well_formed（）中递归时都停止复制整个数组，只需在你返回之前添加更改并在你返回时添加更改，就像你在等待时一样（）函数。

其次，在well_formed（）中，每次对手进行一次增量更改时，都会重新扫描整个数组。这本质上是低效的，相反，你应该寻找机会来保持“状态计数器”。

例如，如果你总是知道你有多少对，你可以轻松检查only_pairs（）。你可以将pair_counter作为数组（或其相关上下文）的一部分，而不是扫描hand（）数组，只要你手动添加一张卡[i]然后检查hand [i] = 2如果它是3，你递增对计数器，你递减它。同样，当您移除一张牌时，如果手牌[j] = 2，则增加对子计数器，但如果它等于1，则减少它。

您可以在很多地方采用此策略，这会对您的表现产生重大影响。

请帮我加快这个麻将算法

3 个答案: