请帮我加快这个麻将算法的速度

Question

我正在用 JavaScript 编写一些与麻将相关的函数。

这是我下面的内容，其中包含测试用例的代码。

请注意，麻将手牌是用数组表示的，其中：

• 元素 0 是这手牌中的牌总数
• 元素 1 到 34 是这手牌中每种类型的牌数
  • 首先是嘎嘎声，然后是点，然后是砰砰声，然后是风，最后是龙

查找等待的函数运行得非常慢。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

您有一个数组来保存手牌的内容，然后您在递归函数中创建一个新数组来保存每次减去一组特定图块的内容。与其创建所有这些数组，不如创建两个数组 - 一个用于容纳正在考虑的手，另一个用于容纳您已经考虑过的手上的牌 - 然后将它们都传递出去。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;

变成这样：

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

你传递 h 和 hc ，并记住要重建整个手牌，你需要添加两个数组。但这可以在考虑 hnd 是否完整时进行。

编辑：这是我更详细的意思：
编辑：我希望现在可以工作......第一次尝试时出现拼写错误。

// 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

我可以看到，在性能方面存在两件事错误。

首先是 David M 已经指出的：每次在 well_formed() 中递归时停止复制整个数组，只需在递归之前添加更改并在返回时取消添加，就像在 waits() 函数中所做的那样。

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，您应该寻找机会保留“状态计数器”。

例如，如果您始终知道自己有多少对，则可以轻松检查 only_pairs() 。您无需扫描 hand() 数组中的任何非对，而是保留pairs_counter作为数组（或其关联上下文）的一部分，每当您向 hand[i] 添加一张牌时，您都会检查 hand[i]=2如果对计数器为 3，则增加它，如果它为 3，则减少它。同样，当您移除一张牌时，如果 hand[j] = 2，您将增加配对计数器，但如果它等于 1，您将减少它。

您可以在很多地方采用此策略，它将对您的绩效产生重大影响。