高尔夫代码: Reversi

Question

好的，这里有一个相当复杂的代码高尔夫挑战:实现一个里弗西游戏(Othello)。

• 游戏应该显示游戏板的当前状态，并允许玩家在一台电脑上交替输入动作。
• 必须捕获不正确的输入和不允许的移动，但可以默默地忽略它们。
• 游戏必须结束时，不能做更多的移动(因为董事会是满的，或因为没有移动将翻转任何部分)。
• 然后游戏必须宣布谁赢了，或者是平局。

在尽可能少的字符中这样做。

会话应该如下所示：

 abcdefgh
1        
2        
3        
4   wb   
5   bw   
6        
7        
8        
b>d3
 abcdefgh
1        
2        
3   b    
4   bb   
5   bw   
6        
7        
8        

Answer 1

Perl，408个字符

这是Perl，现在减少到408个字符。可以用一种更小的方法再砍25个字符来声明获胜者(比如说"B"而不是"Winner: Black\n")。前两个换行符是重要的，其他的是为了可读性而包括的。

sub O{2&$B[$q+=$_]*$%}sub A{grep{$q=$=;$"=''while&O;$B[$q]*O$q=$=}$B[$=]?():@d}
sub F{$B[$q]=$%;O&&&F}sub D{print'
 ',a..h,$/,(map{($e="@B[$_*9+1..$_*9+8]
")=~y/012/ bw/;$_,$e}1..8),@_}
@d=map{$_,-$_}1,8..10;@B=(@z=(0)x40,$%=2,1,(0)x7,1,2,@z);
for$!(%!){$%^=3;for$=(9..80){$=%9*A&&do{{D$%-2?b:w,"> ";
$_=<>;$==(/./g)[1]*9-96+ord;A||redo}F$q=$=for A;last}}}
$X+=/1/-/2/for@B;D"Winner: ",$X<0?White:$X?Black:None,$/

@B拿着棋盘。$B[$i*9+$j]指行$i (1..8)和列$j (1..8)

@d是由8个有效方向组成的列表

O是一种方便的方法。它将$q增量为$_ (当前方向)，如果$B[$q]处的棋子属于当前玩家的对手，则返回非零值。

F处理当前方向的翻转块，$_

A检查当前玩家是否可以在$B[$=]上合法移动，并返回一组可以翻转的方向。

D(@_)绘制板并打印@_

主循环切换$% (当前播放器)并遍历董事会上的位置，为该播放器找到合法的移动。如果发现任何合法移动，请从标准输入读取移动(重复直到输入有效移动)并更新板。

Answer 2

编辑2：

再加上一些调整和技巧，我又把124个字符削减到了752个字符：

using C=System.Console;class O{static int[]b=new int[100];static int c=1;static void Main(){b[44]=b[55]=-1;b[45]=b[54]=1;var g=true;while(g){C.WriteLine(" abcdefgh");for(int i=10;i<90;i++){switch(i%10){case 0:C.Write(i/10);break;case 9:C.WriteLine();break;default:C.Write("w b"[b[i]+1]);break;}}C.Write("w b"[c+1]+">");var l=C.ReadLine();if(l.Length>1){int x=l[0]-96,y=l[1]-48;if(x>0&x<9&y>0&y<9&&b[y*10+x]<1)c*=f(y*10+x,1);}g=false;for(int y=10;y<90;y+=10)for(int x=y;x<y+8;)g|=b[++x]==0&&f(x,0)<0;}int s=0;foreach(int p in b)s+=p;C.WriteLine(s==0?"d":s<0?"w":"b");}static int f(int ofs,int y){var x=1;foreach(int d in new int[]{-11,-10,-9,-1,1,9,10,11}){int l=1,o=ofs;while(b[o+=d]==-c)l++;if(b[o]==c&l>1){x=-1;while(y>0&l-->0)b[o-=d]=c;}}return x;}}

在压实之前：

using Con = System.Console;

class O {

   // Initialise game board, made of ints - 0 is empty, 1 is a black piece, -1 a white.
   // Using a 10x10 board, with the outer ring of empties acting as sentinels.
   static int[] board = new int[100];

   // Color of the current player.
   static int currentColor = 1;

   static void Main() {
     // Set up the four pieces in the middle.
     board[44] = board[55] = -1;
     board[45] = board[54] = 1;
     var legal = true;
     while (legal) {
       // Print game board.
       Con.WriteLine(" abcdefgh");
       for (int i = 10; i < 90; i++) {
         switch (i % 10) {
           case 0: Con.Write(i / 10); break;
           case 9: Con.WriteLine(); break;
           default: Con.Write("w b"[board[i] + 1]); break;
         }
       }
       // Print input, indicating which color's turn it is.
       Con.Write("w b"[currentColor+1] + ">");
       // Parse input.
       var line = Con.ReadLine();
       if (line.Length > 1) {
         int x = line[0] - 96, y = line[1] - 48;
         // Discard malformed input.
         if (x > 0 & x < 9 & y > 0 & y < 9 && board[y * 10 + x] < 1)
           // Check if valid move and if so flip and switch players
           currentColor *= flip(y * 10 + x, 1);
       }
       // See if there are any legal moves by considering all possible ones.
       legal = false;
       for (int y = 10; y < 90; y += 10)
         for (int x = y; x < y + 8;)
           legal |= board[++x] == 0 && flip(x, 0) < 0;
     }
     // Calculate final score: negative is a win for white, positive one for black.
     int score = 0;
     foreach (int piece in board) score += piece;
     Con.WriteLine(score == 0 ? "d" : score < 0 ? "w" : "b");
   }

   // Flip pieces, or explore whether putting down a piece would cause any flips.
   static int flip(int ofs, int commitPutDown) {
     var causesFlips = 1;
     // Explore all straight and diagonal directions from the piece put down.
     foreach (int d in new int[] { -11, -10, -9, -1, 1, 9, 10, 11 }) {
       // Move along that direction - if there is at least one piece of the opposite color next
       // in line, and the pieces of the opposite color are followed by a piece of the same
       // color, do a flip.
       int distance = 1, o = ofs;
       while (board[o += d] == - currentColor) distance++;
       if (board[o] == currentColor & distance > 1) {
         causesFlips = -1;
         while (commitPutDown > 0 & distance-- > 0) board[o -= d] = currentColor;
       }
     }
     return causesFlips;
   }

}

876字符C#版本：

using C=System.Console;class O{static void Main(){int[]b=new int[100];b[44]=b[55]=2;b[45]=b[54]=1;int c=1;while(true){C.WriteLine(" abcdefgh");for(int i=10;i<90;i++){switch(i%10){case 0:C.Write(i/10);break;case 9:C.WriteLine();break;default:C.Write(" bw"[b[i]]);break;}}bool g=false;for(int y=10;y<90;y+=10)for(int x=1;x<9;x++){g|=(b[x+y]==0&&f(x+y,b,c,false));}if(!g)break;C.Write(" bw"[c]+">");var l=C.ReadLine();if(l.Length>1){int x=l[0]-96,y=l[1]-48,ofs;if(x>0&x<9&y>0&y<9&&b[ofs=y*10+x]==0)if(f(ofs,b,c,true))b[ofs]=c;c=3-c;}}int s=0;for(int y=10;y<90;y+=10)for(int x=1;x<9;x++)switch(b[y+x]){case 1:s++;break;case 2:s--;break;}C.WriteLine(s==0?"d":s<0?"w":"b");}static bool f(int ofs,int[]b,int p,bool c){var x=false;foreach(int d in new int[]{-11,-10,-9,-1,1,9,10,11}){int l=1,o=ofs;while(b[o+=d]==3-p)l++;if(b[o]==p&l>1){x=true;if(c)while(l-->1)b[o-=d]=p;}}return x;}}

在压实之前：

using Con = System.Console;

class Othello {

  static void Main() {
    // Initialise game board, made of ints - 0 is empty, 1 is a black piece, 2 a white.
    // Using a 10x10 board, with the outer ring of empties acting as sentinels.
    int[] board = new int[100];
    // Set up the four pieces in the middle.
    board[44] = board[55] = 2;
    board[45] = board[54] = 1;
    // Color of the current player.
    int currentColor = 1;
    while (true) {
      // Print game board.
      Con.WriteLine(" abcdefgh");
      for (int i = 10; i < 90; i++) {
        switch (i % 10) {
          case 0: Con.Write(i / 10); break;
          case 9: Con.WriteLine(); break;
          default: Con.Write(" bw"[board[i]]); break;
        }
      }
      // See if there are any legal moves by considering all possible ones.
      bool legal = false;
      for (int y = 10; y < 90; y += 10) for (int x = 1; x < 9; x++) {
        legal |= (board[x + y] == 0 && flip(x + y, board, currentColor, false));
      }
      if (!legal) break;
      // Print input, indicating which color's turn it is.
      Con.Write(" bw"[currentColor] + ">");
      // Parse input.
      string l = Con.ReadLine();
      if (l.Length > 1) {
        int x = l[0] - 96, y = l[1] - 48;
        int ofs;
        // Discard malformed input.
        if (x > 0 & x < 9 & y > 0 & y < 9 && board[ofs = y * 10 + x] == 0)
          // Check if valid move & flip if it is - if not, continue.
          if (flip(ofs, board, currentColor, true))
            // Put down the piece itself.
            board[ofs] = currentColor;
        // Switch players.
        currentColor = 3 - currentColor;
      }
    }
    // Calculate final score: negative is a win for white, positive one for black.
    int score = 0;
    for (int y = 10; y < 90; y += 10) for (int x = 1; x < 9; x++)
      switch (board[y + x]) {
        case 1: score++; break;
        case 2: score--; break;
      }
    Con.WriteLine(score == 0 ? "d" : score < 0 ? "w" : "b");
  }

  /** Flip pieces, or explore whether putting down a piece would cause any flips. */
  static bool flip(int ofs, int[] board, int playerColor, bool commitPutDown) {
    bool causesFlips = false;
    // Explore all straight and diagonal directions from the piece put down.
    foreach (int d in new int[] { -11, -10, -9, -1, 1, 9, 10, 11 }) {
      // Move along that direction - if there is at least one piece of the opposite color next
      // in line, and the pieces of the opposite color are followed by a piece of the same
      // color, do a flip.
      int distance = 1, o = ofs;
      while (board[o += d] == 3 - playerColor) distance++;
      if (board[o] == playerColor & distance > 1) {
        causesFlips = true;
        if (commitPutDown) while (distance-- > 1) board[o -= d] = playerColor;
      }
    }
    return causesFlips;
  }

}

我使用int100而不是char1010来简化循环和比较。其中有很多小技巧，但除此之外，它基本上与原始Java代码相同。

编辑：

编辑器显示了一些奇怪的"Col“值，但是您应该看看"Ch”值.不是943个字符，是876个.

Answer 3

C++，672个字符

这是我的版本。它的重量为672个字符，包括换行符：

#include <iostream>
#define F for
#define G F(i p=9;p<90;++p)
#define R return
#define C std::cout
typedef int i;i b[100];i f(i p, i s, i d){i t=0;i q[]={-11,-10,-9,-1,1,9,10,11};F(i o=0;o<8;++o){i c=p+q[o];F(;b[c]==-s;c+=q[o]);if(b[c]==s)F(--t;c!=p;c-=q[o],++t)if(d)b[c]=s;}R t;}i h(i s){G if(!b[p]&&f(p,s,0))R 1;R 0;}i main(){F(i p=0;p<100;++p)b[p]=p<10||p>89||!(p%10%9)?2:0;b[44]=b[55]=-1;b[45]=b[54]=1;i s=1;F(;;){C<<" abcdefgh";G C<<(char)(p%10?"w b\n"[b[p]+1]:'0'+p/10);s=h(s)?s:-s;if(!h(s)){s=-34;G s+=b[p];C<<(s<0?'s':s?'b':'d')<<'\n';R 0;}i p=0;F(;p<9||p>90||b[p]||!f(p,s,1);){C<<"w b"[s+1]<<">";std::string m;std::cin>>m;p=m[0]-'`'+(m[1]-'0')*10;}b[p]=s;s=-s;}}