代码高尔夫:康威的生命游戏

Question

的挑战：编写了实现JohnH.Conway生命游戏元胞自动机的最短程序。[链接]

编辑：经过大约一个星期的竞争后，我选择了一个胜利者：pdehaan，用perl以一个字符击败了解决方案。

对于那些没有听说过“生命游戏”的人来说，你需要一个方形细胞的网格(理想的无限)。细胞可以是活的(充满的)或者是死的(空的)。我们通过应用以下规则来确定哪些细胞在下一阶段是活的：

1. 任何活的细胞，如果少于两个活的邻居，就会死亡，就好像是由人口不足引起的。
2. 任何有三个以上活邻居的活牢房都会死，好像是过度拥挤所致。
3. 任何有两三个活邻居的活细胞都会活到下一代。
4. 任何有三个活邻居的死细胞都会变成活细胞，就像通过繁殖一样。

您的程序将读取作为命令行参数指定的40x80字符的ASCII文本文件，以及要执行的迭代次数(N)。最后，经过N次迭代，它将输出到ASCII文件out.txt系统的状态。

下面是一个使用相关文件运行的示例：

in.txt:

................................................................................
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..................................XX............................................
..................................X.............................................
.......................................X........................................
................................XXXXXX.X........................................
................................X...............................................
.................................XX.XX...XX.....................................
..................................X.X....X.X....................................
..................................X.X......X....................................
...................................X.......XX...................................
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迭代100次：

Q:\>life in.txt 100

结式输出(out.txt)

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..................................XX............................................
..................................X.X...........................................
....................................X...........................................
................................XXXXX.XX........................................
................................X.....X.........................................
.................................XX.XX...XX.....................................
..................................X.X....X.X....................................
..................................X.X......X....................................
...................................X.......XX...................................
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The Rules:

• 您需要使用文件I/O来读取/写入文件。
• 您需要接受一个输入文件和作为参数的迭代次数。
• 您需要以指定的格式生成out.txt (如果存在覆盖)。
• 您不需要处理板的边缘(环绕，无限网格.etc)
• 编辑:您的do需要在输出文件中有换行符。

胜利者将由字符计数决定。

祝好运!

Answer 1

Python2.x-210/234个字符

好吧，210个字符的密码有点作弊。

#coding:l1
exec'xÚ=ŽA\nÂ@E÷sŠº1­ƒÆscS‰ØL™Æª··­âî¿GÈÿÜ´1iÖ½;Sçu.~H®J×Þ-‰­Ñ%ª.wê,šÖ§J®d꘲>cÉZË¢V䀻Eîa¿,vKAËÀå̃<»Gce‚ÿ‡ábUt¹)G%£êŠ…óbÒüíÚ¯GÔ/n×Xši&ć:})äðtÏÄJÎòDˆÐÿG¶'.decode('zip')

您可能无法复制和粘贴这些代码并使其工作。它应该是拉丁文1 (ISO-8859-1)，但我认为它在前进的过程中被曲解到了Windows1252中。此外，您的浏览器可能吞下一些非ASCII字符。

因此，如果它不能工作，您可以从普通的7位字符生成文件：

s = """
23 63 6F 64 69 6E 67 3A 6C 31 0A 65 78 65 63 27 78 DA 3D 8E 41 5C 6E C2
40 0C 45 F7 73 8A BA 31 13 AD 83 15 11 11 C6 73 08 63 17 05 53 89 D8 4C
99 C6 AA B7 B7 AD E2 EE BF 47 C8 FF DC B4 31 69 D6 BD 3B 53 E7 75 2E 7E
48 AE 4A D7 DE 90 8F 2D 89 AD D1 25 AA 2E 77 16 EA 2C 9A D6 A7 4A AE 64
EA 98 B2 3E 63 C9 5A CB A2 56 10 0F E4 03 80 BB 45 16 0B EE 04 61 BF 2C
76 0B 4B 41 CB C0 E5 CC 83 03 3C 1E BB 47 63 65 82 FF 87 E1 62 55 1C 74
B9 29 47 25 A3 EA 03 0F 8A 07 85 F3 62 D2 FC ED DA AF 11 47 D4 2F 6E D7
58 9A 69 26 C4 87 3A 7D 29 E4 F0 04 74 CF C4 4A 16 CE F2 1B 44 88 1F D0
FF 47 B6 27 2E 64 65 63 6F 64 65 28 27 7A 69 70 27 29
"""

with open('life.py', 'wb') as f:
    f.write(''.join(chr(int(i, 16)) for i in s.split()))

其结果是一个有效的210字符Python源文件。我在这里所做的就是对原始Python源代码进行压缩压缩。真正的欺骗是我在结果字符串中使用非ASCII字符。它仍然是有效的代码，只是很麻烦。

非压缩版本有234个字符，我认为这仍然是值得尊敬的。

import sys
f,f,n=sys.argv
e=open(f).readlines()
p=range
for v in p(int(n)):e=[''.join('.X'[8+16*(e[t][i]!='.')>>sum(n!='.'for v in e[t-1:t+2]for n in v[i-1:i+2])&1]for i in p(80))for t in p(40)]
open('out.txt','w').write('\n'.join(e))

对于水平滚动很抱歉，但是上面所有的换行符都是必需的，我已经把它们计算为每个字符。

我不会试着读金币密码的。随机选择变量名以实现最佳压缩。是的，我是认真的。更好的格式和注释版本如下：

# get command-line arguments: infile and count
import sys
ignored, infile, count = sys.argv

# read the input into a list (each input line is a string in the list)
data = open(infile).readlines()

# loop the number of times requested on the command line
for loop in range(int(count)):
    # this monstrosity applies the rules for each iteration, replacing
    # the cell data with the next generation
    data = [''.join(

                # choose the next generation's cell from '.' for
                # dead, or 'X' for alive
                '.X'[

                    # here, we build a simple bitmask that implements
                    # the generational rules.  A bit from this integer
                    # will be chosen by the count of live cells in
                    # the 3x3 grid surrounding the current cell.
                    #
                    # if the current cell is dead, this bitmask will
                    # be 8 (0b0000001000).  Since only bit 3 is set,
                    # the next-generation cell will only be alive if
                    # there are exactly 3 living neighbors in this
                    # generation.
                    #
                    # if the current cell is alive, the bitmask will
                    # be 24 (8 + 16, 0b0000011000).  Since both bits
                    # 3 and 4 are set, this cell will survive if there
                    # are either 3 or 4 living cells in its neighborhood,
                    # including itself
                    8 + 16 * (data[y][x] != '.')

                    # shift the relevant bit into position
                    >>

                    # by the count of living cells in the 3x3 grid
                    sum(character != '.' # booleans will convert to 0 or 1
                        for row in data[y - 1 : y + 2]
                        for character in row[x - 1 : x + 2]
                    )

                    # select the relevant bit
                    & 1
                ]

               # for each column and row
                for x in range(80)
            )
            for y in range(40)
    ]

# write the results out
open('out.txt','w').write('\n'.join(data))

抱歉，Pythonistas，对于C-ish括号格式，但是我试图说明每个括号关闭的是什么。

Answer 2

F#，496

我可以减少很多，但我喜欢这个，因为它仍然在大致上和相当可读的。

open System.IO
let mutable a:_[,]=null
let N y x=
 [-1,-1;-1,0;-1,1;0,-1;0,1;1,-1;1,0;1,1]
 |>Seq.sumBy(fun(i,j)->try if a.[y+i,x+j]='X' then 1 else 0 with _->0)
[<EntryPoint>]
let M(r)=
 let b=File.ReadAllLines(r.[0])
 a<-Array2D.init 40 80(fun y x->b.[y].[x])
 for i=1 to int r.[1] do 
  a<-Array2D.init 40 80(fun y x->
   match N y x with|3->'X'|2 when a.[y,x]='X'->'X'|_->'.')
 File.WriteAllLines("out.txt",Array.init 40(fun y->
  System.String(Array.init 80(fun x->a.[y,x]))))
 0

编辑

428

根据请求，下面是我的下一次尝试：

open System
let mutable a,k=null,Array2D.init 40 80
[<EntryPoint>]
let M r=
 a<-k(fun y x->IO.File.ReadAllLines(r.[0]).[y].[x])
 for i=1 to int r.[1] do a<-k(fun y x->match Seq.sumBy(fun(i,j)->try if a.[y+i,x+j]='X'then 1 else 0 with _->0)[-1,-1;-1,0;-1,1;0,-1;0,1;1,-1;1,0;1,1]with|3->'X'|2 when a.[y,x]='X'->'X'|_->'.')
 IO.File.WriteAllLines("out.txt",Array.init 40(fun y->String(Array.init 80(fun x->a.[y,x]))))
 0

这是一个14%的削减与一些基本的高尔夫。我不禁觉得用2D数组/字符串数组(而不是一维数组)输掉了，但现在不想做这种转换。注意如何优雅地读取文件3200次来初始化我的数组:)

Answer 3

以下解决方案使用我自己定制的特定于域的编程语言，我称之为NULL：

3499538

如果您想知道这是如何工作的:我的语言只包含每个程序的一个状态。该语句表示属于代码高尔夫球线程的StackOverflow线程ID。我的编译器将其编译成一个程序，寻找最佳的javascript解决方案(使用SO )，下载它并在web浏览器中运行它。

对于新线程来说，运行时可能更好(出现第一个经过修改的Javascript答案可能需要一些时间)，但从好的方面来说，它只需要很少的编码技巧。