152

挑战

按字符计数的最短代码输入棋盘的 2D 表示,并根据输入输出“真”或“假” 。

该板由 4 种类型的瓷砖制成:

 # - A solid wall
 x - The target the laser has to hit
 / or \ - Mirrors pointing to a direction (depends on laser direction)
 v, ^, > or < - The laser pointing to a direction (down, up, right and left respectively)

只有一束激光,只有一个目标。墙壁必须形成一个任意大小的实心矩形,其中激光和目标放置在里面。“房间”内的墙壁是可能的。

激光射线从它的原点发射并传播到它所指向的方向。如果激光射线击中墙壁,它就会停止。如果激光射到镜子上,它会向镜子指向的方向反射 90 度。镜子是双面的,这意味着两面都是“反射性的”,并且可以以两种方式反射光线。如果激光束击中激光 ( ^v><) 本身,它会被视为一堵墙(激光束会破坏光束器,因此它永远不会击中目标)。

测试用例

输入:
    ##########
    # / \ #
    ##
    # \ X#
    # > / #
    ##########
输出:
    真的

输入:
    ##########
    #vx#
    # / #
    # /#
    # \ #
    ##########
输出:    
    错误的

输入:
    #############
    ###
    # > # #
    ###
    # # X #
    ###
    #############
输出:
    错误的

输入:
    ##########
    #/\/\/\ #
    #\\//\\\ #
    #//\/\/\\#
    #\/\/\/x^#
    ##########
输出:
    真的

代码计数包括输入/​​输出(即完整程序)。

4

28 回答 28

78

Perl,166 160 个字符

Perl,251 248 246 222 214 208 203 201 193 190 180 176 173 170 166 --> 160 个字符。

这场比赛结束时,Solution 有 166 笔,但 A. Rex 找到了几种方法来剃掉另外 6 个字符:

s!.!$t{$s++}=$&!ge,$s=$r+=99for<>;%d='>.^1<2v3'=~/./g;($r)=grep$d|=$d{$t{$_}},%t;
{$_=$t{$r+=(1,-99,-1,99)[$d^=3*/\\/+m</>]};/[\/\\ ]/&&redo}die/x/?true:false,$/

第一行将输入加载到%t棋盘表中,其中$t{99*i+j}包含第 i 行第j的字符。然后,

%d=split//,'>.^1<2v3' ; ($r)=grep{$d|=$d{$t{$_}}}%t

它在 的元素中搜索与或%t匹配的字符,同时将其设置为 0 到 3 之间的值,该值指示激光束的初始方向。> ^ <v$d

在主循环的每次迭代开始时,我们会更新$d光束当前是否在镜子上。XOR'ing 3 给出了\镜像的正确行为,XOR'ing 1 给出了/镜像的正确行为。

$d^=3*/\\/+m</>

接下来,根据$r当前方向更新当前位置。

$r+=(1,-99,-1,99)[$d] ; $_ = $t{$r}

我们将当前位置的字符分配给以$_方便使用匹配运算符。

/[\/\\ ]/ && redo

如果我们在空格或镜像字符上,请继续。否则,true如果我们在目标 ( $_ =~ /x/)上,则终止,否则终止false

限制:可能无法处理超过 99 列的问题。可以以另外 3 个字符为代价来消除此限制,

于 2009-09-26T02:19:42.397 回答
75

Perl,177 个字符

第一个换行符可以去掉;其他两个是强制性的。

$/=%d=split//,' >/^\v';$_=<>;$s='#';{
y/v<^/>v</?do{my$o;$o.=" 
"while s/.$/$o.=$&,""/meg;y'/\\'\/'for$o,$s;$_=$o}:/>x/?die"true
":/>#/?die"false
":s/>(.)/$s$d{$1}/?$s=$1:1;redo}

解释:

$/ = %d = (' ' => '>', '/' => '^', '\\' => 'v');

如果向右移动的光束射入{空白空间,上角镜,下角镜},则它变成{右移光束,上移光束,下移光束}。$/一路初始化——幸运的是“6”不是一个有效的输入字符。

$_ = <>;

将板读入$_.

$s="#";

$s是梁现在坐在上面的任何东西的象征。由于激光发射器将被视为一堵墙,因此首先将其设置为一堵墙。

if (tr/v<^/>v</) {
  my $o;
  $o .= "\n" while s/.$/$o .= $&, ""/meg;
  tr,/\\,\\/, for $o, $s;
  $_ = $o;
}

如果激光束指向除右以外的任何方向,请旋转其符号,然后将整个板旋转到位(同时旋转镜子的符号)。这是一个 90 度的左旋转,通过在调换行和列的同时颠倒行来有效地完成,具有轻微s///e的副作用。y'''在打高尔夫球的代码中, tr 以允许我跳过反斜杠的形式编写一个反斜杠。

die "true\n" if />x/; die "false\n" if />#/;

如果我们击中目标或墙壁,请以正确的信息终止。

$s = $1 if s/>(.)/$s$d{$1}/;

如果激光器前面有空位,请向前移动。如果激光器前面有镜子,向前移动并旋转光束。在任何一种情况下,将“保存的符号”放回旧的梁位置,并将我们刚刚覆盖的东西放入保存的符号中。

redo;

重复直到终止。{...;redo}是小于 2 个字符for(;;){...}和小于 3 个字符while(1){...}

于 2009-09-26T04:25:12.830 回答
39

C89(209 个字符)

#define M(a,b)*p==*#a?m=b,*p=1,q=p:
*q,G[999],*p=G;w;main(m){for(;(*++p=getchar())>0;)M(<,-1)M
(>,1)M(^,-w)M(v,w)!w&*p<11?w=p-G:0;for(;q+=m,m=*q&4?(*q&1?
-1:1)*(m/w?m/w:m*w):*q&9?!puts(*q&1?"false":"true"):m;);}

解释

如果你不理解 C,这个怪物可能很难理解。只是一个预警。

#define M(a,b)*p==*#a?m=b,*p=1,q=p:

这个小宏检查当前字符 ( *p) 是否等于a字符形式 ( *#a) 中的任何内容。如果它们相等,则将移动矢量设置为b( m=b),将此字符标记为墙 ( *p=1),并将起点设置为当前位置 ( q=p)。该宏包括“else”部分。

*q,G[999],*p=G;
w;

声明一些变量。*q是灯的当前位置。*G是作为一维数组的游戏板。*p是填充时的当前读取位置G。*w是板的宽度。

main(m){

很明显mainm是存储运动矢量的变量。(这是main作为优化的参数。)

    for(;(*++p=getchar())>0;)

循环遍历所有字符,G使用p. 跳过G[0]作为优化(无需p在第三部分再次浪费一个字符for)。

        M(<,-1)
        M(>,1)
        M(^,-w)
        M(v,w)

如果可能,请使用上述宏定义激光。 -11分别对应左右,-ww上下。

        !w&*p<11
            ?w=p-G
            :0;

如果当前字符是行尾标记 (ASCII 10),则设置宽度(如果尚未设置)。skippedG[0]允许我们写w=p-G而不是w=p-G+1. 此外,这完成了's的?:链。M

    for(;
        q+=m,

通过移动矢量移动灯光。

        m=
        *q&4
            ?(*q&1?-1:1)*(
                m/w?m/w:m*w
            )

反映运动矢量。

            :*q&9
                ?!puts(*q&1?"false":"true")
                :m
        ;

如果这是一堵墙或x,请退出并显示适当的消息(m=0终止循环)。否则,什么都不做(noop; m=m

    );
}
于 2009-09-26T01:42:30.183 回答
36

我敢打赌人们已经等了很久了。(你的意思是,挑战结束了,没人关心了?)

看哪......我在这里提出了一个解决方案

Befunge-93!

它的重量高达973个字符(如果您足够慈善以忽略空格,则为688个字符,空格仅用于格式化,在实际代码中不执行任何操作)。

警告:不久前,我用 Perl 编写了自己的 Befunge-93 解释器,不幸的是,这就是我真正有时间测试它的全部内容。一般来说,我对它的正确性相当有信心,但它可能对 EOF 有一个奇怪的限制:由于 Perl 的<>运算符在文件末尾返回 undef,因此在数字上下文中将其处理为 0。对于 EOF 具有不同值(例如 -1)的基于 C 的实现,此代码可能不起作用。

003pv   >~v>  #v_"a"43g-!#v_23g03p33v>v
>39#<*v   ::   >:52*-!v   >"rorrE",vg2*
######1   >^vp31+1g31$_03g13gp vv,,<15,
    a#3     >0v       vp30+1g30<>,,#3^@
######p $     0vg34"a"<   >       >vp
^<v>  > ^   p3<>-#v_:05g-!|>:15g-!| $
 >     v^     <   <   <   >^v-g52:< $ 
  v _  >52*"eslaf",,vv|-g53:_      v   
  : ^-"#">#:< #@,,,,<<>:43p0 v0 p34< 
  >">"-!vgv<  ^0p33g31p32-1g3<       
 ^     <#g1|-g34_v#-g34_v#-g34"><v^"<<<<
    v!<^<33>13g1v>03g1-v>03g1+03p$v  $$
>^  _#-v 1>g1-1v>+13pv >03p       v  pp
^_:"^"^#|^g30 <3#   $<           $<>^33
 ^!-"<":<>"v"v^># p#$<>            $^44
^      >#^#_ :" "-#v_ ^   >         ^gg
v  g34$<   ^!<v"/":< >$3p$^>05g43p$ ^55
 >,@   |!-"\"  :_$43g:">"-!|>      ^$32
 *v"x":<      >-^    ^4g52<>:"^" -#v_^
 5>-!#v_"ror"vv$p34g51:<>#|  !-"<":<#|
 ^2,,, ,,"er"<>v      #^^#<>05g43p$$^>^
      >52*"eurt",,,,,@>15g4 3p$$$$  ^#
>:"v"\:"<"\: "^"   -!#^_-!#^_-!      ^
               >                       ^

解释

如果您不熟悉 Befunge 语法和操作,请查看此处

Befunge 是一种基于堆栈的语言,但有一些命令允许人们将字符写入 Befunge 代码。我在两个地方利用了这一点。首先,我将整个输入复制到 Befunge 板上,但位于实际编写的代码下方几行。(当然,当代码运行时,这实际上是不可见的。)

另一个地方在左上角附近:

######
    a#
######

在这种情况下,我在上面突出显示的区域是我存储几个坐标的地方。中间行的第一列是我存储当前“光标位置”的 x 坐标的位置;第二列是我存储 y 坐标的位置;接下来的两列用于存储找到激光束源的 x 和 y 坐标;最后一列(其中包含“a”字符)最终被覆盖以包含当前的光束方向,随着光束的路径被追踪,该方向显然会发生变化。

程序首先将 (0,27) 作为初始光标位置。然后输入一次读取一个字符并放置在光标位置;换行符只会导致 y 坐标增加而 x 坐标返回 0,就像真正的回车一样。最终 undef 被解释器读取,0 字符值用于表示输入结束并继续进行激光迭代步骤。当读取激光字符 [<>^v] 时,它也被复制到内存库(在“a”字符上),并且它的坐标被复制到左侧的列中。

所有这一切的最终结果是,整个文件基本上被复制到 Befunge 代码中,比实际遍历的代码略低一点。

之后,将光束位置复制回光标位置,并执行以下迭代:

  • 检查当前光束方向并适当地增加或减少光标坐标。(我首先这样做是为了避免在第一步就必须处理激光束的角落情况。)
  • 读取该位置的字符。
  • 如果字符是“#”,则将换行符和“false”放在堆栈上,打印并结束。
  • 将其与所有光束字符 [<>^v] 进行比较;如果匹配,也打印“false\n”并结束。
  • 如果字符是空格,则清空堆栈并继续。
  • 如果字符是正斜杠,则将光束方向放到堆栈上并依次与每个方向字符进行比较。当找到一个时,新方向将存储在代码中的同一位置并重复循环。
  • 如果字符是反斜杠,则执行与上述基本相同的操作(反斜杠的正确映射除外)。
  • 如果字符是'x',我们已经击中了目标。打印“true\n”并退出。
  • 如果字符不是这些,打印“error\n”并退出。

如果有足够的需求,我会尝试指出代码中的确切位置。

于 2010-07-22T22:07:12.787 回答
29

F#,36 行,可读性强

好的,只是为了得到一个答案:

let ReadInput() =
    let mutable line = System.Console.ReadLine()
    let X = line.Length 
    let mutable lines = []
    while line <> null do
        lines <- Seq.to_list line :: lines
        line <- System.Console.ReadLine()
    lines <- List.rev lines
    X, lines.Length, lines

let X,Y,a = ReadInput()
let mutable p = 0,0,'v'
for y in 0..Y-1 do
    for x in 0..X-1 do 
        printf "%c" a.[y].[x]
        match a.[y].[x] with 
        |'v'|'^'|'<'|'>' -> p <- x,y,a.[y].[x]
        |_ -> ()
    printfn ""

let NEXT = dict [ '>', (1,0,'^','v')
                  'v', (0,1,'<','>')
                  '<', (-1,0,'v','^')
                  '^', (0,-1,'>','<') ]
let next(x,y,d) =
    let dx, dy, s, b = NEXT.[d]
    x+dx,y+dy,(match a.[y+dy].[x+dx] with
               | '/' -> s
               | '\\'-> b
               | '#'|'v'|'^'|'>'|'<' -> printfn "false"; exit 0
               | 'x' -> printfn "true"; exit 0
               | ' ' -> d)

while true do
    p <- next p    

样品:

##########
#   / \  #
#        #
#   \   x#
# >   /  #
##########
true

##########
#   v x  #
# /      #
#       /#
#   \    #
##########
false

#############
#     #     #
# >   #     #
#     #     #
#     #   x #
#     #     #
#############
false

##########
#/\/\/\  #
#\\//\\\ #
#//\/\/\\#
#\/\/\/x^#
##########
true

##########
#   / \  #
#        #
#/    \ x#
#\>   /  #
##########
false

##########
#  /    \#
# / \    #
#/    \ x#
#\^/\ /  #
##########
false
于 2009-09-26T01:26:42.043 回答
29

Golfscript - 83 个字符(我和 strager 的混搭)

换行符只是在这里换行

:|'v^><'.{|?}%{)}?:$@=?{.[10|?).~)1-1]=$+
:$|=' \/x'?\[.\2^.1^'true''false']=.4/!}do

Golfscript - 107 个字符

为了清楚起见,换行符就在那里

10\:@?):&4:$;{0'>^<v'$(:$=@?:*>}do;
{[1 0&--1&]$=*+:*;[{$}{3$^}{1$^}{"true "}{"false"}]@*=' \/x'?=~5\:$>}do$

这个怎么运作。

第一行计算出初始位置和方向。
每当激光击中镜子时,第二条线就会逐步转向。

于 2009-11-09T01:19:59.253 回答
18

Ruby 中的 353 个字符:

现在有314 277 个字符!

好的,Ruby 中有 256 个字符,现在我完成了。不错的整数。:)

247 个字符。我无法停止。

Ruby 中的223 203 201 个字符

d=x=y=-1;b=readlines.each{|l|d<0&&(d="^>v<".index l[x]if x=l.index(/[>^v<]/)
y+=1)};loop{c=b[y+=[-1,0,1,0][d]][x+=[0,1,0,-1][d]]
c==47?d=[1,0,3,2][d]:c==92?d=3-d:c==35?(p !1;exit):c<?x?0:(p !!1;exit)}

带空格:

d = x = y = -1
b = readlines.each { |l|
  d < 0 && (d = "^>v<".index l[x] if x = l.index(/[>^v<]/); y += 1)
}

loop {
  c = b[y += [-1, 0, 1, 0][d]][x += [0, 1, 0, -1][d]]

  c == 47 ? d = [1, 0, 3, 2][d] :
  c == 92 ? d = 3 - d :
  c == 35 ? (p !1; exit) :
  c < ?x ? 0 : (p !!1; exit)
}

稍微重构:

board = readlines

direction = x = y = -1
board.each do |line|
  if direction < 0
    x = line.index(/[>^v<]/)
    if x
      direction = "^>v<".index line[x]
    end
    y += 1
  end
end

loop do
  x += [0, 1, 0, -1][direction]
  y += [-1, 0, 1, 0][direction]

  ch = board[y][x].chr
  case ch
  when "/"
    direction = [1, 0, 3, 2][direction]
  when "\\"
    direction = 3 - direction
  when "x"
    puts "true"
    exit
  when "#"
    puts "false"
    exit
  end
end
于 2009-09-26T02:37:13.473 回答
17

Python

294 277 253 240 232 个字符,包括换行符:

(第 4 行和第 5 行的第一个字符是制表符,而不是空格)

l='>v<^';x={'/':'^<v>','\\':'v>^<',' ':l};b=[1];r=p=0
while b[-1]:
 b+=[raw_input()];r+=1
 for g in l:
    c=b[r].find(g)
    if-1<c:p=c+1j*r;d=g
while' '<d:z=l.find(d);p+=1j**z;c=b[int(p.imag)][int(p.real)];d=x.get(c,' '*4)[z]
print'#'<c

我忘记了 Python 甚至还有可选的分号。

这个怎么运作

此代码背后的关键思想是使用复数来表示位置和方向。行是假想轴,向下增加。列是实轴,向右增加。

l='>v<^';激光符号列表。选择顺序以使激光方向字符的索引对应于 sqrt(-1) 的幂

x={'/':'^<v>','\\':'v>^<',' ':l};一个转换表,确定当光束离开不同的瓦片时方向如何变化。瓷砖是关键,新的方向是价值。

b=[1];持有董事会。第一个元素是 1(评估为真),因此 while 循环将至少运行一次。

r=p=0 r是输入的当前行号,p是激光束的当前位置。

while b[-1]:当 raw_input 返回空字符串时停止加载板数据

b+=[raw_input()];r+=1将下一行输入附加到板上并增加行计数器

for g in l:依次猜测每个激光方向

c=b[r].find(g)将列设置为激光的位置,如果不在行中(或指向不同的方向),则设置为 -1

if-1<c:p=c+1j*r;d=g如果我们找到了激光,则设置当前位置p和方向dd是其中的字符之一l

将板子装入 后b,当前位置p和方向d已设置为激光源的位置和方向。

while' '<d:space 的 ASCII 值比任何方向符号都低,因此我们将其用作停止标志。

z=l.find(d);字符串中当前方向字符的索引lz稍后用于使用x表格确定新的光束方向,并增加位置。

p+=1j**z;使用 i 的幂增加位置。例如,l.find('<')==2-> i^2 = -1,它将向左移动一列。

c=b[int(p.imag)][int(p.real)];读取当前位置的字符

d=x.get(c,' '*4)[z]在变换表中查找光束的新方向。如果表中不存在当前字符,则设置d为空格。

print'#'<c如果我们停在目标以外的任何东西上,则打印 false。

于 2009-09-26T02:59:48.413 回答
16

F#,255 个字符(并且仍然相当可读!):

好的,经过一晚的休息,我改善了很多:

let a=System.Console.In.ReadToEnd()
let w,c=a.IndexOf"\n"+1,a.IndexOfAny[|'^';'<';'>';'v'|]
let rec n(c,d)=
 let e,s=[|-w,2;-1,3;1,0;w,1|].[d]
 n(c+e,match a.[c+e]with|'/'->s|'\\'->3-s|' '->d|c->printfn"%A"(c='x');exit 0)
n(c,"^<>v".IndexOf a.[c])

让我们一行一行地谈论它。

首先,将所有输入放入一个大的一维数组中(二维数组可能对代码高尔夫不利;只需使用一维数组并在索引中添加/减去一行的宽度即可向上/向下移动一行)。

接下来,我们通过索引到我们的数组来计算输入行的宽度“w”和起始位置“c”。

现在让我们定义“next”函数“n”,它采用当前位置“c”和方向“d”,即上、左、右、下的 0、1、2、3。

index-epsilon 'e' 和 what-new-direction-if-we-hit-a-slash 's' 由表计算。例如,如果当前方向 'd' 为 0(向上),则表的第一个元素为“-w,2”,这意味着我们将索引减少 w,如果我们点击斜线,则新方向为 2 (对)。

现在我们使用 (1) 下一个索引 ("c+e" - current plus epsilon) 和 (2) 新方向递归到下一个函数 'n',我们通过向前查看数组中的内容来计算下一个单元格。如果前瞻字符是斜线,则新方向是“s”。如果是反斜杠,则新方向为 3-s(我们选择的编码 0123 使这项工作有效)。如果它是一个空间,我们只是继续朝着“d”的方向前进。如果是任何其他字符“c”,则游戏结束,如果字符为“x”则打印“true”,否则打印“false”。

首先,我们使用初始位置“c”和起始方向(将方向初始编码为 0123)调用递归函数“n”。

我想我可能仍然可以从它上面刮掉更多的字符,但我对这样的它很满意(255 是一个不错的数字)。

于 2009-09-26T01:55:03.880 回答
16

是Brian 对 C#3 的解决 方案的直接移植,减去了控制台交互。这不是挑战的条目,因为它不是一个完整的程序,我只是想知道他使用的一些 F# 构造如何在 C# 中表示。

bool Run(string input) {
    var a = input.Split(new[] {Environment.NewLine}, StringSplitOptions.None);
    var p = a.SelectMany((line, y) => line.Select((d, x) => new {x, y, d}))
             .First(x => new[] {'v', '^', '<', '>'}.Contains(x.d));
    var NEXT = new[] {
            new {d = '>', dx = 1, dy = 0, s = '^', b = 'v'},
            new {d = 'v', dx = 0, dy = 1, s = '<', b = '>'},
            new {d = '<', dx = -1, dy = 0, s = 'v', b = '^'},
            new {d = '^', dx = 0, dy = -1, s = '>', b = '<'}
        }.ToDictionary(x => x.d);
    while (true) {
        var n = NEXT[p.d];
        int x = p.x + n.dx,
            y = p.y + n.dy;
        var d = a[y][x];
        switch (d) {
            case '/':  d = n.s; break;
            case '\\': d = n.b; break;
            case ' ':  d = p.d; break;
            default: return d == 'x';
        }
        p = new {x, y, d};
    }
}

编辑:经过一些实验,以下相当冗长的搜索代码:

int X = a[0].Length, Y = a.Length;
var p = new {x = 0, y = 0, d = 'v'};
for (var y = 0; y < Y; y++) {
    for (var x = 0; x < X; x++) {
        var d = a[y][x];
        switch (d) {
            case 'v': case '^': case '<': case '>':
                p = new {x, y, d}; break;
        }
    }
}

已被一些更紧凑的 LINQ to Objects 代码取代:

var p = a.SelectMany((line, y) => line.Select((d, x) => new {x, y, d}))
         .First(x => new[] {'v', '^', '<', '>'}.Contains(x.d));
于 2009-09-26T04:19:17.220 回答
11

18203 个字符是一个 Python 解决方案,它可以:

  • 应对“房间”外的镜子
  • 根据 2D 限制计算没有“房间”时的轨迹(规范说了很多关于“房间”中必须有什么但如果房间必须存在则没有)
  • 报告错误

它仍然需要稍微整理一下,我不知道 2D 物理是否规定光束不能自我交叉......

#!/usr/bin/env python
# -*- coding: utf-8 -*-

"""
The shortest code by character count to input a 2D representation of a board, 
and output 'true' or 'false' according to the input.

The board is made out of 4 types of tiles:

# - A solid wall
x - The target the laser has to hit
/ or \ - Mirrors pointing to a direction (depends on laser direction)
v, ^, > or < - The laser pointing to a direction (down, up, right and left
respectively)

There is only one laser and only one target. Walls must form a solid rectangle 
of any size, where the laser and target are placed inside. Walls inside the
'room' are possible.

Laser ray shots and travels from it's origin to the direction it's pointing. If
a laser ray hits the wall, it stops. If a laser ray hits a mirror, it is bounces
90 degrees to the direction the mirror points to. Mirrors are two sided, meaning
both sides are 'reflective' and may bounce a ray in two ways. If a laser ray
hits the laser (^v><) itself, it is treated as a wall (laser beam destroys the
beamer and so it'll never hit the target).
"""



SOLID_WALL, TARGET, MIRROR_NE_SW, MIRROR_NW_SE, LASER_DOWN, LASER_UP, \
LASER_RIGHT, LASER_LEFT = range(8)

MIRRORS = (MIRROR_NE_SW, MIRROR_NW_SE)

LASERS = (LASER_DOWN, LASER_UP, LASER_RIGHT, LASER_LEFT)

DOWN, UP, RIGHT, LEFT = range(4)

LASER_DIRECTIONS = {
    LASER_DOWN : DOWN,
    LASER_UP   : UP,
    LASER_RIGHT: RIGHT,
    LASER_LEFT : LEFT
}

ROW, COLUMN = range(2)

RELATIVE_POSITIONS = {
    DOWN : (ROW,     1),
    UP   : (ROW,    -1),
    RIGHT: (COLUMN,  1),
    LEFT : (COLUMN, -1)
}

TILES = {"#" : SOLID_WALL,
         "x" : TARGET,
         "/" : MIRROR_NE_SW,
         "\\": MIRROR_NW_SE,
         "v" : LASER_DOWN,
         "^" : LASER_UP,
         ">" : LASER_RIGHT,
         "<" : LASER_LEFT}

REFLECTIONS = {MIRROR_NE_SW: {DOWN : LEFT,
                              UP   : RIGHT,
                              RIGHT: UP,
                              LEFT : DOWN},
               MIRROR_NW_SE: {DOWN : RIGHT,
                              UP   : LEFT,
                              RIGHT: DOWN,
                              LEFT : UP}}



def does_laser_hit_target(tiles):
    """
        Follows a lasers trajectory around a grid of tiles determining if it
        will reach the target.

        Keyword arguments:
        tiles --- row/column based version of a board containing symbolic
                  versions of the tiles (walls, laser, target, etc)
    """

    #Obtain the position of the laser
    laser_pos = get_laser_pos(tiles)

    #Retrieve the laser's tile
    laser = get_tile(tiles, laser_pos)

    #Create an editable starting point for the beam
    beam_pos = list(laser_pos)

    #Create an editable direction for the beam
    beam_dir = LASER_DIRECTIONS[laser]

    #Cache the number of rows
    number_of_rows = len(tiles)

    #Keep on looping until an ultimate conclusion
    while True:

        #Discover the axis and offset the beam is travelling to
        axis, offset = RELATIVE_POSITIONS[beam_dir]

        #Modify the beam's position
        beam_pos[axis] += offset

        #Allow for a wrap around in this 2D scenario
        try:

            #Get the beam's new tile
            tile = get_tile(tiles, beam_pos)

        #Perform wrapping
        except IndexError:

            #Obtain the row position
            row_pos = beam_pos[ROW]

            #Handle vertical wrapping
            if axis == ROW:

                #Handle going off the top
                if row_pos == -1:

                    #Move beam to the bottom
                    beam_pos[ROW] = number_of_rows - 1

                #Handle going off the bottom
                elif row_pos == number_of_rows:

                    #Move beam to the top
                    beam_pos[ROW] = 0

            #Handle horizontal wrapping
            elif axis == COLUMN:

                #Obtain the row
                row = tiles[row_pos]

                #Calculate the number of columns
                number_of_cols = len(row)

                #Obtain the column position
                col_pos = beam_pos[COLUMN]

                #Handle going off the left hand side
                if col_pos == -1:

                    #Move beam to the right hand side
                    beam_pos[COLUMN] = number_of_cols - 1

                #Handle going off the right hand side
                elif col_pos == number_of_cols:

                    #Move beam to the left hand side
                    beam_pos[COLUMN] = 0

            #Get the beam's new tile
            tile = get_tile(tiles, beam_pos)

        #Handle hitting a wall or the laser
        if tile in LASERS \
        or tile == SOLID_WALL:
            return False

        #Handle hitting the target
        if tile == TARGET:
            return True

        #Handle hitting a mirror
        if tile in MIRRORS:
            beam_dir = reflect(tile, beam_dir)

def get_laser_pos(tiles):
    """
        Returns the current laser position or an exception.

        Keyword arguments:
        tiles --- row/column based version of a board containing symbolic
                  versions of the tiles (walls, laser, target, etc)
    """

    #Calculate the number of rows
    number_of_rows = len(tiles)

    #Loop through each row by index
    for row_pos in range(number_of_rows):

        #Obtain the current row
        row = tiles[row_pos]

        #Calculate the number of columns
        number_of_cols = len(row)

        #Loop through each column by index
        for col_pos in range(number_of_cols):

            #Obtain the current column
            tile = row[col_pos]

            #Handle finding a laser
            if tile in LASERS:

                #Return the laser's position
                return row_pos, col_pos

def get_tile(tiles, pos):
    """
        Retrieves a tile at the position specified.

        Keyword arguments:
        pos --- a row/column position of the tile
        tiles --- row/column based version of a board containing symbolic
                  versions of the tiles (walls, laser, target, etc)
    """

    #Obtain the row position
    row_pos = pos[ROW]

    #Obtain the column position
    col_pos = pos[COLUMN]

    #Obtain the row
    row = tiles[row_pos]

    #Obtain the tile
    tile = row[col_pos]

    #Return the tile
    return tile

def get_wall_pos(tiles, reverse=False):
    """
        Keyword arguments:
        tiles --- row/column based version of a board containing symbolic
                  versions of the tiles (walls, laser, target, etc)
        reverse --- whether to search in reverse order or not (defaults to no)
    """

    number_of_rows = len(tiles)

    row_iter = range(number_of_rows)

    if reverse:
        row_iter = reversed(row_iter)

    for row_pos in row_iter:
        row = tiles[row_pos]

        number_of_cols = len(row)

        col_iter = range(number_of_cols)

        if reverse:
            col_iter = reversed(col_iter)

        for col_pos in col_iter:
            tile = row[col_pos]

            if tile == SOLID_WALL:
                pos = row_pos, col_pos

                if reverse:
                    offset = -1
                else:
                    offset = 1

                for axis in ROW, COLUMN:
                    next_pos = list(pos)

                    next_pos[axis] += offset

                    try:
                        next_tile = get_tile(tiles, next_pos)
                    except IndexError:
                        next_tile = None

                    if next_tile != SOLID_WALL:
                        raise WallOutsideRoomError(row_pos, col_pos)

                return pos

def identify_tile(tile):
    """
        Returns a symbolic value for every identified tile or None.

        Keyword arguments:
        tile --- the tile to identify
    """

    #Safely lookup the tile
    try:

        #Return known tiles
        return TILES[tile]

    #Handle unknown tiles
    except KeyError:

        #Return a default value
        return

def main():
    """
        Takes a board from STDIN and either returns a result to STDOUT or an
        error to STDERR.

        Called when this file is run on the command line.
    """

    #As this function is the only one to use this module, and it can only be
    #called once in this configuration, it makes sense to only import it here.
    import sys

    #Reads the board from standard input.
    board = sys.stdin.read()

    #Safely handles outside input
    try:

        #Calculates the result of shooting the laser
        result = shoot_laser(board)

    #Handles multiple item errors
    except (MultipleLaserError, MultipleTargetError) as error:

        #Display the error
        sys.stderr.write("%s\n" % str(error))

        #Loop through all the duplicated item symbols
        for symbol in error.symbols:

            #Highlight each symbol in green
            board = board.replace(symbol, "\033[01;31m%s\033[m" % symbol)

        #Display the board
        sys.stderr.write("%s\n" % board)

        #Exit with an error signal
        sys.exit(1)

    #Handles item missing errors
    except (NoLaserError, NoTargetError) as error:

        #Display the error
        sys.stderr.write("%s\n" % str(error))

        #Display the board
        sys.stderr.write("%s\n" % board)

        #Exit with an error signal
        sys.exit(1)

    #Handles errors caused by symbols
    except (OutsideRoomError, WallNotRectangleError) as error:

        #Displays the error
        sys.stderr.write("%s\n" % str(error))

        lines = board.split("\n")

        line = lines[error.row_pos]

        before = line[:error.col_pos]

        after = line[error.col_pos + 1:]

        symbol = line[error.col_pos]

        line = "%s\033[01;31m%s\033[m%s" % (before, symbol, after)

        lines[error.row_pos] = line

        board = "\n".join(lines)

        #Display the board
        sys.stderr.write("%s\n" % board)

        #Exit with an error signal
        sys.exit(1)

    #Handles errors caused by non-solid walls
    except WallNotSolidError as error:

        #Displays the error
        sys.stderr.write("%s\n" % str(error))

        lines = board.split("\n")

        line = lines[error.row_pos]

        before = line[:error.col_pos]

        after = line[error.col_pos + 1:]

        symbol = line[error.col_pos]

        line = "%s\033[01;5;31m#\033[m%s" % (before, after)

        lines[error.row_pos] = line

        board = "\n".join(lines)

        #Display the board
        sys.stderr.write("%s\n" % board)

        #Exit with an error signal
        sys.exit(1)

    #If a result was returned
    else:

        #Converts the result into a string
        result_str = str(result)

        #Makes the string lowercase
        lower_result = result_str.lower()

        #Returns the result
        sys.stdout.write("%s\n" % lower_result)

def parse_board(board):
    """
        Interprets the raw board syntax and returns a grid of tiles.

        Keyword arguments:
        board --- the board containing the tiles (walls, laser, target, etc)
    """

    #Create a container for all the lines
    tiles = list()

    #Loop through all the lines of the board
    for line in board.split("\n"):

        #Identify all the tiles on the line 
        row = [identify_tile(tile) for tile in line]

        #Add the row to the container
        tiles.append(row)

    #Return the container
    return tiles

def reflect(mirror, direction):
    """
        Returns an updated laser direction after it has been reflected on a
        mirror.

        Keyword arguments:
        mirror --- the mirror to reflect the laser from
        direction --- the direction the laser is travelling in
    """

    try:
        direction_lookup = REFLECTIONS[mirror]
    except KeyError:
        raise TypeError("%s is not a mirror.", mirror)

    try:
        return direction_lookup[direction]
    except KeyError:
        raise TypeError("%s is not a direction.", direction)

def shoot_laser(board):
    """
        Shoots the boards laser and returns whether it will hit the target.

        Keyword arguments:
        board --- the board containing the tiles (walls, laser, target, etc)
    """

    tiles = parse_board(board)

    validate_board(tiles)

    return does_laser_hit_target(tiles)

def validate_board(tiles):
    """
        Checks an board to see if it is valid and raises an exception if not.

        Keyword arguments:
        tiles --- row/column based version of a board containing symbolic
                  versions of the tiles (walls, laser, target, etc)
    """

    found_laser = False
    found_target = False

    try:
        n_wall, w_wall = get_wall_pos(tiles)
        s_wall, e_wall = get_wall_pos(tiles, reverse=True)
    except TypeError:
        n_wall = e_wall = s_wall = w_wall = None

    number_of_rows = len(tiles)

    for row_pos in range(number_of_rows):
        row = tiles[row_pos]

        number_of_cols = len(row)

        for col_pos in range(number_of_cols):

            tile = row[col_pos]

            if ((row_pos in (n_wall, s_wall) and
                 col_pos in range(w_wall, e_wall))
                or
                (col_pos in (e_wall, w_wall) and
                 row_pos in range(n_wall, s_wall))):
                if tile != SOLID_WALL:
                    raise WallNotSolidError(row_pos, col_pos)
            elif (n_wall != None and
                  (row_pos < n_wall or
                   col_pos > e_wall or
                   row_pos > s_wall or
                   col_pos < w_wall)):

                if tile in LASERS:
                    raise LaserOutsideRoomError(row_pos, col_pos)
                elif tile == TARGET:
                    raise TargetOutsideRoomError(row_pos, col_pos)
                elif tile == SOLID_WALL:
                    if not (row_pos >= n_wall and
                            col_pos <= e_wall and
                            row_pos <= s_wall and
                            col_pos >= w_wall):
                        raise WallOutsideRoomError(row_pos, col_pos)
            else:
                if tile in LASERS:
                    if not found_laser:
                        found_laser = True
                    else:
                        raise MultipleLaserError(row_pos, col_pos)
                elif tile == TARGET:
                    if not found_target:
                        found_target = True
                    else:
                        raise MultipleTargetError(row_pos, col_pos)

    if not found_laser:
        raise NoLaserError(tiles)

    if not found_target:
        raise NoTargetError(tiles)



class LasersError(Exception):
    """Parent Error Class for all errors raised."""

    pass

class NoLaserError(LasersError):
    """Indicates that there are no lasers on the board."""

    symbols = "^v><"

    def __str__ (self):
        return "No laser (%s) to fire." % ", ".join(self.symbols)

class NoTargetError(LasersError):
    """Indicates that there are no targets on the board."""

    symbols = "x"

    def __str__ (self):
        return "No target (%s) to hit." % ", ".join(self.symbols)

class MultipleLaserError(LasersError):
    """Indicates that there is more than one laser on the board."""

    symbols = "^v><"

    def __str__ (self):
        return "Too many lasers (%s) to fire, only one is allowed." % \
               ", ".join(self.symbols)

class MultipleTargetError(LasersError):
    """Indicates that there is more than one target on the board."""

    symbols = "x"

    def __str__ (self):
        return "Too many targets (%s) to hit, only one is allowed." % \
               ", ".join(self.symbols)

class WallNotSolidError(LasersError):
    """Indicates that the perimeter wall is not solid."""

    __slots__ = ("__row_pos", "__col_pos", "n_wall", "s_wall", "e_wall",
                 "w_wall")

    def __init__(self, row_pos, col_pos):
        self.__row_pos = row_pos
        self.__col_pos = col_pos

    def __str__ (self):
        return "Walls must form a solid rectangle."

    def __get_row_pos(self):
        return self.__row_pos

    def __get_col_pos(self):
        return self.__col_pos

    row_pos = property(__get_row_pos)
    col_pos = property(__get_col_pos)

class WallNotRectangleError(LasersError):
    """Indicates that the perimeter wall is not a rectangle."""

    __slots__ = ("__row_pos", "__col_pos")

    def __init__(self, row_pos, col_pos):
        self.__row_pos = row_pos
        self.__col_pos = col_pos

    def __str__ (self):
        return "Walls must form a rectangle."

    def __get_row_pos(self):
        return self.__row_pos

    def __get_col_pos(self):
        return self.__col_pos

    row_pos = property(__get_row_pos)
    col_pos = property(__get_col_pos)

class OutsideRoomError(LasersError):
    """Indicates an item is outside of the perimeter wall."""

    __slots__ = ("__row_pos", "__col_pos", "__name")

    def __init__(self, row_pos, col_pos, name):
        self.__row_pos = row_pos
        self.__col_pos = col_pos
        self.__name = name

    def __str__ (self):
        return "A %s was found outside of a 'room'." % self.__name

    def __get_row_pos(self):
        return self.__row_pos

    def __get_col_pos(self):
        return self.__col_pos

    row_pos = property(__get_row_pos)
    col_pos = property(__get_col_pos)

class LaserOutsideRoomError(OutsideRoomError):
    """Indicates the laser is outside of the perimeter wall."""

    def __init__ (self, row_pos, col_pos):
        OutsideRoomError.__init__(self, row_pos, col_pos, "laser")

class TargetOutsideRoomError(OutsideRoomError):
    """Indicates the target is outside of the perimeter wall."""

    def __init__ (self, row_pos, col_pos):
        OutsideRoomError.__init__(self, row_pos, col_pos, "target")

class WallOutsideRoomError(OutsideRoomError):
    """Indicates that there is a wall outside of the perimeter wall."""

    def __init__ (self, row_pos, col_pos):
        OutsideRoomError.__init__(self, row_pos, col_pos, "wall")



if __name__ == "__main__":
    main()

显示颜色错误报告的 bash 脚本:

#!/bin/bash

declare -a TESTS

test() {
    echo -e "\033[1m$1\033[0m"
    tput sgr0
    echo "$2" | ./lasers.py
    echo
}

test \
"no laser" \
"    ##########
    #     x  #
    # /      #
    #       /#
    #   \\    #
    ##########"

test \
"multiple lasers" \
"    ##########
    #   v x  #
    # /      #
    #       /#
    #   \\  ^ #
    ##########"

test \
"no target" \
"    ##########
    #   v    #
    # /      #
    #       /#
    #   \\    #
    ##########"

test \
"multiple targets" \
"    ##########
    #   v x  #
    # /      #
    #       /#
    #   \\  x #
    ##########"

test \
"wall not solid" \
"    ##### ####
    #   v x  #
    # /      #
    #       /#
    #   \\    #
    ##########"

test \
"laser_outside_room" \
"    ##########
 >  #     x  #
    # /      #
    #       /#
    #   \\    #
    ##########"

test \
"laser before room" \
" >  ##########
    #     x  #
    # /      #
    #       /#
    #   \\    #
    ##########"

test \
"laser row before room" \
"   >
    ##########
    #     x  #
    # /      #
    #       /#
    #   \\    #
    ##########"

test \
"laser after room" \
"    ##########
    #     x  #
    # /      #
    #       /#
    #   \\    #
    ##########  >"

test \
"laser row after room" \
"    ##########
    #     x  #
    # /      #
    #       /#
    #   \\    #
    ##########
  > "

test \
"target outside room" \
"    ##########
 x  #   v    #
    # /      #
    #       /#
    #   \\    #
    ##########"

test \
"target before room" \
" x  ##########
    #   v    #
    # /      #
    #       /#
    #   \\    #
    ##########"

test \
"target row before room" \
"   x
    ##########
    #   v    #
    # /      #
    #       /#
    #   \\    #
    ##########"

test \
"target after room" \
"    ##########
    #   v    #
    # /      #
    #       /#
    #   \\    #
    ##########   x"

test \
"target row after room" \
"    ##########
    #   v    #
    # /      #
    #       /#
    #   \\    #
    ##########
  x "

test \
"wall outside room" \
"    ##########
 #  #   v    #
    # /      #
    #       /#
    #   \\  x #
    ##########"

test \
"wall before room" \
" #  ##########
    #   v    #
    # /      #
    #       /#
    #   \\  x #
    ##########"

test \
"wall row before room" \
"    #
    ##########
    #   v    #
    # /      #
    #       /#
    #   \\  x #
    ##########"

test \
"wall after room" \
"    ##########
    #   v    #
    # /      #
    #       /#
    #   \\  x #
    ########## #"

test \
"wall row after room" \
"    ##########
    #   v    #
    # /      #
    #       /#
    #   \\  x #
    ##########
  #"

test \
"mirror outside room positive" \
"    ##########
 /  #   / \\  #
    #        #
    #   \\   x#
    # >   /  #
    ########## "

test \
"mirrors outside room negative" \
"    ##########
 \\  #   v x  #
    # /      #
    #       /#
    #   \\    #
    ##########"

test \
"mirror before room positive" \
" \\  ##########
    #   / \\  #
    #        #
    #   \\   x#
    # >   /  #
    ########## "

test \
"mirrors before room negative" \
" /  ##########
    #   v x  #
    # /      #
    #       /#
    #   \\    #
    ##########"

test \
"mirror row before room positive" \
"     \\
    ##########
    #   / \\  #
    #        #
    #   \\   x#
    # >   /  #
    ########## "

test \
"mirrors row before room negative" \
"     \\
    ##########
    #   v x  #
    # /      #
    #       /#
    #   \\    #
    ##########"

test \
"mirror after row positive" \
"    ##########
    #   / \\  #
    #        #
    #   \\   x#
    # >   /  #
    ########## /  "

test \
"mirrors after row negative" \
"    ##########
    #   v x  #
    # /      #
    #       /#
    #   \\    #
    ##########   /  "

test \
"mirror row after row positive" \
"    ##########
    #   / \\  #
    #        #
    #   \\   x#
    # >   /  #
    ########## 
 /  "

test \
"mirrors row after row negative" \
"    ##########
    #   v x  #
    # /      #
    #       /#
    #   \\    #
    ########## 
 /  "

test \
"laser hitting laser" \
"    ##########
    #   v   \\#
    #        #
    #        #
    #x  \\   /#
    ##########"

test \
"mirrors positive" \
"    ##########
    #   / \\  #
    #        #
    #   \\   x#
    # >   /  #
    ########## "

test \
"mirrors negative" \
"    ##########
    #   v x  #
    # /      #
    #       /#
    #   \\    #
    ##########"

test \
"wall collision" \
"    #############
    #     #     #
    # >   #     #
    #     #     #
    #     #   x #
    #     #     #
    #############"

test \
"extreme example" \
"    ##########
    #/\\/\\/\\  #
    #\\\\//\\\\\\ #
    #//\\/\\/\\\\#
    #\\/\\/\\/x^#
    ##########"

test \
"brian example 1" \
"##########
#   / \\  #
#        #
#/    \\ x#
#\\>   /  #
##########"

test \
"brian example 2" \
"##########
#  /    \\#
# / \\    #
#/    \\ x#
#\\^/\\ /  #
##########"

开发中使用的单元测试:

#!/usr/bin/env python
# -*- coding: utf-8 -*-

import unittest

from lasers import *

class TestTileRecognition(unittest.TestCase):
    def test_solid_wall(self):
        self.assertEqual(SOLID_WALL, identify_tile("#"))

    def test_target(self):
        self.assertEqual(TARGET, identify_tile("x"))

    def test_mirror_ne_sw(self):
        self.assertEqual(MIRROR_NE_SW, identify_tile("/"))

    def test_mirror_nw_se(self):
        self.assertEqual(MIRROR_NW_SE, identify_tile("\\"))

    def test_laser_down(self):
        self.assertEqual(LASER_DOWN, identify_tile("v"))

    def test_laser_up(self):
        self.assertEqual(LASER_UP, identify_tile("^"))

    def test_laser_right(self):
        self.assertEqual(LASER_RIGHT, identify_tile(">"))

    def test_laser_left(self):
        self.assertEqual(LASER_LEFT, identify_tile("<"))

    def test_other(self):
        self.assertEqual(None, identify_tile(" "))

class TestReflection(unittest.TestCase):
    def setUp(self):
        self.DIRECTION = LEFT
        self.NOT_DIRECTIO
于 2009-09-27T20:10:16.400 回答
11

红宝石,176 个字符

x=!0;y=0;e="^v<>#x";b=readlines;b.map{|l|(x||=l=~/[v^<>]/)||y+=1};c=e.index(b[y][x])
loop{c<2&&y+=c*2-1;c>1&&x+=2*c-5;e.index(n=b[y][x])&&(p n==?x;exit);c^='  \/'.index(n)||0}

我使用了一个简单的状态机(就像大多数海报一样),没什么特别的。我只是不断地使用我能想到的每一个技巧来减少它。用于改变方向的按位异或(存储为变量中的整数c)比我在早期版本中的条件有了很大的改进。

我怀疑代码会增加x并且y可以缩短。这是执行递增的代码部分:

c<2&&y+=c*2-1;c>1&&x+=(c-2)*2-1

编辑:我能够稍微缩短以上内容:

c<2&&y+=c*2-1;c>1&&x+=2*c-5

激光的当前方向c存储如下:

0 => 向上
1 => 向下
2 => 左
3 => 对

代码依赖于这个事实来增加x正确y的数量(0、1 或 -1)。我尝试重新排列映射到每个方向的数字,寻找一种可以让我做一些按位操作来增加值的排列,因为我有一种唠叨的感觉,它会比算术版本更短。

于 2009-09-28T05:58:22.660 回答
9

C# 3.0

259 个字符

bool S(char[]m){var w=Array.FindIndex(m,x=>x<11)+1;var s=Array.FindIndex(m,x=>x>50&x!=92&x<119);var t=m[s];var d=t<61?-1:t<63?1:t<95?-w:w;var u=0;while(0<1){s+=d;u=m[s];if(u>119)return 0<1;if(u==47|u==92)d+=d>0?-w-1:w+1;else if(u!=32)return 0>1;d=u>47?-d:d;}}

更具可读性:

bool Simulate(char[] m)
{
    var w = Array.FindIndex(m, x => x < 11) + 1;
    var s = Array.FindIndex(m, x => x > 50 & x != 92 & x < 119);
    var t = m[s];
    var d = t < 61 ? -1 : t < 63 ? 1 : t < 95 ? -w : w;
    var u = 0;
    while (0 < 1)
    {
        s += d;
        u = m[s];
        if (u > 119)
            return 0 < 1;
        if (u == 47 | u == 92)
            d += d > 0 ? -w - 1 : w + 1;
        else if (u != 32)
            return 0 > 1;
        d = u > 47 ? -d : d;
    }
}

字符的主要浪费似乎是在寻找地图的宽度和激光源的位置。任何想法如何缩短这个?

于 2009-09-26T13:30:29.287 回答
9

C + ASCII,197 个字符:

G[999],*p=G,w,z,t,*b;main(){for(;(*p++=t=getchar()^32)>=0;w=w|t-42?w:p-G)z=t^86?t^126?t^28?t^30?z:55:68:56:75,b=z?b:p;for(;t=z^55?z^68?z^56?z^75?0:w:-w:-1:1;z^=*b)b+=t;puts(*b^88?"false":"true");}

这个 C 解决方案假定一个 ASCII 字符集,允许我们使用 XOR 镜像技巧。它也非常脆弱——例如,所有输入行的长度必须相同。

它突破了 200 个字符的标记——但是该死的,仍然没有击败那些 Perl 解决方案!

于 2009-09-29T09:48:38.210 回答
9

Golfscript(83 个字符)

你好,咬人!

:\'><v^'.{\?}%{)}?:P@=?{:O[1-1\10?).~)]=P+
:P\=' \/x'?[O.2^.1^'true''false']=.4/!}do
于 2009-11-09T02:15:22.237 回答
9

蟒蛇 - 152

从名为“L”的文件中读取输入

A=open("L").read()
W=A.find('\n')+1
D=P=-1
while P<0:D+=1;P=A.find(">^<v"[D])
while D<4:P+=[1,-W,-1,W][D];D=[D,D^3,D^1,4,5][' \/x'.find(A[P])]
print D<5

要从标准输入中读取,请将第一行替换为

import os;A=os.read(0,1e9)

如果您需要小写真/假,请将最后一行更改为

print`D<5`.lower()
于 2009-11-09T02:30:29.360 回答
7

JavaScript - 265 个字符

更新 IV - 很可能这将是最后一轮更新,通过切换到 do-while 循环并重写运动方程来设法保存更多字符。

更新 III - 感谢 strager 关于删除 Math.abs() 并将变量放在全局名称空间中的建议,再加上变量分配的一些重新排列,使代码减少到 282 个字符。

更新 II - 对代码进行了一些更新以删除使用 != -1 以及更好地使用变量以进行更长的操作。

更新- 通过创建对 indexOf 函数的引用并进行一些更改(感谢 LiraNuna!)并删除不需要的括号。

这是我第一次打代码高尔夫,所以我不确定这能有多好,感谢任何反馈。

完全最小化的版本:

a;b;c;d;e;function f(g){a=function(a){return g.indexOf(a)};b=a("\n")+1;a=g[c=e=a("v")>0?e:e=a("^")>0?e:e=a("<")>0?e:a(">")];d=a=="<"?-1:a==">"?1:a=="^"?-b:b;do{e=d==-1|d==1;a=g[c+=d=a=="\\"?e?b*d:d>0?1:-1:a=="/"?e?-b*d:d>0?1:-1:d];e=a=="x"}while(a!="#"^e);return e}

带评论的原始版本:

character; length; loc; movement; temp;
function checkMaze(maze) {
        // Use a shorter indexOf function
        character = function(string) { return maze.indexOf(string); }
        // Get the length of the maze
        length = character("\n") + 1;
        // Get the location of the laser in the string
        character = maze[loc = temp = character("v") > 0 ? temp :
                               temp = character("^") > 0 ? temp :
                               temp = character("<") > 0 ? temp : character(">")];
        // Get the intial direction that we should travel
        movement = character == "<" ? -1 :
                   character == ">" ? 1 :
                   character == "^" ? -length : length;
        // Move along until we reach the end
        do {
            // Get the current character
            temp = movement == -1 | movement == 1;
            character = maze[loc += movement = character == "\\" ? temp ? length * movement : movement > 0 ? 1 : -1 :
                                               character == "/" ? temp ? -length * movement : movement > 0 ? 1 : -1 : movement];                                   
            // Have we hit a target?
            temp = character == "x";
            // Have we hit a wall?
        } while (character != "#" ^ temp);
        // temp will be false if we hit the target
        return temp;
    }

要测试的网页:

<html>
  <head>
    <title>Code Golf - Lasers</title>
    <script type="text/javascript">
    a;b;c;d;e;function f(g){a=function(a){return g.indexOf(a)};b=a("\n")+1;a=g[c=e=a("v")>0?e:e=a("^")>0?e:e=a("<")>0?e:a(">")];d=a=="<"?-1:a==">"?1:a=="^"?-b:b;do{e=d==-1|d==1;a=g[c+=d=a=="\\"?e?b*d:d>0?1:-1:a=="/"?e?-b*d:d>0?1:-1:d];e=a=="x"}while(a!="#"^e);return e}
    </script>
  </head>
  <body>
    <textarea id="maze" rows="10" cols="10"></textarea>
    <button id="checkMaze" onclick="alert(f(document.getElementById('maze').value))">Maze</button>
  </body>
</html>
于 2009-09-28T19:50:16.640 回答
6

c (K&R) 根据 strager 的更多建议,339 个必要字符。

我的物理学家注意到传播和反射操作是时间反转不变的,所以这个版本从目标发射光线并检查是否到达激光发射器。

实现的其余部分非常直接,并且或多或少完全来自我之前的向前努力。

压缩:

#define R return
#define C case
#define Z x,y
int c,i,j,m[99][99],Z;s(d,e,Z){for(;;)switch(m[x+=d][y+=e]){C'^':R 1==e;
C'>':R-1==d;C'v':R-1==e;C'<':R 1==d;C'#':C'x':R 0;C 92:e=-e;d=-d;C'/':c=d;
d=-e;e=-c;}}main(){while((c=getchar())>0)c==10?i=0,j++:(c==120?x=i,y=j:
i,m[i++][j]=c);puts(s(1,0,Z)|s(0,1,Z)|s(-1,0,Z)|s(0,-1,Z)?"true":"false");}

未压缩(ish):

#define R return
#define C case
#define Z x,y
int c,i,j,m[99][99],Z;
s(d,e,Z)
{
  for(;;)
    switch(m[x+=d][y+=e]){
    C'^': 
      R 1==e;
    C'>': 
      R-1==d;
    C'v': 
      R-1==e;
    C'<': 
      R 1==d;
    C'#':
    C'x':
      R 0;
    C 92:
      e=-e;
      d=-d;
    C'/':
      c=d;
      d=-e;
      e=-c;
    }
}
main(){
  while((c=getchar())>0)
    c==10?i=0,j++:
      (c==120?x=i,y=j:i,m[i++][j]=c);
  puts(s(1,0,Z)|s(0,1,Z)|s(-1,0,Z)|s(0,-1,Z)?"true":"false");
}

没有输入验证,错误的输入会将其发送到无限循环中。输入不大于 99 x 99 时正常工作。需要一个编译器,该编译器将链接标准库而不包括任何头文件。而且我想我已经完成了,即使在他的帮助下, strager 也让我领先了很多。

我宁愿希望有人会展示一种更微妙的方式来完成这项任务。这并没有错,但它不是深奥的魔法。

于 2009-09-26T01:53:48.203 回答
6

镜屋

不是挑战的实际入口,但我基于这个概念编写了一个游戏(不久前)。

它是用 Scala 编写的,开源的,可在此处获得:

它做得更多;处理颜色和各种类型的镜子和设备,但版本 0.00001 正是这个挑战所要求的。不过我已经丢失了那个版本,而且它从来没有针对字符数进行优化。

于 2009-09-26T07:28:54.300 回答
6

红宝石 - 146 个字符

A=$<.read
W=A.index('
')+1
until
q=A.index(">^<v"[d=d ?d+1:0])
end
while d<4
d=[d,d^3,d^1,4,5][(' \/x'.index(A[q+=[1,-W,-1,W][d]])or 4)]
end
p 5>d
于 2009-11-09T05:02:56.230 回答
5

PostScript , 359 字节

第一次尝试,还有很大的改进空间...

/a[{(%stdin)(r)file 99 string readline not{exit}if}loop]def a{{[(^)(>)(<)(v)]{2
copy search{stop}if pop pop}forall}forall}stopped/r count 7 sub def pop
length/c exch def[(>)0(^)1(<)2(v)3>>exch get/d exch def{/r r[0 -1 0 1]d get
add def/c c[1 0 -1 0]d get add def[32 0 47 1 92 3>>a r get c get .knownget
not{exit}if/d exch d xor def}loop a r get c get 120 eq =
于 2009-11-10T23:20:12.187 回答
4

Haskell,395 391 383 361 339 个字符(优化)

仍然使用通用状态机,而不是任何聪明的东西:

k="<>^v"
o(Just x)=x
s y(h:t)=case b of{[]->s(y+1)t;(c:_)->(c,length a,y)}where(a,b)=break(flip elem k)h
r a = f$s 0 a where f(c,x,y)=case i(a!!v!!u)"x /\\"["true",g k,g"v^><",g"^v<>"]of{Just r->r;_->"false"}where{i x y=lookup x.zip y;j=o.i c k;u=j[x-1,x+1,x,x];v=j[y,y,y-1,y+1];g t=f(j t,u,v)}
main=do{z<-getContents;putStrLn$r$lines z}

可读版本:

k="<>^v"    -- "key" for direction
o(Just x)=x -- "only" handle successful search
s y(h:t)=case b of  -- find "start" state
  []->s(y+1)t
  (c:_)->(c,length a,y)
 where (a,b)=break(flip elem k)h
r a = f$s 0 a where -- "run" the state machine (iterate with f)
 f(c,x,y)=case i(a!!v!!u)"x /\\"["true",g k,g"v^><",g"^v<>"] of
   Just r->r
   _->"false"
  where
   i x y=lookup x.zip y -- "index" with x using y as key
   j=o.i c k -- use c as index k as key; assume success
   u=j[x-1,x+1,x,x] -- new x coord
   v=j[y,y,y-1,y+1] -- new y coord
   g t=f(j t,u,v) -- recurse; use t for new direction
main=do
 z<-getContents
 putStrLn$r$lines z
于 2009-09-26T21:58:32.823 回答
3

我相信代码重用,我会使用您的代码之一作为 API :)。

  将 Board.new.validate(input)

32 个字符 \o/... 哇哦

于 2009-09-26T09:10:40.347 回答
3

C++:388 个字符

#include<iostream>
#include<string>
#include<deque>
#include<cstring>
#define w v[y][x]
using namespace std;size_t y,x,*z[]={&y,&x};int main(){string p="^v<>",s;deque<string>v;
while(getline(cin,s))v.push_back(s);while(x=v[++y].find_first_of(p),!(x+1));int 
i=p.find(w),d=i%2*2-1,r=i/2;do while(*z[r]+=d,w=='/'?d=-d,0:w==' ');while(r=!r,
!strchr("#x<^v>",w));cout<<(w=='x'?"true":"false");}

318没有标题)


这个怎么运作:

首先,读取所有行,然后找到激光。以下将评估0为只要还没有找到激光箭头,同时分配到x水平位置。

x=v[++y].find_first_of(p),!(x+1)

然后我们查看找到的方向并将其存储在i. 的偶数值i是顶部/左侧(“减少”),奇数值是底部/右侧(“增加”)。根据该概念,设置了d("direction") 和r("orientation")。我们用方向索引指针数组z,并将方向添加到我们得到的整数上。只有当我们击中斜线时方向才会改变,而当我们击中反斜线时方向保持不变。当然,当我们碰到镜子时,我们总是会改变方向(r = !r)。

于 2009-09-27T23:10:38.907 回答
2

C#

1020 个字符。
1088 个字符(从控制台添加输入)。
925 个字符(重构变量)。
875 个字符(删除了多余的 Dictionary 初始化程序;更改为 Binary & 运算符)

发帖前请注意不要看其他人的。我敢肯定它可能是LINQ了一点。可读版本中的整个 FindLaser 方法对我来说似乎非常可疑。但是,它有效,但已经晚了:)

请注意,可读类包括一个附加方法,该方法在激光移动时打印出当前 Arena。

class L{static void Main(){
A=new Dictionary<Point,string>();
var l=Console.ReadLine();int y=0;
while(l!=""){var a=l.ToCharArray();
for(int x=0;x<a.Count();x++)
A.Add(new Point(x,y),l[x].ToString());
y++;l=Console.ReadLine();}new L();}
static Dictionary<Point,string>A;Point P,O,N,S,W,E;
public L(){N=S=W=E=new Point(0,-1);S.Offset(0,2);
W.Offset(-1,1);E.Offset(1,1);D();
Console.WriteLine(F());}bool F(){
var l=A[P];int m=O.X,n=O.Y,o=P.X,p=P.Y;
bool x=o==m,y=p==n,a=x&p<n,b=x&p>n,c=y&o>m,d=y&o<m;
if(l=="\\"){if(a)T(W);if(b)T(E);if(c)T(S);
if(d)T(N);if(F())return true;}
if(l=="/"){if(a)T(E);if(b)T(W);if(c)T(N);
if(d)T(S);if(F())return true;}return l=="x";}
void T(Point p){O=P;do P.Offset(p);
while(!("\\,/,#,x".Split(',')).Contains(A[P]));}
void D(){P=A.Where(x=>("^,v,>,<".Split(',')).
Contains(x.Value)).First().Key;var c=A[P];
if(c=="^")T(N);if(c=="v")T(S);if(c=="<")T(W);
if(c==">")T(E);}}

可读版本(不是最终的高尔夫版本,但前提相同):

class Laser
{
    private Dictionary<Point, string> Arena;
    private readonly List<string> LaserChars;
    private readonly List<string> OtherChars;

    private Point Position;
    private Point OldPosition;
    private readonly Point North;
    private readonly Point South;
    private readonly Point West;
    private readonly Point East;

    public Laser( List<string> arena )
    {
        SplitArena( arena );
        LaserChars = new List<string> { "^", "v", ">", "<" };
        OtherChars = new List<string> { "\\", "/", "#", "x" };
        North = new Point( 0, -1 );
        South = new Point( 0, 1 );
        West = new Point( -1, 0 );
        East = new Point( 1, 0 );
        FindLaser();
        Console.WriteLine( FindTarget() );
    }

    private void SplitArena( List<string> arena )
    {
        Arena = new Dictionary<Point, string>();
        int y = 0;
        foreach( string str in arena )
        {
            var line = str.ToCharArray();
            for( int x = 0; x < line.Count(); x++ )
            {
                Arena.Add( new Point( x, y ), line[x].ToString() );
            }
            y++;
        }
    }

    private void DrawArena()
    {
        Console.Clear();
        var d = new Dictionary<Point, string>( Arena );

        d[Position] = "*";
        foreach( KeyValuePair<Point, string> p in d )
        {
            if( p.Key.X == 0 )
                Console.WriteLine();

            Console.Write( p.Value );
        }
        System.Threading.Thread.Sleep( 400 );
    }

    private bool FindTarget()
    {
        DrawArena();

        string chr = Arena[Position];

        switch( chr )
        {
            case "\\":
                if( ( Position.X == Position.X ) && ( Position.Y < OldPosition.Y ) )
                {
                    OffSet( West );
                }
                else if( ( Position.X == Position.X ) && ( Position.Y > OldPosition.Y ) )
                {
                    OffSet( East );
                }
                else if( ( Position.Y == Position.Y ) && ( Position.X > OldPosition.X ) )
                {
                    OffSet( South );
                }
                else
                {
                    OffSet( North );
                }
                if( FindTarget() )
                {
                    return true;
                }
                break;
            case "/":
                if( ( Position.X == Position.X ) && ( Position.Y < OldPosition.Y ) )
                {
                    OffSet( East );
                }
                else if( ( Position.X == Position.X ) && ( Position.Y > OldPosition.Y ) )
                {
                    OffSet( West );
                }
                else if( ( Position.Y == Position.Y ) && ( Position.X > OldPosition.X ) )
                {
                    OffSet( North );
                }
                else
                {
                    OffSet( South );
                }
                if( FindTarget() )
                {
                    return true;
                }
                break;
            case "x":
                return true;
            case "#":
                return false;
        }
        return false;
    }

    private void OffSet( Point p )
    {
        OldPosition = Position;
        do
        {
            Position.Offset( p );
        } while( !OtherChars.Contains( Arena[Position] ) );
    }

    private void FindLaser()
    {
        Position = Arena.Where( x => LaserChars.Contains( x.Value ) ).First().Key;

        switch( Arena[Position] )
        {
            case "^":
                OffSet( North );
                break;
            case "v":
                OffSet( South );
                break;
            case "<":
                OffSet( West );
                break;
            case ">":
                OffSet( East );
                break;
        }
    }
}
于 2009-09-26T07:48:27.380 回答
2

Groovy @ 279 个字符

m=/[<>^v]/
i={'><v^'.indexOf(it)}
n=['<':{y--},'>':{y++},'^':{x--},'v':{x++}]
a=['x':{1},'\\':{'v^><'[i(d)]},'/':{'^v<>'[i(d)]},'#':{},' ':{d}]
b=[]
System.in.eachLine {b<<it.inject([]) {r,c->if(c==~m){x=b.size;y=r.size;d=c};r<<c}}
while(d==~m){n[d]();d=a[b[x][y]]()}
println !!d
于 2009-09-27T00:29:31.623 回答
0

Perl 219
我的 perl 版本有392 342 个字符长(我不得不处理光束击中激光的情况):
更新,感谢霍布斯提醒我tr//,现在是 250 个字符:
更新,删除min m//,更改while带来的两个循环一些储蓄;现在只需要一个空间。
L:it;goto L与 长度相同do{it;redo}):

@b=map{($y,$x,$s)=($a,$-[0],$&)if/[<>^v]/;$a++;[split//]}<>;L:$_=$s;$x++if/>/;
$x--if/</;$y++if/v/;$y--if/\^/;$_=$b[$y][$x];die"true\n"if/x/;die"false\n"if
/[<>^v#]/;$s=~tr/<>^v/^v<>/if/\\/;$s=~tr/<>^v/v^></if/\//;goto L

我刮了一些,但它几乎不能与其中的一些竞争,尽管晚了。
它看起来好一点:

#!/usr/bin/perl
@b = map {
    ($y, $x, $s) = ($a, $-[0], $&) if /[<>^v]/;
    $a++;
    [split//]
} <>;
L:
    $_ = $s;
    $x++ if />/;
    $x-- if /</;
    $y++ if /v/;
    $y-- if /\^/;
    $_ = $b[$y][$x];
    die "true\n"  if /x/;
    die "false\n" if /[<>^v#]/;
    $s =~ tr/<>^v/^v<>/ if /\\/;
    $s =~ tr/<>^v/v^></ if /\//;
goto L

@b好吧...老实说,如果您了解是每行中的字符数组,并且可以阅读简单的正则表达式和tr语句,这应该是不言自明的。

于 2009-09-26T02:54:53.650 回答
0

F# - 454(或附近)

游戏有点晚了,但忍不住发布我的 2d 尝试。

更新稍作修改。如果发射器被击中,现在会正确停止。扼杀了布赖恩对 IndexOfAny 的想法(真可惜那行太冗长了)。我实际上还没有设法弄清楚如何让 ReadToEnd 从控制台返回,所以我相信这一点......

我对这个答案很满意,虽然它很短,但它仍然具有相当的可读性。

let s=System.Console.In.ReadToEnd()       //(Not sure how to get this to work!)
let w=s.IndexOf('\n')+1                   //width
let h=(s.Length+1)/w                      //height
//wodge into a 2d array
let a=Microsoft.FSharp.Collections.Array2D.init h (w-1)(fun y x -> s.[y*w+x])
let p=s.IndexOfAny[|'^';'<';'>';'v'|]     //get start pos
let (dx,dy)=                              //get initial direction
 match "^<>v".IndexOf(s.[p]) with
 |0->(0,-1)
 |1->(-1,0)
 |2->(1,0)
 |_->(0,1)
let mutable(x,y)=(p%w,p/w)                //translate into x,y coords
let rec f(dx,dy)=
 x<-x+dx;y<-y+dy                          //mutate coords on each call
 match a.[y,x] with
 |' '->f(dx,dy)                           //keep going same direction
 |'/'->f(-dy,-dx)                         //switch dx/dy and change sign
 |'\\'->f(dy,dx)                          //switch dx/dy and keep sign
 |'x'->"true"
 |_->"false"
System.Console.Write(f(dx,dy))
于 2010-01-19T13:30:41.400 回答