GolfScript-254字符

n%([]:B:$:_"()"@*{:I"()*+|?"[{}/]?[{[[0B$,+:B))\;)]_]+}{B)):ß;:B;qß(:ß;}{8q}{[[0ß0$,)]]+}:8{[[0B-1=:ß)]]+:$q}{ß>$ß<\([0+$,+]\++}:q{[[I$,:ß)]]+}]=~:$}/;{n+[0]:3\{:c;;3:1_:3;{,}{)[$=]_*2/{~\.{c={3|:3}*;}{;.1|1,\:1,<{+0}*;}if}/}/;}/;1$,?)"true""false"if n}%

有点直截了当,第一个循环将regex转换成NFA,第二个循环运行NFA。

Sun Aug 22 00:58:24 EST 2010 (271266)
Sun Aug 22 01:07:11 EST 2010 制造的 []
Sun Aug 22 07:05:50 EST 2010 (265259) 内联进行空状态转换
Sun Aug 22 07:19:21 EST 2010 (259256) 最终状态隐式化
Mon Feb 7 19:24:19 EST 2011 使用 "()""str"*

$ echo "ab*a aa abba abab b"|tr " " "\n"|golfscript regex.gs
true
true
false
false

$ echo "0*1|10 1 10 0110 00001"|tr " " "\n"|golfscript regex.gs
true
true
false
true

$ echo "0*(1|1+0) 1 10 0110 00001"|tr " " "\n"|golfscript regex.gs
true
true
true
true

$ echo "a?b+|(a+b|b+a?)+ abb babab aaa aabba a b"|tr " " "\n"|golfscript regex.gs
true
true
false
true
false
true

$ echo "((A|B|C)+(a|(bbbbb)|bb|c)+)+ ABCABCaccabbbbbaACBbbb ABCABCaccabbbbbaACBbbbb"|tr " " "\n"|golfscript regex.gs
false
true

C(口译), 630 622 617 615 582 576 573 572 554 544 529 502 500 499个字符

#define C char
#define M m(s,o
m(C*s,C*o,C*S,C*p,C*P,C T){C*n=P-1,*q=s,h=*P==41,c=1;for(;h*c;c-=*n--==40)c+=*n==41;
c=*P-42;c=p-P?c-82?T&&c&~1&&c-21?h?2:*S==*P&s<S?M,S-1,p,n,2)||(T&4&&M,S-1,p,P,T|1)):
4:M,T?S:o,p,P-1,T?c&~1?3:7-c:0):T&&s==S||M,o,p,P-1,2):T&&s==S;if(c==2)for(c=4;q<=S;q
++)c|=m(q,S,S,n+h,P-h,2)?M,q,p,n,2)||q<S&T/4&&M,q,p,P,T&6):0;return
c&4?c&1?:T&1&&M,S,p,n,2)||M,o,p,n,0):c;}main(C*w,C**v){C*u;for(w=*++v;*++v;)puts(m(*v
-1,u,u=index(*v,0)-1,w-1,index(w,0)-1,2)?"true":"false");}

用-Wall编译时抱怨argc是char。会在非法模式下崩溃。

argv上的输入,输出 颠倒 正常顺序:

$ ./a.out ab*a aa abba abab b
true
true
false
false

$ ./a.out "0*1|10" 1 10 0110 00001
true
true
false
true

$ ./a.out "0*(1|1+0)" 1 10 0110 00001
true
true
true
true

$ ./a.out "a?b+|(a+b|b+a?)+" abb babab aaa aabba a b
true
true
false
true
false
true

$ ./a.out "((A|B|C)+(a|(bbbbb)|bb|c)+)+" ABCABCaccabbbbbaACBbbb
false
$ ./a.out "((A|B|C)+(a|(bbbbb)|bb|c)+)+" ABCABCaccabbbbbaACBbbbb
true

727 670个字符

这还没有完全达到最低限度,但我想试试看,首先解析regexp是否会对解释它产生影响。确实如此,因为成本更高,尽管解析和匹配都更容易编写/理解。

#define Q struct q
#define C char
#define R return
Q{Q*u,*n,*a;C c,m};Q*P(C*p,C*e){Q*r=calloc(99,1);C*n=p+1,c=1,w;if(p==e)R
r;if(*p==40){for(;c;)c+=(*n==40)-(*n++==41);r->u=P(p+1,n-1);}else
if(*p=='|'){r->a=P(p+1,e);R r;}else r->c=*p;if(n<e){if(*n==43)*n=42,r->n=P(p,e);else
w=*n==42|*n==63,r->n=P(n+w,e),r->m=w?*n:0;r->a=r->n->a;}R r;}M(Q*r,C*s,C*o,C*z){C*p,
e;e=r?r->m==63|r->m==42&&M(r->n,s,o,z)?:*s&&r->c==*s?M(r->m==42?r:r->n,s+1,o,z):2:s
==z;if(e-2)R e;for(p=s,e=0;!r->c&p<=z;p++)e|=M(r->u,s,s,p)&(r->m!=42|p>s)&&M(r->m==
42?r:r->n,p,p,z);R e||r->a&&M(r->a,o,o,z);}main(C
c,C**v){for(;--c>1;)puts(M(P(v[1],index(v[1],0)),v[c],v[c],index(v[c],0))?"true":"false");}

• 要匹配的字符 或 指向要匹配的子模式的结构的指针
• 指向下一个符号结构的指针(如果有)
• 一个\0的乘法器, * 或 ? (pat)+ 被解析为 (pat)(pat)* 马上匹配就容易多了

哈斯克尔:610 627

main=getLine>>=f.words
d=reverse
u=0<1
j=[]
f(r:s)=mapM_(print.any null.c(d$b$'(':r++")"))s
c%(x,y)=(c:x,y)
s _ _ _[]=(j,j)
s n a b (x:y)|n<1&&x==a=(j,x:y)|x==a=f(-1)|x==b=f 1|u=f 0where f k=x%s(n+k)a b y
b r|m==j=r|u=b$d(o++"(("++x)++")("++z++")/)"++w where(c,m)=s 0'|''!'r;_:v=m;(o,_:x)=s 0'('')'$d c;(z,_:w)=s 0')''('v
(!)g f x=f x>>=g
c[]=(:j)
c r=f!c s where(s,f)=i r
p q@(r:s)|r=='('=(s,(:j))|u=(a,f!g)where(w,f)=i q;(a,g)=p w
_?[]=j
c?(h:r)|c==h=[r]|u=j
i(r:q)=maybe(q,(r?))id$lookup r$zip")/*+?"$p q:zip[e,w,w,w][\s->f s++g s,\s->s:l s,l,\s->s:f s]where(w,f)=i q;(e,g)=i w;l s|f s==j=j|u=f s++(f s>>=l)

import Control.Monad
import Data.List

-- (aa|bb|cc)   -->  )|)cc()|)bb()aa((( 

testRegex = "a?b+|(a+b|b+a?)+"

interpret regex = any null . interpret' regex

interpret' regex = compile (rewrite regex)

mapFst f (x, y) = (f x, y)

splitWhileBalanced _ _ _ "" = ("", "")
splitWhileBalanced n b1 b2 (x:xs)
  | n < 1 && x == b1 = ("", x:xs)
  | x == b1   = f (-1)
  | x == b2   = f 1
  | otherwise = f 0
  where
    f k = mapFst (x:) $ splitWhileBalanced (n+k) b1 b2 xs

rewrite regex = reverse $ moveBars $ '(' : regex ++ ")"

moveBars regex
  | mtBar == "" = regex
  | otherwise = moveBars $ reverse (hOpen ++ "((" ++ tOpen) ++ ")(" ++ hClose ++ ")/)" ++ tClose
  where
    (hBar, mtBar) = splitWhileBalanced 0 '|' '!' regex -- '!' is a dummy character
    b:tBar = mtBar
    (hOpen, o:tOpen) = splitWhileBalanced 0 '(' ')' $ reverse hBar
    (hClose, c:tClose) = splitWhileBalanced 0 ')' '(' $ tBar

compile "" = \x -> [x]
compile rs = f <=< compile rs'
  where 
    (rs', f) = compile' rs

paren regex@(r:rs0)
  | r == '('  = (rs0, \x -> [x])
  | otherwise = (rs2, f <=< g)
  where
    (rs1, f) = compile' regex
    (rs2, g) = paren rs1

compile' (r:rs0) = case r of
  '/' -> (rs2, bar)
  '*' -> (rs1, star)
  '+' -> (rs1, plus)
  '?' -> (rs1, mark)
  ')' -> paren rs0
  _   -> (rs0, lit)
  where
    (rs1, f) = compile' rs0
    (rs2, g) = compile' rs1
    bar str = f str ++ g str
    plus str
      | null (f str) = []
      | otherwise = f str ++ (f str >>= plus)
    star str = str : plus str
    mark str = str : f str
    lit = maybe [] (\x -> [x]) . stripPrefix [r]

| 然后反转整个正则表达式。现在操作符是前缀形式的,因此很容易解析。程序像这样解析正则表达式。列表单子用于不确定性。

用法:

> runghc Regex.hs
a?b+|(a+b|b+a?)+ abb babab aaa aabba a b
True
True
False
True
False
True

R=gets.chop;s='';k=[];n=a=0
G={?(=>(A="(a-=1;s<<0)if a>1;")+"k<<[n,a];n=a=0",
Y=?|=>(B="s<<0while 0<a-=1;")+"n+=1",
?)=>B+(C="s<<?|while 0<=n-=1;")+"n,a=k.pop"+F=";a+=1",
?*=>D="s<<c",?+=>D,??=>D}
R.each_char{|c|eval G[c]||A+D+F};eval B+C
def P l,s;l.map{|a|a<<s};end
J={??=>(K="a=k.pop;")+"k<<[{Y=>n=[a[0]]},a[1]<<n]",
?*=>K+(H="P a[1],s={Y=>n=[a[0]]};")+"k<<[s,[n]]",
?+=>K+H+"k<<[a[0],[n]]",
Y=>(I=K+"b=k.pop;")+"k<<[{Y=>[a[0],b[0]]},a[1]+b[1]]",
?\0=>I+"P b[1],a[0];k<<[b[0],a[1]]"}
k=[];s.each_char{|c|eval J[c]||"k<<[{c=>a=[]},[a]]"}
e=k[0];P e[1],R;
p $<.map{|l|s=l.chop;*a=e[0]
s.each_char{|c|eval@f="n=a;a=a.map{|h|h[Y]||[h]}.flatten"while a!=n
a=a.inject([]){|a,h|a+(h[c]||[])}}
eval@f;a.include? R}

(在Ruby1.8中,通过添加下面的别名,还可以使用45个以上的字符)

测试 type testcase.txt | ruby regex.rb

Russ Cox's NFA parser 红宝石色。一个状态被表示为一个散列,其中一个键包含一个下一个状态的数组。

无嗅:

#needed to run on ruby 1.8
class String
  alias :each_char :each_byte 
end  

## Infix to Postfix

R=gets.chop  
p "regexp = #{R}"
k=[] 
s=''  #will hold postfix string
n=a=0 #count braNches and Atoms
postfix = R.each_char{|c|
  case c
    when ?(
        (a-=1;s<<0)if a>1
        k<<[n,a]
        n=a=0
    when ?|
      s<<0while 0<a-=1
      n+=1
    when ?)
      s<<0while 0<a-=1
      s<<?|while 0<=n-=1
      n,a=k.pop;a+=1
    when ?*,?+,??
      s<< c
    else
      (a-=1;s<<0)if a>1
      s<< c
      a+=1
  end
}
s<<0while 0<a-=1
s<<?|while 0<=n-=1

## Postfix to NFA
# State is {char=>s0=[state,state]]
# Frag is [state, [s0,...]]

Y=?|   #choice indicator
X={:true=>[R]} #matcstate

 def patch l,s   #l is list of arrays, s is state.  Put s in each array
   l.map{|a|   a<< s   }
end

k=[]
s.each_char{|c|
 case c
    when ??
      a=k.pop
      s={Y=>n=[a[0]]}
      k<<[s,a[1]<<n]
    when ?*
      a=k.pop
      s={Y=>n=[a[0]]}
      patch(a[1],s)
      k<<[s,[n]]
    when ?+
      a=k.pop
      s={Y=>n=[a[0]]}
      patch(a[1],s)
      k<<[a[0],[n]]
    when ?|
     b=k.pop
     a=k.pop
      s={Y=>[a[0],b[0]]}
      k<<[s,a[1]+b[1]]
    when 0
     b=k.pop
     a=k.pop
     patch(a[1],b[0])
     k<<[a[0],b[1]]
   else
     k<<[{c=>a=[]},[a]]
   end
}
e=k.pop
patch(e[1],X)

#Running the NFA

E=[e[0]] #Evaluator
p $<.map{|l|s=l.chop
  p "evaluating: #{s}" 
  a=E
  s.each_char{|c|
    begin     #skip past split nodes
      s=a.size
      a=a.map{|h|h[?|]||[h]}.flatten  
    end while a.size!=s
    a=a.inject([]){|a,h|
    a+(h[c]||[])}  #add next state or null
  }
  a=a.map{|h|h[?|]||[h]}.flatten
  r = a.include? X  #check for end of pattern
  p r
  r
}

半解释:

• 这是基于在第8章中发现的延续传递风格的解析器组合 高阶Perl .
• 帮助我移除大约70个字符的功劳归于Vincent Pit(VPIT)。
• S { block } sub { block }
• $, 是nil(一个coderef,包含始终匹配的匹配器,不消耗任何内容)
• c 是n元串联(取任意数量的匹配器,如果它们都按顺序成功,则返回一个成功的匹配器)。
• a 是n元交替(取任意数量的匹配器,如果其中任何一个匹配器成功,则返回一个成功的匹配器)。
• A 是regex生成器的助手,它采用连接和传递的交替结构 C 和 一 必要时,返回匹配器。
• k k s() 解析为 s///
• 这个 do @$r 是当前连接列表, @a 是当前替换列表, @p a+ 被视为 aa* ,和 a? 被视为 (a|) 内联输入 b (没有用于 plus 或 maybe
• 这个 S{!length pop}

有关大部分已除雾且注释较多的代码,请参阅 this Gist .

perl regexer.pl 'a?b+|(a+b|b+a?)+' abb babab aaa aabba a b . 代码中没有强制换行符。

use feature':5.12';
sub S(&){pop}
$,=S{goto pop};
sub p{push@{+shift},@_}
sub c{my$l=$,;for my$r(@_){my$L=$l;
$l=S{my($i,$c)=@_;&$L($i,S{&$r(shift,$c)})}}$l}
sub a{my@A=@_;S{my($i,$c,$o)=@_;$o=&$_($i,$c)and return$o for@A;0}}
sub A{$#_?a(map c(@$_),@_):c(@{+pop})}
sub k{my($I,$k)=@_;$k=a c($I,S{&$k}),$,}
$_=shift;$P=do{@a=$r=[];for(/./g){
when('('){p\@p,[@a];@a=$r=[]}
when(')'){$p=A@a;@a=@{pop@p};p$r=$a[-1],$p}
p\@a,$r=[]when'|';
p$r,k pop@$r when'*';
p$r,c $$r[-1],k pop@$r when'+';
p$r,a pop@$r,$,when '?';
my$s=$_;p$r,S{my($_,$c)=@_;s/^\Q$s//&&$_->$c}}A@a};
say&$P($_,S{!length pop})?"true":"false"for@ARGV

// All whitespace is optional
function c(f,p){
    var x=f[0],w=p[0],h="substr",s=f[h](2),b=p[h](1),m=0,t=0,r,g,a=0,l,u="length",y="*";
    switch(f[1]){
        case"+":if(x!=w)return;f=x+y+s;
        case y:return x==w&&c(f,b)||c(s,p);
        case"?":return x==w&&c(s,b)||c(s,p)
    }
    if(x=="("){
        o:for(l=[""];t<f[u];t++){
            switch(f[t]){
                case"|":if(a==1){m=l.push("")-1;continue}break;
                case"(":if(++a==1)continue;break;
                case")":if(!--a)break o
            }
            l[m]+=f[t]
        }
        var v=0,e=t+1;
        return l.some(function(g){
            switch(f[t+1]){
                case y:v=1;
                case"+":e=t+2;
                    for(var q=0;q<f[u];q++)
                        if(c(g+Array(q).join(f[h](0,e))+f[h](e),p))
                            return 1;
                    return;
                case"?":v=1;e=t+2;default:if(c(g+f[h](e),p))return 1;
            }
        })||(v&&c(f[h](e),p))
    }
    return p[u]?(x==w&&c(f[h](1),b)):!f[u]
}

// Make it look nicer
function test(regex, string) { return !!c('(' + regex + ')', string); }

test('a?b+|(a+b|b+a?)+', 'abb') // true
test('a?b+|(a+b|b+a?)+', 'babab') // true

function test(reg, str) {
    console.log('Testing ' + reg + ' against ' + str);
    var c = reg[0], d = str[0];


    switch (reg[1]) {
        case '+': 
            if (c != d) 
                return false;   
            reg = c + '*' + reg.substr(2);
        case '*': 
            return (c == d && test(reg, str.substr(1))) || test(reg.substr(2), str);
        case '?': 
            return (c == d && test(reg.substr(2), str.substr(1))) || test(reg.substr(2), str);
    }

    if (c == '(') {
        var regs = [''];
        o: for (var level = n = i = 0; i < reg.length; i++) {
            //console.log(level + ': ' + n + ': ' + reg[i]);
            switch (reg[i]) {
                case '|': 
                    if (level == 1) { n = regs.push('') - 1; continue; } 
                    break;
                case '(': 
                    if (++level == 1) continue; 
                    break;
                case ')': 
                    if (!--level) break o; 
                    break;
            };
            regs[n] += reg[i];
        }
        //console.log(regs); // An array of alternates (hello|hi) => ['hello', 'hi']        

        var optional = false, end = i+1;
        return regs.some(function(jitem) {
            switch (reg[i+1]) {
                case '*': optional = true; // Fall through
                case '+': 
                    end = i+2;
                    for (var k = 0; k < reg.length; k++)
                        if (test(jitem + Array(k).join(reg.substr(0, i+2)) + reg.substr(i+2), str))
                            return true;
                    return false;

                case '?': optional = true; end = i+2; // Fall through
                default: if (test(jitem + reg.substr(end), str)) return true;       
            }

        }) || (optional && test(reg.substr(end), str));
    }

    if (str == '')
        return reg == '';

    return c == d ? test(reg.substr(1), str.substr(1)) : false;

}

test("hello", "hello") => test("ello", "ello") => test("llo", "llo") => test("lo", "lo") => test("o", "o") => test("", "") 返回true。注:裸露 c 函数不支持隐含的替换。换句话说, hello|hi 不起作用;必须用括号括起来: (hello|hi) .