开发者

Regexercise: factorials

This is an experimental new feature for StackOverlow: exercising your regex muscles by solving various classical problems. There is no one right answer, and in fact we should collect as many right an开发者_如何学Goswers as possible, as long as they offer educational value. All flavors accepted, but please document it clearly. As much as practical, provide testcases/snippets to demonstrate that the pattern "works".

How can we find if a number x is a factorial using regex?

Bonus: if the pattern can determine that x = n!, can it also find n?


Java, with infinite length lookbehind and nested references (see also on ideone.com):

import java.util.regex.*;

class Factorial {
static String assertPrefix(String pattern) {
   return "(?<=(?=^pattern).*)".replace("pattern", pattern);
}
public static void main(String[] args) {
   final Pattern FACTORIAL = Pattern.compile(
      "(?x) (?: inc stepUp)+"
         .replace("inc", "(?=(^|\\1 .))")
         //                      1

         .replace("stepUp", "(?: ^. | (?<=(^.*)) (?=(.*)) (?: notThereYet \\2)+ exactlyThere )")
         //                                2          3

         .replace("notThereYet", "(?:  (?=((?=\\3) .  |  \\4 .)) (?=\\1(.*)) (?=\\4\\5)  )")
         //                                           4                  5

         .replace("exactlyThere", "measure4 measure1")
            .replace("measure4", assertPrefix("\\4(.*)"))
            .replace("measure1", assertPrefix("\\1\\6"))
   );

   for (int n = 0; n < 1000; n++) {
      Matcher m = FACTORIAL.matcher(new String(new char[n]));
      if (m.matches()) {
         System.out.printf("%3s = %s!%n", n, m.group(1).length() + 1);
      }
   }
}
}


With .NET balancing groups, in C# (see also on ideone.com):

var r = new Regex(@"(?xn) 

^(
   (
     ( ^.
     | (?=  (?<temp-n> .)+ )
       (?<= (?<fact>  .+)  )
       (?<n-temp> \k<fact> )+?
       (?(temp) (?!))
     )
     (?<n>)
   )+
 )$

");

for (int x = 0; x < 6000; x++) {
   Match m = r.Match("".PadLeft(x));
   if (m.Success) {
      Console.WriteLine("{0,4} = {1}! ", x, m.Groups["n"].Captures.Count);
   }
}

Note:

The version of .NET used by ideone.com seems to have a bug in the balancing groups that made the reluctant repetition +? necessary in the above snippet. In newer versions, a simple greedy + may suffice. See also: Backtracking a balancing group in a greedy repetition may cause imbalance?

0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜