生成正则表达式的所有有效值

Question

我知道通过使用Xeger，我们可以获得指定模式的随机值。

String regex = "[0-9]{2}"; 
Xeger generator = new Xeger(regex);
String result = generator.generate();

我想知道是否有一种方法可以返回指定正则表达式的所有有效字符串。例如，对于模式：[0-9]{2}，我们可以得到所有的值00来99。

谢谢

编辑：

在这里，我们不考虑+和*等无限输出；我们如何获得有限正则表达式的所有值？

最后编辑：

感谢大家！最后，我不会考虑所有可能的值，因为可能有数千个。我限制一个特定的数字作为减少数量的值的数量。

Answer 1

由于正则表达式是由有限状态机定义的，所以我想知道是否有某种东西可以在此类机器上自动推理，并且是否适合将其用于这项工作……并提供了clojure.core.logic

因此，我查看了正则表达式语法的定义（不幸的是，它缺少{}量词，但它们应该很容易添加到我的代码中）使其适应了Java转义，并得出了以下110行长的clojure程序：

(ns regexp-unfolder.core
  (:require [instaparse.core :as insta])
  (:require [clojure.core.logic :as l])
  (:require [clojure.set :refer [union difference]])
  (:gen-class :methods [#^{:static true} [unfold [String] clojure.lang.LazySeq]])
)

(def parse-regexp (insta/parser 
             "re = union | simple-re?
             union = re '|' simple-re
             simple-re = concat | base-re
             concat = simple-re base-re
             base-re = elementary-re | star | plus
             star = elementary-re '*'
             plus = elementary-re '+'
             elementary-re = group | char | '$' | any | set
             any = '.'
             group = '(' re ')'
             set = positive-set | negative-set
             positive-set = '['  set-items ']'
             negative-set = '[^' set-items ']'
             set-items = set-item*
             set-item = range | char
             range = char '-' char
             char = #'[^\\\\\\-\\[\\]]|\\.'" ))

(def printables (set (map char (range 32 127))))

(declare fns handle-first)

(defn handle-tree [q qto [ type & nodes]]
  (if (nil? nodes)
    [[q [""] qto]]
    ((fns type handle-first) q qto nodes)))

(defn star [q qto node &]
  (cons [q [""] qto]
         (handle-tree q q (first node))))

(defn plus [q qto node &] 
  (concat (handle-tree q qto (first node))
          (handle-tree qto qto (first node))))

(defn any-char [q qto & _] [[q (vec printables) qto]] )

(defn char-range [[c1 _ c2]]
  (let [extract-char (comp int first seq second)]
    (set (map char (range (extract-char c1) (inc (extract-char c2)))))))

(defn items [nodes]
  (union (mapcat
    (fn [[_ [type & ns]]]
      (if (= type :char)
        #{(first ns)}        
        (char-range ns)))
    (rest (second nodes)))))

(defn handle-set [q qto node &] [[q (vec (items node)) qto]])

(defn handle-negset [q qto node &] [[q (vec (difference printables (items node))) qto]])

(defn handle-range [q qto & nodes] [[q (vec (char-range nodes)) qto]])

(defn handle-char [q qto node &] [[q (vec node) qto]] )

(defn handle-concat [q qto nodes] 
  (let [syms (for [x  (rest nodes)] (gensym q))]
    (mapcat handle-tree  (cons q syms) (concat syms [qto] ) nodes)
  ))

(defn handle-first [q qto [node & _]] (handle-tree q qto node))

(def fns {:concat handle-concat, :star star, :plus plus, :any any-char, :positive-set handle-set, :negative-set handle-negset, :char handle-char})

(l/defne transition-membero
  [state trans newstate otransition]
  ([_ _ _ [state trans-set newstate]]
     (l/membero trans trans-set)))

(defn transitiono [state trans newstate transitions]
  (l/conde
   [(l/fresh [f] 
             (l/firsto transitions f)
             (transition-membero state trans newstate f))]
   [(l/fresh [r]
             (l/resto transitions r)
             (transitiono state trans newstate r))])
  )

(declare transitions)

;; Recognize a regexp finite state machine encoded in triplets [state, transition, next-state], adapted from a snippet made by Peteris Erins

(defn recognizeo
  ([input]
     (recognizeo 'q0 input))
  ([q input]
     (l/matche [input] ; start pattern matching on the input
        (['("")]
           (l/== q 'ok)) ; accept the empty string if we are in an accepting state
        ([[i . nput]]
           (l/fresh [qto]
                  (transitiono q i qto transitions) ; assert it must be what we transition to qto from q with input symbol i
                  (recognizeo qto nput)))))) ; recognize the remainder


(defn -unfold [regex] 
  (def transitions 
    (handle-tree 'q0 'ok (parse-regexp regex)))
  (map (partial apply str) (l/run* [q] (recognizeo q))))

用core.logic编写，使其适应正则表达式匹配器应该相当容易

我将可打印字符的范围限制在32到126个ascii之间，否则处理regexp这样的正则表达式会很麻烦[^c]，但是您可以很轻松地扩展它……而且，我还没有实现联合，可选模式和\
w，\ s等字符类的转义符

到目前为止，这是我在clojure中编写的最大的事情，但是基本知识似乎还不错。

regexp-unfolder.core=> (-unfold "ba[rz]")
("bar" "baz")
regexp-unfolder.core=> (-unfold "[a-z3-7]")
("a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "r" "s" "t" "u" "v" "w" "x" "y" "z" "3" "4" "5" "6" "7")
regexp-unfolder.core=> (-unfold "[a-z3-7][01]")
("a0" "a1" "b0" "b1" "c0" "c1" "d0" "d1" "e0" "e1" "f0" "f1" "g0" "g1" "h0" "h1" "i0" "i1" "j0" "j1" "k0" "k1" "l0" "l1" "m0" "m1" "n0" "n1" "o0" "o1" "p0" "p1" "q0" "q1" "r0" "r1" "s0" "s1" "t0" "t1" "u0" "u1" "v0" "v1" "w0" "w1" "x0" "x1" "y0" "y1" "z0" "z1" "30" "31" "40" "41" "50" "51" "60" "70" "61" "71")
regexp-unfolder.core=> (-unfold "[^A-z]")
(" " "@" "!" "\"" "#" "$" "%" "&" "'" "(" ")" "*" "+" "," "-" "." "/" "0" "1" "2" "3" "4" "5" "6" "7" "8" "9" ":" ";" "{" "<" "|" "=" "}" ">" "~" "?")
regexp-unfolder.core=> (take 20 (-unfold "[abc]*"))
("" "a" "b" "c" "aa" "ab" "ac" "ba" "ca" "aaa" "bb" "cb" "aab" "bc" "cc" "aac" "aba" "aca" "baa" "caa")
regexp-unfolder.core=> (take 20 (-unfold "a+b+"))
("ab" "aab" "abb" "abbb" "aaab" "abbbb" "aabb" "abbbbb" "abbbbbb" "aabbb" "abbbbbbb" "abbbbbbbb" "aaaab" "aabbbb" "aaabb" "abbbbbbbbb" "abbbbbbbbbb" "aabbbbb" "abbbbbbbbbbb" "abbbbbbbbbbbb")

自从我开始这种方式以来，我还实现了无限的输出:)

如果有人感兴趣，我在这里上传了

显然，这是一个如何unfold从普通的旧Java 调用的示例：

import static regexp_unfolder.core.unfold;

public class UnfolderExample{
    public static void main(String[] args){
        @SuppressWarnings("unchecked")
        Iterable<String> strings = unfold("a+b+");
        for (String s : strings){
            System.out.println(s);
        }
    }
}