Question

您将获得2个字符串列表 - A和B.找到匹配A中所有字符串的最短正则表达式而不是B中的所有字符串。请注意，此正则表达式可以匹配/不匹配不在A中的其他字符串，而不是B.为简单起见，我们可以假设我们的字母大小只有2个字符 - 0和1.也只允许这些操作符：

* - 0或更多
？ - 0或1
+ - 1或更多
（） - 括号

为简单起见，不允许使用正则表达式运算符。我不知道是否允许或运算符（|）会简化问题。 A和B的路线没有共同的元素。以下是一些例子：

A=[00,01,10]
B=[11]
answer = 1*0+1*

A=[00,01,11]
B=[10]
answer = 0*1*

Answer 1

解决这个问题的一种方法是使用遗传算法。我碰巧有一个genetic solver laying around所以我用以下算法将它应用于你的问题：

从所需的输入中获取不同的标记作为基因
将正则表达式特色添加到基因
用于健身算法
- 确保生成的字符串是有效的正则表达式
- 根据匹配的所需内容获得适合度值它匹配了多少不受欢迎的东西
直到找到成功的正则表达式
- 从不同令牌的数量开始并根据需要递增
- 尝试生成一个通过健身要求的长度的正则表达式

这是我在C＃中的实现

private static void GenerateRegex(IEnumerable<string> target, IEnumerable<string> dontMatch)
{
    string distinctSymbols = new String(target.SelectMany(x => x).Distinct().ToArray());
    string genes = distinctSymbols + "?*()+";

    Func<string, uint> calcFitness = str =>
        {
            if (str.Count(x => x == '(') != str.Count(x => x == ')'))
            {
                return Int32.MaxValue;
            }
            if ("?*+".Any(x => str[0] == x))
            {
                return Int32.MaxValue;
            }
            if ("?*+?*+".ToArray().Permute(2)
                .Any(permutation => str.IndexOf(new string(permutation.ToArray())) != -1))
            {
                return Int32.MaxValue;
            }
            Regex regex;
            try
            {
                regex = new Regex("^" + str + "$");
            }
            catch (Exception)
            {
                return Int32.MaxValue;
            }
            uint fitness = target.Aggregate<string, uint>(0, (current, t) => current + (regex.IsMatch(t) ? 0U : 1));
            uint nonFitness = dontMatch.Aggregate<string, uint>(0, (current, t) => current + (regex.IsMatch(t) ? 10U : 0));
            return fitness + nonFitness;
        };

    for (int targetGeneLength = distinctSymbols.Length; targetGeneLength < genes.Length * 2; targetGeneLength++)
    {
        string best = new GeneticSolver(50).GetBestGenetically(targetGeneLength, genes, calcFitness, true);
        if (calcFitness(best) != 0)
        {
            Console.WriteLine("-- not solved with regex of length " + targetGeneLength);
            continue;
        }
        Console.WriteLine("solved with: " + best);
        break;
    }
}

将其应用于样品的结果：

public void Given_Sample_A()
{
    var target = new[] { "00", "01", "10" };
    var dontMatch = new[] { "11" };

    GenerateRegex(target, dontMatch);
}

输出：

Generation  1 best: 10 (2)
Generation  2 best: 0+ (2)
Generation  5 best: 0* (2)
Generation  8 best: 00 (2)
Generation  9 best: 01 (2)
-- not solved with regex of length 2
Generation  1 best: 10* (2)
Generation  3 best: 00* (2)
Generation  4 best: 01+ (2)
Generation  6 best: 10+ (2)
Generation  9 best: 00? (2)
Generation 11 best: 00+ (2)
Generation 14 best: 0?1 (2)
Generation 21 best: 0*0 (2)
Generation 37 best: 1?0 (2)
Generation 43 best: 10? (2)
Generation 68 best: 01* (2)
Generation 78 best: 1*0 (2)
Generation 79 best: 0*1 (2)
Generation 84 best: 0?0 (2)
Generation 127 best: 01? (2)
Generation 142 best: 0+1 (2)
Generation 146 best: 0+0 (2)
Generation 171 best: 1+0 (2)
-- not solved with regex of length 3
Generation  1 best: 1*0+ (1)
Generation  2 best: 0+1* (1)
Generation 20 best: 1?0+ (1)
Generation 31 best: 1?0* (1)
-- not solved with regex of length 4
Generation  1 best: 1*00? (1)
Generation  2 best: 0*1?0 (1)
Generation  3 best: 1?0?0 (1)
Generation  4 best: 1?00? (1)
Generation  8 best: 1?00* (1)
Generation 12 best: 1*0?0 (1)
Generation 13 best: 1*00* (1)
Generation 41 best: 0*10* (1)
Generation 44 best: 1*0*0 (1)
-- not solved with regex of length 5
Generation  1 best: 0+(1)? (1)
Generation 36 best: 0+()1? (1)
Generation 39 best: 0+(1?) (1)
Generation 61 best: 1*0+1? (0)
solved with: 1*0+1?

第二个样本：

public void Given_Sample_B()
{
    var target = new[] { "00", "01", "11" };
    var dontMatch = new[] { "10" };

    GenerateRegex(target, dontMatch);
}

输出：

Generation  1 best: 00 (2)
Generation  2 best: 01 (2)
Generation  7 best: 0* (2)
Generation 12 best: 0+ (2)
Generation 33 best: 1+ (2)
Generation 36 best: 1* (2)
Generation 53 best: 11 (2)
-- not solved with regex of length 2
Generation  1 best: 00* (2)
Generation  2 best: 0+0 (2)
Generation  7 best: 0+1 (2)
Generation 12 best: 00? (2)
Generation 15 best: 01* (2)
Generation 16 best: 0*0 (2)
Generation 19 best: 01+ (2)
Generation 30 best: 0?0 (2)
Generation 32 best: 0*1 (2)
Generation 42 best: 11* (2)
Generation 43 best: 1+1 (2)
Generation 44 best: 00+ (2)
Generation 87 best: 01? (2)
Generation 96 best: 0?1 (2)
Generation 125 best: 11? (2)
Generation 126 best: 1?1 (2)
Generation 135 best: 11+ (2)
Generation 149 best: 1*1 (2)
-- not solved with regex of length 3
Generation  1 best: 0*1* (0)
solved with: 0*1*

Answer 2

如果这是一个家庭作业问题，那就像“一个家庭作业，在课堂上获得A”类型。我认为在那个问题的某个地方有“或”操作员遗失。

有一个明显的解决方案是A0 | A1 | A2 | ...，但在尝试找到最短的时候似乎更难解决。

我建议使用递归来尝试缩短正则表达式，但这不是一个理想的解决方案。

Answer 3

此项目从给定的单词列表生成正则表达式： https://github.com/bwagner/wordhierarchy

但是，它仅使用“|”，非捕获组“(?:)”和选项“?”。

样本用法：

java -jar dist/wordhierarchy.jar 00 01 10
-> 10|0(?:1|0)

java -jar dist/wordhierarchy.jar 00 01 11
-> 0(?:0|1)|11

java -jar dist/wordhierarchy.jar 000 001 010 011 100 101 110 111
-> 1(?:0(?:0|1)|1(?:0|1))|0(?:1(?:1|0)|0(?:1|0))

java -jar dist/wordhierarchy.jar 000 001 010     100 101 110 111
-> 1(?:0(?:0|1)|1(?:1|0))|0(?:10|0(?:1|0))

Answer 4

“如有疑问，请使用蛮力。”

import re

def learn(ayes, noes, max_size=7):
    def is_ok(rx):
        rx += '$'
        return (all(re.match(rx, s) for s in ayes)
                and not any(re.match(rx, s) for s in noes))
    return find(find(gen_sized(size), is_ok)
                for size in range(max_size + 1))

def find(xs, ok=lambda x: x):
    for x in xs:
        if ok(x):
            return x

def gen_sized(size):
    if 0 == size:
        yield ''
    if 0 < size:
        for rx in gen_sized(size-1):
            yield rx + '0'
            yield rx + '1'
            if rx and rx[-1] not in '*?+':
                yield rx + '*'
                yield rx + '?'
                yield rx + '+'
    if 5 < size:
        for rx in gen_sized(size-3):
            yield '(%s)*' % rx
            yield '(%s)?' % rx
            yield '(%s)+' % rx

这为第一个产生了不同但同样好的答案：0*1?0*。它着眼于1241个试验正则表来解决这两个测试用例（总计）。

搜索有一个大小限制 - 因为这个问题的一般正则表达式版本是NP-hard，所以它的任何程序都会在复杂足够的输入上遇到麻烦。我不会真的想到这个简化的问题。我很乐意看到一些不太明显的答案。

如何从给定的字符串列表中自动生成正则表达式？

4 个答案: