偶数个 0 或奇数个 1 的二进制数的最短正则表达式

Question

Write an expression that contains an even number of 0s or an odd number of 1s

I got it down to:

1*(01*01*)* + 0*10*(10*10*)*

where the first part represents an even number of 0s and the second part an odd number of 1s

However, there's supposed to be a simplified solution that I'm not seeing. Any tips?

Answer 1

Odd-1s part: 0*1(0|10*1)*

Even-0s part, depends:

1. Empty string is correct: (1|01*0)*
2. No-0s is even-0s: (1|01*0)+
3. Must have at least two 0s: 1*(01*01*)+ (as in OP)

old answer: correct under case 1 and 2

(1*(01*0)*)+ | 0*1(0*(10*1)*)*

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Kudos to @OGHaza for helpful comments.

Answer 2

利用偶数长度字符串始终满足您的约束这一事实：

^(([01]{2})*|1*(01*01*)*)$

Answer 3

Define "shortest". If you're looking for the shortest evaluation time (ie fastest) then make sure you do not use capture groups.

here's an example in javascript

^(?:1*(?:01*0)*)+|0*1(?:0*(?:10*1)*)*$

which shows as 20% faster than this expression that uses capture groups but will give you the same answer

^(1*(01*0)*)+|0*1(0*(10*1)*)*$

Answer 4

我发现的最简单的解决方案是：

1+0(0+1)((1+0)(1+0))*

Answer 5

用最少的符号，

1*(01*01*)*

Answer 6

这个表达式怎么样：

(1(11)*+(00)*)