寻找常规语言的补语
你能帮我找到一种语言的补语吗？它以 abab - (a|b)*abab (over an Alphabet {a,b}) 结尾
我想，补语必须包含所有字符串，不以 abab 结尾。 在构建一个 DFA 来补充 (a|b)*abab 后，可以尝试使用 Rij-Algorithm 来完成此操作，但是请帮助我理解在没有 Automaton 和 Rij 的情况下它是如何工作的（因为 Automaton有 5 个状态）。
好吧，单词不能以abab结尾。末尾的 a 和 b 四个字母有 24 种方式。好的，必须删除 abab，这样就有 15 种组合。这是否意味着补语语言是 (a|b)*。（a 和 b 的所有组合的并集没有abab）？但是 (a|b) 还是一开始的样子吗？
请帮助我理解这一点。
Could you help me please to find a complement of a language, which ends with abab - (a|b)*abab (over an alphabet {a,b})
I guess, the complement must contain all string, that don't end with abab.
One can try to do it with Rij-Algorithm after building a DFA for complement of (a|b)*abab, but pleaseee, help me to understand how it works without Automaton and Rij (because that Automaton has 5 states).
Ok, the words are not allowed to end with abab. There are 24 ways for four letters of a's and b's at the end. Okay, abab must be erased so there are 15 combinations. Does it mean, that the complement-language is (a|b)*.(union of all those combinations of a's and b's without abab)? But does (a|b) still stay the same at the beginning?
Help me please to understand this.
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
也许我不太明白你的意思，但这不是更简单吗？我是
(a|b)*(a|bb|aab|bbab)或事件(a|b)*(a|(b|(a|bb)a)b ）？PS 不要忘记还有比
abab短的单词，并且所有这些单词也应该包含在内。即(a|b){0,3}（其中{0,3}表示重复次数 [0; 3]）Maybe I quiet don't understand you, but isn't it much simplier. I'e
(a|b)*(a|bb|aab|bbab)or event(a|b)*(a|(b|(a|bb)a)b)?P.S. Don't forget that there is words shorter than
ababand all of them should be included too. I.e.(a|b){0,3}(where{0,3}denotes amount of repeats [0; 3])