📅 最后修改于: 2020-11-26 10:12:00 🧑 作者: Mango
正则表达式可以如下递归定义-
ε是一个正则表达式，表示包含空字符串的语言。 (L(ε)= {ε})
φ是表示空语言的正则表达式。 (L(φ)= {})
x是一个正则表达式，其中L = {x}
如果X是表示语言L(X)的正则表达式，而Y是表示语言L(Y)的正则表达式，则
X + Y是与语言L(X)∪L(Y)对应的正则表达式，其中L(X + Y)= L(X)∪L(Y) 。
X 。 Y是与语言L(X)相对应的正则表达式。 L(Y)其中L(XY)= L(X)。长(Y)
R *是对应于语言L(R *)的正则表达式，其中L(R *)=(L(R))*
如果我们从1到5多次应用任何规则，它们就是正则表达式。
| Regular Expressions | Regular Set |
|---|---|
| (0 + 10*) | L = { 0, 1, 10, 100, 1000, 10000, … } |
| (0*10*) | L = {1, 01, 10, 010, 0010, …} |
| (0 + ε)(1 + ε) | L = {ε, 0, 1, 01} |
| (a+b)* | Set of strings of a’s and b’s of any length including the null string. So L = { ε, a, b, aa , ab , bb , ba, aaa…….} |
| (a+b)*abb | Set of strings of a’s and b’s ending with the string abb. So L = {abb, aabb, babb, aaabb, ababb, …………..} |
| (11)* | Set consisting of even number of 1’s including empty string, So L= {ε, 11, 1111, 111111, ……….} |
| (aa)*(bb)*b | Set of strings consisting of even number of a’s followed by odd number of b’s , so L = {b, aab, aabbb, aabbbbb, aaaab, aaaabbb, …………..} |
| (aa + ab + ba + bb)* | String of a’s and b’s of even length can be obtained by concatenating any combination of the strings aa, ab, ba and bb including null, so L = {aa, ab, ba, bb, aaab, aaba, …………..} |