Lecture 2 - Regular Expressions¶

Definition :¶

Regular Expression¶

\(a(a\cup b)^nb\) is a regular expression over \(\{a,b\}\)

\(L(R) = \{w \in \{a,b\}^* | w\ is\ a\ string\ that\ starts\ with\ a\ and\ ends\ with\ b\}\)

Automic Regular Expression¶

\(\emptyset\) : \(L(\emptyset) = \emptyset\)
\(a\in \Sigma\) : \(L(a) = \{a\}\)

Composition of Regular Expression¶

\(R_1\cup R_2\) : \(L(R_1\cup R_2) = L(R_1)\cup L(R_2)\)
\(R_1R_2\) : \(L(R_1R_2) = L(R_1)L(R_2)\)
\(R_1^*\) : \(L(R_1^*) = [L(R_1)]^*\)

Precedence of Regular Expression¶

\(* > \cdot > \cup\)

Proof¶

Theorem 1:¶

\(RE \rightarrow NPA\)

We can simply prove that \(RE \rightarrow NPA\) by constructing a NPA from a regular expression.

State Elimination¶

\(NPA \rightarrow RE\)

Prove by using the idea of state elimination and dynamic programming.

Theorem 2:¶

Let \(L\) be a regular language,there must exist an integer \(p\geq 1\) such that every string \(w\in L\) with \(|w|\geq p\) can be written as \(w = xyz\) with

\(|xy|\leq p\)
\(|y|\geq 1\)
for all \(i\geq 0\), \(xy^iz\in L\).

Here \(p\) is called the pumping length.

Proof

If \(L\) is a regular language and it is finite, then we can set \(p\) to be the maximum length +1 of the strings in \(L\).

If \(L\) is a regular language and it is infinite, there exits a NFA \(N\) that accepts \(L\). Let \(N = (Q,\Sigma,\delta,q_0,F)\), where \(Q = \{q_0,q_1,\cdots,q_n\}\).

Suppose \(N\) has \(n\) states, then we can set \(p = n+1\).
consider \(q_0 \ to\ q_p\) [\(q_i\) is state after reading \(a_i\)], then there must \(\exists i,j\) such that \(i\neq j\) and \(q_i = q_j\). Then we can set \(x = w[0,i-1]\), \(y = w[i,j-1]\), \(z = w[j,p-1]\).

Example

\(L = \{0^n1^n | n\geq 0\}\) is not a regular language.

Proof by contradiction
Assume \(L\) is a regular language, then there must exist a pumping length \(p\).
Consider \(w = 0^p1^p\), then \(w = xyz\) with 1. \(|xy|\leq p\) 2. \(|y|\geq 1\) 3. \(xy^iz\in L\).
from i and ii, we can know that \(y = 0^k\) for some \(k\geq 1\).
from iii, we can know that \(xy^2z = 0^{p+k}1^p \notin L\), which is a contradiction.

\(L = \{w\in \{0,1\}^* | w\ contains\ an\ equal\ number\ of\ 0's\ and\ 1's\}\) is not a regular language.

Assume \(L\) is a regular language, then \(L \cap 0^*1^* = \{0^n1^n | n\geq 0\}\) is a regular language, which is a contradiction.

最后更新: 2024年11月11日 23:05:25
创建日期: 2024年9月15日 15:48:04