[英]L = { a^n b^n c^m d^m : n >= 1, m >= 1 } U { a^n b^m c^m d^n : n >= 1, m >= 1 } isRegular?
there is a lot of examples for pumpinglemma proof, but I did not figure out this, can anyone help ? 有很多关于抽水论证的例子,但是我没有弄清楚,有人可以帮忙吗?
L= { a^nb^nc^md^m : n >= 1, m >= 1 } U { a^nb^mc^md^n : n >= 1, m >= 1 } L = {a ^ nb ^ nc ^ md ^ m:n> = 1,m> = 1} U {a ^ nb ^ mc ^ md ^ n:n> = 1,m> = 1}
Consider the regular language R = a*b*cd
. 考虑常规语言R = a*b*cd
。 The intersection of two regular languages must be a regular language. 两种常规语言的交集必须是常规语言。 The intersection of L
and R
is a^nb^n cd
. L
和R
的交点是a^nb^n cd
。 However, this is easily shown not to be regular using the pumping lemma or Myhill-Nerode theorem. 但是,使用抽引引理或Myhill-Nerode定理很容易证明这不是规则的。 This is a contradiction, so L
must not be regular. 这是一个矛盾,因此L
一定不能规则。
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