Checking regularity of a given language.

Question

$L = \left \{ a^nb^mc^p : n+m+p > 5 \right \}.$

Comments

yes n+m+p>5

where 5 is a finite term.

So, a*b*c*-{finite no. of strings}= Regular Language, rt?

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this dfa has a possibility of starting with ba

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ALL ok !!

I think in this way :

L can be written in the following way :

$$\begin{align*} L = \left \{ a^{x}a^{*}b^{y}b^{*}c^{z}c^{*} | x+y+z=6\right \} \end{align*}$$

We have 28 solutions for $\left \{ x+y+z=6\right \} $  So,

$\begin{align*} L = \bigcup \left \{ a^{1}a^{*}b^{1}b^{*}c^{4}c^{*} + a^{1}a^{*}b^{2}b^{*}c^{3}c^{*} + ....etc. \text{all 28 cases} \right \} \end{align*}$

Answer 1

Lets think about its complementary language which is :

anbmcp : n + m + p <= 5

As we know n + m + p <= 5 will have finite no. of solutions which can be found as :

n + m + p + d = 5 [Where d is dummy variable]

So no. of solutions will be : 4-1+5C5 = 8C5 = 56..As this is finite,so is complementary language.

N we know finite language is regular so the complementary language is regular.

Also by closure properties we know that regular languages are closed under complementation.

Hence the original language which is given should be regular.

Answer 2

L={a^nb^mc^p:n+m+p>5}

∑* - {a^nb^mc^p:n+m+p<=5}

regular-finite( which is regular )

regular-regular

regular is closed under intersection and complement . so result is regualar

correct me if i am wrong .

Answer 3

It is a Regular language because It shows all the strings whose length is greater than 5.

Comments

bbaacc does not belong to L

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Yes , u r right. My answer is wrong...

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I think this language contains all the strings in which after 'b' , no 'a' should be come.. and after 'c' , no 'a' and 'b' should be come..