pumping lemma peter linz EX

Question

Let L={a^nb^m: n>=100 , m<=50}

Can you use the pumping lemma to show that L is not regular? Explain your answers

Comments

If we use (a^100)a*(b+€)^50 can we say it's it's regular?

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As Pumping lemma is an negativity test,means if we are not able to find out the any pattern then definitely the language is NOT regular,

but if we found out any pattern then It may or may not be regular.

In the above language as we cant find out any pattern so we can consider that the language is NOT regular,

But as the n,m are independent of each other and also finite so we can construct a FA for the given language,which makes it regular.

Therefore pummping lemma CANT be used for given language.

Correct me if i am wrong!!

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yes given language  is regular so pumping lemma must hold

 but if we try like taking some k =130 then

xy can be a^100 b^30 belongs to L

now assume y=b^20 , z=10, x=a^100

then Wi=x y^i z if we put i=3 the W3=x(b^60)z which does not belong to L

 

Now above like arguments can be  ruled out by  proper understanding of pumping lemma   

In applying the pumping lemma, we must keep in mind what the theorem says. We are guaranteed
the existence of an m as well as the decomposition xyz, but we do not know what they are. We cannot
claim that we have reached a contradiction just because the pumping lemma is violated for some
specific values of m or xyz. On the other hand, the pumping lemma holds for every w ∈ L and every i.
Therefore, if the pumping lemma is violated even for one w or i, then the language cannot be regular.
The correct argument can be visualized as a game we play against an opponent. Our goal is to win
the game by establishing a contradiction of the pumping lemma, while the opponent tries to foil us.
There are four moves in the game.
1. The opponent picks m.
2. Given m, we pick a string w in L of length equal or greater than m. We are free to choose any w,
subject to w ∈ L and |w| ≥ m.
3. The opponent chooses the decomposition xyz, subject to |xy| ≤ m, |y| ≥ 1. We have to assume that
the opponent makes the choice that will make it hardest for us to win the game.
4. We try to pick i in such a way that the pumped string wi, defined in Equation (4.2), is not in L. If
we can do so, we win the game.

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sir pumping lemma is used to prove a non regular language as non regular... and i think if we use it on reg.. language we cant decide that.