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We learn about the Pumping Lemma at the class and I tried to make few examples to understand it...

There I make this example:
Let's say: $L=\{w\in L|w=(0+1)^*1\}$ - i.e. - L is the language of all the words that finish with $1$. (The language is regular of course).

Now, I can take the word: $w=0^{n-1} 1$. $w\in L$ and $|w|\ge n$.
I can take: $u=0^{n-1},v=1,z=\varepsilon$.
$|uv|\le n$ and $|v|>0$, but:
$uv^0z\notin L$ because $uv^0z$ ends with $0$...

What I miss here?
I'd like to understand more where is my mistake...

Raphael
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stud1
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3 Answers3

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The pumping lemma says that, for every regular language there exists some pumping length $p$* such that blah blah blah.

You made two mistakes. First, you tried to choose $p$ based on a specific word, rather than based on the language as a whole. Second, you chose the wrong value of $p$. In particular, for the language you've given, $p=1$ has the property required of the pumping lemma: for every word $w$ in that language, you can write $w=\epsilon y z$ with $|y|=1$, and you'll find that $\epsilon y^n z$ is in the language for all $n\geq 0$.

David Richerby
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The pumping lemma states that "... there exist such $u$, $v$, $z$ that $w = xvz$ ..." but it doesn't state you can divide $w$ into arbitrary three pieces (even provided the other conditions hold).

Anton Trunov
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En tu caso esta algo mal planteado Para verificar con el lema de bombeo se sigue los siguientes pasos: k es un conjunto de estados de A Primero tiene q cumplir L=t(A)y w € A

A es un AFD con |k|=n y |w| >= n

a) w=uvz

b)|uv|=n

c)v ≠¶ (¶ es el vacío)

d) u(v^i)z i>=0 u(v^i )z € T(A)

con eso el AFD es A={{q0,q1,q2},{0,1},d,q1,q3} Con cadena w=0100111 u(v^i )z i >=0 y |uv|<=n u=0 v=1 z=00111 n=3 al incrementar i aun w € T(A)

espero entiendas :)

leny
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