Re: Languages are regualar

On 4月17日, 下午8时03分, 『 』 <tj6...@xxxxxxxxx> wrote:
L={a^nb^m | n/m is an integer}
L=(a^nb^m | n >= 25 m <=25}

Could someone give me some hints on how to do these problems?

Thanks.

Oops, I figured our s=a^pb^p didn't work.

I came up with this:

Let s=a^pab^(p+1)
Then x is some a's or empty, y is the remaining a's of a^p, then pump
down to zero, xy^0z then the numerator( |x|+1 ) will be smaller than
the denominator( p+1) ).
So n/m cannot be an integer.

Is that a good string that I chose?

.

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