# Re: Finite State Automaon Question

## "Vinayak R. Borkar" <vborky@yahoo.com>31 Mar 2005 23:35:13 -0500

From comp.compilers

Related articles
Finite State Automaon Question vborkyREMOVE_THIS@yahoo.com (Vinayak R. Borkar) (2005-03-15)
Re: Finite State Automaon Question torbenm@diku.dk (2005-03-18)
Re: Finite State Automaon Question vborky@yahoo.com (Vinayak R. Borkar) (2005-03-20)
Re: Finite State Automaon Question torbenm@diku.dk (2005-03-24)
Re: Finite State Automaon Question peter.ludemann@gmail.com (2005-03-24)
Re: Finite State Automaon Question nathan.moore@sdc.cox.net (Nathan Moore) (2005-03-25)
Re: Finite State Automaon Question vborky@yahoo.com (Vinayak R. Borkar) (2005-03-31)
Re: Finite State Automaon Question torbenm@diku.dk (2005-04-02)
| List of all articles for this month |

 From: "Vinayak R. Borkar" Newsgroups: comp.compilers Date: 31 Mar 2005 23:35:13 -0500 Organization: Compilers Central References: 05-03-058 05-03-063 05-03-073 05-03-082 Keywords: lex Posted-Date: 31 Mar 2005 23:35:12 EST

Torben Ęgidius Mogensen wrote:
>> > [Method deleted]
> [Alternative method deleted]
>>
>> L1: (ab)*
>> L2: (ab)*abab
>>
>> The DFA for L2 is abab(ab)*
>>
>> By your method, I get L3 to be (ab)*, but this is not right.
>
> Why is it not right?

The problem that I needed to solve was:

Find L3 such that

For any string v in L1 and w in L3, vw is in L2, and
for any string vw in L2, there is a v in L1 and w in L3.

In other words, L2 = L1.L3, where dot means "followed by"

But this is not satisfied by the example above.

Vinayak.

Post a followup to this message