On Dec 11 2000, Hubert Chan wrote:
> Um, I never said that they were the same thing. In fact, I
> specifically said that they different. If you read my post again
> (in fact, the very part that you quoted), you will see that I said
> that which rule you add depends on whether or not you count the
> "Eric" == Eric G Miller writes:
Eric> On Fri, Dec 08, 2000 at 11:16:11AM -0700, Hubert Chan wrote:
>> It all depends on what you mean when you say "word". I used it in the
>> abstract sense, which is just a string of characters. So abcba is a
>> word, even though it is not
> "Rogerio" == Rogerio Brito <[EMAIL PROTECTED]> writes:
Rogerio> On Dec 08 2000, Hubert Chan wrote:
>> Oops. There should be a "| \epsilon" at the end of that (where \epsilon
>> is the empty string) if you count the empty string as a palindrome. If
>> not, then you should ad
ww.kuro5hin.org
> >From karsten Fri Dec 8 19:20:26 2000
> X-Envelope-Sender: [EMAIL PROTECTED]
> Date: Fri, 8 Dec 2000 22:16:03 -0500
> To: debian-user@lists.debian.org
> From: Jim Kroger <[EMAIL PROTECTED]>
> Subject: Re: OT: regular expression question
>
> A
Richard Cobbe wrote:
> Lo, on Friday, December 8, kmself@ix.netcom.com did write:
>
> > Dr. Kroger:
> >
> > Attached are several recent posts you've made to the debian-users
> > mailing list concerning the Debian GNU/Linux operating system
> > (http://www.debian.org/). I think you'll find that in
Lo, on Friday, December 8, kmself@ix.netcom.com did write:
> Dr. Kroger:
>
> Attached are several recent posts you've made to the debian-users
> mailing list concerning the Debian GNU/Linux operating system
> (http://www.debian.org/). I think you'll find that instructions for
> unsubscribing fro
On Dec 08 2000, Hubert Chan wrote:
> Oops. There should be a "| \epsilon" at the end of that (where
> \epsilon is the empty string) if you count the empty string as a
> palindrome. If not, then you should add "| aa | bb".
No, it's not the same thing. The former language generated by
Dr. Kroger:
Attached are several recent posts you've made to the debian-users
mailing list concerning the Debian GNU/Linux operating system
(http://www.debian.org/). I think you'll find that instructions for
unsubscribing from the list are included, as well as a fallback address
for reaching a re
At 10:49 PM +0100 12/8/00, Viktor Rosenfeld wrote:
Hubert Chan wrote:
>
> Viktor> regular. I think, the PDA that recognizes this
language is fairly
> Viktor> easy to construct, but it's late, and I've done
enough theoretical
> Viktor> computer science for today.
>
> For sim
On Fri, Dec 08, 2000 at 11:16:11AM -0700, Hubert Chan wrote:
> It all depends on what you mean when you say "word". I used it in the
> abstract
> sense, which is just a string of characters. So abcba is a word, even though
> it is not an English word.
>
> Strictly, "A Man, a Plan, a Canal, Pana
Hubert Chan wrote:
>
> Viktor> regular. I think, the PDA that recognizes this language is fairly
> Viktor> easy to construct, but it's late, and I've done enough theoretical
> Viktor> computer science for today.
>
> For simplicity, assume that our alphabet is {a,b}. Then the CFG is
> "Eric" == Eric G Miller writes:
Eric> On Thu, Dec 07, 2000 at 05:29:29PM -0700, Hubert Chan wrote:
>> > "Frodo" == Frodo Baggins <[EMAIL PROTECTED]> writes:
>>
Frodo> Take the palindrome w=aaa...abb...b where there are n 'a' and n
>> That's not a palindrome. A pal
> "Hubert" == Hubert Chan <[EMAIL PROTECTED]> writes:
Viktor> regular. I think, the PDA that recognizes this language is fairly
Viktor> easy to construct, but it's late, and I've done enough theoretical
Viktor> computer science for today.
Hubert> For simplicity, assume that o
Viktor Rosenfeld scripsit:
>Frodo Baggins wrote:
>
>> [Pumping Lemma]
>
>> 1 w = xyz
>> 2 y is not empty
>> 3 x has less than n caracters
>
>That would be xy has less than/equal to n characters (|xy| <= n).
>
>> 4 for any integer k, the word w_k = xyy..yz (k times y) is in the
>> langua
On Thu, Dec 07, 2000 at 05:29:29PM -0700, Hubert Chan wrote:
> > "Frodo" == Frodo Baggins <[EMAIL PROTECTED]> writes:
>
> Frodo> Take the palindrome w=aaa...abb...b where there are n 'a' and n
>
> That's not a palindrome. A palindrome is a word in which the first "half" is
> the reverse
> "Viktor" == Viktor Rosenfeld <[EMAIL PROTECTED]> writes:
Viktor> Peter Jay Salzman wrote:
>> i know how to search for palindromes, for instance, a 3 letter
>> palindrome:
>>
>> \(.\)\(.\)\(.\)\3\2\1
>>
>> recently, i started wondering if there was a way, using r
> "Frodo" == Frodo Baggins <[EMAIL PROTECTED]> writes:
Frodo> Take the palindrome w=aaa...abb...b where there are n 'a' and n
That's not a palindrome. A palindrome is a word in which the first "half" is
the reverse of the second "half" (half is in quotes because the word may have
an odd
> "Peter" == Peter Jay Salzman <[EMAIL PROTECTED]> writes:
Peter> i know how to search for palindromes, for instance, a 3 letter
Peter> palindrome: \(.\)\(.\)\(.\)\3\2\1
Well, strictly speaking, that's not really a regular expression. (And I would
also count that as a 6 letter palind
Frodo Baggins wrote:
> [Pumping Lemma]
> 1 w = xyz
> 2 y is not empty
> 3 x has less than n caracters
That would be xy has less than/equal to n characters (|xy| <= n).
> 4 for any integer k, the word w_k = xyy..yz (k times y) is in the
> language (i.e. matches the regex)
MfG Viktor
Peter Jay Salzman scripsit:
>
>recently, i started wondering if there was a way, using regex only, to
>express a palindrome of arbitrary letter length?
>
>i've used some grey matter, and the answer seems to be no, but there's nice
>symmetry here. maybe i'm missing something...
>
Well, maybe you'r
Peter Jay Salzman wrote:
> i know how to search for palindromes, for instance, a 3 letter palindrome:
>
> \(.\)\(.\)\(.\)\3\2\1
>
> recently, i started wondering if there was a way, using regex only, to
> express a palindrome of arbitrary letter length?
>
> i've used some grey matter,
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