Re: Idempotence
- To: mathgroup at smc.vnet.net
- Subject: [mg37830] Re: Idempotence
- From: "Steve Luttrell" <luttrell at _removemefirst_westmal.demon.co.uk>
- Date: Thu, 14 Nov 2002 06:11:43 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Duh! Of course it should have been (list2 // Expand) /. {a^2 -> a, b^2 -> b,
c^2 -> c}.
Steve Luttrell
"News Admin" <news at news.demon.net> wrote in message
news:aqqdt1$sfp$1 at smc.vnet.net...
> You can reduce list2 by expanding everything out, then implementing
> idempotence using replacement rules. In this case all you need are the
rules
> {a^2 -> 1, b^2 -> 1, c^2 -> 1}, but more generally you would need rules to
> deal with higher powers as well.
>
> (list2 // Expand) /. {a^2 -> 1, b^2 -> 1, c^2 -> 1}
>
> yields the result
>
> {-3 + 4 b + 4 a c - 4 a b c, c, 1}
>
> This isn't the same as list1 so I presume there was a typo in your
question.
>
>
> Steve Luttrell
>
> "Tilo Schröder" <tilo.schroeder at unibw-muenchen.de> wrote in message
> news:aqo05b$fpa$1 at smc.vnet.net...
> > Hello,
> >
> > I'm quite new in operating with Mathematica. Solving a problem
concerning
> > network reliability I have to implement an algorithm which uses
> idempotence
> > (e.g.: a*a=a, a^2*b^3=ab).
> > As a result of this fact, the following lists should be the same:
> >
> > list1={b+a c-a b c,c,1}
> > list2={2 b+2 a c-a b c-a c (b+a c-a b c)-b (b+2 a c-a b c-a c (b+a c-a b
> > c)),c,1}
> >
> > The result of "list1===list2" should be true.
> >
> > Does anybody have a hint or an idea how to solve it with Mathematica? I
> > couldn't find anything about it even on Wolfram's webpage.
> >
> > Thank you in advance.
> >
> > Tilo.
> >
> >
> >
>
>
>