Re: Idempotence
- To: mathgroup at smc.vnet.net
- Subject: [mg37773] Re: Idempotence
- From: "News Admin" <news at news.demon.net>
- Date: Tue, 12 Nov 2002 03:13:34 -0500 (EST)
- References: <aqo05b$fpa$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
You can reduce list2 by expanding everything out, then implementing
idempotence using replacement rules. In this case all you need are the rules
{a^2 -> 1, b^2 -> 1, c^2 -> 1}, but more generally you would need rules to
deal with higher powers as well.
(list2 // Expand) /. {a^2 -> 1, b^2 -> 1, c^2 -> 1}
yields the result
{-3 + 4 b + 4 a c - 4 a b c, c, 1}
This isn't the same as list1 so I presume there was a typo in your question.
Steve Luttrell
"Tilo Schröder" <tilo.schroeder at unibw-muenchen.de> wrote in message
news:aqo05b$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.
>
>
>