Re: Idempotence
- To: mathgroup at smc.vnet.net
- Subject: [mg37760] Re: [mg37748] Idempotence
- From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
- Date: Tue, 12 Nov 2002 03:13:08 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Assuming that that's all you want (and, in particular, that you are
happy with the commutativity of multiplication), it is easy:
In[1]:=
a/:a^2=a;
In[2]:=
b/:b^2=b;
In[3]:=
c/:c^2=c;
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};
In[6]:=
list1==list2//Simplify
Out[6]=
True
On Monday, November 11, 2002, at 07:10 PM, Tilo Schröder wrote:
> 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.
>
>
>
>
>
Andrzej Kozlowski
Yokohama, Japan
http://www.mimuw.edu.pl/~akoz/
http://platon.c.u-tokyo.ac.jp/andrzej/