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MathGroup Archive 2004

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Re: How to substitute?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg45543] Re: How to substitute?
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Tue, 13 Jan 2004 04:03:57 -0500 (EST)
  • Organization: Universitaet Leipzig
  • References: <btndpp$4f$1@smc.vnet.net>
  • Reply-to: kuska at informatik.uni-leipzig.de
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,


\!\(\((y\_0 - z\_3)\)\ \((\(-y\_1\) + z\_3)\)\ \((\(-y\_2\) + 
          z\_3)\)\ \((\(-y\_3\) + z\_3)\)\  /. \ \((a_. *y\_i_ + 
          b_. \ \ z\_j_)\) :> \ Q[{a, i}, {b, j}]\)

Regards
  Jens

Steve Gray wrote:
> 
> \!\(\((y\_0 - z\_3)\)\ \((\(-y\_1\) + z\_3)\)\ \((\(-y\_2\) +
>         z\_3)\)\ \((\(-y\_3\) + z\_3)\)\)
> 
>         I have output cells containing various expressions such as
> (y[0]-z[1]) (y[1]-z[2]) (y[4] - z[0]), where I use [x] here to mean
> subscript. The integers can have any small values. For compactness of
> expression, I want to make substitutions which make the above
> expression look like Q[0,1]Q[1,2]Q[4,0]. There are many combinations
> of subscript pairs and I don't want to write explicit substitution
> rules for every possible combination. I tried doing /.(y[i]-z[j]) ->
> Q[i,j] , and /.(y[i_]-z[j_]) -> Q[i,j] but they don't work.
>         Any suggestions will be gratefully received.


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