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MathGroup Archive 2004

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RE: Question on pattern matching

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47803] RE: [mg47765] Question on pattern matching
  • From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
  • Date: Tue, 27 Apr 2004 04:47:42 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com
  • Thread-topic: [mg47765] Question on pattern matching

>-----Original Message-----
>From: Roman Green [mailto:rgreen at mail.ru]
To: mathgroup at smc.vnet.net
>Sent: Monday, April 26, 2004 8:41 AM
>To: mathgroup at smc.vnet.net
>Subject: [mg47803] [mg47765] Question on pattern matching
>
>
>Hi,
>
>I am a very newbie in Mathematica so sorry for possibly stupid 
>question.
>
>In the following Mathematica's output
>In[1]:= SetAttributes[k, Orderless]
>In[2]:= k[a, b, b] /. k[x_, x_, y_] -> 0
>Out[2]= 0
>In[3]:= k[a, b, b] /. k_[x_, x_, y_] -> 0
>Out[3]= k[a, b, b]
>
>I can't understand why Mathematica can find match when 
>applying pattern 
>k[x_, x_, y_], but is unable with more general pattern k_[x_, x_, y_].
>
>Thanks in advance.
>
>---
>
>R. Green
>
>

If you understand Out[2], you recognize that the Attribute Orderless is essential for the pattern to match: 

(first)  the lhs ist brought into normal order:

    k[a, b, b] ---> k[a, b, b]   (no change in this case) 

(then) for the pattern k[x_, x_, y_] it does not suffice to bring it to normal order, as shound not depend on the naming of the patterns variables (k[y_,y_,x_] should give equal matches).
Instead, now all orderings of the the pattern variables are tried, but this shall only be done if the Head k has Attribute Orderless.

Now the pattern k_ or kk_ or whatever, doesn't have this Attribute, hence no match for

In[11]:= k[a, b, b] /. k_[x_, x_, y_] -> -1
Out[11]= k[a, b, b]

but (of course) a match for

In[12]:= k[a, a, b] /. k_[x_, x_, y_] -> 1
Out[12]= 1

as this doesn't need reordering.



If you want to achieve (what I think) you want to do:

In[13]:= 
k[a, b, b] /. expr : kk_[___] :> (expr /. kk[x_, x_, y_] -> 2)
Out[13]= 2

Then the Orderless Attribute of k is exploited (if it just happens to have it).


In[16]:= ClearAttributes[k, Orderless]
In[17]:= 
k[a, b, b] /. expr : kk_[___] :> (expr /. kk[x_, x_, y_] -> 2 )
Out[17]= k[a, b, b]



--
Hartmut Wolf


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