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Re: Re: Question on pattern matching

  • To: mathgroup at
  • Subject: [mg47822] Re: [mg47807] Re: Question on pattern matching
  • From: Andrzej Kozlowski <akoz at>
  • Date: Thu, 29 Apr 2004 00:33:38 -0400 (EDT)
  • References: <>
  • Sender: owner-wri-mathgroup at

On 27 Apr 2004, at 17:47, Bill Rowe wrote:

> On 4/26/04 at 2:40 AM, rgreen at (Roman Green) wrote:
>> 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_].
> The pattern k_ matches an expression named k with any head
> k[a, b, b] is an expression with head k and no name. The two things 
> don't match which is why you get the output you got.
> The pattern _k is any expression with head k. So,
> k[a,b,b]/._k->0
> will result in what you were expecting
> --
> To reply via email subtract one hundred and four
While the above is all true, it seems to me that the key point has to 
do with the Orderless attribute. To start wit,h observe the following:

k[a, b, b] /. k_[x_, y_, y_] -> 0


This matched for obvious reasons.

Now let's give k the attribute Orderless:


Now this matches due to the Orderless attribute:

k[a, b, b] /. k[x_, x_, y_] -> 0


k[a, b, b] /. k_[x_, x_, y_] -> 0


This does not match because k_, unlike k,  "does not have" the 
attribute Orderless (of course this does not strictly make sense since 
only Symbols can have attributes, but I guess you know what i mean): in 
other words no required rearrangement takes place for a match to be 

Andrzej Kozlowski
Chiba, Japan

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