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MathGroup Archive 2004

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Re: Baffled By Underscore Pattern Matching

  • To: mathgroup at smc.vnet.net
  • Subject: [mg46085] Re: Baffled By Underscore Pattern Matching
  • From: drbob at bigfoot.com (Bobby R. Treat)
  • Date: Thu, 5 Feb 2004 04:03:04 -0500 (EST)
  • References: <bvl8su$t50$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Here's a version that selects the largest possible x:

Clear[h]
h[a___, x_, b___, x_, 
    c___] /; x == 
    Max[Select[{a, x, b, c}, 
      Count[{x, a, b, x, c}, 
         #1] > 1 & ]] := 
  hh[x]*h[{a}, {b}, {c}]
h[2, 3, 2, 4, 5, 3]
h[2, 3, 2, 4, 5, 7]
h[8, 3, 8, 4, 5, 3]
h[{2}, {2, 4, 5}, {}]*hh[3]
h[{}, {3}, {4, 5, 7}]*hh[2]
h[{}, {3}, {4, 5, 3}]*hh[8]

Bobby

Harold.Noffke at wpafb.af.mil (Harold Noffke) wrote in message news:<bvl8su$t50$1 at smc.vnet.net>...
> MathGroup:
> 
> In my study of Mathematica 5.0, I have reached "The Mathematica Book >
> Principals of Mathematica > Patterns > 2.3.8 Functions with Variable
> Numbers of Arguments".  The In[1]/Out[1] example I understand, but the
> In[2,3]/Out[3] example (discussed below) has me totally mystified.
> 
> As printed, we have ...
> 
> 	In[2]:= h[a___, x_, b___, x_, c___] := hh[x] h[a, b, c]
> 
> 	In[3]:= h[2, 3, 2, 4, 5, 3]
> 
> 	Out[3]= h[4, 5] hh[2] hh[3]
> 
> Now let's make a change to In[2] ...
> 
> 	In[4]:= Clear[h, hh]
> 	
> 	In[5]:= h[a___, x_, b___, x_, c___] := hh[x] h[{a}, {b}, {c}]
> 	
> 	In[6]:= h[2, 3, 2, 4, 5, 3]
> 	
> 	Out[6]= h[{}, {3}, {4, 5, 3}] hh[2]
> 	
> I did a Trace on this pattern match problem, and found only that
> doublets were pulled out on each iteration.  In order to understand
> what is happening here, I think I need to understand the matching
> process at a level of granularity finer than Trace can supply.  I
> don"t have a clear mental picture of how In[5] manipulates the number
> stream which feeds into it from In[6].  I have no idea of how the
> In[6] lists came to contain the numbers they do.
> 
> Any help, pointers to tutorial papers, or more illuminating examples
> will be greatly appreciated.
> 
> Thanks.
> Harold


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