       Re: Baffled By Underscore Pattern Matching

• To: mathgroup at smc.vnet.net
• Subject: [mg46024] Re: [mg45992] Baffled By Underscore Pattern Matching
• From: "Sseziwa Mukasa,,(978) 536-2359" <mukasa at jeol.com>
• Date: Tue, 3 Feb 2004 03:20:55 -0500 (EST)
• References: <200402021020.FAA29529@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```On Feb 2, 2004, at 5:20 AM, Harold Noffke wrote:

> 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/Out example I understand, but the
> In[2,3]/Out example (discussed below) has me totally mystified.
>
> As printed, we have ...
>
> 	In:= h[a___, x_, b___, x_, c___] := hh[x] h[a, b, c]
>
> 	In:= h[2, 3, 2, 4, 5, 3]
>
> 	Out= h[4, 5] hh hh
>
> Now let's make a change to In ...
>
> 	In:= Clear[h, hh]
>
> 	In:= h[a___, x_, b___, x_, c___] := hh[x] h[{a}, {b}, {c}]
>
> 	In:= h[2, 3, 2, 4, 5, 3]
>
> 	Out= h[{}, {3}, {4, 5, 3}] hh
>
> 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 manipulates the number
> stream which feeds into it from In.  I have no idea of how the
> In lists came to contain the numbers they do.
>
> Any help, pointers to tutorial papers, or more illuminating examples
> will be greatly appreciated.

I think one can understand this function just by carefully tracing the
program.  First of all, one must understand the definition of h.  h
takes an arbitrary number of arguments.  Reading the arguments from
left to right, the first repeated element that's encountered is bound
to x.  a, b, and c are then those arguments which separate the repeated
element.  h is also recursive, calling itself again with the argument
a,b,c.  Your modification on line 5, changed h such that it calls
itself on the arguments {a},{b},{c}, which is three lists.  Given the
original argument 2,3,2,4,5,3, x is bound to 2, the first repeated
argument, a is then empty, b is 3 and c is 4,5,3.  In your modified
version h is thus called with the argument {},{3},{4,5,3}.  None of the
elements are repeated and recursion terminates with your result.  The
original function is called with the argument 3,4,5,3, the empty
argument a being dropped, so x binds to 3, a is empty again, b is 4,5
and c is empty.  On the recursion h is called with 4,5 which has no
repeated element, so recursion terminates with the result seen in
Out.

Hope that's clear enough,

Ssezi

```

• Prev by Date: RE: Drawing specific contours in a 3D Surface Plot?
• Next by Date: Re: Baffled By Underscore Pattern Matching
• Previous by thread: Baffled By Underscore Pattern Matching
• Next by thread: Re: Baffled By Underscore Pattern Matching