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[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. 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[3]. Hope that's clear enough, Ssezi

**References**:**Baffled By Underscore Pattern Matching***From:*Harold.Noffke@wpafb.af.mil (Harold Noffke)

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