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Re: Bug in pattern test, or I did something wrong?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg125911] Re: Bug in pattern test, or I did something wrong?
  • From: Christoph Lhotka <christoph.lhotka at fundp.ac.be>
  • Date: Fri, 6 Apr 2012 06:01:13 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201204050951.FAA13156@smc.vnet.net>

hi,

this is an interesting problem.

The reason why your first replacement rule does not work is hidden in 
the evaluation process.

Define


In[1]:= f[a_ /; MemberQ[{0, 1, 2}, a]] := 0;


As you mention you get:


In[2]:= f[x] /. f[expr_] :> f[-expr]

Out[2]= f[x]


Use Trace to see the reason:


In[3]:= Trace[f[x] /. f[expr_] :> f[-expr]]

Out[3]:= 
{{f[x],{MemberQ[{0,1,2},x],False},f[x]},{{f[expr_],{MemberQ[{0,1,2},expr_],True},0},0:>f[-expr],0:>f[-expr]},f[x]/. 
0:>f[-expr],f[x]}


f[x] evaluates to f[x] while f[expr_] evaluates to 0 thus f[x] /. 0 :> 
f[-expr] does not apply
and you get f[x].

So the problem is the following:


In[4]:= f[expr_]

Out[4]= 0


Which is due to MemberQ


In[5]:= MemberQ[{0, 1, 2}, expr_]

Out[5]= True


The evaluation chain is ok:


In[6]:= 0 /. f[expr_] :> f[-expr]

Out[6]= f[-expr]


By the way, in your definition of g


In[7]:= g[a_?MemberQ[{0, 1, 2}, #] &] := 0


the replacement works


In[8]:= g[x] /. g[expr_] :> g[-expr]

Out[8]= g[-x]


but not the definition:


In[9]:= g[0]

Out[9]= g[0]


If I understand your needs well a simple solution to your problem would be:


In[10]:= Do[h[i] = 0, {i, 0, 2}]


which gives:


In[11]:= {h[0], h[1], h[2], h[x]} /. h[expr_] :> h[-expr]

Out[11]:= {0, 0, 0, h[-x]}




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