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Re: Flat, OneIdentity attributes


Ted,

ClearAll[f];
Attributes[f] = {Flat};
f[2] /. f[n_Integer] :> n + 10

f[2]

I cannot explain this behavior. Further, this is even more unusual

ClearAll[f];
Attributes[f] = {Flat};
f[2] /. f[n_] :> n + 10

f[2] + 10

Mathematica appears to interpret the first case as  
f[2]/.f[n_Integer]:>(n+10)  and the second case as  (f[2]/.f[n_]:>n)+10


Bob Hanlon

In a message dated 1/17/2000 12:15:24 AM, ErsekTR at navair.navy.mil writes:

>For the most part I understand how Flat and OneIdentity are related and
>I
>demonstrate this using Version 4 in the examples below.
>
>In the first example (f) has the attributes Flat and OneIdentity. 
>The pattern matcher treats f[a,2,3] as f[a,f[2,3]] then uses the 
>replacement rule and {1,{2,3}} is returned.
>
>In[1]:=
>ClearAll[f];
>Attributes[f]={Flat,OneIdentity};
>f[1,2,3]//.f[a_,b_]:>{a,b}
>
>Out[3]=
>{1,{2,3}}
>
>---------------------------------------------------
>In the next example the only attribute (f) has is Flat.
>In this case the pattern matcher treats f[1,2,3] as 
>f[f[1],f[f[2],f[3]]] then uses the replacement rule and 
>{f[1],{f[2],f[3]}} is returned.
>
>
>In[4]:=
>ClearAll[f];
>Attributes[f]={Flat};
>f[1,2,3]//.f[a_,b_]:>{a,b}
>
>Out[6]=
>{f[1],{f[2],f[3]}}
>
>OneIdentity the pattern matcher doesn't wrap (f) around a single argument
>when it tries different ways of nesting (f).
>
>--------------------------------
>In the next example (f) has the attributes Flat, OneIdentity and the rule
>is
>used.
>
>In[7]:=
>ClearAll[f]
>Attributes[f]={Flat,OneIdentity};
>f[2]/.f[n_Integer]:>n+10
>
>Out[9]=
>12
>
>--------------------------------
>For reasons I can't understand the rule isn't used in the next example.
>Can
>anyone explain why?
>
>In[10]:=
>ClearAll[f]
>Attributes[f]={Flat};
>f[2]/.f[n_Integer]:>n+10
>
>Out[12]=
>f[2]
>


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