Flat, OneIdentity Again
- To: mathgroup at smc.vnet.net
- Subject: [mg21637] Flat, OneIdentity Again
- From: "Ersek, Ted R" <ErsekTR at navair.navy.mil>
- Date: Tue, 18 Jan 2000 02:35:21 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Thanks to all who answered my question on Flat and OneIdentity. I have more strange behavior on this subject. At http://support.wolfram.com/Kernel/Symbols/System/Flat.html it says in so many words .... If (f) has the Flat attribute you better not have a definition like f[p_]:=p because if you do, then an attempt to evaluate f[1,2] will not work and the kernel will have to quit when the infinite iteration limit is exceeded. In addition I found that you can't even evaluate f[2] in the above case, and it doesn't help if (f) also has the OneIdentity attribute! I wanted to understand just what the kernel is doing to exceed the iteration limit when we try to evaluate f[1,2] or f[2] above. The lines below offer some clues, but also add to the mystery. I wonder if any of you have an explanation. In[1]:= ClearAll[f]; Attributes[f]={Flat}; After the input above (f) has the Flat attribute and no definitions. f[1,2] as f[f[1,2]]. In[3]:= f[1,2]/.f[p_]:>{p} Out[3]= {f[1,2]} The idea that the pattern matcher treats f[1,2] as f[f[1,2]] is sort of verified at Out[4] below. In[4]:= MatchQ[f[1,2],f[_f]] Out[4]= True But if the pattern matcher treats f[1,2] as f[f[1,2]] why doesn't MatchQ return True in Out[4] below ? In[5]:= MatchQ[f[1,2],HoldPattern[f[f[_Integer,_Integer]]]] Out[5]= False Even stranger is the next line where the pattern is much more general! Notice that is a triple blank inside (f). In[6]:= MatchQ[f[1,2],HoldPattern[f[f[___]]]] Out[6]= False All the results above come out the same if (f) has the attributes Flat, OneIdentity. I have a hunch what may be going on here. Perhaps this is a bug. Could it be that the part of the pattern matcher that handles Flat is oblivious to HoldPattern and checks for a match with the patterns f[_Integer,_Integer] and f[___] when it should check for a match with f[f[_Integer,_Integer]] and f[f[___]] respectively in the lines above? I did all this using Version 4. -------------------- Regards, Ted Ersek On 12-18-99 Mathematica tips, tricks at http://www.dot.net.au/~elisha/ersek/Tricks.html had a major update
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