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MathGroup Archive 2000

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Re: Flat, OneIdentity Again

  • To: mathgroup at smc.vnet.net
  • Subject: [mg21659] Re: Flat, OneIdentity Again
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Fri, 21 Jan 2000 04:00:10 -0500 (EST)
  • References: <86190a$l92@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

More examples:

Attributes[f] = {Flat};

f[2, 2, 3] /. f[x_  ..] :> h[x]

        f[h[2], 3]

f[2,2,3] was regarded as  f[f[2,2],3]

--------------------------------------------------
f[2, 2, 3] /. f[x_  .., y_] :> h[x, k[y]]

        h[2, k[f[2, 3]]]

f[2,2,3] was regarded as  f[2,f[2,3]]

----------------------------------------------------
f[2, 2, 3] /. f[x_  .., y__] :> h[x, k[y]]

        h[2, k[2, 3]]

f[2,2,3] was regarded as  f[2,Sequence[2,3]]

Adding the attribute OneIdentity makes no difference

Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565



"Ersek, Ted R" <ErsekTR at navair.navy.mil> wrote in message
news:86190a$l92 at smc.vnet.net...
> 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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