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MathGroup Archive 2000

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Re: Flat, OneIdentity attributes

  • To: mathgroup at smc.vnet.net
  • Subject: [mg21622] Re: Flat, OneIdentity attributes
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Tue, 18 Jan 2000 02:35:11 -0500 (EST)
  • References: <85u3e7$d5j@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com


Ted: below,  I  have added some comments to your text.
I have tried to sort out at what is going with these and other aspects in
the Mathematica in Education and Research, Volume 8, Numbers 3 and 4
(combined), in press.

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>  news:85u3e7$d5j at smc.vnet.net...

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.

ClearAll[f];
Attributes[f] = {Flat, OneIdentity};
f[1, 2, 3] //. f[a_, b_] :> {a, b}

{1, {2, 3}}

(* Begin comment *)
 Follow the evaluation of  ReplaceRepeated (//.) using ReplaceAll (/.):

f[1, 2, 3] /. f[a_, b_] :> {a, b}

{1, f[2, 3]}

To get a match, the pattern matcher regarded f[1, 2, 3] as f[1,f [2, 3]]

% /. f[a_, b_] :> {a, b}

{1, {2, 3}}

The pattern matcher used the rule literally.
(* End comment *)

---------------------------------------------------
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.


ClearAll[f];
Attributes[f] = {Flat};
f[1, 2, 3] //. f[a_, b_] :> {a, b}

{f[1], {f[2], f[3]}}

With OneIdentity the pattern matcher doesn't wrap (f) around a single
argument
when it tries different ways of nesting (f).

(* Begin comment *)
Follow the evaluation

f[1, 2, 3] /. f[a_, b_] :> {a, b}

{f[1], f[2, 3]}

To get a match, the pattern matcher regarded f[1, 2, 3] as f[f[1],f [2, 3]]
,
*not as f[f[1],f[f[2],f[3]]]*.
It looks like the steps may be: find a re-nesting f[1,f[2,3]] that matches
and then, in the absence of OneIdentity,  "level up" to f[f[1],f[2,3]] ", or
it could  be that only such level patterns are checked for matching.

% /. f[a_, b_] :> {a, b}

{f[1], {f[2], f[3]}}

The pattern matcher regards  f[2,3] as f[f[2],f[3]}

Extra example:
 f[1] is regarded as f[f[1]] even when no re-nesting is needed to get a
match..

f[1] /. f[a_] -> {a}

{f[1]}

(* End comment *)

--------------------------------
In the next example (f) has the attributes Flat, OneIdentity and the rule is
used.

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

12

--------------------------------
For reasons I can't understand the rule isn't used in the next example. Can
anyone explain why?

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

f[2]

(* Begin comment *) In the example below shows that the rule is used, but
that f[2] is regarded as f[f[2]]; this explains why in the original example
the rule was not used :-  f[2] does not match n_Integer.

f[2] /. f[n_] :> { n}

{f[2]}

(* End comment *)

---------------------------------------------
Regards,
Ted Ersek

For Mathematica tips, tricks see
http://www.dot.net.au/~elisha/ersek/Tricks.html



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