Re: Pattern Matching
- To: mathgroup at smc.vnet.net
- Subject: [mg23927] Re: Pattern Matching
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Fri, 16 Jun 2000 00:57:06 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Andrzej,
We can modify David's solution to behave in the alternative way:
Module[{u},
x[1] b[1] + x[1]b[2] + x[2] b[1] + x[2] b[2] + 3z /. b[n_]x[n_] -> u /.
n_Integer u -> z]
4 z + b[2] x[1] + b[1] x[2]
or
x[1] b[1] + x[1]b[2] + x[2] b[1] + x[2] b[2] + 3z /. b[n_]x[n_] -> # /.
n_Integer # -> z &[Unique[u]]
4 z + b[2] x[1] + b[1] x[2]
( x[1]b[1] + x[2]b[2] )y + (x[3]b[3] + x[4]b[4])v /. b[n_]x[n_] -> # /.
n_Integer # -> z &[Unique[u]]
v z + y z
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
----- Original Message -----
From: "Andrzej Kozlowski" <andrzej at tuins.ac.jp>
To: mathgroup at smc.vnet.net
<hay at haystack.demon.co.uk>; "David Park" <djmp at earthlink.net>; "Johannes
Ludsteck" <ludsteck at zew.de>; "Wagner Truppel" <wtruppel at uci.edu>
Subject: [mg23927] Re: Pattern Matching
> Dear Hartmut
>
> One problem with questions of this kind is that the author often does not
> give us a really precise idea of what he wants. The result is that a
number
> of potential solutions suggest themselves, which may work in some but not
in
> other situations. For example, take the solution proposed by David Park,
and
> Wagner Truppel:
>
> In[4]:=
> x[1] b[1] + x[1]b[2] + x[2] b[1] + x[2] b[2] /. b[n_]x[n_] -> z /.
> n_Integer z -> z
>
> Out[4]=
> z + b[2] x[1] + b[1] x[2]
>
> is fine, if that is all what Johannes wanted but;
>
> In[5]:=
> x[1] b[1] + x[1]b[2] + x[2] b[1] + x[2] b[2] + 3z /. b[n_]x[n_] -> z /.
> n_Integer z -> z
>
> Out[5]=
> z + b[2] x[1] + b[1] x[2]
>
> may not be quite what he had in mind.
>
> Similarly, with your solution
>
> In[6]:=
> Module[{virgo = True},
> Plus[x[1] b[1], x[1]b[2], x[2] b[1], x[2] b[2]] /.
> b[n_]x[n_] :>
> RuleCondition[ If[virgo, virgo = False; z, Unevaluated[Sequence[]]] ]]
>
> Out[6]=
> z + b[2] x[1] + b[1] x[2]
>
> is fine but
>
>
> In[7]:=
> Module[{virgo = True},
> ( x[1]b[1] + x[2]b[2] )y + (x[3]b[3] + x[4]b[4])v /.
> b[n_]x[n_] :>
> RuleCondition[ If[virgo, virgo = False; z,
Unevaluated[Sequence[]]] ]]
>
> Out[7]=
> y z
>
> while it would seem to me that the answer:
>
> v z + y z
>
> would be more reasonable.
> In both of these cases my "global function" solution:
>
> funct[expr_Plus] := expr /. HoldPattern[Plus[Times[x[_], b[_]] ..] ] :> z;
> funct[(x[i_]*b[j_] /; i != j) + c_] := x[i]*b[j] + funct[c];
> funct[x_] := x
>
>
> seems preferable, since in the above examples it gives:
>
> In[15]:=
> MapAll[funct, ( x[1]b[1] + x[2]b[2] )y + (x[3]b[3] + x[4]b[4])v]
>
> Out[15]=
> v z + y z
>
> In[16]:=
> MapAll[funct, x[1] b[1] + x[1]b[2] + x[2] b[1] + x[2] b[2] + 3z]
>
> Out[16]=
> 4 z + b[2] x[1] + b[1] x[2]
>
> But of course, as I wrote, Johannes did not really make completely clear
> exactly which matches were to be allowed and which not.
>
> Andrzej Kozlowski
>
> --
> Andrzej Kozlowski
> Toyama International University, JAPAN
>
> For Mathematica related links and resources try:
> <http://www.sstreams.com/Mathematica/>
>
>
>
>
> on 00.6.15 5:27 PM, Hartmut Wolf at hwolf at debis.com wrote:
>
> > My comments below at >>>>>>>>
> >
> > Allan Hayes schrieb:
> >>
> >> Johannes,
> >>
> >> Re your first question.
> >>
> >> We get
> >>
> >> x[1] b[1] + x[2] b[2] /.
> >> Plus[Times[x[_], b[_]] ..] -> z
> >>
> >> b[1] x[1] + b[2] x[2]
> >>
> >> possibly because, before the matching and replacement, the left side of
the
> >> rule is evaluated thus:
> >>
> >> Plus[Times[x[_], b[_]] ..]
> >>
> >> b[_] x[_] ..
> >>
> >> So we end up with
> >>
> >> x[1] b[1] + x[2] b[2] /. (*1*)
> >> (x[_] b[_]) .. -> z
> >>
> >> b[1] x[1] + b[2] x[2]
> >>
> >> We can avoid this by using HoldPattern:
> >>
> >> x[1] b[1] + x[2] b[2] /.
> >> HoldPattern[Plus[Times[x[_], b[_]] ..]] -> z
> >>
> >> z
> >>
> >> However it does seem odd that, in spite of (*1*), we get
> >>
> >> x * b[1] + x* b[2] /.
> >> (x* b[_]) .. -> z
> >>
> >> 2 z
> >>
> >> --
> >> 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
> >>
> >
> >>>>>>>>>
> >
> > Dear Allan,
> >
> > as so often you come up with the right questions. You observed (example
> > simplified)
> >
> > In[35]:= h[x, x] /. x .. -> z (*2*)
> > Out[35]= h[z, z]
> >
> > whereas (*1*) doesn't substitute. Before coming back to that, why isn't
> > Out[35]
> > == h[z]?
> >
> > Time to go back to the Book. Although I didn't find it stated, and
although we
> > wouldn't get an error message, all examples given in the book indicate
that
> > the
> > use of Repeated in (*1*) and (*2*) isn't legal. Since Repeat[...] is
> > equivalent
> > to a Sequence the matched object must be enclosed in a head anyways, so
> >
> > In[38]:= h[x, x] /. head_[x ..] -> head[z]
> > Out[38]= h[z]
> >
> > works right, as well as does
> >
> > In[26]:= h[x[1] b[1], x[2] b[2]] /. head_[x[_] b[_] ..] -> head[z]
> > Out[26]= h[z]
> >
> > (And of course the HoldPattern version suggested by you and Andrzej
> > corresponds
> > to this syntax.)
> >
> > However this doesn't yet solve the problem as to replace only for equal
> > indices.
> >
> > In[47]:= x[1] b[1] + x[2] b[2] /. HoldPattern[Plus[Times[x[n_], b[n_]]
..]] ->
> > z
> > Out[47]= z + b[2] x[2]
> >
> > gives only one substitution. I would have liked to restrict the scope of
the
> > pattern variable n_ only to Times[x[n_], b[n_]] and not to the whole
sequence,
> > but I found no way to do that. David Park, and similarily Wagner
Truppel,
> > proposed
> >
> > In[80]:= Plus[x[1] b[1], x[1]b[2], x[2] b[1], x[2] b[2]] /.
b[n_]x[n_] -> z /.
> > n_Integer z -> z
> > Out[80]= z + b[2] x[1] + b[1] x[2]
> >
> > using iterated replacement (which is fine!). But let's try to do that
within
> > one
> > stroke. Since it is easy to simply remove all elements x[n_] b[n_]
> >
> > In[88]:= Plus[x[1] b[1], x[1]b[2], x[2] b[1], x[2] b[2]] /.
> > p : b[n_]x[n_] :> Sequence[]
> > Out[88]= b[2] x[1] + b[1] x[2]
> >
> > we may try just to not remove the first occurance:
> >
> > In[125]:= Module[{virgo = True},
> > Plus[x[1] b[1], x[1]b[2], x[2] b[1], x[2] b[2]] /.
> > b[n_]x[n_] :>
> > RuleCondition[ If[virgo, virgo = False; z, Unevaluated[Sequence[]]] ]]
> > Out[125]=
> > z + b[2] x[1] + b[1] x[2]
> >
> > RuleCondition is not striktly neccessary in this case, but would give a
more
> > pleasing result if the lhs in ReplaceAll is held: then the rhs of
RuleDelayed
> > will be evaluated *after* pattern matching and before substitution (the
> > Trott-Strzebonski-Hayes method; else the rhsides will be evaluated later
when
> > the result of ReplaceAll is evaluated as the final step, which of course
will
> > not occur if held).
> >
> > A less obfuscated method to do the same would be
> >
> > In[127]:= Module[{virgo = True},
> > Plus[x[1] b[1], x[1]b[2], x[2] b[1], x[2] b[2]] /.
> > {b[n_]x[n_] /; First[{virgo, virgo = False}] -> z,
> > b[n_] x[n_] -> Sequence[]}]
> > Out[127]=
> > z + b[2] x[1] + b[1] x[2]
> >
> > where the condition is evaluated at the lhs (of Rule).
> >
> >
> > Finally a remark addressed to Johannes: I certainly would not try to
program
> > that way; it is much easier (and clearer) just to seperate "diagonal"
and
> > "off-diagonal" elements of your sum
> >
> > In[90]:=
> > Plus[x[1] b[1], x[1]b[2], x[2] b[1],
> > x[2] b[2]] /. {{b[n_]x[n_] -> Sequence[]}, {b[n_]x[m_] /; n =!= m ->
> > Sequence[]}}
> > Out[90]=
> > {b[2] x[1] + b[1] x[2], b[1] x[1] + b[2] x[2]}
> >
> > and then proceed with the parts as you like.
> >
> > Kind regards to everyone,
> > Hartmut Wolf
> >
> >
> >
> >> "Johannes Ludsteck" <ludsteck at zew.de> wrote in message
> >> news:8hsu3g$dh4 at smc.vnet.net...
> >>> Dear Group Members,
> >>> I would like to "find" and replace expressions with the simple
> >>> structure x[1] b[1]+x[2] b2]+...+x[n] b[n]
> >>> I tried to use the following replacement rule
> >>> In[27]:= x[1] b[1] + x[2] b[2] /. Plus[Times[x[_], b[_]] ..] -> z
> >>>
> >>> Out[27]= b[1] x[1] + b[2] x[2] + b[3] x[3]
> >>> Which didn't work (Out[27] should be z).
> >>> Why?
> >>> The following FullForm seems to give exactly the structure I used
> >>> in my replacement rule.
> >>>
> >>> In[17]:=
> >>> FullForm[x[1] b[1] + x[2] b[2] + x[3] b[3]]
> >>> Out[17]//FullForm=
> >>> Plus[Times[b[1], x[1]], Times[b[2], x[2]], Times[b[3], x[3]]]
> >>>
> >>> Even if this worked, my pattern wouldn't account for equal indices,
> >>> i.e. it would match x[1] b[500]+x[12] b[3], even if it shouldn't.
> >>>
> >>> Any suggestions?
> >>> Thanks,
> >>> Johannes Ludsteck
> >>>
> >>>
> >>> Johannes Ludsteck
> >>> Centre for European Economic Research (ZEW)
> >>> Department of Labour Economics,
> >>> Human Resources and Social Policy
> >>> Phone (+49)(0)621/1235-157
> >>> Fax (+49)(0)621/1235-225
> >>>
> >>> P.O.Box 103443
> >>> D-68034 Mannheim
> >>> GERMANY
> >>>
> >>> Email: ludsteck at zew.de
> >>>
> >