----- Original Message -----
From: <dp2 at cise.ufl.edu>
To: "Richard Fateman" <fateman at cs.berkeley.edu>
Cc: <macrakis at alum.mit.edu>; <dp2 at cise.ufl.edu>; "Robert Dodier"
<robert.dodier at gmail.com>; <maxima at math.utexas.edu>
Sent: Tuesday, February 07, 2006 8:27 PM
Subject: Re: [Maxima] What is wrong with this rule?
> Can I call that lisp function by using "?".
The function is called m1, but the arguments are not going to look
like macsyma expressions.
So the answer is no.
> Also why the same rule runs if
> we use apply1 simply as I gave example in the previous email. The problem
> is only when apply(apply1...) is used.
I don't know why you are using apply.
If you want to write a program that looks for an operator "*" that
has 3,4, ... arguments, each of the form x(a,b),
and does what you want, it should be possible.
> Do you have any alternative solution for the problem, in which I have to
> apply in this way: x(i,j)*x(k,l)*x(m,n)->some function; it should match
> even the product is x(a,b)*f(a)*x(c,d)*f(d)*x(e,f).
> And similarly for the product of four,five,.... since I am generating
> rules at the time function is called.
> Regards
> Dileep
>
>> The matcher does not backtrack at all.
>> There is one that does, in the integration package
>> but it is not available except from lisp. There are other possibilities
>> (e.g. a Mathematica-style
>> pattern matcher in Lisp) but that would need a lot more syntactic
>> mechanism
>> for the user.
>>
>> RJF
>>
>>
>>
>> Stavros Macrakis wrote:
>>
>>
>>>> if you have a rule like e*f(x)*g(y), then it could try to match
>>>> 3*f(4)*g(5) by matching
>>>> e= 3*f(4) and then failing.
>>>>
>>>> You don't want e to match "anything at all". You want e to match
>>>> something not involving any f( ..).
>>>>
>>>>
>>>
>>> Sure, it might try matching e=3*f(4) to start, but when it fails, it
>>> had better backtrack and try e=3. If it doesn't, that seems like a bug.
>>>
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>>>
>>>
>>
>
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