Can I call that lisp function by using "?". 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.
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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>>
>>
>