Re: [Axiom-mail] Going mad with pattern matching

[Martin Rubey]

Dear Alasdair,

I did now go through your example.  It seems that you hit a bug, but
fortunately, there is an easy workaround.  THe problem is with rules of the
form:

rule ...a...b... | p(a,b) == ...

It seems that in this case, the predicate p is never tested, who knows why.
The workaround is to use the "suchThat" function.  Find the modified ruleset
below.  I know that this looks a bit clumsy, but I do not have the time
currently to debug it completely.

Some comments: 

* rules are applied recursively.  Thus, repeating ztrans on the right hand side
is unnecessary.

(42) -> t := rule a+b == zt(a, b)

   (42)  b + a == 'zt(a,b)
                        Type: RewriteRule(Integer,Integer,Expression Integer)
(43) -> t(a+b+c)

   (43)  zt(a,zt(c,b))
                                                     Type: Expression Integer

* I believe that your second rule (my rule r1) contained a misprint.  Please
  check.

* maybe the invocation of ruleset is unnecessary, and you can simply write the
  suchThat expressions within the rule pile.  I did not try.

All the best,

Martin

zt:=operator 'zt

help(z,a) ==
    tmp := z/(z-1)
    for i in 1..a repeat
        tmp:=-D(tmp,z)
    tmp

myFreeOf? l == freeOf?(l.1, l.2)

r0 := rule zt(f+g,n,z) == zt(f,n,z)+zt(g,n,z)
r1 := suchThat(rule zt(a*f,n,z) == a*zt(f,n,z), [a, n], myFreeOf?)
r2 := suchThat(rule zt(a,n,z) == a*z/(z-1), [a, n], myFreeOf?)
r3 := suchThat(rule zt(a^n,n,z) == z/(z-a), [a, n], myFreeOf?)
r4 := rule zt(0,n,z) == 0
r5 := rule zt(1,n,z) == z/(z-1)
r6 := rule zt(n,n,z) == z/(z-1)^2
r7 := rule zt(n^(a | integer?(a) and a>1),n,z) == help(z, a)

ztransrules := ruleset([r0,r1,r2,r3,r4,r5,r6,r7])$Ruleset(INT, INT, EXPR INT)

ztrans(f,n,z)==ztransrules zt(f,n,z)


Martin Rubey <address@hidden> writes:

> "Alasdair McAndrew" <address@hidden> writes:
> 
> > Thanks, Martin,
> > 
> > As always I am helped by you!  I have two questions; maybe you can answer
> > them:
> > 
> > 1)  How do I include linearity in the pattern matching rules?  At present,
> > the command
> > 
> > ztrans(2+3^n,nz)
> > 
> > which should return the result
> > 
> > 2z/(z-1)+z/(z-3)
> > 
> > produces
> > 
> > (2+3^n)z/(z-1).
> > 
> > That is, the pattern matcher incorrectly applies the rule
> > 
> > zt(a | freeOf?(a,n),n,z) == a*z/(z-1)
> > 
> > even though 2+3^n is not free of n!
> 
> I doubt this.  I'm a little short on time though.  Thus, please do me a 
> favour:
> replace all rules with little local functions -- as I have done with "help".
> Then you can say
> 
> )trace help
> 
> (replace "help" with the name of the functions you want to trace)
> 
> and you should see which rule was invoked.  To see the arguments in a nicer
> form, you can use
> 
> )trace help )mathprint
> 
> but this works only after help was invoked already once.
> 
> 
> > 2)  How do I force answers to be returned in factored form?
> 
> Later...  Short answer: you need to coerce them to an appropriate domain.
> 
> Martin
> 
> 
> 
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