Re: Going mad with pattern matching

Quoting Martin Rubey:

>
> 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. 

Martin, could you please say two words (or a little more?) about
where you found information and examples of how to use suchThat
and ruleset? Sometimes finding out *how* to find out is more
important than the specific answer. 

> Find the modified ruleset below. 
> ... 
> zt:=operator 'zt
>
> help(z,a) ==
>     tmp := z/(z-1)
>     for i in 1..a repeat
>         tmp:=-D(tmp,z)
>     tmp
>

Note that there is a problem in this expression. It does not
correctly compute the ztrans of 'n^a'. 

> 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?)

That's pretty cool! I've never seen that anywhere before... 

> 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)
>

How can one decide on the proper package call to Ruleset?

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

Regards,
Bill Page.