Hello,
I'm using picolisp version 18.12.27 C in a debian 10.10 based distro,
installed as package picolisp 18.12-1
I'm playing with pilog and there're some examples that does not run:
1- the example in doc reference for repeat/0 [1] :
: (be integer (@I) # Generate unlimited supply of integers
(^ @C (box 0)) # Init to zero
(repeat) # Repeat from here
(^ @I (inc @C)) )
-> integer
: (? (integer @X))
@X=1
@X=2
@X=3
@X=4. # Stop
-> NIL
but I got nothing printed, no values for @X var, it keeps doing apparently
nothing until I kill the process
I can make it work if I use the -> functon this way:
: (be integer (@I)
(^ @C (box 0))
(repeat)
(^ @I (inc (-> @C))) )
Is the reference documentation incorrect?
2- the example of factorial in Mia's blog [2] :
: (be factorial (0 1) T)
-> factorial
: (be factorial (@N @X)
(^ @A (dec @N))
(factorial @A @B)
(^ @X (* @N @B)) )
-> factorial
: (? (factorial 5 @X))
@X=120
-> NIL
but when I execute (? (factorial 5 @X)) in my lisp I got:
: (? (factorial 5 @X))
-> NIL
only valid result is when executing (? (factorial 0 @X)) :
: (? (factorial 0 @X))
@X=1
-> NIL
debugging does not help:
: (? factorial (factorial 0 @X))
1 (factorial 0 1)
@X=1
-> NIL
: (? factorial (factorial 5 @X))
2 (factorial 5 @X)
-> NIL
Again I get it working using the -> function this way:
: (be factorial (@N @X)
(^ @A (dec (-> @N)))
(factorial @A @B)
(^ @X (* (-> @N) (-> @B))) )
-> factorial
Now:
: (? (factorial 5 @X))
@X=120
-> NIL
but I cannot get an answer for the inverse query:
: (? (factorial @X 120))
-> NIL
any hint?
I assume the use of -> function is needed but that means picolisp
documentation and Mia's blog is wrong.
Also I would like to know if pilog unification deals with compound
predicates, I mean, in prolog you can write:
p(a).
p(R(y)).
? p( X )
X=a
X=R(y)
Querying for values of X veryfing a p predicate. How can you get that in
pilog?
I've tried this two:
(be p ((R y)))
(be pp ((R (y))))
and it appears to work:
(? (p @X))
@X=(R y)
-> NIL
(? (pp @X))
@X=(R (y))
-> NIL
but I'm not sure it's the same semantics as prolog, because this appears to
me as stating p as a list , not a compound predicate.
In other words, what is the correct translation to this prolog clauses to
pilog?
prolog pilog
p(a). (be p (a))
p(x,y). (be p (x y))
p(r(b)). (be p (r (b))) ? not working (-> NIL)
(be p ((r (b)))) ? gets @X=(r (b))
(be p ((r b))) ? gets @X=(r b)
p(z(a),b) (be p ((z a) b)) ? gets @X=(z a) @Y=b
(be p ((z (a)) b)) ? gets @X=(z (a)) @Y=b
best regards
[1] R (software-lab.de) <https://software-lab.de/doc/refR.html#repeat/0>
[2] How to use Pilog in PicoLisp (picolisp-explored.com)
<https://picolisp-explored.com/how-to-use-pilog-in-picolisp>
