I am experimenting with implementing algorithms from the Red Dragon Book using Clisp 2.49.92. I find that MEMBER doesn't work as I expected when applied to a list of lists. My list is:

; Each production starts with a nonterminal followed by the right side of the production.
; e.g. A -> B + C becomes '(A B + C)

(SETQ GRAMMAR-4-11
'(
(E T EP)
(EP + T EP)
(EP epsilon)
(T F TP)
(TP * F TP)
(TP epsilon)
(F |(| E |)|)
(F id)
)
)

Now if I try (MEMBER '(EP epsilon) GRAMMAR-4-11) I get NIL, whereas I was expecting the third entry to be matched. But

(EQUAL '(EP epsilon) (CADDR GRAMMAR-4-11))

returns T as expected. And if I define

(DEFUN MEMBER-OF (ELEMENT L)
(COND ((NULL L) NIL)
((EQUAL ELEMENT (CAR L)) L)
(T (MEMBER-OF ELEMENT (CDR L)))
)
)

(DEFUN MEMBERP (ELEMENT L)
(COND ((NULL L) NIL)
((EQUAL ELEMENT (CAR L)) T)
(T (MEMBERP ELEMENT (CDR L)))
)
)

these functions find the desired element successfully. So I don't understand why MEMBER can't find the element.

--- Brian McGuinness

Brian McGuinness <b.mcguinness747@gmail.com> writes:

> Now if I try (MEMBER '(EP epsilon) GRAMMAR-4-11) I get NIL, whereas I was expecting the third entry to be matched. But
>
> (EQUAL '(EP epsilon) (CADDR GRAMMAR-4-11))
>
> returns T as expected.

MEMBER uses EQL as test function when none is specified. (MEMBER '(EP
epsilon) GRAMMAR-4-11 :test #'equal) should do what you expect.

--
Laat hulle almal sterf. Ek is tevrede om die wêreld te sien brand en die vallende
konings te spot. Ek en my aasdier sal loop op die as van die verwoeste aarde.

Am 14.12.2022 um 19:12 schrieb Brian McGuinness:

what does:

> (F |(| E |)|)

??

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Jens Kallup <kallup-dev@web.de> writes:

> Am 14.12.2022 um 19:12 schrieb Brian McGuinness:
>
> what does:
>
>> (F |(| E |)|)
>
> ??

|xxx| is a way to write an atomic symbol that contains special charcters
like ( and ):

[1]> (symbol-name '|(|)
"("

--
Ben.

E:\SBCL\tests>sbcl --script test1.lisp
((EP *EPSILON*) (T F TP) (TP * F TP) (TP *EPSILON*) (F ( E )) (F ID))

but i like to expect: (EP *EPSILON*)
How can I do this.

Thanks, Jens

; code running with sbcl
; test1.lisp:
; (defvar *GRAMMAR-4-11*)
(defvar *epsilon*)

(SETQ *GRAMMAR-4-11*
'(
(E T EP)
(EP + T EP)
(EP *epsilon*)
(T F TP)
(TP * F TP)
(TP *epsilon*)
(F |(| E |)|)
(F id)
)
)

(princ (MEMBER '(EP *epsilon*) *GRAMMAR-4-11* :test #'equal))

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ok, I have it done.
But I don't know, if it the right way:

(print (car (member '(T F TP) *GRAMMAR-4-11* :test #'equal)))
(print (car (member '(EP *epsilon*) *GRAMMAR-4-11* :test #'equal)))

This will me display the member of first occurrence.

Jens

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On Thu, 15 Dec 2022 14:11:31 +0100
Jens Kallup <kallup-dev@web.de> wrote:
> ok, I have it done.
> But I don't know, if it the right way:
>
>
> (print (car (member '(T F TP) *GRAMMAR-4-11* :test #'equal)))
> (print (car (member '(EP *epsilon*) *GRAMMAR-4-11* :test #'equal)))
>
>
> This will me display the member of first occurrence.

It's correct but you may find using FIND (no pun intended) more
straightforward.

Am 15.12.2022 um 14:54 schrieb Spiros Bousbouras:

> It's correct but you may find using FIND (no pun intended) more
> straightforward.

in this context:

https://dpaste.com/FQJWV9QMT

how can i combine i1 with "ich", and h1 with "habe", so the german
sentence part "ich habe" will be produced.

Like the Input:
(member '(i1 h1))

give me:
(ich habe)

is that possible ?

Thanks for reading
Jens

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maybe this is better:

(defvar *grammar*)
(defvar *alist*)

(defvar *i0*)
(defvar *i1*)

(setq *alist*
'(
(1 . ich)
(2 . du)

(10 . habe)
(20 . dich)

(30 . gefragt)
)
)

(setq *grammar*
'((1)
(1 10)
(1 10 20)
(1 10 20 30)))

(setq *i0* (assoc '1 *grammar*))

; here (1 . ich)
; but I would test: if *i0* (1) is in alist
; when true, then setq ich, else print error:
; (setq *i1* (assoc '1 *alist* ))

(print *i0* ) ; (1)
(print *i1* ) ; (1 . ICH)

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Ok, thanks. This makes more sense now.

--- Brian McGuinness