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From: HenHanna <HenHanna@devnull.tb>
Newsgroups: comp.lang.lisp
Subject: Re: Given string 'a.bc.' -- replace each dot(.) with 0 or 1
Date: Sat, 25 May 2024 03:06:16 -0700
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On 5/24/2024 5:54 AM, WJ wrote:
> On 5/18/2024, HenHanna wrote:
> 
>>
>> How can i write this function simply?   (in Common Lisp)
>>
>> -- Given a string  'a.bc.'   -- replace each dot(.)  with 0 or 1.
>>
>>         -- So the value is a list of 4 strings:
>>                                   ('a0bc0'  'a0bc1'  'a1bc0'  'a1bc1')
>>
>> -- The order is not important.
>>              If the string has 3 dots, the value is a list of length 8.
>>
>> If the program is going to be simpler,
>>                        pls use, e.g.   (a $ b c $)  rather than  'a.bc.'
> 
> Gauche Scheme:
> 
> (define (dotty s)
>    (define (f r) (dotty (regexp-replace "[.]" s r)))
>    (if (string-scan s #\.)
>      (apply append (map f '("0" "1")))
>      (list s)))
> 
> 
> gosh> (dotty "a.b.c")
> ("a0b0c" "a0b1c" "a1b0c" "a1b1c")
>   
> 


nice...    The description "a Lisp Haiku"  seems appropriate
                            (since i don't fully get how it works)


i've not seen that style before... what else would you write in that 
style?   Permutation?  Combination?  Cartesian-Power?



here's a slight rewrite.  I'd never used map! before today.

(use scheme.list)
(define (dotty x)
   (if (string-scan x #\.)
     (map! (lambda (d) (dotty (regexp-replace "[.]" x d)))
           (list "0" "1"))
     (list x)))

oh..Ok..i still need to do Apply-Append
                 because map! is NOT  (at all like)  Mapcan