Re: map-par slower than map

[Damien Mattei]

Mutex? i do not think code has situation where dead lock could happen, it
is a code about minimalising logic expressions, it uses minterms , minterms
set is a set of minterms :like this:

example:
((1 1 0) (1 1 1)) will be unified : (1 1 x)
because 0 and 1 are replaced by x
the minterms-set could have thousands of pair (mathematic not lisp)
minterms to unify
if there is more than one x as result there is no need to continue so i
escape with a continuation:

minterms-set =
{
((1 0 1 0) (1 1 1 0))
((1 0 1 0) (1 1 0 1))
((1 0 1 0) (1 0 1 1))
((1 0 1 0) (0 1 1 1))
((0 1 1 0) (1 1 1 0))
((0 1 1 0) (1 1 0 1))
((0 1 1 0) (1 0 1 1))
((0 1 1 0) (0 1 1 1))
((0 1 0 1) (1 1 1 0))
((0 1 0 1) (1 1 0 1))
((0 1 0 1) (1 0 1 1))
((0 1 0 1) (0 1 1 1))
((0 0 1 1) (1 1 1 0))
((0 0 1 1) (1 1 0 1))
((0 0 1 1) (1 0 1 1))
((0 0 1 1) (0 1 1 1))
}

replace { } by () to have the list, other example at another level :

minterms-set =
{
((0 x 1 1) (x 1 1 1))
((0 x 1 1) (1 x 1 1))
((0 x 1 1) (1 1 x 1))
((0 x 1 1) (1 1 1 x))
((x 0 1 1) (x 1 1 1))
((x 0 1 1) (1 x 1 1))
((x 0 1 1) (1 1 x 1))
((x 0 1 1) (1 1 1 x))
((0 1 x 1) (x 1 1 1))
((0 1 x 1) (1 x 1 1))
((0 1 x 1) (1 1 x 1))
((0 1 x 1) (1 1 1 x))
((x 1 0 1) (x 1 1 1))
((x 1 0 1) (1 x 1 1))
((x 1 0 1) (1 1 x 1))
((x 1 0 1) (1 1 1 x))
((0 1 1 x) (x 1 1 1))
((0 1 1 x) (1 x 1 1))
((0 1 1 x) (1 1 x 1))
((0 1 1 x) (1 1 1 x))
((x 1 1 0) (x 1 1 1))
((x 1 1 0) (1 x 1 1))
((x 1 1 0) (1 1 x 1))
((x 1 1 0) (1 1 1 x))
((1 0 1 x) (x 1 1 1))
((1 0 1 x) (1 x 1 1))
((1 0 1 x) (1 1 x 1))
((1 0 1 x) (1 1 1 x))
((1 x 1 0) (x 1 1 1))
((1 x 1 0) (1 x 1 1))
((1 x 1 0) (1 1 x 1))
((1 x 1 0) (1 1 1 x))
}

here we see some minterms are already unified

 it is not easy to read even by me because i wrote the code many years ago
and is split in many files, but here it is:

(par-map function-unify-minterms-list minterms-set)

{function-unify-minterms-list <+ (λ (L) (apply
function-unify-two-minterms-and-tag L))}

(define (unify-two-minterms mt1 mt2)
  (function-map-with-escaping-by-kontinuation2
 (macro-function-compare-2-bits-with-continuation) mt1 mt2))

;; (function-map-with-escaping-by-kontinuation2
(macro-function-compare-2-bits-with-continuation)   '(1 1 0 1 0 1 1 0) '(1
1 0 1 1 1 1 1))

;; list1 = (1 1 0 1 0 1 1 0)
;; more-lists = ((1 1 0 1 1 1 1 1))
;; lists = ((1 1 0 1 0 1 1 0) (1 1 0 1 1 1 1 1))
;; clozure = #<procedure:...gos-DrRacket.scm:195:11>

;; #f
;;
;;  (function-map-with-escaping-by-kontinuation2
(macro-function-compare-2-bits-with-continuation)    '(1 1 0 1 0 1 1 0) '(1
1 0 1 1 1 1 0))

;; list1 = (1 1 0 1 0 1 1 0)
;; more-lists = ((1 1 0 1 1 1 1 0))
;; lists = ((1 1 0 1 0 1 1 0) (1 1 0 1 1 1 1 0))
;; clozure = #<procedure:...gos-DrRacket.scm:195:11>

;; '(1 1 0 1 x 1 1 0)
(define (function-map-with-escaping-by-kontinuation2 clozure list1 .
more-lists)
  (call/cc (lambda (kontinuation)
    (let ((lists (cons list1 more-lists))
  (funct-continu ;; this function have the kontinuation in his environment
   (lambda (arg1 . more-args)
     (let ((args (cons arg1 more-args)))
(apply clozure kontinuation args))))) ;; a tester: (apply clozure (cons
conti args))

         ;; (newline)
         ;; (dv list1)
         ;; (dv more-lists)
         ;; (dv lists)
 ;; (dv clozure)
         ;; (newline)

      (apply map funct-continu lists)))))

(define-syntax macro-function-compare-2-bits-with-continuation ;;
continuation version of macro-compare-2-bits
  ;; i need a macro because of external function to the clozure
  (syntax-rules ()
    ((_) (let ((cnt 0)) ;; counter
  (lambda (continuation b1 b2) (if (equal? b1 b2)
 b1
 (begin
   (set! cnt (add1 cnt)) ;; we leave with continuation in case cpt > 1, we
can have used a flag too instead of a counter
   (when (> cnt 1) (continuation #f)) ;; escaping with the continuation
   'x))))))) ;; return x in case of (b1,b2) = (O,1) or (1,0)

what could have caused mutex if in the latter definition above (let ((cnt
0)) ;; counter was defined at top level and shared by all threads!!! yes
there could have be some mutex  but this is not the case, i think even all
function are pure so why is it more slow with // than without?
Damien

On Wed, Oct 12, 2022 at 8:45 PM Maxime Devos <maximedevos@telenet.be> wrote:

> On 12-10-2022 19:19, Damien Mattei wrote:
> > Hello,
> > all is in the title, i test on a approximately 30000 element list , i got
> > 9s with map and 3min 30s with par-map on exactly the same piece of code!?
> >
>  > [...]
>  >
> > translated from Scheme+ to Scheme:
> > (define unified-minterms-set-1 (map function-unify-minterms-list
> > minterms-set)) ;;(par-map function-unify-minterms-list minterms-set))
>
> The definition of 'function-unify-minterms-list' and 'minterms-set' is
> missing.  Without a test case, we can only speculate what's going on.
> (E.g., maybe it grabs a mutex).
>
> Greetings,
> Maxime.
>