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 <maximede...@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. >