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