Exercise 3.39. Which of the five possibilities in the parallel execution shown above remain if we instead serialize execution as follows:
(define x 10) (define s (make-serializer)) (parallel-execute (lambda () (set! x ((s (lambda () (* x x)))))) (s (lambda () (set! x (+ x 1)))))
To simplify description, let's call interleavable pieces of code as follows:
- A1:
(* x x) - A2:
(set! x <ressult of A1>) - B1:
(+ x 1)) - B2:
(set! x <ressult of B1>)
Note that:
- A1, B1 and B2 are not interleaved.
- But A2 might be interleaved into B1 and B2.
Possible orderings are:
- A1 (10 * 10) -> A2 (x = 100) -> B1 (100 + 1) -> B2 (x = 101)
- A1 (10 * 10) -> B1 (10 + 1) -> A2 (x = 100) -> B2 (x = 11)
- A1 (10 * 10) -> B1 (10 + 1) -> B2 (x = 11) -> A2 (x = 100)
- B1 (10 + 1) -> B2 (x = 11) -> A1 (11 * 11) -> A2 (x = 121)
So that 11, 100, 101 and 121 remain.