2020-12-27 20:49:49 +01:00
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(load "util.scm")
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(display "\nex-3.38\n")
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2020-12-28 19:14:07 +01:00
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(display "[answered]\n")
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2020-12-27 20:49:49 +01:00
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2020-12-28 19:14:07 +01:00
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; Peter Paul Mary
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; 110 90 45
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; Peter Mary Paul
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; 110 55 35
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; Mary Peter Paul
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; 50 60 40
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; Mary Paul Peter
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; 50 30 40
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; Paul Mary Peter
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; 80 40 50
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; Paul Peter Mary
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; 80 90 45
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(display "\nex-3.39\n")
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(display "[answered]\n")
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; (define x 10)
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; (define s (make-serializer))
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; (parallel-execute (lambda () (set! x ((s (lambda () (* x x))))))
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; (s (lambda () (set! x (+ x 1)))))
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; 101: P1 sets x to 100 and then P2 increments x to 101.
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; 121: P2 increments x to 11 and then P1 sets x to x times x.
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; 100: P1 accesses x (twice), then P2 sets x to 11, then P1 sets x.
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; 11: P2 accesses x, then P1 sets x to 100, then P2 sets x.
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; My friends on Scheme-Wiki agree with this solution so I feel good about this.
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(display "\nex-3.40\n")
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(display "[answered]\n")
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; (define x 10)
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; (parallel-execute (lambda () (set! x (* x x)))
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; (lambda () (set! x (* x x x))))
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; 10 * 10 = 100
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; 10 * 10 * 10 = 1'000
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; 10 * 1000 = 10'000
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; 10 * 10 * 100 = 10'000
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; 10 * 100 * 100 = 100'000
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; 10^6 = 1'000'000
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; (define x 10)
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; (define s (make-serializer))
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; (parallel-execute (s (lambda () (set! x (* x x))))
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; (s (lambda () (set! x (* x x x)))))
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; 10^2 = 100 -> 100^3 = 1000000
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; 10^3 = 1000 -> 1000^2 = 1000000
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(display "\nex-3.41\n")
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(display "[answered]\n")
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; Ben's concern does not apply. Say, we have a withdraw and a balance check.
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; Depending on the order of execution the balance check would show the balance
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; before or after the withdraw. This behavior is expectected and would not be
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; migated by serializing balance.
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(display "\nex-3.42\n")
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(display "[answered]\n")
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; This should work without issues in all cases. I don't have a good mental
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; model of how the serializer works, yet.
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(display "\nex-3.43\n")
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(display "\nex-3.44\n")
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2020-12-27 20:49:49 +01:00
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