(display "\nex-3.38\n")
(display "[answered]\n")

; Peter Paul Mary
; 110 90 45

; Peter Mary Paul
; 110 55 35

; Mary Peter Paul
; 50 60 40

; Mary Paul Peter
; 50 30 40

; Paul Mary Peter
; 80 40 50

; Paul Peter Mary
; 80 90 45

(display "\nex-3.39\n")
(display "[answered]\n")

; (define x 10)
; (define s (make-serializer))
; (parallel-execute (lambda () (set! x ((s (lambda () (* x x))))))
;                   (s (lambda () (set! x (+ x 1)))))

; 101: 	P1 sets x to 100 and then P2 increments x to 101.
; 121: 	P2 increments x to 11 and then P1 sets x to x times x.
; 100: 	P1 accesses x (twice), then P2 sets x to 11, then P1 sets x.
; 11: 	P2 accesses x, then P1 sets x to 100, then P2 sets x.

; My friends on Scheme-Wiki agree with this solution so I feel good about this.

(display "\nex-3.40\n")
(display "[answered]\n")

; (define x 10)
; (parallel-execute (lambda () (set! x (* x x)))
;                   (lambda () (set! x (* x x x))))

; 10 * 10        = 100
; 10 * 10 * 10   = 1'000
; 10 * 1000      = 10'000
; 10 * 10 * 100  = 10'000
; 10 * 100 * 100 = 100'000
; 10^6           = 1'000'000

; (define x 10)
; (define s (make-serializer))
; (parallel-execute (s (lambda () (set! x (* x x))))
;                   (s (lambda () (set! x (* x x x)))))

; 10^2 = 100  -> 100^3  = 1000000
; 10^3 = 1000 -> 1000^2 = 1000000

(display "\nex-3.41\n")
(display "[answered]\n")

; Ben's concern does not apply. Say, we have a withdraw and a balance check.
; Depending on the order of execution the balance check would show the balance
; before or after the withdraw. This behavior is expected and would not be
; avoided by serializing balance.

(display "\nex-3.42\n")
(display "[answered]\n")

; This should work without issues in all cases. I don't have a good mental
; model of how the serializer works, yet.

(display "\nex-3.43\n")
(display "[answered]\n")

; A swap of account balances maintains each individual balance. Hence, only the
; order of $10, $20, $30 can change and not the value.

;   a      b     c
; $10    $20   $30 (swap a b = 10) (swap a c = 20)
; $20    $10   $30
; $40    $10   $10

; The sum of balances will be preserved because the exchange procedure deposits
; what it withdraws. That works because it calculates `difference` before doing
; the transactions.

; If the individual transactions would not be serialized we could run into a
; situation where (swap a c) overrides a to be $30, for example, as we have
; seen in previous exercises.

(display "\nex-3.44\n")
(display "[answered]\n")

; Louis Reasoner is wrong. Assuming the balance on from-account is high enough
; transfer is going to work. The essential difference is that executing a list
; of transactions of absolute amounts is commutative. Exchange, on the other,
; is not commutative. On exchange has to finish before another one and the
; order matters.

(display "\nex-3.45\n")
(display "[answered]\n")

; When serialized-exchange is call it tries to serialize the deposit and
; withdraw calls that are already serialized by the same serializer. That would
; result in a deadlock because because the outer serialized procedure would
; have to call the inner serialized procedure and they cannot be executed at
; the same time because they are serialized by the same serializer.