Answer till 3.42

ex-3_38-xx.scm

@@ -1,5 +1,78 @@
 (load "util.scm")
 
 (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 expectected and would not be
+; migated 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 "\nex-3.44\n")