Added a bunch of solutions to the exercise problems in the concurrency

section.

diff --git a/src/sicp/ex3_39.rkt b/src/sicp/ex3_39.rkt
new file mode 100644 (file)
index 0000000..47ee525
--- /dev/null
+++ b/src/sicp/ex3_39.rkt
@@ -0,0 +1,11 @@
+#lang racket
+
+#|
+
+Only the below three cases will remain:
+
+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 (121).
+100:   P1 accesses x (twice), then P2 sets x to 11, then P1 sets x.
+
+|#
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diff --git a/src/sicp/ex3_40.rkt b/src/sicp/ex3_40.rkt
new file mode 100644 (file)
index 0000000..9329db8
--- /dev/null
+++ b/src/sicp/ex3_40.rkt
@@ -0,0 +1,41 @@
+#lang racket
+
+#|
+
+(define x 10)
+
+(parallel-execute (lambda () (set! x (* x x)))
+                  (lambda () (set! x (* x x x))))
+
+list all possible values of x.
+
+|#
+
+#|
+
+1. 100
+2. 1000
+3. (expt 100 3) = 1000000
+4. (expt (expt 10 3) 2) = 1000000
+5. (* 10 (* 10 10 10)) = 10000
+6. (* 10 (* 10 10) (* 10 10)) = 100000
+7. (* 10 10 (* 10 10)) = 10000
+
+|#
+
+#|
+
+Which of these possibilities remain if we instead use serialized procedures:
+
+(define x 10)
+
+(define s (make-serializer))
+
+(parallel-execute (s (lambda () (set! x (* x x))))
+                  (s (lambda () (set! x (* x x x)))))
+
+ans:
+
+=> 1000000
+
+|#
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diff --git a/src/sicp/ex3_41.rkt b/src/sicp/ex3_41.rkt
new file mode 100644 (file)
index 0000000..a20f95e
--- /dev/null
+++ b/src/sicp/ex3_41.rkt
@@ -0,0 +1,11 @@
+#lang racket
+
+#|
+
+I don't agree. 'balance' is only read and for a particular account, it is 
+not usually read while another transaction is in progress. Further, even if
+it is read concurrently with a transaction (deposit or withdrawal), one will
+get either the value before the transaction or the one after. Both are 
+acceptable.
+
+|#
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diff --git a/src/sicp/ex3_42.rkt b/src/sicp/ex3_42.rkt
new file mode 100644 (file)
index 0000000..88ba9f5
--- /dev/null
+++ b/src/sicp/ex3_42.rkt
@@ -0,0 +1,7 @@
+#lang racket
+
+#|
+
+Yes, it is safe.
+
+|#
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