Understand orders of growth for count-change

ex1.14.dot

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+digraph {
+  s [label="(cc 11 5)"];
+  s -> sl;
+  s -> sr;
+  sl [label="(cc 11 4)"];
+  sl -> sll;
+  sl -> slr;
+  sll [label="(cc 11 3)"];
+  sll -> slll;
+  sll -> sllr;
+  slll [label="(cc 11 2)"];
+  slll -> sllll;
+  slll -> slllr;
+  sllll [label="(cc 11 1)"];
+  sllll -> slllll;
+  sllll -> sllllr;
+  slllll [label="(cc 11 0)"];
+  sllllr [label="(cc 10 1)"];
+  sllllr -> sllllrl;
+  sllllr -> sllllrr;
+  sllllrl [label="(cc 10 0)"];
+  sllllrr [label="(cc 9 1)"];
+  sllllrr -> sllllrrl;
+  sllllrr -> sllllrrr;
+  sllllrrl [label="(cc 9 0)"];
+  sllllrrr [label="(cc 8 1)"];
+  sllllrrr -> sllllrrrl;
+  sllllrrr -> sllllrrrr;
+  sllllrrrl [label="(cc 8 0)"];
+  sllllrrrr [label="(cc 7 1)"];
+  sllllrrrr -> sllllrrrrl;
+  sllllrrrr -> sllllrrrrr;
+  sllllrrrrl [label="(cc 7 0)"];
+  sllllrrrrr [label="(cc 6 1)"];
+  sllllrrrrr -> sllllrrrrrl;
+  sllllrrrrr -> sllllrrrrrr;
+  sllllrrrrrl [label="(cc 6 0)"];
+  sllllrrrrrr [label="(cc 5 1)"];
+  sllllrrrrrr -> sllllrrrrrrl;
+  sllllrrrrrr -> sllllrrrrrrr;
+  sllllrrrrrrl [label="(cc 5 0)"];
+  sllllrrrrrrr [label="(cc 4 1)"];
+  sllllrrrrrrr -> sllllrrrrrrrl;
+  sllllrrrrrrr -> sllllrrrrrrrr;
+  sllllrrrrrrrl [label="(cc 4 0)"];
+  sllllrrrrrrrr [label="(cc 3 1)"];
+  sllllrrrrrrrr -> sllllrrrrrrrrl;
+  sllllrrrrrrrr -> sllllrrrrrrrrr;
+  sllllrrrrrrrrl [label="(cc 3 0)"];
+  sllllrrrrrrrrr [label="(cc 2 1)"];
+  sllllrrrrrrrrr -> sllllrrrrrrrrrl;
+  sllllrrrrrrrrr -> sllllrrrrrrrrrr;
+  sllllrrrrrrrrrl [label="(cc 2 0)"];
+  sllllrrrrrrrrrr [label="(cc 1 1)"];
+  sllllrrrrrrrrrr -> sllllrrrrrrrrrrl;
+  sllllrrrrrrrrrr -> sllllrrrrrrrrrrr;
+  sllllrrrrrrrrrrl [label="(cc 1 0)"];
+  sllllrrrrrrrrrrr [label="(cc 0 1)"];
+  slllr [label="(cc 6 2)"];
+  slllr -> slllrl;
+  slllr -> slllrr;
+  slllrl [label="(cc 6 1)"];
+  slllrl -> slllrll;
+  slllrl -> slllrlr;
+  slllrll [label="(cc 6 0)"];
+  slllrlr [label="(cc 5 1)"];
+  slllrlr -> slllrlrl;
+  slllrlr -> slllrlrr;
+  slllrlrl [label="(cc 5 0)"];
+  slllrlrr [label="(cc 4 1)"];
+  slllrlrr -> slllrlrrl;
+  slllrlrr -> slllrlrrr;
+  slllrlrrl [label="(cc 4 0)"];
+  slllrlrrr [label="(cc 3 1)"];
+  slllrlrrr -> slllrlrrrl;
+  slllrlrrr -> slllrlrrrr;
+  slllrlrrrl [label="(cc 3 0)"];
+  slllrlrrrr [label="(cc 2 1)"];
+  slllrlrrrr -> slllrlrrrrl;
+  slllrlrrrr -> slllrlrrrrr;
+  slllrlrrrrl [label="(cc 2 0)"];
+  slllrlrrrrr [label="(cc 1 1)"];
+  slllrlrrrrr -> slllrlrrrrrl;
+  slllrlrrrrr -> slllrlrrrrrr;
+  slllrlrrrrrl [label="(cc 1 0)"];
+  slllrlrrrrrr [label="(cc 0 1)"];
+  slllrr [label="(cc 1 2)"];
+  slllrr -> slllrrl;
+  slllrr -> slllrrr;
+  slllrrl [label="(cc 1 1)"];
+  slllrrl -> slllrrll;
+  slllrrl -> slllrrlr;
+  slllrrll [label="(cc 1 0)"];
+  slllrrlr [label="(cc 0 1)"];
+  slllrrr [label="(cc -4 2)"];
+  sllr [label="(cc 1 3)"];
+  sllr -> sllrl;
+  sllr -> sllrr;
+  sllrl [label="(cc 1 2)"];
+  sllrl -> sllrll;
+  sllrl -> sllrlr;
+  sllrll [label="(cc 1 1)"];
+  sllrll -> sllrlll;
+  sllrll -> sllrllr;
+  sllrlll [label="(cc 1 0)"];
+  sllrllr [label="(cc 0 1)"];
+  sllrlr [label="(cc -4 2)"];
+  sllrr [label="(cc -9 3)"];
+  slr [label="(cc -14 4)"];
+  sr [label="(cc -39 5)"];
+
+}

ex1.14.scm

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+ (count-change amount)
+  (cc amount 5))
+
+(define (cc amount kinds-of-coins)
+  (cond ((= amount 0) 1)
+        ((or (< amount 0) (= kinds-of-coins 0)) 0)
+        (else (+ (cc amount
+                     (- kinds-of-coins 1))
+                 (cc (- amount
+                        (first-denomination kinds-of-coins))
+                     kinds-of-coins)))))
+
+(define (first-denomination kinds-of-coins)
+  (cond ((= kinds-of-coins 1) 1)
+        ((= kinds-of-coins 2) 5)
+        ((= kinds-of-coins 3) 10)
+        ((= kinds-of-coins 4) 25)
+        ((= kinds-of-coins 5) 50)))
+
+
+;; The tree-recursive process of count-change is similar in form to the
+;; tree-recursive Fibonacci computation, which requires a number of steps that
+;; grows exponentially with n and space that grows linearly with n. It seems
+;; that count-change requires O(2^amount) steps and O(amount) space.
+
+;; To visualize the tree, I adapted code from
+;; http://tobilehman.com/blog/2013/04/07/visualization-of-sicp-exercise-1-dot-14/. See below.
+;;
+;; I saved output in "ex1.14.dot" and used GraphViz to visualize as
+;; "ex1.14.tree.pdf".
+;;
+;; GraphViz command: "dot -Tpdf ex1.14.dot -o ex1.14.tree.pdf"
+
+(define (cc-graph amount kinds-of-coins)
+
+  (define (display-node label amount kinds-of-coins)
+    (display "  ")
+    (display label)
+    (display " [label=\"")
+    (display `(cc ,amount ,kinds-of-coins))
+    (display "\"];")
+    (newline))
+
+  (define (display-edge a b)
+    (display "  ")
+    (display a)
+    (display " -> ")
+    (display b)
+    (display ";")
+    (newline))
+
+  (define (base-case amount kinds-of-coins)
+    (or (< amount 0)
+        (= kinds-of-coins 0)
+        (= amount 0)))
+
+  (define (left label)
+    (string-append label "l"))
+
+  (define (right label)
+    (string-append label "r"))
+
+  ; label is the unique label of the function invocation, e.g.  "sllrl" is
+  ; reached by traveling (from the root) left then left then right then left.
+  (define (recurse label amount kinds-of-coins)
+    (cond ((base-case amount kinds-of-coins)
+           (display-node label amount kinds-of-coins))
+          (else
+           (display-node label amount kinds-of-coins)
+           (display-edge label (left label))
+           (display-edge label (right label))
+           (recurse (left label) amount (- kinds-of-coins 1))
+           (recurse (right label)
+                    (- amount (first-denomination kinds-of-coins))
+                    kinds-of-coins))))
+
+  (display "digraph {")
+  (newline)
+  (recurse "s" amount kinds-of-coins)
+  (newline)
+  (display "}"))