The Ackermann function is wild

ex1.10.scm

@@ -0,0 +1,44 @@
+(define (A x y)
+  (cond ((= y 0) 0)
+        ((= x 0) (* 2 y))
+        ((= y 1) 2)
+        (else (A (- x 1)
+                 (A x (- y 1))))))
+
+(define (pow base exponent)
+  (cond ((= base 0) 0)
+        ((= exponent 0) 1)
+        (else (* base (pow base (- exponent 1))))))
+
+
+(pow 2 10)
+
+;Value: 1024
+
+(pow 2 16)
+
+;Value: 65536
+
+(A 1 10)
+
+;Value: 1024
+
+(A 2 4)
+
+;Value: 65536
+
+(A 3 3)
+
+;Value: 65536
+
+(define (f n) (A 0 n))
+
+; 2*n
+
+(define (g n) (A 1 n))
+
+; 2^n
+
+(define (h n) (A 2 n))
+
+; 2^2^2^2...^2, with (- n 1) 2's in total

ex1.3.scm

@@ -1,11 +1,18 @@
 ;; Exercise 1.3
 
-(define (sum-of-squares x y) (+ (square x) (square y)))
+(define (sum-of-squares x y)
+  (+ (square x) (square y)))
+
 (define (sum-squares-of-two-largest a b c)
   (cond ((and (> a c) (> b c)) (sum-of-squares a b))
         ((and (> a b) (> c b)) (sum-of-squares a c))
         (else (sum-of-squares b c))))
 
-(= (sum-squares-of-two-largest 3 4 5) (sum-of-squares 4 5))
-(= (sum-squares-of-two-largest 5 4 3) (sum-of-squares 4 5))
-(= (sum-squares-of-two-largest 3 5 4) (sum-of-squares 4 5))
+(= (sum-squares-of-two-largest 3 4 5)
+   (sum-of-squares 4 5))
+
+(= (sum-squares-of-two-largest 5 4 3)
+   (sum-of-squares 4 5))
+
+(= (sum-squares-of-two-largest 3 5 4)
+   (sum-of-squares 4 5))

ex1.6.scm

@@ -2,8 +2,6 @@
   (cond (predicate then-clause)
         (else else-clause)))
 
-(new-if (= 2 3) 0 5)
-
 (new-if (= 1 1) 0 5)
 
 (define (new-sqrt-iter guess x)

ex1.9.scm

@@ -0,0 +1,21 @@
+(define (inc a) (+ a 1))
+(define (dec a) (- a 1))
+
+;; The below procedure generates a recursive process because the process cannot
+;; be summarized at every step by a set of state variable values -- there is
+;; additional "hidden" information maintained by the interpreter that indicates
+;; "where the process is" in negotiating a chain of deferred operations.
+
+(define (plus-recur a b)
+  (if (= a 0)
+      b
+      (inc (plus-recur (dec a) b))))
+
+;; The below procedure generates an iterative process because the state at any
+;; stage in the evolution of the process is captured completely by two state
+;; variables, a and b.
+
+(define (plus-iter a b)
+  (if (= a 0)
+      b
+      (plus-iter (dec a) (inc b))))