Add first exercises from subsection 2.3.3

chapter-2/ex-2.59.scm

@@ -0,0 +1,15 @@
+#lang sicp
+
+(define (element-of-set? x set)
+  (cond ((null? set) false)
+	((equal? x (car set)) true)
+	(else (element-of-set? x (cdr set)))))
+
+; we will simply adjoin elements of set1 to set2 if they are not there
+; and drop them if they are there
+(define (union-set set1 set2)
+  (cond ((null? set1) set2)
+	((null? set2) set1)
+	((element-of-set? (car set1) set2)
+	 (union-set (cdr set1) set2))
+	(else (union-set (cdr set1) (cons (car set1) set2)))))

chapter-2/ex-2.60.scm

@@ -0,0 +1,24 @@
+#lang sicp
+
+; element-of-set? is the exact same
+; O(n)
+(define (element-of-set? x set)
+  (cond ((null? set) false)
+	((equal? x (car set)) true)
+	(else (element-of-set? x (cdr set)))))
+
+; O(1)
+(define (adjoin-set x set)
+  (cons x set)) ; we don't have to check anymore
+
+; O(n), where n is the length of set1
+(define (union-set set1 set2)
+  (append set1 set2))
+
+; intersection-set can still be the exact same
+; O(n^2)
+(define (intersection-set set1 set2)
+  (cond ((or (null? set1) (null? set2)) '())
+	((element-of-set? (car set1) set2)
+	 (cons (car set1) (intersection-set (cdr set1) set2)))
+	(else (intersection-set (cdr set1) set2))))

chapter-2/ex-2.61.scm

@@ -0,0 +1,7 @@
+#lang sicp
+
+(define (adjoin-set x set)
+  (cond ((null? set) (list x))
+	((< x (car set)) (cons x set))
+	((= x (car set)) set)
+	((> x (car set)) (cons (car set) (adjoin-set x (cdr set))))))

chapter-2/ex-2.62.scm

@@ -0,0 +1,11 @@
+#lang sicp
+
+(define (union-set set1 set2)
+  (cond ((null? set1) set2)
+	((null? set2) set1)
+	((< (car set1) (car set2)) (cons (car set1)
+					 (union-set (cdr set1) set2)))
+	((= (car set1) (car set2)) (cons (car set1)
+					 (union-set (cdr set1) (cdr set2))))
+	((> (car set1) (car set2)) (cons (car set2)
+					 (union-set set1 (cdr set2))))))