►zthomae’s SICP solutions
▼2 Chapter 2
2.1 Exercise 2.1
2.2 Exercise 2.2
2.3 Exercise 2.3
2.4 Exercise 2.4
2.5 Exercise 2.5
2.6 Exercise 2.6
2.7 Exercise 2.7
2.8 Exercise 2.8
2.9 Exercise 2.9
2.10 Exercise 2.10
2.11 Exercise 2.11
2.12 Exercise 2.12
2.13 Exercise 2.13
2.14 Exercise 2.14
2.15 Exercise 2.15
2.16 Exercise 2.16
2.17 Exercise 2.17
2.18 Exercise 2.18
2.19 Exercise 2.19
2.20 Exercise 2.20
2.21 Exercise 2.21
2.22 Exercise 2.22
2.23 Exercise 2.23
2.24 Exercise 2.24
2.25 Exercise 2.25
2.26 Exercise 2.26
2.27 Exercise 2.27
2.28 Exercise 2.28
2.29 Exercise 2.29
2.30 Exercise 2.30
2.31 Exercise 2.31
2.32 Exercise 2.32
2.33 Exercise 2.33
2.34 Exercise 2.34
2.35 Exercise 2.35
2.36 Exercise 2.36
2.37 Exercise 2.37
2.38 Exercise 2.38
2.39 Exercise 2.39
2.40 Exercise 2.40
2.41 Exercise 2.41
2.42 Exercise 2.42
2.43 Exercise 2.43
2.44 Exercise 2.44
2.45 Exercise 2.45
2.46 Exercise 2.46
2.47 Exercise 2.47
2.48 Exercise 2.48
2.49 Exercise 2.49
2.50 Exercise 2.50
2.51 Exercise 2.51
2.52 Exercise 2.52
2.53 Exercise 2.53
2.54 Exercise 2.54
2.55 Exercise 2.55
2.56 Exercise 2.56
2.57 Exercise 2.57
2.58 Exercise 2.58
2.59 Exercise 2.59
2.60 Exercise 2.60
2.61 Exercise 2.61
2.62 Exercise 2.62
2.63 Exercise 2.63
2.64 Exercise 2.64
2.65 Exercise 2.65
2.66 Exercise 2.66
2.67 Exercise 2.67
2.68 Exercise 2.68
2.69 Exercise 2.69
2.70 Exercise 2.70
2.71 Exercise 2.71
2.72 Exercise 2.72
2.73 Exercise 2.73
2.74 Exercise 2.74
2.75 Exercise 2.75
2.76 Exercise 2.76
2.77 Exercise 2.77
2.78 Exercise 2.78
2.79 Exercise 2.79
2.80 Exercise 2.80
2.81 Exercise 2.81
2.82 Exercise 2.82
2.83 Exercise 2.83
2.84 Exercise 2.84
2.85 Exercise 2.85
2.86 Exercise 2.86
2.87 Exercise 2.87
2.88 Exercise 2.88
2.89 Exercise 2.89
2.90 Exercise 2.90
2.91 Exercise 2.91
2.92 Exercise 2.92
2.93 Exercise 2.93
2.94 Exercise 2.94
2.95 Exercise 2.95
2.96 Exercise 2.96
2.97 Exercise 2.97
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2.62 Exercise 2.62

Like the new intersection-set, this implementation of union-set is O(n) because we examine the elements of each set one time at most.

(define (union-set set1 set2)
  (cond ((null? set1) set2)
        ((null? set2) set1)
        (else
         (let ((x1 (car set1)) (x2 (car set2)))
           (cond ((= x1 x2)
                  (cons x1 (union-set (cdr set1) (cdr set2))))
                 ((< x1 x2)
                  (cons x1 (union-set (cdr set1) set2)))
                 ((< x2 x1)
                  (cons x2 (union-set set1 (cdr set2)))))))))
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