►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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2.32 Exercise 2.32

The idea behind this algorithm is as follows: The set of all subsets of a set s is the same as the union of the set of subsets of s containing an arbitrary element x from s and the set of subsets of s not containing x. The latter is equivalent to the set of subsets of s minus the set containing x.

There is a bijection between these two subsets of the subsets of s: To go from a subset of s containing x to one not containing x (or vice versa), remove x from (or add it to) it. In other words, if we know the subsets of s not containing x, we can get all of the subsets of s by adding to this set each of its constituent subsets with x added to them.

In the code below, the base case is trying to get the subsets of an empty set, which is just a set containing the empty set ((list nil)). The element x is selected as the first element of the set, and so we then calculate the subsets of (cdr s) in order to add (car s) to each of them and append them to the subsets of (cdr s) to get our answer. The operation of adding (car s) to a subset of (cdr s) is:

(lambda (subset)
(append subset (list (car s))))

The complete procedure is below.

(define (subsets s)
  (if (null? s)
      (list nil)
      (let ((rest (subsets (cdr s))))
        (append
         rest
         (map (lambda (subset) (append subset (list (car s))))
              rest)))))
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