2.2.2 Hierarchical Structures

Printer-friendly version

The representation of sequences in terms of lists generalizes naturally to represent sequences whose elements may themselves be sequences. For example, we can regard the object ((1 2) 3 4) constructed by

(cons (list 1 2) (list 3 4))

as a list of three items, the first of which is itself a list, (1 2). Indeed, this is suggested by the form in which the result is printed by the interpreter. Figure 2.5 shows the representation of this structure in terms of pairs.

Another way to think of sequences whose elements are sequences is as trees. The elements of the sequence are the branches of the tree, and elements that are themselves sequences are subtrees. Figure 2.6 shows the structure in figure 2.5 viewed as a tree.

Recursion is a natural tool for dealing with tree structures, since we can often reduce operations on trees to operations on their branches, which reduce in turn to operations on the branches of the branches, and so on, until we reach the leaves of the tree. As an example, compare the length procedure of section 2.2.1 with the count-leaves procedure, which returns the total number of leaves of a tree:

(define x (cons (list 1 2) (list 3 4)))

(length x)
3
(count-leaves x)
4

(list x x)
(((1 2) 3 4) ((1 2) 3 4))

(length (list x x))
2

(count-leaves (list x x))
8

To implement count-leaves, recall the recursive plan for computing length:

• Length of a list x is 1 plus length of the cdr of x.
• Length of the empty list is 0.

Count-leaves is similar. The value for the empty list is the same:

• Count-leaves of the empty list is 0.

But in the reduction step, where we strip off the car of the list, we must take into account that the car may itself be a tree whose leaves we need to count. Thus, the appropriate reduction step is

• Count-leaves of a tree x is count-leaves of the car of x plus count-leaves of the cdr of x.

Finally, by taking cars we reach actual leaves, so we need another base case:

• Count-leaves of a leaf is 1.

To aid in writing recursive procedures on trees, Scheme provides the primitive predicate pair?, which tests whether its argument is a pair. Here is the complete procedure:[13]

(define (count-leaves x)
  (cond ((null? x) 0)  
        ((not (pair? x)) 1)
        (else (+ (count-leaves (car x))
                 (count-leaves (cdr x))))))

(1 3 (5 7) 9)

((7))

(1 (2 (3 (4 (5 (6 7))))))

(define x (list 1 2 3))
(define y (list 4 5 6))

(append x y)

(cons x y)

(list x y)

(define x (list (list 1 2) (list 3 4)))

x
((1 2) (3 4))

(reverse x)
((3 4) (1 2))

(deep-reverse x)
((4 3) (2 1))

(define x (list (list 1 2) (list 3 4)))

(fringe x)
(1 2 3 4)

(fringe (list x x))
(1 2 3 4 1 2 3 4)

(define (make-mobile left right)
  (list left right))

(define (make-branch length structure)
  (list length structure))

(square-tree
 (list 1
       (list 2 (list 3 4) 5)
       (list 6 7)))
(1 (4 (9 16) 25) (36 49))

(define (square-tree tree) (tree-map square tree))

(define (subsets s)
  (if (null? s)
      (list nil)
      (let ((rest (subsets (cdr s))))
        (append rest (map <??> rest)))))

Mapping over trees

Just as map is a powerful abstraction for dealing with sequences, map together with recursion is a powerful abstraction for dealing with trees. For instance, the scale-tree procedure, analogous to scale-list of section 2.2.1 takes as arguments a numeric factor and a tree whose leaves are numbers. It returns a tree of the same shape, where each number is multiplied by the factor. The recursive plan for scale-tree is similar to the one for count-leaves:

(define (scale-tree tree factor)
  (cond ((null? tree) nil)
        ((not (pair? tree)) (* tree factor))
        (else (cons (scale-tree (car tree) factor)
                    (scale-tree (cdr tree) factor)))))
(scale-tree (list 1 (list 2 (list 3 4) 5) (list 6 7))
            10)
(10 (20 (30 40) 50) (60 70))

Another way to implement scale-tree is to regard the tree as a sequence of sub-trees and use map. We map over the sequence, scaling each sub-tree in turn, and return the list of results. In the base case, where the tree is a leaf, we simply multiply by the factor:

(define (scale-tree tree factor)
  (map (lambda (sub-tree)
         (if (pair? sub-tree)
             (scale-tree sub-tree factor)
             (* sub-tree factor)))
       tree))

Many tree operations can be implemented by similar combinations of sequence operations and recursion.

(1 3 (5 7) 9)

((7))

(1 (2 (3 (4 (5 (6 7))))))

(define x (list 1 2 3))
(define y (list 4 5 6))

(append x y)

(cons x y)

(list x y)

(define x (list (list 1 2) (list 3 4)))

x
((1 2) (3 4))

(reverse x)
((3 4) (1 2))

(deep-reverse x)
((4 3) (2 1))

(define x (list (list 1 2) (list 3 4)))

(fringe x)
(1 2 3 4)

(fringe (list x x))
(1 2 3 4 1 2 3 4)

(define (make-mobile left right)
  (list left right))

(define (make-branch length structure)
  (list length structure))

(square-tree
 (list 1
       (list 2 (list 3 4) 5)
       (list 6 7)))
(1 (4 (9 16) 25) (36 49))

(define (square-tree tree) (tree-map square tree))

(define (subsets s)
  (if (null? s)
      (list nil)
      (let ((rest (subsets (cdr s))))
        (append rest (map <??> rest)))))