SICP Exercise 2.64: Constructing Balanced Trees

The following procedure list->tree converts an ordered list to a balanced binary tree. The helper procedure partial-tree takes as arguments an integer n and list of at least n elements and constructs a balanced tree containing the first n elements of the list. The result returned by partial-tree is a pair (formed with cons) whose car is the constructed tree and whose cdr is the list of elements not included in the tree.

(define (list->tree elements)
  (car (partial-tree elements (length elements))))

(define (partial-tree elts n)
  (if (= n 0)
      (cons '() elts)
      (let ((left-size (quotient (- n 1) 2)))
        (let ((left-result (partial-tree elts left-size)))
          (let ((left-tree (car left-result))
                (non-left-elts (cdr left-result))
                (right-size (- n (+ left-size 1))))
            (let ((this-entry (car non-left-elts))
                  (right-result (partial-tree (cdr non-left-elts)
                                              right-size)))
              (let ((right-tree (car right-result))
                    (remaining-elts (cdr right-result)))
                (cons (make-tree this-entry left-tree right-tree)
                      remaining-elts))))))))

1. Write a short paragraph explaining as clearly as you can how partial-tree works. Draw the tree produced by list->tree for the list (1 3 5 7 9 11).
2. What is the order of growth in the number of steps required by list->tree to convert a list of n elements?

(a) Describe How partial-tree Works
Before we give our short paragraph let's first note what quotient does as it's used here without any description. It's official definition in the Scheme R5RS specification can be found here. However, its functionality can be summarized as returning the result of dividing the first numeric operand by the second numeric operand and, if this is not an integer value, rounding it down to the nearest integer towards zero.

Okay, so given the assumptions about the operands to partial-tree listed above let's now describe how partial-tree works...

In the terminating case, when n=0, partial-tree simply returns a pair containing a representation of an empty binary tree and the unaltered list of elements it was called with (as these elements were not added to the binary tree generated). However, in the non-terminating case partial-tree first calculates the number of elements that should go into the left sub-tree of a balanced binary tree of size n, then invokes partial-tree with the elements and that value which both produces such a sub-tree and the list of elements not in that sub-tree. It then takes the head of the unused elements as the value for the current node, the tail as the potential elements for the right sub-tree and calculates how many elements should go into the right sub-tree of a balanced binary tree of size n. To build the right sub-tree it then invokes partial-tree with the unused elements and the required size of the right sub-tree to build it and get any remaining elements. Finally it builds the balanced tree with the generated components and returns this tree along with any remaining elements.

Note that a more succinct way of putting this would be to say that, for the non-terminating case, partial-tree splits the list into four: elements preceding the (n/2)^th position, the element at the (n/2)^th position, elements between the (n/2)^th position (exclusive) and the n^th position (inclusive) in the list and elements after the n^th position in the list. It then builds binary trees for the first and third of these lists using recursive calls to partial-tree, puts these sub-trees together, along with the element at the (n/2)^th position, to produce a tree and returns this along with the fourth list. However, while this captures the effect of the procedure, it doesn't describe the actual implementation.

I'd like to add a further note to this part of the exercise. The authors state that "list->tree converts an ordered list to a balanced binary tree". I'd like to correct this a little... list->tree will convert any list into a balanced binary tree; however, if the list is ordered then list->tree will convert it into a balanced binary search tree.

As for the tree produced by list->tree for the list (1 3 5 7 9 11), well here's the procedure in action:

> (list->tree '(1 3 5 7 9 11))
'(5 (1 () (3 () ())) (9 (7 () ()) (11 () ())))

If, as in the book, we omit empty lists from the tree diagram, then this gives us the following tree:

(b) Order of Growth
First, lets make the assumption that all of the list and pair operations (i.e. car, cdr, cons and list) are Θ(1) operations. I include list in this as make-tree uses it.

Now, excluding the recursive calls, partial-tree performs a fixed set of these operations for each list it processes. As a result we can say that, excluding recursive calls partial-tree runs in constant time (i.e. Θ(1)). We also know that, due to the way in which partial-tree operates, it makes recursive calls such that uses each index in the range 1..n exactly once. Now if you perform n invocations of an Θ(1) operation that gives you an order of growth of Θ(n).

Finally, list->tree simply makes a single call to partial-tree using the length of the list of elements as n. If we make a further assumption that length is, in the worst case, Θ(n) (i.e. if it has to iterate through the list and count all the elements), then this means that list->tree is Θ(n).

SICP Exercise 2.63: Binary Trees to Ordered Lists

Each of the following two procedures converts a binary tree to a list.

(define (tree->list-1 tree)
  (if (null? tree)
      '()
      (append (tree->list-1 (left-branch tree))
              (cons (entry tree)
                    (tree->list-1 (right-branch tree))))))
(define (tree->list-2 tree)
  (define (copy-to-list tree result-list)
    (if (null? tree)
        result-list
        (copy-to-list (left-branch tree)
                      (cons (entry tree)
                            (copy-to-list (right-branch tree)
                                          result-list)))))
  (copy-to-list tree '()))

1. Do the two procedures produce the same result for every tree? If not, how do the results differ? What lists do the two procedures produce for the trees in figure 2.16?
2. Do the two procedures have the same order of growth in the number of steps required to convert a balanced tree with n elements to a list? If not, which one grows more slowly?

(a) What Do the Procedures Do?
Before we actually look in detail at the internals of the two procedures, let's see how they behave with various trees. It's worth noting that the "Sets as binary trees" section is actually using binary search trees to represent sets, not just binary trees. As a result we'll test the procedures using valid binary search trees:

> (tree->list-1 '())
'()
> (tree->list-2 '())
'()
> (tree->list-1 (make-tree 5 '() '()))
'(5)
> (tree->list-2 (make-tree 5 '() '()))
'(5)
> (tree->list-1 (make-tree 3 (make-tree 2 (make-tree 1 '() '()) '()) '()))
'(1 2 3)
> (tree->list-2 (make-tree 3 (make-tree 2 (make-tree 1 '() '()) '()) '()))
'(1 2 3)
> (tree->list-1 (make-tree 1 '() (make-tree 2 '() (make-tree 3 '() '()))))
'(1 2 3)
> (tree->list-2 (make-tree 1 '() (make-tree 2 '() (make-tree 3 '() '()))))
'(1 2 3)
> (tree->list-1 (make-tree 5
                           (make-tree 3
                                      (make-tree 1
                                                 '()
                                                 (make-tree 2 '() '()))
                                      (make-tree 4 '() '()))
                           (make-tree 6
                                      '()
                                      (make-tree 7 '() '()))))
'(1 2 3 4 5 6 7)
> (tree->list-2 (make-tree 5
                           (make-tree 3
                                      (make-tree 1
                                                 '()
                                                 (make-tree 2 '() '()))
                                      (make-tree 4 '() '()))
                           (make-tree 6
                                      '()
                                      (make-tree 7 '() '()))))
'(1 2 3 4 5 6 7)

As the exercise also asks what the two procedures produce for the trees in figure 2.16, let's try those out too:

> (tree->list-1 (make-tree 7
                           (make-tree 3
                                      (make-tree 1 '() '())
                                      (make-tree 5 '() '()))
                           (make-tree 9
                                      '()
                                      (make-tree 11 '() '()))))
'(1 3 5 7 9 11)
> (tree->list-2 (make-tree 7
                           (make-tree 3
                                      (make-tree 1 '() '())
                                      (make-tree 5 '() '()))
                           (make-tree 9
                                      '()
                                      (make-tree 11 '() '()))))
'(1 3 5 7 9 11)
> (tree->list-1 (make-tree 3
                           (make-tree 1 '() '())
                           (make-tree 7
                                      (make-tree 5 '() '())
                                      (make-tree 9
                                                 '()
                                                 (make-tree 11 '() '())))))
'(1 3 5 7 9 11)
> (tree->list-2 (make-tree 3
                           (make-tree 1 '() '())
                           (make-tree 7
                                      (make-tree 5 '() '())
                                      (make-tree 9
                                                 '()
                                                 (make-tree 11 '() '())))))
'(1 3 5 7 9 11)
> (tree->list-1 (make-tree 5
                           (make-tree 3
                                      (make-tree 1 '() '())
                                      '())
                           (make-tree 9
                                      (make-tree 7 '() '())
                                      (make-tree 11 '() '()))))
'(1 3 5 7 9 11)
> (tree->list-2 (make-tree 5
                           (make-tree 3
                                      (make-tree 1 '() '())
                                      '())
                           (make-tree 9
                                      (make-tree 7 '() '())
                                      (make-tree 11 '() '()))))
'(1 3 5 7 9 11)

Okay, so it looks from the results as if both procedures are performing the equivalent of in-order (or symmetric) tree traversals. Given that these are binary search trees containing numeric values, with the left sub-tree of a node containing only values numerically less than the current node's value and the right-sub-tree of a node containing only values numerically greater than the current node's value, this means that we're expecting both procedures to produce a list containing all of the node values from the tree in numerically ascending order. Let's now look at what each of the two procedures do in turn in order to confirm this.

The first procedure, tree->list-1, processes the child nodes of each node recursively, until it reaches the null children. For any non-null node it appends the tree->list-1 for the left node onto the result of consing the current node's value onto the tree->list-1 for the right node. Assuming that we're dealing with a valid binary search tree then, for any node, this will produce a list containing the current node's value at some index, i such that all the elements in the list before index i are numerically less than the current node's value, while all the elements in the list that are after index i are numerically greater than the current node's value. Given that this is done recursively, we can say that tree->list-1 will indeed produce a list containing all of the node values from the tree in numerically ascending order.

Now let's look at tree->list-1. This procedure (or at least the inner procedure, copy-to-list) is more iterative in nature, building up a result list as it goes along, and stopping once it's run out of tree to process. For any non-null node it recursively calls copy-to-list with the left node as the tree to process, and the result of consing the current node's value onto the list returned by recursively calling copy-to-list with the right node as the tree and the result list as built so far. With the same assumption as before this means that: firstly the head of the list produced by the cons is less than all of the elements in the tail of the list, and secondly all of the elements to be consed onto the head of this list by the call to copy-to-list with the left node as the tree are less than the head of the result list so far. As this is performed recursively, we can also say that tree->list-2 will produce a list containing all of the node values from the tree in numerically ascending order.

How Do the Procedures Grow?
Well, first of all both procedures visit each node once in their traversal of the tree. This means that there's going to be a factor of n in both orders of growth. However, they differ in how they build up the results list...

The first procedure, tree->list-1, uses append to join the sublist produced for the left child node onto the list produced by consing the current node's value onto the sublist produced for the right child node. Now append is an Θ(n) operation, as it must iterate through, in this application, the entirety of the sublist produced for the left child node, and use cons (which we're assuming to be an Θ(1) operation) to append each element in reverse order onto the sublist produced for the right child node. As append is invoked once for each of the n nodes in the tree, tree->list-1 is an Θ(n²) operation.

On the other hand, in each recursive call the second procedure, tree->list-2, uses cons to put the current node's value onto the result list. As noted above we're assuming this to be an Θ(1) operation and, as it's performed once per node, this is performed a total of n times. As a result, tree->list-2 is an Θ(n) operation, and so grows more slowly than tree->list-1.

A Weight off My Shoulders

SICP is quite heavy going... Or at least it's quite heavy going when you're trying to work through it in large chunks on a regular basis and keep a (more than) full-time job going as a software engineer. Not to mention having a life outside work as well...

You may have noticed I've been somewhat lax in keeping up with SICP. I'm not the only one. Several of my colleagues have not been able to keep up with the exercises and, when we reached the end of chapter 2, we took the decision to stop work through the book as a group and instead we'll work through the Berkeley SICP video lectures together. Working through the book is still encouraged, but has been left to individuals to drive their own progress.

So I've been taking a little break from the book. Hey, it's been somewhat busy both at work and home of late, and I've not found enough time to do any of my huge list of personal projects. However, I'm back! I might not manage to maintain the same velocity as I've done previously, but I still intend on working my way through the whole book and posting my solutions to the exercises.

Anyway, I've dropped the progress chart, as that no longer makes sense. And, going forward, the SICP exercise index won't have any week dates on it for the exercises, since they don't make sense either.