The final part of section 2.3.3 asks us to consider a database that contains records, each of which has a key and some data. If the database is represented as an unordered list, a record can be looked up by its key using the following procedure.

(define (lookup given-key set-of-records) (cond ((null? set-of-records) false) ((equal? given-key (key (car set-of-records))) (car set-of-records)) (else (lookup given-key (cdr set-of-records)))))

We can define simple procedures for making a record out of a key and its data, and for extracting the key and data from an existing record in order to test the procedure above.

(define (key record) (car record)) (define (data record) (cdr record)) (define (make-record key data) (cons key data)) (define database (list (make-record 1 'Bill) (make-record 2 'Joe) (make-record 3 'Frank) (make-record 4 'John))) > (lookup 3 database) '(3 . Frank) > (data (lookup 1 database)) 'Bill

**Exercise 2.66**asks us to implement the

`lookup`

procedure for the case where the set of records is structured as a binary tree, ordered by the numerical values of the keys.We can start by including the

`list->tree`

and `partial-tree`

procedures given for exercise 2.64, along with a few required supporting procedures.(define (entry tree) (car tree)) (define (left-branch tree) (cadr tree)) (define (right-branch tree) (caddr tree)) (define (make-tree entry left right) (list entry left right)) (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))))))))

This makes it easier to convert the existing database to one structured as a binary tree.

> (define tree-db (list->tree database)) > tree-db '((2 . Joe) ((1 . Bill) () ()) ((3 . Frank) () ((4 . John) () ())))

Finally, we can write the new implementation of

`lookup`

using `element-of-set?`

as a guide.(define (lookup given-key set-of-records) (cond ((null? set-of-records) #f) ((= given-key (key (car set-of-records))) (car set-of-records)) ((< given-key (key (car set-of-records))) (lookup given-key (left-branch set-of-records))) ((> given-key (key (car set-of-records))) (lookup given-key (right-branch set-of-records))))) > (lookup 3 tree-db) '(3 . Frank) > (lookup 1 tree-db) '(1 . Bill) > (lookup 5 tree-db) #f > (data (lookup 2 tree-db)) 'Joe

For links to all of the SICP lecture notes and exercises that I've done so far, see The SICP Challenge.