Summary: In this laboratory, you will ground your understanding of the basic techniques for naming values and procedures in Scheme,
What are the values of the following
You may use DrScheme to help you answer these questions, but be sure you
can explain how it arrived at its answers.
(let ((tone "fa") (call-me "al")) (list call-me tone "l" tone))
;; Solutions to the quadratic equation x^2 - 5x + 4: (let ((discriminant (- (* -5 -5) (* 4 1 4)))) (list (/ (+ (- -5) (sqrt discriminant)) (* 2 1)) (/ (- (- -5) (sqrt discriminant)) (* 2 1))))
(let ((total (+ 8 3 4 2 7))) (let ((mean (/ total 5))) (* mean mean)))
Write a nested
let-expression that binds a total of five
bound to 9387 and each subsequent name bound to a value twice
as large as the one before it --
beta should be
twice as large as
gamma twice as
beta, and so on. The body of the innermost
let-expression should then compute the sum of the values
of the five names.
let*-expression equivalent to the
let-expression in the previous exercise.
Consider the following procedure that squares all the values in a list.
;;; Procedure: ;;; square-values ;;; Parameters: ;;; lst, a list of numbers of the form (num_1 num_2 ... num_n) ;;; Purpose: ;;; Squares all the values in lst. ;;; Produces: ;;; list-of-squares, a list of numbers ;;; Preconditions: ;;; [Standard] ;;; Postconditions: ;;; list-of-squares has the form (square_1 square_2 ... square_n) ;;; For all i, square_i is the square of num_i (that is num_i * num_i). (define square-values (lambda (lst) (let ((square (lambda (val) (* val val)))) (if (null? lst) null (cons (square (car lst)) (square-values (cdr lst)))))))
a. Verify that
square-values works correctly.
b. Try to execute
square outside of
Explain what happens.
Here is a procedure that takes a non-empty list of lists as an argument and returns the longest list in the list (or one of the longest lists, if there is a tie).
;;; Procedure: ;;; longest-list-in-list ;;; Parameters: ;;; los, a list of lists ;;; Purpose: ;;; Finds one of the longest lists in los. ;;; Produces: ;;; longest, a list ;;; Preconditions: ;;; los is a nonempty list. ;;; every element of los is a list. ;;; Postconditions: ;;; Does not affect los. ;;; Returns an element of los. ;;; No element of los is longer than longest. (define longest-list-in-list (lambda (los) ; If there is only one list, that list must be the longest. (if (null? (cdr los)) (car los) ; Otherwise, take the longer of the first list and the ; longest remaining list. (longer-list (car los) (longest-list-in-list (cdr los))))))
This definition of the
includes a call to the
longer-list procedure, which returns
the longer of two given lists:
;;; Procedure: ;;; longer-list ;;; Parameters: ;;; left, a list ;;; right, a list ;;; Purpose: ;;; Find the longer of left and right. ;;; Produces: ;;; longer, a list ;;; Preconditions: ;;; Both left and right are lists. ;;; Postconditions: ;;; longer is a list. ;;; longer is either equal to left or to right. ;;; longer is at least as long as left. ;;; longer is at least as long as right. (define longer-list (lambda (left right) (if (<= (length right) (length left)) left right)))
Revise the definition of
longest-list-in-list so that the
longer-list is bound to the procedure that it denotes
only locally, in a
Note that there are at least two possible ways to do the previous exercise:
The definiens of
longest--list-in-list can be a
let-expression as its body, or it can be a
let-expression with a
lambda-expression as its
body. That is, it can take the form
(define longest-list-in-list (let (...) (lambda (los) ...)))
or the form
(define longest-list-in-list (lambda (los) (let (...) ...)))
longest-list-in-list in whichever way that you did not
define it for the previous exercise.
b. Does the order of nesting affect what happens when the procedure is invoked? If so, which arrangement is better? Why?
The two definitions you came up with in the previous exercises
are not the only alternatives you have in placing the
longer-list is only needed in the recursive case,
you can place the
(define longest-list-in-list (lambda (los) ; If there is only one list, that list must be the longest. (if (null? (cdr los)) (car los) ; Otherwise, take the longer of the first list and the ; longest remaining list. (let ((longer-list (lambda (left right) (if (<= (length right) (length left)) left right)))) (longer-list (car los) (longest-list -in-list (cdr los))))))
Including the original definition, you've now seen or written four
longest-list-in-list. Which do you prefer?
Extend your favorite version of
that it verifies its preconditions (i.e., that
contains lists and that
los is nonempty). If the
preconditions are not met, your procedure should return
It is perfectly acceptable for you to check each list element in turn to determine whether or not it is a list, rather than to check them all at once, in advance.
(time exp) operation, determine
which of the four versions of
indeed the fastest. You should try a variety of lengths for the
outer list. The inner lists can be mostly 0- or 1-element lists..
No notes yet.
Monday, 2 October 2000 [Samuel A. Rebelsky]
Wednesday, 28 February 2001 [Samuel A. Rebelsky]
Thursday, 1 March 2001 [Samuel A. Rebelsky]
awas too frequently read as an article rather than a name.
Monday, 7 October 2002 [Samuel A. Rebelsky]
Tuesday, 8 October 2002 [Samuel A. Rebelsky]
Wednesday, 9 October 2002 [Samuel A. Rebelsky]
Sunday, 2 February 2003 [Samuel A. Rebelsky]
Thursday, 9 September 2003 [Samuel A. Rebelsky]
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