6.001 Recitation – Oct 2, 2002

RI: Konrad Tollmar

• Pop Quiz

0. ‘(say your name)

1. Warm-up

(define (three) 3)

;Value: "three --> #[compound-procedure 2 three]"

three

;Value: #[compound-procedure 2 three]

(three)

;Value: 3

(define four 4)

;Value: "four --> 4"

four

;Value: 4

(four)

;The object 4 is not applicable.

(+ (three) four)

;Value: 7

(define (add-x x)

(lambda (y) (+ y x)))

;Value: "add-x --> #[compound-procedure 3 add-x]"

(define add-10 (add-x 10))

;Value: "add-10 --> #[compound-procedure 4]"

(add-10 5)

;Value: 15

2. Procedures

Given the procedure,

(define apply-to-3-and-4

(lambda (p) (p 3 4)))

What choices for <exp> in a combination of the form (apply-to-3-and-4 <exp>) will cause the following numbers to be returned?

a. 7 - (apply-to-3-and-4 +)

b. 1 -

(apply-to-3-and-4 (lambda (x y) (- y x))), or (apply-to-3-and-4 (lambda (x y) 1) ;-)

c. 2 -

(apply-to-3-and-4 (lambda (x y) (/ y 2)))

d. 3 -

(apply-to-3-and-4 (lambda (x y) x))

e. 4 -

(apply-to-3-and-4 (lambda (x y) y))

3. Recursive procedures

Someone, by misstake i guess, redefined the standard + and – procedures (among many other procedures). Write a new recursive add procedures by only using the =, inc and dec procedures.

(define (add a b)

(if (= a 0)

b

(inc (add (dec a) b))))

4. List procedure

Write a process version of a procedure called remove that takes a number and a list and returns a new version of the list where that number is no longer an element.

(define (remove elm lst) 


 (if (null? lst)


  nil


 (if (= elm (car lst))


  (remove elm (cdr lst))


  (cons (car lst) (remove elm (cdr lst)))))) 

 

or,

 
(define (remove elm lst) (filter (lambda (x) (not (= x elm))) lst))