CS 320 Homework 4 in Scheme
Name_______________________ Complete these exercises and turn in a hard copy of your solutions in class.
Writing and Evaluating Scheme expressions
1. Give a Scheme expression for a list containing the values such that the number 7 is the first element, the symbol a the second, the number -12 the third, and the symbol ??? the fourth.
2. Scheme has a function symbol? that takes one argument and returns #t or #f depending on whether the argument is a symbol or not. Give a Scheme expression that determines whether the first element of a three-element list is a symbol. (The result of the evaluation should be #t.)
3. Suppose that into the Scheme evaluator we enter the definition: (define dozen 12) Give the value of the Scheme expression: (number? dozen)
4. Give two ways to write a Scheme expression of which the value is a one-element list with the empty list as its element.
5. Give a Scheme definition that will determine the sum of 30, 31, 30, 15 and bind the identifier days-in-semester to this sum. Be as concise as possible.
6. Evaluate this Scheme statement. (Do this by hand first; then check your answer by running it.)
((lambda (x) (+ x x)) 4)
7. Evaluate these Scheme statements. (Do this by hand first; then check your answer by running it.)
(define reverse-subtract
(lambda (x y)
(- y x))) (reverse-subtract 7 10)
1
CS320 Homework 4 in Scheme Name________________________
8. These expressions form repetitions. Trace by hand the following loop and show the set of two lists it produces (in other words, the result will be lists within a list).
(let loop
((numbers ¡¯(3 -2 1 6 -5))
(nonneg ¡¯())
(neg ¡¯()))
(cond ((null? numbers)
(list nonneg neg))
((>= (car numbers) 0)
(loop (cdr numbers)
(cons (car numbers) nonneg)
neg))
(else
(loop (cdr numbers)
nonneg
(cons (car numbers) neg)))))
Output:
9. Here is a function called “deleteall” that strips out all occurrences of a value in a given list. The call to the function: (deleteall 0 ‘(4 0 3 0 2 0 1)) results in the list (4 3 2 1).
(define (deleteall atm list)
(cond
( (NULL? list) ‘() )
( (EQ? atm (CAR list)) (deleteall atm (CDR list)))
( ELSE (CONS (CAR list) (deleteall atm (CDR list))))
))
Write a recursive Scheme procedure called “tally” that counts the number of occurrences of a value in a given list. Do not use a counter variable, but let the function result in the desired count. For example, the value of (tally 0 ‘(4 0 3 0 2 0 1)) will be 3. The procedure will be very similar to deleteall.
Lexical Binding (internal definitions)
The binding construct let defines Scheme block structure. In a let expression, the initial values are computed before any of the variables become bound. Evaluate these expressions by hand showing the effects of lexical binding in Scheme.
A. (let ((x 2) (y 3))
(* x y))
B. (let ((x 2) (y 3))
(let ((foo (lambda (z) (+ x y z)))
(x 7))
; call foo
(foo 4)))
Output:
Output:
2