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Key to CSE431 Sample Final handout #8
1. Scheme expressions.
Part A:
(define a 101)
(define b 103)
(define c 105)
(let ((c (+ a c))
(a (+ b c)))
(list a b c))
'(208 103 206)
Part B:
(define a 101)
(define b 103)
(define c 105)
(let* ((b (+ a b))
(a (+ b c)))
(list a b c))
'(309 204 105)
Part C:
(define x 101)
(define (f n) (+ x y n))
(define y 103)
(define a (f 2))
(define x 105)
(define b (f 2))
(define c a)
(set! a 19)
(list a b c)
'(19 210 206)
Part D:
(define a '(1 2 3))
(define b '(2 3))
(define c (append (cdr b) a))
(define d (list a b))
(list c (eq? a (car d)) (eq? (cdr a) b) (eq? (cdr c) a))
'((3 1 2 3) #t #f #t)
2. One possible solution appears below.
(define (occurrences v lst)
(length (filter (lambda (n) (equal? n v)) lst)))
3. One possible solution appears below.
(define (remove-all v lst)
(filter (lambda (n) (not (equal? n v))) lst))
4. One possible solution appears below.
(define (maximum lst)
(define (explore max lst)
(cond ((null? lst) max)
((> (car lst) max) (explore (car lst) (cdr lst)))
(else (explore max (cdr lst)))))
(if (null? lst)
(error "max of empty list")
(explore (car lst) (cdr lst))))
5. One possible solution appears below.
(define (pattern item)
(cond ((number? item) 'number)
((string? item) 'string)
((symbol? item) 'symbol)
((boolean? item) 'boolean)
((list? item) (map pattern item))
(else item)))
6. One possible solution appears below.
(define (switch-pairs lst)
(cond ((not (list? lst)) (error "not a list"))
((null? lst) lst)
((null? (cdr lst)) lst)
(else (cons (cadr lst) (cons (car lst) (switch-pairs (cddr lst)))))))
7. One possible solution appears below.
(define (overlap lst1 lst2)
(cond ((null? lst1) '())
((null? lst2) '())
((< (car lst1) (car lst2)) (overlap (cdr lst1) lst2))
((> (car lst1) (car lst2)) (overlap lst1 (cdr lst2)))
(else (cons (car lst1) (overlap (cdr lst1) (cdr lst2))))))
8. One possible solution appears below.
(define (histogram lst)
(if (null? lst)
lst
(let ((v (car lst)))
(cons (list v (occurrences v lst))
(histogram (remove-all v lst))))))
9. Ruby expressions.
Part A:
x.each {|n| puts n + 2}
12
13
14
15
16
Part B:
for n in y do
puts n * 3
end
6
9
15
21
33
Part C:
x.map {|n| n % 2 == 0}
[true, false, true, false, true]
10. One possible solution appears below.
def switch_pairs(list)
result = []
i = 0
while i < list.length - 1
result << list[i + 1]
result << list[i]
i += 2
end
if i < list.length
result << list[i]
end
result
end
11. One possible solution appears below.
class Range
def first(n=1)
count = 0
for m in self
break if count >= n
count += 1
yield m
end
end
end
12. One possible solution appears below.
class Array
def pairwise
first = true
for n in self
if first
old = n
first = false
else
if !(yield(old, n))
return false
end
old = n
end
end
true
end
end
13. One possible solution appears below.
class Array
def mode
old = nil
max = 0
max_v = nil
count = 0
for v in self
if v == old
count += 1
else
if count > max
max = count
max_v = old
end
old = v
count = 1
end
end
if count > max
[old, count]
else
[max_v, max]
end
end
end
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