QUIZ #1 SOLUTIONS

1. min

One way:

(define (min sent)
  (cond ((empty? (bf sent))
         (first sent) )
        ((< (first sent) (first (bf sent)))
         (min (se (first sent) (bf (bf sent)))) )
        (else (min (bf sent))) ))

That works, but I think that isn't as clear as:

(define (min sent)
  (define (min-helper sent curmin)
    (cond ((empty? sent)
           curmin)
          ((< (first sent) curmin)
           (min-helper (bf sent)
                       (first sent)) )
          (else (min-helper (bf sent)
                            curmin) )))
  (min-helper (bf sent) (first sent)) )

Or, a slightly briefer version of the above:

(define (min sent)
  (define (min-helper sent curmin)
    (if (empty? sent)
        curmin
        (min-helper (bf sent)
                    (if (< (first sent) curmin)
                        (first sent)
                        curmin) )))
  (min-helper (bf sent) (first sent)) )



2. Given the following:

(define (foo x) (+ x 1))
(define (bar x) (- x 1))
(define (baz x)
  (let ((foo (lambda (x) (* x 2)))
        (bar foo))
    (foo (bar x)) ))

What does the following produce?

(foo 10) returns 11
(bar 10) returns 9
(baz 10) returns 22

If you do not understand why (baz 10) is 22, review the first problem from
Lab 2b.  See also:
    http://www.csua.berkeley.edu/~jameslin/cs61a/week02.shtml