CS150 Notes 7 (7 September 2005)

Upcoming Deadlines

• Friday, 9 September: PS2

1. Be optimistic.
   • Assume you can solve it.
   • If you could, how would you solve a bigger problem.
2. Think of the simplest version of the problem, something you can already solve. (This is the base case.)
3. Combine them to solve the problem.

1. Be very optimistic. Since lists themselves are recursive structures, we should be able to come up with a recursive definition for almost anything we can think of doing with a list.
   1. Assume we can solve a smaller version of the problem.
   2. Break the problem into the original list and its cdr.
2. Think of the simplest version of the problem, something you can solve already. For lists, this is usually the null list. (Sometimes, it might be the length 1 list.)
3. Combine them to solve the problem. For lists, we will usually do combine the result of the recursive evaluation on the cdr with the result of doing something with the car of the list.

(define (listproc lst)
  (if (null? lst)
      [[[ insert base case here ]]]
      ([[[ f ]]] (car lst) (listproc (cdr lst)))))

(define (sumlist lst)
  (if (null? lst) 
      0 
      (+ (car lst) (sumlist (cdr lst)))))

(define (insertl lst f stopval)
  (if (null? lst)
      stopval
      (f (car lst) 
	 (insertl (cdr lst) f stopval))))

(define (sumlist lst) (insertl lst + 0))

(define (productlist lst) (insertl lst * 1))

(define (length lst) 
  (insertl lst (lambda (head rest) (+ 1 rest)) 0))

(define (remove cf lst)
   ;; operands: cf is a comparison function that takes one operand and
   ;;              evaluates to #t or #f
   ;;           lst is a list of values
   ;; result: evaluates to a list containing the only elements of lst for
   ;;            which (cf ) evaluates to #f
   (if (null? lst)  _________________________
       (if (cf (car lst))
 
           ______________________________    ;; discard first element


           ______________________________))) ;; keep first element

Could you define remove using just an insertl application? (Explain how or why it is impossible.)

(define (filter cf lst)
   ;; operands: cf is a comparison function that takes one operand and
   ;;              evaluates to #t or #f
   ;;           lst is a list of values
   ;; result: evaluates to a list containing the only elements of lst for
   ;;            which (cf ) evaluates to #t





   )

Challenge: Define a procedure (intsto n) that evaluates to a list containing the integers less than or equal to n in order. For example, (intsto 3) ==> (1 2 3) and (intsto 0) ==> (). This is pretty tricky. Try your definition in DrScheme.

In mathematics you don't understand things, you just get used to them.
John Von Neumann