ex-3.16.md

Exercise 3.16. Ben Bitdiddle decides to write a procedure to count the number of pairs in any list structure. “It's easy,” he reasons. “The number of pairs in any structure is the number in the car plus the number in the cdr plus one more to count the current pair.” So Ben writes the following procedure:

(define (count-pairs x)
  (if (not (pair? x))
      0
      (+ (count-pairs (car x))
         (count-pairs (cdr x))
         1)))

Show that this procedure is not correct.

This procedure is not correct, because it counts shared data structures many times. For example, consider the following list:

z-->[o][o]
     |  |
     v  v
x-->[o][o]-->[o][o]-->[o][/]
     |        |        |
     v        v        v
    [a]      [b]      [c]

As the above diagram shows, there are 5 pairs in z. But (count-paris z) returns 7, because it counts x twice.

In particular, draw box-and-pointer diagrams representing list structures made up of exactly three pairs for which Ben's procedure would return 3; return 4; return 7; never return at all.

(define z3 (list 'a 'b 'c))

z3-->[o][o]-->[o][o]-->[o][/]
      |        |        |
      v        v        v
     [a]      [b]      [c]

(define z4
  (let ([x (list 'a)])
    (list x x)))

z4-->[o][o]-->[o][/]
      |        |
      |  ,-----'
      |  |
      v  v
 x-->[o][/]
      |
      v
     [a]

(define z7
  (let* ([x (list 'a)]
         [y (cons x x)])
    (cons y y)))

z7-->[o][o]   car=3, cdr=3, total=7
      |  |
      v  v
 y-->[o][o]   car=1, cdr=1, total=3
      |  |
      v  v
 x-->[o][/]   car=0, cdr=0, total=1
      |
      v
     [a]

; Note that make-cycle is defined in Exercise 3.13.
(define z* (make-cycle (list 'a 'b 'c)))

      ,--------------------.
      |                    |
      v                    |
z*-->[o][o]-->[o][o]-->[o][/]
      |        |        |
      v        v        v
     [a]      [b]      [c]