*Noah Friedman*22 Apr 2001 23:27 UTC

Sorry to be bringing up such an old subject, but I ran across this thread as the result of a web search and couldn't help replying. Before quoting the following, let me remind you that Harvey Stein proposed that the "generalized length" of a list should be defined as the number of elements, not the number of times you could apply cdr to the list. On Feb 22 1999, Doug Currie wrote the following: >At 9:35 PM +0200 2/22/99, Harvey J. Stein wrote: >>... >>However, I don't think a generalized length is such a good idea. It >>implies a substantial amount of additional overhead on computing the >>length of a list. This overhead remains no matter what you return for >>circular lists - it's the act of detecting them which adds the >>overhead. > >Not necessarily. The length function must compute a count. At some point >this count will exceed a (implementation dependent) limit, e.g., the size >of the heap, when it becomes clear that the list is circular. For >non-circular lists this adds little overhead, just one integer compare per >iteration (in a clever low level implementation, it may add no overhead if >it's combined with the detection of fixnum overflow to switch to bignum >addition). For circular lists, this may take a while to detect, but on >today's fast processors with lists that fit in cache it can be quite >competitive with more complex approaches that allocate memory or use N*N*N >algorithms. Using these counting techniques a list-length can be provided >competitive in performance with a non-circular-list-detecting version. >However, this approach doesn't provide a result for circular list more >useful than 0 or #f. I think you can compute the number of distinct elements in a circular list in linear time and constant space without resorting to any implemented-defined behavior of the runtime environment. I defined the function below in MIT Scheme and evaluated the following: (define l '(1 2 3 4 5 6 7)) (set-cdr! (list-tail l 6) (list-tail l 3)) (circular-list-length l) ;Value: 7 (circular-list-length '(1 2 3)) ;Value: 3 (circular-list-length '()) ;Value: 0 Here's the function definition, which works for non-circular lists as well.

(define (circular-list-length lst) "Return the number of distinct elements of circular list LST." (let ((tortoise lst) (hare lst) (tortoise-advance #t) (len 0)) ;; Find a member of the list guaranteed to be within the cycle, and ;; compute length if list turns out to be non-circular. (do () ((null? hare)) (set! hare (cdr hare)) (set! len (1+ len)) (set! tortoise-advance (not tortoise-advance)) (if tortoise-advance (set! tortoise (cdr tortoise))) (if (eq? hare tortoise) (begin (set! hare '()) (set! len 0)))) (if (and (not (null? lst)) (zero? len)) (begin ;; Find period of cycle. (set! hare (cdr tortoise)) (set! len 1) (do () ((eq? hare tortoise)) (set! hare (cdr hare)) (set! len (1+ len))) ;; Give hare a head start from the start of the list equal to the ;; loop size. If both move at the same speed they must meet at ;; the nexus because they are in phase, i.e. when tortoise enters ;; the loop, hare must still be exactly one loop period ;; ahead--but that means it will be pointing at the same list ;; element. (set! tortoise lst) (set! hare (list-tail lst len)) (do () ((eq? tortoise hare)) (set! hare (cdr hare)) (set! tortoise (cdr tortoise)) (set! len (1+ len))))) len))