#lang scheme/base
;;; PLT Scheme Inference Collection
;;; towers-alist.ss
;;;
;;; Towers of Hanoi from Artificial Intelligence: Tools, Techniques,
;;; and Applications, Tim O'Shea and Marc Eisenstadt, Harper & Rowe,
;;; 1984, pp.45
;;;
;;; The rules of the game are: (1) move one ring at a time and (2)
;;; never place a larger ring on top of a smaller ring.  The object
;;; is to transfer the entire pile of rings from its starting
;;; peg to either of the other pegs - the target peg.

(require (planet williams/inference/inference))

(define-ruleset towers-rules)

;;; If the target peg hld all the rings 1 to n, stop because according
;;; to game rule (2) they must be in their original order and so the
;;; problem is solved.
(define-rule (rule-1 towers-rules)
    (all (ring ? (on . right)))
  ==>
    (printf "Problem solved!~n")
    (succeed))

;;; If there is no current goal - that is, if a ring has just been
;;; successfully moved, or if no rings have yet to be moved - generate
;;; a goal. In this case the goal is to be that of moving to the
;;; target peg the largest ring that is not yet on the target peg.
(define-rule (rule-2 towers-rules)
    (no (move . ?))
    (ring ?size (on ?peg (not (eq? ?peg 'right))))
    (no (ring (?size-1 (> ?size-1 ?size))
              (on ?peg-1 (not (eq? ?peg-1 'right)))))
  ==>
    (assert `(move (size . ,?size) (from . ,?peg) (to . right))))

;;; If there is a current goal, it can be achieved at once of there is
;;; no small rings on top of the ring to be moved (i.e. if the latter
;;; is at the top of its pile), and there are no small rings on the
;;; peg to which it is to be moved (i.e. the ring to be moved is
;;; smaller that the top ring on the peg we intend to move it to). If
;;; this is the case, carry out the move and then delete the current
;;; goal so that rule 2 will apply next time.
(define-rule (rule-3 towers-rules)
    (?move <- (move (size . ?size) (from . ?from) (to . ?to)))
    (?ring <- (ring ?size (on . ?from)))
    (no (ring (?size-1 (< ?size-1 ?size)) (on . ?from)))
    (no (ring (?size-2 (< ?size-2 ?size)) (on . ?to)))
  ==>
    (printf "Move ring ~a from ~a to ~a.~n" ?size ?from ?to)
    (replace ?ring `(ring ,?size (on . ,?to)))
    (retract ?move))

;;; If there is a current goal but its disc cannot be moved as in rule
;;; 3, set up a new goal: that of moving the largest of the obstructing
;;; rings to the peg that is neither of those specified in the current
;;; goal (i.e. well out of the way of the current goal). Delete the
;;; current goal, so that rule 2 will apply to the new goal next time.
(define-rule (rule-4 towers-rules)
    (?move <- (move (size . ?size) (from . ?from) (to . ?to)))
    (peg (?other (not (memq ?other (list ?from ?to)))))
    (ring (?size-1 (< ?size-1 ?size))
          (on ?peg-1 (not (eq? ?peg-1 ?other))))
    (no (ring (?size-2 (< ?size-1 ?size-2 ?size))
              (on ?peg-2 (not (eq? ?peg-2 ?other)))))
  ==>
    (replace ?move `(move (size . ,?size-1) (from . ,?peg-1) (to . ,?other))))

;;; The main routine:
;;; In a new inference environment:
;;;   Activate the towers rule set.
;;;   Optionally, turn on tracing.
;;;   Create the three pegs - left, middle, and right.
;;;   Create the n rings.
;;;   Start the inference.
;;; The rules will print the solution to the problem.
(define (solve-towers n)
  (with-new-inference-environment
   (activate towers-rules)
   ;(current-inference-trace #t)
   ;; Create pegs.
   (assert '(peg left))
   (assert '(peg middle))
   (assert '(peg right))
   ;; Create rings.
   (for ((i (in-range 1 (+ n 1))))
     (assert `(ring ,i (on . left))))
   ;; Start inferencing.
   (start-inference)))

;;; Test with 6 disks.
(solve-towers 6)