euler 42 pandigitals with sub string divisibility

Mar 25, 2015  

These brute force solutions are getting a bit worrying, but here is another one. (I keep promising myself to get out of my comfort zone, but (unfortunately!) code like this is too easy to write).

Statutory Warning: Spoilers ahead

;; For all 0-9 pandigital numbers, find the ones which have successive
;; substrings of length 3 diisible by 2,3,...,17.

(defparameter *prime-divisors* #(17 13 11 7 5 3 2))

;; A number is 0-9 pandigital if it (a) is 10 digits long, and (b) has every digit from 0-9
(defun is-pandigital (num)
  (declare (type fixnum num))
  (let ((digits-seen (make-array 10 :element-type 'bit :initial-element 0)))
    (do* ((n num (floor (/ n 10)))
          (d (mod n 10) (mod n 10))
          (num-digits 0 (1+ num-digits)))
         ((= n 0) (and (= num-digits 10)
                       (every (lambda (x) (= x 1)) digits-seen)))
      (setf (bit digits-seen d) 1))))

(defun check-divisibility (num div-index)
  (= 0 (mod num (aref *prime-divisors* div-index))))

(defun get-three-digit-num (num)
  (let ((ones (mod num 10))
        (tens (mod (floor (/ num 10)) 10))
        (hundreds (mod (floor (/ num 100)) 10)))
    (+ ones
       (* 10 tens)
       (* 100 hundreds))))

;; Go backwards in groups of three digits and check divisibility
(defun has-divisible-substrings (num)
  (do* ((n num (floor (/ n 10)))
        (div-index 0 (1+ div-index))
        (dividend (get-three-digit-num n) (get-three-digit-num n))
        (div-p (check-divisibility dividend div-index)
               (and div-p (check-divisibility dividend div-index))))
       ((< n 10000) div-p)))

(defun check-substring-pandigital-range (start end)
  (declare (type fixnum start end))
  (let ((candidates '()))
    (loop for num from start to end do
         (if (and (is-pandigital num)
                  (has-divisible-substrings num))
             (push num candidates)))

(defun euler43 ()
  (let ((candidates (check-substring-pandigital-range 1234567890 9876543210)))
    (print candidates)
    (apply #'+ candidates)))

As before, evaluating (euler43) shows the final solution (sum), along with the (six!) candidate numbers making up the sum. This took a whopping 4.3 hours to churn through the 8.5 billion numbers. Maybe I need to create a new constraint for these problems: either pen-and-paper, or something slow (like Python ?! :P), so that brute force is never tempting again.