Sunday, April 10, 2016

Project Euler Problem 48 Common Lisp

Project Euler Problem 48

(defun problem48 ()
  (nth-value 1 (truncate (loop for i from 1 to 1000
                               sum (expt i i))
                         (expt 10 10))))

Project Euler Problem 49 Common Lisp

Project Euler Problem 49

(defun digits-to-list (x &optional (lst nil))
  (multiple-value-bind (a b)
      (truncate x 10)
    (if (zerop a)
      (cons b lst)
      (digits-to-list a (cons b lst)))))

(defun prime? (num)
  (cond ((= 2 num) num)
    ((= 3 num) num)
    ((< num 1) nil)
    ((evenp num) nil)
    ((zerop (mod num 3)) nil)
    (t (loop for i from 5 to (isqrt num) by 6
        if (or (zerop (mod num i))
               (zerop (mod num (+ i 2)))) return nil
        finally (return num)))))

(defun primes (num)
  (append '(2 3)
      (loop for i from 5 to  num by 6
           if (prime? i) collect it
           if (prime? (+ i 2)) collect it)))

(defun perm? (x y z)
  (let ((a (sort (digits-to-list x) #'<))
    (b (sort (digits-to-list y) #'<))
    (c (sort (digits-to-list z) #'<)))
    (and (equal a b) (equal b c))))

(defun four-digit-perms ()
  (loop with primes2 = (delete 1000 (primes 9999) :test #'>)
       for a in primes2
       nconc (loop for b in primes2
          nconc (loop for c in primes2
                 while (> a b c)
                 when (= (- a b) (- b c))
                 when (perm? a b c)
                 collect (list c b a)))))

(defun problem49 ()
  (format nil "The answer to Project Euler problem 49 is ~{~a~}"
      (remove-if-not #'stringp
             (mapcar #'write-to-string
                 (second (four-digit-perms))))))

Project Euler Problem 50 Common Lisp

Project Euler Problem 50

(defun prime? (num)
  (cond ((= 2 num) num)
    ((= 3 num) num)
    ((< num 1) nil)
    ((evenp num) nil)
    ((zerop (mod num 3)) nil)
    (t (loop for i from 5 to (isqrt num) by 6
        if (or (zerop (mod num i))
               (zerop (mod num (+ i 2)))) return nil
        finally (return num)))))

(defun primes (num)
  (append '(2 3)
      (loop for i from 5 to  num by 6
           if (prime? i) collect it
           if (prime? (+ i 2)) collect it)))

(defun prime-sums (num)
  (loop for i in (primes num)
     sum i into j
     unless (< num j)
     collect j))

(defun sum-of-consecutive-primes (num)
  (loop with sums = (prime-sums num)
       for i in sums
       collect (member-if #'prime?
              (loop for j in (cons 0 sums)
                   while (> i j)
                   collect (- i j)))))

(defun longest-sum-of-consecutive-primes (num)
  (let ((x (sort (sum-of-consecutive-primes num)

                 #'>
                 :key #'length)))
    (list (caar x) (length (first x)))))

(defun problem50 ()
  (let ((x (longest-sum-of-consecutive-primes 1000000)))
    (format t "The answer to Problem 50 is ~A with ~A terms."

            (first x) (second x))))

Project Euler Problem 46 Common Lisp

Project Euler Problem 46

Note that all square integers from 1 to some number can be found by summing all the odd numbers, e.g. 2 *2 = 4 = 1 + 3, 3 * 3 = 9 = 1 + 3 + 5, 4 * 4 = 16 = 1 + 3 + 5 + 7.  That's what is happening in function conjecture-false?

(defun prime? (num)
  (cond ((= 2 num) num)
    ((= 3 num) num)
    ((< num 1) nil)
    ((evenp num) nil)
    ((zerop (mod num 3)) nil)
    (t (loop for i from 5 to (isqrt num) by 6
        if (or (zerop (mod num i))
               (zerop (mod num (+ i 2)))) return nil
        finally (return num)))))

(defun conjecture-false? (num)
  (loop for i from 1 to num by 2
       sum i into j
       if (prime? (- num (* j 2))) return nil
       finally (return t)))

(defun problem46 ()
  (loop for i upfrom 1 by 2
       unless (prime? i)
       when (conjecture-false? i) return i))

Monday, February 22, 2016

Project Euler Problem 67 Common Lisp

Project Euler Problem 67

(defparameter *data*
  (with-open-file (input "p067_triangle.txt")
    (loop for line = (read-line input nil nil)
          while line
          collect (read-from-string

                    (concatenate 'string "(" line ")")))))

The above code for converting lines in a text file into a nested list came from here.

(defun bottom-up-max (a b)
  (mapcar #'max
          (mapcar #'+ a b)
          (mapcar #'+ (rest a) b)))

(defun problem67 (lst)
  (first (reduce #'bottom-up-max (reverse lst))))


To get the solution, evaluate:

(problem67 *data*)

Sunday, February 21, 2016

Project Euler Problem 18 Common Lisp

Project Euler Problem 18

The idea for the solution below came from Stackoverflow.  Working from "bottom to top", *data* is reversed in function problem18.

 (defparameter *data*
                   '((75)
                    (95 64)
                    (17 47 82)
                    (18 35 87 10)
                    (20 04 82 47 65)
                    (19 01 23 75 03 34)
                    (88 02 77 73 07 63 67)
                    (99 65 04 28 06 16 70 92)
                    (41 41 26 56 83 40 80 70 33)
                    (41 48 72 33 47 32 37 16 94 29)
                    (53 71 44 65 25 43 91 52 97 51 14)
                    (70 11 33 28 77 73 17 78 39 68 17 57)
                    (91 71 52 38 17 14 91 43 58 50 27 29 48)
                    (63 66 04 68 89 53 67 30 73 16 69 87 40 31)
                    (04 62 98 27 23 09 70 98 73 93 38 53 60 04 23)))

(defun bottom-up-max (a b)
  (mapcar #'max
          (mapcar #'+ a b)
          (mapcar #'+ (rest a) b)))

(defun problem18 (lst)
  (first (reduce #'bottom-up-max (reverse lst))))


To get the solution, evaluate:
 
(problem18 *data*)

Friday, February 19, 2016

Project Euler Problem 11 Common Lisp

Project Euler Problem 11

(defparameter *data*
  '((08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08)
    (49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00)
    (81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65)
    (52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91)
    (22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80)
    (24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50)
    (32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70)
    (67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21)
    (24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72)
    (21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95)
    (78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92)
    (16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57)
    (86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58)
    (19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40)
    (04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66)
    (88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69)
    (04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36)
    (20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16)
    (20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54)
    (01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48)))

(defun multiply-first-4 (lst)
  (if (> 4 (length lst))
      0
      (reduce #'* (subseq lst 0 4))))

(defun all-multiples (lst)
  (maplist #'multiply-first-4 lst))

(defun all-horizontals (lst)
  (mapcan #'all-multiples lst))

(defun all-verticals (lst)
  (all-horizontals (apply #'mapcar #'list lst)))

(defun %all-left-diagonals (lst &optional (n 0))
  (if (< n 4)
      (cons (nthcdr n (car lst))
            (%all-left-diagonals (cdr lst) (1+ n)))
      nil))

(defun all-left-diagonals (lst)
  (mapcon #'(lambda (x) (all-verticals (%all-left-diagonals x)))
          lst))

(defun all-right-diagonals (lst)
  (all-left-diagonals (reverse lst)))

(defun problem11 ()
  (reduce #'max (append (all-horizontals *data*)
                        (all-verticals *data*)
                        (all-left-diagonals *data*)
                        (all-right-diagonals *data*))))



Function multiply-first-4 returns the multiplication of the first 4 elements of a list, or it returns 0 if the list has less than 4 elements--otherwise the function will error out.  Better to just simply return 0 then the code complexity for handling a list with less than 4 elements.

Function all-multiples takes a list of say (1 2 3 4 5 6 7) turning it into ((1 2 3 4 5 6 7) (2 3 4 5 67) (3 4 5 6 7) (4 5 6 7) (5 6 7) (6 7) (7)).  It then takes that and applies multiply-first-4 to each of the lists within, resulting in (24 120 360 840 0 0 0).

Function all-horizontals applies all-multiples to a list of list, like *data* above, returning a flat list of all multiplications of 4 elements from left to right (and right to left simultaneously).

Function all-verticals takes a list of list and transposes it.  For example ((1 2 3) (4 5 6) ( 7 8 9)) becomes ((1 4 7) (2 5 8) (3 6 9)).  all-verticals applies that transposed list to all-horizontals.  The result is a flat list of all multiplications of 4 elements going up and down.  Transposing the list is a trick taken from Stackoverflow.


Function all-left-diagonals uses %all-left-diagonals to turn a list like ((1 2 3) (4 5 6) (7 8 9)) into ((1 2 3) (5 6) (9)) and applies it to all-verticals.  The result is a flat list of all multiplications of 4 elements going diagonal, from upper left to lower right.

Function all-right-diagonals reverses a list of lists, ((1 2 3) (4 5 6) (7 8 9)) becoming ((7 8 9) (4 5 6) (1 2 3)), and applies it to all-left-diagonals.  Thus resulting in a flat list of all multiplications of 4 elements going diagonal, from lower left to upper right.

Function problem11 takes all of these flat lists, appends them, and then finds the maximum.