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(ql:quickload '(fiveam))
(defpackage :day21
(:use :cl :fiveam))
(in-package :day21)
;;22:55
;;23:32
(defun find-start (field)
(destructuring-bind (rows cols)
(array-dimensions field)
(loop for row below rows do
(loop for col below cols do
(when (eq (aref field row col) #\S)
(return-from find-start (list row col)))))))
(defun solver1 (lines step-count)
(let* ((field (make-array (list (length lines) (length (car lines)))
:initial-contents lines))
pending
(end-positions (make-hash-table :test #'equal))
(visited (make-hash-table :test #'equal))
(start (find-start field)))
(destructuring-bind (rows cols)
(array-dimensions field)
(push (cons step-count start) pending)
(setf (gethash start visited) step-count)
(loop while pending
for (steps-left row col) = (pop pending)
;; an even number of steps left means it is possible
;; to go and return to the place in the steps available
;; Thus count to the solution.
when (evenp steps-left)
do (setf (gethash (list row col) end-positions) t)
unless (zerop steps-left)
do (loop
for (dr dc) in '((-1 0) (0 -1) (1 0) (0 1))
for nr = (+ row dr)
for nc = (+ col dc)
for new-pos = (list nr nc)
when (and (< (gethash new-pos visited -1) (1- steps-left))
(< -1 nr rows)
(< -1 nc cols)
(not (eq #\# (aref field nr nc))))
do (setf (gethash new-pos visited) (1- steps-left))
(push (cons (1- steps-left) new-pos) pending))))
(hash-table-count end-positions)))
(test solutions
(is (= 16 (solver1 (uiop:read-file-lines "eg-in") 6)))
(is (= 3574 (solver1 (uiop:read-file-lines "input") 64))))
(defun solver2 (lines step-count)
(let* ((field (make-array (list (length lines) (length (car lines)))
:initial-contents lines))
pending
(end-positions (make-hash-table :test #'equal))
(visited (make-hash-table :test #'equal))
(start (find-start field)))
(destructuring-bind (rows cols)
(array-dimensions field)
(push (cons step-count start) pending)
(setf (gethash start visited) step-count)
(loop while pending
for (steps-left row col) = (pop pending)
;; an even number of steps left means it is possible
;; to go and return to the place in the steps available
;; Thus count to the solution.
when (evenp steps-left)
do (setf (gethash (list row col) end-positions) t)
unless (zerop steps-left)
do (loop
for (dr dc) in '((-1 0) (0 -1) (1 0) (0 1))
for nr = (+ row dr)
for nc = (+ col dc)
for new-pos = (list nr nc)
when (and (< (gethash new-pos visited -1) (1- steps-left))
(not (eq #\# (aref field (mod nr rows) (mod nc cols)))))
do (setf (gethash new-pos visited) (1- steps-left))
(push (cons (1- steps-left) new-pos) pending))))
(hash-table-count end-positions)))
(solver2 (uiop:read-file-lines "eg-in") 6)
(solver2 (uiop:read-file-lines "eg-in") 10)
(solver2 (uiop:read-file-lines "eg-in") 50)
(solver2 (uiop:read-file-lines "eg-in") 100)
;; (time (solver2 (uiop:read-file-lines "eg-in") 500))
(defun solve-part2 ()
(let ((x 202300))
(multiple-value-bind (a b c)
(solve-quadratic 3719 33190 91987)
(+ (* a x x) (* b x) c))))
(defun solve-quadratic (r-one r-two r-three)
;; The covered area grows proportional to the square of steps taken.
;; The requested numbers of steps (= 26501365 (+ (* 202300 131) 65))
;; The function that takes to the the edge of the field g(x) = 65 + x * 131
;; The standard quadratic function q(x) = a*x^2 + b*x + c
;; r₁ = q o g (0) = c
;; r₂ = q o g (1) = a + b + c
;; r₃ = q o g (2) = 4a + 2b + c
;;
;; solving is simple
;; a = (r₃ - 2r₂ + r₁) / 2
;; b = (4r₂ - 3r₁ - r₃) / 2
;; c = r₁
(values (/ (+ (- r-three (* 2 r-two))
r-one)
2)
(/ (- (* 4 r-two) (* 3 r-one) r-three) 2)
r-one))
(test sols
(loop for steps in (list 65 (+ 65 131) (+ 65 (* 131 2)))
for result in '(3719 33190 91987)
do (is (= result (solver2 (uiop:read-file-lines "input") steps))))
(is (= 600090522932119 (solve-part2))))
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