Vector 如何求解方案中的n皇后
我正试图解决方案中的n皇后问题。我的教授告诉我使用一个向量作为棋盘,向量的第I个元素代表棋盘的第I列。该元素的值是皇后所在的行,如果列为空,则为-1。因此,[012-1-1]有两列没有皇后,三列皇后被非法放置。 当我运行这个代码时:(place-n-queens044(-1-1-1-1-1)),我得到了#(0123),显然这四个queens都是非法放置的。我认为问题在于我没有在cond-in-place-queen-on-n中检查足够的内容,但我不确定要添加什么来解决将queen放在同一对角线上的问题Vector 如何求解方案中的n皇后,vector,scheme,n-queens,Vector,Scheme,N Queens,我正试图解决方案中的n皇后问题。我的教授告诉我使用一个向量作为棋盘,向量的第I个元素代表棋盘的第I列。该元素的值是皇后所在的行,如果列为空,则为-1。因此,[012-1-1]有两列没有皇后,三列皇后被非法放置。 当我运行这个代码时:(place-n-queens044(-1-1-1-1-1)),我得到了#(0123),显然这四个queens都是非法放置的。我认为问题在于我没有在cond-in-place-queen-on-n中检查足够的内容,但我不确定要添加什么来解决将queen放在同一对角线上
(define (return-row vector queen)
(vector-ref vector (return-col vector queen)))
(define (return-col vector queen)
(remainder queen (vector-length vector)))
(define (checkrow vector nq oq)
(cond
((= (vector-ref vector nq) -1) #f)
((= (vector-ref vector oq) -1) #f)
(else (= (return-row vector nq) (return-row vector oq)))))
(define (checkcol vector nq oq)
(= (return-col vector nq) (return-col vector oq)))
(define (checkdiagonal vector nq oq)
(cond
((= (vector-ref vector nq) -1) #f)
((= (vector-ref vector oq) -1) #f)
(else (= (abs (- (return-row vector nq) (return-row vector oq)))
(abs (- (return-col vector nq) (return-col vector oq)))))))
(define (checkdiagonalagain vector r c oq)
(= (abs (- r (return-row vector oq)))
(abs (- c (return-col vector oq)))) )
(define (checkrowagain vector r oq)
(= r (return-row vector oq)))
(define (checkinterference vector nq oq)
(or (checkrow vector nq oq) (checkcol vector nq oq) (checkdiagonal vector nq oq)))
(define (place-queen-on-n vector r c)
(local ((define (foo x)
(cond
((checkrowagain vector r x) -1)
((= c x) r)
((checkinterference vector c x) -1)
((map (lambda (y) (eq? (vector-ref vector x) y))
(build-list (vector-length vector) values)) (vector-ref vector x))
((eq? (vector-ref vector x) -1) -1)
(else -1))))
(build-vector (vector-length vector) foo)))
(define (place-a-queen vector)
(local ((define (place-queen collist rowlist)
(cond
((empty? collist) '())
((empty? rowlist) '())
(else (append (map (lambda (x) (place-queen-on-n vector x (car collist))) rowlist)
(try vector (cdr collist) rowlist)))
)))
(place-queen (get-possible-col vector) (get-possible-row (vector->list vector) vector))))
(define (try vector collist rowlist)
(cond
((empty? collist) '())
((empty? rowlist) '())
(else (append (map (lambda (x) (place-queen-on-n vector x (car collist))) rowlist)
(try vector (cdr collist) rowlist)))))
(define (get-possible-col vector)
(local ((define (get-ava index)
(cond
((= index (vector-length vector)) '())
((eq? (vector-ref vector index) -1)
(cons index (get-ava (add1 index))))
(else (get-ava (add1 index))))))
(get-ava 0)))
;list is just vector turned into a list
(define (get-possible-row list vector)
(filter positive? list)
(define (thislist) (build-list (vector-length vector) values))
(remove* list (build-list (vector-length vector) values))
)
(define (place-n-queens origination destination vector)
(cond
((= origination destination) vector)
(else (local ((define possible-steps
(place-n-queens/list (add1 origination)
destination
(place-a-queen vector))))
(cond
((boolean? possible-steps) #f)
(else possible-steps))))))
(define (place-n-queens/list origination destination boards)
(cond
((empty? boards) #f)
(else (local ((define possible-steps
(place-n-queens origination destination (car boards))))
(cond
((boolean? possible-steps) (place-n-queens/list origination destination (cdr boards)))
(else possible-steps))
))))
我们非常感谢您的帮助,让这项工作顺利进行 这很难理解。一般来说,n-queens是通过某种回溯来完成的,我不知道你在哪里回溯。困难的部分是管理使用向量时的副作用。在返回之前,必须将电路板设置为先前的状态
(define (n-queens size)
(let ((board (make-vector size -1)))
(let loop ((col 0) (row 0))
(cond ((= col size) board)
((= row size) ;;dead end
(if (= col 0) ;;if first collumn
#f ;;then no solutions
(begin (vector-set! board (- col 1) -1))
#f)))
;;else undo changes made by previous level and signal the error
((safe? col row board)
(vector-set! board col row)
(or (loop (+ col 1) 0)
;;only precede to next column if a safe position is found
(loop col (+ row 1))))
;; keep going if hit a dead end.
(else (loop col (+ row 1)))))))
书写安全吗?但这取决于你
也不确定为什么要从向量移动到列表。这真的把逻辑搞乱了,所以我跟不上。另外,你应该能够自己轻松地通过向量。在place-queen-on-n
中,您可以在向量上使用构建列表,以便在向量上进行映射。
而某种向量折叠可能更合适。此外,map将始终返回一个列表,该列表始终不为false,这意味着cond中之后的任何代码都不会被命中。这是你的问题吗,我不知道,但这是个问题