你的代码
(defun find! ()
(found? 0 ;; initially show the number 1
'(1 2 3) ;; initial list
'() ;; initially no numbers found
3 ;; numbers list width is 3
) )
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ( (j 1 (1+ j) ) )
( (> j 9) lst)
(unless (some (lambda (x) (= x j)) occupied)
(setf (nth index lst) j)
(push j occupied)
(if (found? (1+ index) lst occupied width) ;; recursion
lst
(setf occupied (remove j occupied)))))
(do ( (j 1 (1+ j) ) )
( (> j 9) lst)
(unless (some (lambda (x) (= x j)) occupied)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+ (mapcar (lambda (x y) (* x y))
'(1000 100 10 1)
(list (third lst) (first lst) (first lst) (second lst))
))))
(if (= lefthnd rghthnd)
lst
'nil))))))
缩进和注释样式:行尾注释使用单个分号,对齐非正文参数,正文缩进两个空格
(defun find! ()
(found? 0 ; initially show the number 1
'(1 2 3) ; initial list
'() ; initially no numbers found
3)) ; numbers list width is 3
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (some (lambda (x) (= x j)) occupied)
(setf (nth index lst) j)
(push j occupied)
(if (found? (1+ index) lst occupied width) ; recursion
lst
(setf occupied (remove j occupied)))))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (some (lambda (x) (= x j)) occupied)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar (lambda (x y) (* x y))
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(if (= lefthnd rghthnd)
lst
'nil))))))
使用更有说服力的谓词:find 或 member。不要将 * 包装在 lambda 中,不做其他任何事情。(我将把 find 放在一边!以后。)
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(push j occupied)
(if (found? (1+ index) lst occupied width) ; recursion
lst
(setf occupied (remove j occupied)))))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(if (= lefthnd rghthnd)
lst
'nil))))))
a 的主体do
不返回任何内容。有很多死代码,我们现在删除:
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(push j occupied)
(unless (found? (1+ index) lst occupied width) ; recursion
(setf occupied (remove j occupied)))))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)))))
我们可以有条件地推送,而不是推送然后有条件地移除:
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(when (found? (1+ index) lst occupied width) ; recursion
(push j occupied))))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)))))
虽然它在性能上有所不同,但将外部条件放入内部主体使其在此处更具可读性:
(defun found? (index lst occupied width)
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(when (and (< index (1- width))
(found? (1+ index) lst occupied width)) ; recursion
(push j occupied)))))
除了数到 9 几次之外,这没有任何作用,这似乎与您的发现一致。
我猜你想从死代码中返回一些东西。您可能想使用return-from
它。
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(push j occupied)
(if (found? (1+ index) lst occupied width) ; recursion
(return-from found? lst)
(setf occupied (remove j occupied)))))
(do ((j 1 (1+ j)))
((> j 9) lst)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(when (= lefthnd rghthnd)
(return-from found? lst)))))))
这返回 (1 2 9),这是错误的。问题似乎是即使您运行超过 9,您也会返回列表,但是您想返回 nil,因为您没有找到任何东西。
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) nil) ; <- nothing found
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(push j occupied)
(if (found? (1+ index) lst occupied width) ; recursion
(return-from found? lst)
(setf occupied (remove j occupied)))))
(do ((j 1 (1+ j)))
((> j 9) nil) ; <- nothing found
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(when (= lefthnd rghthnd)
(return-from found? lst)))))))
这将返回 (9 8 1),这是正确的。现在我似乎明白了你想要做什么,让我们再重构一下。无需从占用的列表中推送和删除,只需在前面暂时创建一个新元素的新列表:
(defun found? (index lst occupied width)
(if (< index (1- width))
(do ((j 1 (1+ j)))
((> j 9) nil)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(when (found? (1+ index) ; recursion
lst
(cons j occupied)
width)
(return-from found? lst))))
(do ((j 1 (1+ j)))
((> j 9) nil)
(unless (find j occupied :test #'=)
(setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(when (= lefthnd rghthnd)
(return-from found? lst)))))))
我认为使用循环而不是 do 使这更具可读性:
(defun found? (index lst occupied width)
(if (< index (1- width))
(loop :for j :from 1 :to 9
:unless (find j occupied :test #'=)
:do (setf (nth index lst) j)
(when (found? (1+ index) ; recursion
lst
(cons j occupied)
width)
(return-from found? lst)))
(loop :for j :from 1 :to 9
:unless (find j occupied :test #'=)
:do (setf (nth index lst) j)
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(when (= lefthnd rghthnd)
(return-from found? lst))))))
由于循环相当复杂,我只想写和读一次,所以把外部条件移到里面:
(defun found? (index lst occupied width)
(loop :for j :from 1 :to 9
:unless (find j occupied :test #'=)
:do (setf (nth index lst) j)
(if (< index (1- width))
(when (found? (1+ index) ; recursion
lst
(cons j occupied)
width)
(return-from found? lst))
(let ((lefthnd (* 111 (reduce #'+ lst)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third lst)
(first lst)
(first lst)
(second lst))))))
(when (= lefthnd rghthnd)
(return-from found? lst))))))
你看到占用只是 lst 的前一个或两个元素,颠倒了吗?我们可以通过递归构建 lst,而不是设置列表元素。我们实际上需要为此返回递归结果,所以这是更好的引用透明性。
(defun find! ()
(found? 0 ; initially show the number 1
'() ; initially no numbers found
3)) ; numbers list width is 3
(defun found? (index part width)
(loop :for j :from 1 :to 9
:unless (find j part :test #'=)
:do (if (< index (1- width))
(let ((solution (found? (1+ index) ; recursion
(cons j part)
width)))
(when solution
(return-from found? solution)))
(let* ((full (cons j part))
(lefthnd (* 111 (reduce #'+ full)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third full)
(first full)
(first full)
(second full))))))
(when (= lefthnd rghthnd)
(return-from found? full))))))
索引和宽度现在只用于计数,所以我们只需要一个数字,我们可以向零计数。这也表明我们应该将基本情况移出循环:
(defun find! ()
(found? '() ; initially no numbers found
3)) ; numbers list width is 3
(defun found? (part count)
(if (zerop count)
(let* ((full part) ; just rename to show that the number is complete
(lefthnd (* 111 (reduce #'+ full)))
(rghthnd (reduce #'+
(mapcar #'*
'(1000 100 10 1)
(list (third full)
(first full)
(first full)
(second full))))))
(when (= lefthnd rghthnd)
(return-from found? full)))
(loop :for j :from 1 :to 9
:unless (find j part :test #'=)
:do (let ((solution (found? (cons j part)
(1- count))))
(when solution
(return-from found? solution))))))
我认为这或多或少是你可以做的,如果你把它保持在一个单一的功能。现在您可能希望将排列的生成与实际代码分开。例如,在广泛使用的库中有一些函数可以处理此类事情alexandria
。