二维下料

aimisiyou · 发表于 2016-6-4 21:39:06

本帖最后由 aimisiyou 于 2016-6-4 23:13 编辑

$F(2n+1,2n) = \begin{pmatrix}
F(n,n)  & F(n+1,n) \\
F(n+1,n) & F(n,n)\end{pmatrix}$
其中$F(n,n)由n*n个F（1,1）组成$
$F(2n,2n-1) = \begin{pmatrix}
F(n,n-1)  & F(n,n) \\
F(n,n) & F(n,n-1)\end{pmatrix}$
构造出其中两个$F（3,2）$

aimisiyou · 发表于 2016-6-4 21:54:18

本帖最后由 aimisiyou 于 2016-6-5 09:48 编辑

$F(2n+1,2n-1) = \begin{pmatrix} F(n,n-1) & F(n+1,n) \\ F(n+1,n) & F(n,n-1)\end{pmatrix}$构造的$F(7,5)$如下

aimisiyou · 发表于 2016-6-4 23:04:18

对于$F（n，1）$则可以采取下楼梯方式构造，如下图$F（4，1）$

aimisiyou · 发表于 2016-6-5 09:50:04

本帖最后由 aimisiyou 于 2016-6-5 14:27 编辑

$F(2n+1,2) = \begin{pmatrix} F(n,1) & F(n+1,1) \\ F(n+1,1) & F(n,1)\end{pmatrix}$构造的$F(5,2)$如下

aimisiyou · 发表于 2016-6-5 10:26:27

$F(4n+1,4) = \begin{pmatrix} F(2n,2) & F(2n+1,2) \\ F(2n+1,2) & F(2n,2)\end{pmatrix}$构造的$F(9,4)$如下

aimisiyou · 发表于 2016-6-5 14:28:28

本帖最后由 aimisiyou 于 2016-6-5 15:14 编辑

$F(2n+1,3) = \begin{pmatrix} F(n,1) & F(n+1,2) \\ F(n+1,2) & F(n,1)\end{pmatrix}$构造的$F(5,3)$如下

aimisiyou · 发表于 2016-6-5 15:17:09

本帖最后由 aimisiyou 于 2016-6-5 15:39 编辑

$F(6n+2,3) = \begin{pmatrix} F(3n,1) & F(3n+2,2) \\ F(3n+2,2) & F(3n,1)\end{pmatrix}$构造的$F(8,3)$、$F(14,3)$分别如下

aimisiyou · 发表于 2016-6-5 15:41:14

$F(6n+4,3) = \begin{pmatrix} F(3n+1,1) & F(3n+3,2) \\ F(3n+3,2) & F(3n+1,1)\end{pmatrix}$构造的$F(10,3)$如下

aimisiyou · 发表于 2016-6-5 15:45:03

本帖最后由 aimisiyou 于 2016-6-5 16:18 编辑

$F(4n+3,4) = \begin{pmatrix} F(2n+1,2) & F(2n+2,2) \\ F(2n+2,2) & F(2n+1,2)\end{pmatrix}$构造的$F(7,4)$如下

aimisiyou · 发表于 2016-6-5 16:21:23

本帖最后由 aimisiyou 于 2016-6-5 16:49 编辑

$F(2n,5) = \begin{pmatrix} F(n,2) & F(n,3) \\ F(n,3) & F(n,2)\end{pmatrix}$构造的$F(8,5)$如下
$F(2n+1,5) = \begin{pmatrix} F(n,2) & F(n+1,3) \\ F(n+1,3) & F(n,2)\end{pmatrix}$构造的$F(9,5)$如下

aimisiyou · 发表于 2016-6-5 16:52:07

本帖最后由 aimisiyou 于 2023-9-12 08:38 编辑

$F(2n,2m+1) = \begin{pmatrix}  F(n,m)  & F(n,m+1) \\  F(n,m+1) & F(n,m)\end{pmatrix}$

$F(2n+1,2m+1) = \begin{pmatrix}  F(n,m)  & F(n+1,m+1) \\  F(n+1,m+1) & F(n,m)\end{pmatrix}$

$F(2n+1,2m) = \begin{pmatrix}  F(n,m)  & F(n+1,m) \\  F(n+1,m) & F(n,m)\end{pmatrix}$

故可以通过递归运算求出任意$F(k_1,k_2)$

aimisiyou · 发表于 2016-6-8 10:24:50

本帖最后由 aimisiyou 于 2016-6-8 11:27 编辑

;;;当k1=n,k2=1情况下作图f(pt n a b)
(defun drawone(pt a b flag)
(if flag
(entmake
   (list
      '(0 . "LWPOLYLINE")
      '(100 . "AcDbEntity")
      '(100 . "AcDbPolyline")
      '(90 . 4)
      '(70 . 1)
      (cons 10 (list (car pt) (cadr pt)))
      (cons 10 (list (+ (car pt) a) (cadr pt) ))
      (cons 10 (list (+ (car pt) a) (- (cadr pt) b) ))
      (cons 10 (list (car pt) (- (cadr pt) b) )  )
      (cons 210 '(0 0 1))
   )
)
(entmake
   (list
      '(0 . "LWPOLYLINE")
      '(100 . "AcDbEntity")
      '(100 . "AcDbPolyline")
      '(90 . 4)
      '(70 . 1)
      (cons 10 (list (car pt) (cadr pt)))
      (cons 10 (list (+ (car pt) b) (cadr pt) ))
      (cons 10 (list (+ (car pt) b) (- (cadr pt) a) ))
      (cons 10 (list (car pt) (- (cadr pt) a) )  )
      (cons 210 '(0 0 1))
   )
)
)
)
(defun f(pt1 n a b)
  (setvar "osmode" 0)
  (setq i 1 x (car pt1) y (cadr pt1))
  (while (<= i (+ 1 n))
(setq pt2 (list (+ a (* (- i 1) b) x) (+ (* (- 1 i) a) y) ))
(setq pt3 (list x (+ (- 0 b) (* (- 1 i) a) y) ))
(drawone pt1 a b t)
(setq pt1 (list (+ x (* i b))  (+ y (* i (- 0 a)) )))
(setq j 0 k 0)
(while (<= j (- n i))
      (drawone pt2 a b nil)
      (setq j (+ 1 j))
      (setq pt2 (polar pt2 0 b))
)
(while (and (< k i) (/= i (+ 1 n)))
      (drawone pt3 a b nil)
      (setq k (+ 1 k))
      (setq pt3 (polar pt3 0 b))
)
(setq i (+ i 1))
  )
(princ)
)
(f '(0 0) 20 30 25)

aimisiyou · 发表于 2016-6-9 00:10:42

本帖最后由 aimisiyou 于 2016-10-21 23:52 编辑

;;;测试下递归能否运行，测试了几个还行
(defun ff(ppt n m a b)
  (if (> (setq tmin (gcd  n m)) 1)
   (ff ppt (/ n tmin) (/ m tmin) (* a tmin) (* b tmin) )
   (cond
         ((= m 1) (f ppt n a b))
         ((and (= (* (rem n 2) (rem m 2) ) 0) (> m 1)) (f ppt (- n m)  a b  ))
         ((and (= (* (rem n 2) (rem m 2) ) 1) (> m 1)) (f ppt (- n m -1) a b ))
   )
)
)
(ff '(0 0) 36 30 30 24)

aimisiyou · 发表于 2016-6-11 21:27:09

(defun drawone(pt a b flag)
(if flag
(entmake
   (list
      '(0 . "LWPOLYLINE")
      '(100 . "AcDbEntity")
      '(100 . "AcDbPolyline")
      '(90 . 4)
      '(70 . 1)
      (cons 10 (list (car pt) (cadr pt)))
      (cons 10 (list (+ (car pt) a) (cadr pt) ))
      (cons 10 (list (+ (car pt) a) (- (cadr pt) b) ))
      (cons 10 (list (car pt) (- (cadr pt) b) )  )
      (cons 210 '(0 0 1))
   )
)
(entmake
   (list
      '(0 . "LWPOLYLINE")
      '(100 . "AcDbEntity")
      '(100 . "AcDbPolyline")
      '(90 . 4)
      '(70 . 1)
      (cons 10 (list (car pt) (cadr pt)))
      (cons 10 (list (+ (car pt) b) (cadr pt) ))
      (cons 10 (list (+ (car pt) b) (- (cadr pt) a) ))
      (cons 10 (list (car pt) (- (cadr pt) a) )  )
      (cons 210 '(0 0 1))
   )
)
)
)
(defun f(pt1 n a b)
  (setvar "osmode" 0)
  (setq i 1 x (car pt1) y (cadr pt1))
  (while (<= i (+ 1 n))
(setq pt2 (list (+ a (* (- i 1) b) x) (+ (* (- 1 i) a) y) ))
(setq pt3 (list x (+ (- 0 b) (* (- 1 i) a) y) ))
(drawone pt1 a b t)
(setq pt1 (list (+ x (* i b))  (+ y (* i (- 0 a)) )))
(setq j 0 k 0)
(while (<= j (- n i))
      (drawone pt2 a b nil)
      (setq j (+ 1 j))
      (setq pt2 (polar pt2 0 b))
)
(while (and (< k i) (/= i (+ 1 n)))
      (drawone pt3 a b nil)
      (setq k (+ 1 k))
      (setq pt3 (polar pt3 0 b))
)
(setq i (+ i 1))
  )
(princ)
)
(defun ff(ppt n m a b)
  (if (> (setq tmin (** n m)) 1)
   (ff ppt (/ n tmin) (/ m tmin) (* a tmin) (* b tmin) )
   (if  (= m 1)
         (f ppt n a b)
         (progn
            (ff ppt (/ n 2) (/ m 2)  a b)
            (ff (polar ppt 0 (+ (* a (/ m 2)) (* b (/ n 2)))) (/ (+ 1 n) 2) (/ (+ 1 m) 2) a b)
            (ff (polar ppt (* 1.5 pi) (+ (* a (/ n 2)) (* b (/ m 2)))) (/ (+ 1 n) 2) (/ (+ 1 m) 2) a b)
            (ff (list (+ (car ppt) (* b (/ (1+ n) 2)) (* a (/ (1+ m) 2))) (- (cadr ppt) (* a (/ (1+ n) 2)) (* b (/ (1+ m) 2)))) (/ n 2) (/ m 2)  a b)
         )
   )
)
(princ)
)
(ff '(0 0) 5 4 28 24)

ynhh · 发表于 2016-7-20 10:28:46

请大师整理出一个完整的源码程序
来测试一下啊

账号		自动登录	找回密码
密码			立即注册

[原创] 二维下料

点评