[原创]二分法求解方程代码 （递归+传递函数）

qjchen

近日写某代码，需用到求解方程，由于之前对把函数当参数传递及递归函数没有仔细研究，故以此为契机做了一下研究，以MIT的“Structure and interpretaion of computer programs”《计算机程序的构造和解释》的lisp方言SCHEME代码为蓝本，撰写如下代码，希望对有兴趣的朋友有用：）
其他的如牛顿迭代法应该也可类似编出
当然程序还很简陋，没有其他容错等，有待细化
myfuntosolve中定义了待求解的方程  -x^3+2x+3=0，求解的时候指定的1.0和2.0需满足能使得-x^3+2x+3的值为一正一负
其他方程亦可使用此方法

2. 
   
3. ;;; By qjchen@gmail.com    http://www.mjtd.com/bbs/post.asp?action=edit&BoardID=3&replyID=31269&ID=80886&star=1
   
4. ;;; The main code is mainly taken from the MIT book "Structure and interpretaion of computer programs"
   
5. ;;; judge whether the initial range is suitable
   
6. (defun halfsolve (f a b / a-value b-value)
   
7.   (setq a-value ((eval f) a)
   
8. b-value ((eval f) b)
   
9.   )  
   
10.   (cond
    
11.     ((and (< a-value 0) (> b-value 0)) (searchhalf f a b))
    
12.     ((and (> a-value 0) (< b-value 0)) (searchhalf f b a))
    
13.     ((= a-value 0) a)
    
14.     ((= b-value 0) b)
    
15.     (T (prompt "The Values maybe not between a and b"))
    
16.   )
    
17. )
    
18. 
    
19. ;;core code of dichotomy
    
20. (defun searchhalf (f neg-point pos-point / test-value midpoint)
    
21.   (setq midpoint (/ (+ neg-point pos-point) 2))
    
22.   (cond
    
23.     ((close-enough? neg-point pos-point) midpoint)
    
24.     (T
    
25.      (setq test-value ((eval f) midpoint))
    
26.      (cond
    
27.        ((> test-value 0) (searchhalf f neg-point midpoint))
    
28.        ((< test-value 0) (searchhalf f midpoint pos-point))
    
29.        (T midpoint)
    
30.      )
    
31.     )
    
32.   )
    
33. )
    
34. 
    
35. ;;judge small enough
    
36. (defun close-enough? (x y)
    
37.   (< (abs (- x y)) 1e-10)
    
38. )
    
39. 
    
40. ;; The equation to be solve, -x^3+2x+3=0
    
41. (defun myfuntosolve (x)
    
42.   (- (* x x x) (* 2 x) 3)
    
43. )
    
44. 
    
45. ;;Main function by qjchen@gmail.com
    
46. (defun c:test()
    
47.   (princ (rtos (halfsolve myfuntosolve 1.0 2.0) 2 14))
    
48.   (princ)
    
49. )
    
50. 
    

<br /> ;;; By qjchen@gmail.com    http://www.mjtd.com/bbs/post.asp?action=edit&BoardID=3&replyID=31269&ID=80886&star=1<br /> ;;; The main code is mainly taken from the MIT book "Structure and interpretaion of computer programs"<br /> ;;; judge whether the initial range is suitable<br /> (defun halfsolve (f a b / a-value b-value)<br />   (setq a-value ((eval f) a)<br /> b-value ((eval f) b)<br />   )  <br />   (cond<br />     ((and (< a-value 0) (> b-value 0)) (searchhalf f a b))<br />     ((and (> a-value 0) (< b-value 0)) (searchhalf f b a))<br />     ((= a-value 0) a)<br />     ((= b-value 0) b)<br />     (T (prompt "The Values maybe not between a and b"))<br />   )<br /> )<br /> <br /> ;;core code of dichotomy<br /> (defun searchhalf (f neg-point pos-point / test-value midpoint)<br />   (setq midpoint (/ (+ neg-point pos-point) 2))<br />   (cond<br />     ((close-enough? neg-point pos-point) midpoint)<br />     (T<br />      (setq test-value ((eval f) midpoint))<br />      (cond<br />        ((> test-value 0) (searchhalf f neg-point midpoint))<br />        ((< test-value 0) (searchhalf f midpoint pos-point))<br />        (T midpoint)<br />      )<br />     )<br />   )<br /> )<br /> <br /> ;;judge small enough<br /> (defun close-enough? (x y)<br />   (< (abs (- x y)) 1e-10)<br /> )<br /> <br /> ;; The equation to be solve, -x^3+2x+3=0<br /> (defun myfuntosolve (x)<br />   (- (* x x x) (* 2 x) 3)<br /> )<br /> <br /> ;;Main function by qjchen@gmail.com<br /> (defun c:test()<br />   (princ (rtos (halfsolve myfuntosolve 1.0 2.0) 2 14))<br />   (princ)<br /> )<br /> <br />

xiaowen

新手学习过程中,多谢分享.