BlogJava-ivaneeo's blog-随笔分类-scheme－语言之母http://www.blogjava.net/ivanwan/category/7554.html自由的力量，自由的生活。zh-cnTue, 27 Feb 2007 10:43:40 GMTTue, 27 Feb 2007 10:43:40 GMT60sin函数的递归过程http://www.blogjava.net/ivanwan/archive/2006/07/31/61020.htmlivaneeoivaneeoMon, 31 Jul 2006 07:02:00 GMThttp://www.blogjava.net/ivanwan/archive/2006/07/31/61020.htmlhttp://www.blogjava.net/ivanwan/comments/61020.htmlhttp://www.blogjava.net/ivanwan/archive/2006/07/31/61020.html#Feedback0http://www.blogjava.net/ivanwan/comments/commentRss/61020.htmlhttp://www.blogjava.net/ivanwan/services/trackbacks/61020.htmlsinxbook-Z-G-D-20.gifx if x is sufficiently small, and the trigonometric identity
    ch1-Z-G-19.gif

to reduce the size of the argument of sin. (For purposes of this exercise an angle is considered ``sufficiently small'' if its magnitude is not greater than 0.1 radians.) These ideas are incorporated in the following procedures:
    (define (cube x) (* x x x))
    (define (p x) (- (* 3 x) (* 4 (cube x))))
    (define (sine angle)
       (if (not (> (abs angle) 0.1))
           angle
           (p (sine (/ angle 3.0)))))



ivaneeo 2006-07-31 15:02 发表评论
]]>得到找零的方法数http://www.blogjava.net/ivanwan/archive/2006/07/30/60864.htmlivaneeoivaneeoSun, 30 Jul 2006 07:51:00 GMThttp://www.blogjava.net/ivanwan/archive/2006/07/30/60864.htmlhttp://www.blogjava.net/ivanwan/comments/60864.htmlhttp://www.blogjava.net/ivanwan/archive/2006/07/30/60864.html#Feedback0http://www.blogjava.net/ivanwan/comments/commentRss/60864.htmlhttp://www.blogjava.net/ivanwan/services/trackbacks/60864.html    美元中有50美分(half-dollars),25美分(quarters),10美分(dimes),5美分(nickels),1美分(pennies).
总共有多少种方式?

分成两步:
    1.计算使用50美分找零的方法数.
    2.上面数目加上除了使用50美分找零的方法数以外的数目.

(define (count-change amount)
  (cc amount 6))

(define (first-denomination kinds-of-coins)
  (cond ((= kinds-of-coins 1) 1)
    ((= kinds-of-coins 2) 2)
    ((= kinds-of-coins 3) 5)
    ((= kinds-of-coins 4) 10)
    ((= kinds-of-coins 5) 20)
    ((= kinds-of-coins 6) 50)))

(define (cc amount kinds-of-coins)
  (cond ((= amount 0) 1)
    ((or (< amount 0)
        (= kinds-of-coins 0))
     0)
    (else (+ (cc (- amount                        ;第一步
            (first-denomination kinds-of-coins))
             kinds-of-coins)
         (cc amount                               ;第二步
             (- kinds-of-coins 1))))))


ivaneeo 2006-07-30 15:51 发表评论
]]>Fibonacci函数和递归树http://www.blogjava.net/ivanwan/archive/2006/07/30/60851.htmlivaneeoivaneeoSun, 30 Jul 2006 05:59:00 GMThttp://www.blogjava.net/ivanwan/archive/2006/07/30/60851.htmlhttp://www.blogjava.net/ivanwan/comments/60851.htmlhttp://www.blogjava.net/ivanwan/archive/2006/07/30/60851.html#Feedback0http://www.blogjava.net/ivanwan/comments/commentRss/60851.htmlhttp://www.blogjava.net/ivanwan/services/trackbacks/60851.htmlch1-Z-G-7.gif

(define (fib n)
  (cond ((= n 0) 0)
        ((= n 1) 1)
        (else (+ (fib (- n 1))
                 (fib (- n 2))))))
递归数如下:
ch1-Z-G-13.gif

Fib(n)非常接近book-Z-G-D-11.gifn/book-Z-G-D-13.gif5

book-Z-G-D-11.gif

同样下面的式子也成立:
ch1-Z-G-15.gif

同样使用线性迭代效率要高的多:

(define (fib n)
  (fib-iter 1 0 n))

(define (fib-iter a b count)
  (if (= count 0)
      b
      (fib-iter (+ a b) a (- count 1))))


ivaneeo 2006-07-30 13:59 发表评论
]]>Ackermann函数http://www.blogjava.net/ivanwan/archive/2006/07/30/60844.htmlivaneeoivaneeoSun, 30 Jul 2006 05:21:00 GMThttp://www.blogjava.net/ivanwan/archive/2006/07/30/60844.htmlhttp://www.blogjava.net/ivanwan/comments/60844.htmlhttp://www.blogjava.net/ivanwan/archive/2006/07/30/60844.html#Feedback0http://www.blogjava.net/ivanwan/comments/commentRss/60844.htmlhttp://www.blogjava.net/ivanwan/services/trackbacks/60844.html 　　　　A（m，0）=A（m-1，0） m=1，2，…
　　　　A（m，n）=A（m-1，A（m，n-1）） m=1，2，… n=1，2，…
　　初始条件为
　　　　A（0，n）=n+1，n=0，1，…

(define (A x y)
  (cond ((= y 0) 0)
        ((= x 0) (* 2 y))
        ((= y 1) 2)
        (else (A (- x 1)
                 (A x (- y 1))))))


ivaneeo 2006-07-30 13:21 发表评论
]]>线性递归和迭代http://www.blogjava.net/ivanwan/archive/2006/07/30/60834.htmlivaneeoivaneeoSun, 30 Jul 2006 03:37:00 GMThttp://www.blogjava.net/ivanwan/archive/2006/07/30/60834.htmlhttp://www.blogjava.net/ivanwan/comments/60834.htmlhttp://www.blogjava.net/ivanwan/archive/2006/07/30/60834.html#Feedback0http://www.blogjava.net/ivanwan/comments/commentRss/60834.htmlhttp://www.blogjava.net/ivanwan/services/trackbacks/60834.html1.线性递归

    ch1-Z-G-7.gif
(define (factorial n)
  (if (= n 1)
      1
      (* n (factorial (- n 1)))))
如图所示的线性的膨胀再线性的缩减,就为线性递归.

2.迭代
    ch1-Z-G-10.gif
(define (factorial n)
  (fact-iter 1 1 n))

(define (fact-iter product counter max-count)
  (if (> counter max-count)
      product
      (fact-iter (* counter product)
                 (+ counter 1)
                 max-count)))

注:试验(factorial 10000),递归堆栈溢出,而迭代则可以运行.


ivaneeo 2006-07-30 11:37 发表评论
]]>Newton's method for cube rootshttp://www.blogjava.net/ivanwan/archive/2006/07/30/60822.htmlivaneeoivaneeoSun, 30 Jul 2006 02:12:00 GMThttp://www.blogjava.net/ivanwan/archive/2006/07/30/60822.htmlhttp://www.blogjava.net/ivanwan/comments/60822.htmlhttp://www.blogjava.net/ivanwan/archive/2006/07/30/60822.html#Feedback0http://www.blogjava.net/ivanwan/comments/commentRss/60822.htmlhttp://www.blogjava.net/ivanwan/services/trackbacks/60822.htmlExercise 1.8.  Newton's method for cube roots is based on the fact that if y is an approximation to the cube root of x, then a better approximation is given by the value
ch1-Z-G-5.gif


ivaneeo 2006-07-30 10:12 发表评论
]]>newton方法求解某数的平方根http://www.blogjava.net/ivanwan/archive/2006/07/29/60791.htmlivaneeoivaneeoSat, 29 Jul 2006 14:12:00 GMThttp://www.blogjava.net/ivanwan/archive/2006/07/29/60791.htmlhttp://www.blogjava.net/ivanwan/comments/60791.htmlhttp://www.blogjava.net/ivanwan/archive/2006/07/29/60791.html#Feedback0http://www.blogjava.net/ivanwan/comments/commentRss/60791.htmlhttp://www.blogjava.net/ivanwan/services/trackbacks/60791.html 例子:求2的平方根

Guess Quotient Average
  
1 (2/1) = 2 ((2 + 1)/2) = 1.5
  
1.5 (2/1.5) = 1.3333 ((1.3333 + 1.5)/2) = 1.4167
  
1.4167 (2/1.4167) = 1.4118 ((1.4167 + 1.4118)/2) = 1.4142
  
1.4142 ... ...

一.给出一个scheme的过程
(define (sqrt-iter guess x)
  (if (good-enough? guess x)
      guess
      (sqrt-iter (improve guess x)
                 x)))
二.改进猜的数(x除以猜的数和猜的数的平均值
    (define (average x y)
      (/ (+ x y) 2))
    (define (improve guess x)
      (average guess (/ x guess)))
三.如何得到比较适合的值.这里做了一个假设,猜的值的平方减去x不小于0.001
    (define (good-enough? guess x)
      (< (abs (- (square guess) x)) 0.001))

注:如何用内部过程来使用上面的部分:
(define (sqrt x)
  (define (good-enough? guess)
    (define (square)
      (* guess guess))
    (< (abs (- (square) x)) 0.001))
  (define (improve guess)
    (define (average y)
      (/ (+ guess y)
     2))
    (average (/ x guess)))
  (define (sqrt-iter guess)
    (if (good-enough? guess)
    guess
    (sqrt-iter (improve guess))))
  (sqrt-iter 1.0))



ivaneeo 2006-07-29 22:12 发表评论
]]>在cygwin中编译使用guile的c代码http://www.blogjava.net/ivanwan/archive/2006/02/18/31317.htmlivaneeoivaneeoFri, 17 Feb 2006 16:32:00 GMThttp://www.blogjava.net/ivanwan/archive/2006/02/18/31317.htmlhttp://www.blogjava.net/ivanwan/comments/31317.htmlhttp://www.blogjava.net/ivanwan/archive/2006/02/18/31317.html#Feedback0http://www.blogjava.net/ivanwan/comments/commentRss/31317.htmlhttp://www.blogjava.net/ivanwan/services/trackbacks/31317.htmlmain.c:
#include <stdio.h>
#include <stdlib.h>
#include <libguile.h>

int main (int argc, char *argv[])
{
	SCM func_symbol;
	SCM func;
	
	scm_init_guile();
	
	// Load the scheme function definitions
	scm_c_primitive_load ("script.scm");	
	
	func_symbol = scm_c_lookup("do-hello");
	func = scm_variable_ref(func_symbol);
	
	scm_call_0 (func);

exit(EXIT_SUCCESS);

}

Makefile:
CC = gcc

GUILE_CFLAGS = `guile-config compile`
GUILE_LIBS = `guile-config link`

all:
	$(CC) $(GUILE_CFLAGS) -c main.c
	$(CC) $(GUILE_LIBS) -o main main.o
如果连接有错误的话,把
	$(CC) $(GUILE_LIBS) -o main main.o
改成
	
	$(CC) main.o $(GUILE_LIBS) -o main

注意:因为这里gcc一般都是把-l参数放在最后一个-o参数之前的.


ivaneeo 2006-02-18 00:32 发表评论
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