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sin函数的递归�q�程

ivaneeo — Mon, 31 Jul 2006 07:02:00 GMT

The sine of an angle (specified in radians) can be computed by making use of the approximation sinxx if x is sufficiently small, and the trigonometric identity

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��N��的方法数

ivaneeo — Sun, 30 Jul 2006 07:51:00 GMT

�?00��元需��N��.
    ��元中有50��分(half-dollars),25��分(quarters),10��分(dimes),5��分(nickels),1��分(pennies).
��d��有多��种方式?

分成两步:
    1.计算使用50��分��N��的方法数.
    2.上面数目加上除了使用50��分��N��的方法数以外的数�?

(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                        ;�W�一�?br />            (first-denomination kinds-of-coins))
             kinds-of-coins)
        (cc amount                               ;�W�二�?br />             (- kinds-of-coins 1))))))

ivaneeo 2006-07-30 15:51 发表评论

ivaneeo — Sun, 30 Jul 2006 05:59:00 GMT

Fibonacci函数定义如下:

(define (fib n) (cond ((= n 0) 0) ((= n 1) 1) (else (+ (fib (- n 1)) (fib (- n 2)))))) 递归数如�?Fib(n)非常接近ⁿ/5

同样下面的式子也成立:

同样使用�U�性�P代效率要高的�?

(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

ivaneeo — Sun, 30 Jul 2006 05:21:00 GMT

Ackermann函数可用递推关系如下定义
　　　　A�Q�m�Q?�Q?A�Q�m-1�Q?�Q?m=1�Q?�Q��?br /> 　　　　A�Q�m�Q�n�Q?A�Q�m-1�Q�A�Q�m�Q�n-1�Q�） m=1�Q?�Q��?n=1�Q?�Q��?br /> 　　初始条�g�?br /> 　　　　A�Q?�Q�n�Q?n+1�Q�n=0�Q?�Q��?br />
(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

ivaneeo — Sun, 30 Jul 2006 03:37:00 GMT

关于n!求解:
1.�U�性递归

(define (factorial n) (if (= n 1) 1 (* n (factorial (- n 1))))) 如图所�C�的�U�性的膨胀再线性的�~�减,��׃ؓ�U�性递归. 2.�q�代 (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))) �?试验(factorial 10000),递归堆栈溢出,而�P代则可以�q�行.

ivaneeo 2006-07-30 11:37 发表评论

Newton's method for cube roots

ivaneeo — Sun, 30 Jul 2006 02:12:00 GMT

Exercise 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

ivaneeo 2006-07-30 10:12 发表评论

newton�Ҏ(gu��)��求解某数的��^�Ҏ(gu��)��

ivaneeo — Sat, 29 Jul 2006 14:12:00 GMT

例子:�?的��^�Ҏ(gu��)��

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.�l�出一个scheme的过�E?br />(define (sqrt-iter guess x) (if (good-enough? guess x) guess (sqrt-iter (improve guess x) x))) �?改进猜的�?x除以猜的数和猜的数的�q�_��?br /> (define (average x y) (/ (+ x y) 2))
(define (improve guess x) (average guess (/ x guess))) �?如何得到比较适合的�?�q�里做了一个假�?猜的值的�q�x��减去x不小�?.001 (define (good-enough? guess x) (< (abs (- (square guess) x)) 0.001)) �?如何用内部过�E�来使用上面的部�? (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��用guile的c代码

ivaneeo — Fri, 17 Feb 2006 16:32:00 GMT

写一个简单的例子:
main.c:

#include 
#include 
#include

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�q�接有错误的�?�?br>	$(CC) $(GUILE_LIBS) -o main main.o
�Ҏ(gu��)��
	
	$(CC) main.o $(GUILE_LIBS) -o main

注意:因�ؓ�q�里gcc一般都是把-l参数攑֜�最后一�?o参数之前�?

ivaneeo 2006-02-18 00:32 发表评论