sicp逐日一题[1.45]

Exercise 1.45

We saw in Section 1.3.3 that attempting to compute square roots by naively finding a fixed point of y->x/y does not converge, and that this can be fixed by  average damping. The same method works for finding cube roots as fixed points of the average-dampedy x/y^2. Unfortunately, the process does not work for fourth roots—a single average damp is not enough to make a fixed-point search for y->x/y3 converge. On the other hand, if we average damp twice (i.e., use the average damp of the average damp of y->x/y3) the fixed-point search does converge. Do some experiments to determine how many average damps are required to compute nth roots as a fixed point search based upon repeated average damping of y->x/y^(n-1). Use this to implement a simple procedure for computing nth roots using fixed-point, average-damp,and the repeated procedure of Exercise1.43. Assume that any arithmetic operations you need are available as primitives.

这道题难度太难了，我最后也没能靠本身做出来。一个是怎么找到要执行频频average-damp，我一开始以为是 n-2，试了几个发现明显不是，又猜测是不是 n/2，结果还是不对，最后上网搜了一下才知道是 log 2(n)，感爱好的可以参考知乎的这个回答；知道了重复执行的次数，在编写代码的时候再次遇到了问题，我对于“把一个过程作为另一个过程的返回值”这个概念明白的还是不到位，没有明白(repeated average-damp n)之后还要给它传一个过程作为 average-damp 的参数，最后上网看了别人的答案才明白过来。下面是我的答案：

1. ; 求 x 和 f(x) 的平均值
2. (define (average-damp f)
3.   (lambda (x) (average x (f x))))
5. ; 对于任意正整数 n，求使得 2^k < n 的最大 k 值
6. (define (max-expt n)
7.   (define (iter k pre)
8.     (if (< n pre)
9.         (- k 1)
10.         (iter (+ k 1) (* 2 pre))))
11.   (iter 1 2))
13. (define (nth-root x n)
14.   (fixed-point ((repeated average-damp (max-expt n))
15.                 (lambda (y) (/ x (expt y (- n 1)))))
16.                1.0))
19. (display (nth-root 2 2))
20. (newline)
21. (display (nth-root 32 5))
22. (newline)
24. ; 结果
25. 1.4142135623746899
26. 2.000001512995761

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