习题2.17,直接利用list-ref和length过程
(define (last
-
pair items) (list (list
-
ref items (
-
(length items)
1
))))
习题2.18,采用迭代法
(define (reverse
-
list items) (define (reverse
-
iter i k) (
if
(
null
?
i) k (reverse
-
iter (cdr i) (cons (car i) k)))) (reverse
-
iter items ()))
习题2.20，如果两个数的奇偶相同，那么他们的差模2等于0，根据这一点可以写出：
(define (same
-
parity a . b) (define (same
-
parity
-
temp x y) (cond ((
null
?
y) y) ((
=
(remainder (
-
(car y) x)
2
)
0
) (cons (car y) (same
-
parity
-
temp x (cdr y)))) (
else
(same
-
parity
-
temp x (cdr y))))) (cons a (same
-
parity
-
temp a b)))
利用了基本过程remainder取模
习题2.21，递归方式：
(define (square
-
list items) (
if
(
null
?
items) items (cons (square (car items)) (square
-
list (cdr items)))))
利用map过程：
(define (square
-
list items) (map square items))
习题2.23，这与ruby中的each是一样的意思，将操作应用于集合的每个元素：
(define (
for
-
each proc items) (define (
for
-
each
-
temp proc temp items) (
if
(
null
?
items) #t (
for
-
each
-
temp proc (proc (car items)) (cdr items)))) (
for
-
each
-
temp proc
0
items))
最后返回true
习题2.24，盒子图就不画了，麻烦，解释器输出：
Welcome to DrScheme, version
360
. Language: Standard (R5RS).
>
(list
1
(list
2
(list
3
4
))) (
1
(
2
(
3
4
)))
树形状应当是这样
.
/\
/ \
/\
/ \ .
/\
/ \习题2.25，
第一个list可以表示为(list 1 3 (list 5 7) 9)
因此取7的操作应当是：
(car (cdr (car (cdr (cdr (list
1
3
(list
5
7
)
9
))))))
第二个list表示为:(list (list 7))
因此取7操作为：
(car (car (list (list
7
))))
第三个list可以表示为：
(list
1
(list
2
(list
3
(list
4
(list
5
(list
6
7
))))))
因此取7的操作为：
(define x (list
1
(list
2
(list
3
(list
4
(list
5
(list
6
7
))))))) (car (cdr (car (cdr (car (cdr (car (cdr (car (cdr (car (cdr x))))))))))))
够恐怖！-_-
习题2.26，纯粹的动手题，就不说了
习题2.27，在reverse的基础上进行修改，同样采用迭代，比较难理解：
(define (deep
-
reverse x) (define (reverse
-
iter rest result) (cond ((null? rest) result) ((
not
(pair? (car rest))) (reverse
-
iter (cdr rest) (cons (car rest) result))) (
else
(reverse
-
iter (cdr rest) (cons (deep
-
reverse (car rest)) result))) )) (reverse
-
iter x ()))
习题2.28，递归，利用append过程就容易了：
(define (finge x) (cond ((pair? x) (append (finge (car x)) (finge (cdr x)))) ((null? x) ()) (
else
(list x))))
习题2.29，这一题很明显出来的二叉活动体也是个层次性的树状结构
1）很简单，利用car,cdr
(define (left
-
branch x) (car x)) (define (right
-
branch x) (car (cdr x))) (define (branch
-
length b) (car b)) (define (branch
-
structure b) (car (cdr b)))
2）首先需要一个过程用于求解分支的总重量：
(define (branch
-
weight branch) (let ((structure (branch
-
structure branch))) (
if
(
not
(pair? structure)) structure (total
-
weight structure)))) (define (total
-
weight mobile) (
+
(branch
-
weight (left
-
branch mobile)) (branch
-
weight (right
-
branch mobile))))
利用这个过程写出balanced?过程：
(define (torque branch) (
*
(branch
-
length branch) (branch
-
weight branch))) (define (balanced? mobile) (
=
(torque (left
-
branch mobile)) (torque (right
-
branch mobile))))
3)选择函数和定义函数提供了一层抽象屏蔽，其他函数都是建立在这两个基础上，因此需要改变的仅仅是selector函数：
(define (right
-
branch mobile) (cdr mobile)) (define (branch
-
structure branch) (cdr branch))
习题2.30：
(define (square
-
tree tree) (cond ((null? tree) tree) ((
not
(pair? tree)) (square tree)) (
else
(cons (square
-
tree (car tree)) (square
-
tree (cdr tree)))))) (define (square
-
tree2 tree) (map (
lambda
(x) (
if
(pair? x) (square
-
tree x) (square x))) tree))
习题2.31，进一步抽象出map-tree，与map过程类似，将proc过程作用于树的每个节点：
(define (tree
-
map proc tree) (cond ((null? tree) tree) ((
not
(pair? tree)) (proc tree)) (
else
(cons (tree
-
map proc (car tree)) (tree
-
map proc (cdr tree)))))) (define (square
-
tree3 tree) (tree
-
map square tree))
习题2.32，通过观察，rest总是cdr后的子集，比如对于(list 1 2 3)，连续cdr出来的是：
(2 3)
(3)
()
其他的5个子集应该是car结果与这些子集组合的结果，因此：
(define (subsets s) (
if
(null? s) (list s) (let ((rest (subsets (cdr s)))) (append rest (map (
lambda
(x) (cons (car s) x)) rest)))))
评论
# re: sicp习题2.2节尝试解答 回复 更多评论
2007-12-26 21:31 by
mabusyao
2.32, 跟楼主思路一样，可惜出不了结果
(define (subsets s)
(if (null? s) (list s)
(let ((rest (subsets (cdr s))))
(append rest (map (lambda (x) (cons (car s) x)) rest)))))
(subsets (list 1 2 3))
Welcome to DrScheme, version 371 [3m].
Language: Advanced Student.
(shared ((-2- (list 3)) (-4- (list 2)) (-6- (cons 2 -2-)))
(list empty -2- -4- -6- (list 1) (cons 1 -2-) (cons 1 -4-) (cons 1 -6-)))
>
文章转自庄周梦蝶 ，原文发布时间5.17
相关资源：SICP 习题答案
