习题2.17,直接利用list-ref和length过程
习题2.18,采用迭代法
习题2.21，递归方式：
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：
习题2.24，盒子图就不画了，麻烦，解释器输出：
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：
因此取7操作为：
第三个list可以表示为：
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：
习题2.28，递归，利用append过程就容易了：
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
2）首先需要一个过程用于求解分支的总重量：
利用这个过程写出balanced?过程：
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：
习题2.30：
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：
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 (last
-
pair items)
(list (list - ref items ( - (length items) 1 ))))
(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 (reverse - iter i k)
( if ( null ? i) k (reverse - iter (cdr i) (cons (car i) k))))
(reverse - iter items ()))
(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
(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)))
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：
( if ( null ? items)
items
(cons (square (car items)) (square - list (cdr items)))))
(define (square
-
list items)
(map square 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
(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))
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
Language: Standard (R5RS).
> (list 1 (list 2 (list 3 4 )))
( 1 ( 2 ( 3 4 )))
.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！-_-
(car (cdr (car (cdr (car (cdr (car (cdr (car (cdr (car (cdr x))))))))))))
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 ()))
(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))))
(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)))
(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))))
(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))))
( * (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))
(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))
(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))
(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)))))
( if (null? s)
(list s)
(let ((rest (subsets (cdr s))))
(append rest (map ( lambda (x) (cons (car s) x)) rest)))))
评论
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-)))
>
(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
