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excercise3.12~3.20.rkt
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excercise3.12~3.20.rkt
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#lang sicp
;; Excercise 3.12:
;; The following procedure for appending lists was introduced
;; in Section 2.2.1:
(define (append x y)
(if (null? x)
y
(cons (car x) (append (cdr x) y))))
;; Append forms a new list by successively consing the elements of
;; x onto y. The procedure append! is similar to append, but it is a
;; mutator rather than a constructor. It appends the lists by splicing
;; them together, modifying the final pair of x so that its cdr is now y.
;; (It is an error to call append! with an empty x.)
(define (append! x y)
(set-cdr! (last-pair x) y)
x)
;; Here last-pair is a procedure that returns the last pair
;; in its argument:
(define (last-pair x)
(if (null? (cdr x))
x
(last-pair (cdr x))))
;; Consider the interaction
(define x (list 'a 'b))
(define y (list 'c 'd))
(define z (append x y))
; z
; => (a b c d)
; (cdr x)
; <response>
(define w (append! x y))
; w
; => (a b c d)
; (cdr x)
; <response>
;; What are the missing <response>s? Draw box-and-poiter diagrams to explain your answer.
;; the first (cdr x) => (b)
;; the second (cdr x) => (b c d)
;; Because z is represent as:
;;x --> [*]----> [*]----> '()
;; | |
;; v v
;; 'a 'b
;;
;;y --> [*]----> [*]----> '()
;; | |
;; v v
;; 'c 'd
;;
;;z --> [*]---->[*]---->[*]---->[*]----> '()
;; | | | |
;; v v v v
;; 'a 'b 'c 'd
;; and for the second one:
;;w------+
;; |
;; |
;; v
;;x --> [*]----> [*]----+
;; | | |
;; v v |
;; 'a 'b |
;; |
;; +--------------+
;; |
;; v
;;y --> [*]----> [*]----> '()
;; | |
;; v v
;; 'c 'd
;;
;;z --> [*]---->[*]---->[*]---->[*]----> '()
;; | | | |
;; v v v v
;; 'a 'b 'c 'd
;; Excercise 3.13:
;; Consider the following make-cycle procedure,
;; which uses the last-pair procedure defined in Exercise 3.12:
(define (make-cycle x)
(set-cdr! (last-pair x) x) x)
;; Draw a box-and-pointer diagram that shows the structure zs created
;; by:
(define zs (make-cycle (list 'a 'b 'c)))
;; What happens if we try to compute (last-pair zs)?
;; it will compute an infinite loop, because:
;; +-----------------------+
;; | |
;; v |
;;zs ----> [*]----> [*]----> [*]----+
;; | | |
;; v v v
;; 'a 'b 'c
;; Excercise 3.14:
;; The following procedure is quite useful, although
;; obscure:
(define (mystery x)
(define (loop x y)
(if (null? x)
y
(let ((temp (cdr x)))
(set-cdr! x y)
(loop temp x))))
(loop x '()))
;; Loop uses the “temporary” variable temp to hold the old value of
;; the cdr of x, since the set-cdr! on the next line destroys the cdr.
;; Explain what mystery does in general. Suppose v is defined by
;; (define v (list ’a ’b ’c ’d)). Draw the box-and-pointer diagram
;; that represents the list to which v is bound. Suppose that we
;; now evaluate (define w (mystery v)). Draw box-and-pointer
;; diagrams that show the structures v and w after evaluating this expression.
;; What would be printed as the values of v and w?
;; 实际上mystery是一个修改版的reverse函数:
;;v --> [*]----> [*]----> [*]----> '()
;; | | |
;; v v v
;; 'a 'b 'c
;; (mystery v)
;; (mystery (list 'a 'b 'c))
;;
;; (loop (list 'a 'b 'c) '())
;;
;; (let ((temp (list 'b 'c)))
;; (set-cdr! (list 'a 'b 'c) '())
;; (loop (list 'b 'c) (list 'a)))
;;
;; (loop (list 'b 'c) (list 'a))
;;
;; (let ((temp (list 'c)))
;; (set-cdr! (list 'b 'c) (list 'a))
;; (loop (list 'c) (list 'b 'a)))
;;
;; (loop (list 'c) (list 'b 'a))
;;
;; (let ((temp '()))
;; (set-cdr! (list 'c) (list 'b 'a))
;; (loop '() (list 'c 'b 'a)))
;;
;; (loop '() (list 'c 'b 'a))
;;
;; (list 'c 'b 'a)
;;v------------------------+
;; |
;; v
;;w --> [*]----> [*]----> [*]----> '()
;; | | |
;; v v v
;; 'c 'b 'a
(define m (list 'a 'b))
(define z1 (cons m m))
(define z2 (cons (list 'a 'b) (list 'a 'b)))
(define (set-to-wow! x)
(set-car! (car x) 'wow) x)
;; Excercise 3.15:
;; Draw box-and-pointer diagrams to explain the effect
;; of set-to-wow! on the structures z1 and z2 above.
;;z1 --> [*][*]
;; | |
;; v v
;; x --> [*][*]--> [*][/]
;; | |
;; v v
;; 'wow! 'b
;;z2 --> [*][*]--> [*][*]--> [*][/]
;; | | |
;; | v v
;; | 'a 'b
;; | ^
;; | |
;; +------> [*][*]--> [*][/]
;; |
;; v
;; 'wow!
;; Excercise 3.16:
;; Ben Bitdiddle decides to write a procedure to count
;; the number of pairs in any list structure. “It’s easy,” he reasons.
;; “The number of pairs in any structure is the number in the car plus
;; the number in the cdr plus one more to count the current pair.” So
;; Ben writes the following procedure:
(define (count-pairs x)
(if (not (pair? x))
0
(+ (count-pairs (car x))
(count-pairs (cdr x))
1)))
;; Show that this procedure is not correct. In particular, draw boxand-
;; pointer diagrams representing list structures made up of exactly
;; three pairs for which Ben’s procedure would return 3; return 4
;; return 7; never return at all.
(define str1 '(foo bar baz))
(count-pairs str1) ; 3
; str1 -> ( . ) -> ( . ) -> ( . ) ->null
; | | |
; v v v
; 'foo 'bar 'baz
(define xx '(foo))
(define yy (cons xx xx))
(define str2 (list yy))
(count-pairs str2) ; 4
; str2 -> ( . ) -> null
; |
; v
; ( . )
; | |
; v v
; ( . ) -> 'null
; |
; v
; 'foo
(define str3 (cons yy yy))
(count-pairs str3) ; 7
; str3 -> ( . )
; | |
; v v
; ( . )
; | |
; v v
; ( . ) -> null
; |
; v
; 'foo
(define str4 '(foo bar baz))
(set-cdr! (cddr str4) str4)
;(count-pairs str4) ; maximum recursion depth exceeded
; ,-------------------,
; | |
; v |
; str4 -> ( . ) -> ( . ) -> ( . )
; | | |
; v v v
; 'foo 'bar 'baz
;; Excercise 3.17:
;; Devise a correct version of the count-pairs procedure
;; of Exercise 3.16 that returns the number of distinct pairs in
;; any structure. (Hint: Traverse the structure, maintaining an auxiliary
;; data structure that is used to keep track of which pairs have
;; already been counted.)
(define (count-real-pairs x)
(length (inner x '())))
(define (false? v)
(if (pair? v)
#f
#t))
(define (inner x memo-list)
(if (and (pair? x) (false? (memq x memo-list)))
(inner (car x)
(inner (cdr x)
(cons x memo-list)))
memo-list))
;; Excercise 3.18:
;; Write a procedure that examines a list and determines
;; whether it contains a cycle, that is, whether a program that
;; tried to find the end of the list by taking successive cdrs would go
;; into an infinite loop. Exercise 3.13 constructed such lists.
;; 网上抄的解法，因为第一次想就已经找到常量空间解法了。
(define (isloop? lst)
(let ((identity (cons '() '())))
(define (iter remain-list)
(cond ((null? remain-list)
#f)
((eq? identity (car remain-list))
#t)
(else
(set-car! remain-list identity)
(iter (cdr remain-list)))))
(iter lst)))
;; Excercise 3.19:
;; Redo Exercise 3.18 using an algorithm that takes
;; only a constant amount of space. (This requires a very clever idea.)
;; 使用快慢步解决这个问题，快的每次走两步，慢的每次走一步，如果有环那么肯定会相遇
(define (loop? lst)
(define (safe-cdr l)
(if (pair? l)
(cdr l)
'()))
(define (iter a b)
(cond [(or (null? a) (null? b)) #f]
[(eq? a b) #t]
[(eq? a (safe-cdr b)) #t]
[else
(iter (safe-cdr a) (safe-cdr (safe-cdr b)))]))
(iter (safe-cdr lst) (safe-cdr (safe-cdr lst))))
;; Excercise 3.20:
;; Draw environment diagrams to illustrate the evaluation
;; of the sequence of expressions
;(define x (cons 1 2))
;(define z (cons x x))
;(set-car! (cdr z) 17)
;(car x)
;17
;; using the procedural implementation of pairs given above
;; (Compare Excercise 3.11.)
;; +------------------------------+
;;global -> | |
;;env | x |
;; +--|---------------------------+
;; | ^
;; | |
;; | +----------+
;; | E1 -> | x: 1 |
;; | | y: 2 |
;; | | |
;; | | set-x! -----> ...
;; | | set-y! -----> ...
;; +--------->dispatch ---> parameters: m
;; | | body: (cond ((eq? m 'car) 'car)
;; +----------+ ((eq? m 'cdr) 'cdr)
;; ((eq? m 'set-car!) 'set-car!)
;; ((eq? m 'set-cdr!) 'set-cdr!)
;; (else
;; (error "..." m)))
;; +-------------------------------------------------------+
;;global -> | |
;;env | z x |
;; +--|---------------------------|------------------------+
;; | ^ | ^
;; | | | |
;; | | | +----------+
;; | | | | x: 1 |
;; | | | | y: 2 |
;; | | | | |
;; | | | | set-x! -----> ...
;; | | | | set-y! -----> ...
;; | | +------->dispatch ---> parameters: m
;; | | | ^ ^ | body: ...
;; | | +--|-|-----+
;; | +----------+ | |
;; | E2 -> | x: ------------------------+ |
;; | | y: --------------------------+
;; | | |
;; | | set-x! -----> ...
;; | | set-y! -----> ...
;; +--------->dispatch ---> parameters: m
;; | | body: (cond ((eq? m 'car) 'car)
;; +----------+ ((eq? m 'cdr) 'cdr)
;; ((eq? m 'set-car!) 'set-car!)
;; ((eq? m 'set-cdr!) 'set-cdr!)
;; (else
;; (error "..." m)))
;; (set-car! (cdr z) 17)有以下两个步骤：
;; 1. 执行(cdr z)返回x
;; 2. 执行(set-car! x 17)，引发表达式((x 'set-car!) 17)执行，然后引发(set-x! 17)
;; 最终x的car被设置为17
;; +-------------------------------------------------------+
;;global -> | |
;;env | z x |
;; +--|---------------------------|------------------------+
;; | ^ | ^
;; | | | |
;; | | | +----------+
;; | | | | x: 17 |
;; | | | | y: 2 |
;; | | | | |
;; | | | | set-x! -----> ...
;; | | | | set-y! -----> ...
;; | | +------->dispatch ---> parameters: m
;; | | | ^ ^ | body: ...
;; | | +--|-|-----+
;; | +----------+ | |
;; | E2 -> | x: ------------------------+ |
;; | | y: --------------------------+
;; | | |
;; | | set-x! -----> ...
;; | | set-y! -----> ...
;; +--------->dispatch ---> parameters: m
;; | | body: (cond ((eq? m 'car) 'car)
;; +----------+ ((eq? m 'cdr) 'cdr)
;; ((eq? m 'set-car!) 'set-car!)
;; ((eq? m 'set-cdr!) 'set-cdr!)
;; (else
;; (error "..." m)))