電腦程式的構造和解釋 練習題2 71

2021-10-06 09:40:28 字數 3801 閱讀 8904

#lang racket

(define (element-of-

set? x set

)(cond (

(null

?set

) #f)

((equal? x (car set

)) #t)

(else

(element-of-

set? x (cdr set))

)))(define (make-leaf symbol weight)

(list 'leaf symbol weight)

)(define (leaf?

object

)(eq?

(car object

) 'leaf)

)(define (symbol-leaf x)

(cadr x)

)(define (weight-leaf x)

(caddr x)

)(define (make-code-tree left right)

(list left

right

(symbols right))(

+(weight left)

(weight right)))

)(define (left-branch tree)

(car tree)

)(define (right-branch tree)

(cadr tree)

)(define (symbols tree)(if

(leaf? tree)

(list (symbol-leaf tree)

)(caddr tree)))

(define (weight tree)(if

(leaf? tree)

(weight-leaf tree)

(cadddr tree)))

(define (decode bits tree)

(define (decode-l bits current-branch)(if

(null

? bits)

'()(

let(

(next-branch

(choose-branch (car bits) current-branch)))

(if(leaf? next-branch)

(cons (symbol-leaf next-branch)

(decode-l (cdr bits) tree)

)(decode-l (cdr bits) next-branch)))

))(decode-l bits tree)

)(define (choose-branch bit branch)

(cond (

(= bit 0

)(left-branch branch))(

(= bit 1

)(right-branch branch))(

else

(error "bad bit -- chosse-branch" bit)))

)(define (adjoin-

set x set

)(cond (

(null

?set

)(list x))(

(<

(weight x)

(weight (car set))

)(cons x set))

(else

(cons (car set

)(adjoin-

set x (cdr set))

))))

(define (make-leaf-

set pairs)(if

(null

? pairs)

'()(

let(

(pair (car pairs)))

(adjoin-

set(make-leaf (car pair)

(cadr pair)

)(make-leaf-

set(cdr pairs)))

)))(define (encode message tree)(if

(null

? message)

'()(encode (cdr message) tree)))

)(define (encode-symbol message tree)

(cond (

(leaf? tree)

null)(

(element-of-

set? message (symbols (left-branch tree)))

(cons 0

(encode-symbol message (left-branch tree)))

)((element-of-

set? message (symbols (right-branch tree)))

(cons 1

(encode-symbol message (right-branch tree)))

)(else

(error "bad bit -- chosse-branch" message)))

)(define (generate-huffman-tree pairs)

(successive-merge (make-leaf-

set pairs)))

(define (successive-merge set)(

if(=(length set)1

)(car set

)(successive-merge (adjoin-

set(make-code-tree (car set

)(cadr set))

(cddr set))

)))(define n5-pairs '((a

16)(b

8)(c

4)(d

2)(e

1)))

(define n10-pairs '((a

512)(b

256)(c

128)(d

64)(e

32)(f

16)(g

8)(h

4)(i

2)(j

1)))

(define n5-tree (generate-huffman-tree n5-pairs)

)(define n10-tree (generate-huffman-tree n10-pairs)

)(define n5-max '(aa

aa a)

)(define n5-min '(ee

ee e)

)(define n10-max '(aa

aaaa

aaa a)

)(define n10-min '(jj

jjjj

jjj j)

)(length (encode n5-max n5-tree)

)(length (encode n5-min n5-tree)

)(length (encode n10-max n10-tree)

)(length (encode n10-min n10-tree)

)

執行結果

5

2010

90

如果是有n個符號進行編碼,那麼,最小的編碼長度位n,最大的編碼長度為n*(n-1),因為最低頻率出現的符號,在樹的最低端,樹一共有n-1層,所以他的編碼長度是n-1,所以一共就是n*(n-1)的長度

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