電腦程式的構造和解釋 練習題2 70

2021-10-06 09:38:24 字數 3774 閱讀 3756

#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 rock-pairs '((na

16)(yip9)

(sha3)

(a2)

(get2)

(job2)

(boom1)

(wah1)

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

)(define rock-code '(

geta

jobsha

nana

nana

nana

nana

geta

jobsha

nana

nana

nana

nana

wahyip

yipyip

yipyip

yipyip

yipyip

sha boom)

)(encode rock-code rock-tree)

執行結果

'(11

1111

1001

1110

1110

0000

0000

1111

1110

0111

1011

1000

0000

0011

0101

0101

0101

0101

0101

0111

0110

11)

不定長編碼為84個二進位制位

如果使用定長編碼表示8個符號,需要定長編碼位3位,一共是36個符號,定長編碼需要108個二進位制位可以表示。

電腦程式的構造和解釋 練習題1 44

平滑函式的公式為 f s x f x dx f x f x dx 3f s x f x dx f x f x dx 3 fs x f x dx f x f x d x 3他的幾何意義應該是將每個點的導數,變化率變小。說簡單點就是函式那塊曲線彎度比較大,加上這個函式就會減小他的彎度,直到他無限接近於一...

電腦程式的構造和解釋 練習題2 19

lang racket define no more?coin values null coin values define except first denomination coin values cdr coin values define first denomination coin va...

電腦程式的構造和解釋 練習題2 61

跟element of set?一樣的道理,當元素大於當時的set表時,就說明元素根本不在這個表裡,所以將元素直接插入到當時表的前面就可以了。最壞的情況還是把錶迴圈了一遍,最好的情況是只迴圈了一次,這樣平均的步數就是n 2次。define adjoin set x set cond null set...