SICP HuffmanCode

本节主要介绍了利用抽象符号来进行Huffman Code

说一下我觉得难的地方,首先,相较于之前,第一次用scheme写个比较完整的功能,可能头绪比较乱,最初的那些construct以及selector的创建是一个难的地方

之后是对于编树将操作一步步分离,又一次见识到了啥叫抽象,像我思考的时候就是一整段功能一起思考,然后一阵头大，具体功能实现不是难点,主要还是

设计整个程序的思路比较难,以下为代码

(define (make-leaf symbol weight)(list 'leaf symbol weight))(define (leaf? object)(eq? (car object) 'leaf))(define (symbol-leaf object)(cadr object))(define (weight-leaf object)(caddr object))(define (make-code-tree left right)(list left right (append (symbols left) (symbols right)) (+ (weight left) (weight right))))(define (left-tree object)(car object))(define (right-tree object)(cadr object))(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)(decode-1 bits tree tree))(define (decode-1 bits tree current-branch)(if (null? bits) ()(let ((next-branch(choose-branch (car bits) current-branch)))(if (leaf? next-branch)(cons (symbol-leaf next-branch)(decode-1 (cdr bits) tree tree))(decode-1 (cdr bits) tree next-branch)))))(define (choose-branch bit branch)(cond((= bit 0) (left-tree branch))((= bit 1) (right-tree branch))(else (error "bad-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 sample-tree (make-code-tree (make-leaf 'A 4) (make-code-tree (make-leaf 'B 2) (make-code-tree (make-leaf 'C 1)  (make-leaf 'D 1)))))(define sample-message '(0 1 1 0 0 1 0 1 0 1 1 1 0))(define (encode message tree)(if (null? message)()(append (encode-symbol (car message) tree) (encode (cdr message) tree))))(define (element-of-set? element set)(cond((null? set) #f)((eq? element (car set)) #t)(else (element-of-set? element (cdr set)))))(define (encode-symbol message tree)(cond((element-of-set? message (symbols tree))(cond((leaf? tree) 0)((element-of-set? message (symbols (left-tree tree)))(if (leaf? (left-tree tree)) (list 0) (cons 0 (encode-symbol message (left-tree tree)))))((element-of-set? message (symbols (right-tree tree)))(if (leaf? (right-tree tree)) (list 1) (cons 1 (encode-symbol message (right-tree tree)))))))(else (error "Not valid value"))))(define (generate-huffman-tree pairs)(successive-merge (make-leaf-set pairs)))(define (successive-merge set)(cond((null? set) ())((null? (cdr set)) set)((null? (cddr set)) (make-code-tree (car set) (cadr set)))(else (make-code-tree (car set) (successive-merge (cdr set))))))(define pairs (list '(a 2) '(na 16) '(boom 1) '(sha 3) '(get 2) '(yip 10) '(job 2) '(wah 1)))(define message1 '(get a job))(define message2 '(sha na na na na na na))(define message3 '(wah yip yip yip yip yip))(define message4 '(sha boom))

然后是假设我们的频率为2的幂,这导致了我们取两个最小的相加,不会产生越位现象,也就是除了最后2位,其他都是很难看的一条线下来,

例如我们举个栗子

1 2 4 8 16

我们的编码树为(16 (8 ( 4 (2 1))))

其中括号层数可以看做树的层数,显然由于没有越位,我们的树看起来非常的简单,然后,它的层数为n-1

之后是要求我们评估encode的时间界

为了方便我们把代码从上面单独拎出来

(define (encode message tree)(if (null? message)()(append (encode-symbol (car message) tree) (encode (cdr message) tree))))(define (element-of-set? element set)(cond((null? set) #f)((eq? element (car set)) #t)(else (element-of-set? element (cdr set)))))(define (encode-symbol message tree)(cond((element-of-set? message (symbols tree))(cond((leaf? tree) 0)((element-of-set? message (symbols (left-tree tree)))(if (leaf? (left-tree tree)) (list 0) (cons 0 (encode-symbol message (left-tree tree)))))((element-of-set? message (symbols (right-tree tree)))(if (leaf? (right-tree tree)) (list 1) (cons 1 (encode-symbol message (right-tree tree)))))))(else (error "Not valid value"))))

由于不同的树,不同的编码信息会导致不同结果我们再次只看一下有哪些操作,首先我们对于一个symbol进行了一次element-of-set的判断,需要N 之后又进行了递归的element-of-set的调用约1/2*N^2，最后我们对M个symbol进行编码,最好情况 O(M*N) 最坏情况 O(m*N^2)
由于只是粗粗估计可能会有错误,欢迎指正