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#lang racket
(require "../../shared/lists.rkt")
(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
(append (symbol-set left)
(symbol-set right))
(+ (weight left)
(weight right))))
(define (left-branch tree)
(car tree))
(define (right-branch tree)
(cadr tree))
(define (symbol-set 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-1 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-1 (cdr bits) tree))
(decode-1 (cdr bits) next-branch)))))
(decode-1 bits tree))
(define (choose-branch bit branch)
(cond ((= bit 0) (left-branch branch))
((= bit 1) (right-branch branch))
(else (error "bad bit -- CHOOSE-BRANCH" bit))))
(define (adjoin-set x mset)
(cond ((null? mset) (list x))
((< (weight x) (weight (car mset))) (cons x mset))
(else (cons (car mset)
(adjoin-set x (cdr mset))))))
(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))))))
;; 67
(define sample-tree
(make-code-tree (make-leaf 'A 4)
(make-code-tree
(make-leaf 'B 2)
(make-code-tree
(make-leaf 'D 1)
(make-leaf 'C 1)))))
(define sample-message '(0 1 1 0 0 1 0 1 0 1 1 1 0))
(decode sample-message sample-tree)
;; '(A D A B B C A)
;; 68
(define (encode message tree)
(if (null? message)
'()
(append (encode-symbol (car message) tree)
(encode (cdr message) tree))))
;; just use id as key
(define (key x)
x)
;; scan for symbol
(define (list-lookup given-key set-of-records)
(cond ((null? set-of-records) false)
((equal? given-key (key (car set-of-records)))
(car set-of-records))
(else (list-lookup given-key (cdr set-of-records)))))
;; without using symbol list, for some reason
;; This is probably not the best way to do it because we cannot short circuit the search using the symbol list that is available at each node.
;; (define (encode-symbol s tree)
;; (define (bits result subtree)
;; (cond ((and (leaf? subtree)
;; (equal? s (car (symbol-set subtree))))
;; result)
;; ((leaf? subtree) '())
;; (else
;; (let ((left (bits
;; (cons 0 result)
;; (left-branch subtree))))
;; (if (null? left)
;; (bits (cons 1 result) (right-branch subtree))
;; left)))))
;; (reverse (bits '() tree)))
(define (encode-symbol s tree)
(cond ((leaf? tree) '())
((list-lookup s (symbol-set (left-branch tree)))
(cons 0 (encode-symbol s (left-branch tree))))
((list-lookup s (symbol-set (right-branch tree)))
(cons 1 (encode-symbol s (right-branch tree))))
(else (error "Symbol not in tree -- ENCODE-SYMBOL" s tree))))
(encode (decode sample-message sample-tree) sample-tree)
(encode '(E) sample-tree)
;; 69
(define (generate-huffman-tree pairs)
(successive-merge (make-leaf-set pairs)))
(define (successive-merge weighted-leafs)
(define (merge lightest leafs)
(if (null? leafs)
lightest
(let ((leafs-with-node
(adjoin-set (make-code-tree
lightest
(car leafs))
(cdr leafs))))
(merge (car leafs-with-node)
(cdr leafs-with-node)))))
(merge (car weighted-leafs)
(cdr weighted-leafs)))
(generate-huffman-tree '((A 4) (B 2) (C 1) (D 1)))
;; 70
(define rock-songs-leafs '((A 2) (BOOM 1)
(GET 2) (JOB 2)
(NA 16) (SHA 3)
(YIP 9) (WAH 1)))
(define rock-songs-tree
(generate-huffman-tree rock-songs-leafs))
(define rock-songs-bits
(encode '(GET A JOB
SHA NA NA NA NA NA NA NA NA
GET A JOB
SHA NA NA NA NA NA NA NA NA
WAH YIP YIP YIP YIP YIP YIP YIP YIP YIP
SHA BOOM)
rock-songs-tree))
;; variable length codes with huffman tree
(length rock-songs-bits)
;; fixed length codes with 8 symbols
(expt 2 (length rock-songs-leafs))
;; 71
;; Most frequent is always a leaf from the root, so only one bit
;; Least frequent requires d bits where d is the depth of the tree,
;; in this case the tree is unbalanced so the depth is linear or n-1.
;; (= (+ 2**(i-2) 2**(i-1)) (- 2**i 1))
;; 72
;; You can do best worst case scenario's for searches and positions in the tree
;; and you can argue average time complexities for different known weight distributions that result in balanced or unbalanced trees.
;; I did them on paper and I'm a messy louise reasoner.
;; Maybe I'll come back someday and write it up.
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