#lang sicp

;;;SECTION 2.3.3

;; representing

(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 (symbols left) (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)))

;; decoding
(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))))

;; sets

(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 'D 1)
                                   (make-leaf 'C 1)))))

(define sample-message '(0 1 1 0 0 1 0 1 0 1 1 1 0))

(define (memq item x)
  (cond ((null? x) false)
        ((eq? item (car x)) x)
        (else (memq item (cdr x)))))

(define (encode-branch symbol tree)
  (let ((left-sym (symbols (left-branch tree)))
        (right-sym (symbols (right-branch tree))))
    (cond ((memq symbol left-sym) (cons 0 (left-branch tree)))
          ((memq symbol right-sym) (cons 1 (right-branch tree)))
          (else (cons '() '())))))
  
(define (encode-symbol symbol tree)
  (if (null? tree) '()
      (let ((code (car (encode-branch symbol tree)))
            (branch (cdr (encode-branch symbol tree))))
        (if (not (null? code))
            (if (leaf? branch)
                (list code)
                (cons code (encode-symbol symbol branch)))
            (error symbol "not in tree")))))
      

(define (encode message tree)
  (if (null? message)
      '()
      (append (encode-symbol (car message) tree)
              (encode (cdr message) tree))))

;; EXERCISE 2.69

(define (successive-merge leaf-set)
  (if (null? (cdr leaf-set))
      (car leaf-set)
      (let ((element1 (car leaf-set))
            (element2 (cadr leaf-set)))
        (successive-merge
         (adjoin-set (make-code-tree element1 element2)
                     (cddr leaf-set))))))
    

(define (generate-huffman-tree pairs)
  (successive-merge (make-leaf-set pairs)))

;; 2.70

(define song-pairs '((a 2) (get 2) (sha 3) (wah 1) (boom 1) (job 2) (na 16) (yip 9)))

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

(define song '(get a job sha na na na na na na na na get a job sha na na na na na na na na sha yip yip yip yip yip yip yip yip yip sha boom))

(define encoded-length (length (encode song song-tree))) ;; 83 bits

;; Song is 36 symobls long.  A fixed-length code would require 3-bits per symbol (8 symbols).  So the
;; song would need 36*3=108 bits.

;; 2.71

(define pairs-5 '((a 1) (b 2) (c 4) (d 8) (e 16)))

(define pairs-10 (append pairs-5
                         '((f 32) (g 64) (h 128) (i 256) (j 512))))

;; To encode the most frequent symbol requires 1 bit.  To encode the most frequent symbol requires
;; n-1 bits (where there are n symbols with weights 1,...,2^{n-1}).  The resulting huffman tree in this
;; case is the most unbalanced tree possible, so is n-1 deep.

;; 2.72
;; To encode the most frequent symbol requires a single lookup into n-1 symbols plus a lookup of
;; 1 symbol.  The tree is only 1 deep for this case, so O(n).
;; For the least frequent symbol, we need to look up from (n-1) symbols and 1 symbol, then (n-2) symbols
;; and one symbol.  This repeats n times, so O(n^2).