Add exercises from section 2.3.4
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67
chapter-2/ex-2.67.scm
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67
chapter-2/ex-2.67.scm
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#lang sicp
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(define (make-leaf symbol weight) (list 'leaf symbol weight))
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(define (leaf? object) (eq? (car object) 'leaf))
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(define (symbol-leaf x) (cadr x))
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(define (weight-leaf x) (caddr x))
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(define (make-code-tree left right)
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(list left
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right
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(append (symbols left) (symbols right))
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(+ (weight left) (weight right))))
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(define (left-branch tree) (car tree))
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(define (right-branch tree) (cadr tree))
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(define (symbols tree)
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(if (leaf? tree)
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(list (symbol-leaf tree))
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(caddr tree)))
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(define (weight tree)
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(if (leaf? tree)
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(weight-leaf tree)
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(cadddr tree)))
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(define (decode bits tree)
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(define (decode-1 bits current-branch)
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(if (null? bits)
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'()
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(let ((next-branch
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(choose-branch (car bits) current-branch)))
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(if (leaf? next-branch)
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(cons (symbol-leaf next-branch)
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(decode-1 (cdr bits) tree))
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(decode-1 (cdr bits) next-branch)))))
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(decode-1 bits tree))
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(define (choose-branch bit branch)
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(cond ((= bit 0) (left-branch branch))
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((= bit 1) (right-branch branch))
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(else (error "bad bit: CHOOSE-BRANCH" bit))))
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(define (adjoin-set x set)
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(cond ((null? set) (list x))
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((< (weight x) (weight (car set))) (cons x set))
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(else (cons (car set)
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(adjoin-set x (cdr set))))))
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(define (make-leaf-set pairs)
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(if (null? pairs)
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'()
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(let ((pair (car pairs)))
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(adjoin-set (make-leaf (car pair) ; symbol
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(cadr pair)) ; frequency
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(make-leaf-set (cdr pairs))))))
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(define sample-tree
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(make-code-tree (make-leaf 'A 4)
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(make-code-tree
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(make-leaf 'B 2)
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(make-code-tree
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(make-leaf 'D 1)
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(make-leaf 'C 1)))))
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(define sample-message '(0 1 1 0 0 1 0 1 0 1 1 1 0))
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; (decode sample-message sample-tree)
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; (A D A B B C A)
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99
chapter-2/ex-2.68.scm
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99
chapter-2/ex-2.68.scm
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#lang sicp
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(define (make-leaf symbol weight) (list 'leaf symbol weight))
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(define (leaf? object) (eq? (car object) 'leaf))
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(define (symbol-leaf x) (cadr x))
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(define (weight-leaf x) (caddr x))
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(define (make-code-tree left right)
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(list left
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right
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(append (symbols left) (symbols right))
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(+ (weight left) (weight right))))
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(define (left-branch tree) (car tree))
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(define (right-branch tree) (cadr tree))
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(define (symbols tree)
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(if (leaf? tree)
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(list (symbol-leaf tree))
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(caddr tree)))
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(define (weight tree)
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(if (leaf? tree)
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(weight-leaf tree)
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(cadddr tree)))
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(define (decode bits tree)
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(define (decode-1 bits current-branch)
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(if (null? bits)
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'()
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(let ((next-branch
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(choose-branch (car bits) current-branch)))
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(if (leaf? next-branch)
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(cons (symbol-leaf next-branch)
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(decode-1 (cdr bits) tree))
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(decode-1 (cdr bits) next-branch)))))
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(decode-1 bits tree))
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(define (choose-branch bit branch)
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(cond ((= bit 0) (left-branch branch))
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((= bit 1) (right-branch branch))
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(else (error "bad bit: CHOOSE-BRANCH" bit))))
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(define (adjoin-set x set)
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(cond ((null? set) (list x))
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((< (weight x) (weight (car set))) (cons x set))
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(else (cons (car set)
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(adjoin-set x (cdr set))))))
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(define (make-leaf-set pairs)
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(if (null? pairs)
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'()
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(let ((pair (car pairs)))
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(adjoin-set (make-leaf (car pair) ; symbol
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(cadr pair)) ; frequency
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(make-leaf-set (cdr pairs))))))
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(define sample-tree
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(make-code-tree (make-leaf 'A 4)
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(make-code-tree
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(make-leaf 'B 2)
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(make-code-tree
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(make-leaf 'D 1)
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(make-leaf 'C 1)))))
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(define sample-message '(0 1 1 0 0 1 0 1 0 1 1 1 0))
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; (decode sample-message sample-tree)
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; (A D A B B C A)
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; until now previous exercise
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; setup for the current one
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(define (encode message tree)
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(if (null? message)
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'()
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(append (encode-symbol (car message) tree)
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(encode (cdr message) tree))))
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; my procedure
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; assumes that a tree is well formed, ie. that if a symbol appears in a
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; parent node's list it must be in one of the child branches
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; not the most efficient since it checks symbol lists at almost every level
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; twice
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(define (encode-symbol symbol tree)
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(cond ((not (member-set symbol (symbols tree)))
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(error "encode-symbol: no seq for the symbol in tree" symbol tree))
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((leaf? tree)
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'())
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((member-set symbol (symbols (left-branch tree)))
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(cons 0 (encode-symbol symbol (left-branch tree))))
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((member-set symbol (symbols (right-branch tree)))
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(cons 1 (encode-symbol symbol (right-branch tree))))))
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(define member-set member)
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; since a set is represented as a list, we can simply use the member procedure
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;test:
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; input: (A D A B B C A)
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; output:(0 1 1 0 0 1 0 1 0 1 1 1 0)
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; it works!
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66
chapter-2/ex-2.69.scm
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66
chapter-2/ex-2.69.scm
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#lang sicp
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(define (make-leaf symbol weight) (list 'leaf symbol weight))
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(define (leaf? object) (eq? (car object) 'leaf))
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(define (symbol-leaf x) (cadr x))
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(define (weight-leaf x) (caddr x))
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(define (make-code-tree left right)
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(list left
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right
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(append (symbols left) (symbols right))
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(+ (weight left) (weight right))))
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(define (left-branch tree) (car tree))
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(define (right-branch tree) (cadr tree))
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(define (symbols tree)
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(if (leaf? tree)
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(list (symbol-leaf tree))
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(caddr tree)))
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(define (weight tree)
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(if (leaf? tree)
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(weight-leaf tree)
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(cadddr tree)))
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(define (decode bits tree)
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(define (decode-1 bits current-branch)
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(if (null? bits)
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'()
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(let ((next-branch
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(choose-branch (car bits) current-branch)))
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(if (leaf? next-branch)
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(cons (symbol-leaf next-branch)
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(decode-1 (cdr bits) tree))
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(decode-1 (cdr bits) next-branch)))))
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(decode-1 bits tree))
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(define (choose-branch bit branch)
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(cond ((= bit 0) (left-branch branch))
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((= bit 1) (right-branch branch))
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(else (error "bad bit: CHOOSE-BRANCH" bit))))
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(define (adjoin-set x set)
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(cond ((null? set) (list x))
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((< (weight x) (weight (car set))) (cons x set))
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(else (cons (car set)
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(adjoin-set x (cdr set))))))
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(define (make-leaf-set pairs)
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(if (null? pairs)
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'()
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(let ((pair (car pairs)))
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(adjoin-set (make-leaf (car pair) ; symbol
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(cadr pair)) ; frequency
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(make-leaf-set (cdr pairs))))))
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; actual exercise
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(define (generate-huffman-tree pairs)
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(successive-merge (make-leaf-set pairs)))
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(define (successive-merge tree-set)
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(if (null? (cdr tree-set)) ; (= (length tree-list) 1)
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(car tree-set)
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(let ((tree1 (car tree-set))
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(tree2 (cadr tree-set)))
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(successive-merge (adjoin-set (make-code-tree tree1 tree2)
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(cddr tree-set))))))
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109
chapter-2/ex-2.70.scm
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109
chapter-2/ex-2.70.scm
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#lang sicp
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(define (make-leaf symbol weight) (list 'leaf symbol weight))
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(define (leaf? object) (eq? (car object) 'leaf))
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(define (symbol-leaf x) (cadr x))
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(define (weight-leaf x) (caddr x))
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(define (make-code-tree left right)
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(list left
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right
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(append (symbols left) (symbols right))
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(+ (weight left) (weight right))))
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(define (left-branch tree) (car tree))
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(define (right-branch tree) (cadr tree))
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(define (symbols tree)
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(if (leaf? tree)
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(list (symbol-leaf tree))
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(caddr tree)))
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(define (weight tree)
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(if (leaf? tree)
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(weight-leaf tree)
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(cadddr tree)))
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(define (decode bits tree)
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(define (decode-1 bits current-branch)
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(if (null? bits)
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'()
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(let ((next-branch
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(choose-branch (car bits) current-branch)))
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(if (leaf? next-branch)
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(cons (symbol-leaf next-branch)
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(decode-1 (cdr bits) tree))
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(decode-1 (cdr bits) next-branch)))))
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(decode-1 bits tree))
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(define (choose-branch bit branch)
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(cond ((= bit 0) (left-branch branch))
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((= bit 1) (right-branch branch))
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(else (error "bad bit: CHOOSE-BRANCH" bit))))
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(define (adjoin-set x set)
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(cond ((null? set) (list x))
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((< (weight x) (weight (car set))) (cons x set))
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(else (cons (car set)
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(adjoin-set x (cdr set))))))
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(define (make-leaf-set pairs)
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(if (null? pairs)
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'()
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(let ((pair (car pairs)))
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(adjoin-set (make-leaf (car pair) ; symbol
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(cadr pair)) ; frequency
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(make-leaf-set (cdr pairs))))))
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(define (generate-huffman-tree pairs)
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(successive-merge (make-leaf-set pairs)))
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(define (successive-merge tree-set)
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(if (null? (cdr tree-set)) ; (= (length tree-list) 1)
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(car tree-set)
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(let ((tree1 (car tree-set))
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(tree2 (cadr tree-set)))
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(successive-merge (adjoin-set (make-code-tree tree1 tree2)
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(cddr tree-set))))))
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; the encode functionality from 2.68
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(define (encode message tree)
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(if (null? message)
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'()
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(append (encode-symbol (car message) tree)
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(encode (cdr message) tree))))
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(define (encode-symbol symbol tree)
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(cond ((not (member-set symbol (symbols tree)))
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(error "encode-symbol: no seq for the symbol in tree" symbol tree))
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((leaf? tree)
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'())
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((member-set symbol (symbols (left-branch tree)))
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(cons 0 (encode-symbol symbol (left-branch tree))))
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((member-set symbol (symbols (right-branch tree)))
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(cons 1 (encode-symbol symbol (right-branch tree))))))
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(define member-set member)
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; since a set is represented as a list, we can simply use the member procedure
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; actual exercise 2.70
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(define sample-tree
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(generate-huffman-tree '((a 2)
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(get 2)
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(sha 3)
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(wah 1)
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(boom 1)
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(job 2)
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(na 16)
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(yip 9))))
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(define sample-message
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'(get a job
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sha na na na na na na na na
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get a job
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sha na na na na na na na na
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wah yip yip yip yip yip yip yip yip yip
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sha boom))
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; (length (encode sample-message sample-tree)) -> 84
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; if we used a fixed length code, it would have to have length of >=3
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; since there are 36 words here, the answer would be 108
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36
chapter-2/ex-2.71.txt
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36
chapter-2/ex-2.71.txt
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(A 1), (B 2), (C 4), ...
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Tree sketch (n=5):
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*
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/ \
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(E 16) *
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/ \
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(D 8) *
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/ \
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(C 4) *
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/ \
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(B 2) (A 1)
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tree sketch (n=10):
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*
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/ \
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(J 512) *
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/ \
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(I 256) *
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/ \
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(H 128) *
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/ \
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(G 64) *
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/ \
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(F 32) *
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/ \
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(E 16) *
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/ \
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(D 8) *
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/ \
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(C 4) *
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/ \
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(B 2) (A 1)
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The most frequent symbol always requires exactly 1 bit.
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The two least frequent symbols will always require exactly n-1 bits.
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12
chapter-2/ex-2.72.txt
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Searching a single node requires on the order of k operations, where
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k is the number of symbols given at that node.
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If the tree is roughly balanced, then the number of symbols listed in
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each node on the way down should decrease exponentially, and the number
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of these searches would be on the order of O(log(n)), where n is the
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number of symbols generally available for encoding.
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If this is true, than each encoding of a symbol would take on the order
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of O(n*log(n)) operations, however if the tree is very unbalanced,
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which is the case in exercise 2.71, the order of growth is roughly
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O(n^2), for the least frequent symbols, and O(n), for the most common
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symbol.
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