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Published byMiracle Mellin Modified about 1 year ago

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Lisp II

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How EQUAL could be defined (defun equal (x y) ; this is how equal could be defined (cond ((numberp x) (= x y)) ((atom x) (eq x y)) ((atom y) nil) ((equal (car x) (car y)) (equal (cdr x) (cdr y)))))

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Some simple list processing examples (defun member (x l) (cond ((atom l) nil) ((equal x (car l)) (cdr l)) (T (member x (cdr l)))) (defun append (l1 l2) (if (null l1) l2 (cons (car l1) (append (cdr l1) l2))))

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Variations on reverse ;; either of these does O(n^2) cons operations (defun reverse (l) (if (null l) nil (append (reverse (cdr l)) (list (car l))))) (defun reverse (l) (and l (append (reverse (cdr l)) (list (car l)))))

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Variations on reverse ;; this tail recursive operation does O(n) conses (defun reverse (l) (reverse1 l nil) (defun reverse (l acc) (if (null l) acc (reverse (cdr l) (cons (car l) acc))))

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Flatten I (defun flatten (l) (cond ((null l) nil) ; empty list do nothing ((atom (car l)) ; cons an atom onto flattend cdr (cons (car l) (flatten (cdr l)))) ; otherwise flatten head & tail and append results (t (append (flatten (car l)) (flatten (cdr l))))))

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Flatten II ;; this version avoids append, which is expensive. (defun flatten (l) (flatten1 l nil)) (defun flatten1 (l acc) (cond ((null list) acc) ; all done ((atom l) ; stick atom on the front of acc (cons l acc)) ; of the accumulator (t (flatten1 (car l) (flatten1 (cdr l) acc)))

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Higher order functions (defun mapcar (f l) (if (null l) nil (cons (apply f (list (car l))) (mapcar f (cdr l)))))

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Mapcar II (defun mapcar1 (f l) (if (null l) nil (cons (apply f (list (car l))) (mapcar f (cdr l))))) (defun mapcar (f &rest args) "Apply f to successive cars of all ARGS. Return the list of results." ;; If no list is exhausted, (if (not (memq 'nil args)) ; apply function to CARs. (cons (apply f (mapcar1 'car args)) (mapcar* f (mapcar1 'cdr args)))))

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