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sicp-study-group | ||
================ | ||
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A lay-led study of [SICP](https://mitpress.mit.edu/sicp/) in Berkeley, CA. | ||
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* http://www.meetup.com/codeselfstudy/events/187351632/ | ||
* http://www.meetup.com/codeselfstudy/events/187758402/ | ||
* ... | ||
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This repo houses my solutions to exercises. |
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;;; Installed Scheme via `brew install mit-scheme` on OS X. | ||
;;; Running Scheme interpreter in Emacs via `M-x run-scheme`. | ||
;;; Donny Winston, 2014-06-10 | ||
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;; MIT/GNU Scheme running under OS X | ||
;; Type `^C' (control-C) followed by `H' to obtain information about interrupts. | ||
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;; Copyright (C) 2014 Massachusetts Institute of Technology | ||
;; This is free software; see the source for copying conditions. There is NO | ||
;; warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR | ||
;; PURPOSE. | ||
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;; Image saved on Friday May 23, 2014 at 5:50:11 PM | ||
;; Release 9.2 || Microcode 15.3 || Runtime 15.7 || SF 4.41 || LIAR/C 4.118 | ||
;; Edwin 3.116 | ||
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;; Exercise 1.1 | ||
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1 ]=> 10 | ||
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;Value: 10 | ||
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1 ]=> (+ 5 3 4) | ||
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;Value: 12 | ||
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1 ]=> (- 9 1) | ||
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;Value: 8 | ||
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1 ]=> (/ 6 2) | ||
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;Value: 3 | ||
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1 ]=> (+ (* 2 4) (- 4 6)) | ||
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;Value: 6 | ||
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1 ]=> (define a 3) | ||
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;Value: a | ||
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1 ]=> (define b (+ a 1)) | ||
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;Value: b | ||
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1 ]=> (+ a b (* a b)) | ||
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;Value: 19 | ||
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1 ]=> (= a b) | ||
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;Value: #f | ||
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1 ]=> (if (and (> b a) (< b (* a b))) | ||
b | ||
a) | ||
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;Value: 4 | ||
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1 ]=> (cond ((= a 4) 6) | ||
((= b 4) (+ 6 7 a)) | ||
(else 25)) | ||
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;Value: 16 | ||
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1 ]=> (+ 2 (if (> b a) b a)) | ||
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;Value: 6 | ||
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1 ]=> (* (cond ((> a b) a) | ||
((< a b) b) | ||
(else -1)) | ||
(+ a 1)) | ||
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;Value: 16 | ||
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1 ]=> |
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;; Exercise 1.2 | ||
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(/ (+ 4 5 (- 2 (- 3 (+ 6 (/ 4 5))))) | ||
(* 3 (- 6 2) (- 2 7))) | ||
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;Value: -37/150 |
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;; Exercise 1.3 | ||
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(define (sum-of-squares x y) (+ (square x) (square y))) | ||
(define (sum-squares-of-two-largest a b c) | ||
(cond ((and (> a c) (> b c)) (sum-of-squares a b)) | ||
((and (> a b) (> c b)) (sum-of-squares a c)) | ||
(else (sum-of-squares b c)))) | ||
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(= (sum-squares-of-two-largest 3 4 5) (sum-of-squares 4 5)) | ||
(= (sum-squares-of-two-largest 5 4 3) (sum-of-squares 4 5)) | ||
(= (sum-squares-of-two-largest 3 5 4) (sum-of-squares 4 5)) |
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;; Exercise 1.4 | ||
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(define (a-plus-abs-b a b) | ||
((if (> b 0) + -) a b)) | ||
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;; If b > 0, return `(+ a b)`. | ||
;; Otherwise, return `(- a b)`. | ||
;; Thus, return `(+ a (abs b))`, | ||
;; where `(abs b)` is the absolute value of b. |
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;; Exercise 1.5 | ||
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;; normal-order evaluation: "fully expand and then reduce" | ||
;; applicative-order evaluation: "evaluate the arguments and then apply" | ||
;; Lisp uses applicative-order evaluation, but normal-order can be useful | ||
;; e.g. for stream processing (to be introduced in Ch 3). | ||
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(define (p) (p)) | ||
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(define (test x y) | ||
(if (= x 0) | ||
0 | ||
y)) | ||
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(test 0 (p)) | ||
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;; With applicative-order evaluation, the evaluation will never return a value | ||
;; because `(p)` will never return. With normal-order evaluation, because the | ||
;; arguments to `test` are not evaluated prior to evaluating the body of `test`, | ||
;; and because `y` is never evaluated (because `(= x 0)` evaluates to true), | ||
;; `(test 0 (p))` will return 0. |