blog/content/post/2008/11/30/2008-11-30-reading-sicp-2.md

80 lines
5.9 KiB
Markdown
Raw Blame History

This file contains ambiguous Unicode characters

This file contains Unicode characters that might be confused with other characters. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.

---
title: Reading SICP 2
author: kazu634
date: 2008-11-30
wordtwit_post_info:
- 'O:8:"stdClass":13:{s:6:"manual";b:0;s:11:"tweet_times";i:1;s:5:"delay";i:0;s:7:"enabled";i:1;s:10:"separation";s:2:"60";s:7:"version";s:3:"3.7";s:14:"tweet_template";b:0;s:6:"status";i:2;s:6:"result";a:0:{}s:13:"tweet_counter";i:2;s:13:"tweet_log_ids";a:1:{i:0;i:4409;}s:9:"hash_tags";a:0:{}s:8:"accounts";a:1:{i:0;s:7:"kazu634";}}'
categories:
- Lisp
---
<div class="section">
<p>
<i><a href="http://d.hatena.ne.jp/asin/0262510871" onclick="__gaTracker('send', 'event', 'outbound-article', 'http://d.hatena.ne.jp/asin/0262510871', 'Structure and Interpretation of Computer Programs (MIT Electrical Engineering and Computer Science)');">Structure and Interpretation of Computer Programs (MIT Electrical Engineering and Computer Science)</a></i>を読む、その2です。今回は
</p>
<ul>
<li>
compound procedure例えば (define (square a) (* a a))しておいてから、(define (sum-of-square a b) (+ (square a) (square b)))と書いた時の「(define (sum-of-square a b) (+ (square a) (square b)))」のこと) <ul>
<li>
substitution model (インタープリタ内部でどのようにcompound proedureを評価するのか。暫定的な説明として、「手続きを置き換え(substitute)てprimitiveな手続きのみの組み合わせに変換する」というモデルが導入される。でも、これは後で破綻するよって書いてあった) <ul>
<li>
Applicative orderインタープリタが行っている方法。引数を先に評価してから、operandを適用する
</li>
<li>
normal order(まずはoperandをどんどんprimitiveな手続きに変換する)
</li>
</ul>
</li>
</ul>
</li>
</ul>
<p>
というようなことが書いてありました(私の理解が間違えていなければ…)。
</p>
<h4>
Excercise (pp 20-21)
</h4>
<h5>
Excercise 1.2
</h5>
<blockquote>
<p>
Translate the following expression into prefix form
</p>
<blockquote>
<p>
<img src="http://d.hatena.ne.jp/cgi-bin/mimetex.cgi?\frac{5~+~4~+~&#40;2~-~&#40;3~-~&#40;6~+~\frac{4}{5}&#41;&#41;&#41;}{3&#40;6~-~2&#41;&#40;2~-~7&#41;}" class="tex" alt="¥frac{5 + 4 + &#40;2 - &#40;3 - &#40;6 + ¥frac{4}{5}&#41;&#41;&#41;}{3&#40;6 - 2&#41;&#40;2 - 7&#41;}" />
</p>
</blockquote>
</blockquote>
<pre class="syntax-highlight">
<span class="synSpecial">(</span><span class="synStatement">/</span> <span class="synSpecial">(</span><span class="synStatement">+</span> <span class="synConstant">5</span> <span class="synConstant">4</span> <span class="synSpecial">(</span><span class="synStatement">-</span> <span class="synConstant">2</span> <span class="synSpecial">(</span><span class="synStatement">-</span> <span class="synConstant">3</span> <span class="synSpecial">(</span><span class="synStatement">+</span> <span class="synConstant">6</span> <span class="synSpecial">(</span><span class="synStatement">/</span> <span class="synConstant">4</span> <span class="synConstant">5</span><span class="synSpecial">)))))</span> <span class="synSpecial">(</span><span class="synStatement">*</span> <span class="synConstant">3</span> <span class="synSpecial">(</span><span class="synStatement">-</span> <span class="synConstant">6</span> <span class="synConstant">2</span><span class="synSpecial">)</span> <span class="synSpecial">(</span><span class="synStatement">-</span> <span class="synConstant">2</span> <span class="synConstant">7</span><span class="synSpecial">)))</span>
</pre>
<h5>
Excercise 1.3
</h5>
<blockquote>
<p>
Define a procedure that takes 3 numbers as arguments and returns the sum of the squares of the two larger numbers.
</p>
</blockquote>
<pre class="syntax-highlight">
<span class="synSpecial">(</span>define <span class="synSpecial">(</span>sum-of-square a b c<span class="synSpecial">)</span>
<span class="synSpecial">(</span>define <span class="synSpecial">(</span>square n<span class="synSpecial">)</span>
<span class="synSpecial">(</span><span class="synStatement">*</span> n n<span class="synSpecial">))</span>
<span class="synSpecial">(</span><span class="synStatement">cond</span> [<span class="synSpecial">(</span><span class="synStatement">and</span> <span class="synSpecial">(</span><span class="synStatement">&#60;</span> a b<span class="synSpecial">)</span> <span class="synSpecial">(</span><span class="synStatement">&#60;</span> a c<span class="synSpecial">))</span> <span class="synSpecial">(</span><span class="synStatement">+</span> <span class="synSpecial">(</span>square b<span class="synSpecial">)</span> <span class="synSpecial">(</span>square c<span class="synSpecial">))</span>] <span class="synComment">; when a is the smallest</span>
[<span class="synSpecial">(</span><span class="synStatement">and</span> <span class="synSpecial">(</span><span class="synStatement">&#60;</span> b a<span class="synSpecial">)</span> <span class="synSpecial">(</span><span class="synStatement">&#60;</span> b c<span class="synSpecial">))</span> <span class="synSpecial">(</span><span class="synStatement">+</span> <span class="synSpecial">(</span>square a<span class="synSpecial">)</span> <span class="synSpecial">(</span>square c<span class="synSpecial">))</span>] <span class="synComment">; when b is the smallest</span>
[<span class="synSpecial">(</span><span class="synStatement">and</span> <span class="synSpecial">(</span><span class="synStatement">&#60;</span> c a<span class="synSpecial">)</span> <span class="synSpecial">(</span><span class="synStatement">&#60;</span> c b<span class="synSpecial">))</span> <span class="synSpecial">(</span><span class="synStatement">+</span> <span class="synSpecial">(</span>square a<span class="synSpecial">)</span> <span class="synSpecial">(</span>square b<span class="synSpecial">))</span>]<span class="synSpecial">))</span> <span class="synComment">; when c is the smallest</span>
</pre>
</div>