Randomize conditionals #201
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hakatashi
requested changes
Dec 7, 2016
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| conditional2-calc = (x) -> | ||
| if x is 5 | ||
| then 10 |
| @@ -51,6 +54,16 @@ sum-of-digits = (n) -> n.to-string!split '' .map (parse-int _, 10) .reduce (+) | |||
| fibonacci-calc = (n, value = 1, prev = 0) -> | |||
| if n is 0 then prev else fibonacci-calc n - 1, value + prev, value | |||
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| conditional1-calc = (x) -> | |||
| if (x % 2) is 0 then 2 * x else 3 * x | |||
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括弧いらない気がする
(LiveScriptではなるべく括弧を殺していきたい)
| expect io .to.satisfy io-spec | ||
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| expect zip io.input, io.output .to.all.satisfy ([input, output]) -> | ||
| output is conditional1-calc input |
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inputが200未満になる旨のテストもあったほうがよさそう
| output: [10, 18, null, null], | ||
| ioGenerator: (random) => { | ||
| const generateAnswer = (input) => { | ||
| if (input === 5) { |
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inputが5か6のテストケースはすでに(固定で)あるので、5か6がランダムに生成されても嘘解法が通りやすくなるだけな気がするけど、どうだろう⋯⋯
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これ、めんどくさくてやってなかっただけなので、普通にやります
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各種、終わりました。レビューお願いします |
hakatashi
reviewed
Dec 8, 2016
| } | ||
| return number + offset; | ||
| }; | ||
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うーん、やりたいことは何となくわかるけど、ちょっと複雑になりすぎな感じがある。ここまで一般的なアルゴリズムを書かなくても、今回の場合は1から200までの数値を含む配列から5と6を除いたものから2個乱択する処理を書けば十分だと思う。
こんな感じ。
const candidates = Array.from({length: 200}, (item, index) => index + 1).filter(n => n !== 5 && n !== 6);
const input1 = candidates[Math.floor(random() * candidates.length)];
const input2 = candidates[Math.floor(random() * candidates.length)];あと、5や6が生まれなくなるのでgenerateAnswer関数もいらないと思う。
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＞ 2個乱択する処理を書けば十分
わかりました・・・
＞ 5や6が生まれなくなるのでgenerateAnswer関数
たしかにそうですね・・・
hakatashi
approved these changes
Dec 8, 2016
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マージします |
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通りました。レビューお願いします