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Dec 29, 2022
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Change gcd code #150

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40 changes: 36 additions & 4 deletions docs/02advanced/04recursion/index.mdx
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Expand Up @@ -214,20 +214,52 @@ $$

<Answer>

ユークリッドの互除法を用いれば簡単にできます。
ユークリッドの互除法を用いれば簡単にできます。$\mathrm{gcd}(a, b)$ を $a$ と $b$ の最大公約数とします。

例：$30$ と $18$ の最大公約数を求める。

- $30$ を $18$ で割った余りは、$12$
- $18$ を $12$ で割った余りは、$6$
- $12$ を $6$ で割った余りは、$0$
$$
\begin{align*}
\mathrm{gcd}(30, 18) &= \mathrm{gcd}(18, 30 - 18\times 1) \\
&= \mathrm{gcd}(18, 12) \\
&= \mathrm{gcd}(12, 18 - 12\times 1) \\
&= \mathrm{gcd}(12, 6) \\
&= \mathrm{gcd}(6, 12 - 6\times 2) \\
&= \mathrm{gcd}(6, 0) \\
&= 6
\end{align*}
$$

よって、最大公約数は $6$

これを使うと、最大公約数を求めるプログラムは次のようになります。

<ViewSource path="/recursion/gcd.ipynb" />

:::note

このプログラムは、はじめの引数が $a < b$ の場合でも動きます。一度目の再帰で、`gcd(b, a % b)` = `gcd(b, a)` が返されるからです。

つまり、$18$ と $30$ の最大公約数を求めるとき、

$$
\begin{align*}
\mathrm{gcd}(18, 30) &= \mathrm{gcd}(30, 18 - 30\times 0) \\
&= \mathrm{gcd}(30, 18) \\
&= \mathrm{gcd}(18, 30 - 18\times 1) \\
&= \mathrm{gcd}(18, 12) \\
&= \mathrm{gcd}(12, 18 - 12\times 1) \\
&= \mathrm{gcd}(12, 6) \\
&= \mathrm{gcd}(6, 12 - 6\times 2) \\
&= \mathrm{gcd}(6, 0) \\
&= 6
\end{align*}
$$

となります。

:::

また、最大公約数 $\mathrm{gcd}(a, b)$ と最小公倍数 $\mathrm{lcm}(a, b)$ の間には次のような関係があります。

$$
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8 changes: 3 additions & 5 deletions static/practice/gcd_recursive.ipynb
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Expand Up @@ -7,12 +7,10 @@
"metadata": {},
"source": [
"def gcd(a, b):\n",
" if a < b:\n",
" a, b = b, a\n",
" if b != 0:\n",
" return gcd(b, a % b)\n",
" else:\n",
" if b == 0:\n",
" return a\n",
" else:\n",
" return gcd(b, a % b)\n",
"\n",
"\n",
"print(gcd(12, 16))"
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2 changes: 0 additions & 2 deletions static/recursion/gcd.ipynb
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Expand Up @@ -7,8 +7,6 @@
"metadata": {},
"source": [
"def gcd(a, b):\n",
" if a < b:\n",
" a, b = b, a\n",
" if b == 0:\n",
" return a\n",
" else:\n",
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14 changes: 6 additions & 8 deletions static/recursion/lcm.ipynb
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Expand Up @@ -6,17 +6,15 @@
{
"metadata": {},
"source": [
"def gcd(m, n):\n",
" if m < n:\n",
" m, n = n, m\n",
" if n == 0:\n",
" return m\n",
"def gcd(a, b):\n",
" if b == 0:\n",
" return a\n",
" else:\n",
" return gcd(n, m % n)\n",
" return gcd(b, a % b)\n",
"\n",
"\n",
"def lcm(m, n):\n",
" return m * n / gcd(m, n)\n",
"def lcm(a, b):\n",
" return a * b / gcd(a, b)\n",
"\n",
"\n",
"print(lcm(30, 18))"
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