学習環境
- Surface 3 (4G LTE)、Surface 3 タイプ カバー、Surface ペン(端末)
- Windows 10 Pro (OS)
- Nebo(Windows アプリ)
- iPad Pro + Apple Pencil
- MyScript Nebo(iPad アプリ)
- 参考書籍
代数系入門 (松坂 和夫(著)、岩波書店)の第1章(整数)、6(同値関係、合同式)、問題3.を取り組んでみる。
よって、
コード(Emacs)
Python 3
#!/usr/bin/env python3
from sympy import pprint, symbols
import random
def mod(a, b, m):
return (a - b) % m == 0
k = symbols('k', integer=True, positive=True)
for _ in range(10):
a = random.randrange(-100, 100)
b = random.randrange(-100, 100)
m = random.randrange(1, 11)
for t in [mod(a, b, m), mod(a, b, m)]:
if t:
print(f'{a} ≡ {b} (mod {m}): {t}')
print(f'{a * k} ≡ {b * k} (mod {m * k}): {mod(a * k, b * k, m * k)}')
print()
入出力結果(Terminal, Jupyter(IPython))
$ ./sample3.py 10 ≡ -47 (mod 1): True 10*k ≡ -47*k (mod k): True 10 ≡ -47 (mod 1): True 10*k ≡ -47*k (mod k): True 36 ≡ 38 (mod 1): True 36*k ≡ 38*k (mod k): True 36 ≡ 38 (mod 1): True 36*k ≡ 38*k (mod k): True -2 ≡ -50 (mod 6): True -2*k ≡ -50*k (mod 6*k): True -2 ≡ -50 (mod 6): True -2*k ≡ -50*k (mod 6*k): True $
HTML5
<pre id="output0"></pre> <input id="a0" type="number" step="1" value="6"> ≡ <input id="b0" type="number" step="1" value="11"> (mod <input id="m0" type="number" step="1" value="5">) <br> k = <input id="k0" type="number" min="1" step="1" value="2"> <button id="run0">run</button> <button id="clear0">clear</button> <script src="sample3.js"></script>
JavaScript
let pre0 = document.querySelector('#output0'),
run0 = document.querySelector('#run0'),
clear0 = document.querySelector('#clear0'),
input_a0 = document.querySelector('#a0'),
input_b0 = document.querySelector('#b0'),
input_m0 = document.querySelector('#m0'),
input_k0 = document.querySelector('#k0'),
inputs = [input_a0, input_b0, input_m0, input_k0],
range = (n) => {
let result = [];
for (let i = 0; i < n; i += 1) {
result.push(i);
}
return result;
},
p = (text) => pre0.textContent += text + '\n',
clear = () => pre0.textContent = '',
mod = (a, b, m) => (a - b) % m === 0,
output = () => {
let a = parseInt(input_a0.value, 10),
b = parseInt(input_b0.value, 10),
m = parseInt(input_m0.value, 10),
k = parseInt(input_k0.value, 10);
p(mod(a, b, m));
p(mod(a * k, b * k, m * k));
};
run0.onclick = output;
clear0.onclick = clear;
inputs.forEach((input) => input.onchange = output);
output();
≡
(mod )
k =
0 コメント:
コメントを投稿