2017年12月27日水曜日

学習環境

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。


  1. a k - b k = a - b k = m n k n = n m k

    よって、

    a k b k m o d m k

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 =

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