Algorithm - Python - 水晶をどうぞ壊してください(壊れた水晶の個数、カウンター、forループ、if文、break文)

問題解決のPythonプログラミング
原著
楽天ブックス(Kobo) Yahoo!

開発環境

• macOS Mojave - Apple (OS)
• Emacs (Text Editor)
• Windows 10 Pro (OS)
• Visual Studio Code (Text Editor)
• Python 3.7 (プログラミング言語)

問題解決のPythonプログラミング ―数学パズルで鍛えるアルゴリズム的思考 (Srini Devadas (著)、黒川 利明 (翻訳)、オライリージャパン)の5章(水晶をどうぞ壊してください)、練習問題(問題2)を取り組んでみる。

コード(Emacs)

Python 3

#!/usr/bin/env python3
def convert_to_decimal(r, d, rep):
    '''
    >>> convert_to_decimal(4, 4, [0, 0, 0, 0])
    0
    >>> convert_to_decimal(4, 4, [1, 2, 3, 3])
    111
    >>> convert_to_decimal(4, 4, [2, 0, 0, 0])
    128
    >>> convert_to_decimal(4, 5, [1, 0, 0, 1, 4]) # 誤植？
    65
    >>> convert_to_decimal(4, 5, [1, 3, 3, 3, 4]) # 誤植？
    127
    >>> convert_to_decimal(4, 4, [1, 0, 0, 1]) # 修正
    65
    >>> convert_to_decimal(4, 4, [1, 3, 3, 3]) # 修正
    127
    >>> convert_to_decimal(4, 4, [1, 2, 0, 1])
    97
    >>> convert_to_decimal(4, 4, [1, 2, 3, 3])
    111
    '''
    return sum([rep[i] * r ** (d - (i + 1)) for i in range(d)])


def how_hard_is_the_crystal(n, d, did_break=None):
    r = 1
    while r ** d <= n:
        r += 1
    print(f'Radix chosen is {r}')

    # while r ** (d - 1) > n:
    #     d -= 1
    ball_number = 1
    num_drops = 0
    floor_no_break = [0] * d
    for i in range(d):
        for j in range(r - 1):
            floor_no_break[i] += 1
            floor = convert_to_decimal(r, d, floor_no_break)
            if floor > n:
                floor_no_break[i] -= 1
                break
            print(f'Broken balls: {ball_number - 1}')
            print(f'Drop ball {ball_number} from floor {floor}')
            # did_break = input('Did the ball break (yes/no)?: ')
            num_drops += 1
            # if did_break == 'yes':
            if did_break[i] == 'yes':
                floor_no_break[i] -= 1
                ball_number += 1
                break
    hardness = convert_to_decimal(r, d, floor_no_break)
    return hardness, num_drops


if __name__ == '__main__':
    # import doctest
    # doctest.testmod()

    def yes_no(n):
        if n % 2 == 0:
            return 'yes'
        return 'no'
    for did_break in [['no'] * 6,
                      ['yes'] * 6,
                      [yes_no(i) for i in range(6)],
                      [yes_no(i + 1) for i in range(6)]]:
        print(how_hard_is_the_crystal(128, 6, did_break))

入出力結果(Terminal, Jupyter(IPython))

$ ./sample2.py
Radix chosen is 3
Broken balls: 0
Drop ball 1 from floor 81
Broken balls: 0
Drop ball 1 from floor 108
Broken balls: 0
Drop ball 1 from floor 117
Broken balls: 0
Drop ball 1 from floor 126
Broken balls: 0
Drop ball 1 from floor 127
Broken balls: 0
Drop ball 1 from floor 128
(128, 6)
Radix chosen is 3
Broken balls: 0
Drop ball 1 from floor 81
Broken balls: 1
Drop ball 2 from floor 27
Broken balls: 2
Drop ball 3 from floor 9
Broken balls: 3
Drop ball 4 from floor 3
Broken balls: 4
Drop ball 5 from floor 1
(0, 5)
Radix chosen is 3
Broken balls: 0
Drop ball 1 from floor 81
Broken balls: 0
Drop ball 1 from floor 108
Broken balls: 1
Drop ball 2 from floor 90
Broken balls: 1
Drop ball 2 from floor 99
Broken balls: 1
Drop ball 2 from floor 102
Broken balls: 2
Drop ball 3 from floor 100
Broken balls: 2
Drop ball 3 from floor 101
(101, 7)
Radix chosen is 3
Broken balls: 0
Drop ball 1 from floor 81
Broken balls: 1
Drop ball 2 from floor 27
Broken balls: 1
Drop ball 2 from floor 54
Broken balls: 1
Drop ball 2 from floor 63
Broken balls: 2
Drop ball 3 from floor 57
Broken balls: 2
Drop ball 3 from floor 60
Broken balls: 2
Drop ball 3 from floor 61
(60, 7)
$