Problem 5: Hailstone (100pts)

Problem

Douglas Hofstadter's Pulitzer-prize-winning book, Gödel, Escher, Bach, poses the following mathematical puzzle.

Pick a positive integer \(x\) as the start.

If \(x\) is even, divide it by 2.

If \(x\) is odd, multiply it by 3 and add 1.

Continue this process until \(x\) is 1.

The number \(x\) will travel up and down but eventually end at 1 (at least for all numbers that have ever been tried -- nobody has ever proved that the sequence will terminate). Analogously, a hailstone travels up and down in the atmosphere before eventually landing on earth.

Breaking News (or at least the closest thing to that in math). There has been a recent development in the hailstone conjecture that shows that almost all numbers will eventually get to 1 if you repeat this process. This isn't a complete proof but a major breakthrough.

Breaking News (or at least the closest thing to that in math). There has been a recent development in the hailstone conjecture that shows that almost all numbers will eventually get to 1 if you repeat this process. This isn't a complete proof but a major breakthrough.

Hailstone sequences can get quite long! Try 27. What's the longest you can find?

道格拉斯·霍夫斯塔特（Douglas Hofstadter）荣获普利策奖的著作《哥德尔、埃舍尔、巴赫》（Gödel, Escher, Bach）提出了以下数学谜题。

选择一个正整数 \(x\) 作为起点。
如果 \(x\) 是偶数，则将其除以 2。
如果 \(x\) 是奇数，则将其乘以 3 并加 1。
重复此过程直到 \(x\) 等于 1。

数字 \(x\) 会上下波动，但最终会以 1 结束（至少对于所有已尝试过的数字来说是如此——没有人证明过这个序列一定会终止）。类似地，冰雹在最终落到地面之前，会在大气层中上下运动。

劲爆消息！（或者至少是数学领域最劲爆的消息）。关于冰雹序列猜想（Hailstone Conjecture）最近有了新的进展，表明如果你重复这个过程，几乎所有数字最终都会达到 1。这并非一个完整的证明，但却是一个重大的突破。

冰雹序列可能会相当长！试试 27。您能找到的最长序列是多长？

def hailstone(x):
    """Print the hailstone sequence starting at x and return its
    length.

    >>> a = hailstone(10)
    10
    5
    16
    8
    4
    2
    1
    >>> a
    7
    """
    "*** YOUR CODE HERE ***"

Hints

你可能需要使用 while 循环。
这个函数只有一个参数，可能不太适合用递归来做（但也不是不能做）
注意如果你在运算中使用了单除号 /，那么结果会变成浮点数！不管能不能整除！
别忘了打印开头（x 自己）和结尾（1）

Solutions

使用 while 循环解决这个问题。注意当 x == 1 时要退出循环。

def hailstone(x):
    # Initialize the sequence length to 1 to count the starting number x.
    length = 1

    # Loop continues as long as the current number x is not 1.
    while x != 1:
        # Print the current number in the sequence.
        print(x)

        # Apply the Hailstone rules:
        if x % 2 == 0:
            # If x is even, the next number is x / 2 (use // for integer division).
            x = x // 2
        else:
            # If x is odd, the next number is 3 * x + 1.
            x = 3 * x + 1

        # Increment the length of the sequence.
        length += 1

    # After the loop, x is 1. Print the final number in the sequence.
    print(x) # print(1) is also correct.

    # Return the total length of the sequence.
    return length