Collatz 循环结构

Question

请帮助我理解我做错了什么。问题是

The Collatz Sequence Write a function named collatz() that has one parameter named number. If number is even, then collatz() should print number // 2 and return this value. If number is odd, then collatz() should print and return 3 * number + 1.

然后编写一个程序，让用户输入一个整数，并在该数字上不断调用 collatz() 直到函数 return 的值为 1。（令人惊讶的是，这个序列实际上适用于任何整数——迟早，使用这个数列，你会得到 1！即使是数学家也不确定为什么。你的程序正在探索所谓的 Collatz 数列，有时被称为“最简单的不可能的数学问题。”）

记得使用 int() 函数将 return 值从 input() 转换为整数；否则，它将是一个字符串值。

提示：如果number %2 == 0，则为偶数，如果number %2 == 1，则为奇数

def collatz(number):
    if number%2==0:
        print(number//2)
    else:
        print(3*number+1)

x = int(input('print num'))
while TRUE:
    if collatz(x)!=1:
        break

Answer 1

重复 -

正如其他人在评论中所述，您的函数 collatz() 必须 return 一个整数也，才能再次输入 collatz() .

在您已经完成的基础上，我们可以让函数 collatz_sequence(x) 重复调用 collatz() 以获得所需的结果：

def collatz(x):
    if x % 2 == 0:
        a = x//2
    else:
        a = 3*x+1
    print(a)
    return a

def collatz_sequence(x):
    print(x)
    while x!=1:
        x=collatz(x)

这是一个示例输出：

>>> collatz_sequence(6)
6
3
10
5
16
8
4
2
1

Answer 2

您必须在函数 collatz(num):

中打印和 return

def collatz(number):
    """prints and returns the next number in the Collatz sequence
    """
    if number % 2 == 0:
        next_collatz_number = number // 2
    else:
        next_collatz_number = 3 * number + 1
    print(next_collatz_number)
    return next_collatz_number

x = int(input('print num'))

while True:
    x = collatz(x)
    if x == 1:
        break

Answer 3

如果你想对抗，我就是这样写的同样的代码

# Write a function named collatz() that has one parameter named number. If
# number is even, then collatz() should print number // 2 and return this value.
# If number is odd, then collatz() should print and return 3 * number + 1


# Then write a program that lets the user type in an integer and that keeps
# calling collatz() on that number until the function returns the value 1.


# Add try and except statements to the previous project to detect whether the
# user types in a noninteger string. Normally, the int() function will raise a
# ValueError error if it is passed a noninteger string, as in int('puppy'). In the
# except clause, print a message to the user saying they must enter an integer.

integernuber = False
while not integernuber:

    try:
        number = int(input('Type a number: '))

        integernumber = True

    except:
        print('Please entere an integer number')

        def collatz():

            global number

            if (number % 2) == 0:
                return number / 2
                number = (number / 2)
            elif (number % 2) == 1:
                return 3 * number + 1
                number = (3 * number + 1)
            else:
                return

        while collatz() <= 1:
            number = int(collatz())
            print(number)

希望对您有所帮助

Answer 4

try:
     number = int(input('Enter a number:'))
     while number != 1:
          number = collatz(number)
except ValueError:
     print('Enter a valid integer')

def collatz(num):
    if num % 2 == 0:
        print(str(num // 2))
        return num // 2
    else:
        print(str(3 * num + 1))
        return (3 * num + 1)

在这段代码中，我已经匹配了他们的约束（整数约束）并返回了值。我在 try 块中使用了简单的递归来获得所需的输出。

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