Rewrite the code so he can do less (now 20 lines)

[Laycee]

Your job is to simplify my program, but to keep it functional.

According to the Collatany number can be translated into a sequence of figures that always ends with units. This sequence is called Unsolved mathematics problems'Cause that hypothesis was never proven. The sequence of words is very simple: there is a function in which we hand over any number. If the number's clear, then we'll split it by two. число / 2 ) If it's not clear, then we multiply it by 3 + 1 (i.e. число * 3 + 1 ) The importance that we have regained our function is once again a function.

    def collatz( number ) :
      if number % 2 == 0 : # чётное
        result = number // 2
      if number % 2 == 1 : # нечётное
        result = 3 * number + 1
      print( result )
      return result
error = True
while error :
  error = True
  try :
    num = int( input( 'Введите число больше нуля: ' ) )
    if num > 0 :
      error = False
  except :
    error = True

firstTime = True
while num != 1 or firstTime :
  firstTime = False
  num = collatz( num )

# если number - чётное, тогда это число надо разделить на 2 без остатка

[simrah]

Your job is to simplify my program, but to keep it functional.

a Your task is to learn to correctly formulate the question : ) or to deal with the issue as a teaching assignment.

Option 1:

def collatz(number):
    result = (number // 2) if number % 2 == 0 else (3 * number + 1)
    print(result)
    if result != 1:
        collatz(result)
num = 0
while num <= 0:

try:

num = int(input('Введите число больше нуля: '))

except:

num = 0
collatz(num)

The function can be redesigned in such a way that the number 1 is immediately processed and the entire combination can be removed from the first.

def collatz(number):

print(number)

if number != 1:

result = (number // 2) if number % 2 == 0 else (3 * number + 1)

collatz(result)

No. print within the function, i.e. the establishment of the list and the withdrawal of the list:

def collatz(number):

res = [number]

if number != 1:

result = (number // 2) if number % 2 == 0 else (3 * number + 1)

res += collatz(result)

return res

num = 0
while num <= 0:

try:

num = int(input('Введите число больше нуля: '))

except:

num = 0
print(*collatz(num), sep='\n')

And the decision is in line 1:

def collatz(number):

return [number] + ([] if number == 1 else (collatz((number // 2) if number % 2 == 0 else (3 * number + 1))))

num = 0
while num <= 0:

try:

num = int(input('Введите число больше нуля: '))

except:

num = 0
print(*collatz(num), sep='\n')

A shorter version of the function:

def collatz(value):

return [value] + ([] if value == 1 else (collatz((3 * value + 1) if value % 2 else (value // 2))))

Now, if there is no need to check the wrong entry, the code can be turned down as follows:

def collatz(i):

return [i] + ([] if i <= 1 else (collatz((3 * i + 1) if i % 2 else (i // 2))))

print(*collatz(int(input('Введите число больше нуля: '))), sep='\n')

- 3 lines:)