Making a collatz program automate the boring stuff

Question

I\'m trying to write a Collatz program using the guidelines from a project found at the end of chapter 3 of Automate the Boring Stuff with Python. I\'m using python 3.

Answer 1

I had a hard time understanding the logic of this exercise. but I don't know what I did wrong to display the error if we have a negative number.

def collatz(number):
if number % 2 == 0:
    print(number // 2)
    return number // 2

elif number % 2 == 1:
    result = 3 * number + 1
    print(result)
    return result

while True:
try:
    entry = input('enter a positive number: ')
    while entry != 1:
        entry = collatz(int(entry))

    # if we have a negative number
    while entry < 1:
        break

except ValueError:
    print('Error, enter valid number')

Answer 2

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


print("Enter a number:")
try:
    n = int(input())
    cyclic(n)
except ValueError:
    print("Unvalied value")

The simplest one

Answer 3

My 17 lines of code for the same exercise that I have came up with.

    def collatz(number):
    """ check if the number is even or odd and performs calculations.
    """
    if number % 2  == 0: # even
        print(number // 2)
        return number //2
    elif number % 2 != 0: # odd 
        result = 3*number+1
        print(result)
        return result

try:
    n = input('Enter number: ') # takes user input
    while n !=1: # performs while loop until 'n' becomes 1
        n = collatz(int(n)) # passes 'n' to collatz() function until it arrives at '1'
except ValueError:
    print('Value Error. Please enter integer.')

Answer 4

You can simply try this

while True:
 number=int(input('Enter next positive number or 0 to quit: '))
 iteration=0
 currentNumber=0
 item=1
 sumOfNumbers=0
 print(number,end='  ')
 if(number):
    while currentNumber !=1 :
        currentNumber=int(number/2) if(number%2==0) else number*3+1
        number=currentNumber
        iteration +=1; item +=1
        sumOfNumbers +=currentNumber
        print(currentNumber,end ='\n' if(item %5==0) else '  ')
    print('\nIt took ',iteration,'iterations to arrive at 1')
    print('The average  is ',round((sumOfNumbers/iteration),2))
 else :
    break

Answer 5

def collatz(number):
    if(number%2==0):
        n=number//2
        print(n)
        return n
    else:
        ev=3*number+1
        print(ev)
        return ev
num1=input("Enter a number: \n")

try:
    num= int(num1)
    if(num==1):
        print("Enter an integer greater than 1")
    elif(num>1):
        a=collatz(num) 
        while(True):
            if(a==1):
                break
            else:
                a=collatz(a)
    else:
        print("Please, Enter a positive integer to begin the Collatz sequence")

except:
    print("please, Enter an integer")

Try to came up with a solution based on up to chapter Function from automate the boring stuff. If need help related to Collatz Problem, then visit here: http://mathworld.wolfram.com/CollatzProblem.html

Answer 6

i am reading the same course and i made a very long solution (improving it when i learn somethign new). i suggest keeping your collatz program up to date as you progress in the chapters, its good training. mine has string manipulation and saving to a \collatzrecords.txt now!

I solved the core problem by using recursion (a method calls itself):

def autocollatz(number):
global spam                     
spam.append(number)             
if number % 2 == 0:             
    autocollatz (int(number/2))
elif number % 2 == 1 and number != 1:
    autocollatz(int(number*3+1))

spam is my list for all the values a number "sees" on its way to 1. as you can see, when the number is even the ethod is called agin with number/2. if the number is even it is called with number*3+1.

modified the number == 1 check a bit. i hope it saves calculating time - im up to 23 000 000 already! (current record is 15 733 191 with 704 steps to get it to 1)