Hi, im a begginer to python, and i've been doing it a small while now, i've been attempting some puzzles of a website, and have met a small problem, and i can't find it. My code is below and i will explain my problem, ask how to sort it, and ask if i am missing something important.

``````a = range(1,1000)
b = []
d = 0
for n in a:
while n != 1:
if n % 2 == 0:
b.append(n)
n = n/2
if n % 2 != 0:
b.append(n)
n = n*3 + 1
else: b.append(n)
if len(b) > d:
d = len(b)
b = []``````

Now the aspects of this code is to find out how long it takes to get to 1, using the 'Collatz Problem' i take the value of n to be inside the range, and if its not 1, i want it to go through the number, decide if its even or odd, then perform a calculation, firstl, im recording the number in a list, so i know the starting number, and then it will perform the calculation, i want it to continue this loop until it reaches 1, im not sure how to do this, i have the else statement at ehd end because i believe that is what i need, i then will take the length of b which is my list, and see if it is higher than my previous list, i then want it to record if so and erase the list, so i can start over at 0, with my next number, but i seem to not be able to make my list work, and i get memory failure of B several times, if someone can help, it would really be apprechiated.

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Last Post by sammyboy289

So in summary (for lazy to read like me), what is the issue? Doesn't work as intended or throws exception? What is the error?

The error is, my b statement doesn't keep resetting which it should, and my list continues even once its reached 1, which i dont understand why :( also, im not sure if my else: statement is in the right place. Sorry i write a lot

Let's see what happens when n is 2. Your first if condition is true, you divide n by 2 and it becomes 1. Now your second if condition is also true! Welcome to the infinite loop.

should it loop, i didn't think it would, because thats while n != 1, so it should stop that loop and move onto the else: statement, or is that incorrect? i can't quite work out, how to make 2 loops, and when one ends, go to the next statement. That's the main problem im dealing with, sorry if im explaining wrong or being really useless

Shorten the list and add a temporary print for testing. See what that will do for you:

``````# shorten list for test
a = range(1, 10)
b = []
d = 0
for n in a:
while n != 1:
if n % 2 == 0:
b.append(n)
n = n/2
elif n % 2 != 0:
b.append(n)
n = n*3 + 1
else: b.append(n)
print b, d, len(b)  # test
if len(b) > d:
d = len(b)
b = []``````

At run time, things happen one step at a time. Let's trace the execution of your program with n being 2

``````while n != 1:
# Python checks that n is not 1, and enters a loop
if n % 2 == 0:
# Python checks that n is even and enters the if-clause
...
n = n/2 # [B]n becomes 1[/B]
if n % 2 != 0:
# Remember, n is 1, an odd number, so python enters the if-clause
...
n = n*3 + 1 # [B]n becomes 4[/B]
# while loop goes into next iteration, n is 4
while n != 1:
# n is 4, enter the loop
if n % 2 == 0:
# n is even, enter the if-clause
...
n = n/2 # n becomes 2
if n % 2 != 0: # n is even, if-clause is not entered.
# Next iteration of while loop. n is 2. Etc, etc, etc``````

wow, that worked, thank you so much, if its not too much to ask, do you mind saying how you changed it, and what it does?, because i notice the elif: statement which im not good with, and one question about the last part, as i will only want one resut to show, will i need to change

``````#
print b, d, len(b)
if len(b) > d:
d = len(b)
b = []``````

to something like

``````print b, d, len(b)
if len(b) > d:
d = len(b)
b = []
if len(b) < d:
b = []``````

and thank you very much

Edited by sammyboy289: n/a

ok, i want to thank nezachem for showing why it went wrong, and i want to thankEne Uran as to showing me what was the way to fix it. I have finally sorted out my code, and i am very thankful to all of you for helping me. You have helped me understand and hopefully i can become better at python :D