cool thanks thats what i changed it to than just now i changed to do  
something else and instead of leaving the program running for 4 or 5  
days as i had been i got my answer in less than a second ^_^"

i realized that i was wasting way to much time checking every  
possible answer because the number is jumping around so much so  
instead i wrote a real quick program infact  i wrote it in  
interactive that just starts at 1 and works the problem backwards for  
how ever long i tell it to :))

thanks for considering the homework, but this time you did'nt have to  
worry i did the homework last week i was just doing this for fun, it  
randomly asked something like "what about a number that takes 100  
steps to get to 1" and i thought what about 1000 :)) now i have one  
running looking for a number that takes 1000000 steps unfortunately i  
dont think ill get any extra credit for doing any of this lol


here is the program sorry copying it from interactive:

 >>> number=1
 >>> count=1000
 >>> count=0
 >>> while count < 1000:
...     if (number*2)%2 == 0:
...             number=number*2
...     else:
...             number=(number-1)/3.0
...     count=count+1
...     print number



On Sep 4, 2007, at 3:00 AM, Alan Gauld wrote:

> Max,
>
>> just wondering if theirs a way to check if a larger number is even
>> or
>> odd
>
> I notice this is homework so I won't give a direct answer.
> But look at the mod operator '%'  - It returns the remainder
> of an integer division. An even number is exactly divisible
> by 2.
>
> HTH,
>
> Alan G.
>
> "max baseman" <[EMAIL PROTECTED]> wrote
>
>> about before ive let it sit for a few days now and i reached a
>> number
>> to high to convert to a decimal by adding 0.0
>
> Thats a terrible way to convert to a float. Just use the float
> function
> (actually a type):
>
> f = float(anInt)
>
> But to determine if its odd/even you don't need to use floats at all.
>
>
>> here's the program:
>>
>> count=1
>> numstart=268549802
>> number=0
>> high=0
>> a=0
>> while 1==1:
>>     numstart=numstart+1
>>     number=numstart
>>     count=1
>>     while number !=1:
>>         if number/2 == (number+0.0)/2:
>>             number=number/2
>>         else:
>>             number=(number*3)+1
>>         count=count+1
>>     if count > a:
>>         a=count
>>         print numstart,":",count
>
>
> Hmm, I have no idea what this is doing...
>
>>
>> thanks
>>
>> _______________________________________________
>> Tutor maillist  -  Tutor@python.org
>> http://mail.python.org/mailman/listinfo/tutor
>>
>
>
> _______________________________________________
> Tutor maillist  -  Tutor@python.org
> http://mail.python.org/mailman/listinfo/tutor

_______________________________________________
Tutor maillist  -  Tutor@python.org
http://mail.python.org/mailman/listinfo/tutor

Reply via email to