Hi Carlos,
I hope this module would help you with the binary conversion part:
def tobits(inte,size = 3):
"""Copy an integer's bits from memory"""
s=''
for i in range(size):
s += str((inte & (1<>i)
#bits are extracted right-to-left
s = s[::-1] #reverse the result string
Hi Again,
Luke, I left it this way now:
x = A_List[i-1]
y = A_List[i]
z = A_List[(i+1) % A_Len]
Probably the other way is shorter but right now I feel comfortable as it
is now :-[ , Danny had already mentioned this method before, but it
didn't sink properly until now. And yes I need it to wrap
Carlos wrote:
> Hello to all,
>
> Ok, after reading your comments I ended up with this:
>
> #Easy_2DCA_01.py
> #A very basic 2D script that lets you play with Wolfram's rule 30
>
> A_List = [0]*10+[1]+[0]*10
> A_Len = len(A_List)
> B_List = []
> print A_List
> Iterations = 5
>
> rule_30 = {
>
"Carlos" <[EMAIL PROTECTED]> wrote
> As you may have noticed the name of the script has changed, I will
> now
> try a 3D CA. I have been thinking that the best way to do this would
> be
> to use a matrix, like those found in SciPy. Do you think this is a
> good
> idea?
Using a matrix is fine
Hello to all,
Ok, after reading your comments I ended up with this:
#Easy_2DCA_01.py
#A very basic 2D script that lets you play with Wolfram's rule 30
A_List = [0]*10+[1]+[0]*10
A_Len = len(A_List)
B_List = []
print A_List
Iterations = 5
rule_30 = {
(1, 1, 1) : 0,
(1, 1, 0) : 0,
Danny Yoo wrote:
>> Do you know a better way to do this?
>> [snip stuff]
>>
>
> Hi Carlos,
>
> Here's one way to handle it. If you're familiar with modulo "clock"
> arithmetic, you might want to apply it above.
>
Also, consider this example, Carlos:
>>> aList = [1,2,3,4,5,6]
>>> a,b,c =
"Danny Yoo" <[EMAIL PROTECTED]> wrote
> Here's one way to handle it. If you're familiar with modulo "clock"
> arithmetic, you might want to apply it above.
Indeed, I should have thought of that.
Good catch Danny.
Alan G.
___
Tutor maillist - Tut
"Carlos" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Do you know a better way to do this?
>
>if i < A_Len-1:
>a = A_List[i-1]
>b = A_List[i]
>c = A_List[i+1]
>else:
>a = A_List[i-1]
>
> Do you know a better way to do this?
>
>if i < A_Len-1:
>
>a = A_List[i-1]
>b = A_List[i]
>c = A_List[i+1]
>
>else:
>
>a = A_List[i-1]
>b = A_List[i]
>c = A_List[0]
Hi Carlos,
Hi again,
Thanks for your comments they are really appreciated.
Alan, yes you are right, there is no need to use S, A_Len works the same
way. Beginner mistake
The dictionary way seems to be way better than my idea, will give it a try.
And Danny, your CA is very nice.
Do you know a better way
"Carlos" <[EMAIL PROTECTED]> wrote
> Here is the code:
> Note: Right now maya is not needed to run the code.
>
> import time
>
> #This is the initial state. You can put as many integers as you wish
> A_List = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1]
>
> A_Len = len(A_List)
> print 'A_List: ',A_Lis
On Mon, 6 Nov 2006, Carlos wrote:
> This is my first script, I mean the first that I'm trying to do on my
> own. Its a very simple Cellular Automata thing, the idea is that the
> initial list A_List is rewritten based on some rules, in this case
> Wolfram's rule 30. You can modify the list le
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