Ah, in the case of looking for all n-digit bit-strings that contain exactly
m-1's, the recursive solution is even nicer. Here's how I think of it:
Base Case(s):
- if you want 0 1's (m==0) then return all 0's.
- if you want all 1's (n==m) then return all 1's.
Otherwise Recursive Case:
- return th
"Luke Paireepinart" <[EMAIL PROTECTED]> wrote
> You could also do this by iterating in base-16 instead of base-10...
I was going to suggest the same but using octal which
has the same property but fewer values to map. :-)
> hexmap =
> {"0":"","1":"0001","2":"0010","3":"0011","4":"0100","5"
Andy Cheesman wrote:
> The code works great, Thanks for the speedy response. The only problem
> which I can see is that the code scales very bad with the size of n.
>
You could also do this by iterating in base-16 instead of base-10...
given a string of hex,
like "59FDE", there is a direct corre
The code works great, Thanks for the speedy response. The only problem
which I can see is that the code scales very bad with the size of n.
So, as I want a small subsection of the data (i.e lines where there are
only 4 1s, number in the code below) for a system where n is large(>20).
The idea is
Another simplest way of doing it is
>>>
from random import *
from sys import *
def bin_list(n):
bin_val = 0
if n >= 0:
bin_val = 2**n - 1
list = []
while bin_val >= 0:
list.append([((bin_val >> y) & 1) for y in range(n-1,-1,-1)])
bin_val -= 1
shuffle(
The 2**n different lists that you are seeking have a direct association to
the binary representation of the integers 0 through (2**n)-1.
You can use this fact and the "repeated division method" for converting
numbers between different bases to generate these lists and form the desired
list of list
Hi people
I am trying to generate an array of all possible combinations of 1, and
zeros (see example data) for a rather nice Kinetic mote Carlo program
which I am writing python. So far, I've been working out for
combinations for 4 or less species by hand as it is quick! but I am
looking to automa
