Calling J from Python

[Nick Craig-Wood]

George Sakkis <george.sakkis at gmail.com> wrote:
>  On Feb 9, 9:20 pm, bearophileH... at lycos.com wrote:
> > This is a bit simpler, but probably there are simpler solutions using
> > modular arithmetic:
> >
> > l = [1]
> > for _ in range(15):
> >     print ''.join(" *"[x] for x in l)
> >     l = [1] + [l[i+1]^l[i] for i in range(len(l)-1)] + [1]
> >
> > Bye,
> > bearophile
> 
>  Here's another one, adapted from the example (in Java) in Wikipedia's
>  entry (http://en.wikipedia.org/wiki/Sierpinski_triangle):
> 
>  N=15
>  for x in xrange(N,0,-1):
>      print ''.join('* '[x&y!=0] for y in xrange(N+1-x))
> 

This is my solution after a few minutes thought.

It uses a different algorithm for generating the triangle.  If python
could output binary numbers it would be more elegant...

>>> n = 1
>>> for i in range(16):
...     print ("%X" % n).replace('0', ' ').replace('1', '*')
...     n = n ^ (n << 4)
... 
*
**
* *
****
*   *
**  **
* * * *
********
*       *
**      **
* *     * *
****    ****
*   *   *   *
**  **  **  **
* * * * * * * *
****************


-- 
Nick Craig-Wood <nick at craig-wood.com> -- http://www.craig-wood.com/nick