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Talk:Van der Corput sequence

From Rosetta Code

Python output[edit]

Looks to me like the base 2 sample output for the Python example is actually base 3?--Tikkanz 08:34, 11 March 2011 (UTC)

Umm, my (very good) maths teachers defence in such situations was to say "Excellent lad, you've found the deliberate mistake"! :-)
I'm at work at the moment but will correct the copy/paste error this evening. Thanks, --Paddy3118 09:53, 11 March 2011 (UTC)
Hmm? The text says base 3, numbers do look like base 2. Edit conflict? --Ledrug 07:03, 10 June 2011 (UTC)

displaying of terms[edit]

In every reference I've looked at, the 2nd term of the van der Corput sequenct (for base two) is
.1
(not) .10000000

I suggest that trailing zeroes illegitimize the terms. Mathematically, of course, .1 is equal to .100 (except to an engineer, where trailing zeroes signify more precision). -- Gerard Schildberger 03:28, 26 March 2012 (UTC)

Generation of the image in the task description[edit]

My windows machine has packed up so I am using Ipython on Ubuntu. I did the following to create the image:

In [211]: from __future__ import division
 
In [212]: def vdc(n, base=2):
...: vdc, denom = 0,1
...: while n:
...: denom *= base
...: n, remainder = divmod(n, base)
...: vdc += remainder / denom
...: return vdc
 
In [213]: plt.plot([(random.random()*0.5, 0.5+vdc(i)*0.5) for i in range(2500)], '.')
Out[213]:
[<matplotlib.lines.Line2D at 0x12c73f2c>,
<matplotlib.lines.Line2D at 0x1311fe4c>]
 
In [214]: plt.title('Distribution: Van der Corput (top) vs pseudorandom')
Out[214]: <matplotlib.text.Text at 0x12ed6fcc>
 
In [215]:

--Paddy3118 21:59, 7 August 2012 (UTC)