Talk:Van der Corput sequence: Difference between revisions
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== Python output == |
== Python output == |
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Looks to me like the base 2 sample output for the Python example is actually base 3?--[[User:Tikkanz|Tikkanz]] 08:34, 11 March 2011 (UTC) |
Looks to me like the base 2 sample output for the Python example is actually base 3?--[[User:Tikkanz|Tikkanz]] 08:34, 11 March 2011 (UTC) |
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: Umm, my (very good) maths teachers defence in such situations was to say "Excellent lad, you've found the deliberate mistake"! :-) |
: Umm, my (very good) maths teachers defence in such situations was to say "Excellent lad, you've found the deliberate mistake"! :-) |
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: I'm at work at the moment but will correct the copy/paste error this evening. Thanks, --[[User:Paddy3118|Paddy3118]] 09:53, 11 March 2011 (UTC) |
: I'm at work at the moment but will correct the copy/paste error this evening. Thanks, --[[User:Paddy3118|Paddy3118]] 09:53, 11 March 2011 (UTC) |
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: Hmm? The text says base 3, numbers do look like base 2. Edit conflict? --[[User:Ledrug|Ledrug]] 07:03, 10 June 2011 (UTC) |
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== displaying of terms == |
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In every reference I've looked at, the 2nd term of the van der Corput sequence (for base two) is |
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'''.1''' |
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(not) '''.10000000''' |
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<br><br>I suggest that trailing zeroes illegitimatizes the terms. |
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Mathematically, of course, '''.1''' is equal to '''.100''' (except to an engineer, where trailing zeroes signify more precision). -- [[User:Gerard Schildberger|Gerard Schildberger]] 03:28, 26 March 2012 (UTC) |
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==Generation of the image in the task description== |
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My windows machine has packed up so I am using Ipython on Ubuntu. I did the following to create the image: |
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<lang python>In [211]: from __future__ import division |
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In [212]: def vdc(n, base=2): |
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...: vdc, denom = 0,1 |
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...: while n: |
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...: denom *= base |
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...: n, remainder = divmod(n, base) |
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...: vdc += remainder / denom |
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...: return vdc |
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In [213]: plt.plot([(random.random()*0.5, 0.5+vdc(i)*0.5) for i in range(2500)], '.') |
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Out[213]: |
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[<matplotlib.lines.Line2D at 0x12c73f2c>, |
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<matplotlib.lines.Line2D at 0x1311fe4c>] |
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In [214]: plt.title('Distribution: Van der Corput (top) vs pseudorandom') |
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Out[214]: <matplotlib.text.Text at 0x12ed6fcc> |
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In [215]: </lang> |
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--[[User:Paddy3118|Paddy3118]] 21:59, 7 August 2012 (UTC) |
Latest revision as of 21:30, 4 October 2020
Python output
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
In every reference I've looked at, the 2nd term of the van der Corput sequence (for base two) is
.1 (not) .10000000
I suggest that trailing zeroes illegitimatizes 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
My windows machine has packed up so I am using Ipython on Ubuntu. I did the following to create the image:
<lang python>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]: </lang> --Paddy3118 21:59, 7 August 2012 (UTC)