dont get you, i am saying sleect numbers in range 1 to 10, starting from 1 with a step of 2.
1 to 10 by 3 = 1 4 7 10
1 to 10 by 5 = 1 6
1 to 10 by 1 = 1 to 10 = 1 2 3 4 5 6 7 8 9 10
Added by Steve Lewis at 3:15pm on November 11, 2013
pen Brep"; I didn't know it worked on flat surfaces. And I think it's only fair to include in your benchmark the considerable time 'SUnion' takes in this example: 21.9 seconds for 121 rings and likely much more with 400 or 1,000+ rings.
Then I noticed the pattern doesn't match. Checked the circles and they are the same. The distance between them, however, is different: 7 instead of 6. When I change that value to 6, the Python fails badly. All the holes and gaps are gone, which destroys the pattern:
I can't do the "two phase" approach on an 11 X 11 grid, but I can do 6 X 6 and 2 X 2 to get a 12 X 12 grid (40 'SUnion' operations) in 28 seconds total. That beats your benchmark of ~37 seconds for an 11 X 11 grid, if you include the 'SUnion' in your code.
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This must be a bug because its true for dividing by all odd numbers:
(i-1)/3
(i-2)/5
(i-3)/7
(i-4)/9
(i-5)/11
....
(i-n)/2n+1
And you can't make it work for even numbers
Added by Danny Boyes at 5:06pm on January 13, 2010
could accomplish this by adding something like '5.0 + Slider * 3.0' into the expression field... (and having an integer slider with domain 0 to 6).
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 10:19am on February 24, 2011
nts me this:
[[0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7]]
this is what I wanted but how to convert this to tree in grasshopper?
In grasshopper I just get:
8x IronPython.Runtime.List…