4 to 9
Is there a way to do this with the list or sequence components other than retrieving individual list items and then lofting between each list item (circle). There must be a more elegant way to do this.
Thanks very much.
please see the attached definition as a sample...…
ee 3)
{5}
0 15
{6}
0 16
And I want to place points at every possible combination of these coordinates, treating Tree 1 as X coordinates, Tree 2 as Y coordinates, and Tree 3 as Z coordinates. Also, I would like the list of points to be a tree with paths corresponding to the coordinates. Wouldn't it be nice if I could plug these trees into a Point XYZ, with a new "branch cross reference" method, and get the following result?
{0:3:5}
0 {10.0, 13.0, 15.0}
{0:3:6}
0 {10.0, 13.0, 16.0}
{0:4:5}
0 {10.0, 14.0, 15.0}
{0:4:6}
0 {10.0, 14.0, 16.0}
{1:3:5}
0 {11.0, 13.0, 15.0}
{1:3:6}
0 {11.0, 13.0, 16.0}
{1:4:5}
0 {11.0, 14.0, 15.0}
{1:4:6}
0 {11.0, 14.0, 16.0}
{2:3:5}
0 {12.0, 13.0, 15.0}
{2:3:6}
0 {12.0, 13.0, 16.0}
{2:4:5}
0 {12.0, 14.0, 15.0}
{2:4:6}
0 {12.0, 14.0, 16.0}
In this form of cross referencing, every combination of individual branches from the different lists is used as separate input, and the output for each combination is put onto a branch in the result whose path is the concatenation of the input branch paths used.…
Added by Andy Edwards at 7:03pm on November 3, 2009
hink you need recursion to modify the random seed; many other ways to accomplish that (use the length of each curve as the seed, for example).
Using multiples of twelve makes it harder for me to grasp the essence of the matter; another way of looking at it is that you want to generate random integers from 2 to 5 (24,36, 48 and 60) and have them add up exactly to curve lengths of 5 (x12=60), 9 (x12=108) or 14 (x12=168).
So you want to generate random numbers until their sum ('Mass Addition') plus 5 is equal to or greater than the curve length (5, 9 or 14). The last number in the series is then not random but just the difference between the two.
For example, for curve length = 5 (x12=60), there are only three possible numbers that can be used as the first in the sequence: 2, 3 or 5. If it's 5, you're done. If it's 2, the second number is 3 (5-2), if it's 3, the second number is 2 (5 - 3). You can't use '4' at all because the remainder, 1 (x12=12) isn't one of your solution options.
There is no point in generating the last number randomly, eh?
P.S. You didn't use 'Internalize data' for the 'Curve (Crv)' param in your GH file.…
Added by Joseph Oster at 2:29pm on September 12, 2015
output will show a tree with 3 branches of 4 integers each that I can pass on to other components. What is the best way to do it?
I have tried creating a tree and using a for loop to do so, but it didn't work.
Thank you for your help.
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