owing:
{0}0. d1. e2. f
3. g4. h5. i
{1}0. a1. b2. c
3. g4. h5. i
{2}0. a1. b2. c
3. e
4. f
5. g
Thought maybe I could use relative Item but I cant figure out how to do an offset that includes multiples.
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tangle. It doesn't need to me something complicated, it can look like this some of these roofs for example:
http://www.google.com/imgres?imgurl=http://www.zigersnead.com/blog/wp-content/uploads/2007/10/southern-cross-station-image-10.jpg&imgrefurl=http://www.zigersnead.com/current/blog/category/architecture/&usg=__0PpBu9zdO4Mg7TvHGIBKNsJPYuA=&h=395&w=640&sz=60&hl=en&start=471&zoom=1&tbnid=AIgPprbX_LhEQM:&tbnh=106&tbnw=172&ei=jsqRTan5BdHs4gaVt9HJAg&prev=/images%3Fq%3Dorganic%2Bshaped%2Broof%26hl%3Den%26sa%3DX%26biw%3D1666%26bih%3D1024%26tbs%3Disch:1&chk=sbg&itbs=1&iact=rc&dur=234&oei=VMqRTaXcNoz14QbRz8yfAg&page=10&ndsp=53&ved=1t:429,r:14,s:471&tx=51&ty=84
Where can I find that kind of Grasshopper file, that will enable me doing this?
Thank you.…
53 → 53 → 63 → 74 → 74 → 84 → 9
As you can see from the above list the connection sequence comes in waves of three, where each group of similar indices on the left is associated with a group of three incrementing indices on the right.
Some combination of Series components will probably generate this list, but it'll only work for the first ring, the second one will need a different connection pattern. It is perhaps better to just encode the integer pairs by hand. But then you cannot change your mind about the number of sides later.…
Added by David Rutten at 10:39am on October 21, 2015
ems in the same way. Lofting was particularly difficult, you had to have a separate loft component for every lofted surface that you wanted to generate because the component would/could only see one large list of inputs. Then came along the data structures in GH v0.6 which allowed for the segregation of multiple input sets.
If you go to Section 8: The Garden of Forking Paths of the Grasshopper Primer 2nd Edition you will find the image above describing the storing of data.
Here you will notice a similarity between the path {0;0;0;0}(N=6) and the pathmapper Mask {A;B;C;D}(i). A is a placeholder for all of the first Branch structures (in this case just 0). B is a place holder for all the second branch structures possibly either 0, 1 or 2 in this case. And so forth.
(i) is a place holder for the index of N. If you think of it like a for loop the i plays the same role. For the example {A;B;C;D}(i) --> {i\3}
{0;0;0;0}(0) --> {0\3} = {0}
{0;0;0;0}(1) --> {1\3} = {0}
{0;0;0;0}(2) --> {2\3} = {0}
{0;0;0;0}(3) --> {3\3} = {1}
{0;0;0;0}(4) --> {4\3} = {1}
{0;0;0;0}(5) --> {5\3} = {1}
{0;0;0;1}(0) --> {0\3} = {0}
{0;0;0;1}(1) --> {1\3} = {0}
{0;0;0;1}(2) --> {2\3} = {0}
{0;0;0;1}(3) --> {3\3} = {1}
{0;0;0;1}(4) --> {4\3} = {1}
{0;0;0;1}(5) --> {5\3} = {1}
{0;0;0;1}(6) --> {6\3} = {2}
{0;0;0;1}(7) --> {7\3} = {2}
{0;0;0;1}(8) --> {8\3} = {2}
...
{0;2;1;1}(8) --> {8\3} = {2}
I'm not entirely sure why you want to do this particular exercise but it goes some way towards describing the process.
The reason for the tidy up: every time the data stream passes through a component that influences the path structure it adds a branch. This can get very unwieldy if you let it go to far. some times I've ended up with structures like {0;0;1;0;0;0;3;0;0;0;14}(N=1) and by remapping the structure to {A;B;C} you get {0;0;1}(N=15) and is much neater to deal with.
If you ever need to see what the structure is there is a component called Param Viewer on the first Tab Param>Special Icon is a tree. It has two modes text and visual double click to switch between the two.
Have a look at this example of three scenarios in three situations to see how the data structure changes depending on what components are doing.
…
o relate to each plane its sections i.e.:
paths=624 - {0;0;2;0}(N=1)
the number of section planes is 70 and somewhere I have 4 section lines and somewhere 5 or 8, is it clear?
1 plane = 1 set of sections
thank you for helping me.…
Added by paul piciorul at 5:35am on October 28, 2010
or each branch goes to col 1, element 2 to col 2 and so on. The issue I am having is how to deal with branches that have less than 5 elements. Notice how in the attachment the number 8 is at row 2 where I want it to be in row 3.…
Added by jon kontuly at 1:03pm on September 30, 2014