r example:
{1} {6} Is this possible to do with Replace Branches?
{2} {5}
{3} => {3}
{4} {4}
{5} {5}
{6} {6}
If it were two lists, then it's very easy with List Replace, but kill me, I can't get what's the logic in Replace Branches...
Please help!
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Added by Artyom Maxim at 10:05am on January 29, 2013
If you right click on the 'Random' component, you will find an option for "Integer Numbers".
You might need to generate random integers of 3, 4, 5 or 6 and multiply that by ten?
a COM error.
I tried to fix the COM with visual studio by updating the references but no luck.
I am currently using rhino 6 because i want to use the new make2d command in grasshopper. But the plug in doesnt work with 5 and 6. same issue.
Have there been any work on this ?
Thank you,
Joey Doherty
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cture, Rhino treats them as a single flat list. For example a surface can have 10 rows and 6 columns of control-points, resulting in a list of 60 points.
But 10 times 6 isn't the only way to get to 60. If you want to make a surface out of a list of 60 points, you'll also have to tell Rhino how those 60 points should be interpreted in terms of a grid. It could be 2*30, 3*20, 4*15, 5*12, 6*10, and all of the aforementioned products the other way around.
Sometimes there's only one way for a number of points to fit into a rectangular grid. For example if you provide 49 points, then 7*7 is the only way to make it work, but these cases are rare so we always demand you give us all the information required to actually make a rectangular grid of control-points from a linear collection.
As for "Why is it, sometimes we need to attach additional value into it?", this is usually because when you divide a domain or a curve into N segments, you end up with N+1 points. For example take the domain {0 to 5}, and divide it into 5 equal subdomains. You end up with {0 to 1}, {1 to 2}, {2 to 3}, {3 to 4} and {4 to 5}. However there are six numbers that mark the transitions between these domains 0, 1, 2, 3, 4 and 5. This is why you often have to add 1 to the UCount, because the number that controls the UCount often results in N+1 actual points.…
Added by David Rutten at 8:30am on December 25, 2014
Thanks RM, somehow this option of yours doesn't work if you have other numbers (i have tried to add a third branch with (12;12;12;5;8) (randomly) and the result goes to 0
{4}-0;3
{5}-6;7
{6}-5;7
{7}-5;6
Here it can be shown that there are two subgraphs containing 0,1,2,3,4 and 5,6,7. How can I use spiderweb (either using scripting or the components) to give me this result when I have many more vertices??
Thanks,
Sam…