a seed, and instead creating a pattern where each color has a seed/control slider for each row? For example, row 1: brown 2, tan 6, yellow 7, purple 3, repeat. row 2: brown 6, tan 1, yellow 4, purple 10, repeat. row 3: yellow 5, purple 1, brown 3, tan 10, repeat. row 4: purple 2, brown 7, tan 3, yellow 4, repeat. Then repeat that sequence up the wall? For each color, the number in the sequence should be adjustable.
Thank you again for your help!…
4}
{0;2;0}
{0;2;1}
{0;2;2}
{0;2;3}
{0;2;4}
You cannot flip this because this is more complex than a rectangular matrix. You're going to have to do the mapping yourself. Try a Path Mapper with the following masks:
{A;B;C}(i) -> {A;B;i}(C)
Which should give you a structure that results in 3 lofts.
--
David Rutten
david@mcneel.com
Poprad, Slovakia
…
Added by David Rutten at 3:18pm on November 27, 2011
middle index, and choose that point with List Item. If even, for example 4 points (0, 1, 2, 3), you'll get 2, so subtract one and choose those two indices, 1 and 2. I only had a few minutes to play with this, so it isn't a fully-baked solution, but it should take you a little further.…
square units. Then you have an integral number of fragments on each side. This means that if all fragments need to have the same surface area, you can only have the following possibilities for side A:
1 fragment = 100 square units
2 fragments = 50 square units each
3 fragments = 33⅓ square units each
4 fragments = 25 square units each
5 fragments = 20 " "
6 fragments = 16⅔ " "
etc.
For side B, the numbers are mostly different
1 fragment = 300 unit²
2 fragments = 150 unit²
3 fragments = 100 unit²
4 fragments = 75 unit²
For side C they are different still. Unless you join fragments across on both sides of the edges of the box, I very much doubt you'll be able to pull this off.
The solution I attached will create fragments as identical as possible, but it's a very boring outcome...
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
ches it with the first branch in Tree B (and then the first branch in Tree C if more than two trees are involved).
I'm planning to add better branch matching logic, but I'm not going to touch it until I have a good idea about what's needed and how it can be accomplished without breaking existing files.
So, the branch "address" is only used to sort the branches in a single tree. Thus, a tree with the following branches is always sorted in the exact same way:
{0;0}
{0;1}
{0;2}
{0;3;0}
{0;3;1}
{1;6}
If you have another tree with different branches:
{0}
{1}
{2}
{3}
{4}
{5}
Then the matching will be:
{0;0} -> {0}
{0;1} -> {1}
{0;2} -> {2}
{0;3;0} -> {3}
{0;3;1} -> {4}
{1;6} -> {5}
As long as people adhere to your advice: "it is best for the addresses of each tree branch to be in the same format", there will be no problem. But it is at the moment extremely difficult to perform complex matchings.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 9:25am on August 11, 2010
vas
Closing and creating a new file (memory resets when this is done) @4:00, 5:57, 6:53
System slow down and crashes @ 8:16 (takes 5 minutes to end the process - perhaps not the most entertaining movie to watch until the end - a good point to turn the kettle on)…