o three parts:
branch 1:
{0;0} N = 3
{0;1} N = 3
branch 2:
{1;0} N = 5
{1;1} N = 5
branch 3:
{2;0} N = 30
{2;1} N = 30
parthmapper won't change the length of branch, explode tree won't give me two branches in one output
…
(0) --> {0}(0) overwrite etc
---
{2;2}(1) --> {2}(1)
{2;3}(0) --> {2}(0)
{2;3}(1) --> {2}(1)
Correct use of index in this case {A;B}(i) --> {A}
but because you don't need to use the index on the right hand side, don't specify it on the left, as it takes the [Path Mapper] longer to process when the index is used.…
Added by Danny Boyes at 2:00pm on September 25, 2011
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
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
{2:2} {2:3}
Each branch has 10 points. I'd like all the points in {0} to draw lines to each sub-branch {0,#}, {1} to {1:#}, etc. I am simply confounded, how would this be accomplished?
Thanks!
…
, ByVal generation As int32, ByRef allboxes As list(Of box))
If generation - 1 < 0 Then
Exit Sub
End If
Dim subboxes As New List(Of Box)
Dim corners As Point3d()=box.getcorners
Dim xvec As Vector3d = (corners(1) - corners(0)) / 3
Dim yvec As Vector3d = (corners(3) - corners(0)) / 3
Dim zvec As Vector3d = (corners(4) - corners(0)) / 3
box.transform(transform.Scale(corners(0), 1 / 3))
For m As int32=0 To 2
For n As Int32=0 To 2
For k As int32=0 To 2
If Not (m * m + n * n = 2) And Not (n * n + k * k = 2) And Not (k * k + m * m = 2) Then
Dim sub_box As Box = box
Dim transvec As Vector3d = xvec * m + yvec * n + zvec * k
sub_box.Transform(transform.Translation(transvec))
allboxes.Add(sub_box)
subD_box(sub_box, generation - 1, allboxes)
' all_boxes.remove(sub_box)
End If
Next
Next
Next
End Sub
'''''''''''''''''''''''''''''''''''''''''…