the one-but-last list [4]. After running out of the n- items avalaible it should continue with the second item of list 0 and so on for all items on all the lists.
Intput, six lists of 30 items each
[0] (n=30)
[1] (n=30)
[2] (n=30)
[3] (n=30)
[4] (n=30)
[5] (n=30)
Output, 18 lists of 10 items each
[0],i=0;[5],i=4; [4],i=7;...
[0],i=1;[5],i=5; [4],i=8;...
...
[5],i=1;[4],i=5; [3],i=0;...
I thought perhaps the weave component or the relative tree item component but didn't manage to figure out how to compose the mask. I couldn't find much on how to use these. I guess it should wrap the lists, but not the items.
Any help would be greatly appreciated.…
Added by Thorsten Lang at 2:27am on January 24, 2011
is passable give different distance in list of point to delete
ex: point 1 - 30
point 2 - 20
point 3 - 10
...........
the point all in one list
i try a lot but didn't found the way ...
icture you can see what I mean, this is just with randome reduce, it should be like this:
->If the last object is (1) then is a 50% chance that the next one is also (1)
-->If the last two objects are (1) then is a 40% chance that the next one is also (1)
--->If the last three objects are (1) then is a 30% chance that the next one is also (1)
---->If the last four objects are (1) then the next object must be (2)
->If the last object is (2) then is a 50% chance that the next one is also (2)
-->If the last two objects are (2) then is a 30% chance that the next one is also (2)
--->If the last three objects are (2) then is a 10% chance that the next one is also (2)
---->If the last four objects are (2) then the next object must be (1)
It would be nice if it is possible to change the number how many objects should be tested and the percent values.
Has anybody an idea how I can do this?
Best Regards
…
ved from the source surface.
3. Send the center points to the VB component.
If you send 15 points to the VB component, it will then return 15 * 4 spirals.…
Added by June-Hao Hou at 9:41pm on December 2, 2010
1+2+3+4+5+6 = 21
1+2+3+4+5+6+7 = 28
1+2+3+4+5+6+7+8 = 36
1+2+3+4+5+6+7+8+9 = 45
Is there a tool, that can do me that job?
How do I get this List {1,3,6,10,15,21,28,36,45}?…
Added by Ahmed Hossam at 2:19pm on September 22, 2013
g to keep the {15} {16} branch structure because its keeping the organization by floor level.
Any insight into how the Curve Plane Intersection is matching the branches for operation would be helpful as well.
Thanks!
…