e
7. True
8. True <-- this one
9. True
10. False
11. True
12. False
13. True
14. True <-- this one
15. True
16. False
17. True
18. False
19. True
20. True <-- this one
21. True
22. False
23. True
24. False
25. True
26. True <-- this one
27. True
28. False
29. True
30. False
31. True
32. True <-- this one
33. True
Any idea how I can solve this?
Thanks!…
a specific domain, for example:
0.) 0 to 1 -----> 11 random values from 0 to 1 (0.245,0.678,0.36,0.78,.28,0.18........)
1.) 1 to 2 -----> 11 random values from 1 to 2 (1.26,1.36,1.01,1.68,1.26,1.96.........)
3.) 2 to 3 -----> 11 random values from 2 to 3 (2.96,2.45,2.78,2.56,2.98,2.10..........)
4.) 3 to 4 and so on where I have a data set containing 11 paths with 11 values and the values fall within the specific domain.
Like my post above I have the correct path but I need to feed it the correct seed to get different values for each number. I tried grafting a series similar to the last post but it scrambles my data. Thanks so much for the help!
…
} (N=11) {0;1} (N=11) {0;2}(N = 11) {0;3}(N = 11) {0;4}(N = 11)
2. I run the Points that are coming out from the Divide Curve Components through the Path Mapper components with this definition:
{A;B} (i) > {A} (i)
3. I run data coming out from Path Mapper component through:
a) Parameter Viewer component and the result is:
{0} N=11 (data with 1 branches)
b) Point > Panel and the result is:
collection of 11 point (N=11) which is the exactly the same as the collection of point belonging to {0;4} (N = 11).
So, here is the question:
why the collection of points coming out from the Path Mapper {A;B} (i) > {A} (i) component is the same as the collection of points belonging to the curve {0;4}(N = 11) ?
Anyway ... It 's the first time I ask a question here... so I would like to thank you for what you do with your work! Thank you! You are really great!…
e now contains (40x3x11=)1320 points with a branch structure of {0-39,0-2} i.e
{0;0}
{0;1}
{0;2}
{1;0}
...
{39;2}
with each branch containing 11 points.
I now want to create lines from the points on line {x;0} to {x;1}, {x;1} to {x;2}, and {x;2} to {x;0}, which will give me a triangular grid on each triangle. How do i do this? I think I need to split the tree 40 times to give me 40 single layered trees but I imagine there's a cleverer way of doing this!
This feel like it should be simple, but I;m having trouble working with the double-layered branch structure. Any help would be greatly appreciated!
Thanks,
Matt…