nd router's external IP (分享器), not your computer's IP (it should look like 192.168.0.xxx).
Here's a quick tutorial to find your "real" IP: https://www.youtube.com/watch?v=g6Eyj8pjrNU…
hink you need recursion to modify the random seed; many other ways to accomplish that (use the length of each curve as the seed, for example).
Using multiples of twelve makes it harder for me to grasp the essence of the matter; another way of looking at it is that you want to generate random integers from 2 to 5 (24,36, 48 and 60) and have them add up exactly to curve lengths of 5 (x12=60), 9 (x12=108) or 14 (x12=168).
So you want to generate random numbers until their sum ('Mass Addition') plus 5 is equal to or greater than the curve length (5, 9 or 14). The last number in the series is then not random but just the difference between the two.
For example, for curve length = 5 (x12=60), there are only three possible numbers that can be used as the first in the sequence: 2, 3 or 5. If it's 5, you're done. If it's 2, the second number is 3 (5-2), if it's 3, the second number is 2 (5 - 3). You can't use '4' at all because the remainder, 1 (x12=12) isn't one of your solution options.
There is no point in generating the last number randomly, eh?
P.S. You didn't use 'Internalize data' for the 'Curve (Crv)' param in your GH file.…
Added by Joseph Oster at 2:29pm on September 12, 2015
is created for each point (25 paths, N=1 for each) which is feed into [Pull Point] for the pull geometry [G].
Correspondingly for the 4 source points a branch is created for each point and duplicated 25 times (4 paths, N=25 for each). This tree then needs to be inverted with [Path Mapper] so the structure will correspond to the format of the pull geometry. The mapping {A;B;C}(i) > {i}(B) produces (25 paths, N=4 for each) the structure to feed into the search point [P].
The [Pull Point] boolean toggle [C] needs to get set to False to obtain all the distances between all search and pull points (4 x 25 = 100 values).
Simultaneously there is also an index being created to correspond to the list of the 4 source points. This index is the integers 0 to 3 which are branched and inverted similar to the source points (25 paths, N=4 for each).
The distance output [D] from [Pull Point] is then sorted synchronously with the source point index for each branch. From the following screengrab branch {0;0} corresponds to a point in the 5 x 5 grid and the shortest distance between that point and a referenced source point index is 5.261. The index of the referenced source point is 3.
For each following sorted branch the first sorted index value will correspond to the closest source point (first [List Item] shown). This index value is then used to select from the original list of duplicated and inverted points and this is done for each of the 25 branches (second List Item shown).
Draw a line or whatever an away we go!…
the next challenge i am confronted with is that all 5 circles of the above-mentioned radi are placed in each point. My aim is to randomly distribute ONE of the created circles in each point, instead of 5 in each point.…
om frame -5 till frame -10 a frame spacing of 100mm is used,
etc.
Frame 0 is located on X=0 mm
Frame -5 will be on X=-500 mm
Frame -6 will be on X=(X of frame -5) -25 = -525 mm
Frame -11 wil be on X= ((X of frame -10) -10 = ?? mm
etc.
Cheers,
Bas…