In the file attached there is an attempt where I highlighted a range of points around 50%, but I got just this far, and I do not know if there is a better way.
Thanks always for the help.
Best regards,
Federico…
ay how many valid permutations exist.
But allow me to guesstimate a number for 20 components (no more, no less). Here are my starting assumptions:
Let's say the average input and output parameter count of any component is 2. So we have 20 components, each with 2 inputs and 2 outputs.
There are roughly 35 types of parameter, so the odds of connecting two parameters at random that have the same type are roughly 3%. However there are many conversions defined and often you want a parameter of type A to seed a parameter of type B. So let's say that 10% of random connections are in fact valid. (This assumption ignores the obvious fact that certain parameters (number, point, vector) are far more common than others, so the odds of connecting identical types are actually much higher than 3%)
Now even when data can be shared between two parameters, that doesn't mean that hooking them up will result in a valid operation (let's ignore for the time being that the far majority of combinations that are valid are also bullshit). So let's say that even when we manage to pick two parameters that can communicate, the odds of us ending up with a valid component combo are still only 1 in 2.
We will limit ourselves to only single connections between parameters. At no point will a single parameter seed more than one recipient and at no point will any parameter have more than one source. We do allow for parameters which do not share or receive data.
So let's start by creating the total number of permutations that are possible simply by positioning all 20 components from left to right. This is important because we're not allowed to make wires go from right to left. The left most component can be any one of 20. So we have 20 possible permutations for the first one. Then for each of those we have 19 options to fill the second-left-most slot. 20×19×18×17×...×3×2×1 = 20! ~2.5×1018.
We can now start drawing wires from the output of component #1 to the inputs of any of the other components. We can choose to share no outputs, output #1, output #2 or both with any of the downstream components (19 of them, with two inputs each). That's 2×(19×2) + (19×2)×(19×2-1) ~ 1500 possible connections we can make for the outputs of the first component. The second component is very similar, but it only has 18 possible targets and some of the inputs will already have been used. So now we have 2×(18×2-1) + (18×2-1)×(18×2-1) ~1300. If we very roughly (not to mention very incorrectly, but I'm too tired to do the math properly) extrapolate to the other 18 components where the number of possible connections decreases in a similar fashion thoughout, we end up with a total number of 1500×1300×1140×1007×891×789×697×...×83×51×24×1 which is roughly 6.5×1050. However note that only 10% of these wires connect compatible parameters and only 50% of those will connect compatible components. So the number of valid connections we can make is roughly 3×1049.
All we have to do now is multiply the total number of valid connection per permutation with the total number of possible permutations; 20! × 3×1049 which comes to 7×1067 or 72 unvigintillion as Wolfram|Alpha tells me.
Impressive as these numbers sound, remember that by far the most of these permutations result in utter nonsense. Nonsense that produces a result, but not a meaningful one.
EDIT: This computation is way off, see this response for an improved estimate.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 12:06pm on March 15, 2013
- valence 3 junctures whose incident edges form 120 degree angles.
The video above shows 20k particles navigating a periodic 64x64x64 voxel grid at approximately 20 fps.…
Added by David Reeves at 1:54am on September 20, 2014
h is known to have the same density in both directions (i.e. W == H) then you can take the square root of the total number and figure out the number of points in both directions.
If however the mesh is not square, there's no easy way to really figure out whether it's supposed to be 20×5 or 5×20, or 10×10 or 25×4 or 4×25. You might be able to figure it out by looking at mesh topology, i.e. which faces are connected by edges, but that takes a lot of coding. You might also be able to figure it out by looking at the shape of the mesh, to try and find a single column of points by their positions, but this will require some pretty heavy duty heuristics.
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David Rutten
david@mcneel.com
Seattle, WA…
Added by David Rutten at 12:07pm on November 22, 2010
make it a square
3. create a slider that shifts the selceted points on each level/circle meanwhile moving in z direction
you understand what I mean?
unfortunaly I do not have rhino here, I can give it a try tomorrow morning
Best Regards
DeDackel
…
of 400 interlocked rings in a 20 X 20 grid.
V1 - A single 'suLoop' component doing 400 'SUnion' operations (20 X 20): 11.6 minutes
V2 - Two phases: 5 X 10 in phase one and 2 X 4 in phase 2, 58 'SUnions' total: ~88 seconds combined
V3 - Two phases: 4 X 5 in phase one and 4 X 5 in phase 2, 40 'SUnions' total: ~104 seconds combined
Again, these Profiler benchmarks don't reflect the whole picture, and might be affected by other things I was doing on the laptop while the code was running.…
Added by Joseph Oster at 12:29pm on March 23, 2017
need at least 5 different panels with a random organization on the surface,it's not my case, because I have funny dimensions,but ideally could work on a square grid.
any idea?…