terior wall. I disconnected this air wall and the model runs fine I ran it for 19 generations for over the period of 7 hours.
2. Earlier, when I ran the model with air wall connected to altConstruction_, I observed that in the middle of this optimization process, one or many zones will disappear altogether from rhino scene and that is when the results became undefined. In the image you attached, if you see in the top left corner, zone Commercial 1F is missing. I witnessed a similar behavior. This disappearance of zones may explain the erratic highs and lows in fitness numbers that you are talking about.
3. I also made a few changes in how you apply glazing ratios. Please find file with changes.
4. My results are attached as well. When I used air wall as altConstruction, the model ran for 7 generations and then started displaying errors as I mentioned in my earlier responses. After removing air wall and making a change in how glazing ratio is applied, the model ran successfully for 19 generations.
Hope this helps,
Devang
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allows you to select every third item between 34 and 55 (for example)... or any other set of branches based on any path mask.
If i only need 2 branches its no big deal (might as well hand/hard code it), but if i need 20 branches out of 300 that are each 7 apart, it saves some math and typing...…
g definition but in diva for grasshopper in material it just appear dusty_med and not metal_railings and metal_treads. How I should write the correct definition?
void brightfunc dusty_med4 dirt dirt.cal -s 101 .25
dusty_med metal metal_railings005 .7 .7 .7 .3 .2
dusty_med metal metal_treads005 .5 .5 .5 .3 .2…
but you could use some combination of components to control the length of the '1' run and value of the repetition.
Inserting lists into other lists is in fact the much more complex. Grasshopper vanilla only provides a component for inserting individual items into a list (although many items can be inserted at once, making the insertion calculation index maths a lot easier).
To do it right you'll have to repeat the '1', '2', '3' as many times as you need to insert it, and then, generate the matching list of insertion indices '3', '3', '3', '7, '7', '7', ... This will involve a fair amount of Series or Duplicate, or Stack, or whatever...…
identical knots on end ends of the curve, so the knot vector will look something like:
0,0,0,1,2,2,2
So one at each end and one in the middle.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 3:47am on January 7, 2013
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
a seed, and instead creating a pattern where each color has a seed/control slider for each row? For example, row 1: brown 2, tan 6, yellow 7, purple 3, repeat. row 2: brown 6, tan 1, yellow 4, purple 10, repeat. row 3: yellow 5, purple 1, brown 3, tan 10, repeat. row 4: purple 2, brown 7, tan 3, yellow 4, repeat. Then repeat that sequence up the wall? For each color, the number in the sequence should be adjustable.
Thank you again for your help!…