rated by "<" symbols. Examples: "2<10", "2<4<10", "Pow(2, 1)<5*Sin(3)<10".
The entered text contains 2 or 3 segments separated by two or more consecutive dots. Examples "2..10", "2..4..10", "Pow(2, 1)....5*Sin(3)..10".
If only two segments are provided, then the initial value will be the same as the minimum value. If a bounds number or a default value is written as a simple number, then the number of decimal places will be harvested. I.e. "2..4..10" is not the same as "2..4..10.00" as the former will result in an integer slider and the latter in a slider with two decimal places.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 10:08am on February 15, 2013
. etc. So it's 80020 which is ~1058. Assuming you're allowed to use the same component more than once.
1058 × 1049 = 10107 total possible algorithms. When talking about big numbers I only have three frames of reference. The distance from us to the edge of the observable universe is roughly 1029 millimeters, the observable universe contains 1080 protons and the volume of the observable universe is roughly 5×10105 cubic nanometers. So you could more or less put a different valid Grasshopper algorithm into every cubic nanometer of this universe.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
x and min values for x,y,z and calculate energy for each optionand collect these results in excel sheet ...
option No. x y z Annual coiling demand(by DIva)
1 10 10 10
2 10 15 20
3 11 10 19
4 12 14 17
5 15 16 15
6 16 11 14
7 18 12 12
.
.
etc
Regards ...
hossam
Hossam.wefki@gmail.com…
Tree:
{0;0;0} N = 2
{0;0;1} N = 1
{0;0;2} N = 3
{0;1;0} N = 5
{0;1;1} N = 8
{0;1;2} N = 10
If we apply the aforementioned mapping to this tree, we'll end up with the following result:
{0;0} N = 6
{0;1} N = 23
Basically {0;0;0}, {0;0;1} and {0;0;2} are combined into a single path {0;0} as we disregard the third index because "C" is no longer present in the target mapping.
Because we only use the Mapper to modify paths, we do not lose any data items, though we might lose some of the paths.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 1:03pm on August 25, 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
I'm trying to retrieve data from a tag heuer timing system through rs 232. Firefly ports available says com ports 1 and 3 are available. Is one of those two the rs 232 9 pin? Thanks for your help.
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