Using Generate Sample Set to avoid the Repeat node


Hat tip to Oleg who posted this tip on the mailing list some time ago.

The basic ideas is that instead of looping over an array with a Repeat with Counter, you use Generate Sample Set to get a data set, and you do everything in a per-generated element context.

GenerateSampleSEt_vs_Repeat

As an example, let’s revisit the problem of taking one array and creating a new array where each element is the sum of the elements before it.

Array_accumulation

The old way, with a Repeat with Counter node, looks like this:
RepeatWithCounter

Using Generate Sample Set, you can work with a data set and use that to access the array elements:
GenerateSampleSet

Generate Sample Set is set up to give an exact number of samples:
GenerateSampleSet-PPG

Saturday Snippet: Getting data from a DataArray2D ICE attribute


I wanted to do this JScript, but I had to do it in Python first, to establish that it was actually possible (with JScript, you’ve got to mess around with VBarrays and such).

# Using one of the CrowdFX sample scenes:
Application.SelectObj("Pedestrian_Mesh.Actor_Copies", None, None);
o = Application.Selection(0)

a = o.ActivePrimitive.Geometry.GetICEAttributeFromName("Materials")
print len(a.DataArray2D)
print len(a.DataArray2D[0] )
print a.DataArray2D[0][0]
for s in a.DataArray2D[0][0]:
    print s

# 1
# 1
# (u'', u'Sources.Materials.PedestrianLib.Shoes', u'Sources.Materials.PedestrianLib.Hair', u'Sources.Materials.PedestrianLib.Legs', u'Sources.Materials.PedestrianLib.Skin', u'Sources.Materials.PedestrianLib.Shirt')
# Sources.Materials.PedestrianLib.Shoes
# Sources.Materials.PedestrianLib.Hair
# Sources.Materials.PedestrianLib.Legs
# Sources.Materials.PedestrianLib.Skin
# Sources.Materials.PedestrianLib.Shirt

After I had it working in Python, I was able to figure it out in JScript:

// Using one of the CrowdFX sample scenes:
SelectObj("Pedestrian_Mesh.Actor_Copies", null, null);
o = Selection(0);

a = o.ActivePrimitive.Geometry.GetICEAttributeFromName("Materials");

x = new VBArray( a.DataArray2D ).toArray();
y = new VBArray( x[0] ).toArray();
for ( var i = 0; i < y.length; i++ )
{
    LogMessage( y[i] );
}

Hexagon tiling with ICE


hexagon_tiling_rr
Here’s a relatively simple ICE tree that arranges hexagons on the XZ plane. The “difficult” part was creating the arrays of position coordinates. For that I used the standard modulo technique. It’s funny, it all makes perfect sense when you’re plugging stuff together, but afterwards it’s hard to make heads or tail of what you did 🙂
hexagon_tiling_1

To understand what’s going on here, it helps to look at the final arrays that are built:
hexagon_tiling_show_values

The math for arranging the hexagon tiles is pretty simple. I started with a simple test point cloud to make sure I understood what I needed to do. After that, it was just a question of setting up the arrays.
hexagon_tiling_testing

If you want to take a look at the ICE trees, here’s some compounds. Note that this isn’t a finished piece of work. It’s more of a draft version. For example, my hexagon has a side length of 5 and that’s hardcoded into the ICE tree right now.
Hexagon_Tiler.xsicompound
Hexagon_Math_Tester.xsicompound

ICE: Building a regular hexagon


Let’s build a regular hexagon!

In this example, I take the vector (5,0,0) and rotate it by 60 degrees, then 120, then 180, and so on, until I have the 6 points of the hexagon. In ICE, that equates to getting an array of rotations, stuffing that into Rotate Vector, and getting an array of positions back.

Notice how the length of each side is the same (5), as is the distance of each point from the local origin.
RegularHexagon

Before I jumped into building the polygon, I first did a quick test with a point cloud, to make sure I understood how to add points in the right places. In general, I think it’s good practice to do some kind of “proof of concept” before you really dive into the details.
RegularHexagon-prelims

Screenshots of the week


LKLightning 2.0 Tutorial 01 – Swirling7

Display debug per-object attribute on distant point cloud
by Fabricio Chamon
DebugAttributeAtBBox

Moving cached geometry
by Alok Gandhi
jhddjdgc

Reflections
by NNois
Reflection

Turbulize Null position
by Helli
nullposition

Splitting edges
by iamVFX, julian johnson
SplitEdgeEqually
SubdivideEdgeByLength

Finding he indices of negative values in an array
by iamVFX
FindIndicesOfNegativeValuesInArray

Splitting an edge into equal-length new edges


Here’s the basic idea of how to split an edge into N equal-length new edges. Notice how I work backwards (in the sense that my split ratio decreases). That way I don’t know have to know (or care) about the new edges. I just keep splitting the same edge, whose EdgeIndex I already know.

SplitEdge0

SplitEdge1

SplitEdge2

SplitEdge3

Based on that observation, here’s a rough draft of an ICE tree that takes an edge of length N, and splits it into N equal-length edges.

SplitEdges