Open Scene Graph - Usage of DrawElementsUInt: Drawing a cloth without duplicating vertices - openscenegraph

I am currently working on simulating a cloth like material and then displaying the results via Open Scene Graph.
I've gotten the setup to display something cloth like, by just dumping all the vertices into 1 Vec3Array and then displaying them with a standard Point based DrawArrays. However I am looking into adding the faces between the vertices so that a further part of my application can visually see the cloth.
This is currently what I am attempting as for the PrimitiveSet
// create and add a DrawArray Primitive (see include/osg/Primitive). The first
// parameter passed to the DrawArrays constructor is the Primitive::Mode which
// in this case is POINTS (which has the same value GL_POINTS), the second
// parameter is the index position into the vertex array of the first point
// to draw, and the third parameter is the number of points to draw.
unsigned int k = CLOTH_SIZE_X;
unsigned int n = CLOTH_SIZE_Y;
osg::ref_ptr<osg::DrawElementsUInt> indices = new osg::DrawElementsUInt(GL_QUADS, (k) * (n));
for (uint y_i = 0; y_i < n - 1; y_i++) {
for (uint x_i = 0; x_i < k - 1; x_i++) {
(*indices)[y_i * k + x_i] = y_i * k + x_i;
(*indices)[y_i * (k + 1) + x_i] = y_i * (k + 1) + x_i;
(*indices)[y_i * (k + 1) + x_i + 1] = y_i * (k + 1) + x_i + 1;
(*indices)[y_i * k + x_i] = y_i * k + x_i + 1;
}
}
geom->addPrimitiveSet(indices.get());
This does however cause memory corruption when running, and I am not fluent enough in Assembly code to decipher what it is trying to do wrong when CLion gives me the disassembled code.
My thought was that I would iterate over each of the faces of my cloth and then select the 4 indices of the vertices that belong to it. The vertices are inputted from top left to bottom right in order. So:
0 1 2 3 ... k-1
k k+1 k+2 k+3 ... 2k-1
2k 2k+1 2k+2 2k+3 ... 3k-1
...
Has anyone come across this specific use-case before and does he/she perhaps have a solution for my problem? Any help would be greatly appreciated.

You might want to look into using DrawArrays with QUAD_STRIP (or TRIANGLE_STRIP because quads are frowned upon these days). There's an example here:
http://openscenegraph.sourceforge.net/documentation/OpenSceneGraph/examples/osggeometry/osggeometry.cpp
It's slightly less efficient than Elements/indices, but it's also less complicated to manage the relationship between the two related containers (the vertices and the indices).
If you really want to do the Elements/indices route, we'd probably need to see more repro code to see what's going on.

Related

Math.Net Exponential Moving Average

I'm using simple moving average in Math.Net, but now that I also need to calculate EMA (exponential moving average) or any kind of weighted moving average, I don't find it in the library.
I looked over all methods under MathNet.Numerics.Statistics and beyond, but didn't find anything similar.
Is it missing in library or I need to reference some additional package?
I don't see any EMA in MathNet.Numerics, however it's trivial to program. The routine below is based on the definition at Investopedia.
public double[] EMA(double[] x, int N)
{
// x is the input series
// N is the notional age of the data used
// k is the smoothing constant
double k = 2.0 / (N + 1);
double[] y = new double[x.Length];
y[0] = x[0];
for (int i = 1; i < x.Length; i++) y[i] = k * x[i] + (1 - k) * y[i - 1];
return y;
}
Occasionally I found this package: https://daveskender.github.io/Stock.Indicators/docs/INDICATORS.html It targets to the latest .NET framework and has very detailed documents.
Try this:
public IEnumerable<double> EMA(IEnumerable<double> items, int notationalAge)
{
double k = 2.0d / (notationalAge + 1), prev = 0.0d;
var e = items.GetEnumerator();
if (!e.MoveNext()) yield break;
yield return prev = e.Current;
while(e.MoveNext())
{
yield return prev = (k * e.Current) + (1 - k) * prev;
}
}
It will still work with arrays, but also List, Queue, Stack, IReadOnlyCollection, etc.
Although it's not explicitly stated I also get the sense this is working with money, in which case it really ought to use decimal instead of double.

Spectrogram - Calculating is wrong

Ok, so basically, I am implementing the following algorithm:
1) Slice signal of size 256 with an overlap of 128
2) Multiply each chunk with the Hanning window
3) Get DFT
4) Compute the abs value sqrt(re*re+im*im)
Plotting these values, as a imshow I get the following result:
This looks ok, it's clearly showing some difference, i.e. the spike where the signal has most amplitude shows. However, in Python I get this result:
I know that I'm doing something right, but, also doing something wrong. I just can't seem to find out where which is making me not think I have done it correctly.
Any rough ideas to where I could be going wrong here? I mean, is plotting the abs value the right way here or not?
Thanks
EDIT:
Result after clamping..
UPDATE:
Code:
for(unsigned j=0; (j < stft_temp[i].size()/2); j++)
{
double v = 10 * log10(stft_temp[i][j].re * stft_temp[i][j].re + stft_temp[i][j].im * stft_temp[i][j].im);
double pixe = 1.5 * (v + 100);
STFT[i][j] = (int) pixe;
}
Typically you might want to use a log magnitude and then scale to the required range, which would usually be 0..255. In pseudo-code:
mag_dB = 10 * log10(re * re + im * im); // get log magnitude (dB)
pixel_intensity = 1.5 * (mag_dB + 100); // offset and scale
pixel_intensity = min(pixel_intensity, 255); // clamp to 0..255
pixel_intensity = max(pixel_intensity, 0);

How to "soften" the edges of a polyline?

Given a line made up of several points, how do I make the line smoother/ curvier/ softer through adding intermediate points -- while keeping the original points completely intact and unmoved?
To illustrate, I want to go from the above to the below in this illustration:
Note how in the above picture, if we start at the bottom there will be a sharper right turn. In the bottom image however, this sharp right turn is made a bit "softer" by adding an intermediate point which is positioned in the middle of the two points, and using averages of the angles of the other lines. (Differently put, imagine the lines a race car would drive, as it couldn't abruptly change direction.) Note how, however, none of the original points was "touched", I just added more points.
Thanks!! For what it's worth, I'm implementing this using JavaScript and Canvas.
with each edge (e1 & e2) adjacent to each 'middle' edge (me) do
let X = 0.5 x length(me)
find 2 cubic bezier control points by extending the adjacent edges by X (see algorithm below)
get midpoint of cubic bezier (by applying formula below)
insert new 'midpoint' between me's two coordinates.
FloatPoint ExtendLine(const FloatPoint A, const FloatPoint B, single distance)
{
FloatPoint newB;
float lenAB = sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
newB.X = B.x - (B.x - A.x) / lenAB * distance;
newB.Y = B.Y - (B.Y - A.Y) / lenAB * distance;
return newB;
}
Edit: Formula for Bezier Curve midpoint: p(0.5) = 0.125(p0) + 0.375(p1) + 0.375(p2) + 0.125(p3)
The following code found elsewhere here does the job for me, in the specific context of JavaScript-Canvas which I'm using -- but please see Angus' answer for a more general approach:
var max = points.length;
context.beginPath();
var i = 0;
context.moveTo(points[i].x, points[i].y);
for (i = 1; i < max - 2; i++) {
var xc = (points[i].x + points[i + 1].x) * .5;
var yc = (points[i].y + points[i + 1].y) * .5;
context.quadraticCurveTo(points[i].x, points[i].y, xc, yc);
}
context.quadraticCurveTo(points[max - 2].x, points[max - 2].y, points[max - 1].x,points[max - 1].y);
context.closePath();
context.stroke();

C# Parallel processing concept

I am working on image processing with C# and implementing integral histogram. I am not getting into the details, but assume that I have MxN matrix and each cell value is the sum of itself and its left and upper neighbor, minus left upper corner neighbor. This works fast but I want to make it faster for large images or for real time image processing performance.
matrix[i,j] += matrix[i-1,j] + matrix[i,j-1] - matrix[i-1,j-1];
The actual implementation is:
for (int i = 0; i < width; i++)
for (int j = 0; j < height; j++)
{
int left = 0, upper = 0, u_l_corner = 0;
if (j - 1 >= 0)
{
left = matrix[i, j - 1];
}
if (i - 1 >= 0)
{
upper = matrix[i - 1, j];
}
if (j - 1 >= 0 && i - 1 >= 0)
u_l_corner = matrix[i - 1, j - 1];
matrix[i, j] += left + upper - u_l_corner;
}
So the calculation is dependent on the previous values of cells. Therefore, it does not look like it can be implemented in parallel(at least to me). But still, just want to make sure before go any further..
Can this algorithm be implemented in parallel using Parallel.For or any other method in C#? If so a simple example is highly appreciated, but if not, I better work on to find a "parallel image histogram algorithm", if any exists.
Thanks in advance.
As far as I'm concerned it is possible to make this algorithm parallel but i find no use of doing so if your matrix is relatively small ( time to process is less than few miliseconds ).
If you are very interested in making this algorithm parallel you could make this task splitted to exactly "j" tasks ( number of items in "y" axis ).
Key to doing so is starting first thread to calculate points in first row of this matrice ( [i, 0] ) then starting second thread delayed - second thread should chase first thread - must never overtake preceding thread.

How do I convert the 2 control points of a cubic curve to the single control point of a quadratic curve?

Having searched the web, I see various people in various forums alluding to approximating a cubic curve with a quadratic one. But I can't find the formula.
What I want is this:
input: startX, startY, control1X, control1Y, control2X, control2Y, endX, endY
output: startX, startY, controlX, controlY, endX, endY
Actually, since the starting and ending points will be the same, all I really need is...
input: startX, startY, control1X, control1Y, control2X, control2Y, endX, endY
output: controlX, controlY
As mentioned, going from 4 control points to 3 is normally going to be an approximation. There's only one case where it will be exact - when the cubic bezier curve is actually a degree-elevated quadratic bezier curve.
You can use the degree elevation equations to come up with an approximation. It's simple, and the results are usually pretty good.
Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2. Then for degree elevation, the equations are:
Q0 = P0
Q1 = 1/3 P0 + 2/3 P1
Q2 = 2/3 P1 + 1/3 P2
Q3 = P2
In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from the equations above:
P1 = 3/2 Q1 - 1/2 Q0
P1 = 3/2 Q2 - 1/2 Q3
If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since it's likely not, your best bet is to average them. So,
P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3
To translate to your terms:
controlX = -0.25*startX + .75*control1X + .75*control2X -0.25*endX
Y is computed similarly - the dimensions are independent, so this works for 3d (or n-d).
This will be an approximation. If you need a better approximation, one way to get it is by subdividing the initial cubic using the deCastlejau algorithm, and then degree-reduce each segment. If you need better continuity, there are other approximation methods that are less quick and dirty.
The cubic can have loops and cusps, which quadratic cannot have. This means that there are not simple solutions nearly never. If cubic is already a quadratic, then the simple solution exists. Normally you have to divide cubic to parts that are quadratics. And you have to decide what are the critical points for subdividing.
http://fontforge.org/bezier.html#ps2ttf says:
"Other sources I have read on the net suggest checking the cubic spline for points of inflection (which quadratic splines cannot have) and forcing breaks there. To my eye this actually makes the result worse, it uses more points and the approximation does not look as close as it does when ignoring the points of inflection. So I ignore them."
This is true, the inflection points (second derivatives of cubic) are not enough. But if you take into account also local extremes (min, max) which are the first derivatives of cubic function, and force breaks on those all, then the sub curves are all quadratic and can be presented by quadratics.
I tested the below functions, they work as expected (find all critical points of cubic and divides the cubic to down-elevated cubics). When those sub curves are drawn, the curve is exactly the same as original cubic, but for some reason, when sub curves are drawn as quadratics, the result is nearly right, but not exactly.
So this answer is not for strict help for the problem, but those functions provide a starting point for cubic to quadratic conversion.
To find both local extremes and inflection points, the following get_t_values_of_critical_points() should provide them. The
function compare_num(a,b) {
if (a < b) return -1;
if (a > b) return 1;
return 0;
}
function find_inflection_points(p1x,p1y,p2x,p2y,p3x,p3y,p4x,p4y)
{
var ax = -p1x + 3*p2x - 3*p3x + p4x;
var bx = 3*p1x - 6*p2x + 3*p3x;
var cx = -3*p1x + 3*p2x;
var ay = -p1y + 3*p2y - 3*p3y + p4y;
var by = 3*p1y - 6*p2y + 3*p3y;
var cy = -3*p1y + 3*p2y;
var a = 3*(ay*bx-ax*by);
var b = 3*(ay*cx-ax*cy);
var c = by*cx-bx*cy;
var r2 = b*b - 4*a*c;
var firstIfp = 0;
var secondIfp = 0;
if (r2>=0 && a!==0)
{
var r = Math.sqrt(r2);
firstIfp = (-b + r) / (2*a);
secondIfp = (-b - r) / (2*a);
if ((firstIfp>0 && firstIfp<1) && (secondIfp>0 && secondIfp<1))
{
if (firstIfp>secondIfp)
{
var tmp = firstIfp;
firstIfp = secondIfp;
secondIfp = tmp;
}
if (secondIfp-firstIfp >0.00001)
return [firstIfp, secondIfp];
else return [firstIfp];
}
else if (firstIfp>0 && firstIfp<1)
return [firstIfp];
else if (secondIfp>0 && secondIfp<1)
{
firstIfp = secondIfp;
return [firstIfp];
}
return [];
}
else return [];
}
function get_t_values_of_critical_points(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y) {
var a = (c2x - 2 * c1x + p1x) - (p2x - 2 * c2x + c1x),
b = 2 * (c1x - p1x) - 2 * (c2x - c1x),
c = p1x - c1x,
t1 = (-b + Math.sqrt(b * b - 4 * a * c)) / 2 / a,
t2 = (-b - Math.sqrt(b * b - 4 * a * c)) / 2 / a,
tvalues=[];
Math.abs(t1) > "1e12" && (t1 = 0.5);
Math.abs(t2) > "1e12" && (t2 = 0.5);
if (t1 >= 0 && t1 <= 1 && tvalues.indexOf(t1)==-1) tvalues.push(t1)
if (t2 >= 0 && t2 <= 1 && tvalues.indexOf(t2)==-1) tvalues.push(t2);
a = (c2y - 2 * c1y + p1y) - (p2y - 2 * c2y + c1y);
b = 2 * (c1y - p1y) - 2 * (c2y - c1y);
c = p1y - c1y;
t1 = (-b + Math.sqrt(b * b - 4 * a * c)) / 2 / a;
t2 = (-b - Math.sqrt(b * b - 4 * a * c)) / 2 / a;
Math.abs(t1) > "1e12" && (t1 = 0.5);
Math.abs(t2) > "1e12" && (t2 = 0.5);
if (t1 >= 0 && t1 <= 1 && tvalues.indexOf(t1)==-1) tvalues.push(t1);
if (t2 >= 0 && t2 <= 1 && tvalues.indexOf(t2)==-1) tvalues.push(t2);
var inflectionpoints = find_inflection_points(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y);
if (inflectionpoints[0]) tvalues.push(inflectionpoints[0]);
if (inflectionpoints[1]) tvalues.push(inflectionpoints[1]);
tvalues.sort(compare_num);
return tvalues;
};
And when you have those critical t values (which are from range 0-1), you can divide the cubic to parts:
function CPoint()
{
var arg = arguments;
if (arg.length==1)
{
this.X = arg[0].X;
this.Y = arg[0].Y;
}
else if (arg.length==2)
{
this.X = arg[0];
this.Y = arg[1];
}
}
function subdivide_cubic_to_cubics()
{
var arg = arguments;
if (arg.length!=9) return [];
var m_p1 = {X:arg[0], Y:arg[1]};
var m_p2 = {X:arg[2], Y:arg[3]};
var m_p3 = {X:arg[4], Y:arg[5]};
var m_p4 = {X:arg[6], Y:arg[7]};
var t = arg[8];
var p1p = new CPoint(m_p1.X + (m_p2.X - m_p1.X) * t,
m_p1.Y + (m_p2.Y - m_p1.Y) * t);
var p2p = new CPoint(m_p2.X + (m_p3.X - m_p2.X) * t,
m_p2.Y + (m_p3.Y - m_p2.Y) * t);
var p3p = new CPoint(m_p3.X + (m_p4.X - m_p3.X) * t,
m_p3.Y + (m_p4.Y - m_p3.Y) * t);
var p1d = new CPoint(p1p.X + (p2p.X - p1p.X) * t,
p1p.Y + (p2p.Y - p1p.Y) * t);
var p2d = new CPoint(p2p.X + (p3p.X - p2p.X) * t,
p2p.Y + (p3p.Y - p2p.Y) * t);
var p1t = new CPoint(p1d.X + (p2d.X - p1d.X) * t,
p1d.Y + (p2d.Y - p1d.Y) * t);
return [[m_p1.X, m_p1.Y, p1p.X, p1p.Y, p1d.X, p1d.Y, p1t.X, p1t.Y],
[p1t.X, p1t.Y, p2d.X, p2d.Y, p3p.X, p3p.Y, m_p4.X, m_p4.Y]];
}
subdivide_cubic_to_cubics() in above code divides an original cubic curve to two parts by the value t. Because get_t_values_of_critical_points() returns t values as an array sorted by t value, you can easily traverse all t values and get the corresponding sub curve. When you have those divided curves, you have to divide the 2nd sub curve by the next t value.
When all splitting is proceeded, you have the control points of all sub curves. Now there are left only the cubic control point conversion to quadratic. Because all sub curves are now down-elevated cubics, the corresponding quadratic control points are easy to calculate. The first and last of quadratic control points are the same as cubic's (sub curve) first and last control point and the middle one is found in the point, where lines P1-P2 and P4-P3 crosses.
Conventions/terminology
Cubic defined by: P1/2 - anchor points, C1/C2 control points
|x| is the euclidean norm of x
mid-point approx of cubic: a quad that shares the same anchors with the cubic and has the control point at C = (3·C2 - P2 + 3·C1 - P1)/4
Algorithm
pick an absolute precision (prec)
Compute the Tdiv as the root of (cubic) equation sqrt(3)/18 · |P2 - 3·C2 + 3·C1 - P1|/2 · Tdiv ^ 3 = prec
if Tdiv < 0.5 divide the cubic at Tdiv. First segment [0..Tdiv] can be approximated with by a quadratic, with a defect less than prec, by the mid-point approximation. Repeat from step 2 with the second resulted segment (corresponding to 1-Tdiv)
0.5<=Tdiv<1 - simply divide the cubic in two. The two halves can be approximated by the mid-point approximation
Tdiv>=1 - the entire cubic can be approximated by the mid-point approximation
The "magic formula" at step 2 is demonstrated (with interactive examples) on this page.
Another derivation of tfinniga's answer:
First see Wikipedia Bezier curve
for the formulas for quadratic and cubic Bezier curves (also nice animations):
Q(t) = (1-t)^2 P0 + 2 (1-t) t Q + t^2 P3
P(t) + (1-t)^3 P0 + 3 (1-t)^2 t P1 + 3 (1-t) t^2 P2 + t^3 P3
Require these to match at the middle, t = 1/2:
(P0 + 2 Q + P3) / 4 = (P0 + 3 P1 + 3 P2 + P3) / 8
=> Q = P1 + P2 - (P0 + P1 + P2 + P3) / 4
(Q written like this has a geometric interpretation:
Pmid = middle of P0 P1 P2 P3
P12mid = midway between P1 and P2
draw a line from Pmid to P12mid, and that far again: you're at Q.
Hope this makes sense -- draw a couple of examples.)
In general, you'll have to use multiple quadratic curves - many cases of cubic curves can't be even vaguely approximated with a single quadratic curve.
There is a good article discussing the problem, and a number of ways to solve it, at http://www.timotheegroleau.com/Flash/articles/cubic_bezier_in_flash.htm (including interactive demonstrations).
I should note that Adrian's solution is great for single cubics, but when the cubics are segments of a smooth cubic spline, then using his midpoint approximation method causes slope continuity at the nodes of the segments to be lost. So the method described at http://fontforge.org/bezier.html#ps2ttf is much better if you are working with font glyphs or for any other reason you want to retain the smoothness of the curve (which is most probably the case).
Even though this is an old question, many people like me will see it in search results, so I'm posting this here.
I would probably draw a series of curves instead of trying to draw one curve using a different alg. Sort of like drawing two half circles to make up a whole circle.
Try looking for opensource Postcript font to Truetype font converters. I'm sure they have it. Postscript uses cubic bezier curves, whereas Truetype uses quadratic bezier curves. Good luck.

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