I have a mesh shown in the picture below. Each segment of the mesh has a positive (np) and negative normal (nn). Each pair of normals belongs to a segment from the mesh.
The red dot in the middle is my observation point. I need to find out which normals are visible to this observation point.
The method I have followed so far is to to calculate a normal vector from the observation point to the middle of the each mesh segment (ob_i). I then do the dot product on the ob_i and np or nn. The result of this is either greater than zero or less than. If it is less than the vector is facing the other way and can be regarded as hidden from the observation point.
With this method I get the following result:
Heres how I calculate the dot product and check for visibility:
vector<vector<int>> calculateShadowingMatrix(vector<point> observationNormals, vector<vector<point>> normals){
vector<vector<int>> shadowMatrix;
for(unsigned ii = 0; ii < normals.size(); ii++){
vector<int> visibilty;
for(unsigned jj = 0; jj < normals[ii].size(); jj++) {
double dot = dotProduct(observationNormals[ii], normals[ii][jj]);
if (dot <= 0.0) {
visibilty.push_back(1);
} else {
visibilty.push_back(0);
}
}
shadowMatrix.push_back(visibilty);
}
return shadowMatrix;
}
double dotProduct(point _u, point _v){
return _u.getX()*_v.getX() - _u.getY()*_v.getY();
}
I am trying to get all of the normal vectors which are facing the observation point to be present and zero out the ones which are facing outwards. I get errors, as can be seen in the second image, some of the vector are facing the wrong direction.
Final result should look as follows:
Your dot product calculation is wrong. It should be X * X + Y * Y.
Related
I have a 2-D array of squares (square shape), Each square has 50 units length and x, y co-ordinates. The distance between the squares is 5 units. The x, y co-ordinates are the bottom left corner of the square. Now, given any point(x,y co-ordinates) how can i find the closest square to this point.
square **sq = new square*[10];
for(int i=0;i<10;++i){
sq[i]=new square[10];
}
int m=0, n=0;
for(int i=0;i<10;++i){
m=0;
for(int j=0;j<10;++j){
sq[i][j].setCoOrdinates(m,n);
m+=55;
}
n+=55;
}
// Given a point (x, y) how can i find the index (i, j) of closest square to this point.
Let's rigorously define the definition of distance from a square to a point: "The minimum of all lengths from the point (x,y) to some point within the rectangle".
This gives some clear rules for defining the distance to a rectangle. For any rectangle, if the point(x,y) is directly above, below, to the left of, or to the right of the sides of the rectangle, the minimum distance to the rectangle is the straight line drawn either vertically or horizontally through the point. For instance, if your point is (40, 90) and your rectangle's bottom-left is (0,0) (and it is a 50x50 square), you can draw a vertical line through your point and the distance is min(90-(0+50), 90-0)
If the point is not directly above, below, left of, or right of the sides of the square, then the closest distance is to the closest of the four corners. You can simply take the min of all the distances to the four corners, which can be found by using the distance formula.
Simply apply this logic to each of your squares in your array and you should be good to go! Time O(NM) where n is the number of rows of squares and M is the number of columns of squares. In other words, O(number of squares you have). Space O(1).
Lets start with the simpler problem: All rectangles are 55 units wide, ie there is no empty space between them...
The best container is the one that you do not need. There is a simple relation between i and j index of a square in the array and its m and n coordinates:
const int distance = 55;
int m(int i) { return i*distance; }
int n(int j) { return m(j); }
As the relation is linear, you can invert it:
int i(int m) { return m / distance; }
int j(int n) { return i(n); }
Using integer arithmetics is fine here, because we get the same result as if we had used floating points and then rounded down.
This is already the full solution for the simpler problem. Given coordinats m and n the closest square is at index i(m),j(n).
Now lets introduce the gap:
cosnt int width = 50;
const int gap = 5;
Now we have to distinguish two possibilities: The given coordinats are inside a square or outside. When it is outside there are two candidates for the cloest square.
int i_with_gap(int m) {
int i = m / distance;
// point is inside the square?
if (m % distance <= width) return i;
// point is closer to square at index i?
if (m % distance <= width+ gap/2.0) return i;
// otherwise i+1 is closer
return i+1;
}
I am applying a 3D Voronoi pattern on a mesh. Using those loops, I am able to compute the cell position, an id and the distance.
But I would like to compute a normal based on the generated pattern.
How can I generate a normal or reorient the current normal based on this pattern and associated cells ?
The aim is to provide a faced look for the mesh. Each cell's normal should point in the same direction and adjacent cells point in different directions. Those directions should be based on the original mesh normals, I don't want to totally break mesh normals and have those points in random directions.
Here's how I generate the Voronoi pattern.
float3 p = floor(position);
float3 f = frac(position);
float id = 0.0;
float distance = 10.0;
for (int k = -1; k <= 1; k++)
{
for (int j = -1; j <= 1; j++)
{
for (int i = -1; i <= 1; i++)
{
float3 cell = float3(float(i), float(j), float(k));
float3 random = hash3(p + cell);
float3 r = cell - f + random * angleOffset;
float d = dot(r, r);
if (d < distance)
{
id = random;
distance = d;
cellPosition = cell + p;
normal = ?
}
}
}
}
And here's the hash function :
float3 hash3(float3 x)
{
x = float3(dot(x, float3(127.1, 311.7, 74.7)),
dot(x, float3(269.5, 183.3, 246.1)),
dot(x, float3(113.5, 271.9, 124.6)));
return frac(sin(x)*43758.5453123);
}
This looks like a fairly expensive fragment shader, it might more sense to bake out a normal map than to try to do this in real time.
It's hard to tell what your shader is doing but I think it's checking every pixel against a 3x3 grid of voronoi cells. One weird thing is that random is a vec3 that somehow gets assigned to id, which is just a scalar.
Anyway, it sounds like you would like to perturb the mesh-supplied normal by a random vector, but you'd like all pixels corresponding to a particular voronoi cell to be perturbed in the same way.
Since you already have a variable called random which presumably represents a random value generated deterministically as a function of the voronoi cell, you could just use that. For example, the following would perturb the normal by a small amount:
normal = normalize(meshNormal + 0.2 * normalize(random));
If you want to give more weight to the random component, just increase the 0.2 constant.
I asked a question about a network which I've been building last week, and I iterated on the suggestions which lead me to finding a few problems. I've come back to this project and fixed up all the issues and learnt a lot more about CNNs in the process. Now I'm stuck on an issue were all of my weights move to massively negative values, which coupled with the RELU ends in the output image always being completely black (making it impossible for the classifier to do it's job).
On two labeled images:
These are passed into a two layer network, one classifier (which gets 100% on its own) and a one filter 3*3 convolutional layer.
On the first iteration the output from the conv layer looks like (images in same order as above):
The filter is 3*3*3, due to the images being RGB. The weights are all random numbers between 0.0f-1.0f. On the next iteration the images are completely black, printing the filters shows that they are now in range of -49678.5f (the highest I can see) and -61932.3f.
This issue in turn is due to the gradients being passed back from the Logistic Regression/Linear layer being crazy high for the cross (label 0, prediction 0). For the circle (label 1, prediction 0) the values are between roughly -12 and -5, but for the cross they are all in the positive high 1000 to high 2000 range.
The code which sends these back looks something like (some parts omitted):
void LinearClassifier::Train(float * x,float output, float y)
{
float h = output - y;
float average = 0.0f;
for (int i =1; i < m_NumberOfWeights; ++i)
{
float error = h*x[i-1];
m_pGradients[i-1] = error;
average += error;
}
average /= static_cast<float>(m_NumberOfWeights-1);
for (int theta = 1; theta < m_NumberOfWeights; ++theta)
{
m_pWeights[theta] = m_pWeights[theta] - learningRate*m_pGradients[theta-1];
}
// Bias
m_pWeights[0] -= learningRate*average;
}
This is passed back to the single convolution layer:
// This code is in three nested for loops (for layer,for outWidth, for outHeight)
float gradient = 0.0f;
// ReLu Derivative
if ( m_pOutputBuffer[outputIndex] > 0.0f)
{
gradient = outputGradients[outputIndex];
}
for (int z = 0; z < m_InputDepth; ++z)
{
for ( int u = 0; u < m_FilterSize; ++u)
{
for ( int v = 0; v < m_FilterSize; ++v)
{
int x = outX + u - 1;
int y = outY + v - 1;
int inputIndex = x + y*m_OutputWidth + z*m_OutputWidth*m_OutputHeight;
int kernelIndex = u + v*m_FilterSize + z*m_FilterSize*m_FilterSize;
m_pGradients[inputIndex] += m_Filters[layer][kernelIndex]*gradient;
m_GradientSum[layer][kernelIndex] += input[inputIndex]*gradient;
}
}
}
This code is iterated over by passing each image in a one at a time fashion. The gradients are obviously going in the right direction but how do I stop the huge gradients from throwing the prediction function?
RELU activations are notorious for doing this. You usually have to use a low learning rate. The reasoning behind this is that when the RELU returns positive numbers it can continue to learn freely, but if a unit gets in a position where the signal coming into it is always negative it can become a "dead" neuron and never activate again.
Also initializing your weights is more delicate with RELU. It appears that you are initializing to range 0-1 which creates a huge bias. Two tips here - Use a range centered around 0, and a range that is much smaller. A normal distribution with mean 0 and std 0.02 usually works well.
I fixed it by downscaling the gradients int the CNN layer, but now I'm confused as to why this works/is needed so if anyone has any intuition as to why this works that'd be great.
Im trying to calculate the angle between two edges in a graph, in order to do that I transfer both edges to origin and then used dot product to calculate the angle. my problem is that for some edges like e1 and e2 the output of angle(e1,e2) is -1.#INDOO.
what is this output? is it an error?
Here is my code:
double angle(Edge e1, Edge e2){
Edge t1 = e1, t2 = e2;
Point tail1 = t1.getTail(), head1 = t1.getHead();
Point u(head1.getX() - tail1.getX(), head1.getY() - tail1.getY());
Point tail2 = t2.getTail(), head2 = t2.getHead();
Point v(head2.getX() - tail2.getX(), head2.getY() - tail2.getY());
double dotProduct = u.getX()*v.getX() + u.getY()*v.getY();
double cosAlpha = dotProduct / (e1.getLength()*e2.getLength());
return acos(cosAlpha);
}
Edge is a class that holds two Points, and Point is a class that holds two double numbers as x and y.
Im using angle(e1,e2) to calculate the orthogonal projection length of a vector like b on to a vector like a :
double orthogonalProjectionLength(Edge b, Edge a){
return (b.getLength()*sin(angle(b, a) * (PI / 180)));
}
and this function also sometimes gives me -1.#INDOO. you can see the implementation of Point and Edge here.
My input is a set S of n Points in 2D space. Iv constructed all edges between p and q (p,q are in S) and then tried to calculate the angle like this:
for (int i = 0; i < E.size(); i++)
for (int j = 0; j < E.size(); j++){
if (i == j)
cerr << fixed << angle(E[i], E[j]) << endl; //E : set of all edges
}
If the problem comes from cos() and sin() functions, how can I fix it? is here other libraries that calculate sin and cos in more efficient way?
look at this example.
the inputs in this example are two distinct points(like p and q), and there are two Edges between them (pq and qp). shouldnt the angle(pq , qp) always be 180 ? and angle(pq,pq) and angle(qp,qp) should be 0. my programm shows two different kinds of behavior, sometimes angle(qp,qp) == angle(pq,pq) ==0 and angle(pq , qp) == angle(pq , qp) == 180.0, and sometimes the answer is -1.#INDOO for all four edges.
Here is a code example.
run it for several times and you will see the error.
You want the projection and you go via all this trig? You just need to dot b with the unit vector in the direction of a. So the final answer is
(Xa.Xb + Ya.Yb) / square_root(Xa^2 + Ya^2)
Did you check that cosAlpha doesn't reach 1.000000000000000000000001? That would explain the results, and provide another reason not to go all around the houses like this.
It seems like dividing by zero. Make sure that your vectors always have 0< length.
Answer moved from mine comment
check if your dot product is in <-1,+1> range ...
due to float rounding it can be for example 1.000002045 which will cause acos to fail.
so add two ifs and clamp to this range.
or use faster way: acos(0.99999*dot)
but that lowers the precision for all angles
and also if 0.9999 constant is too big then the error is still present
A recommended way to compute angles is by means of the atan2 function, taking two arguments. It returns the angle on four quadrants.
You can use it in two ways:
compute the angles of u and v separately and subtract: atan2(Vy, Vx) - atan2(Uy, Ux).
compute the cross- and dot-products: atan2(Ux.Vy - Uy.Vx, Ux.Uy + Vx.Vy).
The only case of failure is (0, 0).
Ok, I'm having a bit of trouble finding a solution for this that seems to be a simple geometry problem.
I have a list of triple coordinates that form a square angle.
Between all these triple-coordinates I want to find a pair that forms up a square.
I believe the best I can do to exemplify is show an image:
and 2. are irrelevant. 3. and 4. are what I'm looking for.
For each triple coordinate I have the midle point, where the angle is, and two other points that describe the two segments that form the angle.
Summing it up, given six points, 2 for the diagonal + 4 other points, how can I find if these make a square?
obs: the two lines that make the angle are consistent but don't have the same size.
obs2:the lines from different triples may not intersect
Thank you for time and any help and insight provided.
If any term I used is incorrect or just plain hard to understand let me know, I'm not a native english speaker.
Edit: The code as it stands.
//for all triples
for (size_t i = 0; i < toTry.size() - 1; i++) {
Vec2i center_i = toTry[i].avg;
//NormalizedDiagonal = ((Side1 - Center) + (Side2 - Center));
Vec2i a = toTry[i].p, b = toTry[i].q;
Vec2f normalized_i = normalizedDiagonal(center_i, toTry[i].p, toTry[i].q);
for (size_t j = i + 1; j < toTry.size(); j++) {
Vec2i center_j = toTry[j].avg;
//Se os pontos sao proximos, nao importam
if (areClose(center_i, center_j, 25))
continue;
Vec2f normalized_j = normalizedDiagonal(center_j, toTry[j].p, toTry[j].q);
line(src, Point(center_i[0], center_i[1]), Point(center_i[0] + 1 * normalized_i[0], center_i[1] + 1 * normalized_i[1]), Scalar(255, 255, 255), 1);
//test if antiparallel
if (abs(normalized_i[0] - normalized_j[0]) > 0.1 || abs(normalized_i[1] - normalized_j[1] > 0.1))
continue;
Vec2f delta;
delta[0] = center_j[0] - center_i[0]; delta[1] = center_j[1] - center_i[1];
double dd = sqrt(pow((center_i[0] - center_j[0]), 2) + pow((center_i[1] - center_j[1]), 2));
//delta[0] = delta[0] / dd;
//delta[1] = delta[1] / dd;
float dotProduct = normalized_i[0] * delta[0] + normalized_i[1] * delta[1];
//test if do product < 0
if (dotProduct < 0)
continue;
float deltaDotDiagonal = delta[0] * normalized_i[0] + delta[1] * normalized_i[1];
menor_d[0] = delta[0] - deltaDotDiagonal * normalized_i[0];
menor_d[1] = delta[1] - deltaDotDiagonal * normalized_i[1];
dd = sqrt(pow((center_j[0] - menor_d[0]), 2) + pow((center_j[1] - menor_d[1]), 2));
if(dd < 25)
[...]
Just to be clear, the actual lengths of the side segments is irrelevant, right? All you care about is whether the semi-infinite lines formed by the side segments of two triples form a square? Or do the actual segments need to intersect?
Assuming the former, a method to check whether two triples form a square is as follows. Let's use the Point3D and Vector3D from the System.Windows.Media.Media3D namespace to define some terminology, since these are decent general-purpose 3d double precision points and vectors that support basic linear algebra methods. These are c# so you can't use them directly but I'd like to be able to refer to some of the basic methods mentioned there.
Here is the basic method to check if two triples intersect:
Define a triple as follows: Center, Side1 and Side2 as three Point3D structures.
For each triple, define the normalized diagonal vector as
NormalizedDiagonal = ((Side1 - Center) + (Side2 - Center));
NormalizedDiagonal.Normalize()
(You might want to cache this for performance.)
Check if the two centers are equal within some linear tolerance you define. If equal, return false -- it's a degenerate case.
Check if the two diagonal vectors are antiparallel within some angular tolerance you define. (I.e. NormalizedDiagonal1 == -NormalizedDiagonal2 with some tolerance.) If not, return false, not a square.
Compute the vector from triple2.Center to triple2.Center: delta = triple2.Center - triple1.Center.
If double deltaDotDiagonal = DotProduct(delta, triple1.NormalizedDiagonal) < 0, return false - the two triples point away from each other.
Finally, compute the distance from the center of triple2 to the (infinite) diagonal line passing through the center triple1. If zero (within your linear tolerance) they form a square.
To compute that distance: distance = (delta - deltaDotDiagonal*triple1.NormalizedDiagonal).Length
Note: deltaDotDiagonal*triple1.NormalizedDiagonal is the projection of the delta vector onto triple1.NormalizedDiagonal, and thus delta - deltaDotDiagonal*triple1.NormalizedDiagonal is the component of delta that is perpendicular to that diagonal. Its length is the distance we seek.
Finally, If your definition of a square requires that the actual side segments intersect, you can add an extra check that the lengths of all the side segments are less than sqrt(2) * delta.Length.
This method checks if two triples form a square. Finding all triples that form squares is, of course, O(N-squared). If this is a problem, you can put them in an array and sort then by angle = Atan2(NormalizedDiagonal.Y, NormalizedDiagonal.X). Having done that, you can find triples that potentially form squares with a given triple by binary-searching the array for triples with angles = +/- π from the angle of the current triple, within your angular tolerance. (When the angle is near π you will need to check both the beginning and end of the array.)
Update
OK, let's see if I can do this with your classes. I don't have definitions for Vec2i and Vec2f so I could get this wrong...
double getLength(Vec2f vector)
{
return sqrt(pow(vector[0], 2) + pow(vector[1], 2));
}
Vec2f scaleVector(Vec2f vec, float scale)
{
Vec2f scaled;
scaled[0] = vec[0] * scale;
scaled[1] = vec[1] * scale;
return scaled;
}
Vec2f subtractVectorsAsFloat(Vec2i first, Vec2i second)
{
// return first - second as float.
Vec2f diff;
diff[0] = first[0] - second[0];
diff[1] = first[1] - second[1];
return diff;
}
Vec2f subtractVectorsAsFloat(Vec2f first, Vec2f second)
{
// return first - second as float.
Vec2f diff;
diff[0] = first[0] - second[0];
diff[1] = first[1] - second[1];
return diff;
}
double dot(Vec2f first, Vec2f second)
{
return first[0] * second[0] + first[1] * second[1];
}
//for all triples
for (size_t i = 0; i < toTry.size() - 1; i++) {
Vec2i center_i = toTry[i].avg;
//NormalizedDiagonal = ((Side1 - Center) + (Side2 - Center));
Vec2i a = toTry[i].p, b = toTry[i].q;
Vec2f normalized_i = normalizedDiagonal(center_i, toTry[i].p, toTry[i].q);
for (size_t j = i + 1; j < toTry.size(); j++) {
Vec2i center_j = toTry[j].avg;
//Se os pontos sao proximos, nao importam
if (areClose(center_i, center_j, 25))
continue;
Vec2f normalized_j = normalizedDiagonal(center_j, toTry[j].p, toTry[j].q);
//test if antiparallel
if (abs(normalized_i[0] - normalized_j[0]) > 0.1 || abs(normalized_i[1] - normalized_j[1] > 0.1))
continue;
// get a vector pointing from center_i to center_j.
Vec2f delta = subtractVectorsAsFloat(center_j, center_i);
//test if do product < 0
float deltaDotDiagonal = dot(delta, normalized_i);
if (deltaDotDiagonal < 0)
continue;
Vec2f deltaProjectedOntoDiagonal = scaleVector(normalized_i, deltaDotDiagonal);
// Subtracting the dot product of delta projected onto normalized_i will leave the component
// of delta which is perpendicular to normalized_i...
Vec2f distanceVec = subtractVectorsAsFloat(deltaProjectedOntoDiagonal, center_j);
// ... the length of which is the distance from center_j
// to the diagonal through center_i.
double distance = getLength(distanceVec);
if(distance < 25) {
}
}
There are two approaches to solving this. One is a very direct approach that involves finding the intersection of two line segments.
You just use the triple coordinates to figure out the midpoint, and the two line segments that protrude from it (trivial). Do this for both triple-sets.
Now calculate the intersection points, if they exist, for all four possible permutations of the extending line segments. From the original answer to a similar question:
You might look at the code I wrote for Computational Geometry in C,
which discusses this question in detail (Chapter 1, Section 5). The
code is available as SegSegInt from the links at that web site.
In a nutshell, I recommend a different approach, using signed area of
triangles. Then comparing appropriate triples of points, one can
distinguish proper from improper intersections, and all degenerate
cases. Once they are distinguished, finding the point of intersection
is easy.
An alternate, image processing approach, would be to render the lines, define one unique color for the lines, and then apply an seed/flood fill algorithm to the first white zone found, applying a new unique color to future zones, until you flood fill an enclosed area that doesn't touch the border of the image.
Good luck!
References
finding the intersection of two line segments in 2d (with potential degeneracies), Accessed 2014-08-18, <https://math.stackexchange.com/questions/276735/finding-the-intersection-of-two-line-segments-in-2d-with-potential-degeneracies>
In a pair of segments, call one "the base segment" and one that is obtained by rotating the base segment by π/2 counterclockwise is "the other segment".
For each triple, compute the angle between the base segment and the X axis. Call this its principal angle.
Sort triples by the principal angle.
Now for each triple with the principal angle of α any potential square-forming mate has the principal angle of α+π (mod 2π). This is easy to find by binary search.
Furthermore, for two candidate triples with vertices a and a' and principal angles α and α+π, the angle of vector aa' should be α+π/4.
Finally, if each of the four segments is at least |aa'|/√2 long, we have a square.