Related
I am trying to quantize surface normals into let's say 8 bins.
For example, when computing features like HOG to quantize 2D gradients [x,y] into 8 bins we just take the angle with the y plane i.e. arctan(y/x) which will give us an angle between 0-360.
My question is, given a 3D direction [x,y,z], a surface normal in this case, how can we histogram it in a similar way? Do we just project onto one plane and use that angle i.e. the dot product of [x,y,z] and [0,1,0] for example?
Thanks
EDIT
I also read a paper recently where they quantized surface normals by measuring angles between normal and precomputed vectors that which are arranged around a right circular cone shape. I have added a link to this paper in the question (section 3.3.2 last paragraph), is this an effective approach? And if so, how do we compute these vectors?
Quantizing a continuous topological space corresponds to partitioning it and assigning labels to each partition. The straightforward standard approach for this scenario (quantizing normals) is as follows.
Choose your favorite uniform polyhedron:
http://en.wikipedia.org/wiki/Tetrahedron (4 faces)
http://en.wikipedia.org/wiki/Cube (6 faces)
http://en.wikipedia.org/wiki/Octahedron (8 faces)
http://en.wikipedia.org/wiki/Dodecahedron (12 faces)
http://en.wikipedia.org/wiki/Icosahedron (20 faces)
In general: http://en.wikipedia.org/wiki/Schl%C3%A4fli_symbol
Develop a mapping function from a normal on the unit sphere to the face of your chosen polyhedron that the normal intersects.
I would advise doing an argmax across polyhedron faces, taking the dot product of your normal and each polyhedron face normal. The one that gives the highest dot product is the face your normal should be binned into.
Use the face normal for each polyhedron face as the label for that face.
Prefer this approach to the approach suggested by others of mapping to spherical coordinates and then binning those. That approach suffers from too much sensitivity near the poles of the sphere.
Edit
In the paper you added to your question, the same idea is being used. There, however, the normals are restricted to a hemisphere - the only surfaces directly visible in an image have surface normals no more than 90 degrees away from the vector from the surface to the viewpoint.
The paper wants to quantize these surface normals into 8 values, represented by 8-bit integers with exactly one bit set to 1 and the rest set to 0. The 8 precomputed normals are computed as:
ntx = cos(a)*cos(t)
nty = cos(a)*sin(t)
ntz = sin(a)
where a = pi/4 and t = 0, pi/4, 2*pi/4, 3*pi/4, ..., 7*pi/4.
Notice
[cos(a)*cos(t)]2 + [cos(a)*sin(t)]2 + [sin(a)]2 = cos2(a)[cos2(t) + sin2(t)] + sin2(a) = cos2(a) + sin2(a) = 1
given a 3D direction [x,y,z], a surface normal in this case, how can
we histogram it in a similar way?
In the first case you quantize the polar orientation theta of the gradients. Now you need to quantize the spherical orientations theta and phi in a 2D histogram.
Do we just project onto one plane and use that angle
The binning of the sphere determines how you summarize the information to build a compact yet descriptive histogram.
Projecting the normal is not a good idea, if theta is more important than phi, just use more bins for theta
EDIT
Timothy Shields points in his comment and his answer that a regular binning of theta and phi won't produce a regular binning over the sphere as the bins will be bunched toward the poles.
His answer gives a solution. Alternatively, the non-regular binning described here can be hacked as follows:
Phi is quantized regularly in [0,pi]. For theta rather than quantizing the range [0,pi], the range [-1,1] is quantized instead;
For each quantized value u in [-1,1], theta is computed as
theta = arcsin(sqrt(1 - u * u)) * sign(u)
sign(u) returns -1 if u is negative, 1 otherwise.
The computed theta along with phi produce a regular quantization over the sphere.
To have an idea of the equation given above look at this article. It describes the situation in the context of random sampling though.
EDIT
In the above hack Timothy Shields points out that only the area of the bins is considered. The valence of the vertices (point of intersection of neighboring bins) won't be regular because of the poles singularity.
A hack for the previous hack would be to remesh the bins into a regular quadrilateral mesh and keep the regular area.
A heuristic to optimize this problem with the global constraints of having the same valence and the area can be inspired from Integer-Grid Maps Quad Meshing.
With the two hacks, this answer is too hacky and a little out of context as opposed to Timothy Shields answer.
A 3-dimensional normal cannot be quantized into a 1-D array as easily as for a 2-D normal (e.g., using arctan). I would recommend histogramming it into a 2-d space with a polar angle and an azimuth angle. For example, use spherical coordinates where the r (radius) value is always 1.0 (since your surface normal is normalized, length 1.0). In this case, you can throw away the r-value and just use polar angle θ (theta), and azimuthal angle φ (phi) to quantize the 3D normal.
I have created a 3D plot (a surface) using wireframe function. I wonder if there is any functions by which I can calculate the volume under the surface in a 3D plot?
Here is a sample of my data plus the wrieframe syntax I used to create my 3D (surface) plot:
x1<-c(13,27,41,55,69,83,97,111,125,139)
x2<-c(27,55,83,111,139,166,194,222,250,278)
x3<-c(41,83,125,166,208,250,292,333,375,417)
x4<-c(55,111,166,222,278,333,389,445,500,556)
x5<-c(69,139,208,278,347,417,487,556,626,695)
x6<-c(83,166,250,333,417,500,584,667,751,834)
x7<-c(97,194,292,389,487,584,681,779,876,974)
x8<-c(111,222,333,445,556,667,779,890,1001,1113)
x9<-c(125,250,375,500,626,751,876,1001,1127,1252)
x10<-c(139,278,417,556,695,834,974,1113,1252,1391)
df<-data.frame(x1,x2,x3,x4,x5,x6,x7,x8,x9,x10)
df.matrix<-as.matrix(df)
wireframe(df.matrix,
aspect = c(61/87, 0.4),scales=list(arrows=FALSE,cex=.5,tick.number="10",z=list(arrows=T)),ylim=c(1:10),xlab=expression(phi1),ylab="Percentile",zlab=" Loss",main="Random Classifier",
light.source = c(10,10,10),drape=T,col.regions = rainbow(100, s = 1, v = 1, start = 0, end = max(1,100 - 1)/100, alpha = 1),screen=list(z=-60,x=-60))
Note: my real data is a 100X100 matrix
Thanks
The data you are feeding to wireframe is a grid of values. Hence one estimate of the volume of whatever underlying surface this is approximating is the sum of the grid values multiplied by the grid cell areas. This is just like adding up the heights of histogram bars to get the number of values in your histogram.
The problem I see with you doing this on your data is that the cell areas are going to be in odd units - percentiles on one axis, phi on the other has unknown units, so your volume is going to have units of loss times units of percentile times units of phi.
This isn't a problem if you want to compare volumes of similar things on exactly the same grid, but if you have surfaces on different grids (different values of phi, or different percentiles) then you need to be careful.
Now, noting that wireframe doesn't draw like a 3d histogram would (looking like square tower blocks) this gives us another way to estimate the volume. Your 10x10 matrix is plotted as 9x9 squares. Divide each of those squares into triangles and then compute the volume of the 192 right truncated triangular prisms (I think this is what they are - they are equilateral triangular prisms with a right angle and one sloping end). The formula for that should be out there somewhere. Probably base area times height to the centroid of the triangle or something.
I thought maybe this would be in the raster package, but it isn't. There's code for computing the surface area but not the volume! I'm sure the raster maintainer would be happy to have some code for this!
If the points are arbitrary (ie, don't follow smooth function), it seems like you're looking for the volume of the convex hull (minimum surface) surrounding these points. One package to help you calculate this is alphashape3d.
You'll need a 3-column matrix of the coordinates to form the right type of object to make the calculation but it seems rather straight-forward.
Imagine you have a grid of sample points of a function z = f(x, y) where 1 < x < N and 1 < y < N. The formula is not given, but just the raw data, that could be for example the grey level of an image.
I would like to find, given a point A, whose x and y coordinates are given (and z is known from the data, so A is a vertex of the surface) a number M of points that lie on the circumference of the circle with center in A and radius R that are a good approximation of a circular "cloth" draped on the imaginary surface described by the data points. Imagine also that the edges of the surface are a triangle mesh.
The biggest constraint in the approximation is that the sum of the length of the edges of the resulting polygon is constantly R * 2 * PI, so that moving the A point across the surface would just change the M points but never the sum of their reciprocal distances. The draping doesn't need to be perfect, it would be nice though to be as close as possible to the surface., or always on one side of the surface, above or below.
Could anybody give me a pointer to something to read about this? Is this a known problem?
I feel that the problem is not completely formulated, I'd already like some help to give a complete description of it.
I have a set of points (with unknow coordinates) and the distance matrix. I need to find the coordinates of these points in order to plot them and show the solution of my algorithm.
I can set one of these points in the coordinate (0,0) to simpify, and find the others. Can anyone tell me if it's possible to find the coordinates of the other points, and if yes, how?
Thanks in advance!
EDIT
Forgot to say that I need the coordinates on x-y only
The answers based on angles are cumbersome to implement and can't be easily generalized to data in higher dimensions. A better approach is that mentioned in my and WimC's answers here: given the distance matrix D(i, j), define
M(i, j) = 0.5*(D(1, j)^2 + D(i, 1)^2 - D(i, j)^2)
which should be a positive semi-definite matrix with rank equal to the minimal Euclidean dimension k in which the points can be embedded. The coordinates of the points can then be obtained from the k eigenvectors v(i) of M corresponding to non-zero eigenvalues q(i): place the vectors sqrt(q(i))*v(i) as columns in an n x k matrix X; then each row of X is a point. In other words, sqrt(q(i))*v(i) gives the ith component of all of the points.
The eigenvalues and eigenvectors of a matrix can be obtained easily in most programming languages (e.g., using GSL in C/C++, using the built-in function eig in Matlab, using Numpy in Python, etc.)
Note that this particular method always places the first point at the origin, but any rotation, reflection, or translation of the points will also satisfy the original distance matrix.
Step 1, arbitrarily assign one point P1 as (0,0).
Step 2, arbitrarily assign one point P2 along the positive x axis. (0, Dp1p2)
Step 3, find a point P3 such that
Dp1p2 ~= Dp1p3+Dp2p3
Dp1p3 ~= Dp1p2+Dp2p3
Dp2p3 ~= Dp1p3+Dp1p2
and set that point in the "positive" y domain (if it meets any of these criteria, the point should be placed on the P1P2 axis).
Use the cosine law to determine the distance:
cos (A) = (Dp1p2^2 + Dp1p3^2 - Dp2p3^2)/(2*Dp1p2* Dp1p3)
P3 = (Dp1p3 * cos (A), Dp1p3 * sin(A))
You have now successfully built an orthonormal space and placed three points in that space.
Step 4: To determine all the other points, repeat step 3, to give you a tentative y coordinate.
(Xn, Yn).
Compare the distance {(Xn, Yn), (X3, Y3)} to Dp3pn in your matrix. If it is identical, you have successfully identified the coordinate for point n. Otherwise, the point n is at (Xn, -Yn).
Note there is an alternative to step 4, but it is too much math for a Saturday afternoon
If for points p, q, and r you have pq, qr, and rp in your matrix, you have a triangle.
Wherever you have a triangle in your matrix you can compute one of two solutions for that triangle (independent of a euclidean transform of the triangle on the plane). That is, for each triangle you compute, it's mirror image is also a triangle that satisfies the distance constraints on p, q, and r. The fact that there are two solutions even for a triangle leads to the chirality problem: You have to choose the chirality (orientation) of each triangle, and not all choices may lead to a feasible solution to the problem.
Nevertheless, I have some suggestions. If the number entries is small, consider using simulated annealing. You could incorporate chirality into the annealing step. This will be slow for large systems, and it may not converge to a perfect solution, but for some problems it's the best you and do.
The second suggestion will not give you a perfect solution, but it will distribute the error: the method of least squares. In your case the objective function will be the error between the distances in your matrix, and actual distances between your points.
This is a math problem. To derive coordinate matrix X only given by its distance matrix.
However there is an efficient solution to this -- Multidimensional Scaling, that do some linear algebra. Simply put, it requires a pairwise Euclidean distance matrix D, and the output is the estimated coordinate Y (perhaps rotated), which is a proximation to X. For programming reason, just use SciKit.manifold.MDS in Python.
The "eigenvector" method given by the favourite replies above is very general and automatically outputs a set of coordinates as the OP requested, however I noticed that that algorithm does not even ask for a desired orientation (rotation angle) for the frame of the output points, the algorithm chooses that orientation all by itself!
People who use it might want to know at what angle the frame will be tipped before hand so I found an equation which gives the answer for the case of up to three input points, however I have not had time to generalize it to n-points and hope someone will do that and add it to this discussion. Here are the three angles the output sides will form with the x-axis as a function of the input side lengths:
angle side a = arcsin(sqrt(((c+b+a)*(c+b-a)*(c-b+a)*(-c+b+a)*(c^2-b^2)^2)/(a^4*((c^2+b^2-a^2)^2+(c^2-b^2)^2))))*180/Pi/2
angle side b = arcsin(sqrt(((c+b+a)*(c+b-a)*(c-b+a)*(-c+b+a)*(c^2+b^2-a^2)^2)/(4*b^4*((c^2+b^2-a^2)^2+(c^2-b^2)^2))))*180/Pi/2
angle side c = arcsin(sqrt(((c+b+a)*(c+b-a)*(c-b+a)*(-c+b+a)*(c^2+b^2-a^2)^2)/(4*c^4*((c^2+b^2-a^2)^2+(c^2-b^2)^2))))*180/Pi/2
Those equations also lead directly to a solution to the OP's problem of finding the coordinates for each point because: the side lengths are already given from the OP as the input, and my equations give the slope of each side versus the x-axis of the solution, thus revealing the vector for each side of the polygon answer, and summing those sides through vector addition up to a desired vertex will produce the coordinate of that vertex. So if anyone can extend my angle equations to handling beyond three input lengths (but I note: that might be impossible?), it might be a very fast way to the general solution of the OP's question, since slow parts of the algorithms that people gave above like "least square fitting" or "matrix equation solving" might be avoidable.
Problem: Suppose you have a collection of points in the 2D plane. I want to know if this set of points sits on a regular grid (if they are a subset of a 2D lattice). I would like some ideas on how to do this.
For now, let's say I'm only interested in whether these points form an axis-aligned rectangular grid (that the underlying lattice is rectangular, aligned with the x and y axes), and that it is a complete rectangle (the subset of the lattice has a rectangular boundary with no holes). Any solutions must be quite efficient (better than O(N^2)), since N can be hundreds of thousands or millions.
Context: I wrote a 2D vector field plot generator which works for an arbitrarily sampled vector field. In the case that the sampling is on a regular grid, there are simpler/more efficient interpolation schemes for generating the plot, and I would like to know when I can use this special case. The special case is sufficiently better that it merits doing. The program is written in C.
This might be dumb but if your points were to lie on a regular grid, then wouldn't peaks in the Fourier transform of the coordinates all be exact multiples of the grid resolution? You could do a separate Fourier transform the X and Y coordinates. If theres no holes on grid then the FT would be a delta function I think. FFT is O(nlog(n)).
p.s. I would have left this as a comment but my rep is too low..
Not quite sure if this is what you are after but for a collection of 2d points on a plane you can always fit them on a rectangular grid (down to the precision of your points anyway), the problem may be the grid they fit to may be too sparsly populated by the points to provide any benefit to your algorithm.
to find a rectangular grid that fits a set of points you essentially need to find the GCD of all the x coordinates and the GCD of all the y coordinates with the origin at xmin,ymin this should be O( n (log n)^2) I think.
How you decide if this grid is then too sparse is not clear however
If the points all come only from intersections on the grid then the hough transform of your set of points might help you. If you find that two mutually perpendicular sets of lines occur most often (meaning you find peaks at four values of theta all 90 degrees apart) and you find repeating peaks in gamma space then you have a grid. Otherwise not.
Here's a solution that works in O(ND log N), where N is the number of points and D is the number of dimensions (2 in your case).
Allocate D arrays with space for N numbers: X, Y, Z, etc. (Time: O(ND))
Iterate through your point list and add the x-coordinate to list X, the y-coordinate to list Y, etc. (Time: O(ND))
Sort each of the new lists. (Time: O(ND log N))
Count the number of unique values in each list and make sure the difference between successive unique values is the same across the whole list. (Time: O(ND))
If
the unique values in each dimension are equally spaced, and
if the product of the number of unique values of each coordinate is equal to the number of original points (length(uniq(X))*length(uniq(Y))* ... == N,
then the points are in a regular rectangular grid.
Let's say a grid is defined by an orientation Or (within 0 and 90 deg) and a resolution Res. You could compute a cost function that evaluate if a grid (Or, Res) sticks to your points. For example, you could compute the average distance of each point to its closest point of the grid.
Your problem is then to find the (Or, Res) pair that minimize the cost function. In order to narrow the search space and improve the , some a heuristic to test "good" candidate grids could be used.
This approach is the same as the one used in the Hough transform proposed by jilles. The (Or, Res) space is comparable to the Hough's gamma space.