In the picture of SVM from Wikipedia, at the lower left corner - pointed by the red arrow, there's b / ||w||. How is that calculated? In other words, why is the line in the picture b / ||w||? Thanks.
The line represents the affine subspace of points x, whose scalar product with the weight vector w yields b. As w is typically not a unit vector (i.e. need not have length 1), one has to divide b by the norm (the "length") of w to get the actual distance from the origin.
More precisely: imagine a vector x starting at the origin and reaching out to a point on the red line and let u be the unit vector in the direction of w, i.e. u = w / ||w||. Then the scalar product of x and u multiplied with u is the projection of x onto the unit vector u and its length corresponds to the distance of the red line from the origin. If you calculate the scalar product of <x,w> (written as x*w in the graphics) instead, you still get a projection on u, which has length b (that's actually how b is defined), so to get back the distance from the origin one has to calculate b/||w||.
I'm studying a course on vision systems and one of the questions posed was;
For the accumulator shown;
Determine the most likely r,θ combination representing the straight line of the greatest strength in the original image.
From my understanding of the accumulator this would be r = 60, θ = 150 as the 41 votes is the highest number of votes in this cluster of large votes. Am I correct with this combination?
And hence calculate the equation of this line in the form y = mx + c
I'm not sure of the conversion steps required to convert the r = 60, θ = 150 to y = mx + c with the information given since r = 60, θ = 150 denotes 1 point on the line.
State the resolution of your answer and give your reasoning
I assume the resolution is got to do with some of the steps in the auscultation and not the actual resolution of the original image since that's irrelevant to the edges detected in the image.
Any guidance on the above 3 points would be greatly appreciated!
Yes, this is correct.
This is asking you what the slope and intercept are of the line given r and theta. r and theta are not one point on the line, they are one point of the accumulator. r and theta describe a line using the line equation in polar coordinates: . This is the cool thing about the hough transform, every line in one space, (i.e. image space) can be described by a point in another space (r, theta). This could be done with m and b from the line equation , but as we all know, m is undefined for vertical lines. This is the reason the polar line equation is used. It is important to note that the line described by the HT r and theta refers to a line from the origin extending to the actual line in the image. This means your image line y = mx + b equation will need to be orthogonal to the polar equation. The wiki article on the HT describes this well and shows examples. I would recommend drawing a diagram of your r and theta extending to a line like this:
Then use trig to get two points on the red line. Two points are enough to give you m and b from the line equation.
I'm not entirely sure what "resolution" refers to in this context. But it does seem like your line estimator will have some precision loss since r is every 20 mm and theta is every 15 degrees. Perhaps it is asking what degree of error you could get given an accumulator of this resolution.
I would like to construct a plane from a list of 3D points in OpenCV. I would like to obtain the result by finding the four parameters in the following form: Ax+By+Cz+D = 0. Would anyone suggest me a way to do it?
If the data does not contain outliers and does not contain more than one plane. Furthermore, all the points lay exactly on a plane (the data is not noisy), it is so simple:
Pick up three random points which are not lay on the same line.
solve this system of linear equations:
x1+by1+cz1+d = 0
x2+by2+cz2+d = 0
x3+by3+cz3+d = 0
then :
A= Choose any number you want in order to match your scale.
B= b*A
C= c*A
D= d*A
If the data is noisy or contains outliers or more than plane (or both) you need then some kind of Robust Estimation techniques. Search for RANSAC as a start.
if you are familar with RANSAC you can see this example here (it is about lines you can simply generlize it to deal with plane)
Answer by leveraging OpenCV
If you want to have equation solved by 3 points, just like follows:
ax + by + cz = 1
Example
you have three points: cv::Point3f p1, p2 and p3, and here is the code:
cv::Matx33f M(p1.x, p1.y, p1.z,
p2.x, p2.y, p2.z,
p3.x, p3.y, p3.z);
cv::Vec3f d(1, 1, 1);
cv::Vec3f coef = M.inv() * d;
Then, a, b, c are coef(0), coef(1), coef(2) sequentially.
I have 3D points and I need to make an 2D orthographic projection of them onto a plane that is defined by the origin and a normal n. The meaning of this is basically looking at the points from the top (given the vertical vector). How can I do it?
What I'm thinking is:
project point P onto the 3D plane: P - P dot n * n
look at the 3D plane from the "back" in respect to the normal (not sure how to define this)
do an ortho projection using max-min coordinates of the points in the plane to define the clipping
I am working with iOS.
One way to do this would be to:
rotate the coordinate system so that the plane of interest lies in the x-y plane, and the normal vector n is aligned with the z-axis
project the points onto the x-y plane by setting their z-components to 0
Set up the coordinate transformation
There are infinitely many solutions to this problem since we can always rotate a solution in the x-y plane to get another valid solution.
To fix this, let's choose a vector v lying in the plane that will line up with the x-axis after the transformation. Any vector will do; let's take the vector in the plane with coordinates x=1 and y=0.
Since our plane intersects the origin, its equation is:
x*n1 + y*n2 + z*n3 = 0
z = -(x*n1 + y*n2)/n3
After substituting x=1, y=0, we see that
v = [1 0 -n1/n3]
We also need to make sure v is normalized, so set
v = v/sqrt(v1*v1 + v2*v2 + v3*v3)
EDIT: The above method will fail in cases where n3=0. An alternative method to find v is to take a random point P1 from our point set that is not a scalar multiple of n and calculate v = P1 - P1 dot n * n, which is the projection of P1 into the plane. Just keep searching through your points until you find one that satisfies (P1 dot n/norm(n)) != P1 and this is guaranteed to work.
Now we need a vector u that will line up with the y-axis after the transformation. We get this from the cross product of n and v:
u = n cross v
If n and v are normalized, then u is automatically normalized.
Next, create the matrix
M = [ v1 v2 v3 ]
[ u1 u2 u3 ]
[ n1 n2 n3 ]
Transform the points
Now given a 3 by N array of points P, we just follow the two steps above
P_transformed = M*P
P_plane = set the third row of P_transformed to zero
The x-y coordinates of P_plane are now a 2D coordinate system in the plane.
If you need to get the 3D spatial coordinates back, just do the reverse transformation with P_space = M_transpose*P_plane.
I've been working on a project where I use Bezier paths to draw the curves I need. Each basic shape in my project consists of three cubic Bezier curves arranged end to end so that the slopes match where they meet.
The basic problem I need to solve is whether the compound curve made of the three Bezier curves intersects with itself. After thinking about it for a while, I've figured out that given the constraints of the curves, I can simplify the task to something else:
The curvature of each of the three Bezier paths should be opposite in curvature direction relative to the curve it's abutted to. In other words, there should be an inflection point where one bezier curve abuts to another one. If that is not the case, I want to reject the parameter set that generated the curves and select a different set.
In any case, my basic question is how to detect whether there is an inflection point where the curves abut each other.
In the illustration, each of the three Bezier curves is shown using a different color. The left black curve curves in the opposite direction from the red curve at the point where they meet, but the right black curve curves in the same direction. There is an inflection point where the red and left black curve meet but not where the red and right black curve meet.
Edit:
Below, I've added another image, showing the polygon enclosing the Bezier path. The crossing lines of the polygon, shown in the black curve, tests for an inflection point, not a loop. I'm guessing that one curve intersecting another can be tested by checking whether the enclosing polygons intersect, as illustrated by the red and blue curves.
P.S. Since there has been some question about the constraints, I will list some of them here:
The left most point and the rightmost point have the same y value.
The x value of the control point of the leftmost point is less than
the x value of the control point of the rightmost point. This keeps
the black and blue curves from intersecting each other.
The slope at the leftmost and rightmost points is within about +/- 10
degrees of horizontal.
The intersection of the black and red curves, and the intersection of
the red and blue curves divide the full curve in approximately
thirds. I don't have exact numbers, but a sample bound would be that
the x value of the left end of the red curve is somewhere between 25%
and 40% of the x value of the rightmost point.
The y value of the points of intersection are +/- some small fraction
of the overall width.
The slope at the intersections is > 0.6 and < 3.0 (positive or
negative).
The equation for curvature is moderately simple. You only need the sign of the curvature, so you can skip a little math. You are basically interested in the sign of the cross product of the first and second derivatives.
This simplification only works because the curves join smoothly. Without equal tangents a more complex test would be needed.
The sign of curvature of curve P:
ax = P[1].x - P[0].x; // a = P1 - P0
ay = P[1].y - P[0].y;
bx = P[2].x - P[1].x - ax; // b = P2 - P1 - a
by = P[2].y - P[1].y - ay;
cx = P[3].x - P[2].x - bx*2 - ax; // c = P3 - P2 - 2b - a
cy = P[3].y - P[2].y - by*2 - ay;
bc = bx*cy - cx*by;
ac = ax*cy - cx*ay;
ab = ax*by - bx*ay;
r = ab + ac*t + bc*t*t;
Note that r is the cross product of the first and second derivative and the sign indicate the direction of curvature. Calculate r at t=1 on the left curve and at t=0 on the right curve. If the product is negative, then there is an inflection point.
If you have curves U, V and W where U is the left black, V is the middle red, and W is the right black, then calculate bc, ac, and ab above for each. The following test will be true if both joins are inflection points:
(Uab+Uac+Ubc)*(Vab) < 0 && (Vab+Vac+Vbc)*(Wab) < 0
The equation for curvature has a denominator that I have ignored. It does not affect the sign of curvature and would only be zero if the curve was a line.
Math summary:
// start with the classic bezier curve equation
P = (1-t)^3*P0 + 3*(1-t)^2*t*P1 + 3*(1-t)*t^2*P2 + t^3*P3
// convert to polynomial
P = P0 + 3*t*(P1 - P0) + 3*t^2*(P2 - 2*P1 + P0) + t^3*(P3 - 3*P2 + 3*P1 - P0)
// rename the terms to a,b,c
P = P0 + 3at + 3btt + cttt
// find the first and second derivatives
P' = 3a + 6bt + 3ctt
P" = 6b + 6ct
// and the cross product after some reduction
P' x P" = ab + act + bctt
One of the deterministic ways to check if a bezier curve has a double-point or self-intersection is to calculate the inverse equation and evaluate the root, since the inversion equation of a bezier curve is always zero at the point of self-intersection.
As detailed in this example by T.W.Sederberg –course notes. Then for detecting whether two bezier curves (red and black in the question) intersect with each other, there are a few methods, binary subdivision (easier to implement) and bezier clipping (very good balance of efficiency and code complexity), implicitization (not worth it).
But a more efficient way to do it probably is to subdivide the curves into small line-segments and find the intersection. Here is one way to do it. Especially when you are interested in knowing whether the path self-intersect, but not in an accurate point of intersection.
If you have well defined assumptions about the locations of the CVs of the piece-wise bezier curves (or poly-bezier as #Pomax mentioned above), you can try out the curvature based method as you have mentioned in your question.
Here is a plot of the curvature on a similar path as in the question. Under similar circumstances, it seems this could give you the solution you are looking for. Also, in case of a cusp, it seems this quick and dirty method still works!
# Victor Engel : Thank you for the clarification. Now the solution is even easier.
In short, look at torques of both curves. If they pull together, the "inflexion" occurs, if they are opposed, the "curvature" continues.
Remark: Words "inflexion" and "curvature" have here a more intuitive nature, than a strict mathematical meaning, therefore apostrophes are used.
As before, a set of points P0,P1,P2,P3 defines the first cubic Bezier curve, and Q0,Q1,Q2 and Q3 the second.
Moreover, 4 vectors: a,b,u,w will be useful.
a = P2P3
b = P1P2
u = Q0Q1
v = Q1Q2
I will omit a C1-continuity checking, it's already done by you. ( This means P3=Q0 and a x u=0)
Finally, Tp and Tq torques will appear.
Tp = b x a
( "x" means a vector product, but Tp on the planar is treated like a simple number, not a vector. }
if Tp=0 {very rarely vectors a, b can be parallel.}
b = P0P1
Tp = b x a
if Tp=0
No! It's can't be! Who straightened the curve???
STOP
endif
endif
Tq = u x v
if Tq=0 {also vectors u, v can be parallel.}
v = Q2Q3
Tq = u x v
if Tq=0
Oh no! What happened to my curve???
STOP
endif
endif
Now the final test:
if Tp*Tq < 0 then Houston! We have AN "INFLEXION"!
else WE CONTINUE THE TURN!
Suppose you have 4 points: P0, P1, P2 and P3, which describes a cubic Bezier's curve (cBc for short) . P0 and P3 are starting and ending points, P1 and P2 are directional points.
When P0, P1 and P2 are collinear, then cBc has an inflection point at P0.
When P1, P2 and P3 are collinear, then cBC has an inflection point at P3.
Remarks.
1) Use the vector product ( P0P1 x P1P2 and P1P2 x P2P3 respectively) to check the collinearity. The result should be a zero-lenght vector.
2) Avoid the situation where P0, P1, P2 and P3 are all collinear, the cBc they create aren't curves in the common sense.
Corollary.
When P1=P2, cBc has inflection points at both sides.