As on the picture:
Could someone help me understand what exactly what delta means in the gradient descent algorithm?
This is a partial derivative with respect to theta_0.
The term is a derivative with respect to the theta 0.
Mark theta as coordinate on X-axis (let it be A)
Find corresponding coordinate on Y-axis (let it be B) so the point belongs to the function J
Draw tangent line to that function at the point (A, B)
The derivative is the slope of this tangent line.
The derivative is used to control two aspects of the cost function (J function) minimization:
direction - sign of the slope tells you in which direction you should move along the X-axis in order to converge J
rate - magnitude of the slope tells you how fast you should move
Related
I am trying to bend a mesh along a spline curve and currently out of ideas … at first I thought I just add spline point vectors to mesh's vertices , but I am looking for more optimized version of it …
so guys …
How can I bend a mesh along a spline, so that mesh, with some forward axis vector follows the spline and bends according to it and also repeat along the spline …
???
I believe there are many ways to do what you want. Some years ago I worked out an approach using Conformal Geometric Algebra. Of course you can do it using conventional 3D math as its described in Instant Mesh Deformation and Deformation styles for spline-based skeletal animation papers.
A simple method is as follows:
Your spline function is a function S(t): R -> R^3, it takes a scalar between [0:1] and give you points in R^3.
Project each mesh vertex on the spline curve. The projection is orthogonal in the sense that it follows the direction of a normal vector to the curve. So your mesh vertex v_i is projected to a point v'_i in the spline where S(t_i) = v'_i. Form a vector p_i = v_i - v'_i (which is normal to the curve) so each mesh vertex can be expressed as:
v_i = S(t_i) + p_i
Compute an orthogonal coordinate system at "each point" of the spline. That coordinate system is known as Frenet-Serret frame. The first vector to determine is the tangent to the curve. It is uniquely defined as the derivative of S(t) so tangent T = S(t)/dt. The other two vectors, the normal N and binormal B, can be computed in different ways, check the above reference papers for that.
Express the vector p_i (from step 1) in terms of the Frenet-Serret frame at point S(t_i). Such that the vector p_i is a linear combination of T, N and B. Create a matrix A with columns T, N and B. You need to find x_i such that:
A x_i = p_i
That can be solved by inverting the matrix A (actually taking the transpose should suffice). So each mesh vertex can be computed as:
v_i = S(t_i) + A x_i
You can store the pair (t_i, x_i) instead of v_i (you don't need to store v_i anymore since you can compute it from t_i and x_i).
To deform the mesh deformation the spline control points must be translated, then you need to recompute the Frenet-Serret frame of each spline point (taking the derivative of the S(t) to compute T and updating the N and B as suggested in above reference papers). Once you have the updated T, N and B, you can define the matrix A and then compute the mesh vertex positions using formula from step 3.
Results can be seen in pictures of the above mentioned papers.
This question and this question both show how to split a cubic Bézier curve at a particular parameterized value 0 ≤ t ≤ 1 along the curve, composing the original curve shape from two new segments. I need to split my Bézier curve at a point along the curve whose coordinate I know, but not the parameterized value t for the point.
For example, consider Adobe Illustrator, where the user can click on a curve to add a point into the path, without affecting the shape of the path.
Assuming I find the point on the curve closest to where the user clicks, how do I calculate the control points from this? Is there a formula to split a Bézier curve given a point on the curve?
Alternatively (and less desirably), given a point on the curve, is there a way to determine the parameterized value t corresponding to that point (other than using De Casteljau's algorithm in a binary search)?
My Bézier curve happens to only be in 2D, but a great answer would include the vector math needed to apply in arbitrary dimensions.
It is possible, and perhaps simpler, to determine the parametric value of a point on the curve without using De Casteljau's algorithm, but you will have to use heuristics to find a good starting value and similarly approximate the result.
One possible, and fairly simple way is to use Newton's method such that:
tn+1 = tn - ( bx(tn) - cx ) / bx'(tn)
Where bx(t) refers to the x component of some Bezier curve in polynomial form with the control points x0, x1, x2 and x3, bx'(t) is the first derivative and cx is a point on the curve such that:
cx = bx(t) | 0 < t < 1
the coefficients of bx(t) are:
A = -x0 + 3x1 - 3x2 + x3
B = 3x0 - 6x1 + 3x2
C = -3x0 + 3x1
D = x0
and:
bx(t) = At3 + Bt2 + Ct + D,
bx'(t) = 3At2 + 2Bt + C
Now finding a good starting value to plug into Newton's method is the tricky part. For most curves which do not contain loops or cusps, you can simply use the formula:
tn = ( cx - x0 ) / ( x3 - x0 ) | x0 < x1 < x2 < x3
Now you already have:
bx(tn) ≈ cx
So applying one or more iterations of Newton's method will give a better approximation of t for cx.
Note that the Newton Raphson algorithm has quadratic convergence. In most cases a good starting value will result in negligible improvement after two iterations, i.e. less than half a pixel.
Finally it's worth noting that cubic Bezier curves have exact solutions for finding extrema via finding roots of the first derivative. So curves which are problematic can simply be subdivided at their extrema to remove loops or cusps, then better results can be obtained by analyzing the resulting section in question. Subdividing cubics in this way will satisfy the above constraint.
Hermite Spline tangent estimation
I'm trying to come up with an algorithm or method that will allow me to estimate the tangent's magnitude (the direction is given) such as the interpolated spline will be the best fit for a given curve (blue dots)
In general points are in 3D space, but even a 2D solution will put me on track.
this is the Hermit Spline form:
Vector3D p0, p1, m0 (tangent at p0), m1 (tangent at p1)
p = (2*t^3 - 3*t^2 + 1)*p0 + (t^3-2*t^2+t)*m0 + (3*t^2 - 2*t^3)*p1 + (t^3-t^2)*m1;
My intuition suggest me some sort of least squares method using m0 and m1 magnitudes as the unknowns (could I apply least squares to each coordinates ecuation?) but somehow it should involve projecting and transforming that spline equation over the vector p1-p0 and instead of returning a 3D point, just the distance of that point to the base p1-p0. But I guess I got lost there :(
Perhaps the solution is easy and I would really love some light into this darkness.
Thanks in advance for your time!
I am reading an article about about cubic Hermite interpolation. In the cardinal spline curve section they give a formula to calculate tangents at end points given by:
Ti = a * ( Pi+1 - Pi-1 )
However, if I have two points P1 and P2 then to find T1
T1 = a*(P2-P0).
I need to calculate this but what should be my P0 point be? Similarly to find T2 I will need to know P3. Can anyone clarify this?
You're right, this formula only makes sense for inner points in your spline which have neighbours on both sides. For endpoints you have to get the tangent from other constraints. The common solutions are:
supply manually chosen tangent points
choose the tangent such that the curvature at the endpoint is zero, this is referred to as natural boundary condition
choose periodic boundary conditions, that is, the tangents of the start- and endpoint are equal. Then you only have to specify one of the tangents. For a closed spline, you can get the last tangent from the natural boundary condition.
These ideas are brought up in the context of cubic splines, which require to solve a system of linear equations in order to get the polynomial coefficients for any part of the spline, because they minimize the total curvature of the whole spline, but in your case of Hermite splines they should be applicable, too.
If I got a polynomial curve, and I want to find all monotonic curve segments and corresponding intervals by programming.
What's the best way to do this...
I want to avoid solving equation like f'(x) = 0;
Using some nice numerical ways to do this,like bi-section, is preferred.
f'(x) expression is available.
Thanks.
Add additional details. For example, I get a curve in 2d space, and its polynomial is
x: f(t)
y: g(t)
t is [0,1]
So, if I want to get its monotonic curve segment, I must know the position of t where its tangent vector is (1,0).
One direct way to resolve this is to setup an equation "f'(x) = 0".
But I want to use the most efficient way to do this.
For example, I try to use recursive ways to find this.
Divide the range [0,1] to four parts, and check whether the four tangents projection on vector (1,0) are in same direction, and two points are close enough. If not, continue to divide the range into 4 parts, until they are in same direction in (1,0) and (0,1), and close enough.
I think you will have to find the roots of f'(x) using a numerical method (feel free to implement any root-seeking algorithm you want, Wikipedia has a list). The roots will be those points where the gradient reaches zero; say x1, x2, x3.
You then have a set of intervals (-inf, x1) (x1, x2) etc, continuity of a polynomial ensures that the gradient will be always positive or always negative between a particular pair of points.
So evaluating the gradient sign at a point within each interval will tell you whether that interval is monotically increasing or not. If you don't care for a "strictly" increasing section, you could patch together adjacent intervals which have positive gradient (as a point of inflection will show up as one of the f'(x)=0 roots).
As an alternative to computing the roots of f', you can also use Sturm Sequences.
They allow counting the number of roots (here, the roots of f') in an interval.
The monotic curve segments are delimited by the roots of f'(x). You can find the roots by using an iterative algorithm like Newton's method.