I want to solve the following support vector machine problem
The soft margin support vector machine solves the following optimization problem:
What does the second term minimize?
What does the first term minimize?
What does xi measure?
The first one maximizes the margin ( Actally margin formula is 2/W , so here minimizing W/2 is equal to maximizing the margin)
The second term minimzes the error from actual class. (In case of where data is not linearly separable, it will try to maximize margin by allowing the border to move with making error on some points), so the second term will decide how much to move border by allowing error.
I'm using the object detection api and tuning the parameters for a SSD task. My question refers to the box coder at https://github.com/tensorflow/models/blob/master/research/object_detection/box_coders/faster_rcnn_box_coder.py.
Why setting these scales factors to [10,10,5,5]? The original paper does not explain it. I suspect that it has to do either assigning a different weight to the 4 components of location error (tx, ty, tw, th) or with some numerical stability issue, but I would like to have a confirmation. Thanks
I find the answer here https://leimao.github.io/blog/Bounding-Box-Encoding-Decoding/, where the variables are used as some sort of Representation Encoding With Variance. The question was also the subject of this issue https://github.com/rykov8/ssd_keras/issues/53
The network predicts changes for each anchor box. That is, for each anchor block, it predicts an offset for x, y position and width, height.
a short description of these parameters can be found, for example to link:
https://medium.com/#smallfishbigsea/understand-ssd-and-implement-your-own-caa3232cd6ad
https://lambdalabs.com/blog/how-to-implement-ssd-object-detection-in-tensorflow/
My problem is explained in the following image
http://i.stack.imgur.com/n6mZt.png
I have a finite (but rather large) amount of such pieces that need to be stacked in a way so that the REMAINING area is the smallest possible. The pieces are locked in the horizontal axis (time) and have fixed height. They can only be stacked.
The remaining area is defined by the maximum point of the stack that depends on which pieces have been selected. The best combination in the example image would be the [1 1 0]. (The trivial [0 0 0] case will not be allowed by other constraints)
My only variables are binaries (Yes or No) for each piece. The objective is a little more complicated than what I am describing, but my greatest problem right now is how to formulate the expression
Max{Stacked_Pieces} - Stacked_Pieces_Profile
in the objective function. The result of this expression is a vector of course (timeseries) but it will be further reduced to a number through other manipulations.
Essentially my problem is how to write
Max{A} - A, where A = 1xN vector
In a way compatible with a linear (or even quadratic) objective. Or am I dealing with a non-linear problem?
EDIT: The problem is like a Knapsack problem the main difference being that there is no knapsack to fill up. i.e. the size of the knapsack varies according to the selected pieces and is always equal to the top of the stacked profile
Thanks everybody!
From what I understand you can basically try to solve it as a normal knapsack problem in multiple iterations, finding the minimal.
Now, finding the height of the knapsack is a problem, which means you need multiple iterations. Because you need to solve the knapsack problem to see if a certain height will work, you need multiple iterations.
Note that you do know an upper and a lower bound for the height. I'm not sure if rotation is applicable, but you can fill in the gaps here:
Min = max(max height of smallest piece, total size / width)
Max = sum(height of all pieces).
Basically solving it means finding the smallest height [Min <= x <= Max] that fits all pieces. The easiest way to do that is by using a 'for' loop, but you can do it better:
Try min, max, half
if half fits -> max = half; iterate (goto 1)
if half doesn't fit -> min = half; iterate (goto 1)
As for solving the knapsack problem, for each iteration, I'd check if all pieces can still be fitted. Use bit-masks and AND/OR/XOR operations if you can to speed things up.
Basically you can do it like this:
Grab bit 'x'. Fill with next block
Check if this leads to a possible solution
Find next bit that can be filled
Note that you might want to use intrinsics in C++ to speed this up. Modern CPU's are quite good with this.
As for code: I've made some code that solves the bedlam cube in the past; I'm pretty sure that if you google for that, you'll find some fast solvers.
Good luck!
I want to detect the best rototraslation matrix between two set of points.
The second set of points is the same of the first, but rotated, traslated and affecteb by noise.
I tried to use least squared method by obviously the solution is usually similar to a rotation matrix, but with incompatible structure (for example, where i should get a value that represents the cosine of an angle i could get a value >1).
I've searched for the Constrained Least Squared method but it seems to me that the constrains of a rototraslation matrix cannot be expressed in this form.
In this PDF i've stated the problem more formally:
http://dl.dropbox.com/u/3185608/minquad_en.pdf
Thank you for the help.
The short answer: What you will need here is "Principal Component Analysis".
Apply this to both sets of points centered at their respective centers of mass. The PCA will effectively give you a rotation matrix for each aligned to the data set principal components. Multiplying the inverse matrix of the original set by the new rotation will give you a matrix that takes the old (centered) set to the new. Inverse translations and translations can similarly be applied to the rotation to create a homogeneous matrix that maps the one set to the other.
The book PRINCE, Simon JD. Computer vision: models, learning, and inference. Cambridge University Press, 2012.
gives, in Appendix "B.4 Reparameterization", some info about how to constrain a matrix to be a rotation matrix.
It seems to me that your problem has also a solution based on SVD: see the Kabsch algorithm also described by Olga Sorkine-Hornung and Michael Rabinovich in
Least-Squares Rigid Motion Using SVD and, more practically, by Nghia Kien Ho in FINDING OPTIMAL ROTATION AND TRANSLATION BETWEEN CORRESPONDING 3D POINTS.
My question is fairly simple. I have two tetrahedra, each with a current position, a linear speed in space, an angular velocity and a center of mass (center of rotation, actually).
Having this data, I am trying to find a (fast) algorithm which would precisely determine (1) whether they would collide at some point in time, and if it is the case, (2) after how much time they collided and (3) the point of collision.
Most people would solve this by doing triangle-triangle collision detection, but this would waste a few CPU cycles on redundant operations such as checking the same edge of one tetrahedron against the same edge of the other tetrahedron upon checking up different triangles. This only means I'll optimize things a bit. Nothing to worry about.
The problem is that I am not aware of any public CCD (continuous collision detection) triangle-triangle algorithm which takes self-rotation in account.
Therefore, I need an algorithm which would be inputted the following data:
vertex data for three triangles
position and center of rotation/mass
linear velocity and angular velocity
And would output the following:
Whether there is a collision
After how much time the collision occurred
In which point in space the collision occurred
Thanks in advance for your help.
The commonly used discrete collision detection would check the triangles of each shape for collision, over successive discrete points in time. While straightforward to compute, it could miss a fast moving object hitting another one, due to the collision happening between discrete points in time tested.
Continuous collision detection would first compute the volumes traced by each triangle over an infinity of time. For a triangle moving at constant speed and without rotation, this volume could look like a triangular prism. CCD would then check for collision between the volumes, and finally trace back if and at what time the triangles actually shared the same space.
When angular velocity is introduced, the volume traced by each triangle no longer looks like a prism. It might look more like the shape of a screw, like a strand of DNA, or some other non-trivial shapes you might get by rotating a triangle around some arbitrary axis while dragging it linearly. Computing the shape of such volume is no easy feat.
One approach might first compute the sphere that contains an entire tetrahedron when it is rotating at the given angular velocity vector, if it was not moving linearly. You can compute a rotation circle for each vertex, and derive the sphere from that. Given a sphere, we can now approximate the extruded CCD volume as a cylinder with the radius of the sphere and progressing along the linear velocity vector. Finding collisions of such cylinders gets us a first approximation for an area to search for collisions in.
A second, complementary approach might attempt to approximate the actual volume traced by each triangle by breaking it down into small, almost-prismatic sub-volumes. It would take the triangle positions at two increments of time, and add surfaces generated by tracing the triangle vertices at those moments. It's an approximation because it connects a straight line rather than an actual curve. For the approximation to avoid gross errors, the duration between each successive moments needs to be short enough such that the triangle only completes a small fraction of a rotation. The duration can be derived from the angular velocity.
The second approach creates many more polygons! You can use the first approach to limit the search volume, and then use the second to get higher precision.
If you're solving this for a game engine, you might find the precision of above sufficient (I would still shudder at the computational cost). If, rather, you're writing a CAD program or working on your thesis, you might find it less than satisfying. In the latter case, you might want to refine the second approach, perhaps by a better geometric description of the volume occupied by a turning, moving triangle -- when limited to a small turn angle.
I have spent quite a lot of time wondering about geometry problems like this one, and it seems like accurate solutions, despite their simple statements, are way too complicated to be practical, even for analogous 2D cases.
But intuitively I see that such solutions do exist when you consider linear translation velocities and linear angular velocities. Don't think you'll find the answer on the web or in any book because what we're talking about here are special, yet complex, cases. An iterative solution is probably what you want anyway -- the rest of the world is satisfied with those, so why shouldn't you be?
If you were trying to collide non-rotating tetrahedra, I'd suggest a taking the Minkowski sum and performing a ray check, but that won't work with rotation.
The best I can come up with is to perform swept-sphere collision using their bounding spheres to give you a range of times to check using bisection or what-have-you.
Here's an outline of a closed-form mathematical approach. Each element of this will be easy to express individually, and the final combination of these would be a closed form expression if one could ever write it out:
1) The equation of motion for each point of the tetrahedra is fairly simple in it's own coordinate system. The motion of the center of mass (CM) will just move smoothly along a straight line and the corner points will rotate around an axis through the CM, assumed to be the z-axis here, so the equation for each corner point (parameterized by time, t) is p = vt + x + r(sin(wt+s)i + cos(wt + s)j ), where v is the vector velocity of the center of mass; r is the radius of the projection onto the x-y plane; i, j, and k are the x, y and z unit vectors; and x and s account for the starting position and phase of rotation at t=0.
2) Note that each object has it's own coordinate system to easily represent the motion, but to compare them you'll need to rotate each into a common coordinate system, which may as well be the coordinate system of the screen. (Note though that the different coordinate systems are fixed in space and not traveling with the tetrahedra.) So determine the rotation matrices and apply them to each trajectory (i.e. the points and CM of each of the tetrahedra).
3) Now you have an equation for each trajectory all within the same coordinate system and you need to find the times of the intersections. This can be found by testing whether any of the line segments from the points to the CM of a tetrahedron intersects the any of the triangles of another. This also has a closed-form expression, as can be found here.
Layering these steps will make for terribly ugly equations, but it wouldn't be hard to solve them computationally (although with the rotation of the tetrahedra you need to be sure not to get stuck in a local minimum). Another option might be to plug it into something like Mathematica to do the cranking for you. (Not all problems have easy answers.)
Sorry I'm not a math boff and have no idea what the correct terminology is. Hope my poor terms don't hide my meaning too much.
Pick some arbitrary timestep.
Compute the bounds of each shape in two dimensions perpendicular to the axis it is moving on for the timestep.
For a timestep:
If the shaft of those bounds for any two objects intersect, half timestep and start recurse in.
A kind of binary search of increasingly fine precision to discover the point at which a finite intersection occurs.
Your problem can be cast into a linear programming problem and solved exactly.
First, suppose (p0,p1,p2,p3) are the vertexes at time t0, and (q0,q1,q2,q3) are the vertexes at time t1 for the first tetrahedron, then in 4d space-time, they fill the following 4d closed volume
V = { (r,t) | (r,t) = a0 (p0,t0) + … + a3 (p3,t0) + b0 (q0,t1) + … + b3 (q3,t1) }
Here the a0...a3 and b0…b3 parameters are in the interval [0,1] and sum to 1:
a0+a1+a2+a3+b0+b1+b2+b3=1
The second tetrahedron is similarly a convex polygon (add a ‘ to everything above to define V’ the 4d volume for that moving tetrahedron.
Now the intersection of two convex polygon is a convex polygon. The first time this happens would satisfy the following linear programming problem:
If (p0,p1,p2,p3) moves to (q0,q1,q2,q3)
and (p0’,p1’,p2’,p3’) moves to (q0’,q1’,q2’,q3’)
then the first time of intersection happens at points/times (r,t):
Minimize t0*(a0+a1+a2+a3)+t1*(b0+b1+b2+b3) subject to
0 <= ak <=1, 0<=bk <=1, 0 <= ak’ <=1, 0<=bk’ <=1, k=0..4
a0*(p0,t0) + … + a3*(p3,t0) + b0*(q0,t1) + … + b3*(q3,t1)
= a0’*(p0’,t0) + … + a3’*(p3’,t0) + b0’*(q0’,t1) + … + b3’*(q3’,t1)
The last is actually 4 equations, one for each dimension of (r,t).
This is a total of 20 linear constraints of the 16 values ak,bk,ak', and bk'.
If there is a solution, then
(r,t)= a0*(p0,t0) + … + a3*(p3,t0) + b0*(q0,t1) + … + b3*(q3,t1)
Is a point of first intersection. Otherwise they do not intersect.
Thought about this in the past but lost interest... The best way to go about solving it would be to abstract out one object.
Make a coordinate system where the first tetrahedron is the center (barycentric coords or a skewed system with one point as the origin) and abstract out the rotation by making the other tetrahedron rotate around the center. This should give you parametric equations if you make the rotation times time.
Add the movement of the center of mass towards the first and its spin and you have a set of equations for movement relative to the first (distance).
Solve for t where the distance equals zero.
Obviously with this method the more effects you add (like wind resistance) the messier the equations get buts its still probably the simplest (almost every other collision technique uses this method of abstraction). The biggest problem is if you add any effects that have feedback with no analytical solution the whole equation becomes unsolvable.
Note: If you go the route of of a skewed system watch out for pitfalls with distance. You must be in the right octant! This method favors vectors and quaternions though, while the barycentric coords favors matrices. So pick whichever your system uses most effectively.