I'm trying to draw about 4000-10000 segments using NSBezierPath on every drawRect of an NSView (about a 300x300 pixel box). This is very resource heavy and is taking a lot time to draw (relatively long).
Can someone suggest a substitute for this? I've tried using a single NSBezierPath for 1000 segments at a time, but it's still too resource heavy.
I'm looking for any possible alternatives. I'm sure OpenGL would be faster, but I don't know if I have to learn a new platform in order to do what I need. I'm open to suggestions.
Not an answer, just test results
I did a simple experiment with Mathematica. This experiment gives us an absolute upper bound for your time, since I used no optimization, no GPU, an interpreted language, etc. So I think much more than one order of magnitude is achievable.
Results:
Generating a 10.000 bezier curves list
b = Table[
{Hue[RandomReal[]],
BezierCurve#RandomReal[{0, 300}, {4, 2}]}, {10000}];
is very quick, because mathematica does not evaluate nothing.
Now rendering:
h1 = AbsoluteTime[]; Print#Graphics[b]; h2 = AbsoluteTime[]; Print[h2 - h1];
Time spent 11.8 secs
Result:
PS: The intention is to set a timing baseline for our mindset.
Related
Suppose that we have several squares of the same size. We want to draw n rectangles (red and yellow rectangles here) to contain these squares.
The goal is to have the least wasted space possible.
In the example below, n = 2 and the solution on the right is preferred because it results in only one wasted space.
Are there any known algorithms already in place to solve these kind of problems?
UPDATE:
The arrangement of the squares is arbitrary and they are always above the X axis!
UPDATE2:
To make the question easier, let's assume that the so called container rectangles are on top of each other! (Red and yellow rectangles here)
A little more complicated case:
let's assume two rectangles are used for this one too. As it can be seen, the 3rd solution results in the least wasted space.
This question is almost identical to a hiring puzzle that ITA Software posed, called "Strawberry Fields" (scroll down for Strawberry Fields; change the greenhouse cost from 10 to 0). I can confirm that integer programming, specifically branch and price where the high-level decisions are whether to put two squares in the same rectangle, works very, very well for this problem. Here's my custom solver, written in C. You'll need to change the greenhouse cost in strawberry_fields.h from 10 to 0.
This type of rectangle cover is hard (NP-hard actually, you can use it to solve the Rectangle Cover Problem), but you can solve this with integer linear programming, as follows:
minimize sum[i] take[i] * area[i]
st
sum[i] take[i] == n
for every filled cell x,y:
sum[rectangle i covers x,y] take[i] == 1
take[i] in { 0, 1 }
Where the lists of rectangles contains only "reasonable" rectangles that you might need. ie only rectangles that cannot be made smaller without uncovering some filled cell, and you can skip certain "interior rectangles" that you can tell can never be part of a solution because they would leave a shape that's harder to cover. Generating those rectangles is a fun exercise in its own right, but generating too many isn't a big problem, just slower. In the solution, any take[i] that is 1 corresponds to a rectangle that you take.
You can throw this into any available solver, such as GLPK (free) or Gurobi (commercial and academic licenses available).
This should generally be faster than brute force, because the linear relaxation (same model as above, but the last constraint converted to 0 <= take[i] <= 1) can be used to guide the search, and various plane cutting tricks can be applied.
More advanced tricks can be found in this paper, such as tricks that use the fractional solution from the linear relaxation.
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 am working on a program that needs to fit numerous cosine waves in order to determine one of the parameters for the function. The equation that I am using is y = y_0 + Acos((4*pi*L)/x + pi) where L is the value that I am trying to obtain from the best fit line.
I know that it is possible to do this correctly by hand for each set of data, but what is the best way to automate this process? I am currently reading in the data from text files, and running a loop with the initial paramiters changing until I have an array of paramater values that have an amplitude similar to the data, then I check the percent difference between points on the center peak and two end peaks to try to pick the best one. It in consistently picking lower values than what I get when fitting by hand (almost exactly one phase off). So is there a way to improve this method, or another method that works better?
Edit: My LabVIEW version has a cos fitting VI which is what I am using, the problem is when I try to automate the fitting by changing the initial parameters using a loop, I cant figure out how to get the program to pick the same best fit line as a human would pick.
Why not just use a Fast Fourier Transform? This should be way faster than fitting a cosine. In the result vector of complex numbers look for the largest peak of in the totals. You're given frequency (position in the FFT result vector), amplitude and phase.
You can evaluate the goodness of the fit by computing the difference between fitting curve and your data. A VI does this in the "Advanced curve fitting" palette. Then all you have to do is pick up the best fit.
As part of my work, I often have to visualize complex 3 dimensional densities. One program suite that I work with outputs the radial component of the densities as a set of 781 points on a logarithmic grid, ri = (Rmax/Rstep)^((i-1)/(pts-1), times a spherical harmonic. For low symmetry systems, the number of spherical harmonics can be fairly large to ensure accuracy, e.g. one system requires 49 harmonics corresponding to lmax = 6. So, to use this data within Mathematica, I would have a sum of up to 49 interpolated functions with each multiplied by a different spherical harmonic. While using v.6 and constructing the interpolated radial functions using Interpolation and setting r = Sqrt(x^2 + y^2 + z^2), I would stop ContourPlot3D after well over an hour without anything displayed. This included reducing both the InterpolationOrder and MaxRecursion to 1.
Several alternatives presented themselves:
Evaluate the density function on a fixed grid, and use ListContourPlot instead.
Or, linearly spline the radial function and use Piecewise to stitch them together. (This presented itself, as I could use simplify to help reduce the complexity of the resulting function.)
I ended up using both, as InterpolatingFunction gives a noticeable delay in its evaluation, and with up to 49 interpolated functions to evaluate, any delay can become noticeable. Also, ContourPlot3D was faster with the spline, but it didn't give me the speed up I desired.
I'll freely admit that I haven't tried Interpolation on v.7, nor I have tried this on my upgraded hardware (G4 v. Intel Core i5). However, I'm looking for alternatives to my current scheme; preferably, one where I can use ContourPlot3D directly. I could try some other form of spline, such as a B-spline, and possibly combine that with UnitBox instead of using Piecewise.
Edit: Just to clarify, my current implementation involves creating a first order spline for each radial part, multiplying each one by their respective spherical harmonic, summing and Simplifying the equations on each radial interval, and then using Piecewise to bind them into one function. So, my implementation is semi-analytical in that the spherical harmonics are exact, and only the radial part is numerical. This is part of the reason why I would like to be able to use ContourPlot3D, so that I can take advantage of the semi-analytical nature of the data. As a point of note, the radial grid is fine enough that a good representation of the radial part is generated and can be smoothly interpolated. While this gave me a significant speed-up, when I wrote the code, it was still to slow for the hardware I was using at the time.
So, instead of using ContourPlot3D, I would first generate the function, as above, then I would evaluate it on an 803 Cartesian grid. It is the data from this step that I used in ListContourPlot3D. Since this is not an adaptive grid, in some places this was too course, and I was missing features.
If you can do without Mathematica, I would suggest you have a look at Paraview (US government funded FOSS, all platforms) which I have found to be superior to everything when it comes to visualizing massive amounts of data.
The core of the software is the "Visualization Toolkit" VTK, and you can find/write other frontends if need be.
VTK/Paraview can handle almost any data-type: scalar and vector on structured grids or random points, polygons, time-series data, etc. From Mathematica I often just dump grid data into VTK legacy format which in then simplest case looks like this
# vtk DataFile Version 2.0
Generated by mma via vtkGridDump
ASCII
DATASET STRUCTURED_POINTS
DIMENSIONS 49 25 15
SPACING 0.125 0.125 0.0625
ORIGIN 8.5 5. 0.7124999999999999
POINT_DATA 18375
SCALARS RF_pondpot_1V1MHz1amu double 1
LOOKUP_TABLE default
0.04709501616121583
0.04135197485227461
... <18373 more numbers> ...
HTH!
If it really is the interpolation of the radial functions that is slowing you down, you could consider hand-coding that part based on your knowledge of the sample points. As demonstrated below, this gives a significant speedup:
I set things up with your notation. lookuprvals is a list of 100000 r values to look up for timing.
First, look at stock interpolation as a basemark
With[{interp=Interpolation[N#Transpose#{rvals,yvals}]},
Timing[interp[lookuprvals]][[1]]]
Out[259]= 2.28466
Switching to 0th-order interpolation is already an order of magnitude faster (first order is almost same speed):
With[{interp=Interpolation[N#Transpose#{rvals,yvals},InterpolationOrder->0]},
Timing[interp[lookuprvals]][[1]]]
Out[271]= 0.146486
We can get another 1.5 order of magnitude by calculating indices directly:
Module[{avg=MovingAverage[yvals,2],idxfact=N[(pts-1) /Log[Rmax/Rstep]]},
Timing[res=Part[avg,Ceiling[idxfact Log[lookuprvals]]]][[1]]]
Out[272]= 0.006067
As a middle ground, do a log-linear interpolation by hand. This is slower than the above solution but still much faster than stock interpolation:
Module[{diffs=Differences[yvals],
idxfact=N[(pts-1) /Log[Rmax/Rstep]]},
Timing[Block[{idxraw,idxfloor,idxrel},
idxraw=1+idxfact Log[lookuprvals];
idxfloor=Floor[idxraw];
idxrel=idxraw-idxfloor;
res=Part[yvals,idxfloor]+Part[diffs,idxfloor]idxrel
]][[1]]]
Out[276]= 0.026557
If you have the memory for it, I would cache the spherical harmonics and radius (or even radius-index) on the full grid. Then flatten the grid caches so you can do
Sum[ interpolate[yvals[lm],gridrvals] gridylmvals[lm], {lm,lmvals} ]
and recreate your grid as discussed here.
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.