I have read about DDA. But I just came across the term symmetric DDA. What is it ? How is it different from DDA ?
The DDA (Digital Differential Analyzer) algorithm is used to find out interpolating points between any given two points, linearly (i.e. straight line). Now since this is to be done on a digital computer - speed is an important factor.
The equation of a straight line is given by m=Δx/Δy eq(i), where Δx = x(2)-x(1) & Δy = y(2)-y(1),now using this equation we could compute successive points that lie on the line. But then this is the discrete world of raster graphics - so we require integral coordinates.
In simple DDA eq(i) is transformed to m=eΔx/eΔy where e, call it the increment factor, is a positive real number. since putting the same number in numerator and denominator does not change anything - but if suitably chosen - it can help us in generating discrete points thereby reducing the overload of having to round off the resultant points.
Basically what we need to do is: increment the coordinates by a fixed small amount, beginning from the starting point, and each time we have a new point progressing towards the end point.
In simple DDA - e is chosen as 1/max(|Δx|,|Δy|) such that one of the coordinate is integral and only the other coordinate has to be rounded. i.e. P(i+1) = P(i)+(1,Round(e*Δy)) here one coordinate is being incremented by 1 and the other by e*Δy
In symmetric DDA - e is chosen such that though both the co-ordinates of the resultant points has to be rounded off, it can be done so very efficiently, thus quickly.
Specifically e is chosen as 1/2^n where 2^(n-1) <= max(|Δx|,|Δy|) < 2^n. In other words the length of the line is taken to be 2^n aligned. The increments for the two coordinates are e*Δx and e*Δy. With suitably chosen initial fraction part of the beginning coordinates: this causes the points to be generated as mixed fractions whose fractional parts are in a cyclic series, i.e. they repeat over a small length. The resultant coordinates can thus easily be rounded off based on two fixed length look-up tables, one for each coordinate.
refer http://w3.msi.vxu.se/~gsu/DAB726-Ht06/Symm-DDA.pdf for an example.
Notice the cyclic repetition in the fractional part of the resultant coordinates.
Related
I have a collection of points in some N-dimensional space, where all I know is the distances between them. Let's say it's an unordered collection of structs like the following:
struct {
int first; // Just some identifier that uniquely specifies a point
int second; // No importance to which point is first or second
float separation; // The distance between the first and second points -- always positive
};
Of course the algorithm doesn't have to be C code. I just wrote the struct in this style to make the problem clear. It rather upsets me that the struct spoils the symmetry between the two end-points, but fixing this just makes things more complicated.
Let's say that the separations are defined by the Pythagorean distance between them, and the space is Euclidean. Let's also specify that the separations are internally consistent. For example, given separations AB, BC and AC, we know that AB + BC >= AC.
I want an algorithm that finds the minimal dimensional space that can contain all the points. Within this algorithm, we can assume that separations that deviate from that defined by the space by less than some specified tolerance can be ignored.
Does anyone know an algorithm that does this? So far, I've only been able to think up non-polynominal algorithms. Can anybody improve on that, or at least make something that is clean and extensible?
Why is this interesting? In Physics there are some low-level theories such as String Theory or Quantum Loop Gravity that do not obviously predict our three dimensional world. This algorithm could be part of a project to find how a 3d world can be emergent.
Thank you everybody who posted ideas here. I now have an answer to my own question. It's not great, in that it executes O(n^3) but at least it's polynomial. Roughly, it works like this:
Represent the problem as a symmetric matrix with zero diagonal -- representing the distances between any two points. This is equivalent to the representation using structs, but much easier to work with.
Assume the ordering of the points implied by the matrix (first column/row = first point) is sensible. (It may be worth pivoting to find a better ordering, but that is todo.)
Now create a rectangular coordinate system to fit the points, starting with the first point, which WLOG we take to be the origin.
Second point defines the x axis
For each subsequent point, we calculate its coordinates one at a time, starting with the x axis. We know the distance from the origin and the distance from point 2. This allows us to calculate the x coordinate, as we end up with two simultaneous equations x^2 + y^2 + ... = s1^2 and (x - x2)^2 + y^2 + ... = s2^2, which allows us to calculate x easily from x2, the x coordinate of point 2, and the distances from points 1 and 2, s1 and s2.
Each new coordinate can be calculated easily, because the matrix of coordinates calculated so far is triangular -- there is only one unknown each time.
The last coordinate for each point is on a new axis -- a dimension that has not yet been used. Calculate its coordinate using Pythagoras on the distance from the origin, as we know all the other coordinates.
It is possible that the coordinate on the new axis will come out imaginary -- a general set of distances cannot always be represented by a coordinate system of any number of dimensions -- at least not with real numbers. If this is the case, I error.
Keep going in this way for each new point, building up a vector of coordinate vectors for each point. In general, this is triangular, but there may be cases where the final coordinate we calculate is near enough to zero that we consider the point's position to be represented by the existing dimensions. I store the coordinates anyway, but keep the number of dimensions the same as the previous point. I also skip these points, as they are not needed for calculating further points (see step 10).
Finally, we have represented all points such that the distances are consistent.
As a final check, I validate that the distances match for all points, including those skipped in step 9.
The number of dimensions needed is the number used for the last point.
If anyone is interested in an implementation of this (in Haskell), it is on my GitHub page at https://github.com/MarcusRainbow/EmergentDimensions/coords.hs.
I have a contour defined by a vector of points i.e. I have a vector of (x_i ,y_i).
Now given any point (a,b) is there a fast way to determine (a,b) lies on the same side as the origin (0,0) or not ?
The vector which defines the contour is a vector of roughly 7000 points. So determination using point by point maybe extremely slow. It would be very kind if someone can give any pointers.
(I am using visual C++ for my computations)
Thanks in advance.
I understand that preprocessing is allowed (otherwise you cannot avoid the "point by point" procedure).
A simple way is by scanning the polyline to decompose it in monotonous sections, i.e. such that the vertices appear by increasing or decreasing x. This takes linear time.
Then when you want to compare a point to the polyline, compare it to every monotonous section independently and find the x interval it faces, by dichotomic search. For a section of length m, this takes Log m operations. When you know the interval, you can immediately tell on what side ot the section you lie.
Repeat for all sections and count the number of "below", the parity of which gives you the answer.
This procedure takes time s Log m for s sections of length m (on geometric average). It is not worst-case optimal but is simple and will behave very decently on most cases.
I am interesting in finding the diameter of two points sets, in 128 dimensions. The first has 10000 points and the second 1000000. For that reason I would like to do something better than the naive approach which takes O(n²). The algorithm will be able to handle any number of points and dimensions, but I am currently very interested in these two particular data sets.
I am very interesting in gaining speed over accuracy, thus, based on this, I would find the (approximate) bounding box of the point set, by computing the min and max value per coordinate, thus O(n*d) time. Then, if I find the diameter of this box, the problem is solved.
In the 3d case, I could find the diameter of the one side, since I know the two edges and then, I could apply the Pythagorean theorem on the other, which is vertical to this side. I am not sure for this however and for sure, I can't see how to generalize it to d dimensions.
An interesting answer can be found here, but it seems to be specific for 3 dimensions and I want a method for d dimensions.
Interesting paper: On computing the diameter of a point set in high dimensional Euclidean space. Link. However, implementing the algorithm seems too much for me in this phase.
The classic 2-approximation algorithm for this problem, with running time O(nd), is to choose an arbitrary point and then return the maximum distance to another point. The diameter is no smaller than this value and no larger than twice this value.
I would like to add a comment, but not enough reputation for that...
I just want to warn other readers that the "bounding box" solution is very inaccurate. Take for example the Euclidean ball of radius one. This set has diameter two, but its bounding box is [-1, 1]^d, which has diameter twice the square root of d. For d = 128, this is already a very bad approximation.
For a crude estimate, I would stay with David Eisenstat's answer.
There is a precision based algorithm which performs very well on any dimension, which is based on computing the dimension of an axial bounding box.
The idea is that it's possible to find the lower and upper boundaries of the axis bounding box length function since it's partial derivatives are limited, and depend on the angle between the axises.
The limit of the local maxima derivatives between two axises in 2d space can be computed as:
sin(a/2)*(1 + tan(a/2))
That means that, for example, for 90deg between axises the boundary is 1.42 (sqrt(2))
Which reduces to a/2 when a => 0, so the upper boundary is proportional to the angle.
For a multidimensional case the formula varies slightly, but still it's easy to compute.
So, the search of local minima convolves in logarithmic time.
The good news is that we can run the search of such local maxima in parallel.
Also, we can filter out both the regions of the search based on the best achieved result so far, as well as the points themselves, which are belo the lower limit of the search in the worst region.
The worst case of the algorithm is where all of the points are placed on the surface of a sphere.
This can be firther improved: when we detect a local search which operates on just few points, we swap to bruteforce for this particular axis. It works fast, because we need only the points which are subject to that particular local search, which can be determined as points actually bound by two opposite spherical cones of a particular angle sharing the same axis.
It's hard to figure out the big O notation, because it depends on desired precision and the distribution of points (bad when most of the points are on a sphere's surface).
The algorithm i use is here:
Set the initial angle a = pi/2.
Take one axis for each dimension. The angle and the axises form the initial 'bucket'
For each axis, compute the span on that axis by projecting all the points onto the axis, and finding min and max of the coordinates on the axis.
Compute the upper and lower bounds of the diameter which is interesting. It's based on the formula: sin(a/2)*(1 + tan(a/2)) and multiplied by assimetry cooficient, computed from the length of the current axis projections.
For the next step, kill all of the points which fall under the lower bound in each dimension at the same time.
For each exis, If the amount of points above the upper bound is less then some reasonable amount (experimentally computed) then compute using a bruteforce (N^2) on the set of the points in question, and adjust the lower bound, and kill the axis for the next step.
For the next step, Kill all of the axises, which have all of their points under the lower bound.
If the precision is satisfactory (upper bound - lower bound) < epsilon, then return the upper bound as the result.
For all of the survived axises, there is a virtual cone on that axis (actually, the two opposite cones), which covers some area on a virtual sphere which encloses a face of the cube. If i'm not mistaken, it's angle would be a * sqrt(2). Set the new angle to a / sqrt(2). Create a whole bucket of new axises (2 * number of dimensions), so the new cone areas would cover the initial cone area. It's the hard part for me, as i have not enough imagination for n>3-dimensional case.
Continue from step (3).
You can paralellize the procedure, synchronizing the limits computed so far for the points from (5) through (7).
I'm going to summarize the algorithm proposed by Timothy Shields.
Pick random point x.
Pick point y furthest from x.
If not done, let x = y, and go to step 2
The more times you repeat, the more accurate the result will be... ??
EDIT: actually this algorithm is not very good. Think about a 2D rectangle with vertices ABCD. There are two maxima: between AC and BD, which are separated by a sizable valley. This algorithm will get stuck at one or the other 50/50. If AC is slightly larger than BD, you'll be getting the wrong answer 50% of the time no matter how many times you iterate. Other regular polygons have the same issue, and in higher dimensions it is even worse.
Given a 3d heightmap (from a laser scanner), how do I find the saddle points?
I.e. given something like this:
I am looking for all points where the curvature is positive in one direction and negative in the other.
(These directions should not need to be aligned with the X and Y axis.
I know how to check whether the curvature in X direction has the opposite sign as the curvature in Y direction, but that does not cover all cases. To make matters worse, the resolution in X is different from the resolution in Y)
Ideally I am looking for an algorithm that can tolerate some amount of noise and only mark "significant" saddle points.
I've been exploring a similar problem for a computational topology class and have had some success with the method outlined below.
First you will need a comparison function that will evaluate the height at two input points and will return < or > (not equal) for any input. One way to do this is that if the points are equal height you use some position-based or random index to find the greater point. You can think of this as adding an infinitesimal perturbation to the height.
Now, for each point, you will compare the height at all the surrounding neighbors (there will be 8 neighbors on a 2D rectangular grid). The lower link for a point will be the set of all neighbors for which the height is less than the point.
If all the neighboring values are in the lower link, you are at a local maximum. If none of the points are in the lower link you are at a local minimum. Otherwise, if the lower link is a single connected set, you are at a regular point on a slope. But if the lower link is two unconnected sets, you are at a saddle.
In 2D you can construct a list of the 8 neighboring point in cyclic order around the point you are checking. You assign a value of +/-1 for each neighbor depending on your comparison function. You can then step through that list (remember to compare the two end points) and count how many times the sign changes to determine the number of connected components in the lower link.
Determining which saddles are "important" is a more difficult analysis. You may wish to look at this: http://www.cs.jhu.edu/~misha/ReadingSeminar/Papers/Gyulassy08.pdf for some guidance.
-Michael
(From a guess at the maths rather than practical experience)
Fit a quadratic to the surface in a small patch around each candidate point, e.g. with least squares. How big the patch is is one way of controlling noise, and you might gain by weighting points depending on their distance from the candidate point. In matrix notation, you can represent the quadratic as x'Ax + b'x + c, where A is symmetric.
The quadratic will have zero gradient at x = (A^-1)b/2. If this not within the patch, discard it.
If A has both +ve and -ve eigenvalues you have a saddle point at x. Since A is only 2x2 and so has at most two eigenvalues, you can ignore the case when it as a zero eigenvalue and so you couldn't invert it at the previous stage.
The parametric boundary of an object can be extract in Matlab by using the bwtraceboundary function. It returns a Q-by-2 matrix B, where Q is the number of boundary pixels for the object and the first and second columns stores the row and column coordinates of the boundary pixels respectively.
What I want to do is to sample this boundary of Q elements by N points that divide the original boundary in segments of equal arch length.
A straightfoward solution that I thought consists in computing the length L of the boundary by summing the distance of all two consecutive boundary pixels. Those distances are either 1 or sqrt(2). Then I divide L by N to find the desired length of the arcs. Finally, I iterate over the boundary again summing the distance of all two consecutive boundary pixels. When the sum is greater or equal the desired arc-length, the current boundary pixel is chosen as one of the N that will compose the sampled boundary.
Is that a good solution? Is there a more efficient/simple solution?
Over the years, I have seen this question a seemingly vast number of times. So I wrote a little tool that will do exactly that. Sample a piecewise linear or even a curvilinear (spline) arc in a general number of dimensions so that the successive points are at a uniform or specified distance along that arc.
In the case of the use of merely piecewise linear arcs, this is rather easy. You sum up the total arc length of the curve, then do an interpolation in arc length, but since that is known to be piecewise linear, it only requires linear interpolation along that length as a function of the cumulative arc length.
In the case of a curved arc, it is most easily done as the solution of a system of ordinary differential equations, watching for events along the way. ODE45 does this nicely.
You can use interparc, as found on the MATLAB Central File Exchange to do this for you, or if you wish to learn to do it yourself for the simple piecewise linear case, read through the first part of the code where I do the piecewise linear arc length interpolation. A nice thing is the linear case is done in a fully vectorized form, so no explicit loops are necessary.