I have one 2d line (it can be a curved line, with loops and so on), and multiple similar paths. I want to compare the first path with the rest, and determine which one is the most similar (in percentage if possible).
I was thinking maybe transforming the paths into bitmaps and then using a library to compare the bitmaps, but that seems like overkill. In my case, I have only an uninterrupted path, made of points, and no different colors or anything.
Can anyone help me?
Edit:
So the first line is the black one. I compare all other lines to it. I want a library or algorithm that can say: the red line is 90% accurate (because it has almost the same shape, and is close to the black one); the blue line is 5% accurate - this percentage is made up for this example... - because it has a similar shape, but it's smaller and not close to the black path.
So the criterion of similarity would be:
how close the lines are one to another
what shape do they have
how big they are
(color doesn't matter)
I know it's impossible to find a library that considers all this. But the most important comparisons should be: are they the same shape and size? The distance I can calculate on my own.
I can think of two measures to express similarity between two lines N (defined as straight line segments between points p0, p1... pr)
M (with straight line segments between q0, q1, ...qs). I assume that p0 and q0 are always closer than p0 and qs.
1) Area
Use the sum of the areas enclosed between N and M, where N and M are more different as the area gets larger.
To get N and M to form a closed shape you should connect p0 and q0 and pr and qs with straight line segments.
To be able to calculate the surface of the enclosed areas, introduce new points at the intersections between segments of N and M, so that you get one or more simple polygons without holes or self-intersections. The area of such a polygon is relatively straightforward to compute (search for "polygon area calculation" around on the web), sum the areas and you have your measure of (dis)similarity.
2) Sampling
Take a predefined number (say, 1000) of sample points O that lie on N (either evenly spaced with respect to the entire line, or evenly spaced
over each line segment of N). For each sample point o in O, we'll calculate the distance to the closest corresponding point on M: the result is the sum of these distances.
Next, reverse the roles: take the sample points from M and calculate each closest corresponding point on N, and sum their distances.
Whichever of these two produces the smallest sum (they're likely not the same!) is the measure of (dis)similarity.
Note: To locate the closest corresponding point on M, locate the closest point for every straight line segment in M (which is simple algebra, google for "shortest distance between a point and a straight line segment"). Use the result from the segment that has the smallest distance to o.
Comparison
Method 1 requires several geometric primitives (point, line segment, polygon) and operations on them (such as calculating intersection points and polygon areas),
in order to implement. This is more work, but produces a more robust result and is easier to optimize for lines consisting of lots of line segments.
Method 2 requires picking a "correct" number of sample points, which can be hard if the lines have alternating parts with little detail
and parts with lots of detail (i.e. a lot of line segments close together), and its implementation is likely to quickly get (very) slow
with a large number of sample points (matching every sample point against every line segment is a quadratic operation).
On the upside, it doesn't require a lot of geometric operations and is relatively easy to implement.
Related
I am creating vertex array for a mesh with given points.So far I was able to create a continuous mesh with thickness.However there is a problem at the intersection of two line segments, the vectors between those segment's sizes needs to be bigger or smaller depending on the situation in order to have a continuous look.
What I have now:
With the given angles theta1 and theta2, how can I calculate the length of red vectors?
What I want:
How I structured my mesh:
You're probably making it more complicated than it needs to be.
Let's start by calculating the red arrows. For any line segment (p_i, p_j), we can calculate the segment's normal with:
dir = normalize(p_j - p_i)
normal = (-dir.y, dir.x) //negate if you want the other direction
At the connection point between two segments, we can just average (and re-normalize) the incident normals. This gives us the red arrows.
The only question that remains is how much you need to shift. The resulting offset for the line segment o_l given an offset of the vertex o_v is:
o_l = o_v * dot(normal_l, normal_v)
This means the following: Both normals are unit vectors. Hence, their dot product is at most one. This is the case when both line segments are parallel. Then, the entire offset of the vertex is transferred to the line. The smaller the angle becomes, the smaller becomes the transferred offset. E.g. if the angle between two consecutive line segments is 120°, then the dot product of normals is 0.5. If you shift the vertex by 1 unit along its normal, both line segment will have a thickness of 0.5.
So, in order to produce a specific line thickness (o_l), we need to shift the vertex by o_v:
o_v = o_l / dot(normal_l, normal_v)
The construction with averaging the line segments' normal for the vertex normal ensures that dot(normal_l1, normal_ v) = dot(normal_l2, normal_v), i.e. the resulting line thickness is equal for both lines in any case.
I need to check if two java.awt.Polygons overlap, but I don't know how.
Area area = new Area(poly1);
area.intersect(new Area(poly2));
return area.isEmpty();
From a geometry perspective, there are conditions that are fulfilled by overlapping polygons.
1) If a segment in A crosses a segment in B, the polygons overlap.
2) If all vertices in B are inside A, or vice-versa, the polygons overlap.
3) If all vertices in A are also vertices of B, the polygons overlap.
Testing for (1) is fairly simple. You just do a little algebra and brute force. If lines are parallel, they don't cross. If the lines are non-parallel, they intersect. If the point of intersection is within either segment, they cross. Maximum iterations is length of A * length of B.
Testing for (2) is a little more complex. One method to tell if a point is inside a poly is this: Pick a point that you know is outside of the poly, often a negative value for x,y works for this. Then, draw a line from the reference point to the test point. If it crosses an odd number of segments, it's inside. It if crosses an even number, or zero segments, it is outside.
I have two ArrayList, Double data type,
1.latitudes
2. longitudes,
each has over 200 elements
say i give a random test coordinates, say (1.33, 103.4), the format is [latitude, longitude]
is there any algorithm to easily find closest point,
or do i have to brute force calculate every possible point, find hypotenuse, and then compare over 200 hypotenuses to return the closest point? thanks
Sort the array of points along one axis. Then, locate the point in the array closest to the required point along this axis and calculate the distance (using whatever metric is appropriate to the problem topology and scale).
Then, search along the array in both directions until the distance to these points is greater than the best result so far. The shortest distance point is the answer.
This can result in having to search the entire array, and is a form of Branch and bound constrained by the geometry of the problem. If the points are reasonably evenly distributed around the point you are searching for, then the scan will not require many trials.
Alternate spatial indices (like quad-trees) will give better results, but your small number of points would make the setup cost in preparing the index much larger than a simple sort. You will need to track the position changes caused by the sort as your other array will not be sorted the same way. If you change the data into a single array of points, then the sort will reorder entire points at the same time.
If your arrays are sorted, you can use binary search to find a position of a requested point in array. After you find index, you should check four near by points to find the closest.
1)Suppose you have two sorted arrays longitudes-wise and latitudes-wise
2)You search first one and find two nearby points
3)Then you search second one and find two more points
4)Now you have from two to four points(results might intersect)
5)These points will form a square around destination point
6)Find the closest point
it's not true that closest lat (or long) value should be choosen to search over the long (or lat) axis, in fact you could stay on a lat (or long) line but far away along the long (or lat) value
so best way is to calculate all distances and sort them
I find out that the GeneralPath class in Java only provides methods to check if a point is inside a general path (to be specific, a polygon with straight line segments). Does someone know how to efficiently check if a point is on the boundary of a general path or not?
Thanks
Dummy solution 1: We can define a circle with radius $\epsilon$ ($\epsilon$ is a very small positive real value). And then, we check a sufficient number of points on the circle to see if one/some of them falls into the general path or not. However, such a dummy method may require a considerable amount of computational effort, which is not very desirable.
Dummy solution 2: We can compute the distances from the point (on the boundary) to every side of the polygon. If the minimal distance is sufficiently small, this point is on the boundary; otherwise, it is not. Again, this method can still be computationally intensive.
I haven't come across a library method ... or potted solution to the problem. I think that the reason is that the problem is fundamentally impossible to solve exactly.
The GeneralPath class inherits a method called getPathIterator from Shape2D. If you look at the javadoc, you will see that the PathIterator object models the path as a sequence of straight line segments. And the getPathIterator method takes a flatness parameter specified as follows:
"flatness - the maximum distance that the line segments used to approximate the curved segments are allowed to deviate from any point on the original curve."
Now, if the shape you are looking at consists of straight line segments, there is a good chance that the path iterator will give you those line segments. But if the shape has curved segments, then the line segments are only an approximation. And it is clearly impossible to test if a point is exactly on a boundary if you don't know what the exact boundary is.
Even assuming that the line segments exactly model the real curve, you still have the problem that (except for special cases), most of the points on the real curve cannot be represented exactly using Java primitive datatypes (int, double, etc). So once again "exactness" is problematic.
I think that the best you can hope for is to test whether your point is within some small delta of the boundary ... and choose a flatness value that is less than that delta, iterate the path line segments, and test the tangential distance of the point from each segment.
Note: if you make flatness very small, you can expect to have to test a very large number of line segments. I don't think there is any way around this computational concern while sticking to the GeneralPath API.
If you restrict the problem to true (i.e. straight-sided) polygons, then you simply need to iterate the line segments, and for each one test to see if the distance from the point to the line is less than some suitable epsilon. This Wikipedia entry gives you the mathematics. Note that exactness is still going to be a concern here ...
You don't have an enormously costly computation, but computing an accurate square-root doesn't come for free, and you have to do it up to N times for an N-sided polygon.
Doing better than that (i.e. getting better than O(N)) is going to be difficult. However, if the polygon is fixed and you are going to be testing a huge number of points against it, then you could consider using a pre-computing a quad-tree data structure to do the resolution. Precomputing the quad-tree will be expensive, but if N is large enough testing a a point will be cheaper. (Roughly O(log(1/epsilon)) in the worst case, instead of O(N) on average. And the further away from the boundary a point is, the cheaper the answer.)
But as I said, quad-trees will only help in limited situations ...
You can use stroked path. BasicStroke has method
public Shape createStrokedShape(Shape s)
So just define thick line BasicStroke, get the stroked shape and check whether the stroked Shape contains your point.
I have a small contest problem in which is given a set of points, in 2D, that form a triangle. This triangle may be subject to an arbitrary rotation, may be subject to an arbitrary translation (both in the 2D plane) and may be subject to a reflection on a mirror, but its dimensions were kept unchanged.
Then, they give me a set of points in the plane, and I have to find 3 points that form my triangle after one or more of those geometric operations.
Example:
5 15
8 5
20 10
6
5 17
5 20
20 5
10 5
15 20
15 10
Output:
5 17
10 5
15 20
I bet it's supposed to apply some known algorithm, but I don't know which. The most common are: convex hull, sweep plane, triangulation, etc.
Can someone give a tip? I don't need the code, only a push, please!
A triangle is uniquely defined (ignoring rotations, flips, and translations) by the lengths of its three sides. Label the vertices of your original triangle A,B,C. You're looking
for points D,E,F such that |AB| = |DE|, |AC| = |DF|, and |BC| = |EF|. The length is given by Pythagoras' formula (but you can save a square root operation at each test by comparing
the squares of the line segment lengths...)
The given triangle is defined by three lengths. You want to find three points in the list separated by exactly those lengths.
Square the given lengths to avoid bothering with sqrt.
Find the square of the distance between every pair of points in the list and only note those that coincide with the given lengths: O(V^2), but with a low coefficient because most lengths will not match.
Now you have a sparse graph with O(V) edges. Find every cycle of size 3 in O(V) time, and prune the matches. (Not sure of the best way, but here is one way with proper big-O.)
Total complexity: O(V^2) but finding the cycles in O(V) may be the limiting factor, depending on the number of points. Spatially sorting the list of points to avoid looking at all pairs should improve the asymptotic behavior, otherwise.
This is generally done with matrix math. This Wikipedia article covers the rotation, translation, and reflection matrices. Here's another site (with pictures).
Since the transformations are just rotation, scaling and mirroring, then you can find the points that form the transformed triangle by checking the dot product of two sides of the triangle:
For the original triangle A, B, C, calculate the dot product of AB.AC, BA.BC and CA.CB
For each set of three points D, E, F, calculate dot product of DE.DF and compare against the three dot products found in 1.
This works since AB.AC = |AB| x |AC| x cos(a), and two lengths and an angle between them defines a triangle.
EDIT: Yes, Jim is right, just one dot product isn't enough, you'll need to do all three, including ED.EF and FD.FE. So in the end, there's the same number of calculations in this as there is in the square distances method.