I have two discs that can move separately with the help of keyboard. The two discs represent two players and I want to code :
If disc1 touches disc2 then the size of disc2 reduces a little bit
Both the discs should not go out of the screen
Given that they're discs, collision detection is actually fairly simple and straightforward. Given two discs with radii R1 and R2, if the distance between the centers of the two objects is less than or equal to R1+R2, then they've collided.
You can compute the distance between the two center points using the Pythagorean theorem: the distance equals the square root of the sum of the delta X squared and delta Y squared.
If you're doing this very often, you probably want to avoid that square root. Fortunately that's pretty easy: square the sum of the two radii, and compare that to the sum of the squares of delta X and delta Y.
Related
Good day.
I have the task of finding the set of points in 2D space for which the sum of the distances to the rectangles is minimal. For example, for two rectangles, the result will be the next area (picture). Any point in this area has the minimum sum of lengths to A and B rectangles.
Which algorithm is suitable for finding a region, all points of which have the minimum sum of lengths? The number of rectangles can be different, they are randomly located. They can even overlap each other. The sides of the rectangles are parallel to the coordinate axes and cannot be rotated. The region must be either a rectangle or a line or a point.
Hint:
The distance map of a rectangle (function that maps any point (x,y) to the closest distance to the rectangle) is made of four slanted planes (slope 45°), four quarter of cones and the rectangle itself, which is at ground level, forming a continuous surface.
To obtain the global distance map, it "suffices" to sum the distance maps of the individual rectangles. A pretty complex surface will result. Depending on the geometries, the minimum might be achieved on a single vertex, a whole edge or a whole face.
The construction of the global map seems more difficult than that of a line arrangement, due to the conic patches. A very difficult problem in the general case, though the axis-aligned constraint might ease it.
Add on Yves's answer.
As Yves described, each rectangle 'divide' plane into 9 parts and adds different distance method in to the sum. Middle part (rectangle) add distance 0, side parts add coordinate distance to that side, corner parts add point distance to that corner. With that approach plan has to be divided into 9^n parts, and distance sum is calculated by adding appropriate rectangle distance functions. That is feasible if number of rectangles is not too large.
Probably it is not needed to calculate all parts since it is easy to calculate some bound on part min value and check is it needed to calculate part at all.
I am not sure, but it seems to me that global distance map is convex function. If that is the case than it can be solved iteratively by similar idea as in linear programming.
I have a map that I can zoom in and out of. Gridlines display the latitude and longitude.
Right now I zoom in by shrinking the borders by a specific amount.
Say top edge is lat 30n, bot edge is lat 28n, left edge is lon -40w, right edge is lon -38w.
So for one zoom in, the new edges are 29.5n, 28.5n, -39.5w,-38.5 respectively.
Zooming in enough times would eventually make the opposite edges equal each other and then flip.
So next zoom in for top and bot edge would be 29n, 29n and then next 28.5n,29.5n which flips the orientation.
I'm looking for a function or a way that I can continuously zoom in, but never have the edges equal each other or flip. In others words something where I can get the edges to continuously get closer and closer to each other, but never touch.
So I have an idea that I'll need some function like
lim x-> infinity of x/(x+1) - 1 + midpoint of the two edges
which basically means, as I zoom in I will approach the midpoint.
If this is valid, I don't know how to implement limits in C++.
This is only one equation too. I have two edges on each axis and so I'd need two equations to approach the midpoint from both sides at an equal rate.
I'm working on an application were I have a set of Contours(each one representing a Potential Line) and I wanna check "How straight" is that contour/shape.
The article I am using as a refrence uses the following technique:
It Matches a "segmented" line crossing the shape like so-
Then grading how "straight" is the line.
Heres an example of the Contours I am working on:
How would you go about implementing this technique?
Is there any other way of checking "How Straight" is a contour\shape?
Regards!
My first guess would be to use a coefficient of determination. That would be, fit a linear line to all your point assuming some reasonable origin where you won't receive rounding errors and calculate R^2.
A more advanced approach, if all contours are disconnected components, would be to calculate the structure model index (the link is for bone morphometry, but they explain the concept and cite the original paper.) This gives you a number that tells you how much your segment is "like a rod". This is just an idea, though. Anything that forms curves or has branches will be less and less like a rod.
I would say that it also depends on what you are using the metric for and if your contours are always generally carrying left to right.
An additional method would be to create the covariance matrix of your points, calculate the eigenvalues from that matrix, and take their ratio (where the ratio is greater than or equal to 1; otherwise, invert the ratio.) This is the basic principle behind a PCA besides the final ratio. If you have a rather linear data set (the data set varies in only one direction) then you will have a very large ratio. As the data set becomes less and less linear (or more uncorrelated) you would see the ratio approach one. A perfectly linear data set would be infinity and a perfect circle one (I believe, but I would appreciate if someone could verify this for me.) Also, working in two dimensions would mean the calculation would be computationally cheap and straight forward.
This would handle outliers very well and would be invariant to the rotation and shape of your contour. You also have a number which is always positive. The only issue would be preventing overflow when dividing the two eigenvalues. Then again you could always divide the smaller eigenvalue by the larger and your metric would be bound between zero and one, one being a circle and zero being a straight line.
Either way, you would need to test if this parameter is sensitive enough for your application.
One example for a simple algorithm is using the dot product between two segments to determine the angle between them. The formula for dot product is:
A * B = ||A|| ||B|| cos(theta)
Solving the equation for cos(theta) yields
cos(theta) = (A * B / (||A|| ||B||))
Since cos(0) = 1, cos(pi) = -1.0 and you're checking for the "straightness" of the lines, a line whose normalization of cos(theta) angles is closest to -1.0 is the straightest.
straightness = SUM(cos(theta))/(number of line segments)
where a straight line is close to -1.0, and a non-straight line approaches 1.0. Keep in mind this is a cursory evaluation of this algorithm and it obviously has edge cases and caveats that would need to be addressed in an implementation.
The trick is to use image moments. In short, you calculate the minimum inertia around an axis, the inertia around an axis perpendicular to this, and the ratio between them (which is always between 0 and 1; since inertia is non-negative)
For a straight line, the inertia along the line is zero, so the ratio is also zero. For a circle, the inertia is the same along all axis so the ratio is one. Your segmented line will be 0.01 or so as it's a fairly good match.
A simpler method is to compare the circumference of the the convex polygon containing the shape with the circumference of the shape itself. For a line, they're trivially equal, and for a not too crooked shape it's still comparable.
I currently have several ellipses. These are defined by a centre point, and then two vectors, one point to the minimum axis and other to the maximum axis.
However, for the program I'm creating I need to be able to deal with these shapes as a polyline. I'm fairly sure there must be formula for generating a set of points from the available data that I have, but I'm unsure how to go about doing it.
Does anyone have any ideas of how to go about this?
Thanks.
(Under assumption that both vectors that represent ellipse axes are parllel to coordinate axes)
If you have a radial ray emanating from the centre of ellipsis at angle angle, then that ray intersects the ellipse at point
x = x_half_axis * cos(angle);
y = y_half_axis * sin(angle);
where x_half_axis and y_half_axis age just the lengths (magnitudes) of your half-axis vectors.
So, just choose some sufficiently small angle step delta. Sweep around your centre point through the entire [0...2*Pi] range with that step, beginning with 0 angle, then delta angle, then 2 * delta angle and so on. For each angle value the coordinates of the ellipse point will be given by the above formulas. That way you will generate your polygonal representation of the ellipse.
If your delta is relatively large (few points on the ellipse) then it should be chosen carefully to make sure your "elliptical polygon" closes nicely: 2*Pi should split into a whole number of delta steps. Albeit for small delta values it does not matter as much.
If your minimum-maximum axis vectors are not parallel to coordinate axes, your can still use the above approach and then transform the resultant points to the proper final position by applying the corresponding rotation transformation.
Fixed-delta angle stepping has some disadvantages though. It generates a denser sequence of polygonal points near the miminum axis of the ellipse (where the curvature is smaller) and a sparser sequence of points near the maximum axis (where the curvature is greater). This is actually the opposite of the desirable behavior: it is better to have higher point density in the regions of higher curvature.
If that is an issue for you, then you can update the algorithm to make it use variadic stepping. The angle delta should gradually decrease as we approach the maximum axis and increase as we approach the minimum axis.
Assuming the center at (Xc,Yc) and the axis vectors (Xm,Ym), (XM,YM) (these two should be orthogonal), the formula is
X = XM cos(t) + Xm sin(t) + Xc
Y = YM cos(t) + Ym sin(t) + Yc
with t in [0,2Pi].
To get a efficient distribution of the endpoints on the outline, I recommend to use the maximum deviation criterion applied recursively: to draw the arc corresponding to the range [t0,t2], try the midpoint value t1=(t0+t2)/2. If the corresponding points are such that the distance of P1 to the line P0P2 is below a constant threshold (such as one pixel), you can approximate the arc by the segment P0P1. Otherwise, repeat the operation for the arcs [t0,t1] and [t1,t2].
Preorder recursion allows you to emit the polyline vertexes in sequence.
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Possible Duplicate:
Finding whether a point lies inside a rectangle or not
There is an interview question that is, "How to determine whether a point lies inside a rectangle"
Note that the rectangle could be rotated as well. So the simple solution of checking point inside the rectangle doesn't stands valid here...
Please share your thoughts on this question..
I found a link on internet, and was trying to understand it, but failed.... Please if any body out here can give complete solution with bit of computer graphics logic, because i have forgotten all the basics....
How to determine if a point is inside rectangle.
Pick a point that's definitely outside the rectangle. Then create a segment from that point to the point in question. Solve the linear equations for intersections between that segment and the segments that make up the rectangle. If you get exactly one intersection, the point is inside the rectangle. Otherwise (0 or 2 intersections), it's outside.
This is trivial to extend to essentially any polygon -- an odd number of intersections means the point is inside the polygon, and an even number means it's outside.
Edit: It may not be immediately obvious, so I'll emphasize that the point we pick outside the rectangle (polygon) is entirely arbitrary. We can pick whatever point we want as long as we're sure it's outside the polygon. To keep our computations easy, what we'll typically do is pick (Px, infinity) (where Px is the x coordinate of the point P that we're testing) -- that is, what we're creating is essentially a vertical ray. That simplifies testing a bit, because we only have to test against one end-point to find an intersection. It also simplifies solving the linear equations to the point that it's barely recognizable as solving linear equations anymore. We really just need to compute the Y coordinate for the line at the Px, and see if it's greater than Py. As such, solving the linear equation breaks down to:
checking whether that X value is within the range of X values for the segment
if it is, plugging the X value into the equation of the line
testing whether the resulting Y value is greater than Py
If those pass, we have an intersection. Also note that the tests can be carried out in parallel (handy if we're doing this on parallel hardware like a GPU).
Simple solution that works in N dimensions for convex polyhedra, of which a 2-dimensional rectangle is a special case:
Represent the polyhedron as the intersection of half-spaces, each defined by a unit normal vector and the distance of the surface hyperplane from the origin along the normal.
For each of these half-spaces, take the dot product of point in question with the defining normal vector. The point is in the half-space if and only if the dot product is less than [or equal to] the defining distance.
The point is inside the polyhedron if and only if it's in every one of the half-spaces.
For a rectangle defined as a counter-clockwise sequence of edges, step 1 amounts to rotating the edges each by 90 degrees clockwise to get the normals, then intersecting the normal line with the line containing the edge to find the distance to the origin.
Assuming step 1 is complete, testing a point takes at most 8 multiplications, 4 additions, and 4 comparisons.
If you want to, you can optimize the process a bit since you have rectangles (and thus opposite sides have opposite normals). Now you're only looking at 2 normals rather than 4, and a range of dot product values which indicate points that lie between the opposite sides. So now you're down to 4 multiplications, 2 additions, and 4 comparisons.
You can also get lucky if the first test you make shows that the point is outside the rectangle, in which case it's just 2 multiplications, 1 addition, and 1-2 comparisons.
This is far from the best solution... But if you have the points in consecutive order, call them a, b, c, and d with an x and a y field, you can use the cross product of the vectors between your point p and each of the consecutive pairs.
If you always get the same sign for the result (i.e., all are positive or all are negative) then you're inside the rectangle; otherwise, you're outside.
Define a new coordinate system with two rectangle sides as unit vectors and transform the coordinate of the point into the new coordinate system. If both coordinates are between 0 and 1, it's inside.
In equations (assuming A,B,C,D are corners of the rectangle, P is the point, _x and _y are the x and y components):
P_x = A_x + x * (B_x - A_x) + y * (D_x - A_x)
P_y = A_y + x * (B_y - A_y) + y * (D_y - A_y)
Solve for x and y and check if they are between 0 and 1
Written as linear equation system (A,B,C,D,P are vectors of length 2):
[ | ] [x] [ ]
[B-A | D-A] * [ ] = [P-A]
[ | ] [y] [ ]
Solving is easy as it has only two dimensions and you can be sure that you are not singular.
You can rotate and move your reference system so it matches position and rotation of the rectangle. Now it is just a matter of simple comparisons between coordinates. This is more a mathematical way, so not the fastest (bellieve #Platinum Azure's one is)
Since the rectangle could be rotated, you might want to consider an algorithm that is used to determine whether a point is interior to a convex polygon.
You could also compute the rotation angle of the rectangle, then transform both the rectangle and the point to axially align the rectangle. Then check to see if the transformed point is inside the axially aligned rectangle.
Finding whether a point lies within a bounded region like rectangle is part of the classic clipping algorithms. Refer to the wikipedia articles on Clipping and Line Clipping to know more about it.
Following the spirit of #Jerry Coffin: create segments from rectangle corners to the point in question. Solve the linear equations. Slope is tan(a). Sum up all seq arctangents diff, if it is 2*PI and each diff < PI - point is inside the rectangle.
Edit Probably enough just check for each sequential difference < Pi...