I am making a reaction game, where you can destroy enemys and earn points, but if the enemy nodes intersect the players node, it reduces the lives by one.
Its a reaction game so I want to give some tolerance points in the intersection. So if they are intersecting, I don't want to see immediately a reduction.
Here is my code:
if (CGRectIntersectsRect(enemy.frame, player.frame)) {
lives--;
}
My goal is a tolerance threshold for the intersection.
I hope someone can help me. (code in objective c please)
From the comments on the question I'm guessing you are not understanding what "inset" means.
If you have a rectangle with origin (10,20), width 100 and height 50 then if you inset the rectangle by 10 on the x-axis and 5 on the y-axis the result is a rectangle with origin (20,25), width 80 and height 40.
Note that the inset rectangle is contained within the original rectangle. #luk2302 suggestion is instead of testing whether the frame rectangles of your two objects intersect you test whether two smaller rectangles intersect. You can generate these smaller rectangles by insetting the frame rectangles by some amount.
You can of course calculate an inset rectangle directly with a few lines of code, but there is an API call CGRectInset (and also NSInsetRect) which will do it for you in one line.
HTH
Related
Suppose that I want to find the 3D position of a cup with its rotation, with image input like this (this cup can be rotated to point in any direction):
Given that I have a bunch of 2D points specifying the top circle and bottom circle like the following image. (Let's assume that these points are given by a person drawing the lines around the cup, so it won't be very accurate. Ellipse fitting or SolvePnP might be needed to recover a good approximation. And the bottom circle is not a complete circle, it's just part of a circle. Sometimes the top part will be occluded as well so we cannot rely that there will be a complete circle)
I also know the physical radius of the top and bottom circle, and the distance between them by using a ruler to measure them beforehand.
I want to find the complete 2 circle like following image (I think I need to find the position of the cup and its up direction before I could project the complete circles):
Let's say that my ultimate goal is to be able to find the closest 2D top point and closest 2D bottom point, given a 2D point on the side of the cup, like the following image:
A point can also be inside of the cup, like so:
Let's define distance(a, b) as a function that find euclidean distance from point a and point b in pixel units.
From that I would be able to calculate the distance(side point, bottom point) / distance(top point, bottom point) which will be a scale number from 0 to 1, if I multiply this number to the physical height of the cup measured by the ruler, then I will know how high the point is from the bottom of the cup in metric unit.
What is the method I can use to find the corresponding top and bottom point given point on the side, so that I can finally find out the height of the point from the bottom of the cup?
I'm thinking of using PnP to solve this but my points do not have correct IDs associated with them. And I don't want to know the exact rotation of the cup, I only want to know the up direction of the cup.
I also think that fitting the ellipse might help somewhat, but maybe it's not the best because the circle is not complete.
If you have any suggestions, please tell me how to obtain the point height from the bottom of the cup.
Given the accuracy issues, I don't think it is worth performing a 3D reconstruction of the cone.
I would perform a "standard" ellipse fit on the top outline, which is the most accurate, then a constrained one on the bottom, knowing the position of the vertical axis. After reduction of the coordinates, the bottom ellipse can be written as
x²/a² + (y - h)²/b² = 1
which can be solved by least-squares.
Note that it could be advantageous to ask the user to point at the endpoints of the straight edges at the bottom, plus the lowest point, instead of the whole curve.
Solving for the closest top and bottom points is a pure 2D problem (draw the line through the given point and the intersection of the sides, and find the intersection points with the ellipse.
I want to find all pixels in an image (in Cartesian coordinates) which lie within certain polar range, r_min r_max theta_min and theta_max. So in other words I have some annular section defined with the parameters mentioned above and I want to find integer x,y coordinates of the pixels which lie within it. The brute force solution comes to mid offcourse (going through all the pixels of the image and checking if it is within it) but I am wondering if there is some more efficient solution to it.
Thanks
In the brute force solution, you can first determine the tight bounding box of the area, by computing the four vertexes and including the four cardinal extreme points as needed. Then for every pixel, you will have to evaluate two circles (quadratic expressions) and two straight lines (linear expressions). By doing the computation incrementally (X => X+1) the number of operations drops to about nothing.
Inside a circle
f(X,Y) = X²+Y²-2XXc-2YYc+Xc²+Yc²-R² <= 0
Incrementally,
f(X+1,Y) = f(X,Y)+2X+1-2Xc <= 0
If you really want to avoid that overhead, you will resort to scanline conversion techniques. First think of filling a slanted rectangle. Drawing two horizontal lines by the intermediate vertices, you decompose the rectangle in two triangles and a parallelogram. Then for any scanline that crosses one of these shapes, you know beforehand what pair of sides you will intersect. From there, you know what portion of the scanline you need to fill.
You can generalize to any shape, in particular your circle segment. Be prepared to a relatively subtle case analysis, but finding the intersections themselves isn't so hard. It may help to split the domain with a vertical through the center so that any horizontal always meets the outline twice, never four times.
We'll assume the center of the section is at 0,0 for simplicity. If not, it's easy to change by offsetting all the coordinates.
For each possible y coordinate from r_max to -r_max, find the x coordinates of the circle of both radii: -sqrt(r*r-y*y) and sqrt(r*r-y*y). For every point that is inside the r_max circle and outside the r_min circle, it might be part of the section and will need further testing.
Now do the same x coordinate calculations, but this time with the line segments described by the angles. You'll need some conditional logic to determine which side of the line is inside and which is outside, and whether it affects the upper or lower part of the section.
I'm drawing squares along a circular path for an iOS application. However, at certain points along the circle, the squares start to go out of the circle's circumference. How do I make sure that the squares stay inside?
Here's an illustration I made. The green squares represent the positions I need the squares to actually be in. The red squares are where they actually appear given the following values for each square's upper-left corner:
x = origin.x + radius * cos(DEGREES_TO_RADIANS(angle));
y = origin.y + radius * sin(DEGREES_TO_RADIANS(angle));
Origin refers to the center of the circle. I have a loop that repeats this for every angle from 1 till 360 degrees.
EDIT: I've changed my design to position the centers of the squares along the circular path rather than their upper left corners.
why not just draw the centers of the squares along a smaller circle inside of the bigger one?
You could do the math to figure out exactly what the radius would have to be to ensure an exact fit, but you could probably trial and error your way there quickly too.
Doing it this way ensures that your objects would end up laid out in an actual circle too, which is not the case if you were merely making sure that one and only one corner of each square touched the larger bounding circle (that would create a slightly octagonal shape instead of a circle)
ryan cumley's answer made me realize how dumb I was all along. I just needed to change each square's anchor point to its center & that solved it. Now every calculated value for x & y would position every square's center exactly on the circular path.
Option 1) You could always find the diameter of the circle and then using Pythagorean Theorem, you could create a square that would fit perfectly within the circle. You could then loop through the square that was just made in the circle to create smaller squares, but I doubt this is what you are aiming for.
Option2) Find out what half of the length of one of the diagonals of the squares should be, and create a ring within the first ring. Then lay down squares at key points (like ever 30 degrees or 15 degrees, etc) along the inner path. Ex: http://i.imgur.com/1XYhoQ0.png
As you can see, the smaller (inner) circle is in the center of each green square, and that ensures that the corners of each square just touches the larger (outer) circle. Obviously my cheaply made picture in paint is not perfect, but mathematically it will work.
I am trying to implement a 2-D fast collision detection with Quad-Tree.
AFAIK, Quad-Tree divides a region into 4 sub-regions, north-west, north-east, south-east and south-west. This dividing works perfectly with a square. But what if the region is a non-square rectangle? In that case, we cannot divide the long edge and the short edge evenly, and the short edge determins how far we can divide.
Am I right on this? Is that meant to be?
Simply take the max of width, height of the bounding box of the region of interest as the side length of the quad tree.
Another solution:
Two quad tree implementtaions that i have seen uses a rectangle internaly, so that would run out of the box, even if the provided root bounds is not a square. They divide both the width and the height of the bounds in each subdivision step. But note that there are ovr 10 different Quadtree types. I am talking about Rectangle Quadtrees.
One implemention explictly uses a a side length which is divided by 2, so that would not work fine for non square root bounds.
However, I still recommend my first sentence, better use a square as root bounds.
This then works for all quad tree types.
How to draw a circle overlapping another circle in the moved phase of touch event,such that no gap is left out between the circles.The circles must be tightly packed to one another,so that even when user moves his hand on the screen faster or lightly,no gap must be present between the circles.
Just two circles? Or many circles? If just two, then detecting if they overlap is simply verifying that their centers are not closer than the sum of their radii. For example, if Circle1's raduis is 10 pixels, and Circle2's radius is 25 pixels, then they overlap if the center of Circle1 is less than 35 pixels from the center of Circle2.
So if you do your calculations in the "moved" phase and find that they're too close, you have to adjust the position of one of them. How you go about that will depend on the specifics of your application. You could:
Keep the y coordinate of the moving circle the same, and calculate the necessary x coordinate to maintain the required distance.
Same as above but swap x and y.
As above, but move the "unmoving" circle away from the "moving" circle.
Some other calculation that makes sense for your application.
NOTE: You should accept some of the answers you've been given.