How to discover if 2 CGPath roteted rects intersect? - ios

Question

I am building 2 rects when view controller is loaded, using CGPath. The rects can be moved with PanGestureRecognizer. The question is how can I know when 2 rects had met? In order to not let them intersect?
MainViewController.m
- (void)viewDidLoad
{
[super viewDidLoad];
for (int i=0; i< 2; i++) {
//create rects of CGPath
CGRectCustomView* randomColorRect =
[[CGRectCustomView alloc]initWithFrame:
CGRectMake(<random place on screen>)];
//random angle
randomColorRect.transform =
CGAffineTransformMakeRotation
(DegreesToRadians([Shared randomIntBetween:0 and:360]));
[self.view addSubview:randomColorRect];
}
}
- (BOOL)areRectsCollide {
???How to find this???
}
CGRectCustomView.m:
- (void)drawRect:(CGRect)rect
{
CGContextRef context = UIGraphicsGetCurrentContext();
CGContextSetLineWidth(context, 8.0);
CGContextStrokePath(context); // do actual stroking
CGContextSetRGBFillColor(context, <green color>, 1);
CGContextFillRect(context, CGRectMake(self.frame.origin.x, self.frame.origin.y, self.frame.size.width, self.frame.size.height));
path = CGContextCopyPath(context);
}
In Apple guide here, there is a function that determines if a path contains point
- (BOOL)containsPoint:(CGPoint)point onPath:(UIBezierPath *)path inFillArea:(BOOL)inFil,
but I have a rectangle which is endless number of points. So what should I do? Breaking my head...

Answer 1

Found it!
I used algorithm described here.
Just one thing: I move the rects around on the screen. So in order for them not to stack after the first collision, I save the last uncollisioned point and if collision accusers, I restore last location.
- (void)handlePan:(UIPanGestureRecognizer *)recognizer {
BOOL rectsColide = NO;
for (RandomColorRect* buttonRectInStoredRects in arreyOfPaths) {
if (buttonRectInStoredRects.tag != recognizer.view.tag) {
if ([self view:buttonRectInStoredRects intersectsWith:recognizer.view]) {
rectsColide = YES;
}
}
}
CGPoint translation = [recognizer translationInView:self.view];
if (!rectsColide) {
lastPoint = recognizer.view.center;
recognizer.view.center = CGPointMake(recognizer.view.center.x + translation.x,
recognizer.view.center.y + translation.y);
}else{
recognizer.view.center = CGPointMake(lastPoint.x ,lastPoint.y);
}
[recognizer setTranslation:CGPointMake(0, 0) inView:self.view];
}
- (void)projectionOfPolygon:(CGPoint *)poly count:(int)count onto:(CGPoint)perp min:(CGFloat *)minp max:(CGFloat *)maxp
{
CGFloat minproj = MAXFLOAT;
CGFloat maxproj = -MAXFLOAT;
for (int j = 0; j < count; j++) {
CGFloat proj = poly[j].x * perp.x + poly[j].y * perp.y;
if (proj > maxproj)
maxproj = proj;
if (proj < minproj)
minproj = proj;
}
*minp = minproj;
*maxp = maxproj;
}
-(BOOL)convexPolygon:(CGPoint *)poly1 count:(int)count1 intersectsWith:(CGPoint *)poly2 count:(int)count2
{
for (int i = 0; i < count1; i++) {
// Perpendicular vector for one edge of poly1:
CGPoint p1 = poly1[i];
CGPoint p2 = poly1[(i+1) % count1];
CGPoint perp = CGPointMake(- (p2.y - p1.y), p2.x - p1.x);
// Projection intervals of poly1, poly2 onto perpendicular vector:
CGFloat minp1, maxp1, minp2, maxp2;
[self projectionOfPolygon:poly1 count:count1 onto:perp min:&minp1 max:&maxp1];
[self projectionOfPolygon:poly2 count:count1 onto:perp min:&minp2 max:&maxp2];
// If projections do not overlap then we have a "separating axis"
// which means that the polygons do not intersect:
if (maxp1 < minp2 || maxp2 < minp1)
return NO;
}
// And now the other way around with edges from poly2:
for (int i = 0; i < count2; i++) {
CGPoint p1 = poly2[i];
CGPoint p2 = poly2[(i+1) % count2];
CGPoint perp = CGPointMake(- (p2.y - p1.y), p2.x - p1.x);
CGFloat minp1, maxp1, minp2, maxp2;
[self projectionOfPolygon:poly1 count:count1 onto:perp min:&minp1 max:&maxp1];
[self projectionOfPolygon:poly2 count:count1 onto:perp min:&minp2 max:&maxp2];
if (maxp1 < minp2 || maxp2 < minp1)
return NO;
}
// No separating axis found, then the polygons must intersect:
return YES;
}
- (BOOL)view:(UIView *)view1 intersectsWith:(UIView *)view2
{
CGPoint poly1[4];
CGRect bounds1 = view1.bounds;
poly1[0] = [view1 convertPoint:bounds1.origin toView:nil];
poly1[1] = [view1 convertPoint:CGPointMake(bounds1.origin.x + bounds1.size.width, bounds1.origin.y) toView:nil];
poly1[2] = [view1 convertPoint:CGPointMake(bounds1.origin.x + bounds1.size.width, bounds1.origin.y + bounds1.size.height) toView:nil];
poly1[3] = [view1 convertPoint:CGPointMake(bounds1.origin.x, bounds1.origin.y + bounds1.size.height) toView:nil];
CGPoint poly2[4];
CGRect bounds2 = view2.bounds;
poly2[0] = [view2 convertPoint:bounds2.origin toView:nil];
poly2[1] = [view2 convertPoint:CGPointMake(bounds2.origin.x + bounds2.size.width, bounds2.origin.y) toView:nil];
poly2[2] = [view2 convertPoint:CGPointMake(bounds2.origin.x + bounds2.size.width, bounds2.origin.y + bounds2.size.height) toView:nil];
poly2[3] = [view2 convertPoint:CGPointMake(bounds2.origin.x, bounds2.origin.y + bounds2.size.height) toView:nil];
return [self convexPolygon:poly1 count:4 intersectsWith:poly2 count:4];
}

Answer 2

If I understand your question, so in order to check if 2 CGRects meet you better use:
/* Return the intersection of `r1' and `r2'. This may return a null rect. */
CG_EXTERN CGRect CGRectIntersection(CGRect r1, CGRect r2)
For example:
CGRect rect1 = CGRectMake(0, 0, 10, 10);
CGRect rect2 = CGRectMake(5, 5, 10, 10);
CGRect rect3 = CGRectMake(20, 20, 10, 10);
CGRect r1 = CGRectIntersection(rect1, rect2); // Returns a CGRect with (0,0,5,5)
CGRect r2 = CGRectIntersection(rect1, rect3); // Returns a CGRect with (Inf,Inf,0,0)
if (CGSizeEqualToSize(r2.size,CGSizeZero)) {
// rect1 rect3 Intersects at CGSizeZero == Do not Intersect
}