how to calculate the dist() from mouseX, mouseY to a rectangle in Processing - processing

Question

If it was the dist to a point it would be
dist(mouseX, mouseY, x, y)
for
point(x,y)
but how can I calculate dist() from the mouse's current position to
rectMode(CORNERS);
rect(x1,y2,x2,y2);
Thanks

Answer 1

Something like this should do it:
float distrect(float x, float y, float x1, float y1, float x2, float y2){
float dx1 = x - x1;
float dx2 = x - x2;
float dy1 = y - y1;
float dy2 = y - y2;
if (dx1*dx2 < 0) { // x is between x1 and x2
if (dy1*dy2 < 0) { // (x,y) is inside the rectangle
return min(min(abs(dx1), abs(dx2)),min(abs(dy1),abs(dy2)));
}
return min(abs(dy1),abs(dy2));
}
if (dy1*dy2 < 0) { // y is between y1 and y2
// we don't have to test for being inside the rectangle, it's already tested.
return min(abs(dx1),abs(dx2));
}
return min(min(dist(x,y,x1,y1),dist(x,y,x2,y2)),min(dist(x,y,x1,y2),dist(x,y,x2,y1)));
}
Basically, you need to figure out if the closes point is on one of the sides, or in the corner. This picture may help, it shows the distance of a point from a rectangle for different positions of the point:

Answer 2

Here's a somewhat interactive program which accomplishes what you're looking for. You can drop it into Processing and run it if you would like.
EDIT: Here's a screenshot:
// Declare vars.
int x_click = -20; // Initializes circle and point off-screen (drawn when draw executes)
int y_click = -20;
float temp = 0.0;
float min_dist = 0.0;
int x1, x2, x3, x4, y1, y2, y3, y4;
// Setup loop.
void setup() {
size(400, 400);
// Calculate the points of a 40x40 centered rectangle
x1 = width/2 - 20;
y1 = height/2 - 20;
x2 = width/2 + 20;
y2 = y1;
x3 = x1;
y3 = height/2 + 20;
x4 = x2;
y4 = y3;
}
// Draw loop.
void draw(){
background(255);
// Draws a purple rectangle in the center of the screen.
rectMode(CENTER);
fill(154, 102, 200);
rect(width/2, height/2, 40, 40);
// Draws an orange circle where the user last clicked.
ellipseMode(CENTER);
fill(204, 102, 0);
ellipse(x_click, y_click, 10, 10);
// Draws black point where the user last clicked.
fill(0);
point(x_click, y_click);
// Draws min dist onscreen.
textAlign(CENTER);
fill(0);
text("min dist = " + min_dist, width/2, height/2 + 150);
}
void mousePressed(){
x_click = mouseX;
y_click = mouseY;
// If the click isn't perpendicular to any side of the rectangle, the min dist is a corner.
if ( ((x_click <= x1) || (x_click >= x2)) && ((y_click <= y1) || (y_click >= y3)) ) {
min_dist = min(min(dist(x1,y1,x_click,y_click),dist(x2,y2,x_click,y_click)), min(dist(x3,y3,x_click,y_click),dist(x4,y4,x_click,y_click)));
} else if( (x_click > x1) && (x_click < x2) && ((y_click < y1) || (y_click > y3)) ) {
// outside of box, closer to top or bottom
min_dist = min(abs(y_click - y1), abs(y_click - y3));
} else if( (y_click > y1) && (y_click < y3) && ((x_click < x1) || (x_click > x2)) ) {
// outside of box, closer to right left
min_dist = min(abs(x_click - x1), abs(x_click - x2));
} else {
// inside of box, check against all boundaries
min_dist = min(min(abs(y_click - y1), abs(y_click - y3)),min(abs(x_click - x1), abs(x_click - x2)));
}
// Print to console for debugging.
//println("minimum distance = " + min_dist);
}

Answer 3

This is what I use. If you are only interested in the relative distance there is probably no need to take the square root which should make it slightly quicker.
- (NSInteger) distanceFromRect: (CGPoint) aPoint rect: (CGRect) aRect
{
NSInteger posX = aPoint.x;
NSInteger posY = aPoint.y;
NSInteger leftEdge = aRect.origin.x;
NSInteger rightEdge = aRect.origin.x + aRect.size.width;
NSInteger topEdge = aRect.origin.y;
NSInteger bottomEdge = aRect.origin.y + aRect.size.height;
NSInteger deltaX = 0;
NSInteger deltaY = 0;
if (posX < leftEdge) deltaX = leftEdge - posX;
else if (posX > rightEdge) deltaX = posX - rightEdge;
if (posY < topEdge) deltaY = topEdge - posY;
else if (posY > bottomEdge) deltaY = posY - bottomEdge;
NSInteger distance = sqrt(deltaX * deltaX + deltaY * deltaY);
return distance;
}