Random point outside of a set bounds? - actionscript-3

Question

I want to generate a point (x,y) that is outside of my screen. The reason for this is I want to tween an object towards the outside of the screen, and the reason for tweening rather than just giving it a random X and Y velocity is because for this instance tweening using greensock is simpler than creating an update loop.
So, what is an elegant way to generate that number? My current solution is crude and doesn't allow for things to tween towards corners.
var quad:int = irand(1, 4); //generates a random integer between and including the min and max
var maxX:Number = frame.width;
var minX:Number = -frame.width;
var maxY:Number = frame.height;
var minY:Number = -frame.height;
if (quad == 1) //top
{
maxY = -frame.height;
minY = -frame.height - 50;
}
else if (quad == 2) //left
{
maxX = -frame.width;
minX = -frame.width - 50;
}
else if (quad == 3) //bottom
{
maxY = frame.height + 50;
minY = frame.height;
}
else if (quad == 4) //right
{
maxX =frame.width + 50;
minX = frame.width;
}
//first param is duration in seconds
//rand generates a number between the given values
objectToTween.tweenTo(rand(1,2), {
x: rand(minX, maxX),
y: rand(minY, maxY),
repeat: -1,
yoyo: false,
scaleX: 1,
scaleY: 1,
delay: rand(0,1),
ease: Quad.easeIn});
So hopefully it's clear that this solution picks a random direction and tweens the object towards that direction. I'm wondering if there is a way to generate a random number between two given ranges. (such as, between -100 and 100, but excluding -50 through 50) and then applying that for 2 dimensions

Answer 1

what you want
So you have a rectangular frame f:
┌───┐
│ │
│ f │
│ │
└───┘
and a rectangular bounding box of an object o to be tweened to the outside of f:
┌─────┐
│ o │
└─────┘
the outside of f is a position of o in which both f and o touch each other but do not overlap.
Some such valid target positions of o are shown below:
┌───┐
│ │
│ f ├─────┐ 1
│ │ o │
└───┴─────┘
┌───┐
│ │
│ f │
│ │ 2
┌───┴─┬─┘
│ o │
└─────┘
┌─────┐
│ o ├───┐
└─────┤ │ 3
│ f │
│ │
└───┘
┌─────┐
│ o │
└┬───┬┘
│ │ 4
│ f │
│ │
└───┘
Due to the contact of o and f, one side of o is always shared with f. For the above cases, that is
f's right is o's left
f's bottom is o's top
f's left is o's right
f's top is o's bottom
In the edge cases (pun intended), two edges are shared, that is, there is one corner point shared (from example 3, shifting o up a little):
┌─────┐
│ o │
└─────┼───┐
│ │
│ f │
│ │
└───┘
o cannot go any further up or to the left. Keep in mind the definition of the outside of f that I stated above which is perfectly adequate and a lot simpler than "the infinity of space excluding the rectangle f".
If you shift o over to the right, you get the maximum of the x position (the right most position):
┌─────┐
│ o │
┌───┼─────┘
│ │
│ f │
│ │
└───┘
I hope this illustrates that at least one dimension is always predetermined by what sides are touching each other and the other is chosen at random from within the limits. At those limits (minimum and maximum) the two frames will only touch each other at the corners.
There's your problem:
x: rand(minX, maxX),
y: rand(minY, maxY),
instead, first set 1 dimension of o randomly to one of the cases represented in the 4 examples above. Either
x is determined by either f's left or
right (also depending on width and registration point of o), or
y is determined by either f's top or bottom (also depending on heihgt and registration point of o)
the other one is chosen at random from with in the limits.
what you need
There's so much randomisation going on that it is hardly justifiable to put that much effort into determining a target position.
Consider an alternative approach. Determine a circle around f, that will guarantee to be outside of f. Then randomly determine an angle (direction) and tween according to that. This will make the tweens appear slightly different, because the duration they are visible varies depending on the direction (angle), but this difference will be barely noticeable.