So divide for one value, and subtract that from the total for the other.)Īnd then, apply the stored separation vectors to each rectangle - similar to what you are already doing. Timeless Treasures business is centered around the most fun category in all cotton quilting prints: Novelties From pigs riding hogs (motorcycles) to.
(Note: when dividing the integers, be aware that you must divide (for example) 5 into 3 and 2. Divide that separation vector evenly (or randomly?) between the two rectangles and replace the stored value for that rectangle if it is greater (or smaller?) in magnitude. When iterating over each pair, determine the minimum separation vector for that pair as explained above. Here's what I would start with:įor each rectangle, store the maximum (or minimum?) separation vector by magnitude. Without trying to implement it myself, I can't say for sure what this would look like. The tricky bit, then, is to make this work for multiple rectangles in a way that can be iterated to a final solution. (This is basically a constrained version of the " separating axis theorem".) This is a directional quantity, so store it as a vector (again: not as a vector3, but as a pair of int). This amount could be split between the two rectangles in any ratio (but in opposite directions). In other words, if you only had two rectangles, this would be the smallest amount you would need to shift them to separate them. Only use a single axis! Your current algorithm is considering both axes at once, based only on position. Autumn Symphony by Timeless Treasures Fabrics Panel size is 23' x 42'.
Autumn Symphony Metallic Cream Harvest Wreath Quilt Panel. Pick the axis with the minimum amount of penetration (so: left edge to right edge, right edge to left edge, top edge to bottom edge, bottom edge to top edge). American Pride by Timeless Treasures Fabrics. You only need to consider each pair of rooms once, so use a loop like this: for(int i = 0 i < count i++)įor(int j = i+1 j < count j++) // <- note how it starts at i+1įor each pair of rectangles, see if they intersect. Because you need to be considering the penetration distance between pairs of rooms (which requires the size). Looking at your Separate() method, you don't seem to be taking into account the size of the rooms. Also, given that you are working with something 2D grid based, you should probably be working with a pair of integers (X and Y axis), rather than a floating-point Vector3. I can't say for sure if it will work, and you'll have to do a bit of legwork to implement this.įirst, a few quick things: What do you define as "lots of iterations"? The TinyKeep demo you linked does it in around 40-50. Please note that I'm only giving a rough starting point for an algorithm here.
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