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    <p>So I've solved this quite some time ago, but to help those who eventually stumble onto this and want some similar functionality:</p> <p>This post is quite some time ago - so let me remember what happened. </p> <p>Creating a Region around a path actually did work but for a very limited set of Paths. What I mean by "Creating a region around a Path" is that for a Path that goes from x1,y1 to x2,y2 create a rectangular Region that covers (for example) x1-50,y1-50 to x2+50,y2+50 where 50 is the pixel weight stroke of the Path. </p> <p>If you can visualise it, it basically creates a rectangular region that covers the Path and it's 50px stroke so you can "fake" Path intersection using Regions. Wherever a Path is, a Region is and so when 2 Paths "intersect", you can check for Region intersection (which you can do but I've forgotten the method names).</p> <p>This however proved to work only for a few Paths. Though I'd like to think my Math proficiency is adequate, I could not get it so that for whichever direction the Path went the Region would work. Different angles, different directions etc. caused the Region not properly drawing under the Path. My above example of using the 50 stroke width would only work for going a particular direction.</p> <p>The solution my parter and I stumbled onto was creating a 2D integer array that mapped onto the screen. Wherever a Path went into a certain direction, we would fill every array cell the Path mapped onto with a specific value (1). We would do the same for the other Path, but with a different value (2). Each move you make you would check the 2D array against the Path co-ordinate to see if it has been occupied or not. There was an important mathematical formula that would extrapolate which cells were visited when you go from x1,y2 to x2,y2 that proved very helpful - I believe it was called something along Brasenheim's formula, or something.</p> <p>It's not the most elegant solution, but it ended up faking Path intersection well. If anyone is interested in a better explanation, you can message me.</p> <p>Good luck!</p>
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