How would one go about finding if two 2D line segments intersect? I know how to do it on paper, but I'm totally baffled how to do it in code. Any suggestions?
How would one go about finding if two 2D line segments intersect? I know how to do it on paper, but I'm totally baffled how to do it in code. Any suggestions?
Last edited by Highland Laddie; 10-13-2005 at 03:31 PM.
Bagpipes – putting the fun back in funeral.
Two lines of code can't intersect, it would drive the compiler nuts.
My best code is written with the delete key.
Are you talking about graphical programming or are you just asking how to code the equation to see if two lines intersect?
If you're talking about the latter, then simply write a program that accepts ( or figures out ) the slopes of both lines, and if they aren't parallel and they are both straight, then they intersect.
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Yeah, I guess it would. I totally ran into that one.
Bagpipes – putting the fun back in funeral.
If this isn't a joke that I don't understand then I'd say you don't understand his question.Originally Posted by Prelude
He's asking how to write code to figure out if two lines on a Cartesian plane intersect.
... atleast I think that's what he's asking.
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>If this isn't a joke that I don't understand then I'd say you don't understand his question.
I guess it's a joke that you don't understand then.
My best code is written with the delete key.
Yeah, I figured that much from his reply.
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Morally, I should tell you to get off your lazy ass and do some research, but since your a fellow celt and I'm feeling lazy myself this morning so here ya goOriginally Posted by Highland Laddie
it's slightly bastardised from existing code so you'll need to make it work properly (i.e. put it in a fucntion or something), but it should get you startedCode:// need to calc line functions for both lines // line func : y = a * x + b // so for line (x1, y1) (x2, y2) // (y2 - y1) (y2 - y1) // a = --------- b = y1 - --------- * x1 // (x2 - x1) (x2 - x1) // calc vals for line from centre to joystick pos int denominator = x2- x1; if (0 == denominator) { // oops divide by zero // should happen for the target line return; } float joyA = (float)((y2 - y1) / (float)denominator); float joyB = y1 - joyA * x1; // calc vals for poly lines int lineDenominator = x2 - x1; if (0 != lineDenominator) { float lineA = (float)((y2 - y1) / (float)lineDenominator); float lineB = y1 - lineA * (float)x1; // intersection point is calculated as // (b2 - b1) (b2 - b1) // xi = --------- yi = a1 * --------- + b1 // (a1 - a2) (a1 - a2) float solvedDenominator = joyA - lineA; if (0.0001f < std::fabs(solvedDenominator)) { float flSolvedX = (lineB - joyB) / solvedDenominator; int solvedX = (int)flSolvedX; int solvedY = (int)(joyA * flSolvedX + joyB); if (solvedX > min(x1, x2) && solvedX < max(x1, x2) && solvedY > min(y1, y2) && solvedY < max(y1, y2)) { int interX = solvedX; int interY = solvedY; } } }
By standard, the variables in slope-intercept is Y = mX + B.
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I stumbled upon another way a while back. Its a good deal simpler, and it does indeed work.
I've modified this so that it is not dependent on my game.Code:struct vertex { float x,z; }; bool intersect (vertex v1, vertex v2, vertex p1, vertex p2); bool counter_clockwise(vertex p1, vertex p2, vertex p3); /*Call this function with your four points *v1,v2 are vertices of one line *p1,p2 are vertices of the other */ bool intersect (vertex v1, vertex v2, vertex p1, vertex p2) { if (counter_clockwise(p1,p2,v1)!=counter_clockwise(p1,p2,v2) && counter_clockwise(v1,v2,p1)!=counter_clockwise(v1,v2,p2)) { return 1; //The lines of collided } else { return 0; } } bool counter_clockwise(vertex p1,vertex p2,vertex p3) { return ((p2.z-p1.z)*(p3.x-p2.x)<(p3.z-p2.z)*(p2.x-p1.x)); }
It should work, and does compile correctly (assuming no Copy/Paste errors).
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Programming Your Mom. http://www.dandongs.com/
true, but I think it might still workOriginally Posted by SlyMaelstrom
Maybe not. As you know, simple inproper uses of variables, be it used in comments, code, or what have you, have a tendency to drive obviously highly intelligent things like home PCs to insanity. Which is the reason that so many PCs have been murdering their end users in thie sleep, lately.
It doesn't fit mine.Originally Posted by CrazyNorman
Be sure to use your right margin, it's a magical thing.
Last edited by SlyMaelstrom; 10-13-2005 at 04:53 PM.
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Well, I think I got something figured out. I have yet to test it and debug it, but I'll get around to it later. Anyways, heres the code.
Code:// Get the slope of the lines double slope1 = End.Y - Start.Y / End.X - Start.X; double slope2 = otherLine.End.Y - otherLine.Start.Y / otherLine.End.X - otherLine.Start.X; // Get the Y - intercepts of the line double yInt1 = slope1 * Start.X - Start.Y; double yInt2 = slope2 * otherLine.Start.X - otherLine.Start.Y; // Return false if the lines are paralell if (slope1 == slope2) { return false; } // Calculate the intersection point double intersectionX = (yInt2 - yInt1) / (slope1 - slope2); double intersectionY = slope1 * intersectionX + yInt1; // Check if the two line segments intersect if (intersectionX < Start.X || intersectionY > Start.Y || intersectionX > End.X || intersectionY < End.Y) { return false; } else { return true; }
Bagpipes – putting the fun back in funeral.
you need to check these lines for division by zeroOriginally Posted by Highland Laddie
"I saw a sign that said 'Drink Canada Dry', so I started"
-- Brendan Behan
Free Compiler: Visual C++ 2005 Express
If you program in C++, you need Boost. You should also know how to use the Standard Library (STL). Want to make games? After reading this, I don't like WxWidgets anymore. Want to add some scripting to your App?