Point inside rect, what is wrong with my logic?

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Original post by SriLumpa Your code does not even read pt, so it couldn't possibly work !

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Original post by alvaro Why do you think that code should work? It's not even close... Perhaps you can try to explain the logic behind your code, and we can try to fix it. Taking the dot product of two positions (DotProduct(v,v1)) is not a well-defined operation, in the sense that the result depends on your choice of origin. `0<=d1' means "the angle formed by v, the origin and v1 is acute", which doesn't even tell you anything about where pt is, so you can't decide that pt is not in the rectangle if this is not true.

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Original post by shultays I tried to explaining in my previous post, here is a code example Vector pt //point to test Vector p1, p2, p3, p4 //corners of rectangle // Vector v1 = p2 - p1; //find edge vectors Vector v2 = p3 - p2; Vector v3 = p4 - p3; Vector v3 = p1 - p4; float f1 = dot(v1, pt-p1); float f2 = dot(v2, pt-p2); float f3 = dot(v3, pt-p3); float f4 = dot(v4, pt-p4); if( (f1 < 0 && f2 < 0 && f3 < 0 && f4 < 0) || (f1 > 0 && f2 > 0 && f3 > 0 && f4 > 0) ){ //if all dot products are negative or positive //inside }else{ //outside }

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Original post by shultays to find if a point inside of a rectangle (for example a b c d) you first find 4 vectors for each side b-a c-b d-c a-d if point is on the same side of all these vectors, it is inside. to find this, you can use dot product all these vectors from the a vector from the second point to our test point dot(b-a, n-a) dot(c-b, n-b) dot(d-c, n-c) dot(a-d, n-d) if all these dot products are negative or positive (it depends on you rotate cw or ccw) your point is inside the rectangle. this works for all kind of shapes, triangles rectangle etc as long as it is a convex polygon

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This works only with rectangles. Try it on a triangle and you will see, it's not working.

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But it is the simplest method for rects anyway.

vector2 diff = p - center;return abs(dot(diff, axis1)) <= extent1 && abs(dot(diff, axis2)) <= extent2;

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Original post by szecs Quote: Original post by shultays This works only with rectangles. Try it on a triangle and you will see, it's not working. But it is the simplest method for rects anyway.

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Original post by shultays Quote: Original post by szecs Quote: Original post by shultays This works only with rectangles. Try it on a triangle and you will see, it's not working. But it is the simplest method for rects anyway. He wanted an algorithm for rectangles. Algorithm will work for any number of sides, as long as it is a convex polygon (since triangles are always convex, it will work) here is an example what does it do. *** image clipped *** you check the angle between a-b and a-n vectors, if it is positive n is on the correct side of a-b. after that you do the same process for all other edges b-c and b-n, and c-a and c-n. if all angles are positive point is inside of the triangle you can increase the edges and it should work edit: forgot to add image

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Original post by jyk Quote: This works only with rectangles. Try it on a triangle and you will see, it's not working. Based on his description of the algorithm, I'm guessing shultays meant to use the perp-dot product rather than the dot product. (Unless I'm missing something, if you substitute perp-dot for dot in the pseudocode that he posted, it will indeed work for any convex polygon, including triangles.) Quote: But it is the simplest method for rects anyway. I think it can get a bit simpler, e.g.: vector2 diff = p - center;return abs(dot(diff, axis1)) <= extent1 && abs(dot(diff, axis2)) <= extent2;

//v1 is the top left of the rectangle//v2 is the bottom right of the rectangleif((p.x >= v1.x && p.x <= v2.x) && (p.y >= v1.y && p.y <= v2.y))return(true);elsereturn(false);

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Original post by szecs The rectangle is rotated. Fuck, I'm spending too much time here....

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Original post by szecs No. It means it is on the correct side of the line, that's perpendicular to a-b, but not a-b itself. So all points will pass the test, which are in a bigger triangle what's sides are perpendicular to the sides of the original triangle. Your statement works with cross products, not dot products. In your drawing, if n is on the opposite side of a-b, it can have the same dot product with the a-b (just mirror n to a-b) So you probably mixed dot and cross products.

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Original post by maspeir How is calculating the dot product several times easier or faster than this: //v1 is the top left of the rectangle//v2 is the bottom right of the rectangleif((p.x >= v1.x && p.x <= v2.x) && (p.y >= v1.y && p.y <= v2.y))return(true);elsereturn(false); This will work for any 2D rectangle and 2D point.