CODEWITHSUNDEEP
algorithm
user631854
user631854

Reputation:

decide the point is in which section of rectangle

I have a rectangle which has known width and height.Now devide rectangle diagonally to form four triangles. Now suppose have a point p(m,n) and i have to decide that the point lies in which triangle out of four triangles. Please suggest me some good algo.

Upvotes: 0

Views: 111

Answers (3)

user631854
user631854

Reputation:

if((x * height < y * width) || ( x*width < y*height )) { test = 1; } else { test = 0; } if((x > width / 2 && y > height / 2)) { test1=1; } else { test1=0; } int section = (test) * 2 + (test1);

                switch(section)
                {
                case 0:
                    Log.d("ABSAR","UP button");

                    break;
                case 1:
                    Log.d("ABSAR","RIGHT button");

                    break;
                case 2:
                    Log.d("ABSAR","LEFT button");

                    break;
                case 3:
                    Log.d("ABSAR","DOWN button");

                    break;
                }

`if point P(x,y) and the rectangle of dimension width * height Then above code works OK...

Upvotes: 0

Petar Minchev
Petar Minchev

Reputation: 47403

Well you know the line equations of the diagonals(because you have two points). Both of the diagonals define two halfplanes.

Diagonal1 equation:

A1 * x + B1 * y + C1 = 0

Diagonal2 equation:

A2 * x + B2 * y + C2 = 0

It becomes:

Substitute A1 * m + B1 * n + C1 to find in which halfplane is the point for the first diagonal. If the number is < 0, then it lies in one halfplane, if it is > 0 then it lies in the other.

Substitute A2 * m + B2 * n + C2 to find in which halfplane is the point for the second diagonal. If the number is < 0, then it lies in one halfplane, if it is > 0 then it lies in the other.

Upvotes: 4

robertrosman
robertrosman

Reputation: 514

Interesting question. I think this algorithm should do it, but I haven't tried it out yet. Assuming that the rectangle is AxB and the point p(m,n).

int section = (m * B < n * A) * 2 + (m > A / 2 || n > B / 2);

The variable section should then be a value between 0 and 3, depending where the point resides. The sections are called:

  0  
2   1
  3

If you wish to name the sections another way feel free to tweak the algo.

Upvotes: 1

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