Recent News First of all, let’s define the rectangles and points in the following structures:struct Point
{
int x;
int y;
};

struct Rectangle
{
Point topLeft; // Denotes the top-left point of the rectangle
Point bottomRight; // Denotes the bottom-right point of the rectangle
};Given two rectangles R1 and R2 . It is easy to visualize that the given two rectangles can not be intersect if one of the following conditions is true.Condition 1: When left edge of R1 is on the right of R2’s right edge. ( That is , R1 is completely on the right of R2).Condition 2: When right edge of R1 is on the left of R2’s left edge. ( That is , R1 is completely on the left of R2).Condition 3: When top edge of R1 is on bottom of R2’s bottom edge ( That is , R1 is completely under R2).Condition 4: When bottom edge of R1 is on top of R2’s top edge ( That is , R1 is completely over R2).Here is the following code based on the above conditions:void RectIntersect( Rectangle R1 , Rectangle Rt2 )
{
if ( ( R1.topLeft.x > R2.bottomRight.x )|| ( R1.bottomRight.x < R2.topLeft.x ) || ( R1.topLeft.y > R2.bottomRight.y ) ||
( R1.bottomRight.y < R2.topLeft.y ) )
{
cout<<” Non – Intersecting”;
}
else
{
cout<<” Intersecting”;
}
}We can also applyDe Morgan’s law:! ( Condition1 || Condition2 || Condition3 || Condition4 )=( ! Condition1 ) && ( ! Condition2 ) && ( ! Condition3 ) && ( ! Condition4 )Here is the following code based on the above conditions:void RectIntersect( Rectangle R1 , Rectangle R2 )
{
if ( ( R1.topLeft.x < R2.bottomRight.x ) && ( R1.bottomRight.x > R2.topLeft.x ) &&
( R1.topLeft.y < R2.bottomRight.y ) &&
( R1.bottomRight.y > R2.topLeft.y ) )
{
cout<<“Intersecting”
}
else
{
cout<<” Non – Intersecting”;
}
} News Reporter