Fast rectangle to rectangle intersection

JavascriptC++Language AgnosticGraphics

Javascript Problem Overview


What's a fast way to test if 2 rectangles are intersecting?


A search on the internet came up with this one-liner (WOOT!), but I don't understand how to write it in Javascript, it seems to be written in an ancient form of C++.

struct
{
    LONG    left;
    LONG    top;
    LONG    right;
    LONG    bottom;
} RECT; 

bool IntersectRect(const RECT * r1, const RECT * r2)
{
    return ! ( r2->left > r1->right
        || r2->right < r1->left
        || r2->top > r1->bottom
        || r2->bottom < r1->top
        );
}

Javascript Solutions


Solution 1 - Javascript

This is how that code can be translated to JavaScript. Note that there is a typo in your code, and in that of the article, as the comments have suggested. Specifically r2->right left should be r2->right < r1->left and r2->bottom top should be r2->bottom < r1->top for the function to work.

function intersectRect(r1, r2) {
  return !(r2.left > r1.right || 
           r2.right < r1.left || 
           r2.top > r1.bottom ||
           r2.bottom < r1.top);
}

Test case:

var rectA = {
  left:   10,
  top:    10,
  right:  30,
  bottom: 30
};

var rectB = {
  left:   20,
  top:    20,
  right:  50,
  bottom: 50
};

var rectC = {
  left:   70,
  top:    70,
  right:  90,
  bottom: 90
};

intersectRect(rectA, rectB);  // returns true
intersectRect(rectA, rectC);  // returns false

Solution 2 - Javascript

function intersect(a, b) {
  return (a.left <= b.right &&
          b.left <= a.right &&
          a.top <= b.bottom &&
          b.top <= a.bottom)
}

This assumes that the top is normally less than bottom (i.e. that y coordinates increase downwards).

Solution 3 - Javascript

This is how the .NET Framework implements Rectangle.Intersect

public bool IntersectsWith(Rectangle rect)
{
  if (rect.X < this.X + this.Width && this.X < rect.X + rect.Width && rect.Y < this.Y + this.Height)
    return this.Y < rect.Y + rect.Height;
  else
    return false;
}

Or the static version:

public static Rectangle Intersect(Rectangle a, Rectangle b)
{
  int x = Math.Max(a.X, b.X);
  int num1 = Math.Min(a.X + a.Width, b.X + b.Width);
  int y = Math.Max(a.Y, b.Y);
  int num2 = Math.Min(a.Y + a.Height, b.Y + b.Height);
  if (num1 >= x && num2 >= y)
    return new Rectangle(x, y, num1 - x, num2 - y);
  else
    return Rectangle.Empty;
}

Solution 4 - Javascript

Another more simple way. (This assumes the y-axis increases downwards).

function intersect(a, b) {
  return Math.max(a.left, b.left) < Math.min(a.right, b.right) &&
          Math.max(a.top, b.top) < Math.min(a.bottom, b.bottom);
}

The 4 numbers (max's and min's) in the condition above also give the intersection points.

Solution 5 - Javascript

This has a Rect type you can use. It is already JavaScript.

https://dxr.mozilla.org/mozilla-beta/source/toolkit/modules/Geometry.jsm

Solution 6 - Javascript

I used a blend of methods, to detect a smaller rectangle inside a large rectangle. This is a nodejs method and uses width/height but can easily be adapted.

            isIntersectingRect: function (r1, r2) {
              var quickCheck = (r1.x <= r2.x + r2.w &&
                      r2.x <= r1.x + r1.w &&
                      r1.y <= r2.y + r2.h &&
                      r2.y <= r1.y + r1.h)
              if (quickCheck) return true;
              var x_overlap = Math.max(0, Math.min(r1.x + r1.w, r2.x + r2.w) - Math.max(r1.x, r2.x));
              var y_overlap = Math.max(0, Math.min(r1.y + r1.h, r2.y + r2.h) - Math.max(r1.y, r2.y));
              var overlapArea = x_overlap * y_overlap;
              return overlapArea == 0;
            }

Solution 7 - Javascript

The current .NET way is simply

    public bool IsEmpty => _width < 0.0;   

    public bool IntersectsWith(Rect rect)
    {
        if (IsEmpty || rect.IsEmpty)
        {
            return false;
        }

        if (rect.Left <= Right && rect.Right >= Left && rect.Top <= Bottom)
        {
            return rect.Bottom >= Top;
        }

        return false;
    }

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionRobin RodricksView Question on Stackoverflow
Solution 1 - JavascriptDaniel VassalloView Answer on Stackoverflow
Solution 2 - JavascriptDS.View Answer on Stackoverflow
Solution 3 - JavascriptSchwarriorView Answer on Stackoverflow
Solution 4 - JavascriptDukeView Answer on Stackoverflow
Solution 5 - JavascriptenglebartView Answer on Stackoverflow
Solution 6 - JavascriptLangerzView Answer on Stackoverflow
Solution 7 - JavascriptSpeedOfSpinView Answer on Stackoverflow