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Given rectangle_A intersecting rectangle_B, which has a union defined such that it is the rectangle containing both rectangles, I want to determine the coordinates of the (not overlapping) rectangles required to add to rectangle_A to create the union of rectangle_A and rectangle_B:

Note: this is just one configuration of the solution set of rectangles. the white rectangles above could be configured differently, as long as they don't overlap.

Is there a simple algorithm for every case of rectangle intersection? I've done a first pass and I miss some corners. Evidently not my forté.

Why? When panning in a UI, I only want to (i) update the new parts of the canvas (ii) keep track of what has been painted as a rectangle (the union of rectangle_A and rectangle_B).

Matt
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jedierikb
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  • Do you always have exactly two rectangles at a time? – ShreevatsaR Jul 14 '10 at 20:03
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    Yes, always just two original rectangles: A and B. These two rectangles can be completely different sizes. – jedierikb Jul 14 '10 at 20:10
  • I find this not clear-in the last image, which rules determine how the resulting rects must be? Your 3. image, would it be also valid to swich the to white rects? Or is "not overlapping" the only restriction? – InsertNickHere Jul 14 '10 at 20:14
  • And will they always intersect? Can it happen that one lies within the other? Can it happen that the x-range of rectangle B is entirely contained with the x-range of A (i.e. B lies "below" A, but with smaller width)? Similarly, can B lie to the right of A (intersecting, but not extending above or below) but with smaller height? – ShreevatsaR Jul 14 '10 at 20:15
  • @InsertNickHere: Good point. The solution rectangles can be in any configuration as long as they are non-overlapping. I will clarify in the OP. – jedierikb Jul 14 '10 at 20:17
  • @ShreevatsaR: For this problem, the perimeter of A will always intersect the perimeter of B. Both of your examples are valid configurations of A and B which require solutions. – jedierikb Jul 14 '10 at 20:20
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    Do you require a solution that presents the fewest possible number of rectangles? If not, a solution that always returns 8 (possibly zero-area) rectangles is trivial. – VeeArr Jul 14 '10 at 20:30
  • @VeeArr: a minimum # of rectangles would be preferable, but I am open to suggestions! – jedierikb Jul 14 '10 at 20:35

3 Answers3

5

If you are not concerned with minimizing the number of rectangles returned, you can simplify the thought process to one that always returns no more than 8 rectangles:

U
+----------+----+-------+
|          |    |       |
|     1    | 2  |  3    |
+----------+----+-------+
|          |    |       |
|     4    | A  |  5    |
|          |    |       |
+----------+----+-------+
|     6    | 7  |  8    |
+----------+----+-------+

U.x1 = min(A.x1,B.x1)
U.x2 = max(A.x2,B.x2)
U.y1 = min(A.y1,B.y1)
U.y2 = max(A.y2,B.y2)
R1.x1 = R4.x1 = R6.x1 = U.x1
R2.x1 = R7.x1 = R1.x2 = R4.x2 = R6.x2 = A.x1
R2.x2 = R7.x2 = R3.x1 = R5.x1 = R8.x1 = A.x2
R3.x2 = R5.x2 = R8.x2 = U.x2
R1.y1 = R2.y1 = R3.y1 = U.y1
R1.y2 = R2.y2 = R3.y2 = R4.y1 = R5.y1 = A.y1
R4.y2 = R5.y2 = R6.y1 = R7.y1 = R8.y1 = A.y2
R6.y2 = R7.y2 = R8.y2 = U.y2

If you wanted, you could then quickly check each rectangle to see if r.x1 == r.x2 || r.y1 == r.y2 (i.e. if it has zero area), and throw it out if so. In most cases, over half of the rectangles can be thrown out this way.

For example, in your three examples, this solution would return 3, 1, and 5 rectangles, and would return 0 in the best case (when B is contained in A) and 8 in the worst case (when A is contained in B).

VeeArr
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    Exactly! The talk about how A and B intersect seems irrelevant. You can always merge 1,2,3 into one and 6,7,8 into another to get no more than 4 rectangles (and that will give optimum if A is completely within the outer rectangle). And if A touches one of the sides, it will give the optimum of 2 or 3 too. –  Jul 14 '10 at 20:54
  • +1 This technique is relatively easy to implement and it handles rectangle A no matter where it is in relation to B. Given the way that the union works, you should always have at 3-5 of these rectangles that you can ignore (they have zero area). Of the remaining rectangles, it should be relatively trivial to combine adjacent rectangles and reduce the number of polygons that you have to deal with. – bta Jul 14 '10 at 20:55
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Say we represent rectangles by a pair of x,y coordinate pairs: x1,y1 for the top-left and x2,y2 for the bottom left corner. Let's also assume y coordinate increase downwards and x coordinates increase left to right.

Now, suppose the rectangle formed by the union of A and B (according to your definition of union) is the rectangle is U.

So, 

U.x1=min(A.x1,B.x1), U.y1=min(A.y1,B.y2) --- top-left corner, take the lowest values
U.x2=max(A.x2,B.x2), U.y2=max(A.y2,B.y2) --- bottom-right corner, take the highest values

Now that we have the larger rectangle U, we can use that to compute the smaller right and bottom rectangles that have to be added to A (the left/top rectangle) to make it U. Lets call them Rt and Bot.

(This time I'm assuming A is the top-left rectangle, if it isn't swap A and B. Also assuming the layout to be similar to that of your picture. If that isn't the case you can adapt this easily).

Rt.x1=A.x2, Rt.y1=A.y1
Rt.x2=A.x2, Rt.y2=B.y2

Bot.x1=A.x1, Bot.y1=A.y2
Bot.x2=A.x2, Bot.y2=B.y2
MAK
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  • I'm quite sure this will only cover the given problem (in the images), but there are many cases where you need more than 2, or just 1 adding rect. The problem is still-how to find the recs in all cases, hardcoding is surely not a good approach here. – InsertNickHere Jul 14 '10 at 20:23
  • @InsertNickHere: As I wrote in my answer, I'm assuming that the layout is what is shown in the diagrams for the last half. You need more than 2 only if B is larger than A and actually covers A and extends on all sides. In that case, U=B, simply deduce the extra rects from that. If less than 2 extra rects are needed, my method returns rects with area=0. Just check for that. – MAK Jul 14 '10 at 20:37
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I'm sorry i cant give a working solution, but...

At first I would try to draw such nice images for every different case that you can imagine. There will be a lot cases, where you need more than 2 rectangles, or just one, right?

I think getting the rect containing the others is trivial-but at this time I can't think of how to proceed. :)

Edit: At this time i'm thinking of a flood fill algorith, just fill up your larger rect. But there are 2 problems with this I can imagine: How to use the flood fill output to generate rects from it? Will it be the right way, or is there a linear algebra solution or something?

InsertNickHere
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