Optimising the drawing of overlapping rectangles

Question

I have a large number of rectangles, and some overlap others; each rectangle has an absolute z-order and a colour. (Each \'rectangle\' is actually the axis-aligned bou

Answer 1

Start by drawing in a random (but correct) order, for example in strict z order. When drawing each frame, either count the number of color changes, or possibly the actual time a complete frame takes. Each frame, try swapping the order of two rectangles. The rectangles to be swapped must not overlap, therefore they can be drawn in any order without violating correctness; aside from that they can be chosen at random, or do a linear pass through the list, or... If doing the swap reduces the number of color changes, keep the new order, if not revert it and try a different swap in the next frame. If doing the swap neither reduces nor increases the number of color changes, keep it with 50% odds. For any rectangles which did not overlap in a previous frame but which start overlapping due to a move, simply exchange them so they are in z order.

This has some relationship to sorting algorithms which swap pairs of items, except that we cannot compare items, we need to go through the whole list and count color changes. This will perform very badly at first but converge to a good order relatively quick, and will adapt to scene changes. I think it is probably not worth it to go through and calculate an optimum order every frame; this will get to, and maintain, a near-optimum order with very little extra work.

Referring to the drawing you have: Initial draw order picked at random: 1,6,2,4,5,8,3,7 (5 color changes). Swap 5,8. New order: 1,6,2,4,8,5,3,7 (4 color changes) => Keep new order.

Answer 2

You are making the very wrong assumption that the performance you will be getting on the desktop browser will somehow determine the performance on your iPhone. You need to understand that the iPhone hardware implements tile-based deferred rendering which means that the fragment shader is used very late in the pipeline anyway. As Apple themselves say (“Do not waste CPU time sorting objects front to back”), Z-sorting your primitives will get you little performance gain.

But here’s my suggestion: if changing the colour is expensive, just don’t change the colour: pass it as a vertex attribute, and merge the behaviours into one super shader so you can draw everything in one or a few batches without even sorting. Then benchmark and determine the optimal batch size for your platform.

Answer 3

Choose colours, not boxes!

At any point in time, one or more boxes will be paintable, i.e. they are able to be painted next without introducing problems (though possibly introducing a cost due to having a different colour from the most recently painted box).

The question at every point is: What colour should we pick to draw next? It's not necessary to think about picking individual paintable boxes to draw, because as soon as you pick a particular box to draw next, you might as well draw all available boxes of the same colour that can be drawn at that time. That's because painting a box never adds constraints to the problem, it only removes them; and choosing not to paint a paintable box when you could do so without changing the current colour cannot make the solution less expensive than it would otherwise be, since you will later have to paint this box and that may require a colour change. This also means it doesn't matter in which order we paint paintable boxes of the same colour, since we will paint all of them at once in a single "block" of box painting operations.

The dependency graph

Start by building a "lies underneath" dependency graph, where each coloured rectangle is represented by a vertex and there is an arc (arrow) from v to u if rectangle v overlaps rectangle u and lies underneath it. My first thought was to use this to build a "must be drawn before" dependency graph by finding the transitive closure, but actually we don't need to do this, since all the algorithms below care about is whether a vertex is paintable or not. Paintable vertices are the vertices that have no predecessors (in-arcs), and taking the transitive closure does not alter whether a vertex has 0 in-arcs or not.

In addition, whenever a box of a given colour has only boxes of the same colour as its ancestors, it will be painted in the same "block" -- since all those ancestors can be painted before it without changing colours.

A speedup

To cut down on computation, notice that whenever all paintable boxes of some particular colour have no different-coloured descendants, painting this colour won't open up any new opportunities for other boxes to become paintable, so we don't need to consider this colour when considering which colour to paint next -- we can always leave it till later with no risk of increasing the cost. In fact it's better to leave painting this colour till later, since by that time other boxes of this colour may have become paintable. Call a colour helpful if there is at least one paintable box of that colour that has a different-coloured descendant. When we get to the point when there are no helpful colours remaining (i.e. when all remaining boxes overlap only boxes of the same colour, or no boxes at all) then we are done: just paint the boxes of each remaining colour, picking colours in any order.

Algorithms

These observations suggest two possible algorithms:

1. A fast but possibly suboptimal greedy algorithm: Choose to paint next the colour that produces the most new paintable vertices. (This will automatically consider only helpful colours.)

2. A slower, exact DP or recursive algorithm: For each possible helpful colour c, consider the dependency graph produced by painting all paintable c-coloured boxes next:

   Let f(g) be the minimum number of colour-changes required to paint all boxes in the dependency graph g. Then

   f(g) = 1 + min(f(p(c, g)))

   for all helpful colours c, where p(c, g) is the dependency graph produced by painting all paintable boxes of colour c. If G is the dependency graph for the original problem, then f(G) will be the minimum number of changes. The colour choices themselves can be reconstructed by tracing backwards through the DP cost matrix.

   f(g) can be memoised to create a dynamic programming algorithm that saves time whenever 2 different permutations of colour choices produce the same graph, which will happen often. But it might be that even after DP, this algorithm could take an amount of time (and therefore space) that is exponential in the number of boxes... I will have a think about whether a nicer bound can be found.

Answer 4

Here's a possibility. You'll have to benchmark it to see if it's actually an improvement.

For all rectangles, back to front:
  If this rectangle has been marked as drawn, skip to the next one
  Set a screen-sized unseen surface to all black
  Call this rectangle's color "the color"
  For rectangles starting with this one and proceeding toward the front
    If (this rectangle's color is the color and
        all the pixels of this rectangle on the unseen are black) then
      Add this rectangle to the to-draw list
    Draw a white rectangle with this rectangle's shape on the unseen surface
    If the unseen surface is more than half white, break
  For all rectangles on the to-draw list:
    Draw the rectangle
    Mark it as drawn

It's not guaranteed to be the most optimal in terms of ordering, but I think it will come pretty close, and it's worst-case quadratic in the pre-drawing step. It does depend on readbacks from the graphics buffer being fast. One trick that might help there is to create a new one pixel surface that is a shrunken version of the area of interest. Its color will be the fraction of the original that was white.